## The basic construction

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 $𝔽\text{.}$ The fact that $𝔽$ is algebraically closed is only for convenience, to avoid the division rings that could arise in the decomposition of $\stackrel{‾}{A}$ just before (4.8) below.

Let $A\subseteq B$ be an inclusion of algebras. Then $B{\otimes }_{𝔽}B$ is an $\left(A,A\right)\text{-bimodule}$ where $A$ acts on the left by left multiplication and on the right by right multiplication. Fix an $\left(A,A\right)\text{-bimodule}$ homomorphism

$ε:B⊗𝔽B⟶A. (4.1)$

The basic construction is the algebra $B{\otimes }_{A}B$ with product given by

$(b1⊗b2) (b3⊗b4)= b1⊗ε (b2⊗b3) b4,for b1, b2,b3,b4∈ B. (4.2)$

More generally, let $A$ be an algebra and let $L$ be a left $A\text{-module}$ and $R$ a right $A\text{-module.}$ Let

$ε:L⊗𝔽R⟶A, (4.3)$

be an $\left(A,A\right)\text{-bimodule}$ homomorphism. The basic construction is the algebra $R{\otimes }_{A}L$ with product given by

$(r1⊗ℓ1) (r2⊗ℓ2)= r1⊗ε (ℓ1⊗r2) ℓ2,for r1,r2∈R and ℓ1,ℓ2∈L. (4.4)$

Theorem 4.18 below determines, explicitly, the structure of the algebra $R{\otimes }_{A}L\text{.}$

Let $N=\text{Rad}\left(A\right)$ and let

$A‾=A/N, L‾=L/NL, and R‾=R/RN. (4.5)$

Define an $\left(\stackrel{‾}{A},\stackrel{‾}{A}\right)\text{-bimodule}$ homomorphism

$ε‾: L‾⊗𝔽R‾ ⟶ A‾ ℓ‾⊗r‾ ⟼ ε(ℓ⊗r)‾ (4.6)$

where $\stackrel{‾}{\ell }=\ell +NL,$ $\stackrel{‾}{r}=r+RN,$ and $\stackrel{‾}{a}=a+N,$ for $\ell \in L,$ $r\in R,$ and $a\in A\text{.}$ 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

$π: R⊗AL ⟶ R‾⊗A‾L‾ r⊗ℓ ⟼ r‾⊗ℓ‾ hasker(π)= R⊗ANL. (4.7)$

The algebra $\stackrel{‾}{A}$ is a split semisimple algebra (an algebra isomorphic to a direct sum of matrix algebras). Fix an algebra isomorphism

$A‾ ⟶∼ ⨁μ∈A^ Mdμ(𝔽) aPQμ ⟵ EPQμ$

where $\stackrel{^}{A}$ is an index set for the components and ${E}_{PQ}^{\mu }$ is the matrix with $1$ in the $\left(P,Q\right)$ entry of the $\mu \text{th}$ block and $0$ in all other entries. Also, fix isomorphisms

$L‾≅ ⨁μ∈A^ A→μ⊗Lμ andR‾≅ ⨁μ∈A^ Rμ⊗A←μ (4.8)$

where $\stackrel{\to }{A}\mu ,\mu \in \stackrel{^}{A},$ are the simple left $\stackrel{‾}{A}\text{-modules,}$ ${\stackrel{←}{A}}^{\mu },\mu \in \stackrel{^}{A},$ are the simple right $\stackrel{‾}{A}\text{-modules,}$ and ${L}^{\mu },{R}^{\mu },\mu \in \stackrel{^}{A},$ are vector spaces. The practical effect of this set-up is that if ${\stackrel{^}{A}}^{\mu }$ is an index set for a basis $\left\{{r}_{Y}^{\mu }\mid Y\in {\stackrel{^}{R}}^{\mu }\right\}$ of ${R}^{\mu },{\stackrel{^}{L}}^{\mu }$ is an index set for a basis $\left\{{\ell }_{X}^{\mu }\mid X\in {\stackrel{^}{L}}^{\mu }\right\}$ of ${L}^{\mu },$ and ${\stackrel{^}{A}}^{\mu }$ is an index set for bases

