Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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}?

share|improve this question
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 '10 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? –  user8617 Aug 19 '10 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 '10 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? –  user8617 Aug 19 '10 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 '10 at 10:09
add comment

1 Answer 1

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.

share|improve this answer
    
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 '10 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 '10 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 '10 at 2:19
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.