The Lie algebra of the generalized unitary group $\{g \in GL_n(\mathbb{C}) : gS\bar{g}^t=S\}$ is $\{XS+S\overline{X}{}^t=0\}$

Question

Let $ S \in M_n(\mathbb{C}) $ be a square matrix and let $ X$ be in the Lie algebra $\mathbb{\mu(S)} $ of the generalized unitary group, $$U(S):=\{g \in GL_n(\mathbb{C}); gS\bar{g}^t=S\} .$$ Show that $$XS+S\overline{X^t}=0 .$$

I need to find the relationship between elements of the Lie algebra and the Lie group

I know that the Lie algebra of a linear Lie group can be characterized in terms of the exponential matrix as follows:

$\mathbb{g} \rightarrow G$ through the map $X \rightarrow \exp(X)=\sum{\frac{X^k}{k!}}$

So in the above case $X \in \mathbb{\mu}$ and $\exp(X)$

Where could I go from here?

Answer 1

It's typically difficult to work directly with the definition of the matrix exponential function. Instead, we can use a standard approach that works with computing the Lie algebra of more or less any matrix Lie group.

Hint As usual, we identify the Lie algebra $\mathfrak u(S)$ with the tangent space $T_I U(S)$. If $X \in \mathfrak u(S)$, there is a curve $\gamma : J \to U(S)$ such that $\gamma(0) = I$ and $\gamma'(0) = X$. Since $\gamma$ has image in $U(S)$, by definition we have $$\gamma(t) S \overline{\gamma(t)}{}^T = S .$$ Now, differentiate.

Answer 2

First, you have the Lie group G =GL(n,C) of invertible complex matrices, whose Lie algebra is the vector space \mathfrak{g}=C^{2n\times n} of all matrices, endowed with the Lie bracket AB-BA. Cartan theorem gives an easy way to compute the Lie subalgebra \mathfrak{h} of a Lie subgroup H. It is the set of elements X of \mathfrak{g} such that \exp(tX) belongs to H for all t. Be careful because \exp is not injective.