The basic construction

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

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 $\bar{A}$ just before (4.8) below.

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

$$\begin{array}{cc}\epsilon :B{\otimes }_{𝔽}B⟶A\text{.}& \text{(4.1)}\end{array}$$

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

$$\begin{array}{cc}(b_1\otimes b_2)(b_3\otimes b_4)=b_1\otimes \epsilon (b_2\otimes b_3)b_4,\text{for}\hspace{0.17em}{b}_1,b_2,b_3,b_4\in B\text{.}& \text{(4.2)}\end{array}$$

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

$$\begin{array}{cc}\epsilon :L{\otimes }_{𝔽}R⟶A,& \text{(4.3)}\end{array}$$

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

$$\begin{array}{cc}(r_1\otimes {\ell }_1)(r_2\otimes {\ell }_2)=r_1\otimes \epsilon ({\ell }_1\otimes r_2){\ell }_2,\text{for}\hspace{0.17em}{r}_1,r_2\in R\hspace{0.17em}\text{and}\hspace{0.17em}{\ell }_1,{\ell }_2\in L\text{.}& \text{(4.4)}\end{array}$$

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

$$\begin{array}{cc}\bar{A}=A/N,\bar{L}=L/NL,\text{and}\bar{R}=R/RN\text{.}& \text{(4.5)}\end{array}$$

Define an $(\bar{A},\bar{A})\text{-bimodule}$ homomorphism

$$\begin{array}{cc}\begin{array}{cccc}\bar{\epsilon }:& \bar{L}{\otimes }_{𝔽}\bar{R}& ⟶& \bar{A}\\ & \bar{\ell }\otimes \bar{r}& ⟼& \overline{\epsilon (\ell \otimes r)}\end{array}& \text{(4.6)}\end{array}$$

where $\bar{\ell }=\ell +NL,$ $\bar{r}=r+RN,$ and $\bar{a}=a+N,$ 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

$$\begin{array}{cc}\begin{array}{cccc}\pi :& R{\otimes }_{A}L& ⟶& \bar{R}{\otimes }_{\bar{A}}\bar{L}\\ & r\otimes \ell & ⟼& \bar{r}\otimes \bar{\ell }\end{array}\text{has}\text{ker}\left(\pi \right)=R{\otimes }_{A}NL\text{.}& \text{(4.7)}\end{array}$$

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

$$\begin{array}{ccc}\bar{A}& \stackrel{\sim }{⟶}& \underset{\mu \in \hat{A}}{⨁}{M}_{d_{\mu }}\left(𝔽\right)\\ a_{PQ}^{\mu }& ⟵& E_{PQ}^{\mu }\end{array}$$

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

$$\begin{array}{cc}\bar{L}\cong \underset{\mu \in \hat{A}}{⨁}{\overrightarrow{A}}^{\mu }\otimes L^{\mu }\text{and}\bar{R}\cong \underset{\mu \in \hat{A}}{⨁}{R}^{\mu }\otimes {\overleftarrow{A}}^{\mu }& \text{(4.8)}\end{array}$$

where $\overrightarrow{A}\mu ,\mu \in \hat{A},$ are the simple left $\bar{A}\text{-modules,}$ ${\overleftarrow{A}}^{\mu },\mu \in \hat{A},$ are the simple right $\bar{A}\text{-modules,}$ and $L^{\mu },R^{\mu },\mu \in \hat{A},$ are vector spaces. The practical effect of this set-up is that if ${\hat{A}}^{\mu }$ is an index set for a basis $\{r_Y^{\mu }\mid Y\in {\hat{R}}^{\mu }\}$ of $R^{\mu },{\hat{L}}^{\mu }$ is an index set for a basis $\{{\ell }_X^{\mu }\mid X\in {\hat{L}}^{\mu }\}$ of $L^{\mu },$ and ${\hat{A}}^{\mu }$ is an index set for bases

$$\begin{array}{cc}\{{\overrightarrow{a}}_Q^{\mu }\mid Q\in {\hat{A}}^{\mu }\}\hspace{0.17em}\text{of}\hspace{0.17em}{\overrightarrow{A}}^{\mu }\text{and}\{{\overleftarrow{a}}_P^{\mu }\mid P\in {\hat{A}}^{\mu }\}\hspace{0.17em}\text{of}\hspace{0.17em}{\overleftarrow{A}}^{\mu }& \text{(4.9)}\end{array}$$

