# BLOG: Heidelberg Laureate Forum

Laureates of mathematics and computer science meet the next generation

As part of this year’s HLF, two alumni – Janelle C. Mason and Jaqueline Godoy Mesquita – interviewed 2019 Abel prize winner Karen Uhlenbeck. The first woman to receive the prize, Uhlenbeck talked about her career and the opportunities she’s had, as well as her work. When asked about her research, she explained how she got interested in the subject of global analysis.

Early in her career, Karen Uhlenbeck was interested in a lot of different kinds of mathematics – “I found almost everything very interesting – it doesn’t matter what it was.” But when she started her MA and PhD studies at Brandeis University with her supervisor Richard Palais, she discovered the new and emerging field of global analysis and found it appealed to her. It crosses between two areas of maths – analysis, the study of functions, and topology, the study of manifolds.

A manifold is roughly an object which, when you look at it close up, looks flat. For example, a sphere is very much not a flat object, but if you zoom in on one part of a sphere, you can see what looks like a flat surface – ‘locally’, as a topologist would say, it looks and behaves like a normal 2-dimensional flat plane.

This can be seen in the way that we map the surface of the earth: some people might use an atlas – a collection of maps or charts, each showing one small flat portion of the surface; others might use a globe to see how the whole thing connects up. Topologists use these same words – they literally describe a manifold as being defined using an ‘atlas’ of ‘charts’ and having a ‘global’ structure.

A sphere is an example of a 2-dimensional manifold (also called a ‘surface’). Other manifolds might locally look like ordinary 3-dimensional space – but globally they might join up in a way that can’t be seen or understood just by looking at one small part. We call a manifold that’s locally n-dimensional, an n-dimensional manifold or n-manifold.

We also require that manifolds have this ‘locally flat’ property everywhere – for example, a circle is a 1-manifold, as every part of it looks like a 1-dimensional piece of straight line, but a figure-of-eight shape isn’t a 1-manifold, as the crossing point in the middle doesn’t look like a straight line. This property allows manifolds to be studied and understood at the simple, local level, even if their global structure is very complex.

In the same way we can define a function on a set of numbers, or on points in a plane, we can define functions on the surface of a manifold, which means we can apply things like calculus to them in order to study how they behave (and this might also tell us something about the manifold itself).

Once we have n-dimensional manifolds, it’s possible to extend this notion to an infinite-dimensional manifold, which would be something that locally behaves like infinite-dimensional space – such as a space of functions. For example, the set of all maps between the interval [0,1] and itself – which map numbers between 0 and 1 onto other numbers between 0 and 1 – can be considered as an infinite dimensional space. Global analysis, which Karen Uhlenbeck found herself studying back in the 1960s, is what happens when infinite dimensional manifolds meet functional analysis.

Describing it as a “faddish and very exciting field”, Uhlenbeck recalls how people discovered that a lot of the calculus you did on finite dimensional manifolds could be done on infinite dimensional manifolds instead. At a time when the field of topology had recently gone through a large amount of development and expansion, people were looking for new ways to apply these ideas, and global analysis used techniques in infinite-dimensional manifold theory to classify the behavior of differential equations – ODEs and PDEs. This allows you to solve optimisation problems, and has applications in physics and quantum field theory.

At the time, this field was emerging, and Uhlenbeck was able to be part of the team which did a lot of the early work in establishing it. “I just got really excited,” she says – and it’s good to know that someone who is now considered one of the founders of modern geometric analysis did so with the glee of getting to play with a subject they’re passionate about.

### Posted by Katie Steckles

is a mathematician based in Manchester, who gives talks and workshops on different areas of maths. She finished her PhD in 2011, and since then has talked about maths in schools, at science festivals, on BBC radio, at music festivals, as part of theatre shows and on the internet. Katie writes blog posts and editorials for The Aperiodical, a semi-regular maths news site.

1. Katie Steckles wrote (02. Oct 2020):
> […] A manifold is roughly an object which, when you look at it close up, looks flat. […] We also require that manifolds have this ‘locally flat’ property everywhere

Does the surface of a cube constitute a (2-dimensional) manifold ? …

2. It’s complicated 🙂 A cube is a topological manifold, because it can be stretched out into e.g. a sphere without cutting or gluing (what we’d call homomorphic to a sphere); however, most topologists are interested in things called ‘smooth manifolds’, where the maps all have to be continuous and satisfy some conditions that mean they’re well-behaved (we’d say something with this property is diffeomorphic to a sphere, like a bigger sphere, or an ellipsoid) – but a cube isn’t, because of the corners and edges where it doesn’t work. Hope that helps!

3. Katie Steckles wrote (02.10.2020, 15:48 o’clock):
> […] Hope that helps!

At least your comment begslends itself to follow up on it. (And sorry that I didn’t get around to do that sooner.)

> [The surface of a] cube is a topological manifold, because it can be stretched out into e.g. a sphere without cutting or gluing

Yes: (the surface of) a cube and a sphere (being the surface of a ball) can be brought into point-to-point correspondence such that they are topologically equivalent (in the sense of a [[homeomorphism]], as far as I understand).
Their distinction is therefore (“only”) metric, with (generally) differing distance ratios between corresponding pairs of points; and by suitably changing many of these distances together (even “always gradually”), one can be deformed into the other.

But: apparently, this is not exactly the justification for calling a (surface of a) cube a “topological manifold”. Instead, there is a dedicated, intrinsic definition for that. And a sphere, on its own, is called a “topological manifold” on the same grounds.

> however, most topologists are interested in things called ‘smooth manifolds’, where the maps all have to be continuous and satisfy some conditions that mean they’re well-behaved (we’d say something with this property is diffeomorphic to a sphere […])

Indeed, in the definition of smooth manifolds (as special cases of differentiable manifolds) an essential (the only?) condition is that all so-called transition maps between any possible two coordinate charts are continuous (see “p.s.” below).

And yes: apparently, a sphere can be covered with coordinate charts whose (mutual) transition maps are all continuous; and there is some “maximal continuous atlas” of all such coverings of a sphere.

> but a cube isn’t, because of the corners and edges where it doesn’t work.

Now, let’s consider a sphere with its “maximal continuous atlas” assigned to its points (in their entirety), and deform it into (the surface of) a cube as considered above: “without cutting or gluing”, “always gradually”. The assignments of coordinates to each point of the (surface of the) cube remain just as they were to the corresponding point of the sphere (before deformation). Then — what exactly “doesn’t work” ?? (Foremost considering only the edges of the cube, if applicable.)

> It’s complicated […]

It’s not rocket science, is it?

p.s. — SciLogs comment $$\LaTeX$$ test:

$$\forall \, \mathcal U_{\alpha}, \mathcal U_{\beta} \subset \mathcal M \, | \, (\mathcal U_{\alpha} \cap \mathcal U_{\beta}) \neq \emptyset :$$
$$\exists \, \phi_{\alpha} : \mathcal U_{\alpha} \longleftrightarrow \mathcal R_{\alpha} \subset \mathbb R^2 \text{ and}$$
$$\exists \, \phi_{\beta} : \mathcal U_{\beta} \longleftrightarrow \mathcal R_{\beta} \subset \mathbb R^2 \text{ and}$$
$$\exists \, \phi_{\beta \alpha} \equiv \phi_{\alpha \beta}^{-1} : (\phi_{\alpha}[ \, (\mathcal U_{\alpha} \cap \mathcal U_{\beta}) \, ]) \subset \mathcal R_{\alpha} \longleftrightarrow (\phi_{\beta}[ \, (\mathcal U_{\alpha} \cap \mathcal U_{\beta}) \, ]) \subset \mathcal R_{\beta} \text{ with}$$
$$\forall \, p \in (\mathcal U_{\alpha} \cap \mathcal U_{\beta}) : \phi_{\beta \alpha}[ \, \phi_{\alpha}[ \, p \, ] \, ] \mapsto \phi_{\beta}[ \, p \, ] \text{ such that}$$
$$\forall \, (x, y) \in (\phi_{\alpha}[ \, (\mathcal U_{\alpha} \cap \mathcal U_{\beta}) \, ]) :$$
$$\exists \, \text{Limit}_{\{ \epsilon \rightarrow 0 \}} \! \! \left[ \frac{(\phi_{\beta \alpha}[ \, (x + \epsilon, \, y) \, ] – \phi_{\beta \alpha}[ \, (x, \, y) \, ])}{\epsilon} \right] \text{ and}$$
$$\exists \, \text{Limit}_{\{ \epsilon \rightarrow 0 \}} \! \! \left[ \frac{(\phi_{\beta \alpha}[ \, (x, \, y + \epsilon) \, ] – \phi_{\beta \alpha}[ \, (x, \, y) \, ])}{\epsilon} \right]$$
etc.

4. @Frank Wappler

It’s somewhat surprising to notice your new explorations in the world of local coordinates. 😉

In case you’re asking whether it is possible to endow the surface of a cube with the structure of a smooth manifold, the answer is yes. To see how this can be done, you may find these notes helpful.

However, for this purpose the surface of the cube has to be considered as an abstract topological manifold and, of course, not as a hypersurface in a Euclidean space.

5. Chrys wrote (12.12.2020, 00:22 o’clock):
> @Frank Wappler It’s somewhat surprising to notice your new explorations in the world of local coordinates.

You seem to mistake a new opportunity to express explorations (in a way that you finally notice) for “new explorations”.

> In case you’re asking whether it is possible to endow the surface of a cube with the structure of a smooth manifold, the answer is yes.

As far as my preceding comment (09.12.2020, 11:48 o’clock) may be construed as asking this question, your reply may be construed as calling into question the applicable parts of Katie Steckles’ comment (02. Oct 2020); not unlike my comment had done (or had at least attempted to do) already.

> […] However, for this purpose the surface of the cube has to be considered as an abstract topological manifold

By considering both as abstract topological manifolds, there appears no remaining distinction between “the surface of a cube” and “a sphere”; Ockham’s razor can and should be used accordingly.
(Btw., I’ve occasionally wondered whether and why it is said that such a topological manifold could be endowed uniquely with the “one corresponding” structure of a smooth manifold …)

> and, of course, not as a hypersurface in a Euclidean space.

Indeed, the above considerations on manifolds and smoothness (or, to acknowledge Katie Steckles’ terminology, “being continuous and well-behaved”) seem to stand quite independently from any “finer details” of distance ratios, ratios of arc lenghts, embeddings, curvature and the like. (The remaining “coarse” requirement would be merely equivalence of, or at least consistency between, the topology which is “abstractly assumed”, and the topology which is induced by measured distance ratios.)

p.s.
Contrariwise concentrating on distance ratios etc., by which a cube can be distinguished from a ball, for instance, we may still speak of “the extrinsic surface of a cube” as the border which an embedded (Euclidean) cube has wrt. ambient (Euclidean) space. We may then consider curves restricted to this “extrinsic surface” and define distance between two points “on this extrinsic surface” as infimum of lengths of curves connecting these two points. But I find it difficult to come up with a matching intrinsic definition of (some metric space constituting) “a surface” which “the extrinsic surface of a cube” would satisfy, too. (And that’s due to “the corners”, not “the edges”.)

6. @Frank Wappler / 14.12.2020, 18:20 o’clock

As you might have seen from Nicolaescu’s notes, the smoothing is carried out in the conceptual framework of abstract manifolds, irrespective of any aspects concerning their possible representation as submanifolds of a Euclidean space.

Clearly, the two surfaces in question are conveniently introduced as subsets of the Euclidean 3-space, but for any given subset $$A \subset \mathbb{R}^3$$ we also have a canonical inclusion map $$\iota_A : A \hookrightarrow \mathbb{R}^3$$. This interpretation of the subset relation allows us to distinguish formally between $$A$$ as an abstract set and its representation $$\iota(A)$$ as a subset, which appears suitable for the situation at hand.

Hence we may now introduce the Euclidean 2-sphere, $$S^2$$, by setting $$\iota_{S^2}(S^2) = \left\{x \in \mathbb{R}^3 \mid \|x\|_2 = 1\right\}$$, and, similarly, the surface of the cube, $$\Sigma^2$$, by something like $$\iota_{\Sigma^2}(\Sigma^2) = \left\{x = x^i e_i \in \mathbb{R}^3 \mid \|x\|_\infty = \max\{|x^i|\} = 1\right\}$$.

Then, having established a smooth structure on $$\Sigma^2$$ the way described in the notes, the remaining distinction you’re apparently asking for is seen from the two inclusion maps: whereas $$\iota_{S^2}$$ is smooth, $$\iota_{\Sigma^2}$$ is still merely continuous.

• p.s. For the sake of correctness, I obviously lost a subscript in $$\iota_A(A)$$ and, moreover, should have emphasized “for any given non-empty subset” … sorry!

7. Chrys wrote (16.12.2020, 12:53 o’clock):
> Clearly, the two surfaces in question are conveniently introduced as subsets of the Euclidean 3-space […]

Sure (see “p.s.” below), and thanks for the follow-up, but …

> […] we also have a canonical inclusion map [for each of these two surfaces …] $$\iota_A : A \hookrightarrow \mathbb{R}^3$$ […]

… sorry: the definition of (how to determine whether a given manifold is a) “smooth manifold” — (which I shall presume Katie Steckles had in mind, unless and until she informs us otherwise. Wouldn’t you?) — is explicitly concerned with
smoothness of any transition maps $$\phi_{\beta\alpha} : \phi_{\alpha}[ \, (\mathcal U_{\alpha} \cap \mathcal U_{\beta}) \, ] \subset \mathbb R^2 \longleftrightarrow \phi_{\beta}[ \, (\mathcal U_{\alpha} \cap \mathcal U_{\beta}) \, ] \subset \mathbb R^2$$ between any overlapping charts (homeomorphisms $$\phi_{\alpha}$$ and $$\phi_{\beta}$$) of open subsets of the manifold under consideration;
and not (or only implicitly) with properties of “its canonical inclusion map” (for whatever might count as “canonical”).

Nevertheless …

> […] whereas [one] is smooth, [the other] is still merely continuous.

… I’d appreciate if you could please spell out which specfic evaluation (defined for any generic $$\iota_A$$, as stated) would lead to the exact conclusion of $$\iota_{S^2}$$ being smooth, and $$\iota_{\Sigma^2}$$ not.

p.s.:

A particular sphere (surface of a ball) would conveniently be introduced as the subset of Euclidean 3-space $$(\mathcal X, d_{\text{E}_3})$$ whose points have all equal (und generally nonzero) distance from one particular selected other point.

Now, to state this formally und explicitly as a set … takes a bit of “gymnastics” … How about:

$$S^2 \equiv \{ x \in \mathcal F_m \subset \mathcal X :$$
$$(\forall a \in \mathcal F_m \, \mid \, d_{\text{E}_3}[ \, x, m \, ] = d_{\text{E}_3}[ \, a, m \, ]) \text{ and }$$
$$(\forall p \in \mathcal X \, | \, (d_{\text{E}_3}[ \, x, m \, ] = d_{\text{E}_3}[ \, p, m \, ]) \Rightarrow (p \in F_m)) \}$$,

where point $$m \in \mathcal X$$ is fixed, of course, and set $$\mathcal F_m \subset \mathcal X$$ is supposed to be fixed as well and thereby turns out to be the same as set $$\mathcal S$$.

It’s also possible (and arguably convenient) to dispense with the additional fixed point $$m$$ and to require instead that the radii of the circumspheres of any four points of set $$\mathcal S$$ are equal (unless these four point happen to be plane wrt. each other, and so on).

A particular surface of a cube $$\Sigma^2$$ could (conveniently) be introduced piecewise:
$$\mathcal T_{abc} \equiv \{ x \in \mathcal F_{abc} \subset \mathcal X :$$
$$(\text{CMD}[ \, a, b, c, x \, ] = 0) \text{ and }$$
$$(\sqrt{ \text{CMD}[ \, a, b, c \, ] } = \sqrt{ \text{CMD}[ \, a, b, x \, ] } + \sqrt{ \text{CMD}[ \, a, x, c \, ] } + \sqrt{ \text{CMD}[ \, x, b, c \, ] }) \text{ and }$$
$$(\forall p \in \mathcal X \, | \, ((\text{CMD}[ \, a, b, c, p \, ] = 0) \text{ and } \( (\sqrt{ \text{CMD}[ \, a, b, c \, ] } = \sqrt{ \text{CMD}[ \, a, b, p \, ] } + \sqrt{ \text{CMD}[ \, a, p, c \, ] } + \sqrt{ \text{CMD}[ \, p, b, c \, ] })) \Rightarrow (p \in F_{abc})) \}$$,

where $$\mathcal T_{abc}$$ denotes one of twelve triangles which make up the surface of the cube, the distance ratios between its three corner points are (without loss of generality)
$$\frac{d_{\text{E}_3}[ \, a, c \, ]}{d_{\text{E}_3}[ \, a, b \, ]} = \frac{d_{\text{E}_3}[ \, a, c \, ]}{d_{\text{E}_3}[ \, b, c \, ]} = \sqrt{2}$$,
$$\text{CMD}$$ denotes the Cayley-Menger-Determinant of all distances between the points given as arguments, and
the given metric space $$(\mathcal X, d_{\text{E}_3})$$ is “Euclidean 3-space” in particular due to the distance ratios being such that the Cayley-Menger-Determinants for any five (or more) points are vanishing, of course.

