I'm following the Bott/Tu textbook on Differential Forms in Algebraic Topology, and while trying to work out one of the exercises there I came across a point I can't fully understand.
If $f: M\rightarrow M$ smooth map and $H^q(f)$ is the induced map on the cohomology $H^q(M)$, I'm trying to understand what is $traceH^q(f)$.
What my intuition tells me is that given $\{{\omega_i}\}$ basis of $H^*(M)$ and $\{\theta_i\}$ the dual basis,
$H^q(f)([\omega]) = [f^*\omega]$ and $H^q(f)[\omega_i] \wedge \theta_i =f^*\omega_i \wedge\theta_i$
So is the expression $traceH^q(f) = \sum_i f^*\omega_i \wedge \theta_i$ correct?
This is the first time I'm going through a graduate text by myself and things begin to get complicated enough to make me feel like I might be getting the definitions wrong, So I figured I might get some help or validation I'm going the right way by asking here.
Best Answer
Given a linear map $f: V \to V$ of a vector space, choosing a basis $\beta = \{v_1, \cdots, v_n\}$, we may identify $f$ with a matrix $f_\beta$ and take the trace of this matrix (the sum along the diagonal). Given a different basis $\beta'$, there is an $n \times n$ matrix $A$ with $\beta = A\beta'$, and then $f_{\beta'} = A^{-1} f_{\beta} A$. Because trace is conjugation invariant, it follows that $\text{tr}(f_\beta)$ does not depend on the choice of basis, so we set $$\text{tr}(f) = \text{tr}(f_\beta)$$ for some choice of basis.
(Nothing here really depends on $V$ being a vector space, but rather on it being free of finite rank. You can give the same definition for, say, an endomorphism of a finitely generated free abelian group.)
To decode this.
You have a vector space $H^q(M)$. The map $H^q(f)$ is defined by $H^q(f) [\omega] = [f^* \omega]$.
You choose a basis $\{\omega_i\}$; as you say, if $\{\theta_i\}$ is a basis of $H^{n-q}(M)$ so that $\int_M \omega_i \wedge \theta_j = \delta_{ij}$ (that is, this integral is $1$ for $i = j$ and $0$ otherwise) --- that is, a dual basis --- then we may compute the matrix of $f^*$ attached to this basis as the matrix with entry $(i,j)$ given by $$\int f^*\omega_i \wedge \theta_j.$$ And you're interested in the sum of the diagonal entries, so that
$$\text{tr}(H^q(f)) = \sum_i \int f^* \omega_i \wedge \theta_i,$$
as you say in your post.
What this comes down to is: Make sure you have a really firm grasp on linear algebra and maybe have a linear algebra text close at hand :) Most of us only truly understood linear algebra with a lot of time and practice and "graduate textbooks" that used it.