# A lie Subgroup of SO(4n)

Hello everyone
thanks to all of you

I have two questions and I hope to get some guide:
1. One of the Lie groups in the Berger's list of holonomy groups of locally irreducible Riemannian manifolds is $Sp(n).Sp(1)\cong Sp(n) \times Sp(1)/(\pm Identity)$, which can be considered as a Lie subgroup of $SO(4n)$. what is the explicit formula for this inclusion?
2. Can be $Sp(n)\times Sp(1).T^1$ a Lie subgroup of $SO(4n)\times T^1$, where $Sp(1).T^1 \cong Sp(1) \times T^1 /(\pm Identity)$?
I think the answer of the second question is negative, but I'm not sure.

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One way to make the subgroup $\mathrm{Sp}(n)\cdot\mathrm{Sp}(1)\subset\mathrm{SO}(4n)$ explicit is to think of $\mathbb{R}^{4n}$ as $\mathbb{H}^n$, i.e. as columns of quaternions of height $n$, where $\mathbb{H}\simeq\mathbb{R}^4$ is the ring of quaternions. Then $\mathrm{Sp}(n)$ can be thought of as the group of $n$-by-$n$ quaternion matrices $A$ that satisfy $A^\ast A = \mathrm{I}_n$, where $A^\ast$ is the conjugate transpose of $A$. This group then acts on $\mathbb{H}^n$ on the left by the usual matrix multiplication $A\cdot x = Ax$, for $A\in\mathrm{Sp}(n)$ and $x\in\mathbb{H}^n$. This action preserves the positive definite quadratic norm $\|x\|^2= x^\ast x$, so $\mathrm{Sp}(n)$ is exhibited as a subgroup of $\mathrm{SO}(4n)$.

This action is irreducible (it acts transitively on the unit sphere in $\mathbb{H}^n$), and its commuting ring is $\mathbb{H}$, thought of as scalar multiplication on the right: $$A\cdot (x q) = (A\cdot x) q$$ for $q\in\mathbb{H}$ (which is true because quaternionic multiplication is associative). Remembering that $\mathrm{Sp}(1)$ is the group of unit quaternions, you can now see how $\mathrm{Sp}(n)\times\mathrm{Sp}(1)$ can act on $\mathbb{H}^n$, just use the rule $$(A,q)\cdot x = A\ x\ \bar q$$ (the conjugation is important, because, otherwise, this won't be a left action). This action is not effective, because $(-A,-q)$ acts the same way as $(A,q)$, so you actually get a faithful representation of $\mathrm{Sp}(n)\cdot\mathrm{Sp}(1)\simeq \bigl(\mathrm{Sp}(n)\times\mathrm{Sp}(1)\bigr)/(\lbrace \pm\mathrm{I}_n,\pm 1\rbrace)$ into $\mathrm{SO}(4n)$.

The commuting ring of this representation is just $\mathbb{R}$, so there is no torus in $\mathrm{SO}(4n)$ that commutes with this action, which answers your second question.

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Thanks a lot for your helpful comments. I guess there is a misunderstanding in the second question. There, I mean $Sp(n)\times (Sp(1).T^1)$ not $(Sp(n)\times (Sp(1)).T^1$. Could you introduce me a good reference for what you,ve used in your comments, for example "commuting ring of an irreducible representation", please. –  Nrd-Math Jan 3 '13 at 11:05
@Nerd-Math: Look up Schur's Lemma; it's basic. Maybe I don't understand what your $\mathrm{Sp}(1).T^1$ means, but I assume that you mean a subgroup of $\mathrm{SO}(4n)$ that commutes with the elements of $\mathrm{Sp}(n)$ and that contains both a copy of $\mathrm{Sp}(1)$ and a separate torus $T^1$ (i.e., a circle). However, the ring of $\mathbb{R}$-linear maps of $\mathbb{H}^n$ to itself that commute with all the elements of $\mathrm{Sp}(n)$ (i.e., its commuting ring) is the linear maps that come from multiplying by elements of $\mathbb{H}$ on the right, so the extra $T^1$ is not possible. –  Robert Bryant Jan 3 '13 at 12:45
Pardon me if I am wrong. Your argument implies $Sp(n)\times Sp(1).T^1$ is not a Lie subgroup of $SO(4n)$. Isn't it? whereas the question is about $SO(4n)\times T^1$. –  Nrd-Math Jan 3 '13 at 16:55
@Nerd-Math: Oh, sorry. I didn't read the question carefully. The answer is still 'no' because such an inclusion would cause an embedding of $\mathrm{Sp}(n)\times\mathrm{Sp}(1)$ as a subgroup of $\mathrm{SO}(4n)\times T^1$, and this is not possible, since it would have to be a subgroup of $\mathrm{SO}(4n)$, which is clearly impossible. –  Robert Bryant Jan 3 '13 at 17:52
I'm sorry for this probably easy question: How does an embedding of $Sp(n)\times Sp(1)$ as a subgroup of $SO(4n)\times T^1$ yield being a subgroup of $SO(4n)$? –  purelymath Jan 11 '13 at 11:10

For all $n,m\ge 0$ we have a homomorphism $Sp(n) \times Sp(m)/(\pm \mathrm{Id})\rightarrow SO(4nm)$. This is just the tensor product of the two defining representations.

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More precisely, it's the 'quaternionic tensor product'. In the usual representation, one has $\mathrm{Sp}(n)\subset\mathrm{SO}(4n)$, so the naïve tensor product would map $\mathrm{Sp}(n)\times\mathrm{Sp}(n)$ into $\mathrm{SO}(4m\cdot 4n)=\mathrm{SO}(16mn)$, which is not what you want. The point is that the commuting ring of $\mathrm{Sp}(n)\subset\mathrm{SO}(4n)$ acting on $\mathbb{R}^{4n}$ is the quaternions $\mathbb{H}$, and one can use this to construct an invariant subspace of dimension $4mn$ in the tensor product $\mathbb{R}^{4m}\otimes \mathbb{R}^{4n}\simeq \mathbb{R}^{16mn}$. –  Robert Bryant Jan 2 '13 at 13:00
(correction) Of course, I meant to type $\mathrm{Sp}(m)\times\mathrm{Sp}(n)$ in my comment above instead of $\mathrm{Sp}(n)\times\mathrm{Sp}(n)$. –  Robert Bryant Jan 2 '13 at 16:48