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In Non-vanishing of L-series of modular forms (easy case?) it was answered that for a cuspidal newform $f$ of weight strictly greater than 2, then $L(f,1)$ is non-zero. (Here the $L$-series is normalized so that the center of the critical strip is given by $s=k/2$.) In particular, for such modular forms, their associated $p$-adic $L$-functions are non-zero. As far as I know the non-vanishing of $p$-adic $L$-functions in the weight 2 case is a highly non-trivial result and relies upon a non-vanishing theorem of Rohrlich on twisted $L$-values. Further, from the non-vanishing of the $p$-adic $L$-function, one can deduce that $L(f,\chi,j)$ is non-zero for all but finitely many pairs $(\chi,j)$ where $\chi$ is a Dirichlet character of $p$-power conductor and $j$ is an integer between $1$ and $k-1$, as long as $p$ is an ordinary prime for $f$.

My questions:

1) Is there a direct argument to prove the non-vanishing of $L(f,\chi,j)$ for all but finitely many $\chi$ and $j$ in the ordinary and weight greater than 2 case (which doesn't use $p$-adic $L$-functions).

2) Is this result known in the non-ordinary case?

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The L-function of $f$ twisted by $\chi$ is the (possibly imprimitive) L-function of some other cuspidal eigenform $f_\chi$, so the answer of the previous mo question addresses the non-vanishing question here. The only problem is the imprimitivity, so you have to account for possible zeroes of the removed Euler factors. These can only occur at $s=(k-1)/2$ (since the roots are Weil numbers), so you get non-vanishing for $0\lt\Re(s)\lt(k-1)/2$ and $\Re(s)\geq (k+1)/2$ for all the twists. This is a rather direct proof away from $s=k/2$. –  Rob Harron Apr 2 '11 at 22:28
    
Thanks Rob H. But from your last answer it does seem that the real content is when $s=k/2$... –  Jupiter Jones Apr 4 '11 at 4:03
    
I am confused. Jupiter: when $k>2$, why is it enough that the L-values at $s=1$ be nonzero to conclude that the $p$-adic L-function does not vanish identically? I believe that in the case, say of a $p$-ordinary cusp form $f$ of weight $k>2$, one could not garantee that the $p$-adic L-function of $f$ does not vanish identically until one had precisely Rohrlich's results (since the non-central critical values are non-zero "for free", as argued by Rob). How can one otherwise exclude the possibility of infinitely many zeros among the values of interpolation? –  monodromy Apr 5 '11 at 0:56
    
@monodromy: When the p-adic L-function, thought of as say a power series in Z_p[[T]], is evaluated at 0 one obtains (essentially) L(f,1). Thus, if L(f,1) is non-zero, this power series is non-zero. But a power series in Z_p[[T]] only has finitely many zeroes (Weierstrauss preparation). Since this power series also interpolates L(f,\chi,j) for j between 1 and k-1, all but finitely many of these values must be non-zero. –  Jupiter Jones Apr 5 '11 at 1:21
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@monodromy: I'm not sure what you're trying to say there. A zero function can't somehow become nonzero at finitely many points. –  David Loeffler Apr 5 '11 at 7:32

3 Answers 3

up vote 8 down vote accepted

So, I just talked to David and he pointed me to his paper L-functions and Division Towers (MR0958262) whose Theorem 1 is the result (and proof) you're looking for. The proof doesn't care whether $p$ is good or bad or whatever. This takes care of the even weight case, the odd weight case having been dealt with in my comment above. (Also, in case you or anyone reading this were interested in the central point for odd weights, David said he had no idea how to do that).

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Excellent! I didn't realize that Rohrlich was at BU. Did you ask him if he was surprised that you get this result "for free" in the ordinary weight greater than 2 case (via p-adic L-functions)? –  Jupiter Jones Apr 4 '11 at 17:40
    
@Jupiter Jones: are you really sure that things are as easy as you suggest? For example are you sure that there is one $p$-adic L-function which is interpolating special values of all twists at all critical points at once, in the higher weight case? I'm no expert but somehow I didn't expect things to be so easy... –  Kevin Buzzard Apr 4 '11 at 18:14
    
@Kevin: There is indeed one p-adic L-function for all the critical twists, see for example section I.14 of Mazur–Tate–Teitelbaum. –  Rob Harron Apr 4 '11 at 22:08
    
@Jupiter Jones: I did not ask him that, though I'd be surprised if he were surprised. –  Rob Harron Apr 4 '11 at 22:39
    
@Kevin: I'm not sure about any of it, but I find the whole thing pretty weird. It does indeed seem that one single p-adic L-function encodes all of those L-values. Thus the non-vanishing of L(f,1) implies the non-vanishing of the whole lot of them (except for finitely many). Back in the 70s (?) when Mazur was conjecturing things like the rank of an elliptic curve over the cyclotomic Zp-extension is finite, was this kind of consideration in his mind at all? Namely, that conjecture (by BSD) is equivalent to the non-vanishing of all but finitely many L(f_E,chi,1) where f_E is the –  Jupiter Jones Apr 5 '11 at 1:14

These arguments on the non-vanishing of $p$-adic $L$-functions are great! I had never seen them before.

