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Let $X$ be a smooth projective variety of dimension $n=\dim X$ over a finite field $\mathbb{F}_q$. As is well known, its zeta function satisfies a functional equation of the form $$Z(X,q^{-n}T^{-1})=\pm q^{n E/2}t^{E}\zeta(X,T),$$ where $E$ is the Euler characteristic of $X$. The question is, what is the sign? The obvious guess is that it only depends on $E$ so the equation may be written in a slightly more precise form as $$Z(X,q^{-n}T^{-1})=(-q^{n/2}t)^{E}\zeta(X,T).$$ This is true for curves, projective spaces $\mathbb{P}^n$ and some other simple examples. But I do not know if it is true in general (and if not, what is really the meaning of the sign).

EDIT. As was pointed out in the comments, the sign was calculated by Deligne in "La Conjecture de Weil, I". (I should have read it before asking, but my French is awful and I never got beyond the first page.) The sign is positive for odd $n$ and is equal to $(-1)^N$ for even $n$, where $N$ is the multiplicity of the eigenvalue $q^{n/2}$ of the Frobenius action in the middle cohomology $H^n(X,\mathbb{Q}_l)$.

Now, we have $(-1)^E=(-1)^{N+N'}$ where $N'$ is the multiplicity of $-q^{n/2}$. (It is due to Poincare duality and the fact that all other eigenvalues come in conjugate pairs.) For the above "guess" to be wrong for some $X$ we need odd (in particular, nonzero) $N'$. I would appreciate an explicit example of this.

Anyway, the formula is true for a quadratic extension of the base field because of a simple identity $$Z(X/\mathbb{F}_q, T)\cdot Z(X/\mathbb{F}_q, -T)=Z(X/\mathbb{F}_{q^2}, T^2).$$

EDIT: The surface (suggested by David E Speyer) $x^2-y^2=z^2+w^2$ over $\mathbb{F}_3$ gives $N=N'=1$, so my guess was off. Also, it illustrates that the sign (unfortunately) has no "geometric meaning", it is an arithmetic invariant. Thanks to all for help.

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    $\begingroup$ The sign is calculated in Deligne Weil I. AFAIR it depends on the middle dimensional cohomology. $\endgroup$
    – user19475
    Commented Apr 15, 2018 at 10:43
  • $\begingroup$ Deligne, Weil I, (2.6): If $d$ is even, let $N$ be the multiplicity of $q^{d/2}$ as an eigenvalue of the Frobenius on $H^d(X,\mathbf{Q}_\ell)$. Then the sign is $1$ if $d$ is odd, and $(-1)^N$ if $d$ is even. $\endgroup$
    – user19475
    Commented Apr 15, 2018 at 11:54
  • $\begingroup$ Thank you. But I am still curious if it is not the same. If I am not mistaken, it would be so if the whole dimension of $H^d(X)$ always has the same parity as $N$. Do you know a counterexample? $\endgroup$
    – Alex Gavrilov
    Commented Apr 15, 2018 at 12:14
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    $\begingroup$ I haven't followed every step of what you've written, but let $X$ be a smooth quadratic surface in $\mathbb{F}_p \mathbb{P}^3$ where Frobenius switches the two rulings; then the action on $H^2$ has one eigenvalue of $p$ and one of $-p$. In equations, I think this should be $x_1^2-x_2^2 = x_3^2-u x_4^2$ with $p$ odd and $u$ a quadratic nonresidue modulo $p$. $\endgroup$
    – David E Speyer
    Commented Apr 15, 2018 at 15:20
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    $\begingroup$ This also gives an example where the sign depends on $p$, since depends on the quadratic character $\left( \tfrac{u}{p} \right)$. $\endgroup$
    – David E Speyer
    Commented Apr 15, 2018 at 15:29

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