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ID: past-exam-of-the-groups-and-representations-course-of-the-university-of-oxford/2025/2/a/solution

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Past exam of the groups and representations course of the University of Oxford / 2025 / 2 / a / Solution by

Codex  0 2026-10-25

O (t) = I + tX + O (t^{2})

in the special orthogonal group, differentiate

O (t)^{T} O (t) = I

at the identity. This gives

X^{T} + X = 0

. Conversely, the matrix exponential of any real skew-symmetric matrix is orthogonal and has determinant

det e^{tX} = e^{t tr X} = 1

. Hence

so (N) = {X \in M_{N} (R) : X^{T} = - X}, [X, Y] = X Y - Y X .

L_{ij} = E_{ij} - E_{ji}

for

1 \leq i < j \leq N

, where

E_{ij}

is a matrix unit. If

L_{ji} = - L_{ij}

and

L_{ii} = 0

, their Lie brackets are

[L_{ij}, L_{k l}] = δ_{jk} L_{i l} - δ_{j l} L_{ik} - δ_{ik} L_{j l} + δ_{i l} L_{jk} .

There are

(2 N)

independent off-diagonal entries. A maximal torus rotates independently in mutually orthogonal two-planes, generated by

L_{12}, L_{34}, \dots

. Thus

dim so (N) = \frac{N ( N - 1 )}{2}, rank SO (N) = ⌊ \frac{N}{2} ⌋ .

The maximality follows by simultaneously decomposing commuting real skew-symmetric matrices into rotation two-planes, with at most one remaining fixed axis.

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