TITLE: Maximal compact subgroup of $\mathrm{SL}(2,\mathbb{H})$
QUESTION [9 upvotes]: $\DeclareMathOperator\Spin{Spin}\DeclareMathOperator\Sp{Sp}\DeclareMathOperator\SL{SL}$The exceptional isomorphism $\Spin(5,1)\simeq \SL(2,\mathbb{H})$ is well-known, and I can find references that say the maximal compact of $\Spin(5,1)$ is $\Spin(5) \simeq \Sp(2)$. So I know the answer to the question, but not the how or why. In particular, is there a proof that $\Sp(2)$ is maximal compact in $\SL(2,\mathbb{H})$ not via the exceptional isomorphisms? Perhaps some sort of analogue of Gram-Schmidt or other explicit factorisation?

REPLY [3 votes]: Here is a different argument from Robert's.
Observation: given a compact Lie group $G$ and a proper closed subgroup $H$ the inclusion $H\to G$ is not a homotopy equivalence, else the homogeneous space $G/H$ would be a closed contractible manifold.
Next, it's a general fact that the inclusion of a maximal compact subgroup $K$ into a semisimple Lie group $G$  is a homotopy equivalence and the quotient space is contractible. Putting these two things together immediately implies that
if $G$ is a semisimple Lie group and $H\subset G$ is a compact subgroup such that $G/H$ is contractible then $H$ is maximal compact in $G$.
It remains to observe that $SL(2,\mathbb H)/Sp(2)$ is contractible.
From Robert's answer the homogeneous (in fact, symmetric) space $SL(2,\mathbb H)/Sp(2)$ can be identified with  the 5-dimensional hyperbolic space which is of course contractible.
Alternatively this can also be seen as follows.
Look at the standard transitive action of $Sp(2)$ on the unit sphere $S^7$ in $\mathbb H^2$. The stabilizer of $(1,0)$ is easily seen to be equal to $Sp(1)=S^3$.
Similarly look at the action of $SL(2,\mathbb H)$ on $\mathbb H^2\setminus (0,0)$ ( which deformation retracts onto   $S^7$). the stabilizer $K$  of $(1,0)$ consists of matrices of the form $\begin{pmatrix} 1&a\\0&b\end{pmatrix}$ where $|b|=1$ and $a\in \mathbb H$ is arbitrary. So $K=S^3\times \mathbb R^4$ topologically.
Comparing long exact  homotopy sequences of the homogeneous space bundles $Sp(1)\to Sp(2)\to Sp(2)/Sp(1)=S^7$ and $K\to SL(2,\mathbb H)\to SL(2,\mathbb H)/K=\mathbb H^2\setminus (0,0)$ and using the natural map between them it now immediately follows by the 5-lemma that the inclusion $Sp(2)\subset SL(2,\mathbb H)$ is a homotopy equivalence. Hence by above  $Sp(2)$ is maximal compact in $SL(2,\mathbb H)$.
By induction this generalizes to show that $Sp(n)$ is maximal compact in $SL(n,\mathbb H)$.