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Given $Sl(2, \mathbb{R})$, how can I define an unitary scalar product on the discrete series representation?

Will the restriction to $SO(2)$ encounter a scaling for each irreducible subrepresentation, or will it embed directly unitarily into the right regular representation of $SO(2)$?

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up vote 2 down vote accepted

On one hand, exactly because a discrete series repn is "discrete series", it is inside $L^2(G)$, and inherits the Hilbert space structure from that.

At the same time, yes, we might like to use an orthogonal basis of $K$-types. One hazard is that the principal series in which discrete series lie are not themselves unitary or unitarizable, so any line of thought that inadvertently depends on this will have problems.

Also, there is the question of normalization of bases of the $K$-types (which here all occur with multiplicities $0$ or $1$). Happily, for holomorphic discrete series, for example, a suitable "raising" operator (in an automorphic forms context called "Maass-Shimura" operators sometimes) sends the $\ell$th $K$-type to the $(\ell+2)$th, giving one choice of normalization of those one-dimensional spaces.

The appropriate raising operator is in the complexified Lie algebra: with the usual $SO(2)$, $\pmatrix{-i & 1 \cr 1 & -i}$ is one choice of raising operator.

There is a computation to do, to see the effect a chosen raising operator has on the $L^2(G)$-norm of the "bottom" $K$-type in a principal series, starting from the fact that $G$ acts unitarily on $L^2(G)$, so the real Lie algebra acts by skew-hermitian operators. Such computations appear in papers of Duke,Iwaniec, et alia, and in papers of Shimura, for example.

Edit: to respond to the comment/question: it's not quite eigenvalues of the raising operator $R$, since, in fact, $v_o, Rv_o, R^2v_o, \ldots$ is an orthogonal basis of the repn space, where $v_o$ is in the "lowest" $K$-type.

Instead, the question is to compute $\langle R^n v_o, R^n v_o\rangle$ where $\langle, \rangle$ is the/a $G$-invariant inner product, in terms of $\langle v_o,v_o\rangle$. Use the fact that the real Lie algebra acts by skew-adjoint operators, since the group acts unitarily, and we also need to keep track of the complexification, since the raising and lowering operators for Cartan in $K$ are not in the real Lie algebra. Indeed, with $x=\pmatrix{0& 1\cr 0&1}$, $y=\pmatrix{0&0\cr 1& 0}$, $h=\pmatrix{1&0\cr 0&-1}$ the usual triple, one choice of raising operator is $R=(x+y-ih)/2$. Thus, its adjoint is $R^*=(-x-y-ih)/2$, which is $-L$, for corresponding lowering operator $L$ with respect to the Cartan in $K$. Thus, starting an induction, $\langle Rv_o,Rv_o\rangle=\langle R^*R v_o,v_o\rangle=\langle -LR v_o,v_o\rangle$. Computing further in the universal enveloping algebra, $$ -LR = (RL-LR) - RL = [R,L] - RL = H - RL $$ where $H$ is the corresponding element in the Lie algebra of $K$ making $H,R,L$ a usual triple (with $[R,L]=H, [H,R]=2R, [H,L]=-2L$). The lowering operator is $0$ on $v_o$, since $v_o$ is a lowest-weight vector in the discrete series, and $H$ acts by $k$ on the weight-$k$ holomorphic discrete series. Thus, $\langle Rv_o,Rv_o\rangle = k\langle v_o,v_o\rangle$.

The computation continues by an induction, which I would likely not get right on the first try, but I hope the idea is clear from the first step, just given.

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I want to rephrase your answer, so that I make sure I understand. The scalling is different than the one for the principal and complementary series reps. The scalling can be computed by analysing the eigenvalue of the lowering and raising operators. In higher rank, $K$-subreps of the same $K$-iso-class (occuring with non-simple multiplicity) can have different scaling on the inner product, right? Thank you. – May 31 '12 at 21:45
Thanks for the hints and the elaboration. – Jun 1 '12 at 17:14
You're welcome. – paul garrett Jun 1 '12 at 17:28

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