such that

$$\begin{array}{cc}{a}_{ST}^{\lambda }{\overrightarrow{a}}_Q^{\mu }={\delta }_{\lambda \mu }{\delta }_{TQ}{\overrightarrow{a}}_S^{\mu }\text{and}{\overleftarrow{a}}_P^{\mu }{a}_{ST}^{\lambda }={\delta }_{\lambda \mu }{\delta }_{PS}{\overleftarrow{a}}_T^{\mu },& \text{(4.10)}\end{array}$$

then

$$\begin{array}{cc}\begin{array}{c}\bar{L}\hspace{0.17em}\text{has basis}\hspace{0.17em}\{{\overrightarrow{a}}_P^{\mu }\otimes {\ell }_X^{\mu }\mid \mu \in \hat{A},P\in {\hat{A}}^{\mu },X\in {\hat{L}}^{\mu }\}\text{and}\\ \bar{R}\hspace{0.17em}\text{has basis}\hspace{0.17em}\{r_Y^{\mu }\otimes {\overleftarrow{a}}_Q^{\mu }\mid \mu \in \hat{A},Q\in {\hat{A}}^{\mu },Y\in {\hat{R}}^{\mu }\}\end{array}& \text{(4.11)}\end{array}$$

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

$$\begin{array}{cc}\epsilon ({\overrightarrow{a}}_Q^{\mu }\otimes {\ell }_X^{\mu }\otimes r_Y^{\mu }\otimes {\overleftarrow{a}}_P^{\mu })={\epsilon }_{XY}^{\mu }{a}_{QP}^{\mu }& \text{(4.12)}\end{array}$$

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

$$\begin{array}{cc}\begin{array}{ccc}\epsilon ({\overrightarrow{a}}_S^{\lambda }\otimes {\ell }_X^{\lambda }\otimes r_Y^{\mu }\otimes {\overleftarrow{a}}_T^{\mu })& =& \epsilon (a_{SQ}^{\lambda }{\overrightarrow{a}}_Q^{\lambda }\otimes {\ell }_X^{\lambda }\otimes r_Y^{\mu }\otimes {\overleftarrow{a}}_P^{\mu }{a}_{PT}^{\mu })\\ & =& a_{SQ}^{\lambda }\epsilon ({\overrightarrow{a}}_Q^{\lambda }\otimes {\ell }_X^{\lambda }\otimes r_Y^{\mu }\otimes {\overleftarrow{a}}_P^{\mu })a_{PT}^{\mu }\\ & =& {\delta }_{\lambda \mu }{a}_{SQ}^{\mu }{\epsilon }_{XY}^{\mu }{a}_{QP}^{\mu }{a}_{PT}^{\mu }={\epsilon }_{XY}^{\mu }{a}_{ST}^{\mu }\text{.}\end{array}& \text{(4.13)}\end{array}$$

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

$$\begin{array}{cc}{E}^{\mu }=\left({\epsilon }_{XY}^{\mu }\right)& \text{(4.14)}\end{array}$$

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 ${\epsilon }_X^{\mu },$

$$\begin{array}{cc}{D}^{\mu }{E}^{\mu }{C}^{\mu }=\text{diag}\left({\epsilon }_X^{\mu }\right)\text{.}& \text{(4.15)}\end{array}$$

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

$$\begin{array}{cc}{\bar{m}}_{XY}^{\mu }=r_X^{\mu }\otimes {\overrightarrow{a}}_P^{\mu }\otimes {\overleftarrow{a}}_P^{\mu }\otimes {\ell }_Y^{\mu },\text{and}{\bar{n}}_{XY}^{\mu }=\sum _{Q_1,Q_2}{C}_{Q_{1}X}^{\mu }{D}_{Y{Q}_2}^{\mu }{\bar{m}}_{Q_1{Q}_2}^{\mu }\text{.}& \text{(4.16)}\end{array}$$

