Consider $G=SO(2n)$ and $K=U(n)$. $(G,K)$ is a symmetric pair. I'm interested in (zonal) spherical functions on $G/K$ which are matrix elements with respect to $K$-fixed vectors in irreducible representations of $G$.

A classical result by E. Cartan says that if $(G, K)$ is a Riemannian symmetric pair then the algebra of integrable functions on $G$ which are bi-invariant under $K$ is commutative, that is, $(G, K)$ is a Gelfand pair.


Then this book cited by Wikipedia says that for a Gelfand pair, the space of $K$-fixed vectors in every irreducible unitary representation of $G$ is at most one-dimensional. But an irreducible spinor representation of $SO(2n)$ has more than one orthogonal $U(n)$-fixed vectors. (I only know how to phrase it in physicists' language: The subgroup $U(n)$ preserves the number of fermions, so it fixes the state with no fermion and the fully filled state.) Is this because a spinor representation really is a representation of $Spin(2n)$ and not $SO(2n)$? Do I miss something here?

EDIT: Assume everything is over $\mathbb{C}$.


There is no contradiction. To use your physical language, the fully filled state is not $K$-fixed as a vector. Rather, the group ${\rm U}(n)$ acts on it by the determinantal character. The subtle point here is that in quantum mechanics, one considers projective representations, so the action by a character cannot be distinguished from the trivial one. The full decomposition of the (half-)spinor representation into $K$-types uses exterior algebra and is well known — see for example Goodman and Wallach's book. In particular, the spinor representation is multiplicity-free: in fact, every $K$-type that occurs has multiplicity 1, and not just the trivial one. This, of course, is a much stronger property.

Incidentally, for odd $n$, the vacuum (1, the state with fermion number $0$) and the fully filled state (the volume form, fermion number $n$) have different parity. So they appear in different halves of the spinor representation.

  • $\begingroup$ Thank you. I see. The trivial $U(n)$-representation (the vacuum) and the $U(n)$-representation on the highest exterior power are not the same, and a subspace being $K$-fixed means that it is actually a trivial representation and not merely any 1-dimensional representation. Is that correct? $\endgroup$ Apr 25 '17 at 2:42
  • $\begingroup$ Yes, although to be completely precise, you should rephrase to "subspace consisting of $K$-fixed vectors means that it is actually a trivial representation". I was glad to have been able to help. $\endgroup$ Apr 25 '17 at 3:13

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