# existence of a ring isomorphism from A to the ring of complex numbers

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 $$\rho=\psi$$.

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 $$A \to \mathbb{C}$$.