Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question

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).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.