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Stream: theory: category theory

Topic: Vector bundles vs internal vector spaces in a topos

Kevin Carlson (Sep 18 2024 at 15:00):

Graham Manuell said:

This suggests that my inclination was correct: https://mathoverflow.net/questions/268836/is-the-theory-of-vector-bundles-just-linear-algebra-done-in-a-suitable-topos

This is pretty interesting, because while the category of vector bundles is not generally constructive, as far as I know the counterexamples are not constructive (eg there’s no kernel of an endomorphism of the trivial line bundle over $\mathbb R$ multiplying by a function which is $0$ on $(-\infty,0]$ and nonzero elsewhere.) The proof that the category of Dedekind real fd vector spaces is constructive, right? So that would show the category of vector bundles is internally abelian?

Mike Shulman (Sep 18 2024 at 15:22):

With B-finite dimensions, I guess the only "exotic" things you'll end up with is vector bundles whose dimensions over different connected components might be different.

Mike Shulman (Sep 18 2024 at 15:25):

And yes, the proof that the category of modules over any ring is an abelian category is constructive. So the category of internal $\mathbb{R}$-modules in any topos is abelian. And since abelianness is a finitary property, this should imply that it's externally abelian as well. The catch is that constructively, not every module over a field is free, while it's only the free internal $\mathbb{R}$-modules that correspond to the usual external notion of "vector bundle".

Kevin Carlson (Sep 18 2024 at 16:27):

Hmm, yeah, let’s see if I can line this all up. The category of $\mathbb R$-modules is internally abelian, hence also externally so. Non-free such modules correspond to certain sheaves that aren’t locally free. Internally finite-dimensional ones are all locally free, maybe because the internal sentence “there exists some $n$ and an isomorphism with the free module of rank $n$”, according to the Kripke-Joyal semantics, means that there’s some cover on which this sentence is true in the external sense. This is making me nervous because I know that a map that’s locally an isomorphism is an isomorphism, but I think it’s fine because we’re not saying a sentence about a given map, but an existentially quantified one about a hom-sheaf.

Kevin Carlson (Sep 18 2024 at 16:31):

And I was also nervous that I could use the argument that “finite” modules are abelian and externalize to get a proof of the false claim that vector bundles are abelian, but I think I was just equivocating between finite-dimensional, ie finite free, and finitely presented.

Mike Shulman (Sep 18 2024 at 17:04):

Yes, I agree with all that, Kevin.

Mike Shulman (Sep 18 2024 at 17:06):

John, I also find Ingo's comment a bit ambiguous as regards free vs locally free. In the purely external point of view, does a "vector bundle" always have a fixed dimension, or could it have different dimensions over different connected components?

John Baez (Sep 18 2024 at 17:08):

Depending on the nuances of the definition it could work either way, but in practice people rarely - never? - want to think about vector bundles with different dimensions on different components.

I guess the most standard definition of 'fiber bundle' starts by fixing a single topological space or manifold as the 'standard fiber', which gets used as the fiber over every component of the base. Similarly, the most standard definition of 'vector bundle' fixes a single vector space as the standard fiber.

Tim Hosgood (Sep 18 2024 at 17:40):

John Baez said:

Depending on the nuances of the definition it could work either way, but in practice people rarely - never? - want to think about vector bundles with different dimensions on different components.

any geometer who wants to think about coherent sheaves likely wants to be secretly doing this :-)

Tim Hosgood (Sep 18 2024 at 17:41):

this conversation is an interesting read to me — i thought i knew about bundles, but reading this i feel like i know nothing! i'm used to seeing statements like "a vector bundle is a 1-truncation of the homotopy limit of the presheaf represented by $\mathrm{GL}_n$ on the Čech nerve" or something like that

John Baez (Sep 18 2024 at 17:42):

Admittedly I was talking about differential geometry, not algebraic geometry, where you might have a moduli space with multiple components of different dimensions, and then its tangent bundle....

Tim Hosgood (Sep 18 2024 at 17:43):

I know that real-analytic geometry also cares about coherent sheaves, but I definitely can't say anything about differential geometry

Mike Shulman (Sep 18 2024 at 18:13):

One could argue that even differential geometers ought to allow manifolds to have connected components of different dimension (e.g. that way the category of manifolds has coproducts), in which case their tangent bundles would have different dimensions on different components. It seems to me kind of like this may be an aspect of the common blindness to disconnected spaces.

John Baez (Sep 18 2024 at 18:23):

It's true that there deserves to be a [[rig category]] of manifolds (with coproducts and products giving the rig structure), thereby forcing manifolds of different dimensions to be in the same category, and manifolds with different components of different dimension. But this is so far down the list of priorities among differential geometers that it doesn't seem to be catching on. A typical paper starts "Let $M$ be an $n$-manifold...."

