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

We let $M_{2}(\mathbb {C} )$ denote the 2x2 matrices over $\mathbb {C}$ , $U$ 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 $\operatorname {Diag} (t,t^{-1})$ . These both act on $M_{2}(\mathbb {C} )$ by (left) matrix multiplication. I wish to find $\mathbb {C} [M_{2}(\mathbb {C} )]^{U}$ and $\mathbb {C} [M_{2}(\mathbb {C} )]^{T}$ , which I think means the polynomials $f:\mathbb {C} [M_{2}(\mathbb {C} )]\to \mathbb {C}$ (i.e. polynomials in the coordinate ring $\mathbb {C} [M_{2}(\mathbb {C} )]$ ) 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 ${\begin{pmatrix}1&t\\0&1\end{pmatrix}}$ ; this takes ${\begin{pmatrix}a&b\\c&d\end{pmatrix}}\mapsto {\begin{pmatrix}a+tc&b+td\\c&d\end{pmatrix}}$ . An f which is invariant under this transformation can be considered (i think) as $f(a,b,c,d)$ rather than $f({\begin{pmatrix}a&b\\c&d\end{pmatrix}})$ , 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

$f(x_{1},x_{2},x_{3},x_{4})=\sum \limits _{i,j,k,l\geq 0}A_{ijkl}x_{1}^{i}x_{2}^{j}x_{3}^{k}x_{4}^{l}=\sum \limits _{i,j,k,l\geq 0}A_{ijkl}(x_{1}+tx_{3})^{i}(x_{2}+tx_{4})^{j}x_{3}^{k}x_{4}^{l}$

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

I then tried looking at this as either a polynomial in $t$ or a polynomial in $t,x_{1},\ldots ,x_{4}$ ; 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 $x_{1}^{i'}x_{2}^{j'}x_{3}^{k'}x_{4}^{l'}$ 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 $\mathbb {C} [c,d,e]$ (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 $\mathbb {C} [x_{1}x_{4},x_{2}x_{4},x_{3}/x_{4}]$ or something like that? Or maybe just any old 3-variable $\mathbb {C} [X,Y,Z]$ 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)[reply]