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Let $k$ be a field. We write $\mathrm{Sym}^g(k^2)$ for the $g$-th symmetric power of the (a?) standard representation of $\mathrm{GL}_2(k)$ ($g\geq 0$ an integer). Here I consider $\mathrm{Sym}^g(k^2)$ as polynomials $P(X,Y)$ that are homogenous of degree equal to $g$ in two variables $X$ and $Y$. Then I equip $\mathrm{Sym}^g(k^2)$ with the right $\mathrm{GL}_2(k)$-action $$ P|_{\gamma}(X,Y) = P(dX-cY,aY-bX),\;\;\;\;\;\;\; \gamma = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in \mathrm{GL_2}(k). $$

Now suppose that $p$ is a prime and $k = \mathbf{F}_p$ is a finite field of size $p$.

Question: Is there a description of $H_0(\mathrm{SL}_2(k), \mathrm{Sym}^g(k^2))$?

To be clear, $H_0$ is the 0-th group homology; it is the largest quotient on which $\mathrm{SL}_2(k)$ acts trivially.

I would be happy with knowing the behavior of the dimension of this space (if that is easier).

I note two similar questions I know the answer to.

If instead we ask for the invariants $H^0(\mathrm{SL}_2(k),\mathrm{Sym}^g(k^2))$ then the answer is given by a theorem of Dickson at the start of the 20th century: any $\mathrm{SL}_2(k)$-invariant is a polynomial in $\theta = X^pY - Y^pX$ and $\psi = (X^{p^2}Y - XY^{p^2})/\theta$. So, you can just count the number of such polynomials of degree $g$ for any given $g$.

I observe though that in characteristic $p$ the $\mathrm{Sym}^g$'s are not self-dual (unlike in characteristic zero, cf. Is $Sym^n (V^*) \cong Sym^n (V)^\ast$ naturally in positive characteristic?). Thus knowing the dimension of the $H^0$ does not give me the dimension of the $H_0$. (And they really are different, e.g. if $g = 6$ and $p=5$.)

The second related computation I know how to do is compute $H_0(H,\mathrm{Sym}^g(k^2))$ where $H \subset \mathrm{SL}_2(k)$ is the upper-triangular matrices. The answer here is that it is $(0)$ for $g \not\equiv 0 \bmod p-1$, and if $g \equiv 0 \bmod p-1$ then it has dimension $1 + \lfloor{g/p\rfloor}$. Explicitly, this space of coinvariants is generated as a vector space by the images of the vectors $(Y^p-X^{p-1}Y)^iY^{g-pi}$ for $i = 0,1,2,\dotsc,\lfloor{g/p\rfloor}$.

This already implies an answer to my question, sometimes: if $g \not\equiv 0 \bmod p-1$, then $H_0(\mathrm{SL}_2(k),\mathrm{Sym}^g(k^2)) = (0)$. Preliminary calculations on a computer suggest that, for $p$ an odd prime and $g \equiv 0 \bmod p-1$, dimension of the $H_0$ grows by $1$ after replacing $g$ by $g + p(p+1)(p-1)/2$. Perhaps someone even sees how to bootstrap the result from $H$ to the whole $\mathrm{SL}_2$, but I don't.

(I added the number theory tag is there are people who study modular symbols that may know the answer.)

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  • $\begingroup$ The span of the vectors $(Y^p-X^{p-1}Y)^iY^{g-pi}$ is better described as the span of the $(X^{p-1})^iY^{g-pi+i}$. $\endgroup$
    – Wilberd van der Kallen
    Commented Sep 22, 2017 at 9:05
  • $\begingroup$ @WilberdvanderKallen Excellent point. It's worth pointing out that my basis naturally arose from considering $\alpha = Y^p-X^{p-1}Y$ as an element of $\mathrm{Sym}^p$. Multiplication by $\alpha$ induces an inclusion $\mathrm{Sym}^g \rightarrow \mathrm{Sym}^{g+p}$ which commutes with the operator $\Delta = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} - 1$. I used this to compute the cokernel of $\Delta$ inductively, which is the first step to calculating the $H_0(H,\mathrm{Sym}^g(k^2))$. Perhaps writing it in your form arises the same way; I need more coffee on Fridays. $\endgroup$
    – tkr
    Commented Sep 22, 2017 at 13:53
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    $\begingroup$ The paper "A study of certain modular representations" (ac.els-cdn.com/0021869378901163/…) proves a lot of results concerning the structure of $\text{Sym}^{g}(k^2)$, among other things that $\text{Sym}^{g+p (p-1)}(k^2) \cong \text{Sym}^g(k^2) \oplus \text{projective module}$. $\endgroup$
    – Kasper Andersen
    Commented Oct 6, 2017 at 9:27

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