Wikipedia:Reference desk/Archives/Mathematics/2012 April 27

= 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

$$f(x_1,x_2,x_3,x_4) = \sum \limits_{i,j,k,l \geq 0} A_{ijkl}x_1^ix_2^jx_3^kx_4^l = \sum \limits_{i,j,k,l \geq 0} A_{ijkl}(x_1+tx_3)^i(x_2+tx_4)^jx_3^kx_4^l $$

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 $$x_1x_4-x_2x_3$$ 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 $$A_{i j k (i+j-k)}x_1^i x_2^j x_3^k x_4^{i+j-k}$$ to be invariant, so would we deduce the invariant ring is $$\mathbb{C}[x_1x_4,x_2x_4,x_3/x_4]$$ 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)