${ a→Qμ∣ Q∈A^μ } of A→μand { a←Pμ∣ P∈A^μ } of A←μ (4.9)$

such that

$aSTλ a→Qμ= δλμ δTQ a→Sμ and a←Pμ aSTλ= δλμ δPS a←Tμ, (4.10)$

then

$L‾ has basis { a→Pμ⊗ ℓXμ∣ μ∈A^,P∈ A^μ,X∈ L^μ } and R‾ has basis { rYμ⊗ a←Qμ ∣ μ∈A^,Q∈ A^μ,Y∈ R^μ } (4.11)$

With notation as in (4.9) and (4.11) the map $\stackrel{‾}{\epsilon }:\stackrel{‾}{L}{\otimes }_{𝔽}\stackrel{‾}{R}\to \stackrel{‾}{A}$ is determined by the constants ${\epsilon }_{XY}^{\mu }\in 𝔽$ given by

$ε ( a→Qμ⊗ ℓXμ⊗ rYμ⊗ a←Pμ ) =εXYμ aQPμ (4.12)$

and ${\epsilon }_{XY}^{\mu }$ does not depend on $Q$ and $P$ since

$ε ( a→Sλ⊗ ℓXλ⊗ rYμ⊗ a←Tμ ) = ε ( aSQλ a→Qλ⊗ ℓXλ⊗ rYμ⊗ a←Pμ aPTμ ) = aSQλε ( a→Qλ⊗ ℓXλ⊗ rYμ⊗ a←Pμ ) aPTμ = δλμ aSQμ εXYμ aQPμ aPTμ= εXYμ aSTμ. (4.13)$

For each $\mu \in \stackrel{^}{A}$ construct a matrix

$Eμ= (εXYμ) (4.14)$

and let ${D}^{\mu }=\left({D}_{ST}^{\mu }\right)$ and ${C}^{\mu }=\left({C}_{ZW}^{\mu }\right)$ be invertible matrices such that ${D}^{\mu }{E}^{\mu }{C}^{\mu }$ is a diagonal matrix with diagonal entries denoted as ${\epsilon }_{X}^{\mu },$

$DμEμCμ= diag(εXμ). (4.15)$

In practice ${D}^{\mu }$ and ${C}^{\mu }$ are found by row reducing ${E}^{\mu }$ to its Smith normal form. The ${\epsilon }_{P}^{\mu }$ are the invariant factors of ${E}^{\mu }\text{.}$

For $\mu \in \stackrel{^}{A},$ $X\in {\stackrel{^}{R}}^{\mu },$ $Y\in {\stackrel{^}{L}}^{\mu },$ define the following elements of $\stackrel{‾}{R}{\otimes }_{\stackrel{‾}{A}}\stackrel{‾}{L}:$

$m‾XYμ= rXμ⊗a→Pμ ⊗a←Pμ⊗ ℓYμ,and n‾XYμ= ∑Q1,Q2 CQ1Xμ DYQ2μ m‾Q1Q2μ. (4.16)$

Since

$rSλ⊗ a→Wλ⊗ a←Zμ⊗ ℓTμ = ( rSλ⊗ a→Pλ aPWλ⊗ a←Zμ⊗ ℓTμ ) = ( rSλ⊗ a→Pλ⊗ aPWλ a←Zμ⊗ ℓTμ ) = δλμ δWZ ( rSλ⊗ a→Pλ⊗ a←Pλ⊗ ℓTλ ) , (4.17)$

the element ${\stackrel{‾}{m}}_{XY}^{\mu }$ does not depend on $P$ and $\left\{{\stackrel{‾}{m}}_{XY}^{\mu }\mid \mu \in \stackrel{^}{A},X\in {\stackrel{^}{R}}^{\mu },Y\in {\stackrel{^}{L}}^{\mu }\right\}$ is a basis of $\stackrel{‾}{R}{\otimes }_{\stackrel{‾}{A}}\stackrel{‾}{L}\text{.}$

