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Stream: learning: questions

Topic: probing with closed vs open balls

Amar Hadzihasanovic (May 19 2021 at 11:52):

The topos of smooth sets is one of the categories of “generalised manifolds” that people consider; the idea is that a generalised manifold is a space “probed by smooth open balls”, which translates to “a sheaf on the site of smooth open balls and smooth maps, with the coverage given by open covers”.
One can then show that both the category of manifolds and smooth maps, and the category of diffeological spaces are full subcategories of the topos of smooth sets.

My question is simply: what if we replace smooth open balls with smooth closed balls (seen as manifolds with boundary) and closed covers?

We should still have a functor from the category of smooth manifolds with boundary, given by $M \mapsto \mathrm{Hom}(-,M)$. Is this fully faithful, or is there something that is “missed” about manifolds by probing with closed balls as opposed to open balls?

Amar Hadzihasanovic (May 19 2021 at 12:02):

For motivation: this is connected to my earlier question on the exterior product.

Because integration of differential forms is done on smooth chains, which are sums of smooth cells, that is, smooth maps from closed balls, it seems that the natural setting for this theory is “spaces probed by smooth closed balls”.

Morgan Rogers (he/him) (May 19 2021 at 12:08):

Amar Hadzihasanovic said:

My question is simply: what if we replace smooth open balls with smooth closed balls (seen as manifolds with boundary) and closed covers?

We should still have a functor from the category of smooth manifolds with boundary, given by $M \mapsto \mathrm{Hom}(-,M)$. Is this fully faithful, or is there something that is “missed” about manifolds by probing with closed balls as opposed to open balls?

It's certainly faithful, since 0-dimensional balls are just points. I'm sure you can prove fullness with just a little work.

Amar Hadzihasanovic (May 19 2021 at 12:26):

Oh no, don't make me do work... :grinning_face_with_smiling_eyes:

Amar Hadzihasanovic (May 19 2021 at 12:31):

I suppose I would just find it surprising if the net effect of using open rather than closed balls were just... to not be able to see boundaries and corners?

Fawzi Hreiki (May 19 2021 at 12:32):

I find it difficult to see (just naively) how you could build all manifolds using colimits of closed balls

Amar Hadzihasanovic (May 19 2021 at 12:36):

I guess the asymmetry between open and closed is given by the fact that through a smooth map, boundary points can map onto interior points, but interior points cannot map onto boundary points.

Amar Hadzihasanovic (May 19 2021 at 13:18):

Anyway, I think Morgan is right that a little work seems to suffice to prove fullness.

Namely, if we have $f: \mathrm{Hom}(-,M) \to \mathrm{Hom}(-,N)$, then the component on points/closed 0-balls fixes what its potential pre-image $\hat{f}: M \to N$ must be, so it suffices to prove that it is smooth.
Now every $x \in M$ has an open neighbourhood with a chart $\varphi$ in a smooth atlas that identifies it either with Euclidean $n$-space or with half Euclidean $n$-space. In both cases we can take a closed $n$-ball $D$ containing $\varphi(x)$, and locally inverting $\varphi$ gives us a smooth map $\psi: D \to M$ almost by definition. Then the composite $\hat{f} \circ \psi: D \to N$ is equal to $f(\psi) \in \mathrm{Hom}(D,N)$, which is smooth by assumption, and that's enough.

Amar Hadzihasanovic (May 19 2021 at 13:21):

So maybe it is the case that people work with open balls only because they don't want any boundaries around.

Reid Barton (May 19 2021 at 14:11):

The issue is that the topology is not subcanonical, so $\mathrm{Hom}(-, M)$ is not a sheaf. For example, $\mathrm{Hom}(-, \mathbb{R})$ is not a sheaf.
The problem is that a function which is a smooth on each member of a closed cover (even if we require the cover to be locally finite or something) need not be smooth. For example, if $f : [0, 2] \to \mathbb{R}$ is smooth on $[0, 1]$ and $[1, 2]$ then there can still be a problem at $1$.

Amar Hadzihasanovic (May 19 2021 at 14:21):

Ah, you're right. Is there a simple fix? Would it work to require a closed cover to restrict, on the interiors, to an open cover of the interior?

Reid Barton (May 19 2021 at 14:27):

I'm not sure actually. I guess the next example would be something like two closed disks in $\mathbb{R^2}$ (I'm imagining two circles of the same radius that overlap, plus their interiors). There are two points where the boundary circles intersect, and I'm not sure whether a function which is smooth on each circle will necessarily be smooth at those intersection points.

Reid Barton (May 19 2021 at 14:27):

You might also need a condition like "the interiors of the boundaries cover the boundary", and so on(?).

Amar Hadzihasanovic (May 19 2021 at 14:38):

Yes, intuitively it seems like the function doesn't even need to be differentiable at the intersection points. I'm picturing a smooth surface over the two disks with two cusps at the intersection points.

Jens Hemelaer (May 19 2021 at 15:49):

I replied earlier, but my reply didn't take into account the important distinction between manifolds with and without boundary here.

A manifold with boundary is locally isomorphic to either $\mathbb{R}^n$ or the halfspace $H^n$ (elements of $\mathbb{R}^n$ that have last coordinate $\geq 0$).

As site for a topos of generalized manifolds with boundary, you can take the category of manifolds with boundary, and smooth maps between them. A sieve is a covering sieve if it contains an open covering. Every manifold admits an open covering by spaces that are each isomorphic to either $\mathbb{R}^n$ or $H^n$ for some natural number $n$. By the Comparison Lemma, this means that the full subcategory on the objects $\mathbb{R}^n$ and $H^n$ form an alternative site for the topos. The Grothendieck topology is the restricted Grothendieck topology, so a sieve is a covering sieve if it generates a covering sieve in the larger category of manifolds with boundary.