8. Frank Wappler wrote (18.12.2020,19:03 o’clock):
> […] A particular surface of a cube $$\Sigma^2$$ could (conveniently) be introduced piecewise:
$$\mathcal T_{abc} \equiv \{ x \in \mathcal F_{abc} \subset \mathcal X :$$
$$(\text{CMD}[ \, a, b, c, x \, ] = 0) \text{ and }$$
$$(\sqrt{ \text{CMD}[ \, a, b, c \, ] } = \sqrt{ \text{CMD}[ \, a, b, x \, ] } + \sqrt{ \text{CMD}[ \, a, x, c \, ] } + \sqrt{ \text{CMD}[ \, x, b, c \, ] }) \text{ and }$$
$$(\forall p \in \mathcal X \, | \, ((\text{CMD}[ \, a, b, c, p \, ] = 0) \text{ and }$$
$$(\sqrt{ \text{CMD}[ \, a, b, c \, ] } = \sqrt{ \text{CMD}[ \, a, b, p \, ] } + \sqrt{ \text{CMD}[ \, a, p, c \, ] } + \sqrt{ \text{CMD}[ \, p, b, c \, ] })) \Rightarrow (p \in F_{abc})) \}$$,

> where $$\mathcal T_{abc}$$ denotes one of twelve triangles which make up the surface of the cube, […]

• p.p.s. For the sake of correctness and completeness, the formula above should end with:
… $$\Rightarrow (p \in F_{abc})) \}$$;
and in the preceding comment (18.12.2020,19:03 o’clock) I’ve erroneously used the distinct symbols $$S^2$$ and $$\mathcal S$$ for denoting what I meant to be the same set.

9. Frank Wappler wrote (18.12.2020, 19:03 o’clock):
> […] … takes a bit of “gymnastics” … […] set $$\mathcal F_m \subset \mathcal X$$ is supposed to be fixed […]

Much easier, and “in hindsight obvious”, is instead to initially fix not only “the midpoint $$m \in \mathcal X$$ but also just one additional point $$r \in \mathcal X$$, $$r \not\equiv m$$, in order to “define the radius $$d_{\text{E}_3}[ \, r, m \, ] \neq 0$$”.

Similarly, in case of defining the surface of a cube, it was sufficient to fix its eight corner points in terms of their mutual distances, and the “additional gymnastics” involving “fixed sets $$\mathcal F_{(\text{triangle})}” turned out to be superfluous from the start. And similarly, a sphere is defined by four initially given points which are not plane wrt. each other. p.s. Chrys schrieb (16.12.2020, 12:53 o’clock): > […] the remaining distinction you’re apparently asking for […] … arises not from consideration of the “ambient distances \(d_{\text{E}_3}$$” but of “intrinsic distances”; i.e. in case of the sphere its “great circle arc lengths” between any two of its points, and in case of the “surface of the cube” the lengths of the shortest “polygon drawn on the surface” (cmp. my comment 14.12.2020, 18:20 o’clock).

10. @Frank Wappler / 18.12.2020, 19:03 o’clock

Well, the surface of a cube is foremostly a geometric shape that had already attracted the attention of ancient Greek scholars. The same can be stated for the sphere, of course.

Relating the cubic shape — intuitively visualized by a surface in 3-space — to the topological concept of manifolds is most easily achieved by specifying a 2-dimensional submanifold according to $$\mathfrak{S}^2_\infty := \left\{x \in \mathbb{R}^3 \mid \|x\|_\infty = 1\right\}$$. Let me briefly refer to this object as the cubic (unit) sphere.

This can analogously be done for the spherical shape, and in that case we obtain what is known as the round (unit) sphere, $$\mathfrak{S}^2_2 := \left\{x \in \mathbb{R}^3 \mid \|x\|_2 = 1\right\}$$.

Note that so far I have merely specified two subsets in $$\mathbb{R}^3$$. But each of these becomes a topological subspace together with the usual subspace topology inherited from the ambient Euclidean space, and then it’s not difficult to see that the map $$h:\mathfrak{S}^2_\infty \to \mathfrak{S}^2_2$$, def. by $$h(x) := x/\|x\|_2$$, is a homeomorphism.

Furthermore, for the round sphere the structure of a submanifold can be deduced from Example 1.2 in Nicolaescu’s notes. That is to say, the structure of a smooth manifold is being established on the preimage of the inclusion map $$\iota_{S^2}$$ such that, in turn, this map becomes a smooth embedding and, therefore, its image (the round sphere) is a submanifold.

For the cubic sphere the construction with the stereographic projections fails. However, we have a homeomorphism $$f := \iota^{-1}_{S^2} \circ h \circ \iota_{\Sigma^2} : \Sigma^2 \to S^2$$, which implies that $$\Sigma^2$$ is a topological manifold and $$\iota_{\Sigma^2}$$ is a topological embedding whose image (the cubic sphere) is thus a submanifold.

What Katie Steckles was apparently trying to explain to you is that the two geometric shapes are homeomorphic as submanifolds in the Euclidean 3-space. But not diffeomorphic, because here your run into trouble at the edges and corners of the cubic shape. The smoothing of the cubic sphere, on the other hand, is performed for abstract manifolds, i.e., for the preimages of the two inclusion maps (not their respective images), whereby no reference to geometric shape is retained. It is just the particular map $$\iota_{\Sigma^2}$$ that produces the cubic shape as a geometric property of its image. And this map, along with its image, remains unchanged, it is not affected by the smoothing procedure.

11. Chrys wrote (19.12.2020, 23:38 o’clock):
> […] we have a homeomorphism $$f := \iota^{-1}_{S^2} \circ h \circ \iota_{\Sigma^2} : \Sigma^2 \to S^2$$

Alright, for a homeomorphism (which had been mentioned already in comment 09.12.2020, 11:48 o’clock, btw.) we’d like to have a bijection $$\Sigma^2 \longleftrightarrow S^2$$;
and making use of your explicit construction of function $$h$$ (regarding the question of its invertibility see below):
$$\Sigma^2 \longleftrightarrow \mathfrak S^2_{\infty} \overset{h}{\longleftrightarrow} \mathfrak S^2_{\2} \longleftrightarrow S^2$$.

(The first and the third/final bijection can and should be spelt out just as explicitly, based on the explicit definitions of $$\Sigma^2$$ and $$S^2$$ as sets with distinctive “additional structures”.)

> the structure of a smooth manifold is being established on the preimage of the inclusion map $$\iota_{S^2}$$ […]

In other words: the structure of a smooth manifold is being established on $$S^2$$; and therefore the structure of a topological manifold must have been established on $$S^2$$ in the first place.

> we have homeomorphism $$f$$ […] which implies that $$\Sigma^2$$ is a topological manifold

I disagree, as I did already above (09.12.2020, 11:48 o’clock):
the structure of a topological manifold can and must have been established on $$\Sigma^2$$ by “first principles” (codified by the definition of a [[topological manifold]] I had linked above), too. Only on that basis we may ask whether, and eventually determine that, by function $$f$$ we do indeed “have a homeomorphism”.

> […] For the cubic sphere the construction with the stereographic projections fails.

Do you mean that there didn’t exist an inverse function to your $$h$$ ?
Well, borrowing the decomposition $$x := x^j \, e_j$$ from your earlier comment

$$h^{(-1)} : \mathfrak S^2_{\2} \to \mathfrak S^2_{\infty}, \qquad h^{(-1)}[ \, x \, ] \mapsto x / \text{Max}[ \, | x^j | \, ]$$

?

> What Katie Steckles was apparently trying to explain to you

… see comment 02.10.2020, 15:48 o’clock …

> is that the two geometric shapes are homeomorphic as submanifolds in the Euclidean 3-space. […]

… this I had essentially acknowledged already in comment 09.12.2020, 11:48 o’clock …

> But not diffeomorphic, because here your run into trouble at the edges and corners of the cubic shape.

That’s still only Katie Steckles’ formulation … which I like to have clarified. I repeat that there is no mentioning of “canonic inclusion maps” in Wikipedia article https://en.wikipedia.org/wiki/Diffeomorphism; and there exist arguably more “well-behaved” assignments of coordinates especially to neighborhoods (open sets) containing pieces of any edge of a cubic surface, than the coordinate assignment implied by set $$\mathfrak S^2_{\infty}$$.

12. Chrys wrote (19.12.2020, 23:38 o’clock):
> […] we have a homeomorphism $$f := \iota^{-1}_{S^2} \circ h \circ \iota_{\Sigma^2} : \Sigma^2 \to S^2$$

Alright, for a homeomorphism (which had been mentioned already in comment 09.12.2020, 11:48 o’clock, btw.) we’d like to have a bijection $$\Sigma^2 \longleftrightarrow S^2$$;
and making use of your explicit construction of function $$h$$ (regarding the question of its invertibility see below):
$$\Sigma^2 \longleftrightarrow \mathfrak S^2_{\infty} \overset{h}{\longleftrightarrow} \mathfrak S^2_{2} \longleftrightarrow S^2$$.

(The first and the third/final bijection can and should be spelt out just as explicitly, based on the explicit definitions of $$\Sigma^2$$ and $$S^2$$ as sets with distinctive “additional structures”, involving values of Euclidean distance, $$d_{\text{E}_3}$$.)

> the structure of a smooth manifold is being established on the preimage of the inclusion map $$\iota_{S^2}$$ […]

In other words: the structure of a smooth manifold is being established on $$S^2$$; and therefore the structure of a topological manifold must have been established on $$S^2$$ in the first place.

> we have homeomorphism $$f$$ […] which implies that $$\Sigma^2$$ is a topological manifold

I disagree, as I did already above (09.12.2020, 11:48 o’clock):
the structure of a topological manifold can and must have been established on $$\Sigma^2$$ by “first principles” (codified by the definition of a [[topological manifold]] I had linked above), too. Only on that basis we may ask whether, and eventually determine that, by function $$f$$ we do indeed “have a homeomorphism”.

> […] For the cubic sphere the construction with the stereographic projections fails.

Do you mean that there didn’t exist an inverse function to your $$h$$ ?
Well, borrowing the decomposition $$x := x^j \, e_j$$ from your earlier comment

$$h^{(-1)} : \mathfrak S^2_{2} \to \mathfrak S^2_{\infty}, \qquad h^{(-1)}[ \, x \, ] \mapsto x / \text{Max}[ \, | x^j | \, ]$$

?

> What Katie Steckles was apparently trying to explain to you

… see comment 02.10.2020, 15:48 o’clock …

> is that the two geometric shapes are homeomorphic as submanifolds in the Euclidean 3-space. […]

… this I had essentially acknowledged already in comment 09.12.2020, 11:48 o’clock …

> But not diffeomorphic, because here your run into trouble at the edges and corners of the cubic shape.

That’s still only Katie Steckles’ formulation … which I like to have clarified. I repeat that there is no mentioning of “canonic inclusion maps” in Wikipedia article https://en.wikipedia.org/wiki/Diffeomorphism; and there are arguably more “well-behaved” assignments of coordinates especially to neighborhoods (open sets) containing pieces of any edge of a cubic surface, than the coordinate assignment implied by set $$\mathfrak S^2_{\infty}$$.

13. Frank Wappler / 22.12.2020, 02:31 o’clock

By definition, $$A \subset \mathbb{R}^3$$ means that $$x \in A \Rightarrow x \in \mathbb{R}^3$$ for all $$x \in A$$. Assuming $$A \neq \emptyset$$ (to avoid trivial pathologies), we can define the canonical inclusion (or injection) map $$\iota_A: A \to \mathbb{R}^3$$ by $$\iota_A(x):= x$$ for all $$x \in A$$. Okay?

Please note that I’m using the mathfrak symbols to denote $$\mathfrak{S}^2_\infty = \iota_{\Sigma^2}(\Sigma^2)$$ and $$\mathfrak{S}^2_2 = \iota_{S^2}(S^2)$$ just for the sake of convenience (to minimize the risk of trivial typos).

Considering both the cubic sphere and the round sphere equipped with the subspace topology induced by the ambient Euclidean space, do you agree that the previously introduced map $$h:\mathfrak{S}^2_\infty \to \mathfrak{S}^2_2$$ is a homeomorphism (of topolgical subspaces)? (And, indeed, it’s inverse given by $$h^{-1}(x) = x/\|x\|_\infty$$.)

Now we adopt the subspace topology for $$S^2$$ and $$\Sigma^2$$, respectively, such that $$\iota_{S^2}: S^2 \to \mathfrak{S}^2_2$$ becomes a homeomorphism, and likewise for $$\Sigma^2$$. So the aforementioned map $$f:\Sigma^2 \to S^2$$ is a homeomorphism. Okay?

An atlas $$\mathfrak{A}$$ on $$S^2$$, consisting of two local charts, is explicitly described by Nicolascu in Example 1.2. By pulling back this atlas to $$\Sigma^2$$ via $$f$$ one gets the atlas $$f^*\mathfrak{A}$$ on $$Sigma^2$$, which shows that this space is at least a topological manifold. The canonical inclusion $$\iota_{\Sigma^2} = h^{-1} \circ \iota_{S^2} \circ f$$ is continuous and injective, hence it is a topological embedding whose image (the cubic sphere) is a topological submanifold of the Euclidean 3-space.

Next you may wish to check yourself that the atlas $$\mathfrak{A}$$ on $$S^2$$ is actually smooth. In addition, observing that the chart mappings are composed of the canonical inclusion and a stereographic projection, we infer that $$\iota_{S^2}$$ is a smooth embedding whose derivative (or tangent map) at $$x$$ is a linear map $$T_x\iota_{S^2}:T_xS^2 \to T_x\mathfrak{S}^2_2$$ from the tangent space at $$x \in S^2$$ onto the tangent plane to the round sphere at the corresponding point $$x = \iota_{S^2}(x) \in \mathfrak{S}^2_2$$.

Finally, as a by-product we see that the pull-back atlas $$f^*\mathfrak{A}$$ on $$Sigma^2$$ is smooth such that, with respect to this differentiable structure, $$\Sigma^2$$ is a smooth manifold and $$f$$ is a diffeomorphism.

As a consequence, you have a tangent space $$T_x\Sigma^2$$ at every $$x \in \Sigma^2$$ and a nice tangent map $$T_x\iota_{\Sigma^2}$$ of rank 2 if $$x$$ corresponds to a point on one of the faces of the cube, but not if (\x\) corresponds to a point on one the edges or the corners. Roughly speaking, that’s because you can’t attach tangent planes to the cubic sphere at these points. So the canonical inclusion map is not a smooth embedding in this case.

• Frank Wappler wrote (23.12.2020,18:32 o’clock):
> […] another atlas consisting of six charts […]

To arrange it perhaps more readably, while correcting some formatting:

$$\mathfrak A_{S^2_6} \equiv \{$$
$$(\mathcal U_{(x^j \gt 0.5)}, \Phi_{(x^j \gt 0.5)} : \mathcal U_{(x^j \gt 0.5)} \longleftrightarrow \{ (x^k, x^l) \})$$,
$$(\mathcal U_{(x^k \gt 0.5)}, \Phi_{(x^k \gt 0.5)} : \mathcal U_{(x^k \gt 0.5)} \longleftrightarrow \{ (x^j, x^l) \})$$,
$$(\mathcal U_{(x^l \gt 0.5)}, \Phi_{(x^l \gt 0.5)} : \mathcal U_{(x^l \gt 0.5)} \longleftrightarrow \{ (x^j, x^k) \})$$,
$$(\mathcal U_{(x^j \lt -0.5)}, \Phi_{(x^j \lt -0.5)} : \mathcal U_{(x^j \lt -0.5)} \longleftrightarrow \{ (x^k, x^l) \})$$,
$$(\mathcal U_{(x^k \lt -0.5)}, \Phi_{(x^k \lt -0.5)} : \mathcal U_{(x^k \lt -0.5)} \longleftrightarrow \{ (x^j, x^l) \})$$ and
$$(\mathcal U_{(x^l \lt -0.5)}, \Phi_{(x^l \lt -0.5)} : \mathcal U_{(x^l \lt -0.5)} \longleftrightarrow \{ (x^j, x^k) \})$$
$$\}$$.

> And I’m under the impression that coordinates assigned to the topological manifold surface $$\Sigma^2$$ quite explicitly as
$$\Phi_{\Sigma^2}[ \, p \, ] \mapsto \Phi[ \, \mathfrak A_{S^2_6} \, ] \circ f^{(-1)}[ \, p \, ]$$
> seem to “work” just fine around edges and corners, too.

14. Chrys wrote (23.12.2020, 00:07 o’clock):
> […] Finally, as a by-product we see that the pull-back atlas $$f * \mathfrak A$$ on $$\Sigma^2$$ is smooth such that, with respect to this differentiable structure, $$\Sigma^2$$ is a smooth manifold and $$f$$ is a diffeomorphism. As a consequence, you have a tangent space $$T_x \Sigma^2$$ at every $$x \in \Sigma^2$$

Very well; pretty much as I thought, too; and in particular including each point of $$\Sigma^2$$ on edges
($$\{ x : 0 = (1 – |x^j|)^2 (1 – |x^k|)^2 + (1 – |x^j|)^2 (1 – |x^l|)^2 + (1 – |x^l|)^2 (1 – |x^l|)^2 \}$$) or corners
($$\{ x : 0 = (1 – |x^j|)^2 + (1 – |x^k|)^2 + (1 – |x^l|)^2 \}$$).

So how does this conclusion agree with comment 02.10.2020, 15:48 o’clock:

called ‘smooth manifold[…]’, where the maps all have to be continuous and satisfy some conditions that mean they’re well-behaved (we’d say something with this property is diffeomorphic to a sphere, like a bigger sphere, or an ellipsoid) – [… the surface of] a cube isn’t

??

> [but] a nice tangent map $$T_x \iota_{\Sigma^2}$$ […] not if $$x$$ corresponds to a point on one the edges or the corners.

Right. But: Is this a property of the manifold (the surface of a cube, with its “natural” topology, and some suitable atlas which clearly exists) ?
Or rather a property of “its inclusion” (how ever “canonical” or conventional it may be), and consequently dependent on coordinates assigned to “ambient space” ?

