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This is probably a silly question.

I was looking at Berger's classification for holonomy groups, and the fourth element is "Quaternion-Kähler manifolds, $\,\dim M=4n, \,\text{Hol}=\text{Sp}(n)\cdot\text{Sp}(1)$".

First I tought $\text{Sp}(n)\cdot\text{Sp}(1)$ meant the direct sum in the group-theoretic sense. But according to de Rham's decomposition theorem, this would mean $M$ is reducible, which contradicts one of the hypothesis of Berger's theorem.

Am I missing something? What does $\text{Sp}(n)\cdot\text{Sp}(1)$ really mean?

rmdmc89
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1 Answers1

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You should think of $Sp(n)\cdot Sp(1)$ as the quaternionic version of being unorientable.

More precisely, $Sp(n)\cdot Sp(1)$ is $(Sp(n)\times Sp(1))/\langle (-I,-1)\rangle$, If you think of $Sp(n)\subset GL(n,\mathbb{H})$ as the group of invertible quaternionic matrices acting on the right-quaternionic vector space $\mathbb{H}^n$ on the left preserving the standard Hermitian form, then $Sp(n)\cdot Sp(1)$ has the additional factor $Sp(1)$ acting by right-multiplication, hence the quotient by central $\pm 1$, and no longer preserve the quaternionic vector space structure (nor the complex vector space structure). You don't get an embedding in $SU(2n)$ or $U(2n)$, because you multiplied both left and right by quaternions.

user10354138
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