Last update: 27 August 2013
The following theorem due to Burnside can be found in Burnside, Proc. London Math. Soc. (2), 3, 1905 p. 430. In his ”Classical Groups”, Weyl also directs the reader to Frobenius, Schur, Siztungsber Preuss. Akad. 1906, p. 209.
Theorem 3.18 (Burnside) Let be an arbitrary algebra and let be an irreducible representation of Then
Without loss, is faithful, since otherwise we may consider the representation which has the same image as
Choose a nonzero By faithfulness of we may choose such that Since is an irreducible Thus, there exists a such that from which it follows that for all In particular, is not nilpotent, so is not in the largest nil ideal. But was an arbitrary nonzero element, so hence is semisimple.
If is not simple, we may write as a direct sum of two-sided ideals. Since is faithful, we may choose such that Then is a submodule of and so irreducibility implies But which contradicts the faithfulness of Thus must be simple, so we must have By 1.23, has a unique irreducible module, namely so
Last time (3.16) we showed that given an arbitrary algebra and a completely decomposable (finite dimensional) representation whose decomposition is given by then the algebra of the representation is given by From the above theorem, writing we have From proposition 3.17, recall that
Suppose is an arbitrary algebra which acts on such that Let be the kernel of the representation of on Then we have a representation which is, by definition, injective. Let by hypothesis, the map is an isomorphism.
Hence, by proposition 3.17, is semisimple, and the irreducible representations of are indexed by and have dimensions Since by the above decomposition where are complex matrices.
Define maps by These are the irreducible representations of but they extend to irreducible representations of via the composition
Suppose and are vector spaces with bases and Then is a vector space with basis
If and any are algebras, then is a vector space which becomes an algebra under the multiplication
If and are modules for and respectively, then is an module with the action defined by
Problem 3.19 Check that the multiplication in is well-defined and that satisfies the axiom of an algebra over Check that the action of on is a well-defined module action.
If and with respect to bases and of and respectively, then If we order the basis lexicographically then the entry of the matrix of is
On the other hand, the matrix is also a matrix, so acts on If we consider the block structure given by and label the rows and columns in the form where indicates the number of the block row or block column and indicates the row or column within that block, then the entry is the entry of namely
Proposition 3.20 If and are algebras with representations and respectively, then is an and the matrix of the representation is given by
Apply these tensor product constructions to the previous situation where is an arbitrary algebra with completely reducible module and where is an arbitrary algebra with
The represent a subset of the isomorphism classes of the irreducible The representations are, similarly, a subset of the irreducible representations of
The algebras and both act on Define an action of on by This is well-defined on since it is the linearization of a bilinear map on Moreover, since and commute on we have Hence, this represents a well-defined module action. This brings us to the most important theorem of the course.
Theorem 3.21 Let and be as above. Then as representations, where are irreducible representations of and are irreducible representations of
Let From the decompositions and we have shown that the representations of and on decompose into irreducible representations as Note that each of the matrices above are block diagonal matrices, with block sizes of as runs over Thus, the action on defined by is represented by the matrices of the form Thus the representation decomposes as
Remarks. Note that has the same center as since both are block diagonal with the same block sizes.
Note also that if we suppose and are semisimple, and that and are representatives of the isomorphism classes of irreducible and modules, respectively, then is semisimple and the set of all are a complete set of irreducible modules for
However, even if is semisimple (as always, is semisimple), the decomposition of into modules contains only those irreducibles of the form There are no summands isomorphic to the irreducible modules of the form for
This is a copy of lectures in Representation Theory given by Arun Ram, compiled by Tom Halverson, Rob Leduc and Mark McKinzie.