Consider transforming the coordinates of (some suitable subset $$\mathcal S$$, containing $$\Sigma^2$$ as well as $$S^2$$, of) “the (Cartesian) $$\mathbb R^3$$ space” under consideration since you introduced “it” here (16.12.2020, 12:53 o’clock):

$$\phi : \mathcal S \subset \mathbb R^3 \rightarrow \mathbb R^3, \qquad \phi[ \, x \, ] \mapsto x / \sqrt{ (\| x \|_2)^2 + (1 – \text{Max}[ \, | x^j | \, ])^2 }$$,

leaving the Euclidean distances (between points to which the coordinate tuples $$\mathcal R^3$$ had been assigned initially) just as they were …

p.s.
> Considering both the cubic sphere and the round sphere equipped with the subspace topology induced by the ambient Euclidean space […]

That’s sufficient to conclude that both the cubic sphere (implied by set $$\mathfrak S^2_{\infty}$$ of Cartesian coordinates assigned to Euclidean space) and the round sphere (implied by set $$\mathfrak S^2_{2}$$ of Cartesian coordinates assigned to Euclidean space) are both individually topological manifolds. There is no need to “show again that $$\Sigma^2$$ is at least a topological manifold” by appealing to $$h$$ and $$f$$ being homeomorphisms.

> [ … $$h : \mathfrak S^2_{\infty} \longleftrightarrow \mathfrak S^2_{2}, \qquad h[ \, x \, ] \mapsto x / \| x \|_2$$ … ]
> it’s inverse given by $$h^{-1}(x) = x / \|x\|_{\infty}$$

I doubt that because, if I understand correctly, $$\|x\|_{\infty}$$ has value $$0$$ on all but six point of set $$\mathfrak S^2_{2}$$;
and I stand by the definition of the inverse of function $$h$$ from above (here it bit more explicit):

$$h^{(-1)} : \mathfrak S^2_{2} \longleftrightarrow \mathfrak S^2_{\infty}, \qquad h^{(-1)}[ \, x \, ] \mapsto x / \text{Max}[ \, \{ | c | : c \in (x^j, x^k, x^l) \equiv x \} \, ]$$.

> […] An atlas $$\mathfrak A$$ on $$S^2$$, consisting of two local charts, is explicitly described by Nicolascu in Example 1.2.

… two charts, each consisting of an open set given as the entire sphere set without one of the two “poles” (or without the other, resp.) and coordinates correspondingly assigned through the stereographic projection “from the removed pole”.

That’s certainly good enough for reaching the conclusion discussed at the beginning of this comment.
However, meanwhile I had thought of another atlas consisting of six charts (in correspondence to “the symmetry of a cube”) which are very explicit and self-explainingly named (in reference to $$\mathfrak S^2_2$$):

$$\mathfrak A_{S^2_6} \equiv \{$$
$$(\mathcal U_{(x^j \gt 0.5)}, \Phi_{(x^j \gt 0.5)} : U_{(x^j \gt 0.5)} \longleftrightarrow \{ (x^k, x^l) \})$$,
$$(\mathcal U_{(x^j \lt -0.5)}, \Phi_{(x^j \lt -0.5)} : U_{(x^j \lt -0.5)} \longleftrightarrow \{ (x^k, x^l) \})$$,
$$(\mathcal U_{(x^k \gt 0.5)}, \Phi_{(x^k \gt 0.5)} : U_{(x^k \gt 0.5)} \longleftrightarrow \{ (x^j, x^l) \})$$,
$$(\mathcal U_{(x^k \lt -0.5)}, \Phi_{(x^k \lt -0.5)} : U_{(x^k \lt -0.5)} \longleftrightarrow \{ (x^j, x^l) \})$$,
$$(\mathcal U_{(x^l \gt 0.5)}, \Phi_{(x^l \gt 0.5)} : U_{(x^l \gt 0.5)} \longleftrightarrow \{ (x^j, x^k) \})$$, and
$$(\mathcal U_{(x^l \lt -0.5)}, \Phi_{(x^l \lt -0.5)} : U_{(x^l \lt -0.5)} \longleftrightarrow \{ (x^j, x^k) \})$$
$$\}$$.

And I’m under the impression that coordinates assigned to the topological manifold surface $$\Sigma^2$$ quite explicitly as
$$\Phi_{\Sigma^2}[ \, p \, ] \mapsto \Phi[ \, A_{S^2_6} \, ] \circ f^{(-1)}[ \, p \, ]$$
seem to “work” just fine around edges and corners, too.

p.p.s.
Happy Holidays!
(The question of the intrinsic distinction between sphere and cubical surface is going to keep at least me entertained …)

• Frank Wappler wrote (23.12.2020,18:32 o’clock):
> […] leaving the Euclidean distances (between points to which the coordinate tuples $$\mathcal R^3$$

… should be: $$\mathbb R^3$$, of course; or perhaps more specificly: the elements of $$\mathbb R^3$$ …

> had been assigned initially) just as they were …

> p.s. […] There is no need to “show again that $$\Sigma^2$$ is at least a topological manifold” by appealing to $$h$$ and $$f$$ being homeomorphisms.

Should have been more explicit:
There is no need to “show again that $$\Sigma^2$$ is at least a topological manifold”;
and especially:
it could not be effected at all by appealing to $$h$$ and $$f$$ being homeomorphisms
(because the topology of $$\Sigma^2$$ needed to been established already in order to claim $$f$$ being a homeomorphism).

15. Frank Wappler wrote (23.12.2020, 18:32 o’clock):
> [… »indeed it’s inverse given by $$h^{−1}(x)=x/\|x\|_{\infty}$$«] I doubt that […]

No, sorry, this doubt of mine was not well-founded.
For one, regardless of how $$\|x\|_{\infty}$$ might be defined in detail (which hasn’t been stated here explicitly yet), it may indeed be intended to equal $$\text{Max}[ \, | x | \, ]$$.
And moreover, as I’ve figured out in the meantime, there’s at least one (“obvious”) explicit definition for $$\|x\|_{\infty}$$ (involving taking a limit, of course) by which it indeed turns out equal to $$\text{Max}[ \, | x | \, ]$$.

• where $$\text{Max}[ \, | x | \, ]$$ should be rather: $$\text{Max}[ \, | x^j | \, ]$$,
or, more explicitly, $$\text{Max}[ \, \{ | c | : c \in (x^j, x^k, x^l) \} \, ]$$.

16. @Frank Wappler / 23.12.2020, 18:32 o’clock ++

A concise overview on the so-called p-norms, $$\|\cdot\|_p \text{with} 1 \le p \le \infty$$, and their properties can be found in this de.wikipedia entry. (I’m slightly surprised that an exactly matching en.wikipedia version doesn’t seem to exist.) Thus, for $$p = 2,\infty$$, the subset $$\mathfrak{S}^2_p$$ is indeed nothing but the unit sphere in the normed vector space $$(\mathbb{R}^3, \|\cdot\|_p)$$.

In your original question you asked how something that is entirely determined by its geometric shape (the surface of a cube) is related to something that, according to its formal definition, does not have a geometric shape at all ( a manifold). So for your question to make sense, it should presumably be understood as “Does the surface of a cube constitute a (2-dimensional) submanifold (of the Euclidean 3-space)?” Because a submanifold of a Euclidean space it is quite naturally associated with geometric shape.

Recall that wikipedia characterizes a submanifold as a subset which has the structure of a manifold and, in addition, for which the inclusion map satisfies certain properties. Hence the answer to your question in my reformulation would be yes—in the sense that the cubic sphere $$\mathfrak{S}^2_\infty \subset \mathbb{R}^3$$, equipped with the subspace topology, is the homeomorphic image of a manifold under the inclusion map $$\iota_{\Sigma^2}:\Sigma^2 \hookrightarrow \mathbb{R}^3$$. This inclusion map thus has the property of being a topological embedding.

Note that up to this point aspects of smoothness were not taken into account. But now we may ask: Is the manifold $$\Sigma^2$$ smoothable? Again the answer is yes. This follows from what Nicolaescu is telling us by applying it to the homeomorphism $$f:\Sigma^2 \to S^2$$ that was introduced earlier.

As a third question we can then ask: Is the inclusion map $$\iota_{\Sigma^2}$$ a smooth embedding of $$\Sigma^2$$, given the smoothed structure? And here the answer is no. In other words, unlike the round sphere, the cubic sphere is not a smooth submanifold of $$\mathbb{R}^3$$. For a smooth embedding it follows as a necessary condition to have a unique tangent plane to the image at each of its points. This condition is clearly not satisfied for the cubic sphere on its edges or corners, in contrast to the case of the round sphere.

17. Chrys wrote (26.12.2020, 15:24 o’clock):

… i.e.

»Does the surface of a cube constitute a (2-dimensional) manifold ?«

in the first comment on this page (02.10.2020, 09:53 o’clock), which was based on a remark by Katie Steckles’ in her above SciLog article, as quoted in my comment …

> you asked how something that is entirely determined by its geometric shape (the surface of a cube)

… right …

> is related to something that, according to its formal definition, does not have a geometric shape at all (a manifold).

Nevertheless, Katie Steckles’ initial remark seemed to appeal to at least some aspect of “geometric shape”, namely:

[…] »that manifolds have this ‘locally flat’ property everywhere«

.

Also, even though manifolds can apparently be discussed without any explicit mentioning of geometry (but instead: of “topology”, and “differential structure”), I’m always curious about how those other notions might be related to, or even founded in, geometry. A certain topology, for instance, may be derived from the geometry of “open balls” as “basis”.

> So for your question to make sense, it should presumably be understood as “Does the surface of a cube constitute a (2-dimensional) submanifold (of the Euclidean 3-space)?” Because a submanifold of a Euclidean space it is quite naturally associated with geometric shape.

Surely this explication of my question, and the technicalities thus introduced, are not purposefully unfair. But as far as the sense of my question needs any further explaination, I’d consider the following formulation fairer:

Does the surface of a cube, given geometrically as metric subspace of Euclidean/flat 3-dimensional metric space, constitute a (2-dimensional) manifold by Katie Steckles’ criterion ?

.

> Recall that wikipedia characterizes a submanifold as a subset which has the structure of a manifold and, in addition, for which the inclusion map satisfies certain properties. Hence the answer to your question in my reformulation would be yes — in the sense that the cubic sphere $$\mathfrak S^2_{\infty} \subset \mathbb R^3$$, equipped with the subspace topology, is the homeomorphic image of a manifold under the inclusion map $$\iota_{\Sigma^2} : \Sigma^2 \hookrightarrow \mathbb{R}^3$$ .

I accept that this is an affirmative response to your reformulation of my question; but I still fail to recognize it proving what had been asked.
If you (want to) claim that function $$\iota_{\Sigma^2}$$ constitutes a homeomorphism you first need to establish that its domain (pre-image) $$\Sigma^2$$ is some specific topological space.

Now (and on this alternative I’m still completely uncertain):

– either you think of $$\Sigma^2$$ as the same as $$\mathfrak S^2_{\infty}$$, together with the topology it “inherits” from “the natural topology” of $$\mathbb R^3$$. Then the topology associated with set $$\Sigma^2$$ is established directly, it is therefore indeed a topological manifold, and any further consideration of map $$\iota_{\Sigma^2}$$ is pointless and trivial. (The choice of symbol, $$\iota$$, reminds me on “identity map” anyways).

– or you think of $$\Sigma^2$$ of a separate set (as I did), “given geometrically” (as I tried to suggest explicitly in comment 18.12.2020, 19:16 o’clock). Then, for what it’s worth, you may spell out map $$\iota_{\Sigma^2}$$ more explicitly (“by suitable means”, for instance

• referring to the distance ratios I used in the mentioned comment,
• declaring some notion of distance on the domain $$\mathbb R^3$$ (Why not “[[Cartesian distance]]” ?),
• require map $$\iota_{\Sigma^2}$$ to be a [[(Scaled) Isometry]].

But all this would still not prove $$\iota_{\Sigma^2}$$ being a homeomorphism; instead you’d have to first consider and establish a topology on set $$\Sigma^2$$ in its own right, perhaps being derived from balls in terms of the distance ratios I suggested.

> Note that up to this point aspects of smoothness were not taken into account. But now we may ask: Is the manifold $$\Sigma^2$$ smoothable?

… this question picks up on Katie Steckles’ reply (02.10.2020, 15:48 o’clock) to my first comment …

> Again the answer is yes.

… and I found the detour involving set $$\mathfrak S^2_{2}$$ and map $$h$$ very helpful on this point; thanks again!

> As a third question we can then ask: Is the inclusion map $$\iota_{\Sigma^2}$$ a smooth embedding of $$\Sigma^2$$, given the smoothed structure?

Alright: There is some map between the (some, any) (smooth) atlas of 2-dimensional coordinate charts associated to the surface of a cube and set $$\mathbb R^3$$.
(I’m hesitant to call this map $$\iota_{\Sigma^2}$$ itself … Rather: “it’s induced by $$\iota_{\Sigma^2}$$”.)

Now: What exactly is “the technique” for (asking and) answering this third question? — especially: Do we (again) need some notion of distance applicable to set $$\mathbb R^3$$ as a whole?, and (also, but separately) some notion of distance applicable to set $$\mathbb R^2$$ as a whole?
Or would it suffice to make appeals to “the natural 1-dimensional metric of $$\mathbb R$$” ?? …

> And here the answer is no. […] a necessary condition to have a unique tangent plane to the image at each of its points. This condition is clearly not satisfied for the cubic sphere on its edges or corners, in contrast to the case of the round sphere.

Well — what exactly is the definition of a “tangent plane (in $$\mathbb R^3$$)” ?, by which to conclude that $$\mathfrak S^2_{\infty}$$ doesn’t provide/constitute such; neither at $$x \equiv (1, 1, 1)$$ nor, say, at $$x \equiv (1, 1, 0)$$.

And a fourth (related) question:
Could ants crawling along the surface of a cube shaped apple (reminiscent of their ancestors on the cover of MTW) even tell whether or not they’re traversing “an edge” ??

18. @Frank Wappler / 28.12.2020, 18:30 o’clock

If I’ve got it right, you have some complaints concerning the use of a phrase like an object that looks locally flat‘ to sketch the idea of a manifold. Well, I see that this informal characterization sounds indeed somewhat misleading. One could instead present a manifold (of dimension $$n$$) heuristically by describing it as an object, or space, that is modeled locally on $$\mathbb{R}^n$$. Commonly this is then loosely rephrased by saying that a manifold looks locally Euclidean‘ or the like.

Let me emphasize that for a topological embedding it is even essential to allow an identification of its preimage with its image as topological spaces. Note that we also have a topological embedding $$h^{-1}\circ \iota_{S^2} : S^2 \hookrightarrow \mathbb{R}^3$$ whose image is $$mathfrak{S}^2_\infty$$. So you can consider $$S^2$$ and the cubic sphere to be “the same” as topological spaces. In this case the edges and vertices of the cubic sphere can obviously not be ascribed to the preimage, so the cubic shape of the image results from the mapping $$h^{-1}$$.

On the other hand, we have $$h^{-1}\circ \iota_{S^2} = \iota_{\Sigma^2} \circ f^{-1}$$ where, in view of the smoothing, $$f^{-1}:S^2 \to \Sigma^2$$ is a diffeomorphism. So here it is not the mapping $$f^{-1}$$ that gives a geometric shape with edges and corners. Therefore, the cubic shape of $$\mathfrak{S}^2_\infty$$ cannot be attributed to the preimage of the inclusion map $$\iota_{\Sigma^2}$$, rather it is to be understood as a feature produced by this mapping. In particular, smoothablity of the manifold $$\Sigma^2$$ does not entail that the topological embedding $$\iota_{\Sigma^2}$$ is a smooth mapping whose image is a smooth submanifold.

Surfaces (i.e., 2-dimensional submanifolds) in Euclidean 3-space arise quite naturally as level sets of real functions defined on $$\mathbb{R}^3$$ or a subset thereof. I have also used this to specify both the round sphere and the cubic sphere, as for $$p = 2,\infty$$ the real functions $$F_p(x) := \|x\|_p$$ have the level sets $$\mathfrak{S}^2_p = F^{-1}_p(\{1\})$$. Then $$F_2 : \mathbb{R}^3 \setminus \{0\} \to \mathbb{R}$$ is a smooth function whose gradient at $$x$$ is the vector $$\nabla F_2(x) = \frac{x}{\|x\|_2} \neq 0$$. Given any $$x_0 \in \mathfrak{S}^2_2$$, the tangent plane to the round sphere at $$x_0$$ is the affine plane
$\left\{ x \in \mathbb{R}^3 \mid \langle \nabla F_2(x_0), x – x_0 \rangle = 0 \right\}$
where, of course, the angle brackets denote the Euclidean scalar product.

For $$F_\infty$$ this approach fails — it is not a smooth function and you’ll find yourself in trouble with the gradient at the edges and corners. Basically, a tangent plane to a surface requires to have a normal vector to the surface at the point in question, and this is not the case for all points of the cubic sphere.

19. Chrys wrote (30.12.2020, 16:01 o’clock):
> Note that we also have a topological embedding [… of] $$S^2$$ whose image is $$\mathfrak S_{\infty}$$.

Indeed (I’m glad you brought that up); let me give it an explicit (and, for what it’s worth, distinctive) name: $$\varsigma_{S^2}$$.
(That’s in order to establish that this particular topological embedding is “just as good”, “just as relevant”, perhaps even “just as canonical” as $$\iota_{S^2}. After all, so far you haven’t provided anything more definitive about “the canonical immersion \(\iota_{S^2}$$” yet, either.)

And likewise there’s a topological embedding of $$\Sigma^2$$ whose image is $$\mathfrak S_{2}$$, let’s call it $$\varsigma_{\Sigma^2}$$.

Apparently therefore, (the relevant part of) the coordinate transformation $$\phi$$ I sketched (“roughly”) in comment 23.12.2020, 18:32 o’clock can be expressed formally as $\phi \equiv \varsigma_{\Sigma^2}^{(-1)} \circ \varsigma_{S^2}$
or likewise $\phi \equiv \iota_{S^2}^{(-1)} \circ \iota_{\Sigma^2}$.

