I'm studying basic representation theory using GTM$42$, encountering a problem on Schur lemma:

Schur Lemma. Given $G$ a group, $V$ a vector space on $\mathbb C$ and an irreducible representation $\rho : G \to Gl(V)$. If a linear map $f : V \rightarrow V$ commutes with all $\rho_g, g\in G$, then $f=\lambda$ for some $\lambda\in \mathbb C$.

Proof of this lemma depends on the fact that $\mathbb C$ is algebraiclly closed. Now I'm asked to give a counterexample of this lemma when $V$ is on $\mathbb Q$ but cannot find one (not familiar with subgroups of $Gl(n,\mathbb Q)$). Moreover I'm wondering if a counterexample exists for any field not algebraiclly closed.

Any help will be appreciated.


1 Answer 1


Schur's lemma relies on the existence of enough eigenvalues. So, take $f$ to not have eigenvalues in the base field. Consider for example rotation of the rational plane $f = \begin{bmatrix} 0&-1\\1&0\end{bmatrix}$. This $f$ has order 4, so I'll take $G = \mathbb{Z}/4\mathbb{Z}$ with $1\mapsto f$.

Now $\mathbb{Q}^2$ is an irreducible (because any proper invariant subspace should be one dimensional) representation of $G$ and $f\colon\mathbb{Q}^2\to\mathbb{Q}^2$ is $G$-linear, but is not scalar multiplication. Incidentally, this also gives an example of an irreducible representation of an abelian group not of dimension 1. (The same applies over any field without a square root of $-1$).

  • 2
    $\begingroup$ This is a nice answer. You can generalize to $G=\mathbb{Z}/n\mathbb{Z}$ acting on the $\mathbb{Q}$-vector space $\mathbb{Q}(\omega_n)$, where $\omega_n$ is a primitive $n$th root of unity. The matrix construction comes from realizing $\mathbb{Q}(\omega_n)\cong \mathbb{Q}[x]/\langle \Phi_n(x)\rangle$, where $\Phi_n$ is the $n$th cyclotomic polynomial. $\endgroup$
    – David Hill
    Oct 21, 2020 at 15:55

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