Group Rings, Group Representation

Group Representation

Let K be a division ring. A homomorphism taking the group G into GLn(K), i.e. the invertible n×n matrices over K, is a representation of G. A faithful representation is a monomorphism.

Every group has a trivial representation, mapping all of G to the identity matrix.

Any subgroup inherits a representation from its parent. The representation remains faithful if the original representation was faithful.

Finite Groups

Since finite groups are isomorphic to permutation groups, every finite group has at least one faithful representation using permutation matrices. This is called the regular representation. This can be rather cumbersome however; for instance, the regular representation of Sn consumes n! rows and columns, while the more traditional representation moves n letters about, and fits inside GLn.

The degree of a representation is the dimension of its matrices. Thus Sn has a faithful representation of degree n. The cyclic group of order l has a faithful representation of degree 1, if you set K to the finite field of order pj, such that pj-1 is divisible by l. Take a primitive element b and raise it to the (pj-1)/l power to find the generator of the cyclic group. Is there such a p and such a j? A theorem, not on my website, says there are infinitely many primes in any arithmetic progression. So you can set j to 1, and find some p that is 1 plus a multiple of l. Or, you can take any prime p that does not divide l, hence p lives in the group of units mod l. In other words, p is torsion in a finite multiplicative group. There is some j such that pj = 1 mod l, and that does the trick.

Why are we Here

I have placed this material under rings, rather than groups, because, as mentioned in the introduction, a representation corresponds to the action of G on the right, when the group ring KG acts on the vector space Kn from the left. The KG module is also a K module, hence the action of anything is K linear, and can be implemented as an n×n matrix over K. Endomorphisms and matrices correspond one to one, though many of these endomorphisms may not be accessible through the action of KG. Since c ∈ G is a unit in KG, its action is invertible, and its matrix is nonsingular. Thus G maps into GLn(K), and we have a representation. Conversely, a representation defines the action of G, and K acts in the usual way; hence Kn becomes a KG module. G representations and KG modules correspond.

Now, if you were brought here from somewhere else, and you don't know anything about group rings, you may still be able to follow the ideas on this page, and the next couple pages, so carry on. It's mostly about groups and matrices.

Equivalent

Two representations are equivalent if a nonsingular matrix Q maps one representation over to the other. That is, Q*r1(G)/Q = r2(G). This map respects the action of G. Review similar matrices for more details.

As described on the aforementioned page, "similar to" is an equivalence relation. Thus equivalent representations of G clump together into equivalence classes. If r1 is equivalent to r2, is equivalent to r3, then r1 is equivalent to r3.

The traditional representation of Sn permutes the coordinates of n space. The x axis might move to the y axis, and the y axis might move to the z axis, and so on. After a change of basis, G permutes n linearly independent vectors, that are no longer the coordinate axes of n space. Still, the representations are equivalent.

Returning to modules over the group ring KG, assume U and V are isomorphic. Of course a KG isomorphism implies a K isomorphism, so a linear transformation f maps U faithfully onto V. Thus U and V have the same dimension, or if you prefer, the representations U and V have the same degree. Since f is an isomorphism it is an invertible matrix. An element of U is a row vector, drawn from K. Apply K on the left, or f on the right, to get a vector in V. So f is a K isomorphism - but it is suppose to be a KG isomorphism. For c in G, apply c, which is a nonsingular matrix over K, then apply f. This has to be the action of G on V. So - if we want to know how G acts on V, it is prescribed. Apply f inverse, then G, then f. As representations, V is equivalent to U. Put this all together and isomorphic KG modules correspond to equivalent representations. In either world, the relation is reflexive, symmetric, and transitive. Collapse equivalence classes down to canonical objects, and each collection of isomorphic, finite dimensional KG modules corresponds to a collection of equivalent representations.

Eigen Values

Since similar matrices have the same eigen values, you can prove two representations are not equivalent by showing r1(c) and r2(c) have different eigen values. (Or the values could be the same, with different multiplicities.) Let's look at some of these eigen values.

The group identity always maps to the identity matrix, and since QI/Q = I, that doesn't change under similarity. The eigen values are all 1's.

Let L be another matrix, the image of c in G, and since G is finite, Lj = I for some j. The eigen values of L, raised to the j, equal 1, and the eigen vectors carry along. New eigen vectors could appear, as when [1,0|0,-1] is squared. Here L has 2 eigenv vectors, and L2 has the entire plane as its eigen space. However, the new vectors are spanned by the original vectors. If this was always the case, we could reverse things, and say that L has n linearly independent eigen vectors, so that Lj can have an n dimensional eigen space. Furthermore, the eigen values of L are all jth roots of 1. There is one fly in the ointment. Let c have order p, and let L = [1,1|0,1] over the field Zp. L has but one eigen vector, yet Lp is the identity matrix, and presents a 2 dimensional eigen space. We can avoid this by insisting that j be a unit in the base field. More generally, the characteristic of the base field does not divide |G|. We'll see this again in the next theorem. Now, convert L to jordan form, and watch what happens as you raise L to the j power. Focus on one block, and note that the diagonal and the subdiagonal commute, hence the binomial theorem applies. The eigen value is raised to the j to get 1, as expected. The subdiagonal becomes the eigen value raised to the j-1, times j. This is nonzero as long as j is nonzero. There is but one eigen vector for this lower triangular block, just as there was one eigen vector for the original block of L. No new vectors appear. Each matrix has n independent eigen vectors, and n eigen values that are various jth roots of 1.

If all the eigen vectors of L are 1 then the diagonalization of L = I, and L is similar to I, and L is already the identity matrix. Thus every matrix, other than the one associated with the group identity, has at least one eigen value that is not 1. In fact, if the eigen values, raised to the k, reach 1, and k < j, then L has order k, and c has order k in G. This is a contradiction; hence the lcm of the exponents needed to bring the eigen values up to 1 is j.

Irreducible / Decomposable

If W is a subspace of Kn, and the matrices in r(G) take W into W, W is a stable subspace of the representation. An irreducible representation has no nontrivial stable subspace. In other words, the corresponding KG module is simple.

A decomposable representation has two nontrivial disjoint subspaces that together span Kn. The two subspaces are called components. Our KG module is the direct sum of two submodules. For example, the permutation matrices corresponding to Sn are decomposable, because the 1 dimensional subspace spanned by 1,1,1…1 is stable, along with the subspace whose vectors have coordinates that sum to 0. (These subspaces happen to be orthogonal.)

The properties defined above depend on the linear transformations implemented by the matrices, and not on the coordinate system of the vector space, so they are not effected by change of basis. If a representation is decomposable, so are all the equivalent representations.

Block Diagonal

Let r(G) be decomposable, with complementary subspaces V and W. Choose a basis for V, and a basis for W. Find an equivalent representation that uses these basis vectors as coordinates. Q is actually the matrix that consists of the basis of V followed by the basis of W, as row vectors. (Remember that Q*r(G)/Q acts on the right, and K acts on the left.) Q is nonsingular, because V and W span the entire space.

Once we have changed basis, V is spanned by the first few components of Kn, and W is spanned by the rest. Every matrix associated with G is block diagonal, such that the upper left block maps V onto V, and the lower right block maps W onto W. Conversely, if the matrices of G are all block diagonal, then the upper left block and the lower right block are stable, and the representation is decomposable.