The basic construction

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

Last update: 9 April 2013

The basic construction

In this section we shall assume that all algebras are finite dimensional algebras over an algebraically closed field 𝔽. The fact that 𝔽 is algebraically closed is only for convenience, to avoid the division rings that could arise in the decomposition of A just before (4.8) below.

Let AB be an inclusion of algebras. Then B𝔽B is an (A,A)-bimodule where A acts on the left by left multiplication and on the right by right multiplication. Fix an (A,A)-bimodule homomorphism

ε:B𝔽BA. (4.1)

The basic construction is the algebra BAB with product given by

(b1b2) (b3b4)= b1ε (b2b3) b4,forb1, b2,b3,b4 B. (4.2)

More generally, let A be an algebra and let L be a left A-module and R a right A-module. Let

ε:L𝔽RA, (4.3)

be an (A,A)-bimodule homomorphism. The basic construction is the algebra RAL with product given by

(r11) (r22)= r1ε (1r2) 2,for r1,r2Rand 1,2L. (4.4)

Theorem 4.18 below determines, explicitly, the structure of the algebra RAL.

Let N=Rad(A) and let

A=A/N, L=L/NL, and R=R/RN. (4.5)

Define an (A,A)-bimodule homomorphism

ε: L𝔽R A r ε(r) (4.6)

where =+NL, r=r+RN, and a=a+N, for L, rR, and aA. Then by basic tensor product relations [Bou1990, Chapter II, Section 3.3 corresponding to Proposition 2 and Section 3.6 corresponding to Proposition 6], the surjective algebra homomorphism

π: RAL RAL r r hasker(π)= RANL. (4.7)

The algebra A is a split semisimple algebra (an algebra isomorphic to a direct sum of matrix algebras). Fix an algebra isomorphism

A μA^ Mdμ(𝔽) aPQμ EPQμ

where A^ is an index set for the components and EPQμ is the matrix with 1 in the (P,Q) entry of the μth block and 0 in all other entries. Also, fix isomorphisms

L μA^ AμLμ andR μA^ RμAμ (4.8)

where Aμ,μA^, are the simple left A-modules, Aμ,μA^, are the simple right A-modules, and Lμ,Rμ,μA^, are vector spaces. The practical effect of this set-up is that if A^μ is an index set for a basis {rYμYR^μ} of Rμ,L^μ is an index set for a basis {XμXL^μ} of Lμ, and A^μ is an index set for bases

{ aQμ QA^μ } ofAμand { aPμ PA^μ } ofAμ (4.9)

such that

aSTλ aQμ= δλμ δTQ aSμ and aPμ aSTλ= δλμ δPS aTμ, (4.10)


Lhas basis { aPμ Xμ μA^,P A^μ,X L^μ } and Rhas basis { rYμ aQμ μA^,Q A^μ,Y R^μ } (4.11)

With notation as in (4.9) and (4.11) the map ε: L𝔽 R A is determined by the constants εXYμ𝔽 given by

ε ( aQμ Xμ rYμ aPμ ) =εXYμ aQPμ (4.12)

and εXYμ does not depend on Q and P since

ε ( aSλ Xλ rYμ aTμ ) = ε ( aSQλ aQλ Xλ rYμ aPμ aPTμ ) = aSQλε ( aQλ Xλ rYμ aPμ ) aPTμ = δλμ aSQμ εXYμ aQPμ aPTμ= εXYμ aSTμ. (4.13)

For each μA^ construct a matrix

Eμ= (εXYμ) (4.14)

and let Dμ=(DSTμ) and Cμ=(CZWμ) be invertible matrices such that DμEμCμ is a diagonal matrix with diagonal entries denoted as εXμ,

DμEμCμ= diag(εXμ). (4.15)

In practice Dμ and Cμ are found by row reducing Eμ to its Smith normal form. The εPμ are the invariant factors of Eμ.

For μA^, XR^μ, YL^μ, define the following elements of RAL:

mXYμ= rXμaPμ aPμ Yμ,and nXYμ= Q1,Q2 CQ1Xμ DYQ2μ mQ1Q2μ. (4.16)


rSλ aWλ aZμ Tμ = ( rSλ aPλ aPWλ aZμ Tμ ) = ( rSλ aPλ aPWλ aZμ Tμ ) = δλμ δWZ ( rSλ aPλ aPλ Tλ ) , (4.17)

the element mXYμ does not depend on P and { mXYμ μA^, XR^μ,Y L^μ } is a basis of RAL.