Perhaps the distinction between domain and range of this coordinate transformation can and should be made more explicit like this:

$phi : {}_{x}\mathbb R^3 \longleftrightarrow {}_{\xi}\mathbb R^3$

and correspondingly we might distinguish
$${}_{x}\| . \|_p$$ from $${}_{\xi}\| . \|_p$$

and declare more explicitly $$\mathfrak S^2_p \equiv {}_{x}\mathfrak S^2_p$$ and
$$h \equiv {}_{x}h$$

to be (“formally”) distinct from $${}_{\xi}\mathfrak S^2_p$$ and $${}_{\xi}h$$, resp.

And last but not least: distinguish
– “Cartesian distance referring to coordinates of the domain of $$\phi$$” from
– “Cartesian distance referring to coordinates of the range of $$\phi$$”.

> […] the cubic shape of $$\mathfrak S_{\infty}$$ […] is to be understood as a feature produced by this mapping.

On which grounds would you attribute “cubic shape” to
$$\mathfrak S_{\infty} \equiv {}_{x}\mathfrak S_{\infty} \equiv \{ x \in {}_{x}\mathbb R^3 : {}_{x}\| x \|_{\infty} = 1 \}$$ ?

And would therefore, too, attribute “cubic shape” to
$${}_{\xi}\mathfrak S_{\infty} \equiv \{ \xi \in {}_{\xi}\mathbb R^3 : {}_{\xi}\| \xi \|_{\infty} = 1 \}$$,
recalling that $${}_{\xi}\mathfrak S_{\infty}$$ is the image of $$S$$ under topological embedding $$\varsigma_{S^2}$$ ?

> […] a tangent plane to a surface requires to have a normal vector […]

A “normal vector” with respect to the coordinates of the domain of coordinate transformation $$\phi$$ ?
Or with respect to the coordinates of the range of coordinate transformation $$\phi$$ ?
Or both ?
Or perhaps some coordinate-free notion of “normal vector” ? …

p.s.
Anyways I’d distinguish
– a “sphere” having constant and non-zero curvature κ, for any four of its points A, B, F, G as maximum of all solutions of
 0 = Det[ { { 0, 1/κ, 1/κ, 1/κ, 1/κ, 1 }, { 1/κ, 0, AB^2, AF^2, AG^2, 1 }, { 1/κ, BA^2, 0, BF^2, BG^2, 1 }, { 1/κ, FA^2, FB^2, 0, FG^2, 1 }, { 1/κ, GA^2, GB^2, GF^2, 0, 1 }, { 1, 1, 1, 1, 1, 0 } } ],

from

– “the surface of a cube” containing flat as well as curved “patches”, and even eight exceptional points in whose “patches” the curvature is not bounded at all (and which in terms of these exceptions resembles less “a surface” but “a body”).

• Please mind the erroneously missing superscripts ($${}^2$$) on some of the $$\mathfrak S$$s in the preceding comment.

• More corrections to the above comment 31.12.2020, 04:59 o’clock (upon reading it as SciLogs comment preview which, unfortunately, is otherwise still not adequately provided):

[…] (the relevant part of) the coordinate transformation $$\phi$$ I sketched (“roughly”) in comment 23.12.2020, 18:32 o’clock can be expressed formally
formally as $\phi \equiv \varsigma_{\Sigma^2} \circ \varsigma_{S^2}^{(-1)}$
or likewise $\phi \equiv \iota_{S^2} \circ \iota_{\Sigma^2}^{(-1)}$.

Perhaps the distinction between domain and range of this coordinate transformation can and should be made more explicit like this:

$\phi : {}_{x}\mathbb R^3 \longleftrightarrow {}_{\xi}\mathbb R^3$

20. More corrections to the above comment 31.12.2020, 04:59 o’clock (upon reading it as SciLogs comment preview which, unfortunately, is otherwise still not adequately provided):

[…] (the relevant parts of) coordinate transformation $$\phi$$ I sketched (“roughly”) in comment 23.12.2020, 18:32 o’clock can be expressed formally
$\phi \equiv \varsigma_{S^2} \circ \iota_{S^2}^{(-1)} \equiv \varsigma_{\Sigma^2} \circ \iota_{\Sigma^2}^{(-1)}$.

Perhaps the distinction between domain and range of this coordinate transformation can and should be made more explicit like this:

$\phi : {}_{x}\mathbb R^3 \longleftrightarrow {}_{\xi}\mathbb R^3$ …

21. @Frank Wappler / 31.12.2020, 04:59 o’clock

I’d think we can safely drop the superscript 2 on the mathfrak{S} symbols for a simplified notation, there seems to be no risk of confusion.

Let me try to draw a commutative diagram of topological spaces and homeomorphisms (to test the local MathJax environment) as follows:
$\begin{matrix} \Sigma^2 & \stackrel{f}{\longrightarrow} & S^2 \\ \quad\Big\downarrow\iota_{\Sigma^2} & & \quad\Big\downarrow\iota_{S^2}\\ \mathfrak{S}_\infty & \stackrel{h}{\longrightarrow} & \mathfrak{S}_2 \\ \end{matrix}$
So $$\varsigma_{\Sigma^2} = \iota_{S^2} \circ f = h \circ \iota_{\Sigma^2} : \Sigma^2 \to \mathfrak{S}_2 \subset \mathbb{R}^3$$ is indeed a smooth embedding, as both $$f$$ and $$\iota_{S^2}$$ are smooth.

Unfortunately, I couldn’t figure out what you are trying to achieve, in the given context, by coordinate transformations of $$\mathbb{R}^3$$. (Especially you should be happy that we don’t need any.)

In elementary geometry, a cube is intuitively presented in 3-space as a regular hexahedron, one of the Platonic solids. Of course, this solid can be freely rotated or displaced in space without altering its geometric shape or size. Once having chosen a formal representation of the cube as a subset of $$\mathbb{R}^3$$, these operations on the cube can analytically be described by what is called a motion of $$\mathbb{R}^3$$. That is to say, a motion of the Euclidean 3-space is an affine transfomation that maps, in particular, the cubic sphere $$\mathfrak{S}_\infty$$ congruently to a geometric object of exactly the same shape and size. A physicist may prefer to take the symmetry axes of a cube as the axes of an attached coordinate frame and then express the motion in a slightly different languege. Anyway, moving a cube or just its boundary in the way just described will not give any new insights with respect to embeddings of $$\Sigma^2$$, because the motion is a congruence mapping and the moved objects are equivalent.

We can also include cubic spheres of different size. For example, with the notation used earlier, we can set $$\mathfrak{S}_\infty(r) := F^{-1}_\infty(\{r\})$$ to define a cubic sphere of “radius” $$r \gt 0$$. However, rescaled spheres can be derived from the unit sphere by a linear transformation. It should be emphasized that geometric shape as well as metric size are extrinsic properties of the submanifold $$\mathfrak{S}_\infty$$, as a result of the embedding. These concepts are not a priori defined for $$\Sigma^2$$, according to the formal definition of a manifold.

As you asked, a normal vector to a plane in Euclidean 3-space simply means a vector that is perpendicular to that plane with respect to the Euclidean scalar procuct. More generally, a normal vector to a surface at a point $$x$$ is then a normal vector to the tangent plane of the surface at $$x$$.

22. Chrys wrote (04.01.2021, 00:34 o’clock):
> Let me try to draw a commutative diagram of topological spaces and homeomorphisms (to test the local MathJax environment) as follows:
$\begin{matrix} \Sigma^2 & \stackrel{f}{\longrightarrow} & S^2 \\ \quad\Big\downarrow\iota_{\Sigma^2} & & \quad\Big\downarrow\iota_{S^2}\\ \mathfrak{S}_\infty & \stackrel{h}{\longrightarrow} & \mathfrak{S}_2 \\ \end{matrix}$

Nice try! …

Allow me to complement it with the following drawing, meant as a mere visual reminder:

$\begin{matrix} & & \Sigma^2 & & \\ & \! \! \! \! \! \! \! \! \Huge \swarrow \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \Large {}^{{}^{\huge \iota_{\large\Sigma^2}}} & \! \! \mid & \! \! \! \! \! \! \! \! \Huge \searrow \! \! \! \! \! \! \! \! \! \! \! \Large {}^{{}^{\huge \varsigma_{\large\Sigma^2}}} & \\ \mathfrak S^2_{\infty} & \! \! \! \stackrel{\curvearrowright}{\tiny \curvearrowleft} {}^{\phi} \! \! \! \! \! \! \! & & \! \! \! \! \! \! \! \! \! \stackrel{\curvearrowleft}{\tiny \curvearrowright} {}^{\phi} & \! \! \! \! \! \! \! \mathfrak S^2_{2} \\ & \! \! \! \! \! \! \! \! \Huge \nwarrow \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \large \varsigma_{\small S^2} & \, {\downarrow}^f & \! \! \! \! \! \! \! \! \Huge \nearrow \! \! \! \! \! \! \! \! \! \! \! \large \iota_{\small S^2} & \\ & & S^2 & & \end{matrix}$

Note the (visually pronounced) distinction between the bijective (invertible) map
$$h : \mathfrak S^2_{\infty} \longleftrightarrow \mathfrak S^2_2$$
which had been explicitly specified (19.12.2020, 23:38 o’clock) as
$$h[ \, x \, ] \mapsto x / \| x \|_2$$

and the bijective (invertible) map
$$\phi : ((\mathbb R^3 \setminus \mathcal X) \cup \mathfrak S^2_2 \cup \mathfrak S^2_{\infty}) \longleftrightarrow ((\mathbb R^3 \setminus \mathcal X) \cup \mathfrak S^2_2 \cup \mathfrak S^2_{\infty})$$,
which, for the time being, I can specify at least partially by
$$\phi[ \, \Sigma^2 \, ] \mapsto S^2$$ and $$\phi[ \, S^2 \, ] \mapsto \Sigma^2$$,
and where $$\mathcal X$$ is some suitable set (“to be excluded from domain and range of $$\phi$$”).

> Unfortunately, I couldn’t figure out what you are trying to achieve, in the given context, by coordinate transformations of $$\mathbb R^3$$. […]

My intention was and remains to point out that so far you had failed to present specifications of the “inclusion maps” $$\iota_{S^2}$$ and $$\iota_{\Sigma^2}$$ since you first mentioned them (16.12.2020, 12:53 o’clock). Therefore I came up with similarly introducing “inclusion maps” $$\varsigma_{S^2}$$ and $$\varsigma_{\Sigma^2}$$ “for contrast”; also illustrating the arbitrariness of distinct homeomorphic coordinate assignments.

You have given nice descriptions of their respective ranges (images), namely the sets $$\mathfrak S^2_2$$ and $$\mathfrak S^2_{\infty}$$, resp., but you haven’t acknowledged any explicit description of their domains besides indicating that these domains are distinct, and accordingly distinctly denoted $$S^2$$ and $$\Sigma^2$$, resp., and calling they distinctly “the Euclidean 2-sphere” and “the surface of the [Euclidean] cube”, resp.; but, time and again, without providing some explicit definition of those notions.

In your latest comment you came up with another clue:

> In elementary geometry, a cube is intuitively presented in 3-space as a regular hexahedron, one of the Platonic solids.

So — might we perhaps at least manage to agree on what’s meant by “3-space”, or more specificly “Euclidean/flat 3-space”?

Do we mean

– a metric space $$(\mathcal P, d_{\text{E}3})$$ consisting of a set $$\mathcal P$$ of “points” and a distance assignment
$$d_{\text{E}3} : \mathcal P \times \mathcal P \longrightarrow \mathbb R_{(\ge 0)}$$,

and do we agree on its further properties being expressible in terms of those distance values ?

> […] geometric shape as well as metric size […] These concepts are not a priori defined for $$\Sigma^2$$, according to the formal definition of a manifold.

The “surface of a cube” which I had brought up is certainly to be thought as having geometric properties (as subset of “Euclidean 3-space” having properties in terms of distance $$d_{\text{E}3}$$, as well as in terms of “intrinsic distance” which is equivalent to the shortest arclength, and thus for any two points the infimum over arcs between these two points restricted to the subset under consideration, of suprema over partial sums of distances $$d_{\text{E}3}$$ of ordered subsets of points of any such arc.

If there is any distinction between “the surface of a cube” and “a sphere” to be made at all, and in particular any difference of their either being a “smooth manifold” (in the sense indicated by Katie Steckles, 02.10.2020, 15:48 o’clock), or not, then it is inevitably founded on their distinction in terms of geometric shape.

I note again that the curvature of a sphere, expressed in terms of its “intrinsic distances” between any four points, has an upper bound (and may be called “smooth” in this respect); while the curvature of the surface of a cube is not bounded (for any four points ever closer to one of its “corners” a.k.a. “points with non-smooth neighborhood”), but exactly zero anywhere else, including for any points on, or arbitrarily close to, one of the “edges”.

p.s.
> As you asked, a normal vector to a plane in Euclidean 3-space simply means […]

Therefore evidently, for the purpose of defining “a normal vector /to a plane in Euclidean 3-space)”, you’re requiring the notion of “a plane in Euclidean 3-space”; presumably a.k.a. “Euclidean 2-space”.

What exactly would be the definition of such an object?,
and importantly (for physicists):
Is it (therefore) invariant under coordinate transformations (such as the coordinate transformation $$\phi$$ sketched above)?

• Frank Wappler wrote (05.01.2021, 03:23 o’clock):
> Allow me to complement with the following drawing, as a mere visual reminder: […] $$\curvearrowleft$$ […]

Unfortunately, this now invalidates https://www.tutorialspoint.com/latex_equation_editor.htm as SciLog comment $$LaTeX$$ previewer …

Let me then offer some substitute:

$\begin{matrix} & & \Sigma^2 & & \\ & \! \! \! \! \! \! \! \! \Huge \swarrow \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \Large {}^{{}^{\huge \iota_{\large\Sigma^2}}} & \, {\mid}^{\, f} & \! \! \! \! \! \! \! \! \Huge \searrow \! \! \! \! \! \! \! \! \! \! \! \Large {}^{{}^{\huge \varsigma_{\large\Sigma^2}}} & \\ \mathfrak S^2_{\infty} & \! \! \! \stackrel{\longrightarrow}{\tiny \leftarrow} {}^{\phi} \! \! \! \! \! \! \! & & \! \! \! \! \! \! \! \! \! \stackrel{\longleftarrow}{\tiny \rightarrow} {}^{\phi} & \! \! \! \! \! \! \! \mathfrak S^2_{2} \\ & \! \! \! \! \! \! \! \! \Huge \nwarrow \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \large \varsigma_{\small S^2} & \, {\downarrow}^f & \! \! \! \! \! \! \! \! \Huge \nearrow \! \! \! \! \! \! \! \! \! \! \! \large \iota_{\small S^2} & \\ & & S^2 & & \end{matrix}$

> Note the (visually pronounced) distinction between $$h$$ […] and $$\phi$$ […]

• Frank Wappler wrote (05.01.2021, 03:23 o’clock):
> […] the bijective (invertible) map
>
$$\phi : ((\mathbb R^3 \setminus \mathcal X) \cup \mathfrak S^2_2 \cup \mathfrak S^2_{\infty}) \leftrightarrow ((\mathbb R^3 \setminus \mathcal X) \cup \mathfrak S^2_2 \cup \mathfrak S^2_{\infty}) > which, for the time being, I can specify at least partially by … \( \phi[ \, \mathfrak S^2_{\infty} \, ] \mapsto \mathfrak S^2_2$$ as well as $$\phi[ \, \mathfrak S^2_2 \, ] \mapsto \mathfrak S^2_{\infty}$$ …

> and where $$\mathcal X$$ is some suitable set (“to be excluded from domain and range of $$\phi$$”).

• Frank Wappler wrote (05.01.2021, 03:23 o’clock):
> […] the bijective (invertible) map
>
$$\phi : ((\mathbb R^3 \setminus \mathcal X) \cup \mathfrak S^2_2 \cup \mathfrak S^2_{\infty}) \leftrightarrow ((\mathbb R^3 \setminus \mathcal X) \cup \mathfrak S^2_2 \cup \mathfrak S^2_{\infty})$$
> which, for the time being, I can specify at least partially by

… $$\phi[ \, \mathfrak S^2_{\infty} \, ] \mapsto \mathfrak S^2_2$$ as well as $$\phi[ \, \mathfrak S^2_2 \, ] \mapsto \mathfrak S^2_{\infty}$$ …

> and where $$\mathcal X$$ is some suitable set (“to be excluded from domain and range of $$\phi$$”).

23. @Frank Wappler / 05.01.2021, 03:23 o’clock

Well, it appears to me that we need some terminological clarification, especially on what is meant by an inclusion map. This originates from the inclusion of sets, i.e., the subset relation $$\subset$$’, which can be interpreted in terms of a mapping, the associated canonical inclusion map. But let me first briefly recall the set theoretic definition of a mapping.

Given two non-empty sets $$A, B$$, a functional relation between $$A$$ and $$B$$ is a subset $$R \subset A\times B$$ with the following properties:
(1) $$\forall x \in A\;\exists y \in B : (x,y) \in R$$, and
(2) $$\forall x\in A : (x,y) \in R \wedge (x,y’) \in R\,\Rightarrow\, y = y’$$.
Then $$R$$ determines a mapping $$f_R : A \to B$$ defined by $$f_R(x) = y\;:\Leftrightarrow\;(x,y) \in R$$. Conversely, every map $$f : A \to B$$ determines a functional relation $$R_f := \left\{(x,f(x)) \mid x \in A\right\} \subset A \times B$$.

