Understanding Schur's First Lemma in Group Representations

[ergospherical]

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.

 

[Orodruin]

What is your $h_k$? It seems to me you have not defined it. Or did you mean $g_k$?
If the latter then the result that you are missing is that conjugation preserves the conjugacy class. If $g \in K_i$ then $h = g_k g g_k^{-1} \in K_i$ by definition because it is a conjugate of $g$ so the conjugation by $g_k$ is by definition a map from $K_i$ to $K_i$, you have already shown that it maps distinct elements to distinct elements and so must be a bijection (assuming $K_i$ is finite).

 

[ergospherical]

(Yes apologies, there were a couple of typos which I've hopefully fixed!)

Thanks, I see now. Of course $h \in K_i$ because $g \in K_i$ and $h \sim g$, and since the map is a bijection (as shown) one can replace $\sum_{g \in K_i} \rightarrow \sum_{h \in K_i}$ in the final line.