Which unimodular lattices $L\subset \mathbb R^2$ minimize $f_t(L):=\sum_{ v\in L} e^{-t \|v\|_2}$? (for parameters $t>0$)

Question

$\DeclareMathOperator\SL{SL}\DeclareMathOperator\SO{SO}$Consider the lattices in $\SL(2,\mathbb R)(\mathbb Z^2)$ up to rotation. The space of such lattices can be identified with the modular surface $\mathcal M:=\SO(2)\backslash \SL(2,\mathbb R)/\SL(2,\mathbb Z)$. We can then define a family of functions $f_t(A):\mathcal M\rightarrow \mathbb R^+$ by

$$f_t(A)=\sum_{ v\in A(\mathbb Z^2)} e^{-t \|v\|_2},$$ for $t>0$, where $\|v\|_2$ denotes the Euclidean norm.

I conjecture that for all $t>0$, $f_t(A)$ has a unique minimum at the lattice which can be tiled by equilateral triangles.

I have computational evidence which suggests that this is true; I have plotted approximations of $f_t(A)$, $$f_{t,N}(A):=\sum_{v\in A(\mathbb Z_N^2)} e^{-t\|v\|_2},$$ where $\mathbb Z_N:= \{m\in\mathbb Z:-N\leq m\leq N\}$, and the conjecture seems to hold. Note that for $t$ small and $N$ small, $f_{t,N}(A)$ is not necessarily minimized by the equilaterally tiled lattice. However, for a fixed $t>0$, it appears that there exists an $N_t>0$ such that for all $N>N_T$, $f_{t,N}(A)$ is minimized by the equilaterally tiled lattice.

Answer 1

The answer is that $f_t(A)$ is uniquely minimized at the hexagonal lattice (up to rotation).

The comment by Marco Golla led me to the following paper by Laurent Bétermin which proves the result in a more general setting: https://arxiv.org/abs/1502.03839