Discrete series representations with non-tempered embedding

We give an example of a semisimple symmetric space G / H and an irreducible representation of G which has multiplicity 1 in L 2 ( G / H ) and multiplicity 2 in C ∞ ( G / H ).


Introduction
Let G be a real reductive group and X = G/H an attached symmetric space.Let further V be a Harish-Chandra module and V ∞ its smooth Fréchet completion of moderate growth.We say that V is H-spherical provided Hom G (V ∞ , C ∞ (X)) = 0 .By Frobenius reciprocity we have Hom G (V ∞ , C ∞ (X)) ≃ (V −∞ ) H with V −∞ = (V ∞ ) ′ the strong dual of V ∞ .We recall that the full multiplicity space (V −∞ ) H is finite dimensional.
Inside of C ∞ (X) we find natural invariant subspaces: the Harish-Chandra Schwartz space C(X) ⊂ L 2 (X) ∩ C ∞ (X) and the space C ∞ temp (X) of tempered functions which lie in L 2+ǫ (X) for all ǫ > 0. Accordingly we obtain subspaces and Hom G (V ∞ , C(X)) respectively.The objective of this paper is to provide an example where To be more specific this happens for X the n-dimensional one-sheeted hyperboloid which is homogeneous for the connected Lorentzian group G = SO 0 (n, 1).Let us briefly introduce the standard notions.
1.1.Notation.Let n ≥ 3 and let G = SO 0 (n, 1) be the identity component of the special Lorentz group SO(n, 1).We denote by H the stabilizer of x 0 := (1, 0, . . ., 0) ∈ R n+1 in G.The entry in the lower right corner of any matrix in SO(n, 1) is non-zero, and SO 0 (n, 1) consists of those matrices for which this entry is positive.From this fact we see that H is the connected subgroup The group G acts transitively on the hyperboloid , and the homogeneous space is a symmetric space.The corresponding involution σ of G is given by conjugation with the diagonal matrix diag(−1, 1, . . ., 1).The subgroup G σ of G of σ-fixed elements is the stabilizer of Rx 0 .This subgroup has two components, one of which is H.For our purpose it is important to use H rather than G σ .The pairs (G, H) and (G, G σ ) differ by the fact that (G, G σ ) is a Gelfand pair, whereas (G, H) is not.In fact it has been shown by van Dijk [8] that X is the only symmetric space of rank one, which is obtained as the homogeneous space of a Gelfand pair.
The regular representation of G on C ∞ (X) decomposes as the direct sum of the invariant subspaces of functions that are even or odd with respect to the G-equivariant symmetry x → −x.The restriction of the regular representation to C ∞ even (X) is isomorphic to the regular representation on C ∞ (G/G σ ), and the non-Gelfandness of (G, H) is therefore caused by the presence of the odd functions.
1.2.Main results.The Lorentzian manifold X carries the G-invariant Laplace-Beltrami operator ∆, which is obtained as the radial part of Let ρ := 1 2 (n − 1) and for each λ ∈ C let E λ (X) be the eigenspace The Laplace-Beltrami operator is a scalar multiple of the Casimir element associated to the Lie group G, and hence every irreducible subspace V of C ∞ (X) is contained in E λ (X) for some λ ∈ C. Apart from the sign, the scalar λ is uniquely determined by the infinitesimal character of V. Conversely, since X = G/H is rank one, ±λ determines the infinitesimal character.Let We can now state our main results.
The manifold X carries a G-invariant measure, which is unique up to scalar multiplication.We denote by L 2 (X) the associated G-invariant space of square integrable functions.
) and E odd λ (X) are irreducible and infinitesimally equivalent.Moreover in this case, ( For n ≥ 4 we have ρ > 1 and it follows that there exists at least one discrete series representation for X = G/H which has multiplicity 1 in C ∞ temp (X), but for which the underlying Harish-Chandra module has multiplicity 2 in C ∞ (X).
The complete Plancherel decomposition for SO 0 (n, 1)/ SO 0 (n − 1, 1) is given in [5].However, this is not needed for the proof of our theorems.

