Variety of conjugacy classes
$begingroup$
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.
ag.algebraic-geometry algebraic-groups invariant-theory reductive-groups conjugacy-classes
$endgroup$
add a comment |
$begingroup$
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.
ag.algebraic-geometry algebraic-groups invariant-theory reductive-groups conjugacy-classes
$endgroup$
1
$begingroup$
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 $q:Gto W/T$, 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
9 hours ago
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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
9 hours ago
3
$begingroup$
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
9 hours ago
add a comment |
$begingroup$
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.
ag.algebraic-geometry algebraic-groups invariant-theory reductive-groups conjugacy-classes
$endgroup$
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.
ag.algebraic-geometry algebraic-groups invariant-theory reductive-groups conjugacy-classes
ag.algebraic-geometry algebraic-groups invariant-theory reductive-groups conjugacy-classes
edited 9 hours ago
YCor
28.2k483136
28.2k483136
asked 10 hours ago
mnrmnr
636315
636315
1
$begingroup$
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 $q:Gto W/T$, 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
9 hours ago
$begingroup$
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
9 hours ago
3
$begingroup$
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
9 hours ago
add a comment |
1
$begingroup$
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 $q:Gto W/T$, 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
9 hours ago
$begingroup$
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
9 hours ago
3
$begingroup$
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
9 hours ago
1
1
$begingroup$
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 $q:Gto W/T$, 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
9 hours ago
$begingroup$
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 $q:Gto W/T$, 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
9 hours ago
$begingroup$
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
9 hours ago
$begingroup$
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
9 hours ago
3
3
$begingroup$
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
9 hours ago
$begingroup$
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
9 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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.
$endgroup$
1
$begingroup$
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
8 hours ago
$begingroup$
Right, you need the traces of all $A^n$, not only the trace of $A$. I will correct this in the answer.
$endgroup$
– ThiKu
7 hours ago
add a comment |
$begingroup$
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 $tin mathbb{C}$, define $epsilon_t:=left(begin{array}{cc}t&-1\1&0end{array}right)$. Take any $Ain SL_2(mathbb{C})$ 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 $pmmathbf{1}$. In the former case, $epsilon_t$ is conjugate to one of $left(begin{array}{cc}pm 1& 1\ 0& pm 1end{array}right)$. Conjugating by $left(begin{array}{cc}n&0\0&frac{1}{n}end{array}right)$ then gives $left(begin{array}{cc}pm 1& frac{1}{n^2}\ 0& pm 1end{array}right)$. Letting $nto infty$ 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).
However, $G/G$, in this generality, is homotopic to the GIT quotient $G/!!/Gcong T/W$, 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.
Examples:
$SU(2)/SU(2)cong [-2,2]$
$SU(3)/SU(3)cong $
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
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.
$endgroup$
1
$begingroup$
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
8 hours ago
$begingroup$
Right, you need the traces of all $A^n$, not only the trace of $A$. I will correct this in the answer.
$endgroup$
– ThiKu
7 hours ago
add a comment |
$begingroup$
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.
$endgroup$
1
$begingroup$
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
8 hours ago
$begingroup$
Right, you need the traces of all $A^n$, not only the trace of $A$. I will correct this in the answer.
$endgroup$
– ThiKu
7 hours ago
add a comment |
$begingroup$
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.
$endgroup$
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.
edited 7 hours ago
answered 9 hours ago
ThiKuThiKu
6,30512137
6,30512137
1
$begingroup$
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
8 hours ago
$begingroup$
Right, you need the traces of all $A^n$, not only the trace of $A$. I will correct this in the answer.
$endgroup$
– ThiKu
7 hours ago
add a comment |
1
$begingroup$
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
8 hours ago
$begingroup$
Right, you need the traces of all $A^n$, not only the trace of $A$. I will correct this in the answer.
$endgroup$
– ThiKu
7 hours ago
1
1
$begingroup$
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
8 hours ago
$begingroup$
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
8 hours ago
$begingroup$
Right, you need the traces of all $A^n$, not only the trace of $A$. I will correct this in the answer.
