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If f is a weight 2 cuspidal newform, then it is common for L(f,1) to vanish. Indeed, the sign of the functional equation of f can force such vanishing. However, if f has weight k>2, then there is no a priori reason why L(f,1) will vanish.

My question: are there known examples where L(f,1)=0 for a newform f of weight strictly greater than 2 or is there some (easy?) reason such examples shouldn't exist?

flag	asked Apr 1 2011 at 22:12 Aaaa 13●2

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How are you normalizing the functional equation? Is $s=1$ meant to be on the critical line? – Stopple Apr 1 2011 at 22:16

Rob Harron · Accepted Answer · 2011-04-02 03:08:19Z UTC

Based on your normalization, $L(s,f)$ is defined as an Euler product for $\Re(s)>\frac{k+1}{2}$, so $L(s,f)$ is non-zero in that right-half plane. Now Jacquet–Shalika MR0432596 showed that that non-zero region extends to the line $\Re(s)=\frac{k+1}{2}$ (for $\mathrm{GL}(n)$, in fact) (this is analogous to the proof of non-vanishing on the line $s=1$ for $\zeta(s)$). Furthermore, $L(s,f)$ satisfies a functional equation $\Lambda(s,f)=i^k\Lambda(k-s,W_Nf)$ where $N$ is the level of $f$, $W_N$ is the Atkin-Lehner operator, so $W_Nf$ is also weight $k$ and level $N$, and $\Lambda(s,f)=N^{s/2}(2\pi)^{-s}\Gamma(s)L(s,f)$. So, $s=\frac{k}{2}$ is the central point (which is indeed $\neq1$ if $k>2$) (stuff like Beilinson–Deligne–Bloch–Kato basically says that vanishing away from the central point should be easier to understand). Now the values of $L(s,f)$ at $0\lt s\leq\frac{k-1}{2}$ are related to values of a possibly different modular form ($W_Nf$) at $k-s$, i.e. in the non-vanishing range $\Re(s)\geq\frac{k+1}{2}$. The only complicating part of this relation is the Gamma factors, but at these $s$, both $s$ and $k-s$ are $>0$, so they are not poles of $\Gamma(s)$. Hence, $L(s,f)\neq0$ at all positive $s$ except possibly in the strip $\frac{k-1}{2}\lt\Re(s)\lt\frac{k+1}{2}$.

GH · Answer 2 · 2011-04-01 23:32:00Z UTC

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The sign of the functional equation of a self-dual cusp form (regardless its weight) can be changed easily by twisting the form with a quadratic character. In particular, the $L$-function of a self-dual cusp form (of any weight) often vanishes at the center. I don't think there is anything special about weight 2. See Theorem 7.6 in Iwaniec: Topics in classical automorphic forms.

answered Apr 1 2011 at 23:32

GH
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Experimental evidence shows that when the sign is $+1$, vanishing is much more common for weight $2$ than for other weights; I think one or two examples are known for weight $4$, and none for weight $\geq 6$. Perhaps William Stein can confirm this? – David Hansen Apr 1 2011 at 23:45

David, very interesting! In my comment I only cared about the sign of the functional equation. – GH Apr 2 2011 at 0:13

Stein has a paper (link below] where they give data for some weight 4 and 6 vanishings of small level. I thought the conjecture was "finitely many" for weight 6 and infinitely for weight 4, though maybe that is for twists. My recollection is that there is no known example beyond elliptic curves that has a triple vanishing. neil-dummigan.staff.shef.ac.uk/dsw_13.dvi – Junkie Apr 2 2011 at 4:27

There is another paper magma.maths.usyd.edu.au/~watkins/papers/heur.pdf and it gives 3 examples of weight 8 forms that vanish (CM) doubly in Table 3 page 23 and Table 5 next page, and says no triple zeros are known in this case. – Junkie Apr 2 2011 at 4:32

Aaaa · Answer 3 · 2011-04-02 00:21:23Z UTC

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I can't comment because I don't have enough points, but my normalization of the $L$-series is $L(f,s) = \sum_n a_n n^{-s}$ where $f=\sum a_n q^n$, and so $1$ is not on the critical line if $k>2$.

answered Apr 2 2011 at 0:21

Aaaa
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Well then this resolves your question, because the Grand Riemann Hypothesis for automorphic $L$-functions says that all the zeroes must occur on the critical line. However, there is no proof of this fact: even zero-free regions for automorphic forms do not eliminate the case of a Siegel zero (if $f$ is self-dual). It is much more common to write the normalisation as $L(s,f) = \sum^{\infty}_{n = 1}{a_f(n) n^{-s}}$, where $f(z) = \sum^{\infty}_{n = 1}{a_f(n) n^{(k - 1)/2} e(nz)}$, as this then ensures that $L(s,f)$ has critical line $\Re(s) = 1/2$. – Peter Humphries Apr 2 2011 at 0:31

@Peter: Actually Hoffstein and Ramakrishnan proved that for GL(2) cusp forms $L(s,f)$ has no Siegel zeros. See imrn.oxfordjournals.org/content/1995/6/… – GH Apr 2 2011 at 1:01

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Why are Siegel zeros relevant here? Aren't these zeroes close to the center of critical strip? Even for k=3, s=1 is pretty from the center. – Aaaa Apr 2 2011 at 1:36

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Siegel zeros are very close to the edge of the critical strip, i.e. very far from the center. In some sense they are the worst (potential) violations of GRH. – GH Apr 2 2011 at 3:12

Non-vanishing of L-series of modular forms (easy case?)

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