# Schur's lemma: why aren't “coordinate change” transformations allowed between equivalent irreducible representations?

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.

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?

• 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. – reuns Dec 22 '19 at 0:42
• @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 – kleingordon Dec 22 '19 at 0:50
• 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. – reuns Dec 22 '19 at 0:57
• I'll think over what you've said, although right now I still don't see how it will ultimately will address my question – kleingordon Dec 22 '19 at 1:00
• 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. – Jyrki Lahtonen Dec 22 '19 at 5:13

• Yes, a change of bases transformation is $$G$$-linear as described in the question.
• 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.