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I feel like I might have made a mistake on this question, 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. Our 2 x 2 matrices $P_{ij}$ act on $\mathbb{C}[X,Y]$ by $X \longmapsto P_{11}X + P_{12}Y$ and $Y \longmapsto P_{21}X + P_{22}Y$, and extending these mappings to $X^i$ and $Y^i$. So, I want to calculate the invariant rings $(i): \mathbb{C}[X,Y]^{GL_2}$ and $(ii): \mathbb{C}[X,Y]^{SL_2}$. My arguments went like this:

$(i)$ Suppose we have some element of $\mathbb{C}[X,Y]^{GL_2}$, say $z = \sum \limits_{i,j \geq 0} A_{i j}X^i Y^j$. Take the matrix $\lambda I$, where $I$ is the identity matrix and $\lambda$ is large. Then $z$ must satisfy $\sum \limits_{i,j \geq 0} A_{i j}X^i Y^j = \sum \limits_{i,j \geq 0} A_{i j}\lambda^{i+j} X^i Y^j$; so clearly $A_{ij}$ must be zero unless $i=j=0$; thus our invariant ring is just $\mathbb{C}$.

$(ii)$ Now our matrices have to have determinant 1. Take the matrix with $A_{11} =A_{22} = 0$ and $A_{12} = \alpha$, $A_{21}=-1/\alpha$ (sorry I can't draw matrices in latex!) Then we must have $A_{ji} = (-1)^{j} \alpha^{i-j}A_{ij}$; so for $i \neq j$, clearly this can't be true for all $\alpha$: thus $A_{ij} = 0$ unless $i=j$. So any element of the invariant ring must be of the form $\sum_i A_i (XY)^i$. But then taking any upper triangular matrix with diagonal elements 1, don't we deduce again that the invariant ring is just $\mathbb{C}$?

The reason I am confused is that these were 2 consecutive problems in the same worksheet, and I didn't expect them to have the same invariant ring. If someone could point out where I've gone wrong or confirm my solutions, I'd be very grateful.

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Do you mean $S^G$ rather than $G^S$? Also, it looks right to me. – anon Apr 26 '12 at 23:36
Yes, sorry. Edited now. – Spyam Apr 27 '12 at 0:39
up vote 6 down vote accepted

Your computations are both correct.

An alternative way to see this is the following: suppose that $f\in\mathbb C[X,Y]$ is invariant under $SL(2,\mathbb C)$. One can easily check that this implies that the function $$\phi:(x,y)\in\mathbb C^2\longmapsto f(x,y)\in\mathbb C$$ is invariant under $SL(2,\mathbb C)$, that is, for each $\begin{pmatrix}a&b\\c&d\end{pmatrix}\in SL(2,\mathbb C)$ we have $$\phi(ax+by,cx+dy)=\phi(x,y), \qquad\forall x,y\in\mathbb C.$$ I claim that this implies that $\phi$ is constant: for example, if $(x,y)$, $(x',y')$ are two non-zero elements of $\mathbb C^2$, there exists a matrix $\begin{pmatrix}a&b\\c&d\end{pmatrix}\in SL(2,\mathbb C)$ such that $x'=ax+by$ and $y'=cx+dy$, so invariance means that $\phi(x',y')=\phi(x,y)$. This tells us that $\phi$ is constant on $\mathbb C^2\setminus\{(0,0)\}$ and, since $\phi$ is continuous, that $\phi$ is constant.

Now check that $\phi$ being constant implies that $f\in\mathbb C$.

Next, suppose that $f\in\mathbb C[X,Y]$ is invariant under $GL(2,\mathbb C)$. Since $SL(2,\mathbb C)\subset GL(2,\mathbb C)$, $f$ is also invariant under $SL(2,\mathbb C)$ and, by what we have just done, $f\in\mathbb C$.

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That's a nice way of thinking about it - thanks for your answer, worries like this are always the problem with second-guessing a problem sheet! – Spyam Apr 27 '12 at 1:18
@hilbert: what I claimed is that given $(x,y)$ and $(x',y')$ there is such a matrix with determinant $1$, but it did it construct it... I'll leave this as an exercise for the interested reader :D – Mariano Suárez-Alvarez Apr 27 '12 at 2:05

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