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I’m trying to check that $\mathfrak{su}(m,n), \mathfrak{sp}(n,\mathbb R), \mathfrak{so}^*(2n)$ are closed under the conjugate transpose as described in Anthony Knapp’s “Lie Groups Beyond an Introduction”. On page 60, Knapp asserts that this follows from the definitions of these groups as

$$\mathfrak{su}(m,n) = \{X\in\mathfrak{sl}(m+n,\mathbb C) \mid X^*I_{m,n} + I_{m,n} X = 0\}$$ $$\mathfrak{sp}(n,\mathbb R) = \{X\in\mathfrak{gl}(2n,\mathbb R) \mid X^tJ_{n,n} + J_{n,n} X = 0\}$$ $$\mathfrak{so}^*(2n) = \{X\in\mathfrak{su}(n,n) \mid X^tI_{n,n}J_{n,n} + I_{n,n}J_{n,n} X = 0\}$$ and the fact that $I_{m,n}^*=I_{m,n}$ and $J_{n,n}^* = - J_{n,n}$.

I thought that the argument should be simple. However, to show for example that $X^*\in \mathfrak{su}(m,n)$ if $X\in \mathfrak{su}(m,n)$ I need to show that $XI_{m,n} + I_{m,n} X^* = 0$ if $X^*I_{m,n} + I_{m,n} X = 0$. Unfortunately, this doesn’t follow from simply taking the conjugate transpose of $X^*I_{m,n} + I_{m,n} X = 0$, which is

\begin{align}0&=(X^*I_{m,n} + I_{m,n} X)^*\\ &= (X^*I_{m,n})^* + (I_{m,n} X)^*\\ &= I_{m,n}^*(X^*)^* + X^*I_{m,n}^* \\ &= I_{m,n}X + X^*I_{m,n},\end{align}

in other words, the same equation. How can I check that $\mathfrak{su}(m,n), \mathfrak{sp}(n,\mathbb R), \mathfrak{so}^*(2n)$ are closed under the conjugate transpose?

Rodrigo
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  • I'm not sure what proof Knapp has in mind but you can prove these individually, relatively easily. For example $\mathfrak{su}(m,n)$ comprises $2\times 2$ block matrices whose block diagonal part is anti-Hermitian and block off-diagonal part is Hermitian. From this, it is clear that that it is preserved under conjugate transpose. – Callum Mar 18 '24 at 23:25

1 Answers1

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The proof may turn out to be easier when making a detour through the associated Lie groups, as it is shown below for the first case.

Let $X \in \mathfrak{su}(m,n)$ be an element of the Lie algebra. Then, $A(t) := e^{tX} \in SU(m,n)$ defines a curve in the corresponding Lie group, in such a way that $A^*I_{m,n}A = I_{m,n}$. Now, let's consider $B(t) := e^{tX^*}$. It satisfies :
$$ B^*I_{m,n}B = e^{tX}I_{m,n}e^{tX^*} = AI_{m,n}A^* = AI_{m,n} \cdot I_{m,n}(I_{m,n}A)^{-1} = AI_{m,n}^2A^{-1}I_{m,n}^{-1} = I_{m,n}, $$ since $I_{m,n}^2 = 1$ and $I_{m,n}^{-1} = I_{m,n}$. In consequence, one concludes that $B \in SU(m,n)$, hence $X^* = \dot{B}(0) \in \mathfrak{su}(m,n)$ in the end.

Abezhiko
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