Proof of the main results
The proof of the two main theorems is divided into several parts.We begin with the analysis on K-types.
We are going to use the diffeomorphism where y = (y 1 , . . ., y n ) ∈ S n−1 and t ∈ R. With the natural action of SO(n) on S n−1 the parameter dependence on y is K-equivariant.
For each j ∈ N 0 we denote by H j ⊂ C ∞ (S n−1 ) the space of spherical harmonics of degree j.We recall that by definition H j consists of the restrictions to S n−1 of all harmonic polynomials on R n , homogeneous of degree j.Equivalently, H j can be defined as the eigenspace where ∆ K is the angular part of the n-dimensional Laplacian is spanned by all functions given in the coordinates (y, t) by Being homogeneous of degree j, the spherical harmonics h ∈ H j satisfy h(−y) = (−1) j h(y) for y ∈ S n−1 .Therefore where parity(j) denotes the parity even or odd of j.Likewise

2.2.
Eigenspaces.With respect to the coordinates (2.1) on X we have (see [7, p. 455]) The K-finite eigenfunctions for ∆ belong to C ∞ (X), and they can be determined as follows.For each j ∈ N 0 we let By separating the variables y and t we see that E λ,j is spanned by the functions f (y, t) = h(y)ϕ(t) for which h ∈ H j and (2.2) This differential equation is invariant under sign change of t.The solution with ϕ(0) = 1 and ϕ ′ (0) = 0 is even, and the solution with ϕ(0) = 0 and ϕ ′ (0) = 1 is odd.Thus the solution space decomposes as the direct sum of the one-dimensional subspaces of even and odd solutions, and we have 2.3.Hypergeometric functions.In fact (2.2) can be transformed into a standard equation of special function theory.We first prepare for the anticipated asymptotic behavior af ϕ by substituting Φ(t) = (cosh t) λ+ρ ϕ(t).This leads to the following equation for Φ(t) Next we change variables.With we replace the limits t = ∞ and t = −∞ by x = 0 and x = 1, respectively.We write Φ(t) = F (x), so that , we arrive at the following equation for the function F (x) solves (2.2).
2.4.L 2 -behavior.In the coordinates (y, t) an invariant measure on X is given by cosh n−1 t dt dy where dt and dy are invariant measures on R and S n−1 , respectively.Hence a function f (y, t) = h(y)ϕ(t) is square integrable if and only if Let φλ,j (t) = ϕ λ,j (−t).By symmetry this function also solves (2.2), and it belongs to L 2 (R, cosh n−1 t dt) if and only if ϕ λ,j does.Lemma 2.1.Let Re λ > 0 and j ∈ N 0 . (1) (2) If j / ∈ λ − ρ + N then ϕ λ,j and φλ,j are linearly independent, and for a sufficiently small ǫ > 0 no non-trivial linear combination belongs to L 2+ǫ (R, cosh n−1 t dt).
Proof.It follows from the definition of ϕ λ,j (t) that Since 2ρ = n − 1 this means ϕ λ,j has the desired L 2 -behavior in the positive direction for all Re λ > 0. The only issue is with the negative direction, or equivalently, with φλ,j (t) for t → ∞.
In particular we note that A = 0 if j / ∈ λ − ρ + N.Under this condition we see that no non-trivial linear combination of ϕ λ,j and φλ,j exhibits L 2+ǫ -behavior in both directions ±∞.
This proves (2) and concludes the proof of the lemma.
Note that with the repeated indices P (λ,λ) l is in fact a Gegenbauer polynomial.

