Let $G$ be a finite group and $X: G\rightarrow\text{GL}_3(\mathbb{C})$ be an irreducible $3$-dimensional complex matrix representation of $G$. Suppose that $$ B=\frac{1}{|G|}\sum_{g\in G}X(g)AX(g)^{-1} $$ where $$ A=\begin{bmatrix} 1&-12&4\\ 0&5&3\\ -2& 1&3 \end{bmatrix}. $$

It is easy to calculate the trace of matrix $B$, that is $\text{Tr}(B)=\text{Tr}(A)=9$. But I don't know how we take advantage of the irreducibility of $X$ and use it to recover the matrix $B$? Any help is appreciated.

  • $\begingroup$ Yeap. $X$ is any irreducible representation. $\endgroup$ – Jay May 9 at 19:06
  • $\begingroup$ Please formulate explicitly a question. For the trivial group we have $B=A$. The question is to show this for all finite groups? $\endgroup$ – dan_fulea May 9 at 19:16
  • 1
    $\begingroup$ Hint: Show this operator $B$ commutes with $X(h)$ for all $h \in G$. $\endgroup$ – Nate May 9 at 19:18
  • $\begingroup$ @Nate Oh, I see. According to Schur's lemma, $B$ should have this form $\alpha I$. Then $\alpha=3$. $\endgroup$ – Jay May 9 at 19:31

I'm going to switch up the notation a little bit, and treat this with a little more generality. Let $\mathsf{k}$ be an algebraically closed field, with $G$ a finite group such that the characteristic of $\mathsf{k}$ does not divide $\left| G \right|$, and suppose that $\rho : G \to \operatorname{GL}(V)$ is a finite dimensional irreducible representation of $G$ on a representation space $V$, and we suppose also that the characteristic of $\mathsf{k}$ does not divide $\operatorname{dim}_{\mathsf{k}}(V)$. Then for $T \in \operatorname{GL}(V)$, we define $\pi_T \in \operatorname{End}(V)$ by

$$ \pi_T = \frac{1}{\left| G \right|}\sum_{g \in G} \rho(g)T\rho(g^{-1}). $$

Then we aim to show that $\rho(h) \circ \pi_T = \pi_T \circ \rho(h)$ for every $h \in G$. Well let $h \in G$, and then

\begin{align*} \rho(h) \circ \pi_T & = \frac{1}{\left| G \right|}\sum_{g \in G} \rho(h)\rho(g)T\rho(g^{-1}) \\ & = \frac{1}{\left| G \right|}\sum_{g \in G}\rho(hg)T\rho(g^{-1}) \\ & = \frac{1}{\left| G \right|}\sum_{g' \in G}\rho(g')T\rho(g'^{-1}h) \\ & = \frac{1}{\left| G \right|}\sum_{g' \in G}\rho(g')T\rho(g'^{-1})\rho(h) \\ & = \pi_T \circ \rho(h). \end{align*}

Thus $\pi_T$ is an $\mathsf{k}$-linear endomorphism of $V$ commuting with the representation $\rho$. Thus $\pi_T \in \operatorname{Hom}_{\mathsf{k}[G]}(V,V)$, and we call such a map an intertwining operator, and we say that $\pi_T$ intertwines $\rho$. Now, since $\mathsf{k}$ is algebraically closed, and $\rho$ is irreducible, Schur's Lemma states that $\operatorname{Hom}_{\mathsf{k}[G]}(V,V)$ is one-dimensional, and in particular every linear endomorphism of $V$ intertwining $\rho$ is a multiple of the identity map on $V$. Thus $\pi_T$ is a multiple of the identity $\operatorname{Id}_V$ on $V$, $\lambda_T$ say. Moreover, as you observed $\operatorname{trace}(\pi_T) = \operatorname{trace}(T)$. But then

$$ \operatorname{trace}(T) = \operatorname{trace}(\pi_T) = \operatorname{trace}(\lambda_T \operatorname{Id}_V) = \operatorname{dim}_{\mathsf{k}}(V)\lambda_T. $$

and so $\lambda_T = \frac{\operatorname{trace}(T)}{\operatorname{dim}_{\mathsf{k}}(V)}$. It follows that $\pi_T = \frac{\operatorname{trace}(T)}{\operatorname{dim}_{\mathsf{k}}(V)} \operatorname{Id}_V$.

Now, using this we conclude that in your example, $B$ is $3I_3$ where $I_3$ is the $3 \times 3$ identity matrix.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.