The following theorem determines the structure of the algebras $R{\otimes }_{A}L$ and $\stackrel{‾}{R}{\otimes }_{\stackrel{‾}{A}}\stackrel{‾}{L}\text{.}$ 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 $\pi :R{\otimes }_{A}L\to \stackrel{‾}{R}{\otimes }_{\stackrel{‾}{A}}\stackrel{‾}{L}$ be as in (4.7) and let $\left\{{k}_{i}\right\}$ be a basis of $\text{ker}\left(\pi \right)=R{\otimes }_{A}NL\text{.}$ Let

$nYTμ∈R⊗AL be such thatπ (nYTμ)= n‾YTμ,$

where the elements ${\stackrel{‾}{n}}_{YT}^{\mu }\in \stackrel{‾}{R}{\otimes }_{\stackrel{‾}{A}}\stackrel{‾}{L}$ are as defined in (4.16).

1. The sets $\left\{{\stackrel{‾}{m}}_{XY}^{\mu }\mid \mu \in \stackrel{^}{A},X\in {\stackrel{^}{R}}^{\mu },Y\in {\stackrel{^}{L}}^{\mu }\right\}$ and $\left\{{\stackrel{‾}{n}}_{XY}^{\mu }\mid \mu \in \stackrel{^}{A},X\in {\stackrel{^}{R}}^{\mu },Y\in {\stackrel{^}{L}}^{\mu }\right\}$ (see (4.16)) are bases of $\stackrel{‾}{R}{\otimes }_{\stackrel{‾}{A}}\stackrel{‾}{L},$ which satisfy $m‾STλ m‾QPμ= δλμ εTQμ m‾SPμ and n‾STλ n‾QPμ= δλμδTQ εTμ n‾SPμ,$ where ${\epsilon }_{TQ}^{\mu }$ and ${\epsilon }_{T}^{\mu }$ are as defined in (4.12) and (4.15).
2. The radical of the algebra $R{\otimes }_{A}L$ is $Rad(R⊗AL)=𝔽-span { ki,nYTμ ∣εYμ=0 or εTμ =0 }$ and the images of the elements $eYTμ= 1εTμ nYTμ, for εYμ≠0 and εTμ ≠0,$ are a set of matrix units in $\left(R{\otimes }_{A}L\right)/\text{Rad}\left(R{\otimes }_{A}L\right)\text{.}$

