- #1
ergospherical
- 1,019
- 1,299
- Homework Statement
- i) Show that the matrix ##B^{\nu}_i = \sum_{g\in K_i} D^{(\nu)}(g)##, consisting of the sum of the matrices of an irreducible ##[n_{\nu}]## representation which correspond to the elements of the conjugacy class ##K_i##, is a multiple of the identity.
- Relevant Equations
- N/A
So long as ##B^{\nu}_i## commutes with all the matrices in the irrep then the result follows from Schur's first lemma. So consider some element ##g_k## and form\begin{align*}
D^{(\nu)}(g_k) B^{\nu}_i D^{(\nu)}(g_k)^{-1} &= \sum_{g \in K_i} D^{(\nu)}(g_k) D^{(\nu)}(g) D^{(\nu)}(g_k)^{-1} \\
&= \sum_{g \in K_i} D^{(\nu)}(g_k g g_k^{-1})
\end{align*}If ##g_k g g_k^{-1} \equiv h \in K_i## then the right hand side would be nothing but ##B^{\nu}_i## (note that if ##g_k g g_k^{-1} = h## and ##g_k g' g_k^{-1} = h##, then it would follow that ##g' = g##, i.e. conjugating with ##g_k## would map each ##g \in K_i## to a distinct element ##h \in K_i##). The bit I can't figure is why ##h## is guaranteed to be in ##K_i##. An equivalence ##g \sim h## implies that there's some group element conjugating the two, but that depends on ##g## and ##h## and surely can't be a single element ##g_k## for every term in the sum? Wonder what I'm overlooking.
D^{(\nu)}(g_k) B^{\nu}_i D^{(\nu)}(g_k)^{-1} &= \sum_{g \in K_i} D^{(\nu)}(g_k) D^{(\nu)}(g) D^{(\nu)}(g_k)^{-1} \\
&= \sum_{g \in K_i} D^{(\nu)}(g_k g g_k^{-1})
\end{align*}If ##g_k g g_k^{-1} \equiv h \in K_i## then the right hand side would be nothing but ##B^{\nu}_i## (note that if ##g_k g g_k^{-1} = h## and ##g_k g' g_k^{-1} = h##, then it would follow that ##g' = g##, i.e. conjugating with ##g_k## would map each ##g \in K_i## to a distinct element ##h \in K_i##). The bit I can't figure is why ##h## is guaranteed to be in ##K_i##. An equivalence ##g \sim h## implies that there's some group element conjugating the two, but that depends on ##g## and ##h## and surely can't be a single element ##g_k## for every term in the sum? Wonder what I'm overlooking.
Last edited: