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I'm using GAP software to find irreducible representations of certain groups, but in many cases the representations are complex and what I need is real representations.

My question is more precisely, given a group $G$ with linear irreducible representation $\rho:G \rightarrow GL_n(\mathbb{C})$ with complex entries in some $\rho(g), g \in G$, is there a straightforward way to obtain a real irreducible representation $\sigma: G \rightarrow GL_m(\mathbb{R})$ corresponding to $\rho$?

I would also be happy with just a computational way of obtaining the representations or a look-up table/resource for usual small groups (order $\leq$ 100).

Thank you in advance.

PSL
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  • The term "linear"in the context of group representations means degree one, so $n=1$ in your notation, but I don't think that's what you mean. – Gerry Myerson May 14 '17 at 00:22
  • For example here there is no requirement of $n=1$. In order to avoid confusion, let me clarify that $n$ here refers to the size of the matrices. – PSL May 14 '17 at 01:06
  • OK, the place I learned it from uses a different definition. – Gerry Myerson May 14 '17 at 02:47

1 Answers1

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Consider the group $C_3$ of three elements. It has exactly three irreducible representations, the trivial one and two that involve nonreal cube roots of unity. You can find real representations of this group (the regular representation is always real, for eaxmple), but other than the trivial one they can't be irreducible.

EDIT: OK, we want real representations that are irreducible over the reals, not necessarily irreducible over the complex numbers. I don't know how to do that for finite groups in general, but for groups of order at most six, here's how:

For the groups of orders 1 and 2, and also for Klein-four and $S_3$, the complex irreducibles are real-valued, so there's nothing to do. The other groups are groups of rotations, so there are representations involving $$\pmatrix{a&-b\cr b&a\cr}$$ where $a$ and $b$ are the sines and cosines of the appropriate angles.

Gerry Myerson
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  • Here it is stated that the degrees of the representations over $\mathbb{R}$ are 1 and 2, so I'm fairly sure there is a way to combine the two representations that are irreducible over $\mathbb{C}$ and obtain one that is irreducible over $\mathbb{R}$. – PSL May 14 '17 at 01:10
  • OK, I see the point; "real irreducible" means real and irreducible over the reals, not real and irreducible over the complex numbers. – Gerry Myerson May 14 '17 at 02:50
  • Thank you for the answer. Actually I realized I was asking for groups of much smaller order than I really need. Updated original question to cover groups of order less or equal to 100. (I was thinking the order as $n$ as in $S_n$ here, which is of course completely wrong). – PSL May 14 '17 at 15:46
  • You do know that there are 267 groups of order 64, right? – Gerry Myerson May 14 '17 at 23:20
  • This doesn't go very far, but might be of interest: https://groupprops.subwiki.org/wiki/Number_of_irreducible_representations_over_reals_equals_number_of_equivalence_classes_under_real_conjugacy – Gerry Myerson May 14 '17 at 23:25
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    See whether this doesn't answer your question: https://math.stackexchange.com/questions/818543/easy-way-to-get-real-irreducible-characters-reps-from-complex-irreducible-char – Gerry Myerson May 14 '17 at 23:27
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    Thank you again for the help. The second link mostly concerns characters, but in the top answer there is a proof which says to replace the complex entries with certain block matrices. This looks like it could be something I'm looking for. However, nothing is mentioned about the irreducibility of the representations obtained in this way. I'll definitely check out the Isaac's book mentioned in the answer. – PSL May 15 '17 at 14:39