The group $\operatorname{SL}\_2(\mathbb Z)$ acts on polynomials in two variables $\mathbb C[x,y]$ via $A\cdot f(x,y)\mapsto f(A^{-1}.(x,y))$ where $(x,y)$ is regarded as a column vector. There are two standard matrices $S=\begin{bmatrix}0&-1\\1&0\end{bmatrix}$ and $T=\begin{bmatrix}1&1\\0&1\end{bmatrix}$ which generate $\operatorname{PSL}\_2(\mathbb Z)$. Let $\epsilon = \begin{bmatrix}-1&0\\0&1\end{bmatrix}\in \operatorname{GL}\_2(\mathbb Z)$. 


In the course of studying the Johnson homomorphism, I defined a space $\Omega\_2(V)$ whose specialization to the case $V=\mathbb C$ in degree $2n$ is defined as follows. Let $R=\mathbb C[x,y]\_{2n}$ and let $\langle a\_1,\ldots, a\_m\rangle=a\_1R+\cdots+a\_mR$.
$$\Omega\_2(\mathbb C)\_{2n}=\dfrac{R}{\langle 1+ST+(ST)^2,1-\epsilon S\rangle+\mathbb C\{x^{2n},y^{2n}\}}.$$


Computer calculations give the dimensions as $0,1,1,2,3,3,4,5,5,6,\ldots$ starting at $n=1$, and I am looking for a proof of the claim that this pattern continues.


If you add the additional relation $\langle 1-\epsilon\rangle$ to this presentation, then it is not hard to see, using for example modular symbols, that the quotient is isomorphic to $H^1(\operatorname{GL}\_2(\mathbb Z);R)$, which is known to be isomorphic to the space of cusp forms for $\operatorname{SL}\_2(\mathbb Z)$ of weight $2n+2$. So we have an exact sequence:
\begin{multline\*}0\to\dfrac{\langle 1-\epsilon\rangle}{\langle 1+(ST)+(ST)^2,1-\epsilon S\rangle\cap \langle 1-\epsilon\rangle+\mathbb C\{x^{2n},y^{2n}\}}\\
\to \Omega\_2(\mathbb C)\_{2n}\to H^1(\operatorname{GL}\_2(\mathbb Z);R)\to 0\end{multline\*}
All the dimensions work out correctly if $$\langle 1+(ST)+(ST)^2,1-\epsilon S\rangle\cap\langle 1-\epsilon\rangle= \langle 1-\epsilon S\rangle\cap\langle 1-\epsilon\rangle=\mathbb C\{x^{2m}y^{2n-2m}-x^{2n-2m}y^{2m}\,:\, 0\leq m\leq n\},$$
but I have not been able to prove this! 


Remarks:


* It might help to use the fact that $\langle
 1+(ST)+(ST)^2\rangle=\ker(1-(ST))$.
* I asked an equivalent version of
this question at [math.stackexhange](https://math.stackexchange.com/questions/1192924/calculating-the-intersection-of-two-spaces-of-polynomials), but have gotten no useful
responses.
* The dimensions of $\Omega\_2(\mathbb C)$ resemble those of the space of cusp forms for the congruence subgroup $\Gamma\_0(3)$, but the polynomial degrees are off by one, and so it seems unlikely that they are related.



Dimensions of a vector space akin to modular symbols
Martin Kassabov sent me an elegant solution to this problem. Start with the observation that $\epsilon S$ and $ST$ generate a copy of the symmetric group $S\_3$. Now decompose $R$ into a direct sum of irreducible $S\_3$-modules. Modding out by $1+ST+(ST)^2$ and $1-\epsilon S$ kill both $1$ dimensional representations and reduce the dimension of the $2$ dimensional representation to $1$. So $\Omega\_{2n}(\mathbb C)$, if you ignore the $x^{2n},y^{2n}$, will have the same dimension as the multiplicity of the two dimensional representations in $R$. It's also not hard to show that further modding out by these two monomials will reduce the dimension by $1$. Now to calculate the $S\_3$ decomposition, calculate the character. It's not too hard to see that $\chi\_R=\begin{bmatrix} 2n+1&1&1\end{bmatrix}, \begin{bmatrix} 2n+1&1&-1\end{bmatrix}$ or $\begin{bmatrix} 2n+1&1&0\end{bmatrix}$ depending on the congruence class of $n$ modulo $3$. From here it follows that the multiplicity of the 2D representation in $R$ is $\frac{2n+1}{3}, \frac{2n+2}{3}$ or $\frac{2n}{3}$ again depending on the congruence class modulo 3. 


