Past exam of the algebra course of the University of Oxford / 2009 / A / 1 / b / i

Table of contents
- Solution i

Solution

For

h \in G

, right-multiplying by

D_{2} (h)^{- 1}

gives

D_{1} (h) T D_{2} (h)^{- 1} = \sum_{g \in G} D_{1} (h g) B D_{2} (g^{- 1} h^{- 1}) .

Make the bijective substitution

k = h g

, so

g^{- 1} h^{- 1} = k^{- 1}

. The sum becomes

\sum_{k} D_{1} (k) B D_{2} (k^{- 1}) = T

. Hence

D_{1} (h) T = T D_{2} (h) (h \in G) .

Thus

T

is an intertwining operator from the second group representation to the first. This is the Schur averaging of rectangular matrices construction.

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