Last update: 9 April 2013
Let be a integral domain and let be an algebra over so that has an
making a ring with identity. Let be the field of fractions of let be the algebraic closure of and set
with multiplication determined by the multiplication in Then is an algebra over
A trace on is a linear map such that
A trace on is nondegenerate if for each there is an such that
Lemma 5.1. Let be a finite dimensional algebra over a field let be a trace on Define a symmetric bilinear form on by for all Let be a basis of Let be the matrix of the form with respect to The following are equivalent:
(2) (1): The trace is degenerate if there is an element such that for all If are such that
for all So exists if and only if the columns of are linearly dependent, i.e. if and only if is not invertible.
(3) (2): Let be the dual basis to with respect to and let be the change of basis matrix from to Then
So the transpose of is the inverse of the matrix So the dual basis to exists if and only if is invertible, i.e. if and only if
Proposition 5.2. Let be an algebra and let be a nondegenerate trace on Define a symmetric bilinear form on by for all Let be a basis of and let be the dual basis to B with respect to
Let be another basis of and let be the dual basis to with respect to Let be the transition matrix from to and let be the inverse of Then
So does not depend on the choice of the basis
The proof of part (b) is the same as the proof of part (a) except with replaced by
Let be an algebra and let be an Define
Theorem 5.3 (Schur’s lemma). Let be a finite dimensional algebra over an algebraically closed field
Let be a nonzero homomorphism. Since is simple, and so is injective. Since is simple, and so is surjective. So is an isomorphism. Thus we may assume that
Since is algebraically closed has an eigenvector and a corresponding eigenvalue Then and so is either 0 or an isomorphism. However, since is not invertible. So So So
Theorem 5.4 (The Centralizer Theorem). Let be a finite dimensional algebra over an algebraically closed field Let be a semisimple and set Suppose that
where is an index set for the irreducible which appear in and the are positive integers.
Index the components in the decomposition of by dummy variables so that we may write
For each let be the isomorphism given by
By Schur's lemma,
Thus each element can be written as
and identified with an element of Since it follows that
(b) As a vector space, is isomorphic to the simple module of column vectors of length The decomposition of as modules follows since
If is an algebra then is the algebra except with the opposite multiplication, i.e.
The left regular representation of is the vector space with action given by left multiplication. Here is serving both as an algebra and as an It is often useful to distinguish the two roles of and use the notation for the i.e. is the vector space
Proposition 5.5 Let be an algebra and let be the regular representation of Then More precisely,
where is given by
Let and let be such that For all
and so Then since
for all and
Theorem 5.6. Suppose that is a finite dimensional algebra over an algebraically closed field such that the regular representation of is completely decomposable. Then is i somorphic to a direct sum of matrix algebras, i.e.
for some set and some positive integers indexed by the elements of
If is completely decomposable then, by Theorem 5.4, is isomorphic to a direct sum of matrix algebras. By Proposition 5.5,
for some set and some positive integers indexed by the elements of The map
where is the transpose of the matrix is an algebra isomorphism. So is isomorphic to a direct sum of matrix algebras.
If is an algebra then the trace tr of the regular representation is the trace on given by
where is the linear transformation of induced by the action of on by left multiplication.
Proposition 5.7. Let The trace of the regular representation is nondegenerate if and only if the integers are all nonzero in In characteristic they could be
As the regular representation
where is the irreducible consisting of column vectors of length For let be the linear transformation of induced by the action of Then the trace tr of the regular representation is given by
where are the irreducible characters of Since the are all nonzero the trace tr is nondegenerate.
Theorem 5.8 (Maschke’s Theorem). Let be a finite dimensional algebra over a field such that the trace of the regular representation of is nondegenerate. Then every representation of is completely decomposable.
Let be a basis of and let be the dual basis of with respect to the form defined by
The dual basis exists because the trace tr is nondegenerate.
Let be an If is irreducible then the result is vacuously true, so we may assume that has a proper submodule Let be a projection onto i.e. and Let
So for all Thus, since tr is nondegenerate,
Let Then for all and so So Let Then for all and so So and as elements of
Note that So
and, by Proposition 5.2(b), is an So is an of which is complementary to By induction on the dimension of and are completely decomposable, and therefore is completely decomposable.
Together, Theorem 5.6, 5.8 and Proposition 5.7 yield the following theorem.
Theorem 5.9 (Artin–Wedderburn Theorem). Let be a finite dimensional algebra over an algebraically closed field Let be a basis of and let be the trace of the regular representation of The following are equivalent:
Remark. Let be an integral domain, and let be an algebra over with basis Then is an element of and in if and only if in In particular, if then is a polynomial. Since a polynomial has only a finite number of roots, for only a finite number of values
Theorem 5.10 (The Tits Deformation Theorem). Let be an integral domain, the field of fractions of the algebraic closure of and the integral closure of in Let be an and let be a basis of For let denote the linear transformation of induced by left multiplication by Let be indeterminates and let
so that is the characteristic polynomial of a “generic” element of
First note that if is another basis of and the change of basis matrix is given by
defines an isomorphism of polynomial rings Thus it follows that if the statements are true for one basis of (or then they are true for every basis of (resp.
(a) Using the decomposition of let be a basis of matrix units in and let be corresponding variables. Then the decomposition of induces a factorization
The polynomial is irreducible since specializing the variables gives
which is irreducible in This provides the factorization of and establishes that By (5.11)
which establishes the last statement.
Any root of is an element of So any root of is an element of and therefore the coefficients of (symmetric functions in the roots of are elements of
(b) Taking the image of the Eq. (5.11), give a factorization of
For the same reason as in (5.12) the factors are irreducible polynomials in
On the other hand, as in the proof of (a), the decomposition of induces a factorization of into irreducibles in These two factorizations must coincide, whence the result.
Applying the Tits deformation theorem to the case where (so that gives the following theorem. The statement in (a) is a consequence of Theorem 5.6 and the remark which follows Theorem 5.9.
Theorem 5.13. Let be a family of algebras defined by generators and relations such that the coefficients of the relations are polynomials in Assume that there is an such that is semisimple. Let be an index set for the irreducible Then
This is an excerpt of a paper entitled Partition algebras, written by Tom Halverson (Mathematics and Computer Science, Macalester College, Saint Paul, MN 55105, United States) and Arun Ram.
This research was supported in part by National Science Foundation Grant DMS-0100975. This research was also supported in part by the National Science Foundation (DMS-0097977) and the National Security Agency (MDA904-01-1-0032).