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Let $\chi$ be a primitive quadratic Dirichlet character of d modulus $m$, and consider the product $$\prod_{\substack{p \text{ prime} \\ \chi(p) = 1}} (1-p^{-2})^{-1}.$$

What can we say about the value of this product? Do we have good upper or lower bounds?

Some observations, ideas, and auxiliary questions

  • When $\chi$ is trivial, it has value $\zeta(2)$.
  • In general, since $\chi(p)$ is equidistributed in the limit, I would "want" the value to be something like

$$\Big(\zeta(2)\prod_{p | m} (1-p^{-2})\Big)^{\frac{1}{2}}.$$

However, if I'm not mistaken, it seems that the error terms in effective forms of this result may cause this to be very far from the truth. We can't ignore what happens before we are close to equidistribution as the tail and the head are both $O(1)$. We can't even control the error term well (without GRH) because of Siegel zeroes.

  • I don't think we can appeal to Dirichlet density versions of CDT since those only tell us things in the limit as $s$ goes to $1$ and here $s = 2$.
  • Is there a way to "Dirichlet character"-ify a proof of $\zeta(2) = \pi^2/6$ to get a formula for this more general case? At least with Euler's proof via Weierstrass factorization, it seems that we would need some holomorphic function which has zeroes whenever $\chi(n) = 1$.

I had a few other ideas but they all seem to run into the same basic problem of "can't ignore the stuff before the limit"... am I missing something?

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    $\begingroup$ The equidistribution of $\chi(p)$ is a special case of Dirichlet's theorem on primes. It was proved about 85 years before Chebotarev's theorem. $\endgroup$
    – GH from MO
    Feb 25 at 10:20
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No need to use Chebotarev or Dirichlet: set $Z(k)=\prod_{p\nmid m}(1-p^{-k})^{-1}$ and $L(k)=\prod_p(1-\chi(p)p^{-k})^{-1}$. Then if $P(k)=\prod_{\chi(p)=1}(1-p^{-k})^{-1}$ we have the recursion $$P(k)=(Z(k)L(k)/Z(2k))^{1/2}P(2k)^{1/2}$$ from which you can deduce bounds that you want.

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  • $\begingroup$ To make sure I understand, the point is that there are closed forms for $L(k)$ and $P(2k)$ goes to $1$ pretty rapidly as $k$ gets large, so with a program (or a lot of paper) we could get an arbitrarily tight numerical approximation, right? $\endgroup$
    – bean
    Feb 25 at 18:59
  • $\begingroup$ I have been doing this programmatically for now, but right now for a lower bound I'm truncating the product for $p \leq x$ and for an upper bound I'm starting at $\zeta(2)$ and removing $p \leq x$ for which $\chi(p) \neq 1$. If I understood right, yours converges much faster. $\endgroup$
    – bean
    Feb 25 at 19:01

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