Since

$$\begin{array}{cc}\begin{array}{ccc}{r}_S^{\lambda }\otimes {\overrightarrow{a}}_W^{\lambda }\otimes {\overleftarrow{a}}_Z^{\mu }\otimes {\ell }_T^{\mu }& =& (r_S^{\lambda }\otimes {\overrightarrow{a}}_P^{\lambda }{a}_{PW}^{\lambda }\otimes {\overleftarrow{a}}_Z^{\mu }\otimes {\ell }_T^{\mu })\\ & =& (r_S^{\lambda }\otimes {\overrightarrow{a}}_P^{\lambda }\otimes a_{PW}^{\lambda }{\overleftarrow{a}}_Z^{\mu }\otimes {\ell }_T^{\mu })\\ & =& {\delta }_{\lambda \mu }{\delta }_{WZ}(r_S^{\lambda }\otimes {\overrightarrow{a}}_P^{\lambda }\otimes {\overleftarrow{a}}_P^{\lambda }\otimes {\ell }_T^{\lambda }),\end{array}& \text{(4.17)}\end{array}$$

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

The following theorem determines the structure of the algebras $R{\otimes }_{A}L$ and $\bar{R}{\otimes }_{\bar{A}}\bar{L}\text{.}$ 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 $\pi :R{\otimes }_{A}L\to \bar{R}{\otimes }_{\bar{A}}\bar{L}$ 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

$$n_{YT}^{\mu }\in R{\otimes }_{A}L\text{be such that}\pi \left(n_{YT}^{\mu }\right)={\bar{n}}_{YT}^{\mu },$$

