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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.

  1. $f(x,y,z)=-f(y,x,z)$
  2. $f(x,y,z)=-f(x,z,y)$
  3. $f(-x,y,z)=f(x,y,z)$
  4. $-f(x,y,z)-f(-y,x+y,z)+f(x,x+y,z)=0$

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:

  1. $f(x,y)=f(y,x)$
  2. $f(x,y)=-f(-x,y)$
  3. $f(x,y)+f(y,-x-y)+f(-x-y,x)=0$

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 $(n)$.

Edit (7/25/2012): There was a mistake in condition 4 as stated, which I have now corrected. Here are some basis elements, as per Kevin Buzzard's suggestion. These are easy enough to generate. I can post more if necessary.

Two variable case ($V=\mathbb R$)

  1. $42 x^5 y^5-25 \left(y^3 x^7+y^7 x^3\right)+4 \left(y x^9+y^9 x\right)$ ($n=10$)
  2. $660 x^7 y^7-539 \left(y^5 x^9+y^9 x^5\right)+245 \left(y^3 x^{11}+y^{11} x^3\right)-36 \left(y x^{13}+y^{13} x\right)$ ($n=14$)
  3. $143 \left(y^7 x^9+y^9 x^7\right)-273 \left(y^5 x^{11}+y^{11} x^5\right)+154 \left(y^3 x^{13}+y^{13} x^3\right)-24 \left(y x^{15}+y^{15} x\right)$ ($n=16$)

Three variable case ($V=\mathbb R$). Let $\sigma$ be the signed sum over all permutations of $3$ letters. Then we have

  1. $\sigma (y^2 x^8 - 3 y^4 x^6)$ ($n=10$)
  2. $\sigma(-11 y^6 x^8+7 y^4x^{10} -2 y^2 x^{12})$ ($n=14$)
  3. $\sigma(-26 y^6 x^{10}-25 y^4 x^{12}-8 y^2 x^{14})$ $(n=16)$
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You may have already seen (and dismissed) this, but your sequence 0,0,0,0,1,0,1,1,1,1,2,1,2 shows up as part of – Barry Cipra Jul 20 '12 at 0:42
@Barry: In the case of dimensions of spaces of cusp forms you have d(n + 12) = d(n) + 1, which is not true of the sequence you linked to. Indeed, this sequence is very simple: the space of modular forms for SL_2(\mathbb{Z}) is generated by Eisenstein series of weights 4 and 6, so the dimension of modular forms for weight n is the number of ways to write n=4a + 6b with a and b nonnegative integers. To get the dimension of the space of cusp forms, subtract 1. – Frank Thorne Jul 20 '12 at 2:22
I'm putting by bet on modular forms and not the A024160 sequence, but it's good to be aware of alternatives! – Jim Conant Jul 20 '12 at 2:38
I'd be tempted to try and write down an isomorphism between your two spaces (and not think about modular forms at all). You've computed dimensions for $n$ small -- but why not compute some explicit bases, and try and spot an isomorphism. If you've already done this, then why not post the polynomials which span the six 1-dimensional spaces in both cases and see if anyone else can spot a link? Perhaps a basis for the first 2-dimensional space in each case might also help. I'm not saying this will definitely work, but it's perhaps worth a try. – Kevin Buzzard Jul 20 '12 at 9:25
This looks like a version of modular symbols to me. – Jeremy Teitelbaum Jul 20 '12 at 11:18

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