# Why is $12$ the smallest weight for which a cusp form exists?

On wikipedia (here) I have read the following:

Twelve is the smallest weight for which a cusp form exists. [...] This fact is related to a constellation of interesting appearances of the number twelve in mathematics ranging from the value of the Riemann zeta function at $−1$ i.e. $\zeta(-1) = -1/12$, the fact that the abelianization of $SL(2,\mathbb{Z})$ has twelve elements, and even the properties of lattice polygons.

I know the dimension formula and the weight formula for modular forms as well as the proof of the weight formula via the residue theorem, but I do not understand how this relates to any of the three concepts mentioned in the quote above. Can anybody tell me more?

• Do you know the valence formula? (I think you might be calling it the weight formula, but I want to be sure) Mar 16 '17 at 1:50
• You need to be careful with such numerology. Some people define weight to be twice the usual value to avoid half integer weights. The values of the zeta function are, in general, comples numbers. Why pick the value at $-1$? I doubt lattice polygons have anything to do with modular forms. I think the answer to your question is that there is no connection. May 9 '17 at 18:40

Regarding the question in the title I must admit that, for me personally, "why" questions are always very hard to answer so I'll simply give some evidence that the (non-) existence of cusp forms can be viewed as a geometric phenomenon. What I am about to relate is very well known and details can be found in many places but I must say that I found the treatment offered in R. Hain's, LECTURES ON MODULI SPACES OF ELLIPTIC CURVES to be particularly instructive.

Let $$\mathcal{M}_{1,1}=\operatorname{SL}_2(\mathbb Z)\backslash\!\!\backslash \mathfrak{H}$$ be the moduli space of complex elliptic curves, where $$\operatorname{SL}_2(\mathbb Z)$$ acts on the upper half-plane $$\mathfrak{H}$$ in the usual way through Möbius transformations. Warning: this is not a complex manifold, rather a more complicated object called a complex orbifold. Nevertheless, all the algebraic invariants that can be associated to complex manifolds (such as fundamental group, homology groups, et cetera) can also be associated to $$\mathcal{M}_{1,1}$$. In particular, there is a good notion of (holomorphic) line bundle over $$\mathcal{M}_{1,1}$$ and, as per usual, we denote by $$\operatorname{Pic}(\mathcal{M}_{1,1})$$ the abelian group of isomorphism classes of such.

Now for our purposes the key statement is that:

$$\operatorname{Pic}(\mathcal{M}_{1,1}) \cong \mathbb Z/12\mathbb Z$$

How does this relate to the (non-) existence of cusp forms? The point is that there exists a line bundle $$\mathcal{L}$$ on $$\mathcal{M}_{1,1}$$ such that global sections of its $$k$$-th tensor power $$\mathcal{L}^{\otimes k}$$, for $$k \in \mathbb Z$$, are so-called weakly holomorphic'' (:=holomorphic on $$\mathfrak{H}$$ and meromorphic at the cusp $$\infty$$) modular forms of weight $$k$$. On the other hand, a line bundle is trivial if and only if there exists a nowhere vanishing global section and since $$\mathcal{L}^{\otimes 12}$$ is necessarily trivial by the above result, there exists a weakly holomorphic modular form of weight $$12$$ that vanishes nowhere on $$\mathfrak{H}$$. The $$k/12$$ formula then forces it to vanish at $$\infty$$ as well and that is your cusp form.

In fact the class of $$\mathcal{L}$$ even generates $$\operatorname{Pic}(\mathcal{M}_{1,1})$$ so that $$12$$ is indeed the smallest weight in which a non-zero cusp form can exist.

A final comment, since you mention $$\zeta(-1)$$ in this context: That is actually equal to the Euler characteristic $$\chi(\mathcal{M}_{1,1})$$ of $$\mathcal{M}_{1,1}$$. More generally Harer and Zagier proved that $$\chi(\mathcal{M}_g)=\zeta(1-2g),$$ for all $$g\geq 2$$, where $$\mathcal{M}_g$$ denotes the orbifold of compact Riemann surfaces of genus $$g$$.

PS: I would have liked to vote up this question but since at the time of my answer it had exactly $$12$$ upvotes already I light-heartedly decided against that.