# Recover a matrix from an irreducible representation.

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.

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