Changing coefficients of cohomology and pullbacks

Question

If I have a compact complex manifold $M$ and a map $f:M\to M$. Let $\mathbb F$ a field. Then $H^k(M;\mathbb F) $ is a $n$-dimensional vector space and chosing a basis, $f^*_{\mathbb F}$ defines a matrix.

Let's say I already know how the matrix $f^*_\mathbb F$ looks for $\mathbb F = \mathbb R$. How do I obtain $f^*_\mathbb C$ from it?
Is it just the same matrix but we think of the entries as complex numbers with vanishing imaginary part?

What about the other way around? If I know $f^*_\mathbb C$, how do I get $f^*_\mathbb R$?
I know that there is a standard way to make a real matrix out of a compley one by replacing every entry $x=a+ ib$ by a block $\begin{pmatrix} a & -b \\ b & a \end{pmatrix} $. But this would change the size of the matrix, so it cannot be right.

Answer 1

Part 1: Relating cohomology with complex and real coefficients

Let's take a look at the universal coefficient theorem for cohomology, which says that for a PID $R$ and $R$-module $M$, there is a natural short exact sequence $$0\to\newcommand\Ext{\operatorname{Ext}}\Ext^1_R(H_{i-1}(X;R),M)\to H^i(X;M)\to \newcommand\Hom{\operatorname{Hom}}\Hom_R(H_i(X;R),M)\to 0,$$ and taking $R=\newcommand\RR{\Bbb{R}}\RR$, we see $H_{i-1}(X;R)$ is free, so $\Ext$ vanishes, so we get natural isomorphisms, $$H^i(X;M)\simeq \Hom_\RR(H_i(X;\RR),M)$$

Now apply this to both $M=\RR$ and $M=\newcommand\CC{\Bbb{C}}\CC$, to get $H^i(X;\RR)\simeq H_i(X;\RR)^*$, where $*$ denotes taking the dual $\RR$-vector space and $$H^i(X;\CC) \simeq \Hom_\RR(H_i(X;\RR),\CC)\simeq H_i(X;\RR)^*\otimes \CC\simeq H^i(X;\RR)\otimes \CC,$$ where this last natural isomorphism follows from this question for example.

Thus we see that cohomology with complex coefficients is the complexification of cohomology with real coefficients.

Part 2: Converting a matrix for $f^*_\RR$ to a matrix for $f^*_\CC$.

Since cohomology with complex coefficients is just the complexification of cohomology with real coefficients, the matrix of the pullback for complex coefficients will be the exact same matrix as the matrix for the pullback with real coefficients, but where we regard the real matrix as now being a complex matrix in the obvious way. (I.e. exactly what you suggested first)

Part 3: The other direction

To go the other way, it's a bit more complicated, and I'll see if I can give the general story for complexifications overall.

Suppose we have a real vector space $V$, and its complexification, $V_\CC=V\otimes\CC$, and another real vector space $W$, and its complexification $W_\CC$, and a map $T : V\to W$, which induces a map $S:V_\CC\to W_\CC$.

Now if we have bases $\{e_i\}$ and $\{f_j\}$ for $V$ and $W$ respectively, and we compute the matrix of $S$ with respect to these (now regarded as bases for $V_\CC$ and $W_\CC$), then you can check that the resulting matrix for $S$ will be real, and will equal the matrix for $T$ computed with respect to $\{e_i\}$ and $\{f_j\}$.

The difficulty arises when we have the matrix of $S$ with respect to what you can think of as "non-real" basis vectors, since then we don't have obvious bases for $V$ and $W$ to compute $T$ with respect to. It's then a bit complicated, so I'll leave my answer here.