The following theorem determines the structure of the algebras RAL and RAL. This theorem is used by W.P. Brown in the study of the Brauer algebra. Part (a) is implicit in [Bro1955, Section 2.2] and part (b) is proved in [Bro1955-2].

Theorem 4.18. Let π:RAL RA L be as in (4.7) and let {ki} be a basis of ker(π)=RANL. Let

nYTμRAL be such thatπ (nYTμ)= nYTμ,

where the elements nYTμRAL are as defined in (4.16).

  1. The sets { mXYμ μA^, XR^μ,Y L^μ } and { nXYμ μA^, XR^μ,Y L^μ } (see (4.16)) are bases of RAL, which satisfy mSTλ mQPμ= δλμ εTQμ mSPμ and nSTλ nQPμ= δλμδTQ εTμ nSPμ, where εTQμ and εTμ are as defined in (4.12) and (4.15).
  2. The radical of the algebra RAL is Rad(RAL)=𝔽-span { ki,nYTμ εYμ=0 orεTμ =0 } and the images of the elements eYTμ= 1εTμ nYTμ, forεYμ0 andεTμ 0, are a set of matrix units in (RAL)/Rad(RAL).


The first statement in (a) follows from the equations in (4.17). If (C-1)μ and (D-1)μ are inverses of the matrices Cμ and Dμ then

X,Y (C-1)XSμ (D-1)TYμ nXY = X,Y,Q1,Q2 (C-1)XSμ CQ1Xμ mQ1Q2 DYQ2μ (D-1)TYμ = Q1,Q2 δSQ1 δQ2T mQ1Q2μ =mSTμ,

and so the elements mSTμ can be written as linear combinations of the nXYμ. This establishes the second statement in (a). By direct computation, using (4.10) and (4.12),

mSTλ mQPμ = ( rSλ aWλ aWλ Tλ ) ( rQμ aZμ aZμ Pμ ) = rSλ aWλε ( aWλ Tλ rQμ aZμ ) aZμ Pμ = δλμ ( rSλ aWλ εTQλ aWZλ Pλ ) = δλμ εTQλ ( rSλ aWλ aWλ Pλ ) = δλμ εTQλ mSPλ,


nSTλ nUVμ = Q1,Q2,Q3,Q4 CQ1Sλ DTQ2λ mQ1Q2λ CQ3Uμ DVQ4μ mQ3Q4μ = Q1,Q2,Q3,Q4 δλμ CQ1Sλ DTQ2λ εQ2Q3μ CQ3Uμ DVQ4μ mQ1Q4μ = δλμ Q1,Q4 δTY εTμ CQ1Sμ DTQ4μ mQ1Q4μ =δλμ δTU εTμ nSVμ.

(b) Let N=Rad(A) as in (4.5). If r1n11, r2n22 RANL with n1Ni for some i>0 then

(r1n11) (r2n22) = r1ε (n11r2) n22 = r1n1ε (1r2) n22RA Ni+1L.

Since N is a nilpotent ideal of A it follows that ker(π)= RANL is a nilpotent ideal of RAL. So ker(π) Rad(RAL).


I=𝔽-span { ki,nYTμ εYμ=0 orεTμ=0 } .

The multiplication rule for the nYT implies that π(I) is an ideal of RAL and thus, by the correspondence between ideals of RAL and ideals of RAL which contain ker(π), I is an ideal of RAL.

If nY1T1μ, nY2T2μ, nY3T3μ { nYTμ εYμ=0 orεTμ=0 } then

nY1T1μ nY2T2μ nY3T3μ= δT1Y2 εY2μ nY1T2μ nY3T3μ= δT1Y2 δT2Y3 εY2μ εT2μ nY1T3μ =0,

since εY2μ=0 or εT2μ=0. Thus any product nY1T1μ nY2T2μ nY3T3μ of three basis elements of I is in ker(π). Since ker(π) is a nilpotent ideal of RAL it follows that I is an ideal of RAL consisting of nilpotent elements. So IRad(RAL).


eYTλ eUVμ= 1εTλ 1εVμ nYTλ nUVμ= δλμ δTY 1εTλεVλ εTλnYVλ =δλμ δTU eYVλ modI,

the images of the elements eYTλ in (4.7) form a set of matrix units in the algebra (RAL)/I. Thus (RAL)/I is a split semisimple algebra and so IRad(RAL).

