Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Problem $1$. Which full rank lattices $\Lambda \subset \mathbb R^d$ have their corresponding theta function $\theta_{\Lambda}(\tau):= \sum_{\bf n \in \Lambda } e^{\pi i \tau ||n||^2} $ zero-free in the complex upper half plane $\mathbb H$?

For example, for $d=1$, the classical theta function $\theta_{\mathbb Z}(\tau):= \sum_{\bf n \in \mathbb Z } e^{\pi i \tau n^2}$ is the ratio of two Dedekind eta functions, so it is the ratio of two infinite products in q, and hence zero-free inside $\mathbb H$. Similarly, any theta function that can be written as an infinite product in a "nice way" is zero-free.

What I would really like is a criterion that, instead of brute-forcing a contour integral to search for zeros, gives us some nice conditions on the lattice $\Lambda$, involving such parameters such as, for example, $vol( \Lambda)$, the successive minima of $\Lambda$ relative to the unit ball, the dual lattice, $Aut(\Lambda)$, etc.

EDIT (on Nov 16, 2012). To consider more examples, fix any dimension $d$ and consider any diagonal positive definite quadratic form $a_1 x_1^2 + \dots + a_d x_d^2$, with $a_j$ any fixed positive real numbers. The corresponding lattice $\Lambda$ is therefore the direct sum of $d$ one-dimensional lattices, namely $\sqrt a_1 \mathbb Z \oplus \cdots \oplus \sqrt a_d \mathbb Z$, so that $\theta_\Lambda$ is the product of $d$ one-dimensional theta functions, and hence $\theta_\Lambda$ is nonzero in $\mathbb H$.

More generally, if $\Lambda$ is reducible in the sense that it is expressible as the direct sum of some lower dim'l lattices, then we call the corresponding theta function reducible. It is therefore natural to ask the following more particular question.

Problem 2. Which irreducible theta functions $\theta_{\Lambda}(\tau)$ are zero-free in $\mathbb H$?

share|improve this question
    
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. –  Felipe Voloch Nov 14 '12 at 3:45
    
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? –  Sinai Robins Nov 14 '12 at 4:04
1  
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. –  François Brunault Nov 14 '12 at 9:48
    
@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? –  Sinai Robins Nov 14 '12 at 10:10
2  
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. :( –  Sinai Robins Nov 14 '12 at 12:00
show 7 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.