TITLE: Number of Fixed Points in a Map from the Torus to itself using Lefschetz Trace
QUESTION [5 upvotes]: Let $f: X \to X$ be a continuous map.  For any fixed point $f(x) = x$ with $x \in X$, we can find the index of that fixed point $i(f,x)$.  The Lefschetz-Hopf formula says:
$$ \sum_{x \in \mathrm{Fix}(f)} i(f,x) = \sum_{k \geq 0} (-1)^k \mathrm{Tr}(f_*|H_k(X,\mathbb{Q}))
 $$
I would like to understand the Lefschetz fixed point formula with an example.
  Let's try $X = S^1 \times S^1$ be a 2-dimensional torus and consider the linear map
\begin{eqnarray*}
x &\mapsto& 3x - y\\
y &\mapsto& x + 3y
\end{eqnarray*}
In the complex plane this would be $z \mapsto (3+i)z$.  Here both equations are taken mod 1.  One can compute the number of fixed points of this map to be $\mathbf{5} = (2+i)(2-i) $, since we solve $z = (3+i)z  \mod \mathbb{Z}[i]$ and get the number of lattice points inside the parallogram.  
How do we compute the traces on each of the elements of the homology?

$H_0(S^1 \times S^1) = \mathbb{Q}$
$H_1(S^1 \times S^1) = \mathbb{Q}\oplus \mathbb{Q}$
$H_2(S^1 \times S^1) = \mathbb{Q}$

How do I get the induced action of $f$ on each of the homology groups and verify the traces?

REPLY [4 votes]: Let $V$ be a finite-dimensional real vector space, let $\Gamma$ be a lattice in $V$, so that $V/\Gamma$ is the corresponding torus, and let $f : V \to V$ be a linear map that preserves $\Gamma$. Then

$f$ naturally gives rise to a map $V/\Gamma \to V/\Gamma$, 
$H_k(V/\Gamma)$ can naturally be identified with $\Lambda^k(\Gamma)$, and in particular the action of $f$ on the former is the same as its action on the latter,
hence the trace of the action of $f$ on $\Lambda^k(\Gamma)$ is, up to sign, the $k^{th}$ coefficient of the characteristic polynomial of $f$.

In this special case the Lefschetz fixed point theorem reduces to the claim that 
$$\det(I - f) = \sum_{k=0}^{\dim V} (-1)^k \text{tr}(\Lambda^k(f))$$
which is a straightforward corollary of the more general fact
$$\det(I - ft) = \sum_{k=0}^{\dim V} (-1)^k t^k \text{tr}(\Lambda^k(f)).$$
(Of course we need to assume that $f$ has isolated fixed points, which as it turns out is equivalent to $I - f$ being invertible.)