Timeline for Same functional equation implies same zero order at s=1/2?
Current License: CC BY-SA 3.0
9 events
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| Mar 24, 2012 at 0:54 | comment | added | GH from MO | @Noam: This is very interesting and useful. Thank you for sharing your knowledge with us. | |
| Mar 24, 2012 at 0:25 | comment | added | Noam D. Elkies | @GH: Indeed we have no proof that any eigenform $g$ yields an $L$-function with a central zero of order $4$ or more. [This does not depend on having some $f$ in the same space whose $L$-function has no central zero.] Averaging can yield upper bounds on the mean analytic rank, but (as far as we know) cannot prove higher-order multiplicities than we can find for individual $g$. (And whatever wisdom I've exhibited in this thread is my teachers', not my own.) | |
| Mar 22, 2012 at 17:34 | comment | added | GH from MO | @Noam: I bow to your superior wisdom (seriously). So is there no example known of $f,g\in S_2(\Gamma_0(N))$ with root numbers $1$ such that the order of vanishing of their $L$-functions at the center is $0$ and at least $4$? If there is no known example (with proof), then can their existence be established by some analytic trick (such as averaging)? | |
| Mar 22, 2012 at 14:40 | comment | added | Noam D. Elkies | @GH: if $L(E,s)$ looks locally like $.00000003 + a_2 (s-1)^2 + \cdots$ then it has two zeros near $s=1$ but without more information you can't tell if they're at $s=1$ or just closer than the radius of the smallest contour you've tried. | |
| Mar 22, 2012 at 9:20 | comment | added | GH from MO | @Noam: I was thinking to use the argument principle in practice to count the order of the zero at the center of a concrete modular $L$-function. We have an integral whose value is a nonnegative integer: it can be approximated with an error of $1/3$, say, and then we have the correct value. So once you have the two candidate modular forms this approach should work. Am I missing something? | |
| Mar 22, 2012 at 2:30 | comment | added | Noam D. Elkies | The problem is not computing $L$ and its derivatives, but recognizing zero. $L(E,1)$ can be computed to within $10^{-n}$ in time polynomial in $n$, but if you know $L(E,1) = .000000003$ to within $10^{-7}$, you still don't know that $E$ has positive arithmetic rank without some further argument. Here this argument is available, as it is for $L'(E,1)$ thanks to Gross-Zagier (which is much harder), but so far not beyond that. So if the analytic rank is at least $4$ we can guess but not prove it. Fortunately for your present purpose ranks $0$ and $2$ suffice. | |
| Mar 21, 2012 at 22:27 | comment | added | GH from MO | @Noam: Thanks for your comment. I was thinking of switching to modular forms and working with them directly. Is obtaining the Taylor expansion of a modular $L$-function about its center computationally difficult? | |
| Mar 21, 2012 at 21:17 | comment | added | Noam D. Elkies | That's true, though the fact that a finite computation proves exact vanishing is not obvious (it comes down to identifying $L(E,1)$ with a rational multiple of the real period whose denominator is bounded by the exponent of the torsion group). | |
| Mar 21, 2012 at 20:57 | history | answered | GH from MO | CC BY-SA 3.0 |