Proof. By definition
The equation (1) follows immediately.Then (2) follows since a Gegenbauer polynomial is even or odd according to the parity of its degree.2.6.K-types in L 2 .For λ ∈ C with Re λ ≥ 0 we define and λ > 0, and by D λ = ∅ otherwise.Furthermore we let It follows from Lemma 2.2 and the fact that the parity of (2.6) follows.Assume from now on that λ − ρ ∈ Z.It then follows from Lemma 2.1(1) that U λ ⊂ L 2 (X).
To complete the proof we will find for each j ∈ N 0 a second solution to (2.2), which is linearly independent from ϕ λ,j , and which is not tempered.There are two cases, depending on the parity of n.
If n is even, then ρ, and hence also λ, is not an integer.In that case we already have a second solution at hand, namely ϕ −λ,j .Since this function ϕ −λ,j does not belong to any L p (R, cosh n−1 t dt) if Re λ ≥ ρ.When 0 < Re λ < ρ it belongs to L 2+ǫ only for ǫ > 2 Re λ ρ−Re λ .We now assume n is odd.Then ρ and λ are positive integers.We need to find a solution linearly independent from F (a, b; c; x) to the hypergeometric equation (2.5) with By the method of Frobenius one finds (see [6, p. 5]) such a solution G(a, b, c; x).It has the form for some explicit power series with a 0 = 1 and b 0 = 0.The corresponding solution to (2.2) is It behaves like (cosh t) −λ−ρ (1 + e 2t ) λ as t → ∞ and as before it is not tempered with respect to the invariant measure.2.7.Irreducibility.Let λ ∈ ρ + Z and assume λ > 0. It follows from Lemma 2.3 that U λ is (g, K)-invariant.We will prove that it is an irreducible (g, K)module by using the infinitesimal element (2.7) as a raising and lowering operator between the functions ϕ λ,j which generate U λ together with K.For this we need to find the derivative of ϕ λ,j .
Then S n−1 ≃ K/M.Let h j ∈ H j be the zonal spherical harmonic.This is the unique function in H j which is M-invariant and has the value 1 at the origin (1, 0, . . ., 0) of S n−1 .Furthermore, let f λ,j ∈ E λ (X) be defined by f λ,j (y, t) = h j (y)ϕ λ,j (t).
With the element T from (2.7) the coordinates (y, t) are determined from and since T is centralized by M the left derivative L T f λ,j by T is again Minvariant.It follows that for each j ∈ N 0 , the function L T f λ,j ∈ E λ,K (X) is a linear combination of the same family of functions f λ,• in E λ,K (X).Since h j (x 0 ) = 1 for all j, the coefficients can be determined from the restriction to {(cosh t, 0, . . ., 0, sinh t) | t ∈ R} ⊂ X, on which L T acts just by d dt , and hence they are given by Lemma 2.4.It follows immediately that U λ has no non-trivial (g, K)-invariant subspaces.
Proof.The assumption on λ implies that D λ = N 0 , and hence U λ is equal to one of the two modules E even λ,K (X) and E odd λ,K (X), depending on the parity of λ−ρ.For simplicity of exposition, let us assume a specific parity, say even, of λ − ρ.Then E odd λ,K (X) = U λ is irreducible as seen in Section 2.7.By Kostant's theorem [4,Thm. 8] an irreducible (g, K)-module, which contains the trivial K-type, is uniquely determined up to equivalence by its infinitesimal character.Hence E odd λ,K (X) is equivalent to the irreducible subquotient of E even λ,K (X) containing E even λ,0 (X).Since E odd λ,K (X) and E even λ,K (X) contain the same K-types, all with multiplicity one, we conclude that this subquotient is equal to E even λ,K (X).The lemma is proved.2.9.Conclusion.Assume Re λ > 0. Then E λ,K (X) ∩ L 2 (X) = U λ by Lemma 2.3.By definition U λ is non-zero if and only if λ − ρ ∈ Z.In Section 2.6 we saw that it consists of even functions on X when λ − ρ is odd, and vice versa.Finally, irreducibility was seen in Section 2.7.Thus the proof of Theorem 1.1 is complete.
Assume λ ∈ ρ+Z and 0 < λ < ρ.Then E even λ,K (X) and E odd λ,K (X) are irreducible and equivalent by Lemma 2.5.One of them equals U λ and belongs to L 2 (X), whereas we have seen in Lemma 2.3 that the other one is non-tempered.This proves Theorem 1.2.
F 1 (a, b; c; x), which with the notation (a) m := m−1 k=0 (a + k) is defined by the Gauss series F (a, b; c; x) = ∞ m=0 (a) m (b) m m! (c) m x m for x ∈ C with |x| < 1, for all a, b, c ∈ C except c ∈ −N 0 .It solves Euler's hypergeometric differential equation (2.5) x(1 − x) w ′′ + (c − (a + b + 1)x) w ′ − ab w = 0.The function F (a, b; c; x) is analytic at x = 0 with the value 1, and unless c is an integer it is the unique solution with this property.With a = λ+ρ+j and b = λ−ρ+1−j as above we have a+b+1 = 2(λ+1).By comparing (2.4) and (2.5) we conclude that for each λ / ∈ −N the function