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 Incoming links: Past exam of the groups and representations course of the University of Oxford / 2024 / 2 / a

Past exam of the groups and representations course of the University of Oxford / 2024 / 2 / b / Solution 2026-10-25

Use the Hermitian

H_{a}

from a to record real weight coordinates:

H_{a} e_{j} = (w_{j})_{a} e_{j}

. Reading off the diagonal entries gives

w_{1} = (1, 0, 0), w_{2} = (- 1, 1, 0), w_{3} = (0, - 1, 1), w_{4} = (0, 0, - 1) .

These are weights in the simple-coroot coordinate system, not coordinates in an orthonormal basis. In the skew-Hermitian convention they mean

(i H_{a}) e_{j} = i (w_{j})_{a} e_{j}

. Intrinsically, the weight on

i diag (h_{1}, h_{2}, h_{3}, h_{4})

is the functional

h \mapsto h_{j}

, with

\sum_{j} h_{j} = 0

.

The conjugate representation acts by

U^{*}

, so a diagonal transformation with entries

e^{i h_{j}}

becomes one with entries

e^{- i h_{j}}

. Thus its four weights are

\overset{w}{ˉ}_{1} = (- 1, 0, 0), \overset{w}{ˉ}_{2} = (1, - 1, 0), \overset{w}{ˉ}_{3} = (0, 1, - 1), \overset{w}{ˉ}_{4} = (0, 0, 1) .

All have multiplicity one. In both representations the sum of the weights is zero, consistent with the vanishing trace of every Lie algebra generator.

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