Now let’s assume that, in addition, $$A \subset B$$. In this case we can consider the diagonal $$\Delta_A = \left\{ (x,y) \in A \times B \mid y = x\right\}$$, which is readily seen to be a functional relation. This gives a mapping $$f_{\Delta_A} : A \hookrightarrow B$$ according to $$f_{\Delta_A}(x) = y \;:\Leftrightarrow\;(x,y) \in \Delta_A$$. Alternatively, $$f_{\Delta_A}$$ is uniquely determined as the mapping that satifies $$\Delta_A = \left\{(x,f_{\Delta_A}(x)) \mid x \in A \right\} \subset A \times B$$. This mapping is canonically defined in that it is implicitly given by the inclusion $$A subset B$$ of sets (and nothing else).

For most purposes it is just fine to set formally $$\iota_A := f_{\Delta_A}$$ and then use the same symbol, $$A$$, to denote the preimage as well as the image of this inclusion map. For our spheres, however, I have introduced distinct symbols to distinguish typographically between the preimages, the manifolds $$S^2, \Sigma^2$$, and their respective images, the submanifolds $$\mathfrak{S}_2, \mathfrak{S}_\infty$$. Simply because I wish to stress that the smoothing procedure entails an equivalence for the former (they are diffeomorphic) which can’t be stated for the latter. And, after all, that is exactly what our discussion is all about.

Right now I’m not sure whether I could make this point sufficiently clear, though it should be clarified before going any futher.

P.S. Concerning my previous remarks on the normal vector, perhaps I should have emphasized that this means a non-zero vector. If you are familiar with the Hesse normal form as an analytic equation for representing a plane in the Euclidean 3-space, there should be no problem anyway.

24. Chrys wrote (06.01.2021, 12:45 o’clock):
> Well, it appears to me that we need some terminological clarification

> especially on what is meant by an inclusion map [… as, per 16.12.2020, 12:53 o’clock, for example $$\iota_{\Sigma^2} : \Sigma^2 \hookrightarrow \mathbb{R}^3$$ …]

> […] Now let’s assume that, in addition, $$A \subset B.$$ In this case we can consider the diagonal […] $$\{(x,y) \in A \times B \mid y = x\}$$, […]

Thanks for clarifying that you are conscious and serious about whar this implies. So, if indeed insist on

$$\iota_{\Sigma^2}[ \, x \, ] \mapsto x,$$

and (therefore) its domain $$\Sigma^2) and its range are indeed one and the same set, \(\Sigma^2 \equiv \mathfrak S^2_{\infty}$$, and thus domain $$\Sigma^2) plainly a subset of \(\mathbb R^3$$,
and of course likewise for $$\iota_{S^2}[ \, x \, ] \mapsto x,$$, then

– the maps $$\varsigma_{\Sigma^2}$$ and $$\varsigma_{S^2}$$ I had suggested are no longer worth discussing, because clearly they have properties so substantially different from $$\iota_{\Sigma^2}$$ or $$\iota_{S^2}$$, and

– I should try to clarify my question (arising in my comment 09.12.2020, 11:48 o’clock), because: starting from the notion of a “(smooth) manifold” as “a set, together with certain additional structure (assigned to this set, or expressed in terms of this set)”, I wish the applicable set to be specified as generic as possible, in order for the relevant “additional structure” to be considered and described all the more explicitly. Especially concerning “structure naturally enjoyed” by certain sets of real numbers, or tuples.

Consequently I’d like to restate my question (foremost to Katie Steckles, in response to her comment 02.10.2020, 15:48 o’clock, but of course gladly also for broader discussion) as follows:

Considering the surface of a cube given (for “elementary intuition”) as (regular) hexahedral subset $$\mathcal H^2 \subset \mathcal E^3$$ of a 3-dimensional flat metric space $$\mathcal E^3, d_{(\text{E3})}$$,
with topology $$\mathfrak T_{(\text{H2})}$$ being the subspace topology induced by the “open ball basis” topology of $$\mathcal E^3, d_{(\text{E3})}$$,
can an atlas $$\mathfrak A_{(\text{H2})}$$ be found and, chart by chart, be assigned to appropriate open subsets of set $$\mathcal H^2$$,
such that set $$\mathcal H^2$$ together with atlas $$\mathfrak A_{(\text{H2})}$$ constututes a smooth manifold ? (And if not, then: Please discuss separately the role of the edges of the cube surface, and the role of its corners, contributing to this conclusion.)

And just because it may be of interest in its own right:

Considering the surface of a cube given intrinsically as a metric space $$\mathcal C^2, d_{(\text{C2})}$$,
such that, upon identifying $$\mathcal C^2 \equiv \mathcal H^2$$,
d_{(\text{C2})}[ \, a, b \, ] \mapsto \text{Infimum of lengths of arcs in set } \mathscr A_{(\text{H2})}[ \, a, b \, ] \)
where set $$\mathscr A_{(\text{H2})}[ \, a, b \, ]$$ contains all and only those arcs which connect “points” $$a$$ and $$b$$ and which are subsets of set $$\mathcal H^2$$,
with topology $$\mathfrak T_{(\text{C2})}$$ induced by the “open ball basis” topology of $$\mathcal C^2, d_{(\text{C2})}$$,
can an atlas $$\mathfrak A_{(\text{C2})}$$ be found and, chart by chart, be assigned to appropriate open subsets of set $$\mathcal C^2$$,
such that set $$\mathcal C^2$$ together with atlas $$\mathfrak A_{(\text{C2})}$$ constututes a smooth manifold (by the definition of the english fragment of Wikipedia) ? (If not: please discuss.)

p.s.
> the submanifolds $$\mathfrak S^2_{2}$$, $$\mathfrak S^2_{\infty}$$ […] are [not] diffeomorphic

Taking once again note of the definition, I wonder whether perhaps you might mean rather more explicitly that the map you introduced above,
$$h : \mathfrak S^2_{\infty} \longleftrightarrow \mathfrak S^2_{2}, \qquad h[ \, x \, ] \mapsto x / \|x\|_2$$,
does not constitute a diffeomorphism. (If that’s not what you meant: please clarify.)

If so, could you sketch a proof which is “as elementary as reasonably achieveable”, i.e. presuming concepts only “up to” differentiating functions of a single real-number variable, please ?

• Frank Wappler wrote (07.01.2021, 01:41 o’clock):
> […] So, if indeed

… you insist on

$$\iota_{\Sigma^2}[ \, x \, ] \mapsto x,$$

and (therefore) its domain and its range are indeed one and the same set, $$\Sigma^2 \equiv \mathfrak S^2_{\infty}$$, and thus domain $$\Sigma^2)$$ is plainly a subset of $$\mathbb R^3$$,
and of course likewise for $$\iota_{S^2}[ \, x \, ] \mapsto x,$$, then …

> […] Consequently I’d like to restate my question

… as follows:

Considering the surface of a cube given (for “elementary intuition”) as (regular) hexahedral subset $$\mathcal H^2 \subset \mathcal E^3$$ of a 3-dimensional flat metric space $$\left(\mathcal E^3, d_{(\text{E3})}\right)$$,
with topology $$\mathfrak T_{(\text{H2})}$$ being the subspace topology induced by the “open ball basis” topology of metric space $$\left(\mathcal E^3, d_{(\text{E3})}\right)$$,
can an atlas $$\mathfrak A_{(\text{H2})}$$ be found and, chart by chart, be assigned to appropriate open subsets of set $$\mathcal H^2$$,
such that set $$\mathcal H^2$$ together with atlas $$\mathfrak A_{(\text{H2})}$$ constitutes a smooth manifold ? (And if not, then: Please discuss separately the role of the edges of the cube surface, and the role of its corners, contributing to this conclusion.)

25. @Frank Wappler / 07.01.2021, 01:41 o’clock

Just think of $$\Sigma^2$$ as an extracted copy of $$\mathfrak{S}_\infty$$. Likewise, if you have a multi-page PDF document, you can extract, say, the first page with some suitable software tool and store it on your HDD as a separate PDF document. Then the content of this new PDF can be identified with a subcontent of your original PDF. And if you imagine an algorithm that associates the content of your new PDF with the corresponding subcontent of the original PDF, you have the analogy of an inclusion map.

To see that the submanifolds $$\mathfrak{S}_\infty$$ and $$\mathfrak{S}_2$$ are not diffeomorphic, you can argue by contradiction: Suppose they were, then for every point $$x \in \mathfrak{S}_2$$ the derivative of the supposed diffeomorphism would provide an isomorphism of the tangent plane to the round sphere at $$\x$$ onto the tangent plane to the cubic sphere at the mapped image of $$x$$. But the cubic sphere does not have a tangent plane at points on the corners and edges. So it doesn’t work for these points, and that’s what Katie Steckles had pointed out in her reply to your question. Hence such a supposed diffeomorphism does not exist.

Note that Nicolaescu’s characterization of smoothable manifolds is an assertion about manifolds, not about embedded submanifolds. The only way to “smooth out the corners” of the (embedded) cubic sphere is to map it homeomorphically to some other embedded object and thus to alter its geometric shape.

• Correction: … at $$x$$ onto …

26. Chrys wrote (08.01.2021, 15:11 o’clock):
> Just think of $$Sigma^2$$ as an extracted copy of $$\mathfrak S^2_{\infty}. Can do! But in doing so, I’m coerced into thinking of \(Sigma^2$$ as a specific set of real number tuples.
And that’s not at all what I had in mind as “the surface of a cube” (which I had already linked from the outset, in my first comment 02.10.2020, 09:53 o’clock, especially because it doesn’t contain any mentioning of “real numbers”, or tuples thereof.).

Therefore (cmp. my comment 07.01.2021, 01:41 o’clock), to avoid any confusion, I urge to denote the set to be considered suitably as $$\mathcal H^2$$ (being reminiscent of the word “hexahedron”) or as $$\mathcal C^2$$ (being reminiscent of the word “cube”),
where I’ve introduced these two distinct symbols for denoting one and the same set $$\mathcal H^2 \equiv \mathcal C^2$$ in order to emphasize and facilitate the distinction between the two different metric spaces $$\left \mathcal H^2, d_{(\text{E3})} \right)$$ and $$\left \mathcal C^2, d_{(\text{C2})} \right)$$.

p.s.
> To see that the submanifolds $$\mathfrak S^2_{\infty}$$ and $$\mathfrak S^2_{2}$$ are not diffeomorphic,

My question (07.01.2021, 01:41 o’clock) had been instead on giving an “elementary” proof that
map $$h : \mathfrak S^2_{\infty} \longleftrightarrow \mathfrak S^2_{2}, \qquad h[ \, x \, ] \mapsto x / \|x\|_2$$

for $$\mathfrak S^2_{\infty} := \(\{ x \in \mathbb R^3 : \|x\|_{\infty} = 1 \}$$
and $$\mathfrak S^2_{2} := \(\{ x \in \mathbb R^3 : \|x\|_{2} = 1 \}$$

constitutes a diffeomorphism, i.e. a map with certain properties.

> you can argue by contradiction: Suppose they were, then for every point $$x \in \mathfrak S^2_{2}$$ the derivative of the supposed diffeomorphism

At this point the attempted proof fails to be as elementary or as explicit as requested:
What is the presumed elementary definition of “the derivative” (of some supposed diffeomorphism); or specificly “the derivative” of map $$h$$, or why specificly would “the derivative” of map $$h$$ not exist ?

27. Chrys wrote (08.01.2021, 15:11 o’clock):
> Just think of $$\Sigma^2$$ as an extracted copy of $$\mathfrak S^2_{\infty}$$.

Can do!
But in doing so, I’m coerced into thinking of $$Sigma^2$$ as a specific set of real number tuples.
And that’s not at all what I had in mind as “the surface of a cube” (which I had already linked from the outset, in my first comment 02.10.2020, 09:53 o’clock, especially because it doesn’t contain any mentioning of “real numbers”, or tuples thereof.).

Therefore (cmp. my comment 07.01.2021, 01:41 o’clock), to avoid any confusion, I urge to denote the set to be considered suitably as $$\mathcal H^2$$ (being reminiscent of the word “hexahedron”) or as $$\mathcal C^2$$ (being reminiscent of the word “cube”),
where I’ve introduced these two distinct symbols for denoting one and the same set $$\mathcal H^2 \equiv \mathcal C^2$$ in order to emphasize and facilitate the distinction between the two different metric spaces $$\left( \mathcal H^2, d_{(\text{E3})} \right)$$ and $$\left( \mathcal C^2, d_{(\text{C2})} \right)$$.

p.s.
> To see that the submanifolds $$\mathfrak S^2_{\infty}$$ and $$\mathfrak S^2_{2}$$ are not diffeomorphic,

My question (07.01.2021, 01:41 o’clock) had been instead on giving an “elementary” proof that
map $$h : \mathfrak S^2_{\infty} \longleftrightarrow \mathfrak S^2_{2}, \qquad h[ \, x \, ] \mapsto x / \|x\|_2$$

for $$\mathfrak S^2_{\infty} := \{ x \in \mathbb R^3 : \|x\|_{\infty} = 1 \}$$
and $$\mathfrak S^2_{2} := \{ x \in \mathbb R^3 : \|x\|_{2} = 1 \}$$

constitutes a diffeomorphism, i.e. a map with certain properties.

> you can argue by contradiction: Suppose they were, then for every point $$x \in \mathfrak S^2_{2}$$ the derivative of the supposed diffeomorphism

At this point the attempted proof fails to be as elementary or as explicit as requested:
What is the presumed elementary definition of “the derivative” (of some supposed diffeomorphism); or specificly “the derivative” of map $$h$$, or why specificly would “the derivative” of map $$h$$ not exist ?

28. Chrys wrote (08.01.2021, 15:11 o’clock):
> Just think of $$\Sigma^2$$ as an extracted copy of $$\mathfrak S^2_{\infty}$$.

Can do!
But in doing so, I’m coerced into thinking of $$\Sigma^2$$ as a specific set of real number tuples.
And that’s not at all what I had in mind as “the surface of a cube” (which I had already linked from the outset, in my first comment 02.10.2020, 09:53 o’clock, especially because it doesn’t contain any mentioning of “real numbers”, or tuples thereof.).

Therefore (cmp. my comment 07.01.2021, 01:41 o’clock), to avoid any confusion, I urge to denote the set to be considered suitably as $$\mathcal H^2$$ (being reminiscent of the word “hexahedron”) or as $$\mathcal C^2$$ (being reminiscent of the word “cube”),
where I’ve introduced these two distinct symbols for denoting one and the same set $$\mathcal H^2 \equiv \mathcal C^2$$ in order to emphasize and facilitate the distinction between the two different metric spaces $$\left( \mathcal H^2, d_{(\text{E3})} \right)$$ and $$\left( \mathcal C^2, d_{(\text{C2})} \right)$$.

p.s.
> To see that the submanifolds $$\mathfrak S^2_{\infty}$$ and $$\mathfrak S^2_{2}$$ are not diffeomorphic,

My question (07.01.2021, 01:41 o’clock) had been instead on giving an “elementary” proof that
map $$h : \mathfrak S^2_{\infty} \longleftrightarrow \mathfrak S^2_{2}, \qquad h[ \, x \, ] \mapsto x / \|x\|_2$$

for $$\mathfrak S^2_{\infty} := \{ x \in \mathbb R^3 : \|x\|_{\infty} = 1 \}$$
and $$\mathfrak S^2_{2} := \{ x \in \mathbb R^3 : \|x\|_{2} = 1 \}$$

does not constitute a diffeomorphism, i.e. a map with certain properties.

> you can argue by contradiction: Suppose they were, then for every point $$x \in \mathfrak S^2_{2}$$ the derivative of the supposed diffeomorphism

At this point the attempted proof fails to be as elementary or as explicit as requested:
What is the presumed elementary definition of “the derivative” (of some supposed diffeomorphism); or specificly “the derivative” of map $$h$$, or why specificly would “the derivative” of map $$h$$ not exist ?

29. @Frank Wappler / 11.01.2021, 11:28 o’clock

By considering $$\Sigma^2$$ as a separate set (as opposed to a subset $$\mathfrak{S}_\infy$$ of the vector space $$\mathhbb{R}^3$$ whose elements are real 3-tuples by definition), its elements are not to be conceived as 3-tuples. Rather, the canonical inclusion $$x \mapsto \iota_{\Sigma^2}(x) = (x^1,x^2,x^3) \in \mathbb{R}^3$$ assigns to every $$x\ in \Sigma^2$$ a real 3-tuple as a label by which $$x$$ can be symbolically identified. Notably your definition of $$\varsigma_{\Sigma^2}$$ gives a different labeling, $$x \mapsto \varsigma_{\Sigma^2}(x) = \iota_{\Sigma^2}(x)/\|\iota_{\Sigma^2}(x)\|_2 =: (\xi^1,\xi^2,\xi^3) \in \mathbb{R}^3$$. So these assigned labels can be exchanged without changing anything substantial for the set $$\Sigma^2$$ or its elements.

Yet another labeling is given by $$x \mapsto r\iota_{\Sigma^2}(x) = (rx^1,rx^2,rx^3) \in \mathbb{R}^3$$ for some “radius” $$r \gt 0$$. These examples suggest that properties like the geometric shape or the metric size of the various label (sub-)sets should in fact not be ascribed to $$\Sigma^2$$.

By a cube we usually mean a 3-dimensional figure which is recognized by its faces, edges, and corners, in combination with certain characteristic symmetries. Hence I’d recommend to use the superscript 3 with a formal representation $$\mathcal{C}^3 = \left\{x \in \mathbb{R}^3 \mid \|x\|_\infty \le 1\right\}$$ of a cubic shape in the Euclidean 3-space $$(\mathbb{R}^3, \langle\cdot,\cdot\rangle)$$. In your original question you had apparently distinguished between a cube and its surface, i.e., the boundary $$\partial\mathcal{C}^3 = \mathfrak{S}_\infty$$. And your question was how this surface might be related to the topological concept of a manifold.