What I'm writing here is neither an answer to your question nor an actual proof of any sort. But I think it at least follows the general theme of what you are asking.

Namely, I tried to use 2-variable p-adic L-functions and the non-vanishing of $p$-adic $L$-functions of higher weight modular forms to deduce Rohrlich's theorem in the weight 2 case (i.e. the non-vanishing of $L(f,\chi,1)$ for $f$ a form of weight 2 for all but finitely many $\chi$ of $p$-power conductor). It didn't actually work as I need to assume the non-vanishing of some mu-invariant which is deep stuff, but I think the argument is amusing enough to present in any case.

Here's the argument: put the original weight 2 form $f$ into a Hida family, and write down the corresponding 2-variable $p$-adic $L$-function. For simplicity, let me assume that the ordinary Hecke algebra in this case is just $\Lambda = {\bf Z}_p[[S]]$. Here one sets $S=\gamma^k-1$ to specialize to weight $k$ where $\gamma$ is some topological generator of ${\bf Z}_p^\times$.

Then the two-variable $p$-adic $L$-function can be thought of as a power series in ${\bf Z}_p[[S,T]]$. Say $$ L_p(S,T) = a_0(S) + a_1(S)T + a_2(S)T^2 + \dots $$

First let me point out that this power series is non-zero. Indeed, it interpolates the $p$-adic $L$-functions of each classical form in the Hida family which have already been observed to be non-zero in weight greater than 2 (without invoking Rohrlich's theorem).

Now let's assume that at least one form in the Hida family has zero $\mu$-invariant. This means there is some weight k such that $$ L_p(f_k,T) = L_p(\gamma^k-1,T) = a_0(\gamma^k-1) + a_1(\gamma^k-1)T + a_2(\gamma^k-1)T^2 + \dots $$ has non-zero $\mu$-invariant. In particular, for some $i \geq 0 $, $a_i(\gamma^k-1)$ is not divisible by $p$. This implies that $a_i(S)$ is a unit in ${\bf Z}_p[[S]]$, and in particular is non-zero. Thus, the $p$-adic $L$-function in weight 2 $$ L_p(f_2,T) = L_p(\gamma^2-1,T) = a_0(\gamma^2-1) + a_2(\gamma^2-1)T + \dots + a_i(\gamma^2-1)T^2 + \dots $$ is non-zero as $a_i(\gamma^2-1)$ is non-zero.

Let me point out that one needs to confront this $\mu$ issue in some way. Possibly the two-variable $p$-adic $L$-function could have looked like $$ L_p(S,T) = (S - (\gamma^2-1)) + 0T + 0T^2 + \dots $$ The specialization of this power series to weight 2 then vanishes. But note, this would mean that every form in this Hida family has positive $\mu$-invariant, and moreover, these $\mu$-invariants blow up as you approach weight 2. (Possibly there is some easy reason why this can't happen, but I can't see one.)

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This does not directly address the question, but I hope it will avoid some misunderstandings.

In general, when $k>2$, the L-values (for any twist by a $p$-power order character) at $s=1$ being nonzero is not enough to conclude that the p-adic L-function does not vanish identically. For example, in the case of a $p$-ordinary cusp form $f$ of even weight $k>2$, one could not garantee that the p-adic L-function of $f$ does not vanish identically until one had precisely Rohrlich's results or equivalent (since the non-central critical values are nonzero "for free", as argued by Rob in his first comment).

In fact, in that case the $p$-adic L-function is given by a power series, in $T$ say, and one does not know that it is nonzero until one can show that is does not interpolate zero infinitely many times (i.e. that it does not have infinitely many distinct zeros), and for that it clearly does not suffice to say that just one of the values of interpolation is nonzero. In order words, and I think this is the confusing point, the formula recovering $L(f,1)$ from the value of the $p$-adic $L$-function at $T=0$ is part of an interpolation problem that we do not know a priori whether it has or not a (nonzero) solution $-$ of course, now we know that it does, but not from the mere fact that $L(f,1)\neq 0$ in those cases $-$, and hence one can not reach the conclusion as in the statement of this question.

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@monodromy: Is the confusion here that you don't believe that a non-zero power series in say Z_p[[T]] has finitely many zeroes? Here's the argument: by (p-adic) Weierstrauss preparation, if f(T) is a non-zero element of Z_p[[T]], one can write f(T) = P(T) U(T) where P(T) is a polynomial and U(T) is a unit power series. Since P(T) is a polynomial it has finitely many zeroes. Since U(T) is a unit, it has no zeroes. That's it. Apart from this, I don't understand your objection. –  Jupiter Jones Apr 5 '11 at 2:33
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it looks to me as if he thought that there were functions that vanish identically but have one non-zero value. –  Chris Wuthrich Apr 5 '11 at 9:43
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@Chris Wuthrich: Of course, this it not what I am thinking. I am thinking that a priori, without Rorhlich's result, one can not rule out the possibility of having (if the weight of $f$ is an even $k>2$) $L(f,1)\neq 0$ and yet infinitely many twisted central critical values $L(f,\phi,k/2)$ being zero. So $p$-adic $L$-functions do not give a proof of Rohrlich's result ``for free'', or at least not by the reasoning suggested above. –  monodromy Apr 7 '11 at 5:02

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