Is a twisted de Rham cohomology always the same as the untwisted one?

Question

I am a physicist studying now some supersymmetric sigma models. My question can, however, be reformulated in a purely mathematical language: A twisted de Rham complex involves $d_W = d + dW \wedge $ where $W$ is any even-dimensional form. In all known for me cases the cohomologies of this complex are the same as for the untwisted one. Can one assert that it is always so ? If yes, where is it proven ?

Answer 1

I hadn't heard of this definition of twisted deRham cohomology before reading this (I had heard of the one in this question, but they seem to be different) so I don't have full confidence in this. I claim that the answer is "yes" to a question which I hope is equivalent:

Let $X$ be a smooth manifold. The usual definition of $H^i_{DR}(X)$ is "closed $i$-forms modulo exact $i$-forms". In other words, you take those $i$-forms killed by $d$ and quotient by $i$-forms of the form $d \theta$, for $\theta$ an $(i-1)$-form.

Here is the analogous definition in your setting. Let $\Omega^{even}$ and $\Omega^{odd}$ be the spaces of even and odd forms. Let $d_W(\alpha) = d \alpha + (d W) \wedge \alpha$. Define a form to be $d_W$-closed if it is killed by $d_W$, and to be $d_W$-exact if it is of the form $d_W(\theta)$ for some $\theta$. I will show that the space of $d_W$-closed forms, modulo the space of $d_W$-exact forms, has dimension independent of $W$.

Set $$e^W = \sum \frac{W^k}{k!}.$$ If $W$ has no $0$-form component, then this sum only has finitely many terms because high enough powers of $W$ are $0$; if there is a $0$-form in $W$, then the sum above is still convergent.

Observe that $d (e^W) = e^W dW$. This crucially uses that $W$ is even, so $W$ and $dW$ commute. (If $W$ were purely odd, this would also be true, but if $W$ were a mix of even and odd parts, then it needn't be true.)

So $$d \left( e^W \alpha \right) = e^W dW \alpha + e^W d \alpha = e^W d_W \alpha \quad (\ast)$$ using that again that $e^W$ is even.

So $\alpha$ is $d_W$-closed if and only if $e^W \alpha$ is $d$-closed and $\alpha$ is $d_W$-exact if and only if $e^W \alpha$ is exact. So we can identify the space of $d_W$-closed forms modulo the space of $d_W$-exact forms with the space of $d_W$ forms modulo the space of $d_W$ forms by multiplying by $e^W$.

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Now, in the ordinary deRham cohomology theory, one shows that the orthogonal complement to the $d$-exact forms is the kernel of $d^{\dagger}$. So, instead of quotienting $\mathrm{Ker}(d)$ by $\mathrm{Image}(d)$, one can instead take the intersection $\mathrm{Ker}(d) \cap \mathrm{Ker}(d^{\dagger})$. Is this also true for $d_W$ and $d_{W}^{\dagger}$ ? In that case, the definition I am working with is the same as yours.