Lie Algebra of $SL_n(\mathbb H)$

Question

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?

Answer 1

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)

Answer 2

The group $\mathrm{SL}(n, \mathbb{H})$ is defined here:

• Wikipedia, GL(n,H) and SL(n,H).

It works like this:

The algebra of quaternions is a subalgebra of the $2 \times 2$ complex matrices in a standard way. This lets us regard $n \times n$ quaternionic matrices as certain special $2n \times 2n$ complex matrices. This in turn lets us define a complex-valued determinant of an $n \times n$ quaternionic matrix, called the Study determinant, and also a complex-valued trace.

The $n \times n$ quaternionic matrices for which the Study determinant is 1 form the group $\mathrm{SL}(n,\mathbb{H})$. The Lie algebra of $\mathrm{SL}(n,\mathbb{H})$ then consists of $n \times n$ quaternionic matrices for which the complex-valued trace is 0.

The Study determinant is the square of the Dieudonné determinant, as shown here:

• Helmer Aslaksen, Quaternionic determinants, The Mathematical Intelligencer 18 (1996), 57–65.

But the Dieudonné determinant is nonnegative, so in fact the Study determinant also takes only nonnegative real values, and $\mathrm{SL}(n,\mathbb{H})$ is also the group of quaternionic $n \times n$ matrices with Dieudonné determinant 1.

Answer 3

The Lie algebra of $\mathrm{GL}_n(\mathbb H)$ is just the full matrix space $M_n(\mathbb H)$ with commutator bracket. Viewed as Lie algebra this way, it's called $\mathfrak{gl}_n(\mathbb H)$).

One can show that this is a reductive Lie algebra. Since its centre $\mathfrak z$ is easily seen to be exactly the real scalar matrices, i.e. $\mathrm{diag}(a,a, ..., a)$ with $a\in \mathbb R$, its derived subalgebra

$$\mathfrak{sl}_n(\mathbb H):= [\mathfrak{gl}_n(\mathbb H), \mathfrak{gl}_n(\mathbb H)]$$

consists of those matrices where the real part of the trace is $0$. (Remember a quaternion $q = a +bi+ cj +dk$ has a real part $\mathfrak{Re}(x) := a$ and a pure quaternion part $bi+cj+dk$).

This has real codimension $1$ in $\mathfrak{gl}_n(\mathbb H)$ and corresponds to the Lie group $SL_n(\mathbb H)$ defined as the kernel of the Dieudonné determinant (1, 2).

This obviously avoids the problem in Mariano Suárez-Álvarez' answer. For example, the computation $[ \begin{pmatrix}i&0\\0&-i\end{pmatrix}, \begin{pmatrix}j&0\\0&-j\end{pmatrix}] = \begin{pmatrix}2k&0\\0&2k\end{pmatrix}$ happens without problem in $\mathfrak{sl}_2(\mathbb H)$. (While the commutator of two matrices with trace $0$ need not have trace $0$ necessarily, the commutator of two matrices whose traces have real part $0$ again has trace with real part $0$, and that's all we need.)

This also corresponds to the construction in the Wikipedia article cited in John Baez' answer, but with the correction alluded to in my comment: When we write a matrix $A \in M_n(\mathbb H)$ as matrix $$\tilde{A}= \pmatrix{X & -\overline{Y}\\Y & \overline{X}} \in M_{2n}(\mathbb C)$$ (with $X, Y \in M_n(\mathbb C)$), then the condition to be in $\mathfrak{sl}_n(\mathbb H)$ is not $Tr(X) = 0$ as claimed there, but rather $\mathfrak{Re}(Tr(X))=0$ or equivalently $Tr(\tilde{A})=0$.

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Note 1: It's worthwhile to check all this already in the non-trivial basic case $n=1$. We have $\mathrm{GL}_1(\mathbb H) = \mathbb H^*$, $\mathrm{SL}_1(\mathbb H) =$ the norm-1-quaternions (which is a real Lie group $\simeq SU_2$), and $\mathfrak{gl}_n(\mathbb H) = \mathbb H$ which splits, stable under commutators, as

$$ \mathbb H = \underbrace{\mathbb R}_{\text{centre}} \oplus \underbrace{\text{ pure quaternions }}_{\text{real Lie algebra } \simeq \mathfrak{su}_2 (\simeq \mathfrak{so}_3)}$$

And when written as complex matrices $\pmatrix{\alpha & -\bar{\beta}\\\beta& \bar{\alpha}}$, the subspace of pure quaternions is indeed that with $\alpha + \bar\alpha = 0$ or equivalently $\mathfrak{Re}(\alpha)=0$, not just the ones with $\alpha=0$.

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Note 2: All this generalizes in the best way possible to a base field $k$ (to be safe, $char(k)=0$ although much goes through at least if for $char(k)$ does not divide $n$) and a $k$-central division algebra $D$ (of degree $d:= \sqrt{\dim_k D}$ say). Then again one defines the Lie algebra $\mathfrak{sl}_n(D)$ as the derived subalgebra of the full matrix algebra $\mathfrak{gl}_n(D)$, which is of $k$-codimension $1$. It turns out to be an absolutely simple $k$-Lie algebra of type $A_m$, where $m = dn-1$, of $k$-rational rank $n-1$, with Satake-Tits diagram

For more details, see sections 4.4 and 4.5.2 of my thesis. (Note the errata for page 93.)

It matches the group case as presented in Tits, Jacques (1966), "Classification of algebraic semisimple groups", Algebraic Groups and Discontinuous Subgroups (Proc. Sympos. Pure Math., Boulder, Colo., 1965), Providence, R.I.: American Mathematical Society, pp. 33–62

And of course for the Hamilton quaternions, we just have $d=2$ and the above collapses to the second row in Onishchik / Vinberg's table 4 (they just exclude the compact form $\mathfrak{sl}_1(\mathbb H)$, which in their notation would be $p=0$ and as said above is just $\simeq \mathfrak{su}_2$, with diagram one black dot):