Basic constructions for AB

Let AB be an inclusion of algebras. Let ε1:BA be an (A,A) bimodule homomorphism and use the (A,A)-bimodule homomorphism

ε: B𝔽B A b1b2 ε1(b1b2) (4.19)

and (4.2) to define the basic construction BAB. Theorem 4.28 below provides the structure of BAB in the case where both A and B are split semisimple.

Let us record the following facts:

(4.20a) If pA and pAp=𝔽p then (p1) (BAB) (p1) =𝔽·(p1).
(4.20b) If p is an idempotent of A and pAp=𝔽p then ε1(1)𝔽.
(4.20c) If pA, pAp=𝔽p and if ε1(1)0, then 1ε(1)(p1) is a minimal idempotent in BAB.

These are justfied as follows. If pA and pAp=𝔽p and b1,b2B then (p1) (b1b2) (p1) = (pε1(b1)b2) (p1) = pε1(b1)ε1 (b2p) = pε1(b1)ε1 (b2)p1 =ξp1, for some constant ξ𝔽. This establishes (a). If p is an indempotent of A and pAp=𝔽p then pε1(1)p = ε1(p2) = (ε1(1·p)) = ε1(1)p and so (b) holds. If pA and pAp=𝔽p then (p1)2=ε1(1)(p1) and so, if ε1(1)0, then 1ε(1)(p1) is a minimal idempotent in BAB.

Assume A and B are split semisimple. Let

A^ be an index set for irreducibleA-modules Aμ, B^ be an index set for irreducibleB-modules Bλ, and let A^μ= {Pμ} be an index set for a basis of the simpleA -moduleAμ,

for each μA^ (the composite Pμ is viewed as a single symbol). We think of A^μ as the set of “paths to μ" in the two-level graph

Γwith vertices on level A: A^, vertices on level B: B^,and mμλedgesμ λifAμ appears with multiplicitymμλ inResaB (Bλ). (4.21)

For example, the graph Γ for the symmetric group algebras A=S3 and B=S4 is

A^: B^:

If λB^ then

B^λ= { Pμλ μA^,Pμ A^μandμ λis an edge inΓ } (4.22)

is an index set for a basis of the irreducible B-module Bλ. We think of B^λ as the set of paths to λ in the graph Γ. Let

{ aPQμ μA^, Pμ,Qμ A^μ } and { bPQμνλ λB^, Pμλ,Qνλ B^λ } , (4.23)

be sets of matrix units in the algebras A and B, respectively, so that

aPQμ aSTν = δμν δQS aPTμ and bPQμγλ bSTτνσ = δλσ δQS δγτ bPTμνλ, (4.24)

and such that, for all μA^,P,QA^μ,

aPQμμ= μλ bPQμμλλ (4.25)

where the sum is over all edges μλ in the graph Γ.

Though is not necessary for the following it is conceptually helpful to let C=BAB, let C^=A^, and extend the graph Γ to a graph Γ^ with three levels, so that the edges between level B and level C are the reflections of the edges between level A and level B. In other words,

Γ^has vertices on level C: C^,and an edgeλμ,λB^, μC^,for each edge μλ,μA^,λ B^. (4.26)

For each νC^ define

C^ν= { Pμλν | μA^, λB^, νC^, PμA^μ and μλand λνare edges in Γ^ } , (4.27)

so that C^ν is the set of “paths to ν" in the graph Γ^. Continuing with our previous example, Γ^ is

A^: B^: C^:

Theorem 4.28. Assume A and B are split semisimple, and let the notation and assumption be as in (4.21)–(4.25).

