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The Lie algebra of $SL_n(\mathbb C)$ are the matrices where the trace is $0$. But what is the Lie algebra of $SL_n(\mathbb H)$ where $\mathbb H$ is the quaternions?

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up vote 4 down vote accepted

The obvious canditate for $\mathfrak{sl}_2(H)$ is the space of $2\times 2$ matrices $\begin{pmatrix}a&b\\c&d\end{pmatrix}$ with quaternion entries such that $a+d=0$, with bracket the commutator of matrices, but... that is not a Lie algebra.

For example, the trace of the commutator of $\begin{pmatrix}i&0\\0&-i\end{pmatrix}$ and $\begin{pmatrix}j&0\\0&-j\end{pmatrix}$ is not zero.

The big problem, really, is that you have to decide what you mean by $SL_2(H)$. There is no determinant... (There is the Dieudonné determinant, though)

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I think the easiest way to define $SL_n(\mathbb H)$ is as the group generated by the upper and lower unitriangular matrices with entries in $\mathbb H$ (for $n=1$ one should throw in the multiplicative commutators, i.e., norm $1$ quaternions, I think). This should coincide with the definition by the Dieudonné determinant, and in any case produce a group of real codimension $1$ in $GL_n(\mathbb H)$. – Marc van Leeuwen Jan 7 '12 at 13:28

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