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de Rahm Rham cohomology and flat vector bundles |
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I was wondering whether there is some notion of "vector bundle de Rham cohomology". To be more precise: the k-th de Rham cohomology group of a manifold $H_{dR}^{k}(M)$ is defined as the set of closed forms in $\Omega^k(M)$ modulo the set of exact forms. The coboundary operator is given by the exterior derivative. Let now $E \rightarrow M$ be a vector bundle with connection $\nabla^E$ over $M$, and consider the $E$-valued $k$-forms on $M$: $\Omega^k(M,E)=\Gamma(\Lambda^k TM^\ast \otimes E)$. $$H_{dR}^{k}(M,E) := \frac{ker \quad d^{\nabla^E}|{\Omega^k(M,E)}}{im d^{\nabla^E}|_{\Omega^k(M,E)}}{im \quad d^{\nabla^E}|{\Omega^{k-1}(M,E)}}$$d^{\nabla^E}|_{\Omega^{k-1}(M,E)}}$$ So my question: Is this somehow useful? I mean can one use this definition to make some statements about $M$ or $E$ or whatever? Or is the restriction of $E$ to be a flat vector bundle somehow disturbing? Or is this completely useless? |
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de Rahm cohomology and flat vector bundlesI was wondering whether there is some notion of "vector bundle de Rham cohomology". To be more precise: the k-th de Rham cohomology group of a manifold $H_{dR}^{k}(M)$ is defined as the set of closed forms in $\Omega^k(M)$ modulo the set of exact forms. The coboundary operator is given by the exterior derivative. Let now $E \rightarrow M$ be a vector bundle with connection $\nabla^E$ over $M$, and consider the $E$-valued $k$-forms on $M$: $\Omega^k(M,E)=\Gamma(\Lambda^k TM^\ast \otimes E)$. $$H_{dR}^{k}(M,E) := \frac{ker \quad d^{\nabla^E}|{\Omega^k(M,E)}}{im \quad d^{\nabla^E}|{\Omega^{k-1}(M,E)}}$$ So my question: Is this somehow useful? I mean can one use this definition to make some statements about $M$ or $E$ or whatever? Or is the restriction of $E$ to be a flat vector bundle somehow disturbing? Or is this completely useless?
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