Lectures in Representation Theory

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia

Last update: 27 August 2013

Lecture 19

3Centralizer Algebras

My Favorite Game
—A. Ram

Definition 3.1 If A is an algebra, let Md(A) denote the d×d matrices with entries in A.

Definition 3.2 Let A be a subset of Md(). The centralizer or commutant of A is the set A= {bMd()|ab=baaA} . Note that AA, although the inclusion may be proper.

Proposition 3.3 For any subset AMd(), A is an algebra over .


The set A inherits a ring structure from Md(), since for all b1,b2A, and all aA (b1+b2)a = a(b1+b2) (b1b2)a = a(b1b2). Moreover, cI=Z(A) and so must lie in the center of A. Hence cI acts as scalars on A which gives the vector space and algebra structures.

Proposition 3.4 If A1A2Md() are subsets then A1A2.

Proposition 3.5 A=A.


Since AA, the previous proposition implies that AA. But A(A)=A.

Proposition 3.6 Md()= Z(Md())= Id, Id= Md()

Let V be a representation of an algebra A, so V:AMd(). The image of A is the algebra of the representation V(A) is a subalgebra of Md(). We are interested in V(A), the centralizer algebra of the representation V.

From the module perspective, suppose V is an A-module and let End(V) be the set of all linear transformations of V. Then V(A) EndA(V)=def { bEnd(V)| ba(v)=ab(v) aA,vV } .

Example 3.7 Let A be an arbitrary algebra and let W be an irreducible representation of A. We claim W(A)=I.


Let BW(A). Then for all aA BW(a)=W(a)B and so by Schur’s lemma, B=cI for some cC.

Example 3.8 Let A be an (associative) algebra. Recall the left regular representation A given by a1a2=a1a2. What is EndA(A)?

The associative law a(bc)=(ab)c shows that EndA(A)RA, the algebra of transformations given by right multiplication by elements of A.

Let BEndA(A), and let bA be defined by B1=b. Then for all aA, Ba=Ba1=a B1=ab= ab. Hence B=Rb acts on A as right multiplication by b, and so EndA(A)=RA.

We are interested in the centralizer algebras of arbitrary semisimple algebras, AλMdλ(), so we will develop some notation for block matrix algebras.

Definition 3.9 Let P be a d1×d2 matrix and Q be a r1×r2 matrix, both over the same field (or algebra). The tensor product of P and Q, PQ is the d1r1×d2r2 matrix with entries in described by the block structure PQ= ( p1,1Q p1,2Q p1,d2Q p2,1Q p2,2Q p2,d2Q pd1,1Q pd1,2Q pd1,d2Q ) For example, Ei,jQ is zero other than the r1×r2 block in the (i,j) position, where it agrees with Q.

Problem 3.10 Suppose P1 and P2 are d1×d2 and d2×d3 matrices, respectively, and suppose also that Q1 and Q2, respectively, are r1×r2 and r2×r3 (so that we may form the products P1P2 and Q1Q2).

Check that (P1Q1) (P2Q2) = P1P2 Q1Q2.

Let A be a subalgebra of Md(). We defined Ms(A) to be s×s matrices with entries in A, so elements of Ms(A) are of the form 1i,jsEijai,j, where the ai,j are matrices in A. Thus, Ms(A) is a subalgebra of Msd(). Let us compute the centralizer Ms(A), which also lies inside Msd().

Suppose BMs(A), and write B=1i,jsEi,jbi,j where each bi,j is a d×d complex matrix. Then B commutes with all elements of Ms(A), i.e. for all choices of s2 matrices ak,A, (1i,jsEi,jbi,j) (1k,sEk,ak,)= (1k,sEk,ak,) (1i,jsEi,jbi,j).

If we multiply this out, the left hand side becomes i,k, Ei, bi,k ak,= i, Ei, kbi,k ak,. Performing the same calculation on the right hand side and setting it equal to the previous computation gives i, Ei, kbi,k ak,= k,j Ek,j iak,i bi,j. Comparing coefficients of Em,n, we see that kbm,k ak,n=i am,ibi,n (3.11) for all choices of s2 elements ai,jA, for each 1m,ns.

Since A is a subalgebra, we know 0 and I lie in A. Fix indices mn and choose s2 elements given by am,m=I and all others zero. Then bm,n=i am,i bi,n= kbm,k ak,n=0.

Make a second choice of s2 elements, with mn, where am,n=I and all other ai,j=0. Equation 3.11 becomes bm,m= bm,m am,n= am,n bn,n= bn,n.

Hence B= 1i,js Ei,jbi,j = k=1s Ek,kb is block diagonal. Moreover, 3.11 becomes bam,n= bm,m am,n= am,n bn.n= am,nb and so bA. Thus BIsA, hence Ms(A)IsA (where we mean the set of matrices of the form Isb for bA). The reverse inclusion is left as an exercise, and we claim

Proposition 3.12 If A is a subalgebra of Md() then Ms(A)=IsA.

Notes and References

This is a copy of lectures in Representation Theory given by Arun Ram, compiled by Tom Halverson, Rob Leduc and Mark McKinzie.

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