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defining a super-domain as a pair $(U\subset\mathbb{R}^n, C^\infty(U)\otimes\bigwedge[\theta^1\cdots\theta^m])$, a supermanifold is usually defined as a topological space $M$ endowed with a sheaf of super-algebras (call it $C^\infty(\mathcal{M})$) which is locally isomorphic to a superdomain. My question is: how can I formulate this local isomorphism here? In one side, there is a given sheaf constructed in a topological space, in the other side there is an object constructed in the euclidean space.

BVquantization
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1 Answers1

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The local isomorphism is a local isomorphism of ringed spaces, i.e. a homeomorphism $f: U\to V\subset M$ where $V$ and $U$ are open and an isomorphism of sheaves $F:f_*(C^\infty(\mathbb{R}^n)\otimes \bigwedge[\theta^1,\cdots, \theta^m])\to C^\infty(\mathcal{M})\vert_{V}$.

Edit: (I mixed up pullback and pushforward, resulting in something which did not make sense)

J.V.Gaiter
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  • How do you define $f^*(\theta^i)$? – BVquantization Aug 10 '23 at 17:20
  • In principle, $f$ is just a map between topological spaces, and there is no information of the grassmann algebra there. Therefore I think what you wrote: $f_*(C^\infty(R^n)\otimes\bigwedge[\theta^1\cdots\theta^n] )$ is not well defined – BVquantization Aug 10 '23 at 18:40
  • This is the pushforward of sheaves, which is always defined for a continuous map. – J.V.Gaiter Aug 10 '23 at 19:15
  • You should define the morphism using the pullback map $f^*:C^\infty(M)|_V \to \left(C^\infty(U)\otimes \bigwedge [\theta^\bullet]\right)$. It is part of the definition of the morphism of ringed spaces, and cannot be deduced from the homeomorphism $f:U\to V$. – Johannes Moerland Feb 17 '25 at 09:03