The claim is that you can also take as a site the full subcategory on the closed balls of radius $1$. I will denote the closed ball of dimension $n$ by $B^n$. It's a full subcategory so the maps are again the smooth maps, and as Grothendieck topology we take the restricted Grothendieck topology: a sieve on this category with one object is a covering sieve if and only if it generates a covering sieve in the larger category of $n$-manifolds, which is the case if and only if it contains an open covering.

To show that this is a site for the topos, we can again use the Comparison Lemma: it is enough to show that both $\mathbb{R}^n$ and $H^n$ admit a covering sieve generated by the inclusion of closed balls. For $\mathbb{R}^n$ you can take a covering by open balls and then replace each open ball by its closure. This generates a covering sieve because it contains (as sieve) an open covering. The situation for $H^n$ is more difficult. Here I think you can take closed rectangles (with rounded corners), such that each of these closed rectangles is isomorphic to $B^n$. Now cover $H^n$ with these rectangles such that each rectangles either lies completely in the interior of $H^n$, or one of the edges lines up with the boundary of $H^n$. I believe it is important that every point on the boundary of $H^n$ is contained in the interior of an edge of a square. Then this leads to a covering sieve on $H^n$.

The result is that as a site for the topos of generalized manifolds with boundaries, you can take a site consisting of the objects $B^n$ for each natural number $n$. The Grothendieck topology is the restricted Grothendieck topology, but it is not clear to me yet what the covering sieves look like in practice.

Jens Hemelaer (May 19 2021 at 16:05):

I think this is a more concrete description of the restricted Grothendieck topology:
a family $\{ f_i : M_i \to M \}_{i \in I}$ generates a covering sieve if and only if for each point $x \in M$ there is an open subset $U \subseteq M$, $x \in U$ such that there is a section $s : U \to M_i$.

(If the maps $f_i$ are topological embeddings, then the existence of a section $U \to M_i$ is equivalent to the condition $U \subseteq f_i(M_i)$.)

Jens Hemelaer (May 19 2021 at 16:09):

Reid Barton said:

You might also need a condition like "the interiors of the boundaries cover the boundary", and so on(?).

It should come down to this condition. For manifolds with boundaries, I don't think you need the "so on", but you probably need it if you are working with the more general manifolds with corners.

John Baez (May 19 2021 at 19:23):

Do y'all know about Chen spaces? Here our "probes" are arbitrary convex subsets of $\mathbb{R}^n$'s.

John Baez (May 19 2021 at 19:24):

In the linked paper, Alex Hoffnung and I describe a lot of properties of the category of Chen spaces.

John Baez (May 19 2021 at 19:25):

Of course closed half-spaces of $\mathbb{R}^n$ are convex, and so are quarter-spaces and so on.

Amar Hadzihasanovic (May 19 2021 at 20:21):

Thank you Jens and John!

Amar Hadzihasanovic (May 20 2021 at 11:10):

I think in fact if we work with manifolds with corners, instead of manifolds with boundaries, we can get a nice site with a simpler comparison by taking the full subcategory on the (closed) cubes $I^n$.

A manifold with corners has an open covering with opens that are diffeomorphic to

$$H_i^n := \{(x_1,\ldots,x_n) \in \mathbb{R}^n \mid x_{i+1}, \ldots, x_n \geq 0\}$$

for some $n \in \mathbb{N}$ and $i \in \{0,\ldots,n\}$, so to apply the comparison lemma we just need a covering of $H_i^n$ by closed cubes that generates a covering sieve. I believe that, fixing some small $\varepsilon > 0$,

$$\{[k_1,k_1+1+\varepsilon] \times \ldots \times [k_n, k_n+1+\varepsilon] \mid k_1,\ldots,k_i \in \mathbb{Z}, k_{i+1},\ldots,k_n \in \mathbb{N}\}$$

is such a covering.

Amar Hadzihasanovic (May 20 2021 at 11:17):

(Tbh manifolds with corners is what I had in mind initially, but having had topologist goggles on for too long, I missed at first that they are a different thing in the smooth world :sweat_smile: )

Amar Hadzihasanovic (May 20 2021 at 11:55):

Oh, I only just realised by reading John and Alex Hoffnung's paper that in fact manifolds with corners already sit inside the topos of smooth sets (because they sit inside diffeological spaces); I had naively assumed that smooth maps from open balls would not “see” the boundary or corners, but that's of course not true (e.g. a smooth curve can go wherever it wants).

So in fact this topos that we're defining is just the topos of smooth sets, and to apply the comparison lemma it suffices to cover $\mathbb{R}^n$.

John Baez (May 20 2021 at 18:30):

Sounds right. By the way, there's a smooth one-to-one and onto map from the real line to the L-shaped set

$\{(x,y) \in \mathbb{R}^2: x \ge 0 \text{ and } y \ge 0 \text{ and either } x \text{ or } y \text{ is } 0 \}$

John Baez (May 20 2021 at 18:31):

In other words, the image of a smooth injection from the real line into the plane doesn't need to be a smooth submanifold.

John Baez (May 20 2021 at 18:35):

Another good thing to know is that any subset of a diffeological space naturally acquires the structure of a diffeological space.

John Baez (May 20 2021 at 18:36):

So not only is a manifold with corners a diffeological space, so is the Cantor set.

But the Cantor set is rather dull as a diffeological space since they only smooth maps from $\mathbb{R}^n$ into it are constant!

John Baez (May 20 2021 at 18:38):

It gets the "discrete diffeology".