 Solution

ID: past-exam-of-the-lie-algebras-and-representation-theory-of-symmetric-groups-course-of-the-university-of-oxford/2008/a/1/c/solution

Induct on

i

, proving the claim for all

j

simultaneously. The case

i = 1

[L, L_{j}] = L_{j + 1}

. If it holds for

i

, Jacobi gives

[L_{i + 1}, L_{j}] = [[L, L_{i}], L_{j}] \subseteq [L, [L_{i}, L_{j}]] + [L_{i}, [L, L_{j}]] \subseteq [L, L_{i + j}] + [L_{i}, L_{j + 1}] \subseteq L_{i + j + 1} .

i

, so the induction is valid. Thus

[L_{i}, L_{j}] \subseteq L_{i + j}

for all positive indices.

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