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Complex, contact, Riemannian, pseudo-Riemannian and Finsler geometry, relativity, gauge theory, global analysis.
39
votes
6
answers
4k
views
Advantages of diffeological spaces over general sheaves
I have been playing with/thinking about diffeological spaces a bit recently, and I would like understand something rather crucial before going further. First a little background:
Diffeological spaces …
25
votes
What is an $(\infty,1)$-topos, and why is this a good setting for doing differential geometry?
Ok, so, I will try to answer this best I can. first, I'll tell you a skewed-perspective of what an infinity topos is (or ought-to-be). As for how you can "do differential geometry"- this is a bold sta …
22
votes
3
answers
1k
views
Applications of topological and diferentiable stacks
What are some examples of theorems about topology or differential geometry that have been proven using topological/differentiable stacks, or, some examples of proofs made easier by them? I'm well awar …
21
votes
Geometric imagination of differential forms
A one-form assigns to each vector tangent to a manifold a real number in a linear way. You may think of a vector tangent to a manifold as being determined by two points on the manifold that are "infin …
15
votes
1
answer
2k
views
When does a leaf space admit a (non-Hausdorff) manifold structure?
If $f:M \to N$ is a submersion with connected fibers, then the fibers of $f$ foliate $M$. This is called a simple foliation of $M$ and the leaf space can be identified with $N$. Suppose that a foliati …
14
votes
Synthetic vs. classical differential geometry
In theory, most anything can be expressed with SDG, and there has been some work in expressing some of GR in this context, but I am not sure if much has been done beyond proof-of-concept. You can goog …
14
votes
2
answers
1k
views
Are all manifolds affine?
There is a classical result which says that the assignment $$M \mapsto C^{\infty}\left(M\right)$$ is an embedding of the category of (paracompact Hausdorff) smooth manifolds into the opposite category …
12
votes
1
answer
535
views
Is every representable map a submersion?
Recall that a morphism $f:C \to D$ in a category $\mathscr{C}$ is representable if for all maps $g:E \to D$ in $\mathscr{C},$ the pullback $C \times_{D} E$ exists.
Let now $\mathscr{C}$ be the catego …
12
votes
2
answers
783
views
Generalized smooth spaces and infinite dimensional manifolds
There is a theorem due to Losik which shows that the category of Frechet manifolds embeds fully-faithfully into diffeological spaces. (Diffeological spaces are concrete sheaves on the site of (Euclide …
10
votes
Accepted
Based loop groups as stacks?
$\Omega G$ won't be a differentiable stack unless you are willing to go to infinite dimensions. Provided you are considering $S^1$ as a smooth manifold, Nerses answer above is correct- it gives you a …
10
votes
3
answers
1k
views
Connections on principal bundles via stacks?
Let G be a Lie group and M a smooth manifold. Suppose that P is a principal G-bundle over M. Then by Yoneda, this corresponds to a smooth map $p:M \to [G]$, where $[G]$ is the differentiable stack ass …
9
votes
What is meant by smooth orbifold?
To be quick, just like manifolds, orbifolds have a fixed dimension. This does not vary point to point. This is also true of their tangent spaces. This is actually true for any etale differentiable sta …
9
votes
Accepted
Stacks over diffeologies
I will show that stacks over diffeological spaces are "the same" (in the sense of equivalence of 2-categories) as ordinary stacks on manifolds.
The Grothendieck pre-topology in question is the Grothe …
7
votes
1
answer
1k
views
When does a submersion have connected fibers?
Can we characterize when a submersion $F:M \to N$ between two smooth manifolds has connected fibers? If this is too hard, what are some sufficient conditions?
7
votes
4
answers
5k
views
Cotangent bundle of a differentiable stack
If you ever wanted to construct the tangent bundle of a differentiable stack, it's relatively simple:
First, if $\mathbf{X}$ is a stack coming from a Lie groupoid $\mathcal{G}$, you could just say $\ …