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April 27

Invariant theory: matrices and invariants under upper triangular matrices

Hi all, I'm learning some invariant theory for rings and I'm getting myself a bit confused with this question - feel like I might have made a mistake, and would appreciate some feedback from someone more experienced than me.

If $G$ acts on $S$, we write $S^G=\{s\in S:gs=s\mathrm{\forall }g\in G\}$ for the invariant ring. In particular, if an element g acts on some finite dimensional vector space W over $ℂ$, then for an element $f:W\to ℂ$ of the coordinate ring $ℂ[W]$, we have an action on f by $g:f(\cdot )\mapsto f(g\cdot )$.

We let $M_2(ℂ)$ denote the 2x2 matrices over $ℂ$, $U$ denote the upper triangular unipotent matrices (that is, matrices with '1's on the diagonal, a zero below and anything above), and $T$ denote the matrices of the form $\mathrm{Diag}(t,t^{-1})$. These both act on $M_2(ℂ)$ by (left) matrix multiplication. I wish to find $ℂ[M_2(ℂ){]}^U$ and $ℂ[M_2(ℂ){]}^T$, which I think means the polynomials $f:ℂ[M_2(ℂ)]\to ℂ$ (i.e. polynomials in the coordinate ring $ℂ[M_2(ℂ)]$) which are fixed under the map $f(A)\mapsto f(MA)$ for any matrix M in $U,T$ respectively.

So a matrix in U looks like $\left(\begin{array}{cc}1& t\\ 0& 1\end{array}\right)$; this takes $\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\mapsto \left(\begin{array}{cc}a+tc& b+td\\ c& d\end{array}\right)$. An f which is invariant under this transformation can be considered (i think) as $f(a,b,c,d)$ rather than $f(\left(\begin{array}{cc}a& b\\ c& d\end{array}\right))$, so that $f(a,b,c,d)=f(a+tc,b+td,c,d)$: so the question is essentially asking us, if I understand correctly, to find the polynomials f which are invariant under that transformation. What can we say about such polynomials? I tried expanding it as

$f(x_1,x_2,x_3,x_4)=\sum _{i,j,k,l\ge 0}{A}_{ijkl}{x}_1^i{x}_2^j{x}_3^k{x}_4^l=\sum _{i,j,k,l\ge 0}{A}_{ijkl}(x_1+t{x}_3{)}^i(x_2+t{x}_4{)}^j{x}_3^k{x}_4^l$

for all i, j, k, l, t.

I then tried looking at this as either a polynomial in $t$ or a polynomial in $t,x_1,\dots ,x_4$; we know that all coefficients of $t^r$ are always zero for $r>0$, and these coefficients are polynomials in the $x_m$ and the $A_{ijkl}$; what I really wanted to do is show that these coefficients are necessarily nonzero polynomials in the $x_m$, unless we assume all $A_{ijkl}$ involved are zero; i.e. our polynomial can only be a polynomial in the latter two variables, otherwise it is not fixed for every t.

However, when I tried to determine the coefficient of $t^r$ as a poly in the $x_m$, I find that multiple terms in $(*)$ can contribute to the same term $x_1^{i^{\prime }}{x}_2^{j^{\prime }}{x}_3^{k^{\prime }}{x}_4^{l^{\prime }}$ in the coefficient of $t^r$, and in fact the function $x_1{x}_4-x_2{x}_3$ satisfies the requirements but is obviously a function of all 4 variables (note that this is effectively the determinant, though I don't know if that has any significance). Indeed, functions such only in the latter 2 variables are *included* in our class of possible functions, but they don't make up the whole thing. I'm not sure what more we can say about the class; by choice of t I think we can deduce $f(a,b,c,d)=h(ad-bc,c,d)$ for some h, but I'm not sure where to go from here.

What is the invariant ring exactly? Is it just $ℂ[c,d,e]$ (where e happens to equal ad-bc)? And likewise with the diagonal matrices, I think we get that all terms in the polynomial must be of the form $A_{ijk(i+j-k)}{x}_1^i{x}_2^j{x}_3^k{x}_4^{i+j-k}$ to be invariant, so would we deduce the invariant ring is $ℂ[x_1{x}_4,x_2{x}_4,x_3/x_4]$ or something like that? Or maybe just any old 3-variable $ℂ[X,Y,Z]$ since we effectively have 3 variables? Sorry for the long question, I've just started learning invariant theory and I'm still finding it a bit confusing. Thank you for your help :) 86.26.13.2 (talk) 08:14, 27 April 2012 (UTC)