 Solution

ID: past-exam-of-the-representation-theory-of-semisimple-lie-algebras-course-of-the-university-of-oxford/2010/1/b/solution

For the intended module, index a basis by

v_{k} = a_{l - 2 k}

for

0 \leq k \leq l

, and use

H v_{k} = (l - 2 k) v_{k}, f v_{k} = v_{k + 1}, e v_{k} = k (l + 1 - k) v_{k - 1},

with

v_{- 1} = v_{l + 1} = 0

. This makes the basis endpoints distinct, identifies the lowering operator, and includes its missing raising action at

k = l

. These are the necessary indexing corrections to the printed display. The same definition for

l = 0

supplies the trivial module needed for classification.

Directly on

v_{k}

[H, f] v_{k} [H, e] v_{k} [e, f] v_{k} = - 2 f v_{k}, = 2 e v_{k}, = ((k + 1) (l - k) - k (l + 1 - k)) v_{k} = (l - 2 k) v_{k} .

These hold at the endpoints too, proving the three defining

sl_{2}

relations. Finally

Δ v_{k} = (4 k (l + 1 - k) + (l - 2 k)^{2} + 2 (l - 2 k)) v_{k} = l (l + 2) v_{k} .

Thus the Casimir element acts by that scalar on the whole module.

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