A question about adjoint representation of a compact and connected Lie group.

Question

I have already uploaded the question before, but since it seems that the previous question is so verbose, I rewrite it. My question: Let $G$ be a compact and connected Lie group, and consider an adjoint representation of such Lie group, $Ad(g)$ whose matrix component is denoted by $(\alpha_{ij}(g))$. Then why the determinant of adjoint representation matrix is only $+1$,i.e $det(Ad(g))=+1$.

Answer 1

Recall that a compact connected Lie group $G$ has a real Lie algebra $\mathfrak{g}$, and the adjoint action of G is naturally acting on $\mathfrak{g}$ by $Ad(g): \mathfrak{g} \rightarrow \mathfrak{g}$, $ X \mapsto gXg^{-1}$ for $g \in G, X \in \mathfrak{g}$.

Suppose we have a norm on $\mathfrak{g}$, for example, $\langle X,X \rangle = Tr(X X)$ (trace form), then $\langle Ad(g)X, Ad(g)X \rangle = Tr(gXg^{-1} gXg^{-1}) = Tr(X X) = \langle X,X \rangle$.

Since $\mathfrak{g}$ is real, this means $Ad(g)$ must be orthogonal transformation, so the determinant of $Ad(g)$ has to be $1$ or $-1$.