# Inequality regarding sum of gaussian on lattices

When S is a subset of an inner product space, let d(S) denote ${\sum\limits_{s \in S} e^{- \langle s,s \rangle}}$

Suppose L is a discrete additive subgroup of $\mathbb{R^n}$, M is a subgroup of L, and V is a subspace of $\mathbb{R^n}$.

Does this hold in general: ${\frac{d(L)} {d(L \cap V)}} \geq {\frac{d(M)}{d(M \cap V)}}$?

Why I am interested in this: There are analogies (analogue of the Riemann Roch theorem for example) which imply it makes sense to think of metrized lattices as sheaves and the log of d as the dimension of the global sections of a sheaf(by "metrized lattice" I mean a lattice with an inner product). So then given a map between metrized lattices, say f: A -> B, an analogous way to define "rank of induced map on global sections" would be: log(d(A)) - log(d(ker(f))) (from the rank-nullity theorem). I will denote this quantity as rank(f). So then if this rank function actually behaves as "rank of induced map on global sections", we would expect that $rank(g \circ f)$ is no greater than either rank(f) or rank(g). I am trying to show this for homomorphisms between metrized lattices that don't increase the norm at any point. $rank(g \circ f) \leq rank(f)$ is trivial; $rank(g \circ f) \leq rank(g)$ is equivalent to this problem I have posed.

I am interested in this rank function because it seems to be possible to do derived functor cohomology without an actual left-exact functor but with just the ranks of the maps that a left-exact functor would induce. Fully explaining myself here would result in an unreasonably big MO question.

$\textbf{Update:}$ The ideas mentioned in the previous paragraph, which are what originally motivated me to consider this question, are now fully explained here. Oded and Noah's result is applied in section 7.

• Interesting question. I don't remember encountering such an inequality, and don't readily see a proof or counterexample. Have you tried some examples with $n=2$ and $n=3$ to see if it seems to work numerically in all cases? Mar 16, 2014 at 7:37
• Is $V$ an $L$-rational subspace?
– Asaf
Mar 16, 2014 at 7:54
• Since you can make dilatations of L, your question is really about evaluations of θ−series $\sum_{\lambda\in L}q^{\langle \lambda,\lambda\rangle}$ of lattices at reals in $(0,1)$. (You get of course equality at q=0, your inequality should thus break down at negative evaluations. And you get divergent quantities for $q\geq 1$.) Mar 16, 2014 at 10:56
• @Noam I have made a python script that randomly generates three vectors in $\mathbb{R^3}$, call them a, b, and c, and the lattices they generate A, B, and C. It then checks if ${\frac{d(A + B + C)} {d(A + B)}} \geq {\frac{d(A + C)}{d(A)}}$ . Unless you coincidentally have linear dependence among these vectors, this is going to be a special case of the problem. I have checked several thousand random cases with this script without having found a counterexample. Also, I have proven: when we have two lattices D and E whose intersection is {0}, $\frac{d(D + E)} {d(D)} \geq d(E)$. Mar 16, 2014 at 17:33
• @Asaf I'm not 100% sure what you mean by L-rational subspace. If you mean that the dimension of $L \cap V$ is the same as the dimension of V, then it shouldn't make a difference since replacing V by the span of $L \cap V$ is L-rational by this definition and gives the same quantities in the inequality. Mar 16, 2014 at 17:41