Timeline for Algebraic relation amongst an elliptic function and its convolution
Current License: CC BY-SA 4.0
14 events
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| Apr 11, 2021 at 2:11 | history | edited | Stefano | CC BY-SA 4.0 |
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| Apr 10, 2021 at 23:10 | history | edited | Stefano | CC BY-SA 4.0 |
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| Apr 10, 2021 at 22:52 | comment | added | Stefano | @AlexandreEremenko: and to Paul too, I edited the question following your comments. I was cavalier by not considering the effect of complex poles when defining the periodicity. So the context of my question is reduced to the case in which $h(z)$ is periodic with an imaginary period $\omega_3$ and $f(z)$ is elliptic with periods $\omega_1$ (real) and $\omega_3$. | |
| Apr 10, 2021 at 22:48 | history | edited | Stefano | CC BY-SA 4.0 |
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| Apr 10, 2021 at 20:59 | comment | added | paul garrett | @AlexandreEremenko, ah, indeed. Unclear. Probably the intent/context is also unclear for the asker. Unless I am mistaken, a doubly-periodic function with simple poles is locally $L^1$, so an integral on a (real-) two-torus would have a sense. But certainly not for general elliptic functions. Hard to guess the intent... | |
| Apr 10, 2021 at 20:53 | comment | added | Alexandre Eremenko | @paul garrett: what OP wrote is a 1-dimensonal integral:-) 2 dimensional of a doubly periodic function will be divergent. So it is not clear what the question means. | |
| Apr 10, 2021 at 19:20 | comment | added | François Brunault | It is a general fact that if $f,g$ are two rational functions on an algebraic curve, with minimal polynomial $P(f,g)=0$, then the degree of $P$ with respect to $f$ (resp. $g$) is less than or equal to the degree of $g$ (resp. $f$), with equality if $f,g$ generate the function field. | |
| Apr 10, 2021 at 17:54 | comment | added | paul garrett | @AlexandreEremenko, I'm a little confused: if I construe "convolution" as the obvious real-two-dimensional integral, at a simple pole for example (so, locally integrable), I don't see how to get extra things to prevent double periodicity of the convolution. After all, the whole set-up could be construed as concerning (complex-valued) functions on a two-torus $\mathbb T^2$, no? | |
| Apr 10, 2021 at 17:43 | comment | added | Alexandre Eremenko | @paul garrett: when you go around a pole, something is added (the residue) so a convolution is not a single-valued function. It can be elliptic (single valued) only under very special conditions on $f$ and $g$. | |
| Apr 10, 2021 at 17:39 | comment | added | Alexandre Eremenko | On the other hand, answering the question about $M,N$ in the last (corrected) formula is easy, if one knows that $f,c$ are both elliptic. | |
| Apr 10, 2021 at 17:15 | comment | added | paul garrett | Yes, although the general idea makes some sense, the biggest obstacle, as @AlexandreEremenko comments, is the meaning of convolution when applied to a meromorphic function. E.g., even heuristically, does it smear out poles? What would that mean? (Saying "principal value" does not seem to be enough?) E.g., elliptic functions with double poles are not locally $L^1$... Clarify? | |
| Apr 10, 2021 at 16:21 | history | edited | Stefano | CC BY-SA 4.0 |
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| Apr 10, 2021 at 16:19 | comment | added | Alexandre Eremenko | With convolution, it is problematic. Elliptic function has poles. If you integrate on the real line, and there happens to be a pole on your way, the integral will diverge. Anyway, you want your "convolution" to be defined for complex arguments, and it will not be a function since it will be multivalued. So is is not correct to say that "convolution of an elliptic function with anything is elliptic". Also, in your last eq. you have c in the LHS and g in the RHS. This must be a misprint. | |
| Apr 10, 2021 at 15:05 | history | asked | Stefano | CC BY-SA 4.0 |