Past exam of the lie Groups and Differentiable Manifolds course of the University of Oxford / 2008 / A / 1 / b / i

Here the orthogonal group is the real group $O_n(\mathbb{R})$. Differentiating $g(t{)}^{T}g(t)=I$ at zero gives $X^T+X=0$. Conversely a real skew-symmetric matrix satisfies $(e^{tX}{)}^T{e}^{tX}=e^{-tX}{e}^{tX}=I$. ThereforeFor $n\ge 2$ this is not a complex Lie subalgebra of $M_n(\mathbb{C})$: multiplying a nonzero real member by $i$ leaves this real subspace. For $n=1$ the zero algebra is the trivial complex Lie algebra as well. The complex orthogonal group would be a different group from the real orthogonal group used here.