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We define the skew-symmetric hermitian inner product $$\phi(x,y)=\bar{x}^tjy$$ over $\mathbb{H}^2$ and are asked to calulate the Lie algebra of the group $G\le GL(2,\mathbb{H})$ of automorphisms of $\phi$.

My solution is the following:

We need to find the Lie algebra of $\{X\in GL(2,\mathbb{H})|\bar{X}^tjX=X\}$ so if we take $a(t)\in G$ with $a(0)=1$ and differentiate then we get the condition that:


So the Lie algebra $\mathfrak{g}=\{X\in GL(2,\mathbb{H})|\bar{X}^t=jXj\}$.

I am then asked to calculate the real dimension of this which is where I am having trouble.

If we take $X=\left ( \begin{array}{cc} a & b \\ c & d \end{array} \right )=\left ( \begin{array}{cc} j\bar{a}j & j\bar{c}j \\ j\bar{b}j & j\bar{d}j \end{array} \right )$.

If I now want to calculate the real dimension of this what do I do?

Thanks for any help.

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First of all, original poster has misleading ideas about GL (general linear group) and possibly confuses it with U (unitary group). Automorphisms of ℂn that preserve product give U(n) that contains much less transformations than GL(n, ℂ) has (exercise: check $n = 1$ to feel the difference).

Second, it might be reasonable to define quaternionic inner product as ${\bar x}^{\mathsf T}y$, but a skew-symmetric bilinear form ${\bar x}^{\mathsf T}jy$ over quaternions is degenerate for $n>1$: check $$n = 2,\quad x = y =\begin{pmatrix}1\\k\end{pmatrix}.$$ Respective group of automorphisms hardly makes any sense, and cannot be neither GL(2, ℍ) nor some hypothetical quaternionic unitary group.

Moreover, the form ${\bar x}^{\mathsf T}qy$ for any $q∈{\mathbb H}$ such that $\operatorname{Re}q = 0$ is degenerate as well (exercise).

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