For a Matrix Lie group, consists of derivatives of differentiable curves in with , viewed inside as a real vector space. At the identity, multiplying two such curves adds their derivatives, and reparametrizing a curve multiplies its derivative by any real scalar. Thus is a real vector subspace.
If and , conjugating a curve with derivative proves . Let have and . Differentiating the curve inside the finite-dimensional, hence closed, vector subspace givesThe matrix commutator is real bilinear, antisymmetric, and satisfies the Jacobi identity by associativity of matrix multiplication. This proves that is a real Lie algebra.
The exponential-map theorem for a linear Lie group says whenever . Applying it to every real multiple of gives for every real . Conversely that curve has derivative at zero, so . Hence
Here the orthogonal group is the real group . Differentiating at zero gives . Conversely a real skew-symmetric matrix satisfies . ThereforeFor this is not a complex Lie subalgebra of : multiplying a nonzero real member by leaves this real subspace. For 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.
Differentiating the affine block form givesConversely the exponential of such a block matrix has bottom row and invertible upper-left block , so it remains in the affine group. Its commutator corresponds toThe space is closed under multiplication by and its bracket is complex bilinear. Thus it is a complex Lie algebra of dimension , and its real dimension is .
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