1
$\begingroup$

Let $E\rightarrow X$ be a $\mathbb{Z}_2$-vector bundle (or superbundle for connoisseurs) and consider the superconnection $$A=\nabla + B$$ where $\nabla$ is a connection on $E$ and $B\in\Gamma(End(E))^-$ simply an odd endomorphism. I would like to understand how to compute the curvature $A^2$: I can of course write $$A^2=\nabla^2 + \nabla B + B\nabla + B^2$$ but I would like to understand whether it makes sense to write $$A^2(X,Y)\in \Gamma(End(E))$$ just like for connections. The problem is that $A^2$ is the sum of a 2-form, a 1-form and a simple endomorphism. Hence it makes no sense a priori to write $A^2(X,Y)$. Nevertheless I wonder based on the fact one can always write $$A_XA_Ys := (\nabla_X+B)(\nabla_Y+B)s$$ whether it's possible to make a canonical sense of $A^2(X,Y)$.

$\endgroup$
1
  • $\begingroup$ The curvature form of a superconnection is an even element of the de Rham complex valued in the endomorphism bundle. Its components are (odd) differential forms of all possible degrees (e.g., in your case, you get degrees 0, 1, and 2). You can project onto odd differential forms of degree 2, if you want, but most of the curvature will be discarded. $\endgroup$ Nov 11, 2020 at 17:48

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.