User sigurdur helgason - MathOverflow

Answer by Sigurdur Helgason for Symmetric spaces, Horocycle spaces and intertwining operators

2012-03-25T12:44:08Z

An explicit form is in my paper "A duality ... Advan. Math. 1970 and in my book Geometric Analysis on Symmetric Spaces AMS 2008, pp.554-556. Helgason.

Answer by Sigurdur Helgason for Exceptional isomorphisms of Lie groups

2012-02-25T16:31:49Z

The question goes back to paper A: Cartan's 1914 classification of real simple Lie algebras (Collected papers, Part I, Vol.1) with essentially complete answers for the Lie algebras . A complete discussion together with explicit isomorphisms for all the cases (both the Lie algebras and the Lie groups ) is in my book B: Diff. Geometry, Lie Groups and Symmetric Spaces, pp. 517- 528.

Comments: On the first page of A (bottom) Cartan states that real forms of the same simple Lie algebra over C are in general commpletely determined by the signature of the Killing form. As shown in B p. 517 this is contradicted by so*(18) and so(12,6) as well as by su*(14) and su(9,5). Cartan's phrase "general" clearly and deliberately meant "almost all the time". However, at the end of the paper (p. 352-355) he identifies real forms on the basis of the signature statement (which would require some additional caution). As mentioned in B, p. 520 the isomorphism so*(8) ~ so(6,2) does not occur in Cartan's original list although in a later paper he shows that the corresponding symmetric spaces are isometric.

Generally the group isomorphism in B are computational although some have geometric interpretation like the isomorphism so(3)~su(2) which comes from the standard stereographic projection.

S. Helgason

Answer by Sigurdur Helgason for When are the eigenspaces of the Laplacian on a compact homogeneous space irreducible representations?

2012-02-19T03:46:24Z

For G/K symmetric the joint eigenspaces of the G-invariant differential operators on G/K are all irreducible. Also each irreducible subspace of H has multiplicity bounded by one. For this see my "Groups and Geometric Analysis?" Ch. V Theorems 4.3 and 3.5. Concerning the Laplace Beltrami operator L, the Casimir operator on G (if semisimple) does induce L on G/K (loc. cit. p.331). If G/K is two point homogeneous the G-invariant differential operators on G/K are all polynomial in L (loc. cit. p/288) so for these spaces the answer to Dicks question is yes. For G/K not symmetric Theorem 3.5 p. 533 still gives a decomposition of H into spaces spanned by representation coefficients which are eigenfunctions of the Casimir operator.

Assuming the metric on G/K, (G semisimple) is coming from the Killing form Riemannian structure on G it is still true that the geodesics through the origin in G/K are orbits of one parameter subgroups of G. It seems to me that the argument for Problem A4 p.568 should still show that the Casimir operator on G will induce the Laplace Beltrami operator on G/K. Therefore the decomposition in Theorem 3.5 p.533 should still be a decomposition into eigenfunctions of the Laplacian. But there is no reason to expect irreducibility.

Answer by Sigurdur Helgason for A bounded homogeneous space which fails to be symmetric?

2012-02-19T02:33:56Z

E.Cartan proved in 1936 that for dimension 1 and 2 bounded homogeneous spaces are symmetric. For dimension 3 he did not publish the proof considering it loo long in comparison to the interest of the result. This has now changed with P-Sapiro's example for dimension 4. So the proof for dimension 3 is presumably somewhere in E.Cartan's Nachlass, unpublished.