Consider the space of newforms $S^{\mathrm{new}}_k(\Gamma_1(q))$ of weight $k$ and level $q$ for the congruence subgroup $\Gamma_1(q)$ of $\mathrm{SL}_2(\mathbb{Z})$; for simplicity's sake, let's assume that $q$ is prime. Then for $k \geq 2$, it is known via Riemann-Roch that $$\dim S^{\mathrm{new}}_k(\Gamma_1(q)) = \frac{k - 1}{24} (q^2 - 1) + E(q,k)$$ for an error term $E(q,k)$. This error term can be calculated explicitly (though not particularly neatly): see Theorem 13 of http://www.math.ubc.ca/~gerg/papers/downloads/DSCFN.pdf. So for $k \geq 2$, it is certainly possible to determine $\dim S^{\mathrm{new}}_k(\Gamma_1(q))$ precisely.

For $k = 1$, on the other hand, no such precise equations seem to exist, as the method used to prove the $k \geq 2$ case breaks down. Instead, it is conjectured (see Conjecture 2.1 of http://arxiv.org/pdf/0906.4579v1) that $$\dim S^{\mathrm{new}}_1(\Gamma_1(q)) = \frac{q - 2}{2} h(K_q) + O_{\varepsilon}(q^{\varepsilon}),$$ for any $\varepsilon > 0$ with the error term is uniform in $q$, and where $h(K_q)$ is the class number of $\mathbb{Q}(\sqrt{-q})$; here the leading term comes from the dihedral modular forms, while the error term is due to the others (icosahedral etc.).

Now note that the leading term in the formula for $S^{\mathrm{new}}_k(\Gamma_1(q))$ for $k \geq 2$ vanishes when $k = 1$, so if that formula where to be valid for $k = 1$, we would be left with the error term $E(q,k)$, which we can explicitly compute.

**Question**: Is there a reason why we should not expect $\dim S^{\mathrm{new}}_1(\Gamma_1(q)) = E(q,1)$? Obviously a quick check on Magma or Sage should prove that this is not the case, but unfortunately I don't have either installed.

If not, is there any chance that we will one day find a closed form for $\dim S^{\mathrm{new}}_1(\Gamma_1(q))$?

isa closed form for this number ;-) There are no infinite sums involved, for example ;-) $\endgroup$ – Kevin Buzzard Apr 11 '11 at 7:30