$endgroup$
– ThiKu
7 hours ago
$begingroup$
Right, you need the traces of all $A^n$, not only the trace of $A$. I will correct this in the answer.
$endgroup$
– ThiKu
7 hours ago
add a comment |
$begingroup$
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 $tin mathbb{C}$, define $epsilon_t:=left(begin{array}{cc}t&-1\1&0end{array}right)$. Take any $Ain SL_2(mathbb{C})$ 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 $pmmathbf{1}$. In the former case, $epsilon_t$ is conjugate to one of $left(begin{array}{cc}pm 1& 1\ 0& pm 1end{array}right)$. Conjugating by $left(begin{array}{cc}n&0\0&frac{1}{n}end{array}right)$ then gives $left(begin{array}{cc}pm 1& frac{1}{n^2}\ 0& pm 1end{array}right)$. Letting $nto infty$ 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).
However, $G/G$, in this generality, is homotopic to the GIT quotient $G/!!/Gcong T/W$, 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.
Examples:
$SU(2)/SU(2)cong [-2,2]$
$SU(3)/SU(3)cong $
$endgroup$
add a comment |
$begingroup$
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 $tin mathbb{C}$, define $epsilon_t:=left(begin{array}{cc}t&-1\1&0end{array}right)$. Take any $Ain SL_2(mathbb{C})$ 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 $pmmathbf{1}$. In the former case, $epsilon_t$ is conjugate to one of $left(begin{array}{cc}pm 1& 1\ 0& pm 1end{array}right)$. Conjugating by $left(begin{array}{cc}n&0\0&frac{1}{n}end{array}right)$ then gives $left(begin{array}{cc}pm 1& frac{1}{n^2}\ 0& pm 1end{array}right)$. Letting $nto infty$ 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).
However, $G/G$, in this generality, is homotopic to the GIT quotient $G/!!/Gcong T/W$, 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.
Examples:
$SU(2)/SU(2)cong [-2,2]$
$SU(3)/SU(3)cong $
$endgroup$
add a comment |
$begingroup$
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 $tin mathbb{C}$, define $epsilon_t:=left(begin{array}{cc}t&-1\1&0end{array}right)$. Take any $Ain SL_2(mathbb{C})$ 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 $pmmathbf{1}$. In the former case, $epsilon_t$ is conjugate to one of $left(begin{array}{cc}pm 1& 1\ 0& pm 1end{array}right)$. Conjugating by $left(begin{array}{cc}n&0\0&frac{1}{n}end{array}right)$ then gives $left(begin{array}{cc}pm 1& frac{1}{n^2}\ 0& pm 1end{array}right)$. Letting $nto infty$ 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).
However, $G/G$, in this generality, is homotopic to the GIT quotient $G/!!/Gcong T/W$, 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.
Examples:
$SU(2)/SU(2)cong [-2,2]$
$SU(3)/SU(3)cong $
$endgroup$
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 $tin mathbb{C}$, define $epsilon_t:=left(begin{array}{cc}t&-1\1&0end{array}right)$. Take any $Ain SL_2(mathbb{C})$ 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 $pmmathbf{1}$. In the former case, $epsilon_t$ is conjugate to one of $left(begin{array}{cc}pm 1& 1\ 0& pm 1end{array}right)$. Conjugating by $left(begin{array}{cc}n&0\0&frac{1}{n}end{array}right)$ then gives $left(begin{array}{cc}pm 1& frac{1}{n^2}\ 0& pm 1end{array}right)$. Letting $nto infty$ 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).
However, $G/G$, in this generality, is homotopic to the GIT quotient $G/!!/Gcong T/W$, 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.
Examples:
$SU(2)/SU(2)cong [-2,2]$
$SU(3)/SU(3)cong $
edited 3 hours ago
answered 4 hours ago
Sean LawtonSean Lawton
3,38122142
3,38122142
add a comment |
add a comment |
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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 $q:Gto W/T$, 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.
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– Jason Starr
9 hours ago
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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 $mathbb{C}$).
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– Jason Starr
9 hours ago
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A single example such as $mathrm{SL}_2(mathbb{C})$ shows that the quotient by conjugacy is not Hausdorff. What have you tried?
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– YCor
9 hours ago