# Why use the Lefschetz Zeta function?

Given a compact, triangulable space $X$ and a continuous function $f: X\rightarrow X$, then we define the Lefschetz number $\Lambda_{f}$ by $$\Lambda_{f} = \sum_{k\geq0}(-1)^{k}Tr(f_{\ast}\vert H_{k}(X,\mathbb{C}))$$

If $f$ has only finitely many fixed points, then by the Lefschetz-Hopf theorem $$\Lambda_{f}=\sum_{x=f(x)}i(f,x)$$ where $i(f,x)$ is the fixed point degree of the fixed point $x$. The fixed point degree at a point can loosely be thought of as the multiplicity of the map at that point.

We define the Lefschetz Zeta function by $$\zeta_{f}(z)=\exp\left(\sum_{n=1}^{\infty}\Lambda_{f^n}\frac{z^n}{n}\right)$$ and we see using the Lefschetz Hopf theorem that $$\zeta_{f}(z)=\prod_{i=0}^{n}\det(1-zf_{\ast}\vert H_{i}(X,\mathbb{C}))^{(-1)^{i+1}}$$

Now this is all very well and good, but as far as I can tell there is nothing new to be gained from $\zeta_{f}(z)$. It seems to be specifically constructed so that the Lefschetz Hopf theorem can simplify it. Am I wrong? Does $\zeta_{f}(z)$ actually help us retireive informatin about $f$? Do the coefficients of $\zeta_{f}(z)$ correspond to anything? When would I actually learn anything by constructing this function?

• Two quick comments that don't really merit a full answer: 1) It's a rational function in nice cases by the last product expansion you give; and 2) It's a kind of zeta function and therefore generally counts things nicely. The last bit is admittedly very vague. For a more concrete example, take $f$ to be the Frobenius map. For an algebraic variety $V$ over $\overline{\mathbb{F}}_q$, the fixed points of $f$ are exactly the $\mathbb{F}_q$-rational points. For an example of why this isn't just convenient bookkeeping, see the Weil conjectures. – anomaly Nov 27 '14 at 18:33