“Quasi-unitary” representations of non-compact Lie groups

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Let $G$ be a connected, non-compact Lie group and let $rho: G to GL_n(mathbb C)$ be a complex representation of $G$.



Definition: $rho$ is quasi-unitary if there exists some positive hermitian matrix $A in GL_n(mathbb C)$ such that beginequationsup_g in G||rho(g) cdot A cdot rho(g)^*|| < C endequation for some $C > 0$, where $rho(g)^*$ denotes the conjugate-transpose of $rho(g)$, and $|| cdot ||$ is some matrix-norm.



If $rho$ is unitary, one can chose a positive $A in GL_n(mathbb C)$ such that $rho(g) cdot A cdot rho(g)^* = A$ for all $g in G$, hence $rho$ is quasi-unitary.



Question: Is every quasi-unitary representation unitary ?



I believe that this is the case, but I can't quite prove it. My idea is to consider the conjugate-action of $G$ on $GL_n(mathbb C)$ induced by a quasi-unitary representation $rho$, given by $g.B := rho(g) cdot B cdot rho(g)^*$. Let $A in GL_n(mathbb C)$ be such that $sup_g in G ||g.A|| < C$. Then $G.A$ (the $G$-orbit of $A$) is bounded (hopefully with compact closure). Now it is somehow intuitive to assume that a fixed point $C in overlineG.A$ of the $G$-action must exist. Unfortunaltey, all fixed point Theorems that I know do not quite apply to this situation.




representation-theory lie-groups lie-algebras




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    up vote
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    down vote

    favorite












    Let $G$ be a connected, non-compact Lie group and let $rho: G to GL_n(mathbb C)$ be a complex representation of $G$.



    Definition: $rho$ is quasi-unitary if there exists some positive hermitian matrix $A in GL_n(mathbb C)$ such that beginequationsup_g in G||rho(g) cdot A cdot rho(g)^*|| < C endequation for some $C > 0$, where $rho(g)^*$ denotes the conjugate-transpose of $rho(g)$, and $|| cdot ||$ is some matrix-norm.



    If $rho$ is unitary, one can chose a positive $A in GL_n(mathbb C)$ such that $rho(g) cdot A cdot rho(g)^* = A$ for all $g in G$, hence $rho$ is quasi-unitary.



    Question: Is every quasi-unitary representation unitary ?



    I believe that this is the case, but I can't quite prove it. My idea is to consider the conjugate-action of $G$ on $GL_n(mathbb C)$ induced by a quasi-unitary representation $rho$, given by $g.B := rho(g) cdot B cdot rho(g)^*$. Let $A in GL_n(mathbb C)$ be such that $sup_g in G ||g.A|| < C$. Then $G.A$ (the $G$-orbit of $A$) is bounded (hopefully with compact closure). Now it is somehow intuitive to assume that a fixed point $C in overlineG.A$ of the $G$-action must exist. Unfortunaltey, all fixed point Theorems that I know do not quite apply to this situation.




    representation-theory lie-groups lie-algebras




    share|cite|improve this question





















      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      Let $G$ be a connected, non-compact Lie group and let $rho: G to GL_n(mathbb C)$ be a complex representation of $G$.



      Definition: $rho$ is quasi-unitary if there exists some positive hermitian matrix $A in GL_n(mathbb C)$ such that beginequationsup_g in G||rho(g) cdot A cdot rho(g)^*|| < C endequation for some $C > 0$, where $rho(g)^*$ denotes the conjugate-transpose of $rho(g)$, and $|| cdot ||$ is some matrix-norm.



      If $rho$ is unitary, one can chose a positive $A in GL_n(mathbb C)$ such that $rho(g) cdot A cdot rho(g)^* = A$ for all $g in G$, hence $rho$ is quasi-unitary.



      Question: Is every quasi-unitary representation unitary ?



      I believe that this is the case, but I can't quite prove it. My idea is to consider the conjugate-action of $G$ on $GL_n(mathbb C)$ induced by a quasi-unitary representation $rho$, given by $g.B := rho(g) cdot B cdot rho(g)^*$. Let $A in GL_n(mathbb C)$ be such that $sup_g in G ||g.A|| < C$. Then $G.A$ (the $G$-orbit of $A$) is bounded (hopefully with compact closure). Now it is somehow intuitive to assume that a fixed point $C in overlineG.A$ of the $G$-action must exist. Unfortunaltey, all fixed point Theorems that I know do not quite apply to this situation.




      representation-theory lie-groups lie-algebras




      share|cite|improve this question











      Let $G$ be a connected, non-compact Lie group and let $rho: G to GL_n(mathbb C)$ be a complex representation of $G$.



      Definition: $rho$ is quasi-unitary if there exists some positive hermitian matrix $A in GL_n(mathbb C)$ such that beginequationsup_g in G||rho(g) cdot A cdot rho(g)^*|| < C endequation for some $C > 0$, where $rho(g)^*$ denotes the conjugate-transpose of $rho(g)$, and $|| cdot ||$ is some matrix-norm.



      If $rho$ is unitary, one can chose a positive $A in GL_n(mathbb C)$ such that $rho(g) cdot A cdot rho(g)^* = A$ for all $g in G$, hence $rho$ is quasi-unitary.



      Question: Is every quasi-unitary representation unitary ?



      I believe that this is the case, but I can't quite prove it. My idea is to consider the conjugate-action of $G$ on $GL_n(mathbb C)$ induced by a quasi-unitary representation $rho$, given by $g.B := rho(g) cdot B cdot rho(g)^*$. Let $A in GL_n(mathbb C)$ be such that $sup_g in G ||g.A|| < C$. Then $G.A$ (the $G$-orbit of $A$) is bounded (hopefully with compact closure). Now it is somehow intuitive to assume that a fixed point $C in overlineG.A$ of the $G$-action must exist. Unfortunaltey, all fixed point Theorems that I know do not quite apply to this situation.





      representation-theory lie-groups lie-algebras





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      asked Jul 26 at 8:38









      Berni Waterman

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