Non-vanishing of p-adic L-functions - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T12:36:45Z http://mathoverflow.net/feeds/question/60387 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/60387/non-vanishing-of-p-adic-l-functions Non-vanishing of p-adic L-functions Jupiter Jones 2011-04-02T21:03:27Z 2011-04-11T17:54:35Z <p>In <a href="http://mathoverflow.net/questions/60326" rel="nofollow">http://mathoverflow.net/questions/60326</a> 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 $f$.</p> <p>My questions:</p> <p>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).</p> <p>2) Is this result known in the non-ordinary case?</p> http://mathoverflow.net/questions/60387/non-vanishing-of-p-adic-l-functions/60572#60572 Answer by Rob Harron for Non-vanishing of p-adic L-functions Rob Harron 2011-04-04T16:27:04Z 2011-04-04T16:27:04Z <p>So, I just talked to David and he pointed me to his paper <em>L-functions and Division Towers</em> (<a href="http://www.ams.org/mathscinet-getitem?mr=958262" rel="nofollow">MR0958262</a>) 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).</p> http://mathoverflow.net/questions/60387/non-vanishing-of-p-adic-l-functions/60635#60635 Answer by monodromy for Non-vanishing of p-adic L-functions monodromy 2011-04-05T02:18:37Z 2011-04-05T12:07:54Z <p>This does not directly address the question, but I hope it will avoid some misunderstandings.</p> <p>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). </p> <p>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.</p> http://mathoverflow.net/questions/60387/non-vanishing-of-p-adic-l-functions/61313#61313 Answer by Robert Pollack for Non-vanishing of p-adic L-functions Robert Pollack 2011-04-11T17:54:35Z 2011-04-11T17:54:35Z <p>These arguments on the non-vanishing of $p$-adic $L$-functions are great! I had never seen them before.</p> <p>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.</p> <p>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.</p> <p>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 ${\bf Z}_p^\times$. </p> <p>Then the two-variable $p$-adic $L$-function can be thought of as a power series in ${\bf Z}_p[[S,T]]$. Say $$L_p(S,T) = a_0(S) + a_1(S)T + a_2(S)T^2 + \dots$$</p> <p>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).</p> <p>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.</p> <p>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.)</p>