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This is quite an elementary question, but I'm confused over terminology and whether $H_{dR}^k(M)$ is a module, or vector space (this confusion is exacerbated when it's often called the de Rham "group").

Since $\Omega^k(M)$ is the space of $(k,0)-$ anti-symmetric tensor fields on $M$, it is in fact a $C^{\infty}(M)$ module. We then define $B^k(M) = \{\omega\in\Omega^k(M) \::\: \exists \alpha \in\Omega^{k-1}(M) \text{ such that } d\alpha = \omega\}$ (the space of "exact forms"), and $Z^k(M) = \{\omega \in \Omega^k(M) \::\: d\omega = 0\}$ (the space of "closed" forms). It's not difficult to check that these are submodules of $\Omega^k(M)$.

Then we define $H^k_{dR}(M) = Z^k(M)/B^k(M)$, and I'm told this is a quotient vector space. Why is it not a $C^\infty(M)$ module? When did it become a vector space? If it were defined as $Z^k(T_p(M))/B^k(T_pM)$ (exact and closed forms on $T_p(M)$), this would make a lot more sense, but then it would be a different vector space at every point that we could potentially "glue" together like the tangent bundle. Is this the case?

Then when I go to calculate $H^1_{dR}(S^1)$, it's again treated as a vector space, and I'm supposed to show it's isomorphic to $\mathbb{R}$, but shouldn't it be $\Omega^1(\mathbb{R})$ since this is a $C^{\infty}(\mathbb{R})$ module?

Any elucidation would be greatly appreciated.

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The de Rham cohomology groups are the cohomology of the complex with $k$-th term $\Omega^n(M)$, consisting of the differential $k$-forms on the manifold $M$, with differential $d$, the exterior derivative. Thus $H^k_{dr}(M)=Z^k/B^k$ where $Z^k$ is the kernel of $d:\Omega^k(M)\to \Omega^{k+1}(M)$ and $B^k$ is the image of $d:\Omega^{k-1}(M)\to \Omega^k(M)$. The $\Omega^k(M)$ are vector spaces over $\Bbb R$ and $d$ is $\Bbb R$-linear. Therefore each of $Z^k$, $B^k$ and $H^k_{dr}(M)$ are vector spaces over $\Bbb R$.

Also the $\Omega^k(M)$ are modules for the ring $C^\infty(M)$. But $d$ is not a homomorphism of $C^\infty(M)$-modules. We'd need $d(f\omega)=f\, d\omega$ for that (when $f\in C^\infty(M)$) but actually $d(f\omega)=df\wedge \omega+f\, d\omega$. In general then, $Z^k$ and $B^k$ won't be $C^\infty(M)$-modules.

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  • $\begingroup$ This clears it up nicely. The expression $d(f\omega) = f\, d\omega$ holds if and only if $df \wedge \omega = 0$ for all $\omega$ so we need $df = 0$ so if our manifold is connected, $f$ is constant so we simply take scalar coefficients on $\mathbb{R}$, thus we have an $\mathbb{R}$ vector space. Is this correct? $\endgroup$ – Nico May 2 '18 at 16:42
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You can always find the real numbers inside of $C^\infty(\mathbb{R})$ as the constant functions. So for any $C^\infty(\mathbb{R})$-module $M$, we get an $\mathbb{R}$-action by defining $\lambda \cdot m = c_\lambda m$, where $c_\lambda$ is the constant function with value $\lambda$.

Hence, all $C^\infty(\mathbb{R})$-modules become real vector spaces by forgetting most of the structure, and just remembering that you can multiply by constant functions.

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$B^k(M)$ and $Z^k(M)$ are not submodules of $\Omega^k(M)$, since the exterior derivative is only $\mathbb{R}$-linear, not $C^\infty(M)$-linear. For instance, when $k=0$, for $Z^k(M)$ to be a submodule would mean that if $df=0$ then $d(fg)=0$ for any other function $g$. This is obviously false: $df=0$ means that $f$ is locally constant, but if $f$ is constant and nonzero then $fg$ won't be locally constant unless $g$ is.

Since the exterior derivative is only $\mathbb{R}$-linear, $B^k(M)$ and $Z^k(M)$ are only $\mathbb{R}$-vector spaces, not $C^\infty(M)$-modules, and so the same is true of $H^k_{dR}(M)$.

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