Let $\chi$ and $\psi$ be two quadratic Dirichlet characters and let $L(s,\chi)$ and $L(s,\psi)$ their associated Dirichlet $L$-functions.

Is there a realtion between these two Dirichlet $L$-functions: $L(s,\chi)\times L(s,\psi)$ and $L(s,\chi\times \psi)$?

I mean can we derive an expression which relates these two functions? Thanks in advance.

  • 2
    $\begingroup$ I suppose $\times$ means multiplication here? I would highly suggest using $\cdot$ for that, since $\times$ is usually used for other kinds of operations in mathematics. $\endgroup$
    – Wojowu
    May 25, 2016 at 9:51
  • 1
    $\begingroup$ The DIrichlet series of $\zeta(s) L(s,\chi) L(s,\psi) L(s, \chi \times \psi)$ has all non-negative coefficients (it's also the zeta function of a biquadratic field), which among other things is important in the proof of Siegel's theorem on exceptional zeroes of L-functions. $\endgroup$
    – Terry Tao
    May 25, 2016 at 15:29
  • 4
    $\begingroup$ @TerryTao: A small point of warning is that one needs to replace $\chi$, $\psi$, $\chi\times\psi$ by the underlying primitive characters to obtain the zeta function of the corresponding biquadratic field (and of course these three characters need to be non-principal). $\endgroup$
    – GH from MO
    May 25, 2016 at 16:11
  • $\begingroup$ Ah, right (I always forget about these pesky local factors!). $\endgroup$
    – Terry Tao
    May 25, 2016 at 16:22
  • $\begingroup$ An affirmative answer to the following question can give a divisibility relation between the objects in the question. Let $k$ be an even integer and $p$ a prime number such that $p-1|k$. Suppose that $p$ does not divide $L(1-k,\chi)$, where $\chi$ is a quadratic character. Can we find another quadratic character $\psi$ such that $p$ does not divide $L(1-k,\psi)$ and $L(1-k,\chi.\psi)$ ? Thanks of any comment $\endgroup$
    – user92196
    Jun 6, 2016 at 8:39

2 Answers 2


Using the perspective of "pretentious" multiplicative number theory, one can see that these $L$ functions are correlated with each other through the Deuring-Heilbronn repulsion phenomenon, but there is not a simple identity connecting the two (unless one is willing to introduce zeta functions of biquadratic fields, as mentioned in my comment above).

This is easiest to explain through formal calculations at $s=1$. From formal manipulation of the Euler product one has

$$ L(1,\chi) = \prod_p (1 - \frac{\chi(p)}{p})^{-1}$$ $$ \approx \exp( \sum_p \frac{\chi(p)}{p} ).$$

Thus, we expect Siegel zero type behaviour (unusually small values of $L(1,\chi)$) precisely when the (conditionally convergent) sum $\sum_p \frac{\chi(p)}{p}$ is unusually large and negative, or more informally if the primes $p$ are unusually biased towards the set $\{ n: \chi(n)=-1\}$ rather than $\{ n: \chi(n)=+1\}$ (e.g. if they are unusually biased to be quadratic non-residues with respect to some fixed modulus). (In the language of pretentious number theory, this implies that the Liouville function $\lambda(n)$ (or Mobius function $\mu(n)$) ``pretends'' to be $\chi(n)$ in that the two are positively correlated on primes.)

Now the repulsion phenomenon manifests itself in the simple (heuristic) observation that if $\chi(p)$ is biased to be $-1$, and $\psi(p)$ is also biased to be $-1$, then $\chi(p) \psi(p)$ should be biased to be $+1$. (To put it another way: if $\chi$ and $\psi$ both pretend to be $\lambda$, then $\chi \psi$ should pretend to be $\lambda^2 = +1$.) This can be quantified through the elementary inequality $$ 1 + \chi(p) + \psi(p) + \chi(p) \psi(p) \geq 0$$ which is the logarithmic form of the fact mentioned in my previous comment that $\zeta(s) L(s,\chi) L(s,\psi) L(s, \chi \psi)$ has non-negative Dirichlet coefficients. This can be used to quantify various versions of the repulsion phenomenon, which roughly speaking asserts that at most one of $L(1,\chi), L(1,\psi), L(1,\chi \psi)$ can be exceptionally small. For instance, one precise inequality one can obtain from the above discussion is that $$ \zeta(\sigma) L(\sigma,\chi) L(\sigma,\psi) L(\sigma,\chi \psi) \geq 1$$ for any real $\sigma > 1$. One can compare this with Mertens' famous inequality $$ \zeta(\sigma)^3 |\zeta(\sigma+it)|^4 |\zeta(\sigma+2it)| \geq 1$$ that is commonly used to prove the prime number theorem. (Indeed, if one takes a suitably "adelic" viewpoint then the two inequalities are both consequences of the general theory of Rankin-Selberg products.)

Note however that the repulsion effect only provides inequalities between the behaviour of $L(s,\chi), L(s,\psi), L(s,\chi\psi)$, rather than identities. For instance, if one knew that $\chi(p)$ was $-1$ $60\%$ of the time (for some range of $p$), and $\psi(p)$ was $-1$ $70\%$ of the time, then this lets one bound the proportion of $p$ for which $\chi(p) \psi(p)=+1$ to be between $30\%$ and $90\%$, but does not pin things down precisely. To achieve the latter one would need to understand the precise behaviour of the quantity $1 + \chi(p) + \psi(p) + \chi(p) \psi(p)$, which is captured by the zeta function of a biquadratic field (modulo some local factors, and assuming that $\chi, \psi, \chi \psi$ are non-principal). Unfortunately, the main way we know of to understand the latter is to break things back up into Dirichlet L-functions...

At some point there is going to be a monograph by Granville and Soundararajan on the pretentious approach, but until then, one can look for instance at

Andrew Granville and K. Soundararajan, MR 2276774 Large character sums: pretentious characters and the Pólya-Vinogradov theorem, J. Amer. Math. Soc. 20 (2007), no. 2, 357--384.

and various followup papers for more precise discussions of these topics.


It follows from the functorial properties of characters of representations (1-dimensional as in the case of Dirichlet or not) that:

$$L(s,\chi) \times L(s,\psi)=L(s,\chi+\psi)$$

This basically follows from the fact that we are dealing with a direct sum of representations, $\rho_\chi \oplus \rho_\psi$.

On the other hand, $L(s,\chi \times \psi)$ is essentially a Rankin-Selberg convolution, that is, the L-function of $\rho_\chi \otimes \rho_\psi$.

I'm not aware of any natural relation between those two operations that would allow to express one in terms of the other. Perhaps it would be easier to find a representation-theoretic identity, and then translate it back to L-functions.

  • 4
    $\begingroup$ The OP might know only about Dirichlet L-functions, not other kinds (including not higher-dim. characters) and therefore might be confused since "$\chi + \psi$" is not itself a Dirichlet character anymore. $\endgroup$
    – KConrad
    May 25, 2016 at 14:38
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    $\begingroup$ Perhaps $\chi \oplus \psi$ would be less ambiguous than $\chi + \psi$, then? $\endgroup$
    – Terry Tao
    May 25, 2016 at 15:28
  • $\begingroup$ @KConrad : from the identity of $L$ functions, I guess $\chi + \psi$ means the multiplicative function whose value at the primes is $f(p^k) = \sum_{m=0}^k \chi(p)^m \psi(p)^{k-m}$ ? and $\rho_{\chi} \otimes \rho_\psi$ is supposed to be related with some underlying automorphic forms ? $\endgroup$
    – reuns
    May 28, 2016 at 2:43

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