# Could a motivic spectrum have a “zeta function”?

I'm currently learning about zeta functions, so I apologize in advance if this is riddled with nonsense. Suppose you have a sequence $$E=(E_0,E_1,...)$$ of motivic spaces along with structure maps $$s_i:\mathbb{P}^1\wedge E_i\to E_{i+1}$$ that induce weak equivalences $$\Sigma_{\mathbb{P}^1}E_i\simeq E_{i+1}$$, so that $$E$$ is a motivic spectrum. Is there a sequence of zeta functions $$\zeta_E=(\zeta_{E_0},\zeta_{E_1},...)$$ that respects the structure maps?

There are several pieces of this question that I'm not sure about:

1. If $$E_i$$ is a projective variety over $$\mathbb{F}_q$$, then we can assign a local zeta function $$\zeta_{E_i}$$ to it. Is it possible to do the same for a more general motivic space?
2. If $$X$$ and $$Y$$ are motivic spaces that have zeta functions $$\zeta_X$$ and $$\zeta_Y$$, is there a zeta function $$\zeta_{X\wedge Y}$$ for their smash product? If so, is there any way to express $$\zeta_{X\wedge Y}$$ in terms of $$\zeta_X$$ and $$\zeta_Y$$?
3. If $$X$$ and $$Y$$ are motivic spaces that have zeta functions $$\zeta_X$$ and $$\zeta_Y$$, and if $$f:X\to Y$$ is a sufficiently nice morphism, is there a way to make sense of $$f_*\zeta_X$$ (for example, would this be $$\zeta_{f(X)}$$)? If so, is there a way to relate $$f_*\zeta_X$$ and $$\zeta_Y$$?

Etale cohomology factors through the stable $$\mathbb A^1$$ homotopy type. So you should simply consider your local zeta factor in terms of traces of frobenius on etale cohomology. Since smashing with $$\mathbb P^1$$ corresponds to tensoring with $$\mathbb Q_l(q)$$ the eigenvalues that give you the local factor will be multiplied by $$q$$ each time. This means that the local factor will undergo the transformation $$f(t) \mapsto f(qt)$$. In terms of the global zeta function, this should correspond to translating by $$1$$.

• I suppose this is answered by your first sentence, but does this mean that two weakly equivalent spaces will have the same global zeta function? I was worried that contracting $\mathbb{A}^1$ would change the set of points over which the Euler product is taken. – Stephen McKean Jan 17 at 17:00
• @StephenMcKean Of course $\mathbb A^1$ and $*$ don't have the same zeta function. Two proper spaces that are weakly equivalent have the same zeta function. The zeta function depends on the compactly supported cohomology, which does not factor through the $\mathbb A^1$ homotopy category. But for proper varieties, compactly supported cohomology agrees with ordinary cohomology, hence factors through the homotopy category. – user1092847 Jan 17 at 19:53
• Great, thanks for your clear explanations! – Stephen McKean Jan 17 at 20:00