Timeline for Does there exist a half-integer weight theta function which is is equivalent to 1 modulo 4?
Current License: CC BY-SA 3.0
15 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Jul 29, 2013 at 10:16 | history | edited | Ricardo Andrade | CC BY-SA 3.0 |
fixed errors introduced by preceding edit (x^16 is fine in text but not in latex); added some links while I was at it
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| S Jul 29, 2013 at 9:57 | history | suggested | BlackAdder | CC BY-SA 3.0 |
Put into latex form.
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| Jul 29, 2013 at 9:30 | review | Suggested edits | |||
| S Jul 29, 2013 at 9:57 | |||||
| Mar 27, 2013 at 22:52 | comment | added | paul Monsky | Alternatively, if p=5 mod 8 and q=7 mod 8 then the number of ways of writing pq as s_1+2s_2+8s_3 with the s_i squares is the same as the number of ways of writing pq as s_1+2s_2+8s_3 with the s_i non-zero squares. So the coefficients of x^pq in g^11 and in (1+g)^11 are the same. | |
| Mar 27, 2013 at 21:38 | comment | added | paul Monsky | @Sarah. That's right. Another way to say it--instead of using (1/2)*(phi-(E_4)^m) in my edited comment, use (1/2)*(phi+(E_4)^m). | |
| Mar 27, 2013 at 14:54 | comment | added | stl | Ah, I think I understand it now. We can just add an appropriate power of $1 + 2 \sum q^{n^2} \equiv 1 \pmod{2}$ to $g(x)$ to get $a(x)$. | |
| Mar 27, 2013 at 5:04 | comment | added | stl | I'm convinced that this works for the element $a(x) = 1 + x + x^4 + x^9 + \dots \in Z/2Z[[x]]$, but I think removing the 1 to get $g(x)$ might mess it up. In the MO question about sums of 11 squares, everything is about $a(x)$. In particular, I'm not convinced that you can go between $x_0^2 + x_1^2 + \cdots + x_{10}^2$ and $x_0^2 + 2 x_1^2 + 8 x_3^2$ when you restrict to $x_i > 0$ instead of $x_i \geq 0$. The principal / lemma that Greg Kuperberg brought up in his answer if only for unipotent formal power series. | |
| Mar 26, 2013 at 3:08 | vote | accept | stl | ||
| Mar 25, 2013 at 11:48 | comment | added | paul Monsky | Here's the argument that if u is the mod 2 reduction of an element of Z[[x]] that is the expansion at infinity of some modular form phi of weight w for gamma_0 (N) then the space spanned by the image of u under the formal Hecke operators "T_q" ,q prime, has finite dimension. For fixed N and w the Z-module of such elements of Z[[x]] has finite rank and is stable under the Hecke T_q where (N,q)=1. So the image of this module under mod 2 reduction contains u and has finite Z/2 dimension. The T_q with (2N,q) reduce to "T_q". These "T_q" stabilize the image. And only finitely many q divide N. | |
| Mar 25, 2013 at 0:48 | history | edited | paul Monsky | CC BY-SA 3.0 |
Complete answer given.
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| Mar 22, 2013 at 3:49 | comment | added | paul Monsky | @Will-My proof? is more or less a version of this idea, as I think my conjectures when the level is odd should follow from results of Igusa on the modular curve. But I have no feeling for what happens when the level is even. (I took Katz and Mazur out of the library last year, but it went unread--it's not for amateurs). | |
| Mar 22, 2013 at 3:00 | comment | added | Will Sawin | $g$ satisfies the equation $g^4+ g= \Delta$, where $\Delta$ is the reduction mod $2$ of the $\Delta$ function and generator of the forms of level $1$ mod $2$. This is a Galois extension of Galois group $\mathbb Z/2 \times \mathbb Z/2$. Maybe one could look for this extension inside coverings of the modular curve? | |
| Mar 21, 2013 at 22:28 | history | edited | paul Monsky | CC BY-SA 3.0 |
comment incorporated into answer, slight elaboration
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| Mar 21, 2013 at 5:33 | history | answered | paul Monsky | CC BY-SA 3.0 |