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I have heard quite a few times that "purely odd directions don't carry topological obstructions", but I can't find/understand the exact statement. The question I want to understand is the following: suppose I am given a graded vector bundle $E \rightarrow X$ where the typical fiber $F$ is concentrated in degree 1 and $X$ is an ordinary smooth manifold. Is it safe to say that $E = X \times F$?

AbuuzArbuz
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  • The statement means roughly that the purely odd directions are contractibly fibred over the underlying topological space. In particular, Batchelor's theorem states that any supermanifold $X=(\underline X, \mathcal O X)$ is non-canonically isomorphic to a fibrewise parity reversed _vector bundle $\Pi E \to \underline X$, where $\underline X$ is given the standard smooth structure inherited from $\underline X$. – Johannes Moerland Feb 17 '25 at 08:50
  • In other words, the purely odd directions themselves do not carry any (non-trivial) topology. As an example, any purely odd supermanifold manifold is of the form $\mathbb R^{0|d}$ – Johannes Moerland Feb 17 '25 at 08:54

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I've just looked up the definitions, and maybe I'm missing something, but $E$ = Mobius strip** and $X = S^1$ seems like a counterexample to this conjecture.

** i.e., the unique $\mathbb R^1$ bundle over $S^1$ with nontrivial clutching function in $O(1)$

John Hughes
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