# Two equivalent irreducible representations given by integer matrices

Let $$G$$ be a finite group, and $$\rho_1, \rho_2: G\to GL_n(\mathbb C)$$ be two representations. Suppose that $$\rho_1$$ and $$\rho_2$$ are equivalent (i.e. conjugate over $$\mathbb C$$), and suppose that both groups $$\rho_1(G)$$, $$\rho_2(G)$$ belong to $$GL(n,\mathbb Z)$$. Is it true that these two groups are conjugate in $$GL(n,\mathbb Z)$$?

If not, is this at least true in the case when $$G$$ is a symmetric group $$S_n$$ and the representation $$\rho$$ is irreducible? The motivation for this question is the following: I know that all complex irreducible representations of $$S_n$$ can be defined over integers. I wonder whether there is somehow a canonical choice.

• Specht modules afford the irreducible complex characters and are defined over the integers. But in general Specht modules are not self-dual (even though their characters are real valued) so even here the choice is not entirely canonical. In general they are many different $\mathbb{Z}$-forms inside an integral Specht module: see for instance wildonblog.wordpress.com/2016/10/30/…. – Mark Wildon Jan 15 at 12:28

The smallest counterexample involving irreducible representations of symmetric groups is the $$2$$-dimensional irreducible module for $$\mathbb{C}S_3$$. It can be defined over the integers as the submodule $$U = \langle e_2-e_1, e_3-e_1\rangle_\mathbb{Z}$$ of the natural integral permutation module $$\langle e_1, e_2, e_3 \rangle_\mathbb{Z}$$. Then $$U \otimes_\mathbb{Z} \mathbb{C}$$ is irreducible and affords the ordinary character labelled $$(2,1)$$. The dual $$U^\star = \mathrm{Hom}_{\mathbb{Z}}(U,\mathbb{Z})$$ is isomorphic to the quotient of $$\langle e_1,e_2,e_3 \rangle$$ by the trivial submodule $$\langle e_1+e_2+e_3\rangle$$. The corresponding homomorphisms $$\rho, \rho^\star : S_3 \rightarrow \mathrm{GL}_2(\mathbb{Z})$$ are such that $$\rho(S_3)$$ and $$\rho^\star(S_3)$$ are conjugate in $$\mathrm{GL}_2(\mathbb{C})$$ but not in $$\mathrm{GL}_2(\mathbb{Z})$$.

To prove the final claim: if the representations are $$\mathbb{Z}$$-equivalent then the modules $$U \otimes_\mathbb{Z} \mathbb{F}_3$$ and $$U^\star \otimes_\mathbb{Z} \mathbb{F}_3$$ are isomorphic. The first has a trivial submodule spanned by

$$(e_2-e_1) +(e_3-e_1) = e_1+e_2+e_3;$$

the quotient by this submodule is the sign module. The second is its dual, with the factors in the opposite order. Since both are indecomposable, they are not isomorphic.

• Thank you @Jeremy Rickard – Mark Wildon Jan 15 at 16:01

No, it is not even true for matrices, e.g., $$\left(\begin{array}{rr}0 & 1\\1&0\end{array}\right)$$ and $$\left(\begin{array}{rr}1 & 0\\0&-1\end{array}\right)$$.

• Thanks! Do you think however that for irreducible representations of the symmetric group $S_n$ this is true? – aglearner Jan 15 at 11:47
• No, I see no reason for that. I can see an example below already.... – Bugs Bunny Jan 15 at 12:47

An instructive example (for general $$G$$, not for symmetrc groups) is provided by the case that $$G$$ is a dihedral group with eight elements.

Then $$G$$ has a unique complex irreducible character $$\chi$$ which may be expressed as $${\rm Ind}_{U}^{G}(\lambda)$$ and $${\rm Ind}_{V}^{G}(\mu)$$, where $$U$$ and $$V$$ are the two Klein $$4$$-subgroups of $$G$$, and $$\lambda, \mu$$ are non-trivial linear characters of $$U,V$$ respectively.

These representations exhibit $$G$$ as an absolutely irreducible subgroup of $${\rm GL}(2, \mathbb{Q})$$ with all matrix entries in $$\mathbb{Z}$$ (even in $$\{0,1,-1\}$$). The two given representations are equivalent over $$\mathbb{C}$$, but they are not equivalent as integral representations.

One way to explain this is via J.A. Green's theory of vertices and sources : both these integral representations are indecomposable on reduction ( mod $$2$$). One of the reductions has vertex $$U$$ and one has vertex $$V$$. Since $$U \lhd G$$ and $$V \lhd G$$ , but $$U \neq V$$, we see that $$U$$ and $$V$$ are not $$G$$-conjugate.

Since (by Green's theory) the vertex of an indecomposable module is unique up to conjugacy, these two modules are not isomorphic on reduction (mod $$2$$), so they are certainly not isomorphic as $$\mathbb{Z}G$$-modules