3
$\begingroup$

Let $A=(a_{ij})\in\operatorname{SL}_2(\mathbb{F}_p)$. Consider the ring map $A:\mathbb{F}_p[x,y]\to\mathbb{F}_p[x,y]$ defined by

$$A(x)=a_{11}x+a_{21}y$$ $$A(y)=a_{12}x+a_{22}y$$

and extended multiplicatively. Are there non-constant polynomials $f(x_1,x_2)\in\mathbb{F}_p[x,y]$ such that $f(x,y)=A(f(x,y))$ for all $A\in\operatorname{SL}_2(\mathbb{F}_p)$?

I tried to solve this problem by writing $f(x,y)=\sum\lambda_{ij}x^iy^j$, and discovering different restrictions on the $\lambda_{ij}$ for particularly nice choices of $A$. This quickly got complicated and so I'm hoping this has been studied before and there is a more elegant solution.

Edit: After working through this for some small primes, I believe that the following two polynomials are fixed by all $A$:

$$f(x,y)=x^py-xy^p$$ $$g(x,y)=\sum_{i=0}^px^{(p-i)(p-1)}y^{i(p-1)}$$

However, I can't prove that $g(x,y)$ is fixed for all $p$, and it's not clear to me that these two polynomials generate the entire subalgebra of $\mathbb{F}_p(x,y)$ invariant under the action of $\operatorname{SL}_2(\mathbb{F}_p)$. Is there an invariant $h(x,y)$ not in the subalgebra generated by $f$ and $g$?

$\endgroup$
  • $\begingroup$ If instead of $\Bbb{F}_p$ we had an algebraically closed field of the same characteristic the representation theory of $SL_2$ would give a lot of answers. I need to think about this case, but the following observations can be made immediately. 1) Subspaces $V_n$ of homogeneous polynomials (of a fixed degree $n$) are stable under this action, so it suffices to find the homogeneous invariants $f(x,y)=\sum_{i=0}^n\lambda_ix^{n-i}y^i$. 2) A diagonal matrix with entries $(a,1/a)$ acts on $x^{n-i}y^i$ by a scalar $a^{n-2i}$. For this scalar to be $=1$ for all $a\in\Bbb{F}_p^*$ it is necessary $\endgroup$ – Jyrki Lahtonen Sep 26 '15 at 7:25
  • $\begingroup$ (cont'd) and sufficient that $n-2i$ is a multiple of $p-1$ whenever $\lambda_i\neq0$. This narrows down the search considerably. IIRC trivial representations (=invariants) appear as composition factors of $V_n$ for some $n$, but not necessarily as submodules, which is what you are looking for. I gotta rush now. Hopefully more later. $\endgroup$ – Jyrki Lahtonen Sep 26 '15 at 7:28
  • $\begingroup$ @JyrkiLahtonen: Thanks for your suggestions. If you're still interested, note the edit. $\endgroup$ – Jared Sep 27 '15 at 1:17
4
$\begingroup$

These are the Dickson's polynomials. If we let $$h(x,y)=x^{p^2}y-y^{p^2}x$$ then $h$ is also invariant under $SL(2,p)$ action. It then follows that $$g(x,y)=\frac{h(x,y)}{f(x,y)}$$ is invariant.


For a general $n$, let

$$L_{n,s}=\left|\begin{array}{cccc} x_1&x_2&\cdots&x_n\\ x_1^{p}&x_2^{p}&\cdots&x_n^p\\ \cdots&\cdots&\cdots&\cdots\\ \widehat{x_1^{p^s}}&\widehat{x_2^{p^s}}&\cdots&\widehat{x_n^{p^s}}\\ \cdots&\cdots&\cdots&\cdots\\ x_1^{p^n}&x_2^{p^n}&\cdots&x_n^{p^n} \end{array}\right|$$

where the $p^s$ row is omitted. Then the Dickson's invariant for $SL(n,p)$ are $\dfrac{L_{n,s}}{L_{n,n}}$.


And you are correct, these two generate the whole invariant algebra. I don't know the details but they are likely in Dickson's works.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.