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 ?
Is a twisted de Rham cohomology always the same as the untwisted one?
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differential-geometry
physics
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1Are 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? – 2012-04-09
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1I 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 ? – 2012-04-09
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0Ahh I understand the question better now. Unfortunately, I don't know the answer. – 2012-04-09