 Solution

ID: past-exam-of-the-groups-and-representations-course-of-the-university-of-oxford/2023/3/b/solution

Partition

M

into the same five-by-five blocks as

R

. The equation for block

(i, j)

M_{ij} R_{j} (g) = R_{i} (g) M_{ij}

, so each block is an intertwiner. By Schur's lemma, blocks between the two inequivalent irreducible representations vanish, whereas blocks between identical copies are arbitrary scalar multiples of the relevant identity matrix. Thus the commutant of a semisimple representation is

M = a_{11} I_{n_{1}} a_{21} I_{n_{1}} a_{31} I_{n_{1}} 00 a_{12} I_{n_{1}} a_{22} I_{n_{1}} a_{32} I_{n_{1}} 00 a_{13} I_{n_{1}} a_{23} I_{n_{1}} a_{33} I_{n_{1}} 00 000 b_{11} I_{n_{2}} b_{21} I_{n_{2}} 000 b_{12} I_{n_{2}} b_{22} I_{n_{2}}, a_{ij}, b_{ij} \in C .

Equivalently,

M = (A \otimes I_{n_{1}}) \oplus (B \otimes I_{n_{2}})

, where

A \in M_{3} (C)

and

B \in M_{2} (C)

. Substitution proves every such matrix commutes with every

R (g)

. There is no invertibility requirement on

M

, and its solution space has complex dimension

3^{2} + 2^{2} = 13

, even if

n_{1} = n_{2}

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