Introduction

$$S^1=\mathbb{R}/2\pi \mathbb{Z}$$

Fourier analysis tells us that the function space on $S^1$ is topologically spanned by

$$\{f_k (x)=e^{\sqrt{-1} k x} ; k\in \mathbb{Z}\}.$$

Let us consider the Laplacian $\Delta=-(\partial/\partial x)^2$ .

$$\Delta( f_k(x))=k^2 f_k(x).$$

$$\Delta= {\mathrm{diag}}(\dots,3^2,2^2,1^2,0^2,1^2,2^2,3^2,\dots)$$

The 0 -eigenspace of the Laplacian corresponds to the space of constants $H^0(S^1,\mathbb{R})$ . We omit it and consider somewhat reduced matrix:

$$\tilde \Delta= {\mathrm{diag}}(\dots,3^2,2^2,1^2,1^2,2^2,3^2,\dots).$$

We may then consider its $-s/2$ -th power:

$$\tilde \Delta^{-s/2} = {\mathrm{diag}}(\dots,3^{-s},2^{-s},1^{-s},1^{-s},2^{-s},3^{-s},\dots).$$

Its trace is equal to the Riemann Zeta function (up to the multiplicative constant $2$ .)

$$\operatorname{tr}\tilde \Delta^{-s/2} = 2 \sum _{n=1}^\infty \frac{1}{n^s}= 2 \zeta(s).$$

REMARK 01.1   For any compact Riemannian manifold $M$ , we may mimic the argument. Namely,

1. There are Laplacians $\Delta_i$ on the space $\Omega^i(M,\mathbb{C})$ of $i$ -forms for $i=0,1,2,\dots, \dim M$ .
2. For each $i$ , the 0 -eigenspace of $\Delta_i$ is equal to the $i$ -th cohomology
3. For each $i$ , we may consider reduced Laplacian $\tilde \Delta_i$ . It is an self adjoint operator with positive eigenvalues.
4. The trace of the power $\Delta_i ^{-s/2}$ may be considered as the ``$i$ -th Zeta function of $M$ ''.

REMARK 01.2   Eigen vectors of the Laplacians are ``particles'' on the manifold $M$ . The zeta functions are then considered to be ``generating functions of the number of particles on $M$ ''.

In general, we employ the following principle:

The zeta functions are ``generating functions of number of particles''

The meaning of the term ``particles'' may vary.