Generalized matrix algebra structure
Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au
Last updates: 5 November 2010
Generalized matrix algebras
Let
A be an algebra and fix
a∈A. The
homotope algebra Aa is the algebra
A with a new multiplication given by
x⋅y=xay,for x,y∈A.
If
p,q are invertible elements of
A then the map
Apaq→Aax↦qxpis an algebra isomorphism.
The radical of a homotope algebra
Let
R be a PID and let
A=MnR
and let
ϵ∈A. The
smith normal form says that there exist
p,q∈GLnRsuch thatpϵq=diagϵ1,…,ϵk,0,,0,…,0,withϵ1∣ϵ2∣⋯∣ϵk.
Thus,
Mnϵ≅Mnδ,whereδ=diagϵ1,…,ϵk,0,,0,…,0,
and
RadMnδ=x∈Mn∣if xST≠0 then S>k or T>k,Rad2Mnδ=x∈Mn∣if xST≠0 then S>k and T>k,andRad3Mnδ=0.
RadAa=x∈A∣axa∈RadA
and
Rad3Aa⊂RadA.
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Proof.
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The set
I=x∈A∣axa∈RadA
is an ideal in A since, if y∈A then
ax⋅ya=axay∈RadAa. Why and when is I=RadA??? Or do I care?
□
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A⊆B, both split semisimple
Assume
A⊆B is an inclusion of algebras and that
A and
B are split semisimple. Let
Aˆ be an index set for the irreducible A-modulesAμ,Bˆ be an index set for the irreducible B-modulesBλ,and letAˆμ=P→μ be an index set for a basis of the simple A-module Aμ,
for each
μ∈Aˆ
(the composite
P→μ is viewed as a single symbol). Let
Γ be the two level graph
vertices on level A:Aˆ,vertices on level B:Bˆ,andmμλ edges μ→λ if Aμ appears with multiplicity mμλ in ResABBλ.
| 1.1 |
If
λ∈Bˆ
then
Bˆλ=P→μ→λ∣μ∈Aˆ,P→μ∈Aˆμ and μ→λ an edge in Γ
| 1.2 |
is an index set for a basis of the irreucible
B-module
Bλ. We think of
Bˆλ
as the set of paths to
λ and
Aˆμ
as the "set of paths to
μ" in the graph
Γ. For example, the graph
Γ for the symmetric group algebras
A=ℂS3 and
B=ℂS4 is
picture goes here
Since A and B are split semisimple there exist sets of matrix units in the algebras A and B,
{aP Qμ∣μ∈Aˆ,P→μ,Q→μ∈Aˆμ}and{bP Qμ νλ∣λ∈Bˆ,P→μ→λ,Q→ν→λ∈Bˆλ}
| 1.3 |
respectively, so that
aP QμaS Tν=δμνδQSaP TμandbP Qμ γλbS Tτ νσ=δλσδQSδγτbP Tμ νλ,
| 1.4 |
and such that
aP QμbS Tμ τλ=δQ Sμ γbP Tμ τλandbS Tσ τλ aP Qμ=δT Pτ μbS Qσ μλ.
| 1.5 |
Then
and
aP Qμ=1⋅aP Qμ⋅1=(∑bR Rρ ρλ)aP Qμ∑(bS Sσ σγ)=∑bP Qμ μλ
| 1.7 |
where the sum is over all edges
μ→λ in the graph
Γ.
Now assume that B is a subalgebra of an algebra C and there is an element e∈C such that for all b∈B,
-
ebe=ϵ1b, with
ϵ1b∈A, and
-
ϵ1a1ba2=a1ϵ1ba2 for all a1,a2∈A, and
-
ea=ae, for all a∈A.
Note that the map
ϵ1:B⊗B→Ab1⊗b2↦ϵ1b1b2is an A,A bimodule homomorphism.
Though it is not necessary for the following it is conceptually helpful to let
C=BeB,
let
Cˆ=Aˆ
and extend the graph Γ to a graph
Γˆ
with three levels, so that the edges between level B and level C are reflections of the edges between level A and level B. In other words,
Γˆ
has
vertices on level C:Cˆ,andan edge λ→μ,λ∈Bˆ,μ∈Cˆ, for each edge μ→λ,μ∈Aˆ,λ∈Bˆ.
| 1.8 |
For each
ν∈Cˆ
define
Cˆν=P→μ→λ→ν∣μ∈Aˆ,λ∈Bˆ,ν∈Cˆ,P→μ∈Aˆ and μ→λ and λ→μ are edges in Γˆ,
| 1.9 |
so that
Cˆν
is te set of paths to
ν in the graph
Γˆ.
