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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 …
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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 …
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  • 247
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 …
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  • 247
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 …
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  • 247
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 …
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