A consequence of Schur's lemma for two equivalent irreducible representations of a finite group $G$ on a complex vector space $V$, $r^1(G,V)$ and $r^2(G,V$), is that linear $G$-morphisms between $r^1$ and $r^2$ must be of the form $cI$ where c is a complex scalar.

But what goes wrong if we try to use a "change of coordinate" matrix as the $G$-morphism?

As an explicit example, consider any two-dimensional irreducible representation. Then consider a standard rotation matrix $T_r$ in terms of some angle $\theta$. Through conjugation, I can map the matrix representation of any element of $G$ in $r^1$ to another matrix, and let that be the matrix representation of the same group element for $r^2$ (I'm constructing $r^2$ this way). So by construction we have $r^2 T_r = T_r r^1$.

And I would think that this same matrix $T_r$ could be used for all group elements, because conceptually it amounts to redrawing the coordinate axes, which shouldn't affect the way group actions behave.

But clearly there's a mistake somewhere in my reasoning here. An explicit counter-example could be very helpful.

Edited addition:

I'm reframing my question by using the explicit example of $G = S_3$ and $V = \mathbb{C}^2$.

I may define $\rho^1$ as

$e \mapsto \begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}$, $(123) \mapsto\begin{bmatrix} -1/2 & -\sqrt{3}/2\\ \sqrt{3}/2 & -1/2\end{bmatrix}$, $(132) \mapsto \begin{bmatrix} -1/2 & \sqrt{3}/2\\ -\sqrt{3}/2 & -1/2\end{bmatrix}$,

$(12) \mapsto \begin{bmatrix}1 & 0\\0 & -1\end{bmatrix}$, $(13) \mapsto \begin{bmatrix} -1/2 & \sqrt{3}/2\\ \sqrt{3}/2 & 1/2\end{bmatrix}$, $(23) \mapsto \begin{bmatrix} -1/2 & -\sqrt{3}/2\\ -\sqrt{3}/2 & 1/2\end{bmatrix}$

Then I define $f = \begin{bmatrix}\cos{\theta} & \sin{\theta}\\ -\sin{\theta} & \cos{\theta}\end{bmatrix}$ for any real value of $\theta$, and define $\rho^2$ as

$$ \rho^2(g)= f \, \rho^1(g) \, f^{-1}, \qquad g \in S^3.$$

I've checked that $\rho^2$ is still a representation of $S^3$ , i.e. it obeys the group element multiplication table for $S^3$. I also know $\rho^1$ and $\rho^2$ are irreducible. So, it seems to me I am in conflict with part (2) of Schur's lemma, which I copy below from the textbook by Serre:

Let $\rho^1: G \to \mathbf{GL}(V_1)$ and $\rho^2: G \to > \mathbf{GL}(V_2)$ be two irreducible representations of $G$, and let $f$ be a linear mapping of $V_1$ into $V_2$ such that $\rho_s^2 \circ$ $f$ = $f \circ \rho_s^1$ for all $s \in G$. Then:

(1) If $\rho_1$ and $\rho_2$ are not isomorphic, we have $f = 0$

(2) If $V_1 =V_2$ and $\rho^1 = \rho^2$, $f$ is a homothety (i.e., a scalar multiple of the identity).

How is there no conflict with (2), given the example I posed above?

  • $\begingroup$ What is saying Schur's lemma ? If $(r_i,V_i)$ are irreducible $G$-modules then for any $v_i\in V_i-0$, the $r_i(g)v_i,g\in G$ contain a basis of $V_i$, and a $G$-module homomorphism $T:V_1\to V_2$ will be of the form $T(r_1(g) v_1)= r_2(g)v_2$, if $T$ is well-defined then the two modules are isomorphic and given $v_1$, any $v_2\in V_2$ works. $\endgroup$ – reuns Dec 22 '19 at 0:42
  • $\begingroup$ @reuns I'm sorry, I don't understand. My question is about $T$, and I don't see how this comment demonstrates why the $T$ I proposed is not an acceptable $G$-morphism in all cases. My understanding of Schur's lemma is that for the conditions I stated, $T = c I$. I have seen proofs of Schur's lemma, but I don't understand what goes wrong in the example I posed $\endgroup$ – kleingordon Dec 22 '19 at 0:50
  • $\begingroup$ Do you understand that irreducible implies the $r_i(g) v_i$ form a basis of $V_i$ ? If so then pick some $v_1,v_2$ and define your $T$ in the corresponding basis. $\endgroup$ – reuns Dec 22 '19 at 0:57
  • $\begingroup$ I'll think over what you've said, although right now I still don't see how it will ultimately will address my question $\endgroup$ – kleingordon Dec 22 '19 at 1:00
  • $\begingroup$ Part (2) in the yellow box does state that If $V_1=V_2$ and $\rho^1=\rho^2$, then $f$ is a homotethy. In your example $V_1=V_2$, but the representations are different. $\endgroup$ – Jyrki Lahtonen Dec 22 '19 at 5:13

A summary of the comments/chat:

  • Yes, a change of bases transformation is $G$-linear as described in the question.
  • But then you also change (one of) the representation(s), so the conditions of part (2) no longer hold.
  • Schur's lemma survives in the form that the space of $G$-linear mappings between two irreducible complex representations is $1$-dimensional. In the OP's example the space of $G$-linear transformations between $\rho^1$ and $\rho^2$ consists of scalar multiples of that rotation.

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