Timeline for The loss of double periodicity (ellipticity)
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Apr 4, 2021 at 13:56 | comment | added | Lelouch | @მამუკაჯიბლაძე Ok in my mind, the integration contour is fixed to be $[0,1]_{\mathfrak{a}_2}$. But as $\mathfrak{a}_1$ changes from 0 to $\tau$, the poles start moving, and some poles cross the $[0,1]_{\mathfrak{a}_2}$. I guess you are saying that the residues of these crossed-over poles precisely cause the failure of ellipticity? | |
| Apr 4, 2021 at 13:53 | comment | added | Lelouch | @AlexandreEremenko Ah I see. I thought that as $\mathfrak{a}_1$ moves from 0 to $\tau$, the poles of $\mathfrak{a}_2$ inside the fundamental parallelogram start moving and will eventually cross the integration contour $[0,1]_{\mathfrak{a}_2}$, and I thought those residues will cancel each other. So is this picture incorrect? | |
| Apr 4, 2021 at 13:17 | comment | added | reuns | $\theta(\tau)$ is a modular form not an elliptic function, did you mean $\theta(\tau,.)$ for fixed $\tau$. Then $\int_0^1 \frac1{\wp_i(z+t)-\wp_i(1/4)}dt$ is well-defined and analytic and doubly periodic for $\Im(z)\not \in \Bbb{Z}$ but its analytic continuation has logarithmic branch points at $1/4+\Bbb{Z}/2+i\Bbb{Z}$. | |
| Apr 4, 2021 at 12:41 | comment | added | Alexandre Eremenko | In other words, your integral is not well-defined: for some $a_1$, the function $a_2\mapsto f(a_1,a_2)$ will have a pole on $[0,1]$, and this will make your integral divergent. If you try to exclude those $a_1$, this can make your integral multi-valued. | |
| Apr 4, 2021 at 12:04 | comment | added | მამუკა ჯიბლაძე | Do they? Definitely if you take just a single elliptic function, results of its integration along all possible paths with the same endpoints fill a whole affine lattice (shift of the period lattice). Two different paths with the same endpoints result in the same value of the integral if and only if the loop they form together is cohomologous to zero in $H^1$ of the corresponding Riemann surface (complex 1-torus in this case). | |
| Apr 4, 2021 at 10:10 | history | edited | Lelouch | CC BY-SA 4.0 |
clarification
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| Apr 4, 2021 at 10:09 | comment | added | Lelouch | @მამუკაჯიბლაძე Indeed, $\tau$-shift moves the $\mathfrak{a}_2$-integration path, and it does cross poles. But thanks to the ellipticity, I think their residues precisely cancel. | |
| Apr 4, 2021 at 10:05 | comment | added | Lelouch | @მამუკაჯიბლაძე $\mathfrak{b}$ is a free parameter that resolves a double pole into simple pole. | |
| Apr 4, 2021 at 9:18 | comment | added | მამუკა ჯიბლაძე | Also, what is $\mathfrak b$? | |
| Apr 4, 2021 at 9:14 | comment | added | მამუკა ჯიბლაძე | First thought - since there are poles around, the result of integration is ambiguous, it depends on the path. It might thus be that it is still elliptic if along with the $\tau$-shift you also shift the path accordingly. | |
| Apr 4, 2021 at 6:51 | history | asked | Lelouch | CC BY-SA 4.0 |