Note that the only motivation for introducing an “extracted copy” $$\Sigma^2$$ of the surface $$\mathfrak{S}_\infty$$ is to bring in an object which is then formally shown to be a manifold. And aside from being a set, the only structure we had to assume for the “ectracted copy” is a topology such that the inclusion map $$\iota_{\Sigma^2}:\Sigma^2 \to \mathfrak{S}_\infty$$ becomes a homeomorphism of topological spaces, where the latter carries the subspace topology induced from the ambient Euclidean space.

Under these minimal assumption I’d say that your question allows a reasonable answer: Yes, $$\Sigma^2$$ is a manifold, it’s even a smoothable manifold, but the corresponding topological submanifold $$\mathfrak{S}_\infty$$ is nonetheless not smooth. Can you agree that this is indeed an answer to your question?

Concerning your desire for an “elementary” argument, the most elementary you can hope for is Multivariable Calculus. Observe that the map $$h^{-1}$$ can be extended to $$\rho:\mathbb{R}^3\setminus\{0\} \to \mathbb{R}^3$$ which then retracts the punctured Euclidean 3-space continuously to $$\mathfrak{S}_\infty$$, according to $$x \mapsto x/\|x\|_\infty$$. Then you can treat $$\rho$$ as a function of 3 variables with values in $$\mathbb{R}^3$$ and ask if this map is differentiable. The derivative of $$\rho$$, if it exists, is given by the Jacobian, and essentially you have to deal with the partial derivatives $$\partial \rho^i/\partial x^j$$. Is this elementary enough? And then, is $$\rho$$ differentiable?

• Correction: … as opposed to a subset $$\mathfrak{S}_\infty$$ of the vector space $$\mathbb{R}^3$$ …

30. Chrys wrote (13.01.2021, 13:39 o’clock):
> By considering $$\Sigma^2$$ as a separate set (as opposed to a subset $$\mathfrak S^2_{\infty}$$ of the vector space $$\mathbb R^3$$ whose elements are real 3-tuples by definition),

This was my understanding from the outset.
But unfortunately, it seems that this understanding of $$\Sigma^2$$ had not been kept up in our comments above, and perhaps you didn’t have this understanding of $$\Sigma^2$$ when you first introduced it 16.12.2020, 12:53 o’clock.
So we have a choice of either considering the symbol $$\Sigma^2$$ “burnt” and find and agree on some replacement(s), where the meaning of $$\mathfrak S^2_{\infty}$$ is of course unchanged and undisputed,
or of (re)defining the meaning of $$\Sigma^2$$ “on the fly” (which, to me, seems to be an imposition especially on you. But if you don’t mind … ).

> its elements are not to be conceived as 3-tuples.

Its elements would generically be called “points”, regardless of possible additional “structure”. In other words: the set of real number triples $$\mathfrak S^2_{\infty}$$ (together with a suitable distance definition) could be a specific instance of the generic set with its defining properties.

> Rather, the canonical inclusion $$x \mapsto \iota_{\Sigma^2}(x) = (x^1, x^2, x^3) \in \mathbb R^3$$ assigns to every x inΣ2 a real 3-tuple as a label by which x can be symbolically identified.

Yes: to each point of generic set $$\Sigma^2$$ there may be a real number tuple assigned; and map $$\iota_{\Sigma^2}$$ denotes one specific, “canonical” such assignment.

Some remarks on notation: you seem to use the “mapsto” arrow differently than I do. I find my usage advantageous, and I’ll stick to it. Also: I find that your choice of variable names doesn’t express the distinction between “point itself” and “coordinate tuple assigned”. May I suggest instead:
$$P \in \Sigma^2$$ and $$\iota_{\Sigma^2}[ \, P \, ] \mapsto (x_P, y_P, z_P) \equiv \mathbf x_P.$$

> Notably your definition of $$\varsigma_{\Sigma^2}$$ gives a different labeling

Yes: map $$\varsigma_{\Sigma^2}$$ differs from map $$\iota_{\Sigma^2}$$ by construction:
$$\varsigma_{\Sigma^2} := h \circ \iota_{\Sigma^2}$$,
where you had given an explicit definition of $$h$$ (being different from identity).

But: in keeping with the understanding of $$\Sigma^2$$ as a generic set (of points) with additional properties (I’d say: as a specific metric space) we still lack an explicit full definition of
“the canonical map” $$\iota_{\Sigma^2} : \Sigma^2 \longleftrightarrow \mathfrak S^2_{\infty}$$,
and consequently I haven’t given a full definition of “the map variant”
$$\varsigma_{\Sigma^2} : \Sigma^2 \longleftrightarrow \mathfrak S^2_{2}$$, either. (See however some attempts below.)

> […] the Euclidean 3-space $$\left(\mathbb R^3, \langle \cdot, \cdot\rangle \right)$$

… which I read as: “the set of all real triples, together with a (suitable) inner product”.

This again lacks distinction between “points” and “labels”.
And, again, I’m pointing out the notion of “the Euclidean 3-space” as a metric space, $$\left( \mathcal E, d_{(\text{E}3)}\right)$$.

Its properties can be characterized more completely than I managed above:

For each element $$P \in \mathcal E$$ the corresponding set of pairs $$\{ (A, P) : A \in \mathcal E \}$$ shall constitute a 3-dimensional vector space $$\mathfrak V_P$$ wrt. the field of real numbers, with inner product
$$\langle (J, P), (K, P) \rangle_P := (1/2) \, (d_{(\text{E}3)}[ \, J, P \, ])^2 + (d_{(\text{E}3)}[ \, K, P \, ])^2 – (d_{(\text{E}3)}[ \, J, K \, ])^2$$,

and these inner products of the distinct vector spaces must be consistent:

$$\langle ((K, P) – (J, P)), ((T, P) – (S, P)) \rangle_P = \langle ((K, J), ((T, J) – (S, J)) \rangle_J$$.

> […] examples suggest that properties like the geometric shape or the metric size of the various label (sub-)sets should in fact not be ascribed to $$\Sigma^2$$.

But “the surface of a cube”, as a metric space consisting of generic set $$\Sigma^2$$ together with suitable distances (either “inherited $$d_{(\text{E}3)}$$”, or “intrinsic, or likewise by embedding in 3-dimensional Euclidean space: $$d_{(\text{C}2)}$$) does have geometry (corners, edges, faces) itself.

> […] Yes, $$\Sigma^2$$ is a manifold, it’s even a smoothable manifold,

Why even bother with the labellings $$\mathfrak S^2$$ ??

p.s.
> […] The derivative of $$\rho$$, if it exists, is given by the Jacobian,

… which in turn consists of “simple single variable derivatives”. Fine.
Now let’s try this out on another example:

Is the identity map
$$I_{\mathfrak S^2_{\infty}} : \mathfrak S^2_{\infty} \longleftrightarrow \mathfrak S^2_{\infty}, \qquad I_{\mathfrak S^2_{\infty}}[ \, x \, ] \mapsto x$$
therefore a diffeomorphism ?

31. @Frank Wappler / 15.01.2021, 14:12 o’clock

Sorry for the delayed reply. Unfortunately, I’m still under the impression that you’re missing the point which makes the difference between submanfold’ and manifold’ with respect to the question of smoothing out corners. And I had introduced the different symbols in the hope to prevent any possible ambiguity on what I’m stating for which of these objects, the submanifold (subset, topological subspace) $$\mathfrak{S}_\infty$$ or the manifold (set, topological space) $$\Sigma^2$$. But I have serious doubts that this was successful.

Please note that by considering $$\Sigma^2$$ as what I had informally called an “extracted copy” of the cubic sphere $$\mathfrak{S}_\infty$$ most of the resemblance to a cube is stripped off. It is primarily just a set where each element $$x \in \Sigma^2$$ can be imagined to be marked with a symbolic reference to the “original” vector $$(x^1,x^2,x^3) \in \mathfrak{S}_\infty$$ from which it was “copied”. This is formally expressed by the assignment $$x \mapsto \iota_{\Sigma^2}(x) = (x^1,x^2,x^3)$$ for each $$x\in \Sigma^2$$. — And yes, that’s the way how it is used by convention: the symbol on the left of the mapsto arrow denotes the element to which the value on the right is assigned. Alternatively, I could write $$\iota_{\Sigma^2} : x \mapsto (x^1,x^2,x^3)$$ to express the same assignment.

The only additional structure on $$\Sigma^2$$ adopted from $$\mathfrak{S}_infty$$ is the topology by which the “extracted copy” becomes a topological space homeomorphic to the “original” cubic sphere in $$\mathbb{R}^3$$ endowed with its subspace topology. However, the topology does neither give a geometric shape to the “copy”, nor a concept of metric distance.

Do you expect your considerations on metric distance to provide some insight with respect to your original question on manifolds? Or is it just a new game you have started? It’s not clear to me what you are trying to do with metric distances .

And, finally, do you think that $$\mathfrak{S}^2_\infty$$ can be diffeomorphic to anything at all?

32. Chrys schrieb (23.01.2021, 17:36 o’clock):
> Sorry for the delayed reply.

Much rather: Thank you for following up!

> […] Do you expect your considerations on metric distance

… a pleonasm ? …

> to provide some insight with respect to your original question on manifolds?

My original question (02.10.2020, 09:53 o’clock), as well as Katie Steckles’ initial assertion quoted there, had been concerning the relation (if any) between “being a manifold” and “when you look at it close up, looks flat […] everywhere” (or in other words, in the relevant case: whether there is a finite infimum of all radii of circumspheres of any four points, or not).

The characterization of the object(s) under consideration as metric space(s) appears essential on both grounds:

– providing a topology (by default: through the “open ball basis”), by which to decide which assignments of real nubers are “homeomorphic” wrt. this topology, and which are not, as well as

– facilitating comparisons of radii of circumspheres.

Concerning the “surface of a cube” which I had introduced in my original question in the assumption that it presents a counterexample to Katie Steckles’ initial assertion
we have surely sufficienty established that indeed “it is a manifold” (or more precisely, that a smooth atlas may be assigned to it).
It may still remain to clarify that (and in which specific sense) the “surface of a cube” I referred to does not “look flat everywhere, when you look at it close up”;
and therefore does indeed present a counterexample to Katie Steckles’ initial assertion.

(And as indicated before: I find interesting, too, that there arise certain distinctions between consideration of “intrinsic” vs. “extrinsic” distances; especially concerning the radii of circumspheres of four points “near an edge”.)

> Unfortunately, I’m still under the impression that you’re missing the point which makes the difference between submanfold’ and `manifold’ with respect to the question of smoothing out corners. […]

The conception of the object under consideration (“surface of a cube”, “sphere (being the surface of a ball)”, …) as subset of 3-dimensional flat metric space is of course useful for motivation and for deriving the corresponding “intrinsic distance (ratios)”.
But as pointed out before: the notion of “submanifold” does not appear in the definition of “manifold”, nor specificly in the definition of “smooth manifold” at all.

p.s.
> And, finally, do you think that $$\mathfrac S^2_{\infty}$$ can be diffeomorphic to anything at all?

As suggested above, I’d like to check whether the identity map

$$I_{\mathfrac S^2_{\infty}} : \mathfrac S^2_{\infty} \longleftrightarrow \mathfrac S^2_{\infty}, \qquad I_{\mathfrac S^2_{\infty}}[ \, x \, ] \mapsto x$$

constitutes a diffeomorphism (by the definition you provided, 13.01.2021, 13:39 o’clock).

But I may suggest an even simpler “problem” which may help us to resolve what I consider the relevant issue:

Consider the function

$$p : \{ x \in \mathbb R: a \le x \le b \} \longrightarrow \mathbb R, \qquad p[ \, x \, ] \mapsto \frac{1}{b – a}.$$

Do you consider this function $$p$$ “smooth in $$a$$” (as well as “smooth in $$b$$”), or not ?

(Note the relation — similarities as well as distinction — to “the continuous uniform probability distribution function”.)

• p.s.
> And, finally, do you think that $$\mathfrak S^2_{\infty}$$ can be diffeomorphic to anything at all?

As suggested above, I’d like to check whether the identity map

$$I_{\mathfrak S^2_{\infty}} : \mathfrak S^2_{\infty} \longleftrightarrow \mathfrak S^2_{\infty}, \qquad I_{\mathfrac S^2_{\infty}}[ \, x \, ] \mapsto x$$

constitutes a diffeomorphism (by the definition you provided, 13.01.2021, 13:39 o’clock).

But I may suggest an even simpler “problem” which may help us to resolve what I consider the relevant issue:

Consider the function

$$p : \{ x \in \mathbb R: a \le x \le b \} \longrightarrow \mathbb R, \qquad p[ \, x \, ] \mapsto \frac{1}{b – a}.$$

Do you consider this function $$p$$ “smooth in $$a$$” (as well as “smooth in $$b$$”), or not ?

(Note the relation — similarities as well as distinction — to “the continuous uniform probability distribution function”.)

• p.s.
> And, finally, do you think that $$\mathfrak S^2_{\infty}$$ can be diffeomorphic to anything at all?

As suggested above, I’d like to check whether the identity map

$$I_{\mathfrak S^2_{\infty}} : \mathfrak S^2_{\infty} \longleftrightarrow \mathfrak S^2_{\infty}, \qquad I_{\mathfrak S^2_{\infty}}[ \, x \, ] \mapsto x$$

constitutes a diffeomorphism (by the definition you provided, 13.01.2021, 13:39 o’clock).

But I may suggest an even simpler “problem” which may help us to resolve what I consider the relevant issue:

Consider the function

$$p : \{ x \in \mathbb R: a \le x \le b \} \longrightarrow \mathbb R, \qquad p[ \, x \, ] \mapsto \frac{1}{b – a}.$$

Do you consider this function $$p$$ “smooth in $$a$$” (as well as “smooth in $$b$$”), or not ?

(Note the relation — similarities as well as distinction — to “the continuous uniform probability distribution function”.)

33. @Frank Wappler / 26.01.2021, 10:46 o’clock

To me it’s obvious that just giving an intuitive impression of what manifolds are, in layman’s terms, is all Katie Steckles had in mind when she sketched them roughly as objects which “look locally flat if you zoom in”. For instance, a picture I can get from her wording is that of an embedded surface in 3-space which allows to attach a tangent plane to each of its points. Zooming in to a point would give a local view of the surface where it comes closer and closer to the tangent plane at this point.

At any rate, her focus was clearly not on piecewise flat manifolds. Could that possibly be the class of objects you are especially looking for? (Related terms are “piecewise linear” or “piecewise smooth“.)

By the way, the closed interval $$I := [a,b] \subset \mathbb{R}$$ is a 1-dimensional manifold with boundary, $$\partial I= \{a,b\}$$. That’s yet another class of objects.

34. Chrys schrieb (31.01.2021, 16:05 o’clock):
> […] roughly as objects which “look locally flat if you zoom in”.

I certainly admit that such phrasing triggers my pet-peeve reflex, quite independent from further technicalities about manifolds.

The curvature of a given sphere, for instance, is constant; regardless of the size of the patch (as long as its not “size zero”, nor “of too low dimension”) on which its value is determined, patch by patch.
And the value is determined not by “looking”, but by evaluating vanishing Cayley-Menger determinants (in terms of “extrinsic distances”, solving for the unknown “R^2”) or vanishing Gram determinants (in terms of “intrinsic distances”, solving for the unknown “√{ κ }”).

> For instance, a picture I can get from her wording is that of an embedded surface in 3-space which allows to attach a tangent plane to each of its points.

Presumably you’re thereby thinking of any embedded surface in (flat) 3-space which “at every point” (i.e. in the limit of vanishing patch size) has (finite) extrinsic curvature; in other words: such that the smallest circumspheres for any four points [1] of any patch will have radii not below some limiting nonzero value.
Clearly such surfaces are deserving of interest and deserving of an exclusive short-hand name for themselves. But: it is not “(2-dimensional) manifold”. (Using everyday language I’d call them “smooth surfaces”, “without creases, or cusps”.)

> Zooming in to a point would give a local view of the surface where it comes closer and closer to the tangent plane at this point.

The maximal (flat 3-space) distance between some point “of the surface patch under consideration” and some tangent plane which touches the surface in some other point of this patch decreases with patch size; and indeed generally and eventually over-proportionally (quadratically) so.
The smaller the patch, the better the surface is approximated by any plane touching the patch; absolutely so (anyways), and even relatively so (with distance normalized to the patch size).

But such considerations obviously presume the notion of a “plane (itself)”, as if the prerequisits for defining “a plane” were any different from the prerequisits for defining the notion of any other surface of constant nonzero curvature “itself”. (I’m in favor of being explicit about prerequisits, rather than silently harboring “intuitions”.)

> […] focus was clearly not on piecewise flat manifolds.

Granted. But writing that

“manifolds are roughly” objects of a certain description

doesn’t convey quite the same meaning as writing that

.

Moreover, it remains to be resolved whether there exist Diffeomorphisms between any two of those piecewise flat manifolds, or between any one of them and, say, a sphere (as may be of particular interest to topologists, by Katie Steckles comment 02.10.2020, 15:48 o’clock).

