Let $\mathcal{C}$ denote a category with a zero object. Let $\mathrm{N}(X)$ denote the norm of an object $X$ of a category $\mathcal{C}$, defined as $\mathrm{N}(X) = \,|\mathrm{Hom}(X,X)\,|$. From this terminology, we say a non-zero object $X$ is *finite* if $\mathrm{N}(X)$ is finite. Let $\mathrm{P}(\mathcal{C})$ denote the isomorphism classes of all finite simple objects and $\mathrm{M}(\mathcal{C})$ denote the isomorphism classes of all finite semisimple objects. For the former, letting $[X] \in \mathrm{P}(\mathcal{C})$, the norm $\mathrm{N}([X])$ is well-defined so we choose a representative of an isomorphism class $[X]$ so we have $\mathrm{N}([X]) = \mathrm{N}(X)$. According to [1], the *categorical zeta function* of our category $\mathcal{C}$ is defined as


$$\zeta\_\mathcal{C}(s) = \prod\_{[X] \in \mathrm{P}(\mathcal{C})} \frac{1}{1-\mathrm{N}(X)^{-s}}.$$


As an example, if $\mathcal{C}$ denotes the category of $\mathbb{Z}$-modules, we have $\zeta\_{\mathcal{C}}(s) = \zeta(s)$, the Riemann zeta function. Generalizing this to $O\_K$-modules, where $O\_K$ denotes the ring of integers of a number field $K$, we have $\zeta\_{\mathcal{C}}(s) = \zeta\_K(s)$, the Dedekind zeta function. From the definition of a semisimple object, it seems to me we ought to have the relationship


$$\prod\_{[X] \in \mathrm{P}(\mathcal{C})} \frac{1}{1-\mathrm{N}(X)^{-s}} = \sum\_{[X] \in \mathrm{M}(\mathcal{C})} \frac{1}{\mathrm{N}(X)^s},$$
since we can write semisimple objects into a coproduct (direct sum, in this case) of simple objects but I'm not quite sure if I can exactly treat $\mathrm{N}$ as if it was multiplicative. I'm not finding other sources that consider the above.



> 
> **Question:** Is it true, for any category with a zero object, we have the relationship $$\prod\_{[X] \in \mathrm{P}(\mathcal{C})} \frac{1}{1-\mathrm{N}(X)^{-s}} = \sum\_{[X] \in \mathrm{M}(\mathcal{C})} \frac{1}{\mathrm{N}(X)^s} \;?$$
> 
> 
> 


**References:**


1.) N. Kurokawa. Zeta Functions of Categories. *Proc. Japan Acad.* **Vol 72A** No. 10, 221-222 (1996).



Properties of categorical zeta function
Consider the category $\def\Set{\mathbf{Set}}\Set\_\*$ of pointed sets.


Kurokawa defines the simple objects to be those nonzero $X$ for which the only morphisms $X\to Y$ are monomorphisms or zero, so the only simple object of $\Set\_\*$ is the two-element set $P = \{\*, 0\}$, whose norm is $N(P) = 2$.
Therefore,
$$
\zeta\_{\Set\_\*}(s) = \frac1{1-2^{-s}}
$$


On the other hand, the finite pointed sets are the semisimples, and $N(X) = (n+1)^n$ if $|X|=n+1$.


Therefore,
$$
\sum\_{[X]\in M(\Set\_\*)}\frac1{N(X)^s} = \sum\_{n=0}^\infty\frac1{(n+1)^{ns}}
$$
which does not agree with the zeta function for $\Set\_\*$.




---


I imagine this even fails when $\def\cC{\mathcal{C}}\cC = \def\Z{\mathbb{Z}}\def\Mod{\mathbf{Mod}}\Z\Mod$.
The finite semisimple $\Z$-modules are precisely the direct sums of cyclic groups of prime order, hence there is a unique semisimple $\Z$-module of every finite order.


The issue is that the norm of such a module is not its order, unlike with the simples.
Indeed, even consider $\Z/3\oplus\Z/3$.
In this case, the endomorphism ring has order $3^4$ (rather than the $3^2$ you would want).


In fact, you can prove the following "almost multiplicity" of the norm for an abelian category:


**Claim 1.** If $\cC$ is abelian, then for simple objects $X$ and $Y$, we have
$$
N(X\times Y) = \begin{cases} N(X)\times N(Y), & \text{if $X\not\cong Y$} \\ N(X)^4, & \text{otherwise} \end{cases}
$$



> 
> *Proof.* Finite products and coproducts coincide for abelian categories, so
> $$
> \begin{align\*}
> N(X\times Y) &= |\def\Hom{\operatorname{Hom}}\Hom(X\times Y, X\times Y)| \\
> &= |\Hom(X\sqcup Y, X\times Y)| \\
> &= |\Hom(X, X)|\times|\Hom(X, Y)|\times|\Hom(Y, X)|\times|\Hom(Y, Y)|
> \end{align\*}
> $$
> Now, by Schur's lemma, $\Hom(X, Y) = \{0\}$ if and only if $X\not\cong Y$ for simples $X$ and $Y$.
> Therefore, in one case, you get $N(X)\times N(Y)$, and in the other case, you get $N(X)^4$. $\blacksquare$
> 
> 
> 


In particular, the series $\sum\_{[X]\in M(\Z\Mod)}\frac1{N(X)^s}$ would not agree with $\sum\_{n=1}^\infty\frac1{n^s}$.




---


**Edit.** ($\times2$) After some thought, I believe there is a good "fix" for the norm that should still be satisfying.


**Definition.** For a more general object $Y$ of $\cC$, define its **norm** to be
$$
\bar N(Y) := \prod\_{[X]\in P(\cC)}|\Hom(Y, X)|
$$


In the context of a category where Schur's lemma applies, this will be an extension of the original definition of a norm for simple objects---in that $N(X) = \bar N(X)$ for $X$ simple---but this definition has the advantage of being multiplicative:


**Claim 2.** In an abelian category, for all semisimple objects $X$ and $Y$, we have $\bar N(X\oplus Y) \cong \bar N(X)\times \bar N(Y)$.


The proof of the above claim is similar to Claim 1, but now with this multiplicative norm, one can show (analogous to the usual Euler product identity for the Riemann zeta function) that
$$
\zeta\_\cC(s) = \sum\_{[X]\in M(\cC)}\frac1{\bar N(X)^s}
$$
for $\cC$ an abelian category.




---


I edited my answer to map *into* simples because this actually makes this identity true also for $\Set\_\*$!


We have for any pointed set $X$ of cardinality $n+1$ that $\bar N(X) = |\Hom(X, \{\*, 1\})| = 2^n$, and so
$$
\sum\_{[X]\in M(\Set\_\*)}\frac1{\bar N(X)^s} = \sum\_{n=0}^\infty\frac1{2^{ns}} = \frac1{1-2^{-s}} = \zeta\_{\Set\_\*}(s)
$$


In fact, we can now prove the following


**Claim 3.** In any category $\cC$ with a zero object and finite coproducts such that finite simple objects satisfy Schur's lemma,
$$
\prod\_{[X]\in P(\cC)}\frac1{1-N(X)^{-s}} = \sum\_{[X]\in M(\cC)}\frac1{\bar N(X)^s}
$$
where "Schur's lemma" is the statement that any nonzero map $X\to Y$ between finite simple objects is an isomorphism (which holds also for $\Set\_\*$), and $M(\cC)$ is the set of isomorphism classes of finite coproducts of finite simple objects.


