Let be a finite group acting transitively on a set . An important object in representation theory, much used by Burnside and his successors to study the action of , is the permutation module , defined to be the vector space of all formal linear combinations of the elements of , over a chosen field . Thus has as a canonical basis.

It’s easy to fall into the trap of thinking that the canonical basis of a permutation module is in some sense special, or even unique. The latter is certainly wrong in all but the smallest cases. For example, whenever has characteristic zero, or the characteristic of does not divide , the linear map defined on the canonical basis by

is an isomorphism of -modules. Generally, if has rank in its action on then , and a generic linear combination of these -linearly independent homomorphisms is an automorphism of sending the canonical basis to a basis of permuted by .

The motivation for this post is an example where has a permutation basis not obtained in this way. Take acting on . Let be the set of lines in . (Each line has a unique non-zero point; but we use the line notation to makes clear the distinction between and .) The canonical basis for is therefore

For each plane contained in , let . Clearly , for , so the are permuted by . The duality gives a bijection between the lines and planes. Ordering basis vectors as above, the for non-zero have the coefficients shown in the matrix below.

The determinant of is , so whenever has characteristic or has prime characteristic , the form a permutation basis for . Since acts -transitively on , the endomorphism algebra of is -dimensional, spanned by the identity and the ‘complementing’ map defined above. Therefore this alternative basis is not obtained by an endomorphism of .

Now for the surprising duality. When has characteristic , the three linear relations whose first is

show that . It is clear from the final four rows of that , , , are linearly independent and span a -dimensional submodule of . The sum of these vectors, namely , spans the unique trivial submodule of . Since is odd, it splits off as a direct summand. Observe from rows and of the matrix that

A convenient basis for a complement to the trivial submodule is therefore

We have shown that is a -dimensional -submodule of . In the special case , the representation homomorphism is even an isomorphism. Looking at the definitions of , it might seem almost obvious that is isomorphic to the natural representation of .

This is not the case. There are two -dimensional irreducible -modules, namely the natural module and its dual . These are distinguished by the action of the matrices in of order . Let

be the companion matrix for the irreducible cubic . Computing the action of on we find that

Similar calculations show that and . Reordering basis vectors as , we see that in its action on , is represented by the matrix

This is the companion matrix of the irreducible cubic , which lies in the other conjugacy class of matrices of order to . Therefore .

Since , it follows that where is an indecomposable -dimensional -module with and . As expected, the summand is self-dual, but I think it is a mild surprise that it has and not in its socle.