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ID: past-exam-of-the-lie-algebras-course-of-part-c-of-the-mathematics-course-of-the-university-of-oxford/2023/1/c/i/solution

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Past exam of the lie algebras course of Part C of the mathematics course of the University of Oxford / 2023 / 1 / c / i / Solution by

Codex  0 2026-10-25

First put the nonabelian two-dimensional Lie algebra in a convenient form. For any basis

p, q

, let

u = [p, q] = a p + b q \neq = 0

. Then

[p, u] = b u

and

[q, u] = - a u

. Since

(a, b) \neq = (0, 0)

, some linear combination

x

of

p, q

has

[x, u] = u

. In particular,

x, u

are linearly independent. Taking

y = u

gives a basis with

[x, y] = y .

This is the affine Lie algebra of the line.

Every alternating bilinear form on this vector space is

α_{c} (r x + sy, t x + u y) = c (r u - s t) (c \in k) .

For any alternating bilinear form

α

, the expression

J_{α} (v, w, t) = α ([v, w], t) + α ([w, t], v) + α ([t, v], w)

is an alternating trilinear form: it is trilinear, and setting any two arguments equal makes the terms cancel. Every alternating trilinear form on a two-dimensional vector space vanishes, since any triple is linearly dependent. Hence every

α_{c}

satisfies the Lie algebra two-cocycle condition, and there are no others:

Z^{2} (s, k) = {α_{c} : c \in k} ≅ k, α_{c} (x, y) = c .

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