Variety of conjugacy classes












2












$begingroup$


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.










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$endgroup$








  • 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










  • $begingroup$
    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
















2












$begingroup$


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.










share|cite|improve this question











$endgroup$








  • 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










  • $begingroup$
    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














2












2








2





$begingroup$


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.










share|cite|improve this question











$endgroup$




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






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share|cite|improve this question













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share|cite|improve this question








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














  • 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










2 Answers
2






active

oldest

votes


















3












$begingroup$

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.






share|cite|improve this answer











$endgroup$









  • 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



















2












$begingroup$

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 $



SU(3)/SU(3)






share|cite|improve this answer











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    2 Answers
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    active

    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    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.






    share|cite|improve this answer











    $endgroup$









    • 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
















    3












    $begingroup$

    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.






    share|cite|improve this answer











    $endgroup$









    • 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














    3












    3








    3





    $begingroup$

    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.






    share|cite|improve this answer











    $endgroup$



    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.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    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














    • 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











    2












    $begingroup$

    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 $



    SU(3)/SU(3)






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      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 $



      SU(3)/SU(3)






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        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 $



        SU(3)/SU(3)






        share|cite|improve this answer











        $endgroup$



        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 $



        SU(3)/SU(3)







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        edited 3 hours ago

























        answered 4 hours ago









        Sean LawtonSean Lawton

        3,38122142




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