In the previous example
Γˆ
is
PICTURE GOES HERE
The element of A given by
ϵ1(bP Qμ τλ)=ϵ1(aP PμbP Qμ τλaQ Qτ)=aP Pμϵ1(bP Qμ τλ)aQ Qτ
is zero unless μ=τ and
ϵ1(bP Qμ μλ)=ϵ1(aP RμbR Rμ μλaR Qμ)=aP Rμϵ1(bR Rμ μλ)aR Qμ=ϵμλaP Qμ
| 1.10 |
for some constant
ϵμλ
which does not depend on
P or
Q (since it depends only on
R which can be chosen freely). The element of
C give by
bP Rμ ρλebT Qτ νσ=bP Rμ ρλaR RρebT Qτ νσ=bP Rμ ρλeaR RρbT Qτ νσ
is zero unless
R=T and
ρ=τ and
bP Rμ ρλebR Qρ νσ=bP Sμ ρλaS RρebR Qρ νσ=bP Sμ ρλeaS RρbR Qρ νσ=bP Sμ ρλebS Qρ νσ
does not depend on the choice of
R. If
cP Qμ νλ σγ=bP Tμ γλebT Qγ νσ
| 1.11 |
then
cP Qμ νλ σγcR Sτ ξρ ηπ=(bP Tμ γλebT Qγ νσ)(bR Xτ πρebX Sπ ξη)=δQ Rν τσ ρbP Tμ γλϵ1(bT Xγ πσ)ebX Sπ ξη=δQ Rν τσ ρbP Tμ γλδγπϵγσaT XγebX Sπ ξη=δQ Rν τσ ρδπγϵγσbP Xμ γλebX Sγ ξη=δQ Rν τσ ρδπγϵγσcP Sμ ξλ ηγ.
Define
eP Qμ νλ σγ=1ϵγσcP Qμ νλ σγ,whenver ϵγσ≠0.TheneP Qμ νλ σγeR Sτ ξρ ηπ=δQ Rν τσ ρδγπeP Sμ ξλ ηγ
| 1.12 |
so that these are matrix units. Furthermore
bP Qμ νλeR Sτ ξρ ηπ=δQ Rν τλ ρeP Sμ ξλ ηπandeP Qμ νλ σγbR Sτ ξρ=δQ Rν τσ ρeP Sμ ξλ ργ
| 1.13 |
so that the
bs are related to the
es in the same way that the
as are related to the
b s. Then
e=1⋅e⋅1=(∑bR Rρ ρλ)e(∑bS Sσ σγ)=∑bR Rρ ρλebR Rρ ρλ=∑cR Rρ ρλ γρ=ϵργeR Rρ ρλ γρ.
| 1.14 |
In summary
ebP Qμ τλe=δμτϵμλaP QμeP Qμ νλ σγ=1ϵγσbP Tμ γλebT Qγ νσbP Qμ νλ=∑λ→γeP Qμ νλ λγ
and
eeP Aμ νλ σγ=δμγϵγλ∑γ→τ→γeP Qγ ντ σγ.
R⊗AL, for A semisimple.
Fix isomorphisms
L‾≅⨁μ∈AˆA→μ⊗LμandR‾≅⨁μ∈AˆRμ⊗A←μ
| 1.15 |
where
A→μ,μ∈Aˆμ are the simple left
A‾-modules,
A←μ,
μ∈Aˆ,
are the simple right
A‾-modules, and
Lμ,Rμ,μ∈Aˆ
are vector spaces. In other words, if
A has matrix units
{aP Qμ∣μ∈Aˆ,P→μ,Q→μ∈Aˆμ}thenLhas a basislPX∣P∈Aˆμ,X∈LˆμRhas a basisrYQ∣Q∈Aˆμ,Y∈Lˆμ
such that
aP QμlR Xν=δμνδQRlP XμandrY SνaP Qμ=δμνδSPlY Qμ.
| 1.16 |
The map
ϵ:L‾⊗𝔽R‾→A‾
is determined by the constants
ϵXYμ∈𝔽
given by
ϵ(lQ Xμ⊗rY Pμ)=ϵXYμaQ Pμ
| 1.17 |
and
ϵXYμ
does not depend on
Q and
P since
ϵ(lS Xλ⊗rY Pμ)=ϵ(aS QλlQ Xλ⊗rY PμaP Tμ)=aS Qλϵ(lQ Xλ⊗rY Pμ)aP Tμ=δλμaS QμϵXYμaQ PμaP Tμ=ϵXYμaS Tμ.