 Proof. The first statement in (a) follows from the equations in (4.17). If ${\left({C}^{-1}\right)}^{\mu }$ and ${\left({D}^{-1}\right)}^{\mu }$ are inverses of the matrices ${C}^{\mu }$ and ${D}^{\mu }$ then $∑X,Y (C-1)XSμ (D-1)TYμ n‾XY = ∑X,Y,Q1,Q2 (C-1)XSμ CQ1Xμ m‾Q1Q2 DYQ2μ (D-1)TYμ = ∑Q1,Q2 δSQ1 δQ2T m‾Q1Q2μ =m‾STμ,$ and so the elements ${\stackrel{‾}{m}}_{ST}^{\mu }$ can be written as linear combinations of the ${\stackrel{‾}{n}}_{XY}^{\mu }\text{.}$ This establishes the second statement in (a). By direct computation, using (4.10) and (4.12), $m‾STλ m‾QPμ = ( rSλ⊗ a→Wλ⊗ a←Wλ⊗ ℓTλ ) ( rQμ⊗ a→Zμ⊗ a←Zμ⊗ ℓPμ ) = rSλ⊗ a→Wλ⊗ε ( a←Wλ⊗ ℓTλ⊗ rQμ⊗ a→Zμ ) a←Zμ⊗ ℓPμ = δλμ ( rSλ⊗ a→Wλ⊗ εTQλ a‾WZλ⊗ ℓPλ ) = δλμ εTQλ ( rSλ⊗ a→Wλ⊗ a←Wλ⊗ ℓPλ ) = δλμ εTQλ m‾SPλ,$ and $n‾STλ n‾UVμ = ∑Q1,Q2,Q3,Q4 CQ1Sλ DTQ2λ m‾Q1Q2λ CQ3Uμ DVQ4μ m‾Q3Q4μ = ∑Q1,Q2,Q3,Q4 δλμ CQ1Sλ DTQ2λ εQ2Q3μ CQ3Uμ DVQ4μ m‾Q1Q4μ = δλμ ∑Q1,Q4 δTY εTμ CQ1Sμ DTQ4μ m‾Q1Q4μ =δλμ δTU εTμ n‾SVμ.$ (b) Let $N=\text{Rad}\left(A\right)$ as in (4.5). If ${r}_{1}\otimes {n}_{1}{\ell }_{1},{r}_{2}\otimes {n}_{2}{\ell }_{2}\in R{\otimes }_{A}NL$ with ${n}_{1}\in {N}^{i}$ for some $i\in {ℤ}_{>0}$ then $(r1⊗n1ℓ1) (r2⊗n2ℓ2) = r1⊗ε (n1ℓ1⊗r2) n2ℓ2 = r1⊗n1ε (ℓ1⊗r2) n2ℓ2∈R⊗A Ni+1L.$ Since $N$ is a nilpotent ideal of $A$ it follows that $\text{ker}\left(\pi \right)=R{\otimes }_{A}NL$ is a nilpotent ideal of $R{\otimes }_{A}L\text{.}$ So $\text{ker}\left(\pi \right)\subseteq \text{Rad}\left(R{\otimes }_{A}L\right)\text{.}$ Let $I=𝔽-span { ki,nYTμ ∣εYμ=0 or εTμ=0 } .$ The multiplication rule for the ${\stackrel{‾}{n}}_{YT}$ implies that $\pi \left(I\right)$ is an ideal of $\stackrel{‾}{R}{\otimes }_{\stackrel{‾}{A}}\stackrel{‾}{L}$ and thus, by the correspondence between ideals of $\stackrel{‾}{R}{\otimes }_{\stackrel{‾}{A}}\stackrel{‾}{L}$ and ideals of $R{\otimes }_{A}L$ which contain $\text{ker}\left(\pi \right),$ $I$ is an ideal of $R{\otimes }_{A}L\text{.}$ If ${\stackrel{‾}{n}}_{{Y}_{1}{T}_{1}}^{\mu },{\stackrel{‾}{n}}_{{Y}_{2}{T}_{2}}^{\mu },{\stackrel{‾}{n}}_{{Y}_{3}{T}_{3}}^{\mu }\in \left\{{\stackrel{‾}{n}}_{YT}^{\mu }\mid {\epsilon }_{Y}^{\mu }=0 \text{or} {\epsilon }_{T}^{\mu }=0\right\}$ then $n‾Y1T1μ n‾Y2T2μ n‾Y3T3μ= δT1Y2 εY2μ n‾Y1T2μ n‾Y3T3μ= δT1Y2 δT2Y3 εY2μ εT2μ n‾Y1T3μ =0,$ since ${\epsilon }_{{Y}_{2}}^{\mu }=0$ or ${\epsilon }_{{T}_{2}}^{\mu }=0\text{.}$ Thus any product ${\stackrel{‾}{n}}_{{Y}_{1}{T}_{1}}^{\mu }{\stackrel{‾}{n}}_{{Y}_{2}{T}_{2}}^{\mu }{\stackrel{‾}{n}}_{{Y}_{3}{T}_{3}}^{\mu }$ of three basis elements of $I$ is in $\text{ker}\left(\pi \right)\text{.}$ Since $\text{ker}\left(\pi \right)$ is a nilpotent ideal of $R{\otimes }_{A}L$ it follows that $I$ is an ideal of $R{\otimes }_{A}L$ consisting of nilpotent elements. So $I\subseteq \text{Rad}\left(R{\otimes }_{A}L\right)\text{.}$ Since $eYTλ eUVμ= 1εTλ 1εVμ nYTλ nUVμ= δλμ δTY 1εTλεVλ εTλnYVλ =δλμ δTU eYVλ mod I,$ the images of the elements ${e}_{YT}^{\lambda }$ in (4.7) form a set of matrix units in the algebra $\left(R{\otimes }_{A}L\right)/I\text{.}$ Thus $\left(R{\otimes }_{A}L\right)/I$ is a split semisimple algebra and so $I\supseteq \text{Rad}\left(R{\otimes }_{A}L\right)\text{.}$ $\square$

Basic constructions for $A\subseteq B$

Let $A\subseteq B$ be an inclusion of algebras. Let ${\epsilon }_{1}:B\to A$ be an $\left(A,A\right)$ bimodule homomorphism and use the $\left(A,A\right)\text{-bimodule}$ homomorphism

$ε: B⊗𝔽B ⟶ A b1⊗b2 ⟼ ε1(b1b2) (4.19)$

and (4.2) to define the basic construction $B{\otimes }_{A}B\text{.}$ Theorem 4.28 below provides the structure of $B{\otimes }_{A}B$ in the case where both $A$ and $B$ are split semisimple.