  1. The elements of BAB is given by bPTμγλ bTQγνσ do not depend on the choice of TγA^γ and form a basis of BAB.
  2. For each edge μλ in Γ define a constant εμλ𝔽 by ε1 (bPPμμλ) =εμλ aPPμ. (4.29) Then εμλ is independent of the choice of PμA^μ and ( bPTμγλ bTQγνσ ) ( bRXτπρ bXSπξη ) = δγπ δQR δντ δσρ εγσ ( bPTπμγ bTSγξη ) . Rad(BAB) has basis { bPTμγλ bTQγνσ | εμλ=0 or ενσ=0 } , and the images of the elements ePQμνλσγ =(1εγσ) ( bPTμγλ bTQγνσ ) ,such that εμλ0 and ενσ0, form a set of matrix units in (BAB)/ Rad(BAB).
  3. Let trB:B𝔽 and trA:A𝔽 be traces on B and A, respectively, such that trA(ε1(b))= trB(b), for allbB. (4.30) Let χAμ,μA^, and χBλ,λB^, be the irreducible characters of the algebras A and B, respectively. Define constants trAμ,μA^, and trBλ,λB^, by the equations trA=μA^ trAμχAμ and trB=λB^ trBλχBλ, (4.31) respectively. Then the constants εμλ defined in (4.29) satisfy trBλ= εμλ trAμ.
  4. In the algebra BAB, 11= P μ λ γ bPPμμλ bPPμμγ
  5. By left multiplication, the algebra BAB is a left B-module. If Rad(BAB) is a B-submodule of BAB and ι:B (BAB)/ Rad(BAB) is a left B-module homomorphism then ι(bRSτβπ) =πγ eRSτβππγ .


By (4.11) and (4.25),

B μA^ AμLμ bPQμνλ aPμ Qμνλμ and B νA^ RνAν bPQμνλ rPμν aQν (4.32)

as left A-modules and as right A-modules, respectively. Identify the left and right hand sides of these isomorphisms. Then, by (4.17), the elements of C=BAB given by

mPQμνλσγ = rPμγλγ aTγ aTγ Qγνσγ = bPTμγλ bTQγνσ (4.33)

do not depend on TγA^γ and form a basis of BAB.

(b) By (4.12), the map ε:B𝔽BA is determined by the values

εTQγτλσμμ 𝔽given by εTQγτλσμμ aPPμ=ε ( aPμ Tμγλμ rQτμσμ aPμ ) , (4.34)


εTQγτλσμμ aPPμ = ε ( bPTμγλ bQPτμσ ) = ε1 ( bPTμγλ bQPτμσ ) = δTQγτλσ ε1(bPPμμλ) = δTQγτλσ ε1 ( bPPμμλ bPPμμλ ) = δTQγτλσ εPPμμμλλμ aPPμ.

The matrix Eμ given by (4.14) is diagonal with entries εμλ given by (4.29) and, by (4.17), εμλ is independent of PμA^μ. By Theorem 4.18(a),

mPQμνλσγ mRSτξρηπ= δγπ εQRντσργ mPSμξληγ= δγπ δQRντσρ εγσ mPSμξληγ

in the algebra C. The rest of the statements in part (b) follow from Theorem 4.18(b).

(c) Evaluating the equations in (4.31) and using (4.29) gives

trBλ= trB (bPPμμλ) =trA (ε1(bPPμμλ)) =εμλtrA (aPPμ)= εμλ trAμ. (4.35)

(d) Since

1=Pμλ bPPμμλ in the algebraB,

it follows from part (b) and (4.16) that

11 = ( Pμλ bPPμμλ ) ( Qνγ bQQννγ ) = PμλQνγ δPQδμν ( bPPμμλ bQQννγ ) = P μ λ γ mPPμμλγμ,

giving part (d).

(e) By left multiplication, the algebra BAB is a left B-module. If εγλ0 and εγσ0 then

bRSτβπ ePQμνλσγ =(1εγσ) bRSτβπ ( bPTμγλ bTQγνσ ) =(1εγσ) δSPβμπλ ( bRTτγλ bTQγνσ ) = δSPβμπλ eRQτνπσγ .

Thus, if ι:B (BAB)/ Rad(BAB) is a left B-module homomorphism then

ι(bRSτβπ) = ι (bRSτβπ) ·1 = bRSτβπ Pμλγ ePPμμλλγ = Pμλγ δSPβμπλ eRPτμπλγ = πγ eRSτβππγ.

Notes and References

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).

page history