For each
μ∈Aˆ
construct a matrix
and use row reduction (Smith normal form) to find invertible matrices
Dμ=DSTμandCμ=CZWμsuch thatDμℰμCμ=diagϵXμ
| 1.19 |
is a diagonal matrix with diagonal entries denoted
ϵXμ.
The
ϵPμ
are the
invariant factors of the matrix
ℰμ.
Using notation as in (???) define elements of
R⊗AL
by
mX Yμ=rX Pμ⊗lP Yμ,andnX Yμ=∑Q1,Q2CQ1XμDYQ2μmQ1 Q2μ
| 1.20 |
where
μ∈Aˆ,X∈Rˆμ,Y∈Lˆμ.
-
The sets
{mX Yμ∣μ∈Aˆ,X∈Rˆμ,Y∈Lˆμ}and{nX Yμ∣μ∈Aˆ,X∈Rˆμ,Y∈Lˆμ}
are bases of
R‾⊗A‾L‾,
which satisfy
mS TλmQ Pμ=δλμϵTQμmS PμandnS TλnQ Pμ=δλμδTQϵTμnS Pμ,
where
ϵTQμ
and
ϵTμ
are as defined in (1.17) and (1.19).
-
The radical of the algebra
R⊗AL
is
RadR⊗AL=𝔽-span{nY Tμ∣ϵYμ=0 or ϵTμ=0}
and the images of the elements
eY Tμ=1ϵTμnY Tμ,for ϵYμ≠0 or ϵTμ≠0,
are a set of matrix units in
R⊗AL/RadR⊗AL.
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Proof.
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-
Since
(rS Wλ⊗lZ Tμ)=(rS PλaP Wλ⊗lZ Tμ)=(RS Pλ⊗aP WλlZ Tμ)=δλμδWZ(rS Pλ⊗lP Tλ)
the element
mX Yμ
doe not depend on P and
{mX Yμ∣μ∈Aˆ,X∈Rˆμ,Y∈Lˆμ}is a basis ofR⊗AL
| 1.22 |
and hence
R⊗AL
is a direct sum of generalized matrix algebras. If
C-1μ
and
D-1μ
are the inverses of the matrices Cμ and Dμ then
∑x,YC-1XSμD-1TYμnX Yμ=∑X,Y,Q1,Q2C-1XSμCQ1XμmQ1 Q2μDYQ2μD-1TYμ=∑Q1,Q2δSQ1δQ2TmQ1 Q2μ=mS Tμ,
and so the elements
mS Tμ
can be written as a linear combination of the
nX Yμ.
Thus
{nX Yμ∣μ∈Aˆ,X∈Rˆμ,Y∈Lˆμ}is a basis ofR⊗AL.
| 1.22 |
By direct computation,
mS TμmQ Pλ=(rS Wλ⊗lW Tλ)(rQ Zμ⊗lZ Pμ)=ϵ(lW Tλ⊗rQ Zμ)lZ Pμ=δλμ(rS Wλ⊗ϵTQλaW Zλ lZ Pλ)=δλμϵTQλ(rS Wλ⊗lW Pλ)=δλμϵTQλmS Pλ,
and
nS TλnU Vμ=∑Q1,Q2,Q3,Q4CQ1SλDTQ2λmQ1 Q2λCQ3UμDVQ4μmQ3 Q4μ=∑Q1,Q2,Q3,Q4δλμCQ1SλDTQ2λϵQ1Q2μCQ3UμDVQ4μmQ3 Q4μ=δλμ∑Q1,Q4δTUϵTμCQ1SμDVQ4μmQ1 Q4μ=δλμδTUϵTμnS Vμ.
| 1.23 |
-
Let
I=𝔽-span{nY Tμ∣ϵYμ=0 or ϵTμ=0}
The multipliction rule for the
nY Tμ
implies that I is an ideal of
R⊗AL.
If
nY1 T1μ,nY2 T2μ,nY3 T3μ∈{nY Tμ∣ϵYμ=0 or ϵTμ=0}
then
nY1 T1μnY2 T2μnY3 T3μ=δT1Y2ϵY2μnY1 T1μnY3 T3μ=δT1Y2δT2Y3ϵY2μϵT2μnY1 T3μ,
since
ϵY2μ=0
or
ϵT2μ=0.
