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Complex, contact, Riemannian, pseudo-Riemannian and Finsler geometry, relativity, gauge theory, global analysis.
2
votes
1
answer
184
views
Parity reversed tangent bundle as a supermanifold
I encountered an example in a paper telling that $\underline{SM}(\mathbb{R}^{0|1},X)\cong \pi TX $, where $X$ is some fixed ordinary Riemannian manifold, $\pi TX $ is the supermanifold with base manif …
1
vote
Accepted
Parity reversed tangent bundle as a supermanifold
My advisor told me the answer; I was just one step away from it.
A morphism $\varphi:S\to \pi TX$ is determined by the natural transformation between the structural sheaves, which is locally determine …
2
votes
0
answers
89
views
Parity reversed tangent bundle of a supermanifold and the corresponding Q-structure
I asked this question in MathStackExchange 10 days ago but get no response (not a vote nor a comment), so I'm copying it here below. The link to the original question is:
https://math.stackexchange.co …
2
votes
1
answer
133
views
One-to-one correspondence between super morphisms $\varphi:S\to TX$ and pairs $(f:C^\infty(X...
I'm trying to show that for an ordinary manifold $X$ and a supermanifold $S$, supermanifold morphisms $\varphi:S\to TX$ are one-to-one to the pairs $(f,F) $ where $f:C^\infty(X)\to C^\infty(S)$ is a s …
3
votes
1
answer
330
views
How to show that $\text{Man}(M,\mathbb{R}^n)\cong \mathbb{R}\text{-Alg}(C^\infty(\mathbb{R}^...
I'm trying to show that manifolds are affine, i.e. $\text{Man}(M,N)\cong \mathbb{R}\text{-Alg}(C^\infty(N),C^\infty(M)) $. If I could show this for $N=\mathbb{R}^n$, then I know how to do the rest usi …