[1: Looking at the Introduction to piecewise flat manifolds you had linked above led me to recall what I had understood long ago but failed to appreciate more recently: concerning surfaces or spaces of larger dimension, $$n \ge 2$$, we should consider the radii of circumspheres of not just any $$n + 2$$ points of some patch, but only for “selected $$n + 2$$ points”; e.g. with $$n + 1$$ points constrained to a sphere which has the remaining point as its center. ]

p.s.
> Btw. […]

Let me go ahead and try to calculate some entries into the Jacobian determinant for the suggested function,

$$I_{\mathfrak S^2_{\infty}} : \mathfrak S^2_{\infty} \longleftrightarrow \mathfrak S^2_{\infty}, \qquad I_{\mathfrak S^2_{\infty}}[ \, x \, ] \mapsto x$$,

at least for some selected points (of its domain); for starters at coordinate point $$\mathbf x \equiv (x, y, z) := (0, 0, 1)$$. I get:

$$J_{xx}[ \, I_{\mathfrak S^2_{\infty}} \, ]_{(0, 0, 1)} =$$
$$\frac{\partial}{\partial x} \left[ \phantom{\frac{\partial}{\partial x} } \! \! \! \! \! \! I_{\mathfrak S^2_{\infty}} \left[ \phantom{|} (x, y, z) \, \right]:x \right]_{(0, 0, 1)} =$$
$$\text{lim}_{\{ \epsilon \rightarrow 0 \}} [ \, \frac{ I_{\mathfrak S^2_{\infty}}[ \, (0 + \epsilon, 0, 1) \, ]:x – I_{\mathfrak S^2_{\infty}}[ \, (0, 0, 1) \, ]:x}{\epsilon} \, ] =$$
$$\text{lim}_{\{ \epsilon \rightarrow 0 \}} [ \, \frac{ (0 + \epsilon) – 0}{\epsilon} \, ] =$$
$$\text{lim}_{\{ \epsilon \rightarrow 0 \}} [ \, 1 \, ] =$$
$$1$$.

$$J_{xy}[ \, I_{\mathfrak S^2_{\infty}} \, ]_{(0, 0, 1)} =$$
$$\frac{\partial}{\partial x} \left[ \phantom{\frac{\partial}{\partial y} } \! \! \! \! \! \! I_{\mathfrak S^2_{\infty}} \left[ \phantom{|} (x, y, z) \, \right]:x \right]_{(0, 0, 1)} =$$
$$\text{lim}_{\{ \epsilon \rightarrow 0 \}} [ \, \frac{ I_{\mathfrak S^2_{\infty}}[ \, (0, 0 + \epsilon, 1) \, ]:x – I_{\mathfrak S^2_{\infty}}[ \, (0, 0, 1) \, ]:x}{\epsilon} \, ] =$$
$$\text{lim}_{\{ \epsilon \rightarrow 0 \}} [ \, \frac{ 0 – 0}{\epsilon} \, ] =$$
$$\text{lim}_{\{ \epsilon \rightarrow 0 \}} [ \, 0 \, ] =$$
$$0$$.

Likewise:
$$J_{yy}[ \, I_{\mathfrak S^2_{\infty}} \, ]_{(0, 0, 1)} = 1$$
and
$$J_{yx}[ \, I_{\mathfrak S^2_{\infty}} \, ]_{(0, 0, 1)} = 0$$.

And
$$J_{zx}[ \, I_{\mathfrak S^2_{\infty}} \, ]_{(0, 0, 1)} = 0$$,
and
$$J_{zy}[ \, I_{\mathfrak S^2_{\infty}} \, ]_{(0, 0, 1)} = 0$$

Further:
$$J_{xz}[ \, I_{\mathfrak S^2_{\infty}} \, ]_{(0, 0, 1)} =$$
$$\frac{\partial}{\partial z} \left[ \phantom{\frac{\partial}{\partial z} } \! \! \! \! \! \! I_{\mathfrak S^2_{\infty}} \left[ \phantom{|} (x, y, z) \, \right]:x \right]_{(0, 0, 1)} =$$
$$\text{lim}_{\{ \epsilon \rightarrow 0 \}} [ \, \frac{ I_{\mathfrak S^2_{\infty}}[ \, (0, 0, 1 + \epsilon ) \, ]:x – I_{\mathfrak S^2_{\infty}}[ \, (0, 0, 1) \, ]:x}{\epsilon} \, ] =$$

???

For any values $$\epsilon$$ except $$\epsilon := 0$$ or $$\epsilon := -2$$ is the tripel $$(0, 0, 1 + \epsilon)$$ not in the domain of function $$I_{\mathfrak S^2_{\infty}}$$.
Now what ? …

p.p.s.
The “Introduction to piecewise flat manifolds” linked above I do find quite interesting in its own right. Since it refers to “abstract line segments called edges” and “abstract triangles called faces”, would it be possible to consider whether or not some such “face” and some such “edge” (which doesn’t belong to the boundery of the “face” under consideration) have (nevertheless) one or more “(abstract) point” in common ? (If that’s conceivable, but not explicitly excluded, I could think of some “interesting examples”. …) Also, notably: The notion of “curvature at a vertex” introduced there (Def. 6) is very different from my understanding of curvature (“at a point”) discussed above.

35. Chrys schrieb (31.01.2021, 16:05 o’clock):
> […] roughly as objects which “look locally flat if you zoom in”.

I certainly admit that such phrasing triggers my pet-peeve reflex, quite independent from further technicalities about manifolds.

The curvature of a given sphere, for instance, is constant; regardless of the size of the patch (as long as its not “size zero”, nor “of too low dimension”) on which its value is determined, patch by patch.
And the value is determined not by “looking”, but by evaluating vanishing Cayley-Menger determinants (in terms of “extrinsic distances”, solving for the unknown “R^2”) or vanishing Gram determinants (in terms of “intrinsic distances”, solving for the unknown “√{ κ }”).

> For instance, a picture I can get from her wording is that of an embedded surface in 3-space which allows to attach a tangent plane to each of its points.

Presumably you’re thereby thinking of any embedded surface in (flat) 3-space which “at every point” (i.e. in the limit of vanishing patch size) has (finite) extrinsic curvature; in other words: such that the smallest circumspheres for any four points [1] of any patch will have radii not below some limiting nonzero value.
Clearly such surfaces are deserving of interest and deserving of an exclusive short-hand name for themselves. But: it is not “(2-dimensional) manifold”. (Using everyday language I’d call them “smooth surfaces”, “without creases, or cusps”.)

> Zooming in to a point would give a local view of the surface where it comes closer and closer to the tangent plane at this point.

The maximal (flat 3-space) distance between some point “of the surface patch under consideration” and some tangent plane which touches the surface in some other point of this patch decreases with patch size; and indeed generally and eventually over-proportionally (quadratically) so.
The smaller the patch, the better the surface is approximated by any plane touching the patch; absolutely so (anyways), and even relatively so (with distance normalized to the patch size).

But such considerations obviously presume the notion of a “plane (itself)”, as if the prerequisits for defining “a plane” were any different from the prerequisits for defining the notion of any other surface of constant nonzero curvature “itself”. (I’m in favor of being explicit about prerequisits, rather than silently presuming “intuitions”.)

> […] focus was clearly not on piecewise flat manifolds.

Granted. But writing that

“manifolds are roughly” objects of a certain description

doesn’t convey quite the same meaning as writing that

.

Moreover, it remains to be resolved whether there exist Diffeomorphisms between any two of those piecewise flat manifolds, or between any one of them and, say, a sphere (as may be of particular interest to topologists, by Katie Steckles comment 02.10.2020, 15:48 o’clock).

[1: Looking at the Introduction to piecewise flat manifolds you had linked above led me to recall what I had understood long ago but failed to appreciate more recently: concerning surfaces or spaces of larger dimension, $$n \ge 2$$, we should consider the radii of circumspheres of not just any $$n + 2$$ points of some patch, but only for “selected $$n + 2$$ points”; e.g. with $$n + 1$$ points constrained to a sphere which has the remaining point as its center. ]

p.s.
> Btw. […]

Let me go ahead and try to calculate some entries into the [[Jacobian determinant]] for the suggested function,

$$I_{\mathfrak S^2_{\infty}} : \mathfrak S^2_{\infty} \longleftrightarrow \mathfrak S^2_{\infty}, \qquad I_{\mathfrak S^2_{\infty}}[ \, x \, ] \mapsto x$$,

at least for some selected points (of its domain); for starters at coordinate point $$\mathbf x \equiv (x, y, z) := (0, 0, 1)$$. I get:

$$J_{xx}[ \, I_{\mathfrak S^2_{\infty}} \, ]_{(0, 0, 1)} =$$
$$\frac{\partial}{\partial x} \left[ \phantom{\frac{\partial}{\partial x} } \! \! \! \! \! \! I_{\mathfrak S^2_{\infty}} \left[ \phantom{|} (x, y, z) \, \right]:x \right]_{(0, 0, 1)} =$$
$$\text{lim}_{\{ \epsilon \rightarrow 0 \}} [ \, \frac{ I_{\mathfrak S^2_{\infty}}[ \, (0 \, + \, \epsilon, 0, 1) \, ]:x \, – \, I_{\mathfrak S^2_{\infty}}[ \, (0, 0, 1) \, ]:x}{\epsilon} \, ] =$$
$$\text{lim}_{\{ \epsilon \rightarrow 0 \}} [ \, \frac{ (0 \, + \, \epsilon) \, – \, 0}{\epsilon} \, ] =$$
$$\text{lim}_{\{ \epsilon \rightarrow 0 \}} [ \, 1 \, ] =$$
$$1$$.

$$J_{xy}[ \, I_{\mathfrak S^2_{\infty}} \, ]_{(0, 0, 1)} =$$
$$\frac{\partial}{\partial x} \left[ \phantom{\frac{\partial}{\partial y} } \! \! \! \! \! \! I_{\mathfrak S^2_{\infty}} \left[ \phantom{|} (x, y, z) \, \right]:x \right]_{(0, 0, 1)} =$$
$$\text{lim}_{\{ \epsilon \rightarrow 0 \}} [ \, \frac{ I_{\mathfrak S^2_{\infty}}[ \, (0, 0 \, + \, \epsilon, 1) \, ]:x \, – \, I_{\mathfrak S^2_{\infty}}[ \, (0, 0, 1) \, ]:x}{\epsilon} \, ] =$$
$$\text{lim}_{\{ \epsilon \rightarrow 0 \}} [ \, \frac{ 0 \, – \, 0}{\epsilon} \, ] =$$
$$\text{lim}_{\{ \epsilon \rightarrow 0 \}} [ \, 0 \, ] =$$
$$0$$.

Likewise:
$$J_{yy}[ \, I_{\mathfrak S^2_{\infty}} \, ]_{(0, 0, 1)} = 1$$
and
$$J_{yx}[ \, I_{\mathfrak S^2_{\infty}} \, ]_{(0, 0, 1)} = 0$$.

And
$$J_{zx}[ \, I_{\mathfrak S^2_{\infty}} \, ]_{(0, 0, 1)} = 0$$,
and
$$J_{zy}[ \, I_{\mathfrak S^2_{\infty}} \, ]_{(0, 0, 1)} = 0$$.

Further:
$$J_{xz}[ \, I_{\mathfrak S^2_{\infty}} \, ]_{(0, 0, 1)} =$$
$$\frac{\partial}{\partial z} \left[ \phantom{\frac{\partial}{\partial z} } \! \! \! \! \! \! I_{\mathfrak S^2_{\infty}} \left[ \phantom{|} (x, y, z) \, \right]:x \right]_{(0, 0, 1)} =$$
$$\text{lim}_{\{ \epsilon \rightarrow 0 \}} [ \, \frac{ I_{\mathfrak S^2_{\infty}}[ \, (0, 0, 1 \, + \, \epsilon ) \, ]:x \, – \, I_{\mathfrak S^2_{\infty}}[ \, (0, 0, 1) \, ]:x}{\epsilon} \, ] =$$

???

For any values $$\epsilon$$ except $$\epsilon := 0$$ or $$\epsilon := -2$$ is the tripel $$(0, 0, 1 \, + \, \epsilon)$$ not in the domain of function $$I_{\mathfrak S^2_{\infty}}$$.
Now what ? …

p.p.s.
The “Introduction to piecewise flat manifolds” linked above I do find quite interesting in its own right. Since it refers to “abstract line segments called edges” and “abstract triangles called faces”, would it be possible to consider whether or not some such “face” and some such “edge” (which doesn’t belong to the boundery of the “face” under consideration) have (nevertheless) one or more “(abstract) point” in common ? (If that’s conceivable, but not explicitly excluded, I could think of some “interesting examples”. …)
Also, notably: The notion of “curvature at a vertex” introduced there (Def. 6) is very different from my understanding of curvature (“at a point”) discussed above.

• Frank Wappler wrote (03.02.2021, 11:19 o’clock):
> […] Let me go ahead and try to calculate some entries into the [[Jacobian determinant]] for the suggested function, […]
> […] $$J_{xy}[ \, I_{\mathfrak S^2_{\infty}} \, ]_{(0, 0, 1)} =$$

… should have been:

$$\frac{\partial}{\partial y} \left[ \phantom{\frac{\partial}{\partial y} } \! \! \! \! \! \! I_{\mathfrak S^2_{\infty}} \left[ \phantom{|} (x, y, z) \, \right]:x \right]_{(0, 0, 1)} =$$
$$\text{lim}_{\{ \epsilon \rightarrow 0 \}} [ \, \frac{ I_{\mathfrak S^2_{\infty}}[ \, (0, 0 \, + \, \epsilon, 1) \, ]:x \, – \, I_{\mathfrak S^2_{\infty}}[ \, (0, 0, 1) \, ]:x}{\epsilon} \, ] =$$
$$\text{lim}_{\{ \epsilon \rightarrow 0 \}} [ \, \frac{ 0 \, – \, 0}{\epsilon} \, ] =$$
$$\text{lim}_{\{ \epsilon \rightarrow 0 \}} [ \, 0 \, ] =$$
$$0$$.

> […] Likewise

… but here spelt out for comparison:

$$J_{yx}[ \, I_{\mathfrak S^2_{\infty}} \, ]_{(0, 0, 1)} =$$
$$\frac{\partial}{\partial x} \left[ \phantom{\frac{\partial}{\partial x} } \! \! \! \! \! \! I_{\mathfrak S^2_{\infty}} \left[ \phantom{|} (x, y, z) \, \right]:y \right]_{(0, 0, 1)} =$$
$$\text{lim}_{\{ \epsilon \rightarrow 0 \}} [ \, \frac{ I_{\mathfrak S^2_{\infty}}[ \, (0 \, + \, \epsilon, 0, 1) \, ]:y \, – \, I_{\mathfrak S^2_{\infty}}[ \, (0, 0, 1) \, ]:y}{\epsilon} \, ] =$$
$$\text{lim}_{\{ \epsilon \rightarrow 0 \}} [ \, \frac{ 0 \, – \, 0}{\epsilon} \, ] =$$
$$\text{lim}_{\{ \epsilon \rightarrow 0 \}} [ \, 0 \, ] =$$
$$0$$.

p.s.
I just noticed that the comment I had submitted on 03.02.2021, 11:10 o’clock, which happened to contain three links (as became painfully clear to me only after my attempt at submitting it was answered with the infamous »Dein Kommentar befindet sich in Moderation.«) now appears on this webpage, too.

I can think of more tolerable and reliable modes for submitting, moderating and presenting comments … And I call again for the applicable comment policy to be declared publicly.

36. @Frank Wappler / 03.02.2021, 11:10 o’clock

To give you a comparable textbook example, let me quote from Donald W. Kahn, Introduction to Global Analysis, AP, 1980:

A manifold is a nice topological space that in the small is just like Euclidean space (a rigorous definition will follow). To do mathematical analysis on manifolds, one imposes some condition of smoothness or differentiability concerning the relation of any two nearby pieces, each of which is like Euclidean space. […] A typical example would be the ordinary sphere (the surface of the earth). A nice small region—such as a polar ice cap—is virtually indistinguishable from a small region in the plane, while the entire manifold, the sphere, is fundamentally different from the plane.

That’s not too far away from what Katie Steckles has written here, isn’t it? Would your question concerning the surface of a cube have been answered by Donald Kahn’s informal exposition?

John M. Lee, who has authored several books on the subject matter, suggests that a physicist’s idea of an n-dimensional manifold is just that of an object with n degrees of freedom. Well, this would allow a very concise answer to your question.

The restriction of identity map in $$mathbb{R}^3$$ to the cubic sphere, $$I_{\mathfrak{S}_\infty}:\mathfrak{S}_\infty \to \mathbb{R}^3$$, is smooth at every $$x = x^i e_i$$ with $$|x^i| = 1$$ for exactly one index $$i$$, that is, if $$x$$ is not a vertex or on an edge of the cube. As long as you stay away from the edges and corners you’re not in trouble, so we can say this map is piecewise smooth.

At those $$x$$ where it exists, the derivative, or tangent map, $$T_x I_{\mathfrak{S}_\infty}$$, is a linear map $$T_x\mathfrak{S}_\infty \cong\mathbb{R}^2 \to T_x\mathbb{R}^3 \cong \mathbb{R}^3$$ which can be expressed by a familiar Jacobian matrix. In particular, at $$x = (0,x^2,1)$$ we find
$(T_{(0,x^2,1)}I_{\mathfrak{S}_\infty})_{ij} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 0 \\ \end{pmatrix},$
and likewise, at $$x = (0,1,x^3)$$,
$(T_{(0,1,x^3)}I_{\mathfrak{S}_\infty})_{ij} = \begin{pmatrix} 1 & 0 \\ 0 & 0 \\ 0 & 1 \\ \end{pmatrix}.$
Thus we get
$\lim_{x^2 \to 1}T_{(0,x^2,1)}I_{\mathfrak{S}_\infty} \neq \lim_{x^3 \to 1}T_{(0,1,x^3)}I_{\mathfrak{S}_\infty},$
which implies that $$I_{\mathfrak{S}_\infty}$$ is not smooth at $$x = (0,1,1)$$ and, consequently, is not a smooth diffeomorphism of the cubic sphere onto itself.

37. Chrys wrote (08.02.2021, 09:45 o’clock):
> To give you a comparable textbook example, let me quote from Donald W. Kahn, Introduction to Global Analysis, AP, 1980:

A manifold is a nice topological space that in the small is just like Euclidean space (a rigorous definition will follow).

The comparison required here can obviously only consider topological properties of Euclidean space (i.e. more specificly: [[Euclidean topology]], or perhaps equivalent: [[Box topology]] of real n-tuples); without direct appeal to metric properties of Euclidean space (especially the vanishing of all Cayley-Menger determinants of distances between sufficiently many — $$n + 2$$ — points). I trust that the promised »rigorous definition« does not involve any mentioning of »distance«.

