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

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 ?

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Are you asking if the de Rham complex $(\Omega_X^\cdot,d)$ is homotopic to the twisted de Rham complex $(\Omega_X^\cdot, d+d_\omega)$, where $d_{\omega} := d+ d\omega \wedge$? I don't see how $d_\omega$ maps $\Omega_X^i$ to $\Omega^{i+1}_X$. Could you elaborate? –  Harry Apr 9 '12 at 9:55
I do not know well the mathematical language that you use. Let me explain it in my own terms. For the usual de Rham complex there are some number of even-dimensional forms that are annihilated by both d and d^\dagger. Call this number \beta_{even}. One define analogously \beta_{odd}. The difference \beta_{even} - \beta{odd} is the Euler characteristics of the manifold. For the twisted complex, this difference is not changed. For simple manifolds like S^n I understand how to show that \beta_{even} and \beta_{odd} are not changed SEPARATELY. The question is whether it is true in general ? –  Andrei Smilga Apr 9 '12 at 12:08
Ahh I understand the question better now. Unfortunately, I don't know the answer. –  Harry Apr 9 '12 at 17:50

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 $e^W$.

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.

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Is there any intuition for thinking that studying $e^W$ would be useful for this problem? I mean, I can see how it basically solved almost the whole thing, but I'm wondering what made you think "Looking at $e^W$ might be beneficial!" Is it just that finding forms satisfying $d\omega = 0$ but $\omega \neq d\eta$ for any $\eta$ is essentially solving differential equations, and exponential functions are often helpful in this context? –  Jason DeVito Apr 9 '12 at 19:19
Two things. (1) The kind of twisted cohomology I know uses $d+\alpha \wedge$, where $\alpha$ is a $1$-form. If $\alpha = d w$ for a scalar function $w$, then this is ordinary cohomology, by this argument. (2) I started trying to write down a class $\alpha$ in the kernel of $d_W$ for $X=S^n$ and $W$ a $2$-form. We must have $d \alpha_0$, so $\alpha_0$ is constant -- let's make it $1$. So $d \alpha_2 + dW=0$ -- it seems like a good bet that $\alpha_2 = -W$. So $d \alpha_4 - W dW=0$. I triple checked that $d(W \wedge W) =2 W dW$ for $W$ a $2$-form. So it seems likely that $\alpha_4 = W^2/2$. –  David Speyer Apr 9 '12 at 19:45
So I had $1-W+W^2/2+\cdots$, and at that point there was an obvious guess. –  David Speyer Apr 9 '12 at 19:46
That is the actual history. A clever way to think of it is: We want $d \eta+W \wedge \omega=0$, so we want $\omega = \mathrm{Constant} \cdot e^W$, and when we say "constant", we actually mean "in the kernel of $d$". –  David Speyer Apr 9 '12 at 19:48
Actually, I have found the reference [Mathai + Wu, arXiv:0810.4204], where the statement about the equivalence of the twisted de Rham complexes $d + H$ and $d + H +dB$ ($H$ being a closed 3-form) was proven in the same way as David did (with invoking the exponential $e^B$). I would not be surprised though, if somebody noticed this before them... –  Andrei Smilga Apr 11 '12 at 12:52