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

1. The sets $\{{\bar{m}}_{XY}^{\mu }\mid \mu \in \hat{A},X\in {\hat{R}}^{\mu },Y\in {\hat{L}}^{\mu }\}$ and $\{{\bar{n}}_{XY}^{\mu }\mid \mu \in \hat{A},X\in {\hat{R}}^{\mu },Y\in {\hat{L}}^{\mu }\}$ (see (4.16)) are bases of $\bar{R}{\otimes }_{\bar{A}}\bar{L},$ which satisfy $${\bar{m}}_{ST}^{\lambda }{\bar{m}}_{QP}^{\mu }={\delta }_{\lambda \mu }{\epsilon }_{TQ}^{\mu }{\bar{m}}_{SP}^{\mu }\text{and}{\bar{n}}_{ST}^{\lambda }{\bar{n}}_{QP}^{\mu }={\delta }_{\lambda \mu }{\delta }_{TQ}{\epsilon }_T^{\mu }{\bar{n}}_{SP}^{\mu },$$ 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 $$\text{Rad}\left(R{\otimes }_{A}L\right)=𝔽\text{-span}\{k_i,n_{YT}^{\mu }\mid {\epsilon }_Y^{\mu }=0\hspace{0.17em}\text{or}\hspace{0.17em}{\epsilon }_T^{\mu }=0\}$$ and the images of the elements $$e_{YT}^{\mu }=\frac{1}{{\epsilon }_T^{\mu }}{n}_{YT}^{\mu },\text{for}\hspace{0.17em}{\epsilon }_Y^{\mu }\ne 0\hspace{0.17em}\text{and}\hspace{0.17em}{\epsilon }_T^{\mu }\ne 0,$$ 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 $$\begin{array}{ccc}\sum _{X,Y}{\left(C^{-1}\right)}_{XS}^{\mu }{\left(D^{-1}\right)}_{TY}^{\mu }{\bar{n}}_{XY}& =& \sum _{X,Y,Q_1,Q_2}{\left(C^{-1}\right)}_{XS}^{\mu }{C}_{Q_{1}X}^{\mu }{\bar{m}}_{Q_1{Q}_2}{D}_{Y{Q}_2}^{\mu }{\left(D^{-1}\right)}_{TY}^{\mu }\\ & =& \sum _{Q_1,Q_2}{\delta }_{S{Q}_1}{\delta }_{Q_{2}T}{\bar{m}}_{Q_1{Q}_2}^{\mu }={\bar{m}}_{ST}^{\mu },\end{array}$$ and so the elements ${\bar{m}}_{ST}^{\mu }$ can be written as linear combinations of the ${\bar{n}}_{XY}^{\mu }\text{.}$ This establishes the second statement in (a). By direct computation, using (4.10) and (4.12), $$\begin{array}{ccc}{\bar{m}}_{ST}^{\lambda }{\bar{m}}_{QP}^{\mu }& =& (r_S^{\lambda }\otimes {\overrightarrow{a}}_W^{\lambda }\otimes {\overleftarrow{a}}_W^{\lambda }\otimes {\ell }_T^{\lambda })(r_Q^{\mu }\otimes {\overrightarrow{a}}_Z^{\mu }\otimes {\overleftarrow{a}}_Z^{\mu }\otimes {\ell }_P^{\mu })\\ & =& r_S^{\lambda }\otimes {\overrightarrow{a}}_W^{\lambda }\otimes \epsilon ({\overleftarrow{a}}_W^{\lambda }\otimes {\ell }_T^{\lambda }\otimes r_Q^{\mu }\otimes {\overrightarrow{a}}_Z^{\mu }){\overleftarrow{a}}_Z^{\mu }\otimes {\ell }_P^{\mu }\\ & =& {\delta }_{\lambda \mu }(r_S^{\lambda }\otimes {\overrightarrow{a}}_W^{\lambda }\otimes {\epsilon }_{TQ}^{\lambda }{\bar{a}}_{WZ}^{\lambda }\otimes {\ell }_P^{\lambda })\\ & =& {\delta }_{\lambda \mu }{\epsilon }_{TQ}^{\lambda }(r_S^{\lambda }\otimes {\overrightarrow{a}}_W^{\lambda }\otimes {\overleftarrow{a}}_W^{\lambda }\otimes {\ell }_P^{\lambda })={\delta }_{\lambda \mu }{\epsilon }_{TQ}^{\lambda }{\bar{m}}_{SP}^{\lambda },\end{array}$$ and $$\begin{array}{ccc}{\bar{n}}_{ST}^{\lambda }{\bar{n}}_{UV}^{\mu }& =& \sum _{Q_1,Q_2,Q_3,Q_4}{C}_{Q_{1}S}^{\lambda }{D}_{T{Q}_2}^{\lambda }{\bar{m}}_{Q_1{Q}_2}^{\lambda }{C}_{Q_{3}U}^{\mu }{D}_{V{Q}_4}^{\mu }{\bar{m}}_{Q_3{Q}_4}^{\mu }\\ & =& \sum _{Q_1,Q_2,Q_3,Q_4}{\delta }_{\lambda \mu }{C}_{Q_{1}S}^{\lambda }{D}_{T{Q}_2}^{\lambda }{\epsilon }_{Q_2{Q}_3}^{\mu }{C}_{Q_{3}U}^{\mu }{D}_{V{Q}_4}^{\mu }{\bar{m}}_{Q_1{Q}_4}^{\mu }\\ & =& {\delta }_{\lambda \mu }\sum _{Q_1,Q_4}{\delta }_{TY}{\epsilon }_T^{\mu }{C}_{Q_{1}S}^{\mu }{D}_{T{Q}_4}^{\mu }{\bar{m}}_{Q_1{Q}_4}^{\mu }={\delta }_{\lambda \mu }{\delta }_{TU}{\epsilon }_T^{\mu }{\bar{n}}_{SV}^{\mu }\text{.}\end{array}$$ (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 $i\in {\mathbb{Z}}_{>0}$ then $$\begin{array}{ccc}(r_1\otimes n_1{\ell }_1)(r_2\otimes n_2{\ell }_2)& =& r_1\otimes \epsilon (n_1{\ell }_1\otimes r_2)n_2{\ell }_2\\ & =& r_1\otimes n_1\epsilon ({\ell }_1\otimes r_2)n_2{\ell }_2\in R{\otimes }_A{N}^{i+1}L\text{.}\end{array}$$ 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 $$I=𝔽\text{-span}\{k_i,n_{YT}^{\mu }\mid {\epsilon }_Y^{\mu }=0\hspace{0.17em}\text{or}\hspace{0.17em}{\epsilon }_T^{\mu }=0\}\text{.}$$ The multiplication rule for the ${\bar{n}}_{YT}$ implies that $\pi \left(I\right)$ is an ideal of $\bar{R}{\otimes }_{\bar{A}}\bar{L}$ and thus, by the correspondence between ideals of $\bar{R}{\otimes }_{\bar{A}}\bar{L}$ 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 ${\bar{n}}_{Y_1{T}_1}^{\mu },{\bar{n}}_{Y_2{T}_2}^{\mu },{\bar{n}}_{Y_3{T}_3}^{\mu }\in \{{\bar{n}}_{YT}^{\mu }\mid {\epsilon }_Y^{\mu }=0\hspace{0.17em}\text{or}\hspace{0.17em}{\epsilon }_T^{\mu }=0\}$ then $${\bar{n}}_{Y_1{T}_1}^{\mu }{\bar{n}}_{Y_2{T}_2}^{\mu }{\bar{n}}_{Y_3{T}_3}^{\mu }={\delta }_{T_1{Y}_2}{\epsilon }_{Y_2}^{\mu }{\bar{n}}_{Y_1{T}_2}^{\mu }{\bar{n}}_{Y_3{T}_3}^{\mu }={\delta }_{T_1{Y}_2}{\delta }_{T_2{Y}_3}{\epsilon }_{Y_2}^{\mu }{\epsilon }_{T_2}^{\mu }{\bar{n}}_{Y_1{T}_3}^{\mu }=0,$$ since ${\epsilon }_{Y_2}^{\mu }=0$ or ${\epsilon }_{T_2}^{\mu }=0\text{.}$ Thus any product ${\bar{n}}_{Y_1{T}_1}^{\mu }{\bar{n}}_{Y_2{T}_2}^{\mu }{\bar{n}}_{Y_3{T}_3}^{\mu }$ 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 $$e_{YT}^{\lambda }{e}_{UV}^{\mu }=\frac{1}{{\epsilon }_T^{\lambda }}\frac{1}{{\epsilon }_V^{\mu }}{n}_{YT}^{\lambda }{n}_{UV}^{\mu }={\delta }_{\lambda \mu }{\delta }_{TY}\frac{1}{{\epsilon }_T^{\lambda }{\epsilon }_V^{\lambda }}{\epsilon }_T^{\lambda }{n}_{YV}^{\lambda }={\delta }_{\lambda \mu }{\delta }_{TU}{e}_{YV}^{\lambda }\text{mod}\hspace{0.17em}I,$$ 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 $(A,A)$ bimodule homomorphism and use the $(A,A)\text{-bimodule}$ homomorphism

