Timeline for Zero-free theta functions in the upper half plane
Current License: CC BY-SA 3.0
17 events
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| Nov 16, 2012 at 1:17 | history | edited | Sinai Robins | CC BY-SA 3.0 |
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| Nov 15, 2012 at 17:35 | history | edited | Sinai Robins | CC BY-SA 3.0 |
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| Nov 15, 2012 at 16:59 | history | edited | Sinai Robins | CC BY-SA 3.0 |
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| Nov 15, 2012 at 16:40 | comment | added | Sinai Robins | @Francois: Thanks for your interesting thoughts here. I'll edit the problem statement to incorporate the idea that reducible lattices allow the possibility of many non-zero theta functions for each dimension $d$. | |
| Nov 15, 2012 at 12:13 | comment | added | François Brunault | In fact, since we are dealing with positive definite lattices, there is a unique decomposition of a given lattice into indecomposable ones. In particular $\Lambda \oplus \Lambda'' \cong \Lambda' \oplus \Lambda''$ implies $\Lambda \oplus \Lambda'$, so the Grothendieck group doesn't seem to be of much help here :( | |
| Nov 15, 2012 at 7:54 | comment | added | François Brunault | (...) but he considers only unimodular lattices which are possibly indefinite. I don't know what is the structure of the Grothendieck group in the setting of positive definite lattices which are not necessarily unimodular. | |
| Nov 15, 2012 at 7:52 | comment | added | François Brunault | @Sinai : We can define the following equivalence relation on integral lattices : $\Lambda \sim \Lambda'$ if there exists $\Lambda''$ such that $\Lambda \oplus \Lambda'' \cong \Lambda' \oplus \Lambda''$. This makes a commutative monoid, which we can extend to an abelian group (called Grothendieck group) $K$ as in Serre, Cours d'arithmétique, p. 89. The map $\Lambda \mapsto \theta_\Lambda$ factors through $K$. The set of $x \in K$ such that $\theta_x$ doesn't vanish on the upper half-plane is then a subgroup of $K$. In loc. cit. Serre describes the structure of the Grothendieck group (...) | |
| Nov 14, 2012 at 16:00 | comment | added | Sinai Robins | @Francois: Sounds interesting-can you expand on this point? I'm not that familiar with the Grothendieck group. But it seems that we need to define a reducible theta function to be one that comes from a reducible lattice, versus the corresponding definition for an irreducible theta function. So for reducible $d$-dim'l theta functions $\theta_\Lambda$ that arise from lattices that are direct sums of lower dimensional lattices whose theta functions are in turn zero-free, we still have that $\theta_\Lambda$ is zero-free. It's now natural to ask when the irreducible theta functions are zero-free. | |
| Nov 14, 2012 at 15:20 | comment | added | François Brunault | Good point. The non-vanishing property seems to depend only of the class of the lattice in some Grothendieck group. | |
| Nov 14, 2012 at 12:00 | comment | added | Sinai Robins | Wait - some of the remarks above appear to be contradictory - what about the integer lattice $\mathbb Z^d$? We can decompose its theta function since it has a diagonal quadratic form in the exponent: $\theta_{\mathbb Z^d}(\tau) = \theta^d_{\mathbb Z}(\tau)$, and since $\theta_{\mathbb Z}$ is zero free in $\mathbb H$, so is $\theta^d_{\mathbb Z}$, independent of how large $d$ is. :( | |
| Nov 14, 2012 at 11:48 | comment | added | François Brunault | @Sinai : The following sounds too easy, but in the case $d$ is odd you can consider $\theta_\Lambda^2$ to reduce to the case $d$ is even. You can also reduce to the case the lattice is even by considering $2\Lambda$ but this might not be the answer you're looking for. | |
| Nov 14, 2012 at 10:10 | comment | added | Sinai Robins | @Francois: Yes, thank you, I've tried this before too, and I guess your suggestion might work out to give an optimal bound on the dimension, hopefully, though it's not clear to me how to find the order of vanishing at the cusps from the data. The wt=$d/2$ for a d-dim'l lattice, and the level $N \leq vol(\Lambda)$,by a known theorem. So I guess this might give a partial answer for large enough volume of the lattice vol(Λ) and large enough dimension d, together with the assumption that the lattice Λ is an even integral lattice. What about odd dimensions, or lattices which are not even integral? | |
| Nov 14, 2012 at 9:48 | comment | added | François Brunault | You can work out the number of zeroes (counted with multiplicities) of a modular form with given weight and level from the Riemann-Roch theorem. Assuming you can find an upper bound for the order of vanishing of $\theta_\Lambda$ at the cusps, you could then deduce that $\theta_\Lambda$ must vanish when the weight and the level are large enough. | |
| Nov 14, 2012 at 4:04 | comment | added | Sinai Robins | Well, I think it's true that it's a form for even integral lattices, which means the following. If we let the lattice be $\Lambda:= A(\mathbb Z^d)$, for a matrix $A$, then $A^t A$ should be an integer matrix, with all of its diagonal elements even integers. Then the associated theta function is a modular form on some $\Gamma_0(N)$. Do you know how large $d$ has to be to insure that it does indeed have zeros ? How about any general information about their location? | |
| Nov 14, 2012 at 3:45 | comment | added | Felipe Voloch | If the lattice is a sublattice of the integer lattice then the theta function is a modular form for some congruence subgroup of the modular group and, for $d$ large and even, it will have zeros. | |
| Nov 14, 2012 at 1:25 | history | edited | Sinai Robins | CC BY-SA 3.0 |
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| Nov 13, 2012 at 16:04 | history | asked | Sinai Robins | CC BY-SA 3.0 |