Is it possible to use EichlerSelberg trace formula to compute the dimension of modular forms of weight $k$ for $SL(2,\mathbb Z)$? This was computed by classical methods such as RiemannRoch.

2$\begingroup$ Yes, the dimension of the space is just the trace of the first Hecke operator. Use Theorem 2 in people.mpimbonn.mpg.de/zagier/files/google/estracesl2z/… with $m=1$ there. $\endgroup$– LuciaJan 7, 2014 at 3:31

$\begingroup$ but somehow theorem 2 can't be computed effectively to reach the same formula given by RiemannRoch. $\endgroup$– 7adicJan 7, 2014 at 3:53

2$\begingroup$ The only relevant values of $t$ there are $t=0$, $t=\pm 1$ and $t=\pm 2$. And you just need to know $H(0)$, $H(3)$ and $H(4)$ which are provided in the previous page. $\endgroup$– LuciaJan 7, 2014 at 3:58

1$\begingroup$ Lucia: Your comments look like a perfect answer to me, why don't you put them in the answer box? $\endgroup$– GH from MOJan 7, 2014 at 14:19
2 Answers
Yes; here's an elaboration of my comments.
The EichlerSelberg formula (see Theorem 2 of Zagier's http://people.mpimbonn.mpg.de/zagier/files/google/estracesl2z/fulltext.pdf which appears in Lang's book on Modular Forms) states that the trace of the $m$th Hecke operator on the space of cusp forms of weight $k$ for $SL_2({\Bbb Z})$ equals $$ \frac 12 \sum_{t=\infty}^{\infty} P_k(t,m) H(4mt^2) \frac 12 \sum_{dd^{\prime}=m} \min(d,d^{\prime})^{k1}. $$ Here $H(n)=0$ if $n<0$, $H(0)=\frac{1}{12}$ and for positive $n$, $H(n)$ is a weighted class number for positive definite binary quadratic forms of discriminant $n$. Note that $H(3)=\frac 13$ and $H(4)=\frac 12$. The quantity $P_k(t,m)$ (for $t\le 2\sqrt{m}$) is defined as follows: find $\rho$ such that $\rho=\sqrt{m}$ and the real part of $\rho$ is $t/2$. Then (with the obvious interpretation if $\rho$ is real) $$ P_k(t,m) = \frac{\rho^{k1}\overline{\rho}^{k1}}{\rho\overline{\rho}}. $$
The dimension of the space of cusp forms of weight $k$ corresponds to the $m=1$ case of the above formula. Note that $H(4t^2)=0$ for $t>2$ and so the formula gives, using the values for $H(0)$, $H(3)$ and $H(4)$ $$ \frac 12 + \frac{1}{12} P_k(2,1) \frac{1}{3} P_k(1,1)  \frac 14 P_k(0,1). $$ It's a simple matter to compute that $P_k(2,1)=(k1)$, $P_k(1,1) = \sin(\pi(k1)/3)/\sin(\pi/3)$ and $P_k(2,1) = \sin (\pi(k1)/2)$. This is the desired dimension formula.
For the sake of completeness, it is worth pointing out that the Selberg trace formula establishes an analogous formula for a weighted trace of Hecke eigenvalues of Maass forms. Here one obtains class numbers and regulators of real quadratic fields rather than the imaginary fields above. From the trace formula one can obtain an asymptotic for the number of Maass forms with eigenvalue up to some point $T$ which is the analog of the dimension formula above. Establishing such a formula was one of Selberg's motivations in developing the trace formula.

2$\begingroup$ A technical addendum: this computation is fairly delicate, and in the first edition (or so) of Lang's book there was something wrong in Zagier's computation. By now I don't remember exactly what, something about notabsolute convergence and rearrangement (!), but it certainly gave an incorrect result (as opposed to a merely flawed computation of the correct). This was corrected perhaps in the second edition or so. $\endgroup$ Feb 15, 2016 at 0:15
An adelic reference for $GL(2, \mathbb{Z})$ is also "Traces of Hecke operators" by Knightly and Li. You choose matrix coefficent at the infinite place and the characteristic function $GL_2(\mathbb{Z}_p)$ times the center, and plug it into the Arthur trace formula. They do that almost for weight $k \geq 3$, but they work with matrix coefficient. This gives the dimension formula. This is likely to be more fruitful for generalization to higher rank or deeper level, where not always a classical treatment is available.

1$\begingroup$ I meant to say one wants to work with pseudo matrix coefficients where KL use matrix coefficients:/ $\endgroup$ Feb 6, 2014 at 14:54