S. D. V. Stephens

notes / Studies in Algebra and Group Theory / chapter 9: Representation Theory / From my book

B14.4 The Character Table of

By the splitting criterion for conjugacy classes in (Proposition 11.4.3), the conjugacy classes of are as follows. The transposition commutes with , and commutes with , so the double transpositions and the -cycles each remain a single class of . The centralizer of in has order (Proposition 3.2.13), so it is , and the twenty-four -cycles split into two classes of twelve. Conjugating by the odd permutation gives , which lies in the other class: if with , then the odd permutation would commute with . So has five conjugacy classes:

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It therefore has five irreducible representations.

Letting only the even permutations act turns every representation of into a representation of , its restriction (restriction is studied systematically in Chapter 15). Elements of act trivially on , so restricts to the trivial representation, and for any representation of , the restrictions of and to have the same character, hence are isomorphic. So it suffices to restrict , , and :

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Computing inner products over , and , so and remain irreducible. However, so on , is a direct sum of two non-isomorphic irreducible representations. Neither is , or , since So on , where and are the two remaining irreducible representations of . Their dimensions satisfy and , so .

14.4.1 Determining and

We know . Write ; the first entry is . Since and are not isomorphic to , or , the class function is orthogonal to , and , which are real-valued: Adding the first two equations gives , and with the third, and then . So for some . Since ,

This does not yet determine . But every character of is real-valued. Indeed, every element of is conjugate in to its inverse: a -cycle or a double transposition has the same cycle type as its inverse, and these classes of do not split in ; and a -cycle has inverse , which is its conjugate by the even permutation . So for every representation of and every , by Proposition 13.4.3(iii). Hence is real and . The sign only decides which of the two representations we call ; taking , we get , and similarly for :

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The golden ratio appears naturally. The representations and are exchanged by the automorphism of , for a transposition : letting act on as does defines a new representation with character . Conjugation by fixes the classes of , and and swaps the two classes of -cycles, so this character is . In particular, this automorphism is not conjugation by an element of , which would preserve every conjugacy class.

Note. Without the fact that the characters of are real, the computation above leaves two possibilities, because the dual is again a -dimensional irreducible representation, so it is or . Either and , the characters are real, and ; or , so that , and . Geometry also decides between them. The group is the rotation group of the regular icosahedron (Theorem 11.3.7), and the inclusion is a -dimensional representation of by real matrices, so its character is real. Since the irreducible representations of have dimensions , a -dimensional representation is , or , and this one is nontrivial, so it is or . Its character confirms the table: a rotation by the angle has trace , and rotations by , , and have traces , , and .

The irrational character values have a geometric consequence: there is no regular icosahedron (or dodecahedron) in all of whose vertices have rational coordinates. If there were, we could translate it so that its center, the average of its vertices, is at the origin, keeping the coordinates rational. Then in a basis consisting of three linearly independent vertices, each rotation of the icosahedron, which permutes the vertices, would have a matrix with rational entries, and all character values would be rational. But this representation is or , and both characters take the irrational values on -cycles. This holds in any coordinate system, not just an orthonormal one.