John Baez (Sep 18 2024 at 18:31):

One reason is that in differential geometry and differential topology, the interesting questions about manifolds depend heavily on the dimension. Manifolds of dimension 0 and 1 are more or less beneath contempt (with apologies to the amazing work on 1-dimensional chaos). Manifolds of dimension 2 are "well-understood" but still turn up interesting questions. Manifolds of dimension 3 are incredibly fascinating, with questions like the Poincare conjecture and Thurston's geometrization program. Manifolds of dimension 4 are incredibly fascinating in a completely different way, with questions like the study of exotic $\mathbb{R}^4$'s interacting with Yang-Mills theory, Seiberg-Witten theory and so on. Then comes "high-dimensional manifold theory", where homotopy theory becomes increasingly powerful as you hit 5 or 6 dimensions.

John Baez (Sep 18 2024 at 18:31):

Many people choose to specialize in just one of these realms!

John Baez (Sep 18 2024 at 18:37):

So if I them that the category of manifolds should have coproducts, that will go over about as well as if I went to a biology conference and told them I'd discovered a new kind of animal that's the disjoint union of a giraffe and a flea!

Tobias Fritz (Sep 18 2024 at 18:58):

Mike Shulman said:

One could argue that even differential geometers ought to allow manifolds to have connected components of different dimension (e.g. that way the category of manifolds has coproducts), in which case their tangent bundles would have different dimensions on different components. It seems to me kind of like this may be an aspect of the common blindness to disconnected spaces.

There are some interesting subtleties here for infinitely many connected components, because of which one can argue that "manifold" should be defined such that the connected components have bounded dimension.

Tobias Fritz (Sep 18 2024 at 18:59):

That's because if you have unbounded dimension, then the module of vector fields is no longer finitely generated, and in particular not a dualizable object in the symmetric monoidal category of modules over the algebra of smooth functions. This means that tensor calculus in its usual form, where vectors and covectors form each others' dual modules, no longer works! Having this dualizability is crucial for tensor calculus in the subsequent development of Riemannian geometry, e.g. the definition of the curvature tensor.

Tobias Fritz (Sep 18 2024 at 19:01):

So as long as you want tensor calculus in its usual form, you can't have both infinite coproducts and (finite) products in your category of manifolds. (Because for any nontrivial manifold $M$, the coproduct $\coprod_n M^{\times n}$ would have components of unbounded dimension.)

Mike Shulman (Sep 18 2024 at 19:13):

I was really only thinking about finite coproducts. Do we really want to allow manifolds with infinitely many connected components at all?

Tobias Fritz (Sep 18 2024 at 19:19):

I'd like to be able to say that $\mathbb{Z}$ is a Lie group, so yes :yum: And that's fine for the purposes of differential geometry as long as the dimension is bounded.

John Onstead (Sep 18 2024 at 19:20):

Hi, I don't mean to be impolite (and I'm not sure if this is against the rules or not) but would it by any chance be possible to move the above discussion to a different thread (perhaps starting with, I think, Graham Manuell's comment from yesterday?) I was kind of hoping to use this thread as "notes" to help me (and potentially others) understand fiber bundles from the ground up that I could refer back to when needed. Thanks.

Notification Bot (Sep 18 2024 at 21:04):

37 messages were moved here from #learning: questions > Deepening Understanding of Fiber Bundles by Kevin Carlson.

Kevin Carlson (Sep 18 2024 at 21:07):

@John Onstead I tried to split out the stuff I'm guessing you don't care about and leave your interleaved messages and John B's responses back at the old thread. Hope I didn't blow anything up too badly.

John Onstead (Sep 18 2024 at 22:39):

@Kevin Carlson Thanks so much!

Mike Shulman (Sep 18 2024 at 23:46):

@Tobias Fritz, can you explain more about how the failure of dualizability is a problem? I'm just thinking concretely about tensor fields and Riemannian metrics, and having trouble seeing why it would be a problem to have the same thing happening at unbounded dimensions on different components. Why doesn't everything to do with metrics and curvature just happen as usual on each component?

Tobias Fritz (Sep 19 2024 at 07:19):

Sure. I guess you can use the usual definitions of every component even if the dimension is unbounded, but the problem I'm getting at is that these definitions are then no longer equivalent to the existing fully algebraic formulations, which are convenient to work with.

Tobias Fritz (Sep 19 2024 at 07:24):

To explain why dualizability is important, let me write $V$ for the module of vector fields over the algebra of smooth functions $C^\infty(M)$. Throughout the following, by "linear" I mean with respect to scalar multiplication by $C^\infty(M)$.

Tobias Fritz (Sep 19 2024 at 07:26):

Then dualizability already pops up in the very definition of a (pseudo-)Riemannian metric, which in its purely algebraic formulation is a symmetric bilinear form on $V$ together with another one on the dual module $V^*$ such that these two satisfy the zig-zag identities; this is the algebraic version of the usual pointwise non-degeneracy condition. So, being a (symmetric) dualizability witness is the very definition of (pseudo-)Riemannian metric!

