Consider a reductive group $G$ over an algebraically closed field $K$ of characteristic $0$. I would like to consider the space $X$ of all $G$-conjugacy classes in $G$. Does the space $X$ have some nice geometric structure? (For instance, is it a variety? And/or if I take $K = \mathbb{C}$ is the space $X(\mathbb{C})$ Hausdorff? (It seems not to me, but I would like to ask)

Beware: I have not included the condition that I look at semi-simple conjugacy classes.

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    $\begingroup$ The spectrum of the $K$-subalgebra of invariants of $k[G]$ for the conjugation action is canonically isomorphic to the (finite) quotient of a maximal torus $T$ by the induced conjugation action of the Weyl group $W(G,T):=N_G(T)/T$. For the induced quotient map $q:G\to W/T$, every fiber of a regular element equals the corresponding conjugacy class. Some fibers contain many conjugacy classes. My recollection is that every fiber contains a unique conjugacy class of semisimple elements. $\endgroup$ – Jason Starr Mar 11 at 16:30
  • $\begingroup$ The topological quotient is Hausdorff if and only if $G$ is Abelian (i.e., multiplicative). The map from the topological quotent to $W/T$ should be the initial continuous map to a Hausdorff topological space (when $K$ equals $\mathbb{C}$). $\endgroup$ – Jason Starr Mar 11 at 16:33
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    $\begingroup$ A single example such as $\mathrm{SL}_2(\mathbb{C})$ shows that the quotient by conjugacy is not Hausdorff. What have you tried? $\endgroup$ – YCor Mar 11 at 16:43
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    $\begingroup$ In fact, refusing to restrict to semisimple conjugacy classes is essentially demanding that the quotient not be Hausdorff; the closure of an arbitrary class contains the class of its semisimple part. $\endgroup$ – LSpice Mar 12 at 23:13

As already noted, the quotient need not be even $T_1$. For example, $SL_2(\mathbb{C})/SL_2(\mathbb{C})$ is homeomorphic to $\mathbb{C}$ with double points at $\pm 2$. This quotient is not $T_1$ since two of those points are not closed.

Here is a proof:

For any $t\in \mathbb{C}$, define $\epsilon_t:=\left(\begin{array}{cc}t&-1\\1&0\end{array}\right)$. Take any $A\in SL_2(\mathbb{C})$ so that its trace is not equal to $\pm 2.$ Then it has two distinct eigenvalues determined by its trace (use the quadratic formula on the characteristic polynomial to see this). Consequently, there is a matrix $g_A$ such that $g_A A g_A^{-1}=\epsilon_t$. Consequently, each of these conjugation orbits is closed. If its trace is $\pm 2$, then $A$ is either conjugate to $\epsilon_t$ or equals $\pm\mathbf{1}$. In the former case, $\epsilon_t$ is conjugate to one of $\left(\begin{array}{cc}\pm 1& 1\\ 0& \pm 1\end{array}\right)$. Conjugating by $\left(\begin{array}{cc}n&0\\0&\frac{1}{n}\end{array}\right)$ then gives $\left(\begin{array}{cc}\pm 1& \frac{1}{n^2}\\ 0& \pm 1\end{array}\right)$. Letting $n\to \infty$ we see that $\pm \mathbf{1}$ is in the closure of the orbit of $\epsilon_t$. $\Box$

In general, $G/G$ will not be ``nice'' for complex reductive groups $G$ (unless $G$ is abelian), since the semisimple points are exactly the points with closed conjugation orbits.

However, $G/G$, in this generality, is homotopic to the GIT quotient $G/\!\!/G\cong T/W$, where $T$ is a maximal torus in $G$ and $W$ is the Weyl group. And in turn $G/\!\!/G$ is homotopic to the corresponding quotient $K/K$ for a maximal compact $K$ in $G$.

And, if $K$ is simply connected, its Weyl alcove is homeomorphic to $K/K$. Therefore, $K/K$ is homeomophic to a closed ball in this case.


$SU(2)/SU(2)\cong [-2,2]$

$SU(3)/SU(3)\cong $



What you look it at is exactly the quotient of the representation variety $$Hom({\mathbf Z},G)$$ of representations from the integers ${\mathbf Z}$ to $G$, by the $G$-action via conjugation.

The quotient of the representation variety is in general not a variety, however you can look at the GIT-quotient, which is the character variety $$X({\mathbf Z},G)=Hom({\mathbf Z},G)//G.$$ It is the Hausdorffification of the actual quotient. In the case $G=GL(n,{\mathbf C})$ it is constructed by identifying two representations when all their traces agree. In your case this just means that you identify matrices if the traces of all their powers agree.

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    $\begingroup$ Your last sentence is only true for $\mathrm{SL}_2(\mathbf{C})$. In general, you maybe mean the characteristic polynomial in lieu of traces. $\endgroup$ – YCor Mar 11 at 18:17
  • $\begingroup$ Right, you need the traces of all $A^n$, not only the trace of $A$. I will correct this in the answer. $\endgroup$ – ThiKu Mar 11 at 18:47

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