# Number of Fixed Points in a Map from the Torus to itself using Lefschetz Trace

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?

• That isn't the number of fixed points. That's the number of points $f$ sends to $0$. Commented May 9, 2014 at 17:45
• I may be wrong but I thought this would have only five fixed points. Here $(3+i)z\equiv z \pmod{\Lambda}$, iff $(2+i)z\equiv 0$. And this should have $(2+i)(2-i)=5$ cosets as solutions. Also $tr(f_*\vert H_2)=\det(f) = 10$, $tr(f_*\vert H_1)=6$ (manifestly), and $tr(f_*\vert H_0)=1$. Thus Lefschetz also gives $10-6+1=5$. Commented May 9, 2014 at 17:46
• You could think of the action of $f_*$ on the homology as being the identity on $H_0$ (the unique path component is mapped to itself). On $H_1$ the matrix of $f_*$ is (I hope) almost trivially that of multiplication by $3+i$ on $\Bbb{C}$ viewed as a 2-d vector space over $\Bbb{R}$. After all $H_1$ is "the free abelian group on the line segments $0\to1$ and $0\to i$. As a mapping of the complex plane $f$ stretches the area of all rectangles by a factor of $N(3+i)=10$. Thus the image of the 2-cell wraps around itself 10-fold. Commented May 9, 2014 at 17:58

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.)

• The benefit of working at this level of generality (not picking a particular $f$, not picking a particular dimension, and not picking a particular decomposition of my torus as a product of circles) is that it gets harder for me to say the wrong thing because I'm implicitly requesting more functoriality. The exterior power is almost the only thing that works. Commented May 9, 2014 at 17:50
• Here $f$ is my linear map $\left( \begin{array}{cc} 3 & -1 \\ 1 & 3 \end{array}\right)$, $V = \mathbb{R}^2$, $\Gamma = \mathbb{Z}^2$. In this case, $f$ extends to an action on the wedge products $\Lambda^0(\mathbb{R}^2),\Lambda^1(\mathbb{R}^2),\Lambda^2(\mathbb{R}^2)$, globally. Commented May 9, 2014 at 19:05
• Commented Aug 14, 2022 at 4:12