Let us record the following facts:

(4.20a) If $p\in A$ and $pAp=𝔽p$ then $\left(p\otimes 1\right)\left(B{\otimes }_{A}B\right)\left(p\otimes 1\right)=𝔽·\left(p\otimes 1\right)\text{.}$
(4.20b) If $p$ is an idempotent of $A$ and $pAp=𝔽p$ then ${\epsilon }_{1}\left(1\right)\in 𝔽\text{.}$
(4.20c) If $p\in A,$ $pAp=𝔽p$ and if ${\epsilon }_{1}\left(1\right)\ne 0,$ then $\frac{1}{\epsilon \left(1\right)}\left(p\otimes 1\right)$ is a minimal idempotent in $B{\otimes }_{A}B\text{.}$

These are justfied as follows. If $p\in A$ and $pAp=𝔽p$ and ${b}_{1},{b}_{2}\in B$ then $\left(p\otimes 1\right)\left({b}_{1}\otimes {b}_{2}\right)\left(p\otimes 1\right)=\left(p\otimes {\epsilon }_{1}\left({b}_{1}\right){b}_{2}\right)\left(p\otimes 1\right)=p\otimes {\epsilon }_{1}\left({b}_{1}\right){\epsilon }_{1}\left({b}_{2}p\right)=p{\epsilon }_{1}\left({b}_{1}\right){\epsilon }_{1}\left({b}_{2}\right)p\otimes 1=\xi p\otimes 1,$ for some constant $\xi \in 𝔽\text{.}$ This establishes (a). If $p$ is an indempotent of $A$ and $pAp=𝔽p$ then $p{\epsilon }_{1}\left(1\right)p={\epsilon }_{1}\left({p}^{2}\right)=\left({\epsilon }_{1}\left(1·p\right)\right)={\epsilon }_{1}\left(1\right)p$ and so (b) holds. If $p\in A$ and $pAp=𝔽p$ then ${\left(p\otimes 1\right)}^{2}={\epsilon }_{1}\left(1\right)\left(p\otimes 1\right)$ and so, if ${\epsilon }_{1}\left(1\right)\ne 0,$ then $\frac{1}{\epsilon \left(1\right)}\left(p\otimes 1\right)$ is a minimal idempotent in $B{\otimes }_{A}B\text{.}$

Assume $A$ and $B$ are split semisimple. Let

$A^ be an index set for irreducible A-modules Aμ, B^ be an index set for irreducible B-modules Bλ, and let A^μ= {P→μ} be an index set for a basis of the simple A -module Aμ,$

for each $\mu \in \stackrel{^}{A}$ (the composite $P\to \mu$ is viewed as a single symbol). We think of ${\stackrel{^}{A}}^{\mu }$ as the set of “paths to $\mu \text{"}$ in the two-level graph

$Γ with vertices on level A: A^, vertices on level B: B^,and mμλ edges μ→ λ if Aμ appears with multiplicity mμλ in ResaB (Bλ). (4.21)$

For example, the graph $\Gamma$ for the symmetric group algebras $A=ℂ{S}_{3}$ and $B=ℂ{S}_{4}$ is

$\stackrel{^}{A}: \stackrel{^}{B}:$

If $\lambda \in \stackrel{^}{B}$ then

$B^λ= { P→μ→λ∣ μ∈A^,P→μ∈ A^μ and μ ∈λ is an edge in Γ } (4.22)$

is an index set for a basis of the irreducible $B\text{-module}$ ${B}^{\lambda }\text{.}$ We think of ${\stackrel{^}{B}}^{\lambda }$ as the set of paths to $\lambda$ in the graph $\Gamma \text{.}$ Let

