## irreducible subgroup of SL(n,R)

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Suppose a subgroup of SL(n,R) is irreducible; i.e. R^n contains no proper invariant real subspaces except {0}. Then is it irreducible as a subgroup of SL(n,C)? i.e. Does C^n contain no proper invariant complex subspace except {0}?

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4
This is false: the cyclic group with three elements acts on $R^3$, permuting the coordinates and fixing the subspace $V$ whose coordinates sum to zero. The representation $V$ is irreducible over the reals, but not over the complex numbers. – damiano Aug 19 2010 at 8:57
Thanks for your answer. If we assume the subgroup of SL(n,R) is finitey generated infinite group, is there a counter example? – unknown (google) Aug 19 2010 at 9:08

4
The integers act on $R^2$ via $1 \mapsto \begin{pmatrix}\cos(n) & \sin(n) \cr -\sin(n) & \cos(n) \end{pmatrix}$; this action has no non-trivial invariant subspaces over the real numbers, but decomposes into a sum of two one-dimensional representations over the complex numbers. – damiano Aug 19 2010 at 9:20
If we assume the subgroup of SL(n,R) is finitely generated with more than 2 generators and infinite, is there a counter example? – unknown (google) Aug 19 2010 at 9:47

3
Let $G$ be any subgroup of $SO(2,R)$. This group has a natural real representation of dimension two that is irreducible with only a couple of exceptions. The same representation is not irreducible over the complex numbers. Note that among the various choices for $G$ there are finitely generated infinite groups with any finite number of generators. – damiano Aug 19 2010 at 10:09

## 1 Answer

A representation $\rho$ over a field $K$ is called absolutely irreducible if for any algebraic field extension $L/K$, the representation $\rho\otimes_K L$ obtained by extension of scalars is irreducible (over $L$). It is enough to check this for the algebraic closure. As damiano's examples in the comments show, this is a much stronger property than irreducibility. Serre's "Linear representations of finite groups" contains a criterion for a real representation of a finite group to be absolutely irreducible.

Here is a way in which non absolutely irreducible representations typically arise. Let $L/K$ be a finite separable field extension of degree $d>1$ and $\sigma$ be an irreducible $n$-dimensional representation of $G$ over $L.$ By restriction of scalars, we obtain an $nd$-dimensional representation $\rho$ of $G$ over $K.$ (In the language of linear group actions, the representation space, which is a vector space over $L,$ is viewed as a vector space over $K$). The representation $\rho$ is not absolutely irreducible because $\rho\otimes_K L$ is isomorphic to the direct sum of $d>1$ Galois conjugates of $\sigma.$ Yet $\rho$ is frequently irreducible. For example, under the restriction of scalars from $\mathbb{C}$ to $\mathbb{R}$, the group $U(1,\mathbb{C})$ becomes $SO(2,\mathbb{R}).$ Therefore, any one-dimensional complex unitary representation (i.e. a character) $\sigma$ of a group $G$ gives rise to a two-dimensional real orthogonal representation $\rho$ whose complexification splits into a direct sum of two representations. This is the construction behind damiano's second and third examples.

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Evidently, my effort was in vain. In the absence of any reaction, this answer will be deleted soon. Really, I shouldn't respond to random question of various unknown (google), since past experience shows that it's just a waste of time. – Victor Protsak Aug 19 2010 at 22:15
@VP: you've gotten three upvotes (including mine).  That indicates that three people got something out of your answer.  For the sake of the other two, I hope you don't delete it (I'll still be able to see it).   – Pete L. Clark Aug 20 2010 at 2:04
Thanks, Pete! I won't delete it now, but I sometimes get frustrated with the vote system. A while ago I posted a technical and detailed answer which was completely ignored for more than a week. I am considering adopting a policy of deleting ignored answers after, say 12 hours. – Victor Protsak Aug 20 2010 at 2:19
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