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Hi friends,

I have some questions concerning the critical values of motives, in the sense of Deligne. I will only look at motives of the form $h^i(X)$ where $X$ is a smooth projective algebraic variety over $\mathbb{Q}$. If I understand correctly, the notion of critical value depends only on the Hodge numbers.

Introduce the notations:

$$ \Gamma_\mathbb{C}(s):=2(2\pi)^{-s}\Gamma(s), \quad \Gamma_\mathbb{R}(s):=\pi^{-s/2}\Gamma(\frac{s}{2}) $$ where $\Gamma$ is the usual Gamma function. Then one define the $L$-factor at infinity as follows.

Let us first consider the case where $i$ is odd. Then:

$$ L_\infty(h^i(X), s)=\prod_{p < q} \Gamma_{\mathbb{C}}(s-p)^{h^{p, q}} $$ where $p+q=k$ and $h^{p, q}$ is the corresponding Hodge number.

Example: If $i=3$ and $h^{3, 0}=0$ (for instance $X$ is an hypersurface in $\mathbb{P}^4$), then

$L_\infty(h^3(X), s)=\Gamma_{\mathbb{C}}(s-1)^{h^{2, 1}}$

When $i$ is even, the definition is more involved, as one has also to consider the action of complex conjugation on $H^{p, p}$, which decomposes this space as $H^{p^+}\oplus H^{p^{-}}$. Let

$$ h^{p^{\pm}}:=\dim_{\mathbb{C}} H^{p^{\pm}}. $$

Then $$ L_\infty(h^i(X), s)=\prod_{p < q} \Gamma_{\mathbb{C}}(s-p)^{h^{p, q}}\cdot \Gamma_{\mathbb{R}}(s-\frac{i}{2})^{h^{i/2+}} \Gamma_{\mathbb{R}}(s-\frac{i}{2}+1)^{h^{i/2-}} $$

After having introduced this: an integer $n$ is said to be critical for $M=h^i(X)$ if $L_\infty(M, s)$ has no pole at $s=n$.

In his paper Deligne also asks that $L_\infty(\hat{M}, 1-s)$ has no pole at $n$. Is that necessary or it is a consequence of the former provided one has a functional equation?

Anyway, I would like to know if for my example the critical integers are all integers $n \geq 2$.

Question 2 What about a $K3$ surface? Can one determine the critical integers for the transcedental lattice? In that case there is a $\Gamma(s)$ coming from $h^{2,0}=1$. What about the real Gamma factor?

Thanks for your help!

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The factor $L_\infty(M,s)$ is holomorphic and non-vanishing for $\operatorname{Re}(s)$ large enough, so it is definitely necessary to also ask that $L_\infty(\hat{M},1-s)$ has no pole at the given integer. As an example, for the Riemann zeta function $\zeta(s)$, only the even integers $n \geq 2$ are critical.

I haven't done the computation of critical integers for $L$-functions of $K3$ surfaces, but there should be no difficulty in this computation as the answer depends only on the behaviour of the motive at infinity, which in this case just means (the cohomology of) the complex variety.

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Merci François! Do you have an idea of how to compute the action of complex conjugation on $H^{1,1}$ of a K3 surface, at least in some examples? – critval Mar 21 '13 at 17:55
I'm not sure, but when $X$ is a K3 surface the motive $M=h^2(X)$ has rank 22 and there are many trivial Euler factors so you may want to consider a direct factor of $M$ instead. The only cases I know of give you a motive of the form $M(f)$ where $f$ is a newform of weight 3. This is a rank 2 motive whose critical integers are only $s=1,2$. In this case the functional equation relates $L(f,s)$ and $L(f,3-s)$. – François Brunault Mar 21 '13 at 19:33
Yes, $h^2(X)$ decomposes as $\mathbb{L}^\rho \oplus t(X)$ where $\rho$ is the Picard number of $X$ and you want to consider the submotive $t(X)$. Do you have references for the cases you mention? Are there such examples for $\rho<20$? Thanks – critval Mar 21 '13 at 19:46
I'm not an expert but I think these cases have been completely classified, see the recent work of Schütt and Elkies-Schütt. – François Brunault Mar 22 '13 at 8:34

In your example (with Hodge structure of weight 3 concentrated in bidegrees (2, 1) and (1, 2)) there is only one critical value, at s = 2. At all other integer values of $s$, either $L_\infty(M, s)$ or $L_\infty(M^\vee, 1-s)$ will have a pole.

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Dear David, thanks for your answer! What do you use to compute $L_\infty(M^\vee, 1-s)$? I guess one has $M^\vee=M(3)$ by Poincaré duality, so $L_\infty(M^\vee, 1-s)=L_\infty(M, 4-s)$. But then $s=3$ would also be a critical value. Where is the mistake? – critval Mar 21 '13 at 17:54
I'm pretty sure that, the only integer $s$ such that $s \ge 2$ and $4-s \ge 2$ is $s = 2$. – David Loeffler Mar 21 '13 at 18:58
Lol, you are right. I forgot the -1 in my own computation Thanks – critval Mar 21 '13 at 19:13

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