Thus any product
nY1 T1μnY2 T2μnY3 T3μ
of three basis elements of I is 0. So I is an ideal of
R⊗AL
consisting of nilpotent elements and so
I⊆RadR⊗AL.
Since
eY TλeU Vμ=1ϵTλ1ϵVμnY TλnU Vμ=δλμδTU1ϵTλϵVλϵVλnY Vλ=δλμδTUeY Vλmod I,
the images of the elements
eY Tλ
in (????) form a set of matrix units in the algebra
R⊗AL/I.
Thus
R⊗AL/I
is a split semisimple algebra and so
I⊇RadR⊗AL.
□
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The structure of Zϵ
Let
ϵ:L⊗DR→Cbe a C,C bimodule homomorphism.
The
left radical Lϵ and the
right radical Rϵ of
ϵ are defined by
Lϵ=l∈L∣ϵl⊗r∈RadC,for all r∈R,Rϵ=r∈R∣ϵl⊗r∈RadC,for all l∈L,
The map
ϵ is
nondegenerate if
RadC=0,
Lϵ=0, and
Rϵ=0. Let
C‾=C/RadC,L‾=L/Lϵ,R‾=R/Rϵ,andφ:R⊗CL→R‾⊗C‾L‾r‾⊗l‾↦r⊗l‾
Then
ker φ is generated by
R⊗CLϵ
and
Rϵ⊗CL,
and we have that
ker φ⋅R⋅Rϵ and
L⋅ker φ⊂Lϵ. THne
I=RadC+LC+ker φis a nilpotent ideal of Aϵ,
and
AϵI≅Aϵ‾where the mapϵ‾:L‾⊗DR‾→C‾
is a nondegenerate
(C‾,C‾)
bimodule homomorphism.
If
ϵ:L⊗DR→C
is nondegenerate and R is a projective C-module then there is a
(D,C)
bimodule homomorphism
τ:R→∼L*r↦λr:L→Cl↦ϵl⊗rso thatϵ=ev∘id⊗τ
and
Aϵ≅A(evL).
If C,D,L,R are finite dimensional vector spaces over 𝔽 and D=𝔽 then
ϵ=ϵ0⊕evP:L0⊕P*⊗DR0⊕P→C,
with P projective and
im ϵ0⊆RadC.
If ϵ=ϵ0⊕evP with
P finitely generated and projective then
Aϵ-mod→∼Aϵ0-modM↦eMwheree=1-∑ipi⊗αi.
If
im ϵ⊆RadC
then
RadAϵ0=I=RadC⊕RadD⊕L0⊕R0⊗CL0
and
Aϵ0RadAϵ0≅CRadC⊕DRadD.
Duals and Projectives
Let
L be a
C-module and let
Z=EndCL
so that
L is a
(C,Z)
bimodule. The
dual module to
L is the
(Z,C)
bimodule
L*=HomCL,C.
The
evaluation map is the
(C,C)
bimodule homomorphism
ev:L⊗ZL*→Cλ⊗l↦λl
and the
centralizer map is the
(Z,Z)
bimodule homomorphism
ξ:L*⊗CL→Zλ⊗l↦zλ,l:L→Lm↦λml
Recall that [
Bou, Alg. II §4.2 Cor.]
-
If L is a projective C-module if and only if
1∈im ξ,
-
If L is a projective C-module then ξ is injective,
-
If L is a finitely generated projective C-module then ξ is bijective,
-
If L is a finitely generated free module then
ξ-1z=∑ibi*⊗zbi,
where
b1,…,bd
is a basis of L and
b1*,…,bd*
is the dual basis in M*.
Statement (a) says that
L is projective if and only if there exist
bi∈L and
bi*
such that
if l∈Lthenl=∑ibi*lbiso thatξ∑ibi*⊗bi=1.
References
[Bou]
N. Bourbaki,
Groupes et Algèbres de Lie,
Masson, Paris, 1990.
[GW1]
F. Goodman and H. Wenzl,
The Temperly-Lieb algebra at roots of unity, Pacific J. Math. 161 (1993), 307-334.
MR1242201 (95c:16020)
[GL1]
J. Graham and G. Lehrer,
Diagram algebras, Hecke algebras and decomposition numbers at roots of unity, Ann. Sci. École Norm Sup. (4) 36 (2004), 479-524.
MR2013924 (2004k:20007)
[GL2]
J. Graham and G. Lehrer,
The two-step nilpotent representations of the extended affine Hecke algebra of type A, Compositio Math. 133 (2002), 173-197.
MR1923581 (2004d:20004)
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