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.

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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 $\mathbb{C}$.