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Let $p$ and $q$ be integers. The group $K=SO(p) \times SO(q)$ can be naturally seen as a subgroup of $G=SO(p+q)$. The quotient space $G/K$ is identified with the space of oriented $p$-dimensional subspaces of $\mathbb R^{p+q}$, and carries an action of $G$ (by left multiplication).

A spherical function is a $C^\infty$ function on $G/K$ which is also invariant under left multiplication by $K$, which takes value $1$ at $eK$ and which is an eigenfunction of all $G$-invariant differential operators on $K/U$. I am not familiar with this subject (all I know comes from my reading of Helgason's book "Groups and geometric analysis: integral geometry, invariant differential operators, and spherical functions"), and I am wondering whether all the spherical functions are explicitely described somewhere. I am mainly interested in the case when $q=p$ or $q=p+1$.

I know that the spherical functions are described in term of the irreducible finite-dimensional spherical representations of $G$, by the formula $g\mapsto \int_K \chi(g^{-1} k) dk$, where $\chi$ is the character of an irreducible finite-dimensional spherical representation of $G$. If I understood correctly these representations are explicitely described in terms of their higher weight, and Weyl's formulas give the dimension and the character of the representation in term of this weight.

Is it possible to be more explicit? In the last pages of Helgason's book, the rank one case is treated. Hoogenboom made computations for $G=SU(p+q)$ and $K=SU(p)\times SU(q)$, but I was not able to find a reference for the real case.

I am already interested in the case $p=q=2$.

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For $p=q=2$, is it true that the exceptional isomorphism $SO(4)\simeq (SU(2)\times SU(2))/C_2$ induces a decomposition of the Grassmannian as $(S^2\times S^2)/C_2$? If yes, this basically reduces the description of spherical functions to the one on the sphere $S^2$. – Alain Valette May 4 '11 at 20:59
It's true (and the oriented Grassmannian is $S^2 \times S^2$). – Henry Cohn May 4 '11 at 23:22
Thanks Henry! Indeed, noticing that $SO(2)\times SO(2)$ is a maximal torus in $SO(4)$, it must correspond in the above isomorphism to maximal tori in each of the simple factors. – Alain Valette May 5 '11 at 12:59
Thanks for your comments and answer. Just to be sure I got it right, and before I work enough to understand the genral case: I know that the spherical functions on $S^2$ are given by the Legendre polynomials $P_n$ (= the Jacobi polynomial of degree $n$ with parameters $(0,0)$). And the dimension of the corresponding spherical representation is $(2n+1)$. Does it imply that in the case $p=q=2$, the spherical functions are parametrized by integers $n$ and $m$, and are given as the functions $(s,t)\mapsto P_n(s)P_m(t)$? And the dimension is $(2n+1)(2m+1)$? – Mikael de la Salle May 5 '11 at 14:00
up vote 5 down vote accepted

In the unoriented case, the spherical functions are computed in the following article (they turn out to be multivariate Jacobi polynomials):

A. T. James and A. G. Constantine, Generalized Jacobi polynomials as spherical functions of the Grassmann manifold, Proc. London Math. Soc. (3) 29 (1974), 174-192.

Unfortunately, I don't know a good reference where the oriented case is worked out, but it just amounts to taking a double cover. All the spherical functions for the unoriented Grassmannian lift to the oriented Grassmannian, and this gives half of the spherical functions in the oriented case.

I am more or less sure the remaining spherical functions can be obtained by just a small modification, but I haven't worked it out.

For comparison, the spherical functions on $S^2$ are Jacobi polynomials with parameters $(0,0)$, and those for $\mathbb{R}\mathbb{P}^2$ are Jacobi polynomials with parameters $(0,-1/2)$. Using the notation $P_k^{(\alpha,\beta)}$ for the degree $k$ Jacobi polynomial with parameters $(\alpha,\beta)$, we have $$ P_{2k}^{(0,0)}(t) = P_k^{(0,-1/2)}(2t^2-1) $$ and $$ P_{2k+1}^{(0,0)}(t) = t P_k^{(0,1/2)}(2t^2-1). $$ The former equation just says the spherical functions on $\mathbb{R}\mathbb{P}^2$ lift to those of even degree on $S^2$, and the latter says the ones of odd degree on $S^2$ come from Jacobi polynomials with similar parameters. Presumably the multivariate case works similarly, but I don't know for sure.

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