Consider the $C_3$-representation $\rho$ on $\mathbb{R}^2$ that sends a generator of $C_3$ to the counter-clockwise rotation of the plane by $120$ degrees. Consider the morphisms $A = Hom_{C_3}(\rho,\rho)$ of this representation to itself and think of it as a ring with multiplication being composition. Show that we have an $\mathbb{R}$-linear ring isomorphism from A to $\mathbb{C}$.

For any $T\in A, g\in C_3, T \rho_g = \rho_g T$. We need to find a bijective map $f$ that sends $T\in A$ to some complex number $c$ so that $f(T+U) = f(T)+f(U), f(1) = 1, f(TU) = f(T)f(U)$. Now $C_3$ is an abelian group, but I don't think $\rho$ is necessarily irreducible (e.g. any equilateral triangle in $\mathbb{R}^2$ is a $C_3$-invariant subspace). I'm not sure if it'll be useful to use a corollary of Schur's theorem, which states that for any two irreducible representations $\rho$ and $\psi$, if $T \in Hom(\rho,\psi)$, then T is a scalar multiple of the identity if $\rho=\psi$.


1 Answer 1


The representation $\rho$ is irreducible since the linear map which rotates the plane by $120$ degrees does not have any eigenvectors. Equilateral triangles in the plane are not subspaces!

Now, let $x$ be a generator for $C_3$. With respect to the standard basis, the matrix for $\rho_x$ is $$X = \begin{bmatrix} -1/2 & -\sqrt{3}/2 \\ \sqrt{3}/2 & -1/2 \end{bmatrix}.$$ By direct computation, matrices which commute with $X$ are of the form $\begin{bmatrix} a & -b \\ b & a \end{bmatrix}$. So $A$ consists of linear maps of the form $aI + bT$, where $I$ is the identity and $T$ is rotation by $90$ degrees. The map $aI + bT \mapsto a + bi$ defines an isomorphism $A \to \mathbb{C}$.


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