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This question is related to my previous question. The answers to that question inspired a new question, namely

For a complex manifold $M$, why is $H_{DR}^p(M,\mathbb{C})\cong H_{DR}^p(M,\mathbb{R})\otimes_\mathbb{R}\mathbb{C}$

Where $$H_{DR}^p(M,\mathbb{C})=\frac{\ker d:A^p(M,\mathbb{C})\to A^{p+1}(M,\mathbb{C})}{dA^{p-1}(M,\mathbb{C})}$$ And $$A^p(M,\mathbb{C})=\Gamma(\wedge^p(T^*M\otimes_\mathbb{R} \mathbb{C}_M))$$ and $\mathbb{C}_M$ is the trivial bundle over $M$ with fibre $\mathbb{C}$. And $H_{DR}^p(M,\mathbb{R})$ is the usual de Rham cohomology of the underlying real manifold, with spaces of forms $A^p(M)$. The answers to my previous question suggest that we can use the universal coefficient theorem here, but I don't see how. We have two complexes:

$$\to A^{p-1}(M)\to A^p(M)\to A^{p+1}(M)\to$$ and $$\to A^{p-1}(M,\mathbb{C})\to A^p(M,\mathbb{C})\to A^{p+1}(M,\mathbb{C})\to$$

But here it is not clear that $A_p(M)\otimes \mathbb{C}\cong A_p(M,\mathbb{C})$, since this is the statement that $$\Gamma(\wedge^p T^*M)\otimes\mathbb{C}\cong \Gamma(\wedge^p T^*M\otimes \mathbb{C})$$ which is not obvious to me. And if this is true, its a second matter of confirming that the $d$ on our complex-valued complex corresponds to $d\otimes\mathbb{1}$ under this isomorphism, although that is probably true.

In conclusion I do not see how to apply the universal coefficient theorem here. Any help on how to approach this would be much appreciated

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  • In words, "Every global complex-valued differential form decomposes uniquely into real and imaginary parts, and the exterior derivative $d$ respects this decomposition." (The plausibility of these assertions may depend on your definitions, which is why I'm commenting rather than answering.) – Andrew D. Hwang Apr 23 '16 at 13:12
  • @AndrewD.Hwang my problem is that everywhere I find this result it is just stated in vague terms like this. What I'm looking for is a rigorous proof of this fact, but no one seems to bother... – user2520938 Apr 23 '16 at 13:17
  • In that case, it may be best to work locally and concretely: A complex-valued differential form is a complex-valued section of the space of real-multilinear functions on $TM$ (adapt as needed for your definition of differential forms), so it obviously decomposes into real and imaginary parts. And then $d(\alpha + i\beta) = d\alpha + i, d\beta$, perhaps by definition, though this definition is imposed by the requirements of linearity and the Leibniz rule over wedge products (together with the fact "$i$" is constant). – Andrew D. Hwang Apr 23 '16 at 13:32

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