To do mathematical analysis on manifolds, one imposes some condition of smoothness or differentiability concerning the relation of any two nearby pieces, each of which is like Euclidean space.

Again, the comparison surely only involves topology; and therefore does not claim that the »pieces« under consideration had to be flat as Euclidean space.

A typical example would be the ordinary sphere (the surface of the earth).

This example, which arguably appears outright as a particular metric space, surely is to be understood “together with the obvious suitable manifold topology”; similar to how I introduced “the surface of a cube” above.

A nice small region—such as a polar ice cap—is virtually indistinguishable from a small region in the plane, while the entire manifold, the sphere, is fundamentally different from the plane.

As far as the comparisons are meant to be refer to topology, Kahn’s statements can be understood as being correct: “indistinguishability in the small” referring to the rigorous definition ([[homeomorphism]]), and “fundamental difference on the wholes” referring to a sphere being a compact topological space, while an entire infinite plane is not.

However, if Donald W. Kahn (1935 – 2015) meant to suggest that some small segment of a sphere had a curvature radius different from the nonzero curvature radius of the sphere as a whole, I’d have to disagree.

> That’s not too far away from what Katie Steckles has written here, isn’t it?

Kahn may have considered only those surfaces »typical examples (of 2-dim. manifolds)« whose curvature is “everywhere bounded”.
But that is still far from saying that surfaces whose curvature is not “everywhere bounded” could therefore not be (given a topology of) a manifold, as Katie Steckles appearently suggested in the above SciLogs article, if that’s what she intended.

> Would your question concerning the surface of a cube have been answered by Donald Kahn’s informal exposition?

The surface of a cube would stand as an atypical example to Kahn’s supposed classification of manifolds. In contrast, to Katie Steckles’ apparent claim of “what is a manifold” it stands as counterexample (if indeed it’s nevertheless called manifold).

> The restriction of identity map in $$\mathbb R^3$$ to the cubic sphere, $$I_{\mathfrak{S}_\infty} : \mathfrak{S}_\infty \to \mathbb{R}^3$$ […]
> […] where it exists, the derivative, or tangent map, $$T_x I_{\mathfrak{S}_\infty}$$ is a linear map $$T_x\mathfrak{S}_\infty \cong\mathbb{R}^2 \to T_x\mathbb{R}^3 \cong \mathbb{R}^3$$ which can be expressed by a familiar Jacobian matrix. In particular […] we find $(T_{(0,y,1)}I_{\mathfrak{S}_\infty})_{ij} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 0 \\ \end{pmatrix},$

How exactly did “we find” the two values $$0$$ in the third line ?? …

And if for some function … $$\beta : \mathfrak{S}^2_\infty \rightarrow \mathbb R$$ … happens to hold:

$$\forall -1 \lt z \lt 1 \, \exists \text{Limit}_{\{x \, \rightarrow \, 1 \}}\partial_x \beta[ \, (x, 1, z) \, ]$$,

$$\forall -1 \lt z \lt 1 \, \exists \text{Limit}_{\{y \, \rightarrow \, 1 \}}\partial_y \beta[ \, (1, y, z) \, ]$$,

and

$$\forall -1 \lt z \lt 1 \, \text{Limit}_{\{x \, \rightarrow \, 1 \}}\partial_z \beta[ \, (x, 1, z) \, ] = \text{Limit}_{\{y \, \rightarrow \, 1 \}}\partial_z \beta[ \, (1, y, z) \, ]$$,

then why not assign

$$\partial_x \beta[ \, (1, 1, z) \, ] := \text{Limit}_{\{x \, \rightarrow \, 1 \}}\partial_x \beta[ \, (x, 1, z) \, ]$$,

$$\partial_y \beta[ \, (1, 1, z) \, ] := \text{Limit}_{\{y \, \rightarrow \, 1 \}}\partial_x \beta[ \, (1, y, z) \, ]$$,

along with

$$\partial_z f\beta[ \, (1, 1, z) \, ] := \text{Limit}_{\{x \, \rightarrow \, 1 \}}\partial_x \beta[ \, (x, 1, z) \, ]$$ ??

Anyways — let me affirm (since I was not always have been perfectly clear above) that I do not intend to claim that set $$\mathfrak S^2_{\infty} \equiv \{ x \in \mathbb R^3 : \|x\|_{\infty} = 1 \}$$ is a differentiable manifold at all. It is a 2-dimensional topological manifold (as, I believe, we’ve established and agreed); but the neighborhoods of all points on the edges, incl. vertices, don’t have 2-dimensional coordinate charts assigned (but, obviously, always genuinely 3-dimensional coordinate charts). A few times above I had mistakenly mentioned [[diffeomorphism]], which of course presumes differentiable manifolds from the outset, while I was merely interested in how to possibly define and evaluate differentiability of functions on $$\mathfrak S^2_{\infty}$$.

Instead, the p.s. of my comment 23.12.2020 showed an atlas $$\mathfrak A_{S^2_6}$$ of six overlapping coordinate patches assigned to all points of a sphere $$S^2$$, such that $$S^2, \mathfrak A_{S^2_6}$$ constitutes a differentiable manifold;
and, by using the inverse of function $$f : \Sigma^2 \leftrightarrow S^2$$ discussed earlier, also gave an implicit construction of an atlas $$\mathfrak A_{\Sigma^2_6}$$ of six overlapping coordinate patches assigned to all points of the surface of a cube, $$\Sigma^2$$, by which $$\Sigma^2, \mathfrak A_{\Sigma^2_6}$$ constitutes a differentiable manifold.

38. Chrys wrote (08.02.2021, 09:45 o’clock):
> To give you a comparable textbook example, let me quote from Donald W. Kahn, Introduction to Global Analysis, AP, 1980:

A manifold is a nice topological space that in the small is just like Euclidean space (a rigorous definition will follow).

The comparison required here can obviously only consider topological properties of Euclidean space (i.e. more specificly: [[Euclidean topology]], or perhaps equivalent: [[Box topology]] of real n-tuples); without direct appeal to metric properties of Euclidean space (especially the vanishing of all Cayley-Menger determinants of distances between sufficiently many — $$n + 2$$ — points). I trust that the promised »rigorous definition« does not involve any mentioning of »distance«.

To do mathematical analysis on manifolds, one imposes some condition of smoothness or differentiability concerning the relation of any two nearby pieces, each of which is like Euclidean space.

Again, the comparison surely only involves topology; and therefore does not claim that the »pieces« under consideration had to be flat as Euclidean space.

A typical example would be the ordinary sphere (the surface of the earth).

This example, which arguably appears outright as a particular metric space, surely is to be understood “together with the obvious suitable manifold topology”; similar to how I introduced “the surface of a cube” above.

A nice small region—such as a polar ice cap—is virtually indistinguishable from a small region in the plane, while the entire manifold, the sphere, is fundamentally different from the plane.

As far as the comparisons are meant to be refer to topology, Kahn’s statements can be understood as being correct: “indistinguishability in the small” referring to the rigorous definition ([[homeomorphism]]), and “fundamental difference on the wholes” referring to a sphere being a compact topological space, while an entire infinite plane is not.

However, if Donald W. Kahn (1935 – 2015) meant to suggest that some small segment of a sphere had a curvature radius different from the nonzero curvature radius of the sphere as a whole, I’d have to disagree.

> That’s not too far away from what Katie Steckles has written here, isn’t it?

Kahn may have considered only those surfaces »typical examples (of 2-dim. manifolds)« whose curvature is “everywhere bounded”.
But that is still far from saying that surfaces whose curvature is not “everywhere bounded” could therefore not be (given a topology of) a manifold, as Katie Steckles appearently suggested in the above SciLogs article, if that’s what she intended.

> Would your question concerning the surface of a cube have been answered by Donald Kahn’s informal exposition?

The surface of a cube would stand as an atypical example to Kahn’s supposed classification of manifolds. In contrast, to Katie Steckles’ apparent claim of “what is a manifold” it stands as counterexample (if indeed it’s nevertheless called manifold).

> The restriction of identity map in $$\mathbb R^3$$ to the cubic sphere, $$I_{\mathfrak{S}_\infty} : \mathfrak{S}_\infty \to \mathbb{R}^3$$ […]
> […] where it exists, the derivative, or tangent map, $$T_x I_{\mathfrak{S}_\infty}$$ is a linear map $$T_x\mathfrak{S}_\infty \cong\mathbb{R}^2 \to T_x\mathbb{R}^3 \cong \mathbb{R}^3$$ which can be expressed by a familiar Jacobian matrix. In particular […] we find $(T_{(0,y,1)}I_{\mathfrak{S}_\infty})_{ij} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 0 \\ \end{pmatrix},$

How exactly did “we find” the two values $$0$$ in the third line ?? …

And if for some function … $$\beta : \mathfrak{S}^2_{\infty} \rightarrow \mathbb R$$ … happens to hold:

$$\forall z \in (\text{-}1] … [1) \, \exists \text{Limit}_{\{x \, \rightarrow \, 1 \}}\partial_x \beta[ \, (x, 1, z) \, ]$$,

$$\forall z \in (\text{-}1] … [1) \, \exists \text{Limit}_{\{y \, \rightarrow \, 1 \}}\partial_y \beta[ \, (1, y, z) \, ]$$,

and

$$\forall z \in (\text{-}1] … [1) \, \text{Limit}_{\{x \, \rightarrow \, 1 \}}\partial_z \beta[ \, (x, 1, z) \, ] = \text{Limit}_{\{y \, \rightarrow \, 1 \}}\partial_z \beta[ \, (1, y, z) \, ]$$,

then why not assign for $$z \in (\text{-}1] … [1)$$ (i.e. “along the edge, excluding its ends”):

$$\partial_x \beta[ \, (1, 1, z) \, ] := \text{Limit}_{\{x \, \rightarrow \, 1 \}}\partial_x \beta[ \, (x, 1, z) \, ]$$,

$$\partial_y \beta[ \, (1, 1, z) \, ] := \text{Limit}_{\{y \, \rightarrow \, 1 \}}\partial_y \beta[ \, (1, y, z) \, ]$$,

along with

$$\partial_z \beta[ \, (1, 1, z) \, ] := \text{Limit}_{\{x \, \rightarrow \, 1 \}}\partial_z \beta[ \, (x, 1, z) \, ]$$ ??

(And so on for derivatives “at the vertices” of $$\mathfrak{S}^2_{\infty}$$ …)

Anyways — let me affirm (since I was not always perfectly clear above) that I do not intend to claim that set $$\mathfrak S^2_{\infty} \equiv \{ x \in \mathbb R^3 : \|x\|_{\infty} = 1 \}$$ is a differentiable manifold at all. It is a 2-dimensional topological manifold (as, I believe, we’ve established and agreed); but the neighborhoods of all points on the edges, incl. vertices, don’t have 2-dimensional coordinate charts assigned (but, obviously, always genuinely 3-dimensional coordinate charts). A few times above I had mistakenly mentioned [[diffeomorphism]], which of course presumes differentiable manifolds from the outset, while I was merely interested in how to possibly define and evaluate differentiability of functions on $$\mathfrak S^2_{\infty}$$.

Instead, the p.s. of my comment 23.12.2020 showed an atlas $$\mathfrak A_{S^2_6}$$ of six overlapping coordinate patches assigned to all points of a sphere $$S^2$$, such that $$(S^2, \mathfrak A_{S^2_6})$$ constitutes a differentiable manifold;
and, by using the inverse of function $$f : \Sigma^2 \leftrightarrow S^2$$ discussed earlier, also gave an implicit construction of an atlas $$\mathfrak A_{\Sigma^2_6}$$ of six overlapping coordinate patches assigned to all points of the surface of a cube, $$\Sigma^2$$, by which $$(\Sigma^2, \mathfrak A_{\Sigma^2_6})$$ constitutes a differentiable manifold.

• Frank Wappler wrote (10.02.2021, 18:33 o’clock):
> {…] And if for some function … $$\beta : \mathfrak S^2_{\infty} \rightarrow \mathbb R$$ … happens to hold: […]

Sorry — my subsequent argument was not quite to the point. (I’d like to retract it for now; perhaps returning to some variant later.)

Let me instead revisit a 1-dimensional case once again, in order to seek agreement on terminology and, let’s say, on the nature of my difficulty:

Consider therefore two somewhat related functions, say $$\alpha$$ and $$\beta$$ (I’m recycling this function name, since gradually we’re running out of innocuous function names), defined (just for the sake of my following question) as:

$\alpha : \mathbb R \rightarrow ([0 … 1]), \qquad \alpha[ \, x \, ] \mapsto \begin{cases} 0 & \text{if } x \leq 0 \\ x & \text{if } 0 \leq x \leq 1 \\ 1 & \text{if } 1 \leq x \end{cases}$

and

$\beta : ([0 … 1]) \rightarrow ([0 … 1]), \qquad \beta[ \, x \, ] \mapsto x$.

At argument values $$x := 0$$ and $$x := 1$$ function $$\alpha$$ surely is continuous, and not differentiable; and the reasons for this assessment may be spelt out in detail.

But what about function $$\beta$$ ? — Is there any reason to conclude that $$\beta$$ is not differentiable ?

39. @Frank Wappler / 10.02.2021, 18:33 o’clock

Donald Kahn does not say what exactly it means for a topological space to be “in the small is just like Euclidean space“. Is then the surface of a cube near a vertex just like Euclidean space? That’s far from being obvious, but one should not expect answers to this kind of questions from brief motivating introductions.

If you compare the various definitions of differentiability in literature, you will notice that different authors may use slightly differing assumptions. For $$x_0 \in \Omega \subset \mathbb{R}^n$$, one may define a real function $$f:\Omega \to \mathbb{R}$$ to be differentiable at $$x_0$$ under the assumption that $$x_0$$ is an interior point of $$\Omega$$. For example, you’ll find this in your Bronstein or, at least for the case $$n=1$$, at en.wikipedia under Differentiability of real functions. Basically the same you will also see if you follow the “Deutsch” link on that page.

Clearly, to say that $$f \in C^k(\Omega)$$ then requires the domain $$\Omega$$ to be open, although one may be interested in functions which, along with their derivatives up to order $$k$$, can be continuously extended to the closure $$\overline{\Omega}$$, i.e., in functions which are said to belong to $$C^k(\overline{\Omega})$$.

Especially in the case $$n=1$$, some authors prefer to assume that $$x_0 \in \Omega$$ is a cluster point of $$\Omega$$ and not necessarily an interior point. However, they usually do not follow this strategy for $$n \gt 1$$ and then switch again to interior points. Of course, for your example $$\beta:[0,1] \to \mathbb{R}$$, these authors would have the one-sided derivatives at the end points of the interval already included in their general definition, whereas an additional argument is needed here if differentiability is formally defined at interior points only. That is to say, $$\beta$$ is right differentiable at $$0$$ because we can certainly find an open neighborhood $$0 \in U \subset \mathbb{R}$$ and a function $$\tilde{\beta}:U \to \mathbb{R}$$ which is differentiable at $$x_0 = 0$$ such that $$\beta(x) = \tilde{\beta}(x)$$ for all $$x \in U \cap [0,1]$$. By an analogous argument, $$\beta$$ is left differentiable at the right end of the interval.

To conclude that your function $$\alpha$$ is not differentiable at $$0$$ it is sufficient to note that there is a jump discontinuity for the derivative $$\alpha^{\prime}$$ at this point, $$\lim_{x \nearrow 0} \alpha^{\prime}(x) \neq \lim_{x \searrow 0} \alpha^{\prime}(x)$$. Of course, you can argue that the left and right derivatives of $$\alpha$$ exist but differ at $$0$$, but we would not have to introduce the one-sided derivatives for this purpose.

And similarly, $$I_{\mathfrak{S}_\infty}$$ is seen to be not differentiable on on an edge of the cube because the Jacobian has a jump discontinuity there.

40. Chrys schrieb (15.02.2021, 12:08 o’clock):
> […] Especially in the case $$n = 1$$, some authors prefer to assume that $$x_0 \in \Omega$$ is a cluster point of $$\Omega$$ […]

Indeed. (Or rather: even the notion “cluster point” seems to have several interpretations … see for instance https://math.stackexchange.com/questions/853526/definition-of-cluster-point.)

Relevant is that typically for each point of a domain $$\Omega$$ there are several distinct “approaching paths” within domain $$\Omega$$; which allows to make comparisons of “things calculated along the way” (especially: limits, as far as they exist).

Quantitative comparisons (e.g. whether “the left and the right derivatives” of a particular function wrt. one and the same domain variable are equal, or not) between the results from distinct approaches should only be made if the results being compared are qualitatively the same for the distinct approaches; with the foremost condition that both exist and can be evaluated along the respective paths.

Unlike, for instance, difference quotients wrt. $$z$$ of a function and for an approach path, where the domain provides only exactly one $$z$$ value (e.g. value $$z_{\text{(x,y) face}}$$) for any point of any path within one of the two $$(x,y)$$ faces of a cube surface);
and in decisive qualitative distinction to difference quotients wrt. $$z$$, of the same function, for an approach path with varying $$z$$ values, within one of the two $$(x,z)$$ faces of a cube surface, or just as well within one of the two $$(y,z)$$ faces of a cube surface, which do exist.

> […] different authors may use slightly differing assumptions […]

An excellent opportunity to express my thanks, declare victory, and wander off (for the Weekend ;) …

41. Yes. Manifolds are differentiable/topological structures upon which one can do calculus (generally used in differential geometry); tangent space and geodesic properties are usually the target or a target of study. Varieties are algebraic structures that arise from the intersections of a set of equations, which can be realized geometrically. So, if you imagine wrapping your couch in a blanket, you can get an idea of what a manifold might be. If you shoot an arrow straight through a basketball, you’ll get two puncture points, which give you an algebraic variety (a real algebraic variety, as the intersections only involve real numbers).