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 $c_1, c_2?$ – Igor Rivin Dec 15 '15 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 '15 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 '15 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... – David E Speyer Dec 16 '15 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? – paul garrett Dec 16 '15 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.