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Baruch's paper (Annals of Math. 158 (2003), 207-252) proves the following statement, originally claimed by Kirillov.

Theorem. Let $G=\mathrm{GL}_n(K)$ where $K$ is either $\mathbb{R}$ or $\mathbb{C}$, and let $P=P_n(K)$ be the subgroup of matrices in $\mathrm{GL}_n(K)$ consisting of matrices whose last row is $(0,0,\dots,0,1)$. If $\pi$ is an irreducible unitary representation of $G$ on a Hilbert space $H$ then $\pi|P$ is irreducible.

Representation theory is not my expertise, so the following question might be very basic. I might add further questions here if that is acceptable.

Question 1. On page 208 the author considers an arbitrary bounded linear operator $R:H\to H$ which commutes with $\pi(p)$ for all $p\in P$. He also considers, for any $f\in C_c^\infty(G)$, the distribution $\Lambda_R(f)=\mathrm{trace}(R\pi(f))$. Then he says: "It is easy to see that $\Lambda_R$ is an eigendistribution with respect to the center of the universal enveloping algebra associated to $G$." What is the meaning of "eigendistribution" here, and why is the statement true?

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    $\begingroup$ I have not looked at the paper, but I guess an eigendistribution is justa distribution which is an eigenvalue. That is, when you act on it by a central element in the enveloping algebra, it gets multiplied by a scalar. $\endgroup$ Jan 12, 2012 at 2:59
  • $\begingroup$ Thanks, Mariano, that would make sense. I hope someone can clarify this with proof. $\endgroup$
    – GH from MO
    Jan 12, 2012 at 3:29
  • $\begingroup$ Let us take R=Identity. Do we agree that trace( pi(f)) is eigendistribution ? It seems it is character. Center of U(g) should act on it by scalars. $\endgroup$ Jan 12, 2012 at 6:54
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    $\begingroup$ And welcome to the 10k club, GH! $\endgroup$ May 9, 2012 at 5:46
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    $\begingroup$ Dear David Roberts: Thank you, that was very kind of you! $\endgroup$
    – GH from MO
    May 9, 2012 at 9:07

3 Answers 3

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Since $\pi$ is irreducible, the center of the enveloping algebra acts by scalars (by Schur's lemma). Hence, for $X\in Z(\mathcal{U}G)$ we have $$\Lambda_R(Xf)=Tr(R\pi(Xf))=Tr(Rd\pi(X)\pi(f))=d\pi(X)Tr(R\pi(f))=d\pi(X)\Lambda_R(f)$$ where $d\pi$ is the derived representation extended to the enveloping algebra.

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    $\begingroup$ And for this to make sense, $\pi(f)$ is trace-class, which is not obvious. $\endgroup$ Jan 12, 2012 at 13:44
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    $\begingroup$ @Paul: Right, this is really not trivial. On the other hand, the OP took it for granted. $\endgroup$ Jan 12, 2012 at 16:40
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    $\begingroup$ I don't think "Schur's lemma" is completely obvious in this setting either, especially when $G = GL_n \mathbb R$ is disconnected. $\endgroup$
    – Faisal
    Jan 12, 2012 at 18:21
  • $\begingroup$ @Alain: Thank you. At the moment I only see that $X$ commutes with $\pi(g)$ for $g\in\exp(G)$. Also, I think $\Lambda_R(Xf)=d\pi(-X)\Lambda_R(f)$. Indeed, the left hand side is the $t$-derivative at $t=0$ of the operator $\int_G f(e^{tX}g)\pi(g)\ dg$, using that $X$ commutes with $\pi(g)$, and the last integral equals $\int_G f(g)\pi(e^{-tX}g)\ dg$. Am I missing something? $\endgroup$
    – GH from MO
    Jan 13, 2012 at 5:18
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    $\begingroup$ @Faisal: You made a very good point: I overlooked the fact that $GL_n(\mathbb{R})$ is not connected (and connectedness is important, as GH's computation shows). If the restriction of $\pi$ to $GL_n^+(\mathbb{R})$ (=matrices with positive determinant) is still irreducible, then my argument goes through. If $n$ is odd, this will be the case (since $GL_n(\mathbb{R})$ is then a direct product). The case of even $n$ remains unclear to me. $\endgroup$ Jan 13, 2012 at 8:50
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The following answer to Question 1 is based on Alain Valette's response, it is too long for a comment.

For $X\in Z(\mathcal{U}\mathfrak{g})$ the actions of $X$ and $G^0=\mathrm{GL}^+_n(\mathbb{R})$ commute, hence we are done by Schur's lemma when $\pi|G^0$ is irreducible. Otherwise $\pi|G^0$ is the sum of two irreducible representations $(\rho^\pm,H^\pm)$ which are conjugates of each other by the matrix $i:=\mathrm{diag}(-1,1,\dots,1)\in P$. In particular, $\pi(i)$ interchanges $H^\pm$. By Schur's lemma, $X$ acts on $H^{\pm}$ by some scalars $c^\pm(X)\in\mathbb{C}$. We are done if $c^+(X)=c^-(X)$.

Assume that $c^+(X)\neq c^-(X)$, and let $f\in C_0^\infty(G)$ be supported in a connected component of $G$. As $R$ commutes with $\pi(i)$, we have $$\mathrm{trace}(R\pi(Xf))=\mathrm{trace}(R\pi(i)\pi(Xf)\pi(i)).$$ Note that $\pi(f)$ either leaves the spaces $H^\pm$ invariant or it maps one into the other, depending on whether $f$ is supported in $G^0$ or in $G^0i$. In either case the previous equation together with $$\Lambda_R(f)=\mathrm{trace}(R\pi(f))=\mathrm{trace}(R\pi(i)\pi(f)\pi(i))$$ yields $$c^+(X)\Lambda_R(f)=c^-(X)\Lambda_R(f).$$ This shows that $\Lambda_R(f)=0$, hence $\Lambda_R=0$ as a whole.

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    $\begingroup$ Nice use of the extra assumption that $\pi|_P$ commutes with $R$. $\endgroup$ Jan 14, 2012 at 13:00
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Here is a complement on GH's answer, too long for a comment. Assume $n=2$, and write $GL_2^+(\mathbb{R})$ as $SL_2(\mathbb{R})\times\mathbb{R}$, so that $\mathcal{U}(G)\simeq\mathcal{U}(SL_2(\mathbb{R})\otimes\mathcal{U}(\mathbb{R})$ and $Z(\mathcal{U}(G))\simeq Z(\mathcal{U}(SL_2(\mathbb{R}))\otimes\mathcal{U}(\mathbb{R})$. Now $\mathcal{U}(\mathbb{R})=\mathbb{C}[X]$ and, as is well-known, $Z(\mathcal{U}(SL_2(\mathbb{R}))=\mathbb{C}[\Omega]$, where $\Omega$ is the Casimir operator. Now the adjoint action by the diagonal matrix $i=diag(-1,1)$ fixes $\Omega$, so that (using GH's notations) $c^+=c^-$, without appealing to the extra assumption on $\pi$. I'm not sure that this generalizes to other even $n$'s.

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  • $\begingroup$ Nice observation! $\endgroup$
    – GH from MO
    Jan 15, 2012 at 15:22

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