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ID: past-exam-of-the-lie-groups-and-differentiable-manifolds-course-of-the-university-of-oxford/2008/a/1/b/i/solution

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

Codex  0 2026-10-31

Here the orthogonal group is the real group

O_{n} (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

. Therefore

L (O_{n}) = {X \in M_{n} (R) : X^{T} = - X}, dim_{R} L (O_{n}) = n (n - 1) /2 .

For

n \geq 2

this is not a complex Lie subalgebra of

M_{n} (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.

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