# what is the equivalent of the Euler constant for higher dimensional lattices

Let $\Lambda$ be a unimodular lattice in $\mathbb R^d$. Then there are constants such that

$$\sum_{\substack{\gamma\in \Lambda\\0<|\gamma|<R\\}} \frac{1}{|\gamma|^d} = c_1 \log R + c_2 + o(1).$$

My questions are: Does $c_2$ depend on the lattice ? If yes, how ?

• What is a reference for the existence of $c_1, c_2?$ Dec 15, 2015 at 4:56
• Well. I don't have a reference. However, if you take a fundamental domain $D$ for the lattice, assuming it is compact, and invariant by $x\to -x$, then one can compare each tem with the integral of $|x|^{-d}$ over some $D+\gamma$. Now thanks to the symmetry of $D$, this involves an integral remainder with only the hessian of $|x|^{-d}$, that decreases fast enough for the sum to be convergent. So up to a constant and $o(1)$, the sum is the integral of the function over the reunion of $\gamma+D$. (continued in next comment)
– user84131
Dec 15, 2015 at 12:55
• Then this is almost the integral over a big ball of radius $R$, up to $o (1)$ minus the integral over $D$ (for $\gamma =0$), hence the formula.
– user84131
Dec 15, 2015 at 12:55
• Added automorphic-forms tag, since we are talking about a function on $SL_d(\mathbb{R})$ which is clearly invariant for the right $SL_d(\mathbb{Z})$ action and the left $SO_d(\mathbb{R})$ action. If only we had something like a holomorphicity condition... Dec 16, 2015 at 16:28
• Do you in fact want what you literally asked, or do you actually want something about the Laurent expansion of the corresponding generalized Epstein zeta function at the leading pole? Dec 16, 2015 at 22:42

I work out the case $d=2$ below. I didn't check everything carefully, so hopefully there are no errors.

Up to homothety, any lattice is equivalent to one generated by the complex numbers $1,z$ with $z \in \mathbb{H}$. In fact, $z$ can be chosen to lie in the standard fundamental domain for $SL_2(\mathbb{Z}) \backslash \mathbb{H}$. To make such a lattice unimodular, we simply re-scale by a scalar $\lambda > 0$ to get $\lambda, \lambda z$ with $\lambda = y^{-1/2}$. Here $z= x +i y$, $y>0$. Then any element of the lattice $\Lambda$ may be written uniquely as $\lambda cz + \lambda d$ with $(c,d) \in \mathbb{Z}^2$, $(c,d) \neq (0,0)$. The sum in question, in this notation, is then $$\sum_{0 < |\lambda cz + \lambda d | < R} |\lambda cz + \lambda d |^{-2}.$$ By a Perron-type formula, we can evaluate such a sum asymptotically by a contour integral of the form $$\lim_{T \rightarrow \infty} \frac{1}{2 \pi i} \int_{\sigma - iT}^{\sigma + iT} R^s F(s) \frac{ds}{s},$$ where $$F(s) = \sum_{(c,d) \neq (0,0)} |\lambda cz + \lambda d |^{-2-s}.$$ In practice, all that matters is the analytic behavior of $F(s)$ near $s=0$.

Now $F(s)$ is closely related to the Eisenstein series $E(z,s)$ defined by $$E(z,s) = \frac{1}{2} \sum_{\gcd(c,d) =1 } \frac{y^s}{|cz+d|^{2s}} = \frac12 \frac{1}{\zeta(2s)} \sum_{(c,d) \neq (0,0)} \frac{y^s}{|cz+d|^{2s}}.$$ Unless I made a mistake, a short calculation (pulling out a gcd to give the zeta function) gives $F(s) = 2 \zeta(2+s) E(z,1+\frac{s}{2}).$

The constant $c_1$ only depends on the residue of $F(s)$ at $s=0$, which one can surely calculate quite easily; it does not depend on the lattice of course. To get $c_2$ one needs to calculate the next term in the Laurent expansion of $F(s)$, which I believe equals a constant minus $\log y^{1/2} |\eta(z)|^2$. Here this function $f(z)=\log y^{1/2} |\eta(z)|^2$ is $SL_2(\mathbb{Z})$-invariant. Now $f(z)$ depends on $z$, and so yes $c_2$ depends on the lattice in a rather interesting way.

I wager that for $d \geq 3$ one needs to find the relevant Eisenstein series and its Laurent expansion.

$\def\RR{\mathbb{R}}\def\ZZ{\mathbb{Z}}$This is essentially the constant term in the Epstein $\zeta$-function. Given a lattice $\Lambda$ in $\RR^d$, the Epstein $\zeta$ function is $$Z(\Lambda, s) = \sum_{g \in \Lambda \setminus \{ 0 \}} \frac{1}{(g^T g)^s}.$$ $Z$ has a simple pole at $d/2$ with residue $\tfrac{\pi^{d/2}}{\sqrt{\det \Lambda} \ \Gamma(d/2)}$, and no other poles on $\mathrm{Re}(s)\geq d/2$. Set $$Z(s) = \frac{\pi^{d/2}}{(\det \Lambda) \ \Gamma(d/2) (s-d/2)}+c(\Lambda) + O(s-d/2)$$

There are standard tools to convert Dirichlet series estimates to partial sum estimates. If I didn't drop any constants, then $$c_1 = \frac{2 \pi^{d/2}}{(\det \Lambda) \ \Gamma(d/2)} \quad c_2 = c(\Lambda).$$ Theorem 4 of Terras, "Bessel Series Expansions of the Epstein Zeta Function and the Functional Equation", Trans. AMS, Vol. 183 (Sep., 1973), pp. 477-486 gives a formula for $c(\Lambda)$ in terms of other functions, but I don't feel competent to summarize it.