${ aPQμ ∣μ∈A^, P→μ,Q→μ∈ A^μ } and { bPQμνλ ∣λ∈B^, P→μ→λ,Q→ν→λ ∈B^λ } , (4.23)$

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 $\mu \in \stackrel{^}{A},P,Q\in {\stackrel{^}{A}}^{\mu },$

$aPQμμ= ∑μ→λ bPQμμλλ (4.25)$

where the sum is over all edges $\mu \to \lambda$ in the graph $\Gamma \text{.}$

Though is not necessary for the following it is conceptually helpful to let $C=B{\otimes }_{A}B,$ let $\stackrel{^}{C}=\stackrel{^}{A},$ and extend the graph $\Gamma$ to a graph $\stackrel{^}{\Gamma }$ 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,

$Γ^ has vertices on level C: C^,and an edge λ→μ,λ∈B^, μ∈C^, for each edge μ→λ,μ∈A^,λ ∈B^. (4.26)$

For each $\nu \in \stackrel{^}{C}$ define

$C^ν= { P→μ→λ→ν | μ∈A^, λ∈B^, ν∈C^, P→μ∈A^μ and μ→λ and λ→ν are edges in Γ^ } , (4.27)$

so that ${\stackrel{^}{C}}^{\nu }$ is the set of “paths to $\nu \text{"}$ in the graph $\stackrel{^}{\Gamma }\text{.}$ Continuing with our previous example, $\stackrel{^}{\Gamma }$ is

$\stackrel{^}{A}: \stackrel{^}{B}: \stackrel{^}{C}:$

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 $B{\otimes }_{A}B$ is given by $bPTμγλ ⊗ bTQγνσ$ do not depend on the choice of $T\to \gamma \in {\stackrel{^}{A}}^{\gamma }$ and form a basis of $B{\otimes }_{A}B\text{.}$
2. For each edge $\mu \to \lambda$ in $\Gamma$ define a constant ${\epsilon }_{\mu }^{\lambda }\in 𝔽$ by $ε1 (bPPμμλ) =εμλ aPPμ. (4.29)$ Then ${\epsilon }_{\mu }^{\lambda }$ is independent of the choice of $P\to \mu \in {\stackrel{^}{A}}^{\mu }$ and $( bPTμγλ ⊗ bTQγνσ ) ( bRXτπρ ⊗ bXSπξη ) = δγπ δQR δντ δσρ εγσ ( bPTπμγ ⊗ bTSγξη ) . Rad(B⊗AB) has basis { bPTμγλ ⊗ bTQγνσ | εμλ=0 or ενσ=0 } ,$ and the images of the elements $ePQμνλσγ =(1εγσ) ( bPTμγλ ⊗ bTQγνσ ) ,such that εμλ≠0 and ενσ≠0,$ form a set of matrix units in $\left(B{\otimes }_{A}B\right)/\text{Rad}\left(B{\otimes }_{A}B\right)\text{.}$
3. Let ${\text{tr}}_{B}:B\to 𝔽$ and ${\text{tr}}_{A}:A\to 𝔽$ be traces on $B$ and $A,$ respectively, such that $trA(ε1(b))= trB(b), for all b∈B. (4.30)$ Let ${\chi }_{A}^{\mu },\mu \in \stackrel{^}{A},$ and ${\chi }_{B}^{\lambda },\lambda \in \stackrel{^}{B},$ be the irreducible characters of the algebras $A$ and $B,$ respectively. Define constants ${\text{tr}}_{A}^{\mu },\mu \in \stackrel{^}{A},$ and ${\text{tr}}_{B}^{\lambda },\lambda \in \stackrel{^}{B},$ by the equations $trA=∑μ∈A^ trAμχAμ and trB=∑λ∈B^ trBλχBλ, (4.31)$ respectively. Then the constants ${\epsilon }_{\mu }^{\lambda }$ defined in (4.29) satisfy $trBλ= εμλ trAμ.$
4. In the algebra $B{\otimes }_{A}B,$ $1⊗1= ∑ P ↓ \mu ↓ ↓ \lambda \gamma bPPμμλ⊗ bPPμμγ$
5. By left multiplication, the algebra $B{\otimes }_{A}B$ is a left $B\text{-module.}$ If $\text{Rad}\left(B{\otimes }_{A}B\right)$ is a $B\text{-submodule}$ of $B{\otimes }_{A}B$ and $\iota :B\to \left(B{\otimes }_{A}B\right)/\text{Rad}\left(B{\otimes }_{A}B\right)$ is a left $B\text{-module}$ homomorphism then $ι(bRSτβπ) =∑π→γ eRSτβππγ .$

