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Let $\mathfrak{gl}_n(\mathbb{R})$ be the Lie algebra of matrices with real entries and $GL_n(\mathbb{R})$ its associated Lie group. Recall that a linear subgroup $G \subseteq GL_n(\mathbb{R})$ acts by conjugation on $\mathfrak{gl}_n(\mathbb{R})$, that is, for $g \in G$ its action on $A \in \mathfrak{gl}_n(\mathfrak{R})$, is defined by

$$g(A) = g^{-1}Ag.$$

Definition: Let $G \subseteq GL_n(\mathbb{R})$ be a subgroup. A polynomial $f \in\mathbb{R}[(X_{ij})_{1\leq i,j\leq n}]$ is called invariant on $\mathfrak{gl}_n(\mathbb{R})$ with respect to conjugation by elements in $G$ iff

$$\forall g \in G, \forall A \in \mathfrak{gl}_n(\mathbb{R}): f(g^{-1}Ag) = f(A)$$

We will denote by $G'$ the set of all invariant polynomials.

Question: What is known about $G'$?

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    $\begingroup$ This is very well known. The generic matrix is diagonalizable, so the continuous invariant functions are functions of the eigenvalues. The eigenvalues can be permuted freely by conjugation, so the continuous invariant functions are functions of the permutation invariants of the eigenvalues, which are the elementary symmetric functions of the eigenvalues, or in other words the coefficients of the characteristic polynomial. It is then easy to see that these polynomials freely generate the ring of invariant polynomials, and generate the rings of smooth and of continuous invariant functions. $\endgroup$
    – Ben McKay
    Commented Sep 10, 2015 at 16:30
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    $\begingroup$ Sorry, I meant it is very well known for $G=GL(n,R)$. For the orthogonal group, the result is the same by Zariski density. For the special orthogonal group, the same except for the Pfaffian, which is independent. $\endgroup$
    – Ben McKay
    Commented Sep 10, 2015 at 16:42
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    $\begingroup$ Mehta, 1988, Basic sets of invariants for finite reflection groups, deals with the adjoint representations of simple Lie groups, proving that the invariants are generated by the usual suspects, freely. $\endgroup$
    – Ben McKay
    Commented Sep 10, 2015 at 16:44
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    $\begingroup$ @BenMcKay Why not leave this as an answer? $\endgroup$
    – Yemon Choi
    Commented Sep 10, 2015 at 17:25
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    $\begingroup$ @BenMcKay Aren't we dealing with the action on the Lie algebra of $GL_n$, not the action on the Lie algebra of $G$, and thus not dealing with the adjoint representation? $\endgroup$
    – Will Sawin
    Commented Sep 10, 2015 at 18:05

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For the action of the upper unipotent group, which we shall denote by $G$ here, one can continue in the following way: Let $g_r = \pmatrix{1 & r \\ 0 & 1}\in G$. Consider a $2\times 2$ real matrix $M=\pmatrix{a & b \\ c & d }$. Then $g_rMg_r^{-1} = \pmatrix{a+rc & b-r(a-d) - r^2c \\ c & d-rc}$. Let us write $x=a+d$ and $y = a-d$. Then $x$ and $c$ are invariants under the action of $G$, and the action of $g_r$ on the coordinate functions $y$ and $b$ is given by $g_r\cdot y = y+2rc$ and $g_r\cdot b = b -ry - r^2c$. A direct calculation shows that the polynomial $bc+4y^2$ is also $G$-invariant (it is a linear combination, in fact, of the trace and determinant polynomials). I claim that $\mathbb{R}[a,b,c,d]^G = \mathbb{R}[bc+4y^2,x,c]$.

To see why this is true, consider the localization $\mathbb{R}[a,b,c,d]_c$. Since the action of $G$ on $c$ is trivial, we have an action of $G$ on this ring. We can then write $\mathbb{R}[a,b,c,d]_c = \mathbb{R}[x,y,c,b+4y^2/c]_c$. In this presentation the action of $G$ on 3 of the generators is trivial, and one can show by an easy induction that if $G$ acts trivially on $p(x,y,c,b+4y^2/c)$ then no nontrivial power of $y$ appear in $p$. We get that $\mathbb{R}[a,b,c,d]_c^G = \mathbb{R}[x,c,b+4y^2/c]_c$. The ring of invariants in $\mathbb{R}[a,b,c,d]$ will just be the intersection of the ring of invariants in $\mathbb{R}[a,b,c,d]_c$ with $\mathbb{R}[a,b,c,d]$, and one can easily show that this ring is exactly $\mathbb{R}[c,x,bc+4y^2] = \mathbb{R}[c,tr,det]$.

About the embedding of $GL(1,\mathbb{C})$ in $GL(2,\mathbb{R})$: We denote the group by $G$ again. We need to find the invariants in $\mathbb{R}[a,b,c,d]$. We will find the invariants in the extension of scalars: $\mathbb{C}[a,b,c,d]$. The advatage is that now the action of $x+iy\in \mathbb{C}^{\times}$ is given by one dimensional representations: on $a+d$ and $c-b$ the action is trivial, while on $(c+b) + (d-a)i$ and $(c+b) + (a-d)i$ the action is given by $(x-iy)^2/(x^2+y^2)$ and $(x+iy)^2/(x^2+y^2)$. Since these two weights are of infinite order, and they are the inverses of each other, we get easily that the ring of invariants is now spanned by $a+d, b+c$ and $((c+b) + (d-a)i)((c+b) - (d-a)i) = (c+b)^2 + (a-d)^2$. Another set of generators will then indeed be $a+d,c-b, a^2+b^2+c^2+d^2$.

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  • $\begingroup$ I am wondering why it doesn't compile the tex code $\endgroup$
    – Mike Cocos
    Commented Sep 12, 2015 at 18:54

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