In the course of doing some calculations on a project I am working on, I came across the following presentation of a vector space. It is generated by homogenous polynomials of even degree $n$ over a vector space $V\oplus V\oplus V$, satisfying the following conditions.
- $f(x,y,z)=-f(y,x,z)$
- $f(x,y,z)=-f(x,z,y)$
- $f(-x,y,z)=f(x,y,z)$
- $-f(x,y,z)-f(x+y,-x,z)+f(x,y-x,z)=0$
Computer calculations indicate that when $V$ is $1$-dimensional, the dimensions of this space, starting with $n=2$ are $0,0,0,0,1,0,1,1,1,1,2,1,2,$ which looks like the dimensions of the spaces of classical cusp forms. I'm curious if anyone can see whether there really is an isomorphism.
There is a similar problem for polynomials over $V\oplus V$ which I do know how to solve. In this case, the polynomial is supposed to satisfy:
- $f(x,y)=f(y,x)$
- $f(x,y)=-f(-x,y)$
- $f(x,y)+f(y,-x-y)+f(-x-y,x)=0$
Over a general $V$ the answer is a bit complicated, but when $\dim V=1$, you do get classical cusp forms. Basically one uses the Eichler-Shimura isomorphism $H^1_{cusp}(SL_2(\mathbb Z),Sym^k(\mathbb C^2))\cong\mathcal S_{k+2}\oplus \overline{\mathcal S_{k+2}}$, where $\mathcal S_{k+2}$ denotes the space of cusp forms of weight $k+2$. Examining the group cohomology chain complex, one can derive the isomorphism. It might be enlightening to have a more direct computation of the dimension.
Anyway, in summary, I'm looking for a proof that the above presentation of polynomials over $V\oplus V\oplus V$ gives classical cusp forms when $\dim V=1$ and any further ideas about what happens as $\dim V$ increases would be helpful too! It's clear that the answer will decompose as a direct sum of Schur functors $\mathbb S_{\lambda}(V)$ where $\lambda$ is a partition of $n$ with $\leq 3$ rows. Presumably the multiplicities will be given by modular form spaces. The $\dim V=1$ computation is picking up the partition $(n)$.