Timeline for Non-vanishing of p-adic L-functions
Current License: CC BY-SA 2.5
20 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
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| Apr 11, 2011 at 17:54 | answer | added | sibilant | timeline score: 6 | |
| Apr 8, 2011 at 14:48 | comment | added | Rob Harron | @monodromy: It is surprising that this comes for free, but I can't see anything wrong with Jupiter's reasoning. I've talked to Rob Pollack and David Rohrlich about this and they are both quite surprised by this argument, but neither can see a problem with it. So, it's understandable that you don't believe it. It's just that, so far, your anti-arguments haven't been correct. | |
| Apr 6, 2011 at 22:00 | comment | added | monodromy | @Rob: Thanks for your kind comments. But then you agree with Jupiter Jones that if $f$ is a ($p$-ordinary) cusp form of weight $k>2$, then the mere fact that $L(f,1)\neq 0$ automatically implies (as I understand is being claimed in the question) that "In particular, for such modular forms, their associated $p$-adic $L$-functions are non-zero"? I don't believe this is true, and I was trying to convey a reason why. | |
| Apr 6, 2011 at 5:20 | comment | added | Rob Harron | (cont'd) However, in higher weights, what the alliteratively-named Jupiter Jones has pointed out, is that you no longer need to worry about the central points. This makes the argument much easier (basically trivial in the weight $\geq4$ case and still easier in the weight 3 case). | |
| Apr 6, 2011 at 5:17 | comment | added | Rob Harron | @monodromy: BUT, the $p$-adic interpolation property is not inconsistent. It seems like you're trying to argue that maybe one doesn't even know that the $p$-adic $L$-function exists. But it does. Typically, the construction is to first create an element of the Iwasawa algebra, then show it has the required interpolation property. After that you're likely to be interested to know if you've constructed the zero function. At this point, the $p$-adic $L$ being non-zero is equivalent to it having finitely many zeros. This is a tricky question in weight 2 because all you have are BSD-related points. | |
| Apr 6, 2011 at 2:21 | comment | added | monodromy | @ Loeffler: I am trying to say: suppose for example the weight of $f$ is an even $k>2$ and imagine a situation in which there do exist infinitely many twisted $L$-values of interpolation which are zero. Then there would obviously not exist a nonzero $p$-adic $L$-function in the Iwasawa algebra interpolating all those values, and yet, the desired interpolation property would include the nonzero value $L(f,1)$ among the values that we want the $p$-adic $L$−function tointerpolate. That is: the $p$-adic interpolation problem could well be inconsistent! | |
| Apr 5, 2011 at 7:32 | comment | added | David Loeffler | @monodromy: I'm not sure what you're trying to say there. A zero function can't somehow become nonzero at finitely many points. | |
| Apr 5, 2011 at 2:18 | answer | added | monodromy | timeline score: 0 | |
| Apr 5, 2011 at 1:45 | comment | added | monodromy | that we de not know a priori whether it has or not a (nonzero) solution, and hence one can not conclude as you did. | |
| Apr 5, 2011 at 1:45 | comment | added | monodromy | @Jupiter: If the $p$-adic L-function is given by a power series, one does not know that it is non-zero 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 point that is confusing you, the formula involving $L(f,1)$ that you mention is part of an interpolation problem | |
| Apr 5, 2011 at 1:30 | history | edited | Jupiter Jones | CC BY-SA 2.5 |
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| Apr 5, 2011 at 1:21 | comment | added | Jupiter Jones | @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. | |
| Apr 5, 2011 at 0:56 | comment | added | monodromy | 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? | |
| Apr 4, 2011 at 17:41 | vote | accept | Jupiter Jones | ||
| Apr 4, 2011 at 16:27 | answer | added | Rob Harron | timeline score: 8 | |
| Apr 4, 2011 at 14:20 | history | edited | Jupiter Jones |
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| Apr 4, 2011 at 4:03 | comment | added | Jupiter Jones | Thanks Rob H. But from your last answer it does seem that the real content is when $s=k/2$... | |
| Apr 2, 2011 at 22:28 | comment | added | Rob Harron | 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 $s=k/2$. | |
| Apr 2, 2011 at 21:03 | history | asked | Jupiter Jones | CC BY-SA 2.5 |