Tobias Fritz (Sep 19 2024 at 07:33):

Of course this doesn't show that tensor calculus in general doesn't make sense without dualizability. So my actual main point is this: one can define a tensor of rank $(m,n)$ as a linear map $V^{\otimes n} \to V^{\otimes m}$, but one often wants to move some or all of the tensor factors to the other side by turning them into a $V^*$, even in the absence of a metric. For example, tensor contraction amounts to applying the trace, so the symmetric monoidal subcategory of modules formed by the powers of $V$ needs to be traced. (Maybe one can get away with an abstract trace to some extent without dualizability, but I'm not sure how far this will go.)

Tobias Fritz (Sep 19 2024 at 07:39):

Also Penrose graphical notation for differential geometry is basically string diagrams with a few extra bells and whistles. Since there's no distinction between inputs and outputs in these diagrams, they implicitly assume dualizability. Usually the diagrams are even drawn with undirected wires, which amounts to having made a particular choice of metric.

Tobias Fritz (Sep 19 2024 at 07:57):

BTW I recently wrote a paper on doing (pseudo-)Riemannian geometry in purely algebraic terms. (That may sound a bit unoriginal and partly that's true, but for a first paper involving anything geometrical I'm happy with it.) The gist is to observe that a lot of differential geometry makes sense fully generally in its algebraic form over any commutative algebra whose module of derivations is dualizable, and I've pushed this up to the development of geodesics. One of the nice things about this is that one automatically gets a fibred/parametric version of differential geometry like this by considering $C^\infty(M)$ as an algebra over $C^\infty(N)$ for any smooth submersion $M \to N$.

Mike Shulman (Sep 19 2024 at 15:43):

Yeah, I realize all that. I guess maybe my question would have been better phrased as "why do you prefer to work with the algebraic formulation?"

Mike Shulman (Sep 19 2024 at 15:57):

I also wonder, given that everything seems to work fine pointwise, whether there could be some reformulation of the algebraic approach, e.g. some "local" notion of duality, that would still work.

Tobias Fritz (Sep 19 2024 at 18:39):

Mike Shulman said:

Yeah, I realize all that. I guess maybe my question would have been better phrased as "why do you prefer to work with the algebraic formulation?"

Ah, I see :sweat_smile: So here are the main reasons why I prefer the algebraic formulation:

1. Greater generality: we're not confined to the dogma that smooth functions should be the default class of functions in differential geometry. For example, the book Geometric Theory of Generalized Functions with Applications to General Relativity develops differential geometry based on versions of the Colombeau algebra. This flavour of differential geometry is also an instance of the algebraic framework, and the authors of that book could have spared themselves a good deal of work if they had realized that. Of course this is not an argument against a potential pointwise/local algebraic formalism, but I haven't thought about such a thing because I personally don't see any use case for it. In algebraic geometry, we want to have access to the local information due to the scarcity of regular functions on a variety; but because smooth functions on a manifold are in ample supply, this reason doesn't seem to apply to differential geometry.
2. Conceptual simplicity: The algebraic formulation is just easier to state than the standard manifold one if you start from first principles.
3. Physical intuition: In thinking about general relativity, I've come to view this theory as secretly algebraic and point-free in spirit. The reason is that when you solve the Einstein field equations, then you're typically not fixing a manifold to begin with, but rather determining the topology in the process jointly with the metric. What is involved in this process is not easy to formalize in terms of manifolds, but in algebraic terms it's more clear and quite nice. To illustrate this, let's consider a (1+1)-dimensional universe with a time coordinate $t$ and a space coordinate $x$, so that a typical ansatz for the metric would be $g = a(t) \, dt^2 - dx^2$, where the function $a$ tells us how the size of the universe evolves with time. A priori we don't know what topology we want, and it could e.g. be $\mathbb{R} \times \mathbb{R}$ or $S^1 \times \mathbb{R}$ (a temporally cyclic universe). But algebraically, we're just starting with the polynomial ring in variables $t$, $x$ and $a$, and throw in an $a^{-1}$ in order for our $g$ to have an inverse. We then plug this into the Einstein equations and get an additional differential relation between $a$ and $t$, and this defines our actual spacetime. This is basically how cosmology is treated in the textbooks. (There's something nice that happens with regards to the constants of integration when solving the equations, and I could argue further that doing physics within a given spacetime is similarly secretly algebraic, but I'll skip these things in order to wrap up.) So this is how I've personally come to think of general relativity. I don't want to consider this as a necessary or objective fact, but more like an intuition; an intuition which should have its uses in a physics context by providing a new perspective on old things.

Is this more like an answer to your question now?

Mike Shulman (Sep 19 2024 at 19:20):

Yes, thanks, that's very interesting!

Mike Shulman (Sep 19 2024 at 19:21):

I still wonder whether there's some reformulation of the algebrac approach that could handle manifolds with components of unbounded dimension. Like maybe some kind of filtered or graded rings/modules where degree $n$ sees only dimensions $\le n$, so you can get all the dimensions included without spurious "infinite-dimensional" stuff.