Dimension of the space of cusp forms

Question

I guess this is rather basic, and as much a reality check as a question.

We have the classical equations for the dimension of the space of modular forms of weight $k$ for $\Gamma(1)$:

$$\dim M_{12j+r} = \left\{ \begin{array}{lr} j+1 & r=0,4,6,8,10\\ j & otherwise \end{array} \right. $$

Now, in Daniel Bump's "Automorphic Forms and Representations", it says:

$$\dim M_{k}=\dim S_{k}+1$$ for $k \geq 4$. I guess that is a mistake, since the leading discussion makes clear that the extra modular form is an Eisenstein series, and those only exist for even $k \geq 4$. Or am I perhaps missing something?

Computing $\dim M_{k}$ for $k \leq 16$ gives me:

$0,0,0,1,0,1,0,1,0,1,0,2,1,1,1,2$

By the preceding reasoning, this means that the first cusp shows up in $k=12$, the very famous $\Delta$. The second would be $k=16$ (which one is that?).

Or to put it another way, $\dim S_{k}$ would be:

$0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,1$

* My question is, is this correct?

Answer 1

Yes, this is correct (although your list of dimensions of $M_k$ seems to start with $k = 0$, while your list for $S_k$ starts with $k = 1$). There are no nonzero modular forms of level 1 and odd weight, either Eisenstein or cuspidal. I guess the restriction to even $k \ge 4$ must have been accidentally omitted in Bump's book.

Answer 2

The second one would be the product $E_4 \; \Delta$ which is an eigenform (multiplicative coefficients, completely non-obvious at first) because the cuspidal subspace is 1-dimensional.