Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 1 / 19I (Representation Theory)

Schur lemma over $\mathbb{C}$ states that an intertwining operator $T:V\to W$ between finite-dimensional irreducible complex representations is either zero or an isomorphism. In particular, every intertwining endomorphism of $V$ is a scalar multiple of the identity.

Indeed, $\ker T\subseteq V$ and $\mathrm{im}T\subseteq W$ are invariant subspaces. If $T\ne 0$, irreducibility gives $\ker T=0$ and $\mathrm{im}T=W$, so $T$ is an isomorphism. For an endomorphism $T:V\to V$, the fundamental theorem of algebra gives an eigenvalue $\lambda \in \mathbb{C}$. Then $T-\lambda I$ is an intertwiner with nonzero kernel, so the first part forces $T-\lambda I=0$. Hence $T=\lambda I$, completing the Proof of Schur lemma.

Let $G=C_3=⟨g⟩$ and let $e_0,e_1,e_2$ be the standard basis of its real regular representation, with $g{e}_j=e_{j+1}$ modulo three. SetBoth are invariant, they are orthogonal, andThe line $U$ is the trivial representation. On the plane $W$, the generator acts as a rotation through $2\pi /3$. Such a rotation has no real eigenline, so $W$ has no nonzero proper invariant real subspace and is irreducible.

The two irreducible real modules $U$ and $W$ are nonisomorphic, so the kernel-image argument from part (a), which is valid over any field, givesEvery endomorphism of the one-dimensional trivial module is scalar:

Identify the oriented plane $W$ with $\mathbb{C}$, so that $g$ acts by multiplication by $e^{2\pi i/3}$. A real-linear map commutes with this rotation exactly when it is multiplication by a complex number. Equivalently,where $J$ is rotation through $\pi /2$. Taking $J$ gives a nonscalar intertwining endomorphism of an irreducible real representation. This is the Failure of the scalar conclusion of Schur lemma over the real numbers.

Let $e_0,\dots ,e_{n-1}$ be the standard basis and let the generator $g$ send $e_j$ to $e_{j+1}$ modulo $n$. The constant vectorspans a trivial module. Since $n$ is even, the alternating vectorspans a sign module on which $g$ acts by $-1$.

For each $1\le k<n/2$, define the real discrete Fourier modesThe plane $W_k=\mathrm{span}\{c_k,s_k\}$ is invariant, and $g$ acts on it by rotation through $2\pi k/n$. Since this angle is neither zero nor $\pi$, $W_k$ is irreducible over $\mathbb{R}$. The traces $2\cos (2\pi k/n)$ are distinct for $1\le k<n/2$, so these modules are pairwise nonisomorphic. Orthogonality of the discrete Fourier transform givesThe dimensions sum to $1+1+2(n-2)/2=n$, so this is the whole regular module: two nonisomorphic one-dimensional summands and $(n-2)/2$ nonisomorphic two-dimensional summands.

When $n$ is odd there is no alternating sign mode. The same cosine-sine construction gives invariant irreducible planes $W_k$ foragain pairwise nonisomorphic because their generator traces are distinct. The real Fourier decomposition isIts dimensions sum to $1+2(n-1)/2=n$. Thus the real regular representation of a finite cyclic group has one one-dimensional irreducible summand and $(n-1)/2$ nonisomorphic two-dimensional irreducible summands.