# How do functional equations for zeta functions arise from the structure of a homology group?

I have read in various disparate sources that certain zeta functions satisfy functional equations as a consequence of some structure on some homology group. Here is an example of a quote in this spirit, from the Wikipedia on functional equations for L-functions.

There are also functional equations for the local zeta-functions, arising at a fundamental level for the (analogue of) Poincaré duality in étale cohomology

Let me be clear that I do not understand this quote.

Can someone elaborate on what the author of this quote is talking about here? Are there other simple examples of this that don't come from number theory? Thanks!

• The Weil conjectures (starting with elliptic curves over finite fields) Oct 19, 2019 at 2:41

There are examples that don't come from number theory, although it's not much simpler. Specifically, the Lefschetz zeta function.

Let $$X$$ be a compact manifold of dimension $$d$$ and let $$f: X\to X$$ be a map. Let $$L(f^n)$$ be the number of fixed points of $$f^n$$, counted with appropriate multiplicity. Then if we define

$$\zeta_f(t) = e^{ \sum_{n=1}^{\infty} L(f^n) \frac{t^n}{n} }$$

by the Lefschetz fixed point formula we have

$$\zeta_f(t) =e^{ \sum_{n=1}^{\infty} L(f^n) \frac{t^n}{n} } = e^{ \sum_{n=1}^{\infty} \sum_{i=1}^{d} (-1)^i \operatorname{tr} ( f^n, H^i(X,\mathbb Q))\frac{t^n}{n} }=\prod_{i=1}^{d} \left( e^{ \sum_{n=1}^{\infty} \operatorname{tr} ( f^n, H^i(X,\mathbb Q))\frac{t^n}{n} }\right)^{(-1)^i}$$

$$\prod_{i=1}^{d}\left( \det ( 1 - t f, H^i(X,\mathbb Q)) \right)^{(-1)^i}$$

Now using the fact that the eigenvalues of $$f$$ on $$H^i$$ are $$\deg f$$ divided by the eigenvalues of $$f$$ on $$H^{d-i}$$, which comes from Poincare duality, by substituting each term we can prove the functional equation (if $$d$$ is even)

$$\zeta_f ((\deg f)^{-1} t^{-1} ) = \prod_{i=1}^{d}\left( \det ( 1 - (\deg f)^{-1} t^{-1} f, H^i(X,\mathbb Q)) \right)^{(-1)^i}= \prod_{i=1}^{d}\left( \det ( 1 - t^{-1} f^{-1} , H^{d-i} (X,\mathbb Q)) \right)^{(-1)^i}= \prod_{i=1}^{d}\left( \det ( 1 - t^{-1} f^{-1} , H^{i} (X,\mathbb Q)) \right)^{(-1)^i}$$

$$=\prod_{i=1}^{d}\left( \det ( tf - 1, H^i(X,\mathbb Q)) \right)^{(-1)^i} t^{ (-1)^{i+1} \dim H^i(X,\mathbb Q)} / \det(f,H^i(X,\mathbb Q)) =\prod_{i=1}^{d}\left( \det ( 1- tf 1, H^i(X,\mathbb Q)) \right)^{(-1)^i} t^{ -\chi(M)} C = C t^{-\chi(M)} \zeta_f(t)$$ for some constant $$C$$.

The number theory one is proved exactly the same way.

• Typo: if $(\lambda_i)$ are the eigenvalues of $f$ acting on $H^i$, the eigenvalues of $f$ acting on $H^{d-i}$ are $\ \deg(f)^d/\lambda _i$ .
– abx
Oct 19, 2019 at 6:01
• @abx Thanks! That is not quite right either - I meant $(\deg f)$. Oct 19, 2019 at 11:51
• Oops! Yes, of course.
– abx
Oct 19, 2019 at 14:06