 Proof. By (4.11) and (4.25), $B ⟶∼ ⨁μ∈A^ A→μ⊗Lμ bPQμνλ ⟼ a→Pμ⊗ ℓQμνλμ and B ⟶∼ ⨁ν∈A^ Rν⊗A←ν bPQμνλ ⟼ rPμν⊗ a←Qν (4.32)$ as left $A\text{-modules}$ and as right $A\text{-modules,}$ respectively. Identify the left and right hand sides of these isomorphisms. Then, by (4.17), the elements of $C=B{\otimes }_{A}B$ given by $m∼PQμνλσγ = rPμγλγ ⊗ a←Tγ ⊗ a→Tγ ⊗ ℓQγνσγ = bPTμγλ ⊗ bTQγνσ (4.33)$ do not depend on $T\to \gamma \in {\stackrel{^}{A}}^{\gamma }$ and form a basis of $B{\otimes }_{A}B\text{.}$ (b) By (4.12), the map $\epsilon :B{\otimes }_{𝔽}B\to A$ is determined by the values $εTQγτλσμμ ∈𝔽given by εTQγτλσμμ aPPμ=ε ( a→Pμ⊗ ℓTμγλμ⊗ rQτμσμ⊗ a←Pμ ) , (4.34)$ since $εTQγτλσμμ aPPμ = ε ( bPTμγλ⊗ bQPτμσ ) = ε1 ( bPTμγλ⊗ bQPτμσ ) = δTQγτλσ ε1(bPPμμλ) = δTQγτλσ ε1 ( bPPμμλ bPPμμλ ) = δTQγτλσ εPPμμμλλμ aPPμ.$ The matrix ${E}^{\mu }$ given by (4.14) is diagonal with entries ${\epsilon }_{\mu }^{\lambda }$ given by (4.29) and, by (4.17), ${\epsilon }_{\mu }^{\lambda }$ is independent of $P\to \mu \in {\stackrel{^}{A}}^{\mu }\text{.}$ By Theorem 4.18(a), $m‾PQμνλσγ m‾RSτξρηπ= δγπ εQRντσργ m‾PSμξληγ= δγπ δQRντσρ εγσ m‾PSμξληγ$ in the algebra $C\text{.}$ 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 $1=∑P→μ→λ bPPμμλ in the algebra B,$ it follows from part (b) and (4.16) that $1⊗1 = ( ∑P→μ→λ bPPμμλ ) ⊗ ( ∑Q→ν→γ bQQννγ ) = ∑P→μ→λQ→ν→γ δPQδμν ( bPPμμλ⊗ bQQννγ ) = ∑ P ↓ \mu ↓ ↓ \lambda \gamma m‾PPμμλγμ,$ giving part (d). (e) By left multiplication, the algebra $B{\otimes }_{A}B$ is a left $B\text{-module.}$ If ${\epsilon }_{\gamma }^{\lambda }\ne 0$ and ${\epsilon }_{\gamma }^{\sigma }\ne 0$ then $bRSτβπ ePQμνλσγ =(1εγσ) bRSτβπ ( bPTμγλ⊗ bTQγνσ ) =(1εγσ) δSPβμπλ ( bRTτγλ⊗ bTQγνσ ) = δSPβμπλ eRQτνπσγ .$ Thus, if $\iota :B\to \left(B{\otimes }_{A}B\right)/\text{Rad}\left(B{\otimes }_{A}B\right)$ is a left $B\text{-module}$ homomorphism then $ι(bRSτβπ) = ι (bRSτβπ) ·1 = bRSτβπ ∑P→μ→λ→γ ePPμμλλγ = ∑P→μ→λ→γ δSPβμπλ eRPτμπλγ = ∑π→γ eRSτβππγ.$ $\square$

## 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).