$$\begin{array}{cc}\begin{array}{cccc}\epsilon :& B{\otimes }_{𝔽}B& ⟶& A\\ & b_1\otimes b_2& ⟼& {\epsilon }_1\left(b_1{b}_2\right)\end{array}& \text{(4.19)}\end{array}$$

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 $(p\otimes 1)\left(B{\otimes }_{A}B\right)(p\otimes 1)=𝔽·(p\otimes 1)\text{.}$
(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 $\frac{1}{\epsilon \left(1\right)}(p\otimes 1)$ 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 $(p\otimes 1)(b_1\otimes b_2)(p\otimes 1)=(p\otimes {\epsilon }_1\left(b_1\right)b_2)(p\otimes 1)=p\otimes {\epsilon }_1\left(b_1\right){\epsilon }_1\left(b_{2}p\right)=p{\epsilon }_1\left(b_1\right){\epsilon }_1\left(b_2\right)p\otimes 1=\xi p\otimes 1,$ for some constant $\xi \in 𝔽\text{.}$ This establishes (a). If $p$ is an indempotent of $A$ and $pAp=𝔽p$ then $p{\epsilon }_1\left(1\right)p={\epsilon }_1\left(p^2\right)=\left({\epsilon }_1(1·p)\right)={\epsilon }_1\left(1\right)p$ and so (b) holds. If $p\in A$ and $pAp=𝔽p$ then ${(p\otimes 1)}^2={\epsilon }_1\left(1\right)(p\otimes 1)$ and so, if ${\epsilon }_1\left(1\right)\ne 0,$ then $\frac{1}{\epsilon \left(1\right)}(p\otimes 1)$ is a minimal idempotent in $B{\otimes }_{A}B\text{.}$

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

$$\begin{array}{c}\hat{A}\hspace{0.17em}\text{be an index set for irreducible}\hspace{0.17em}A\text{-modules}\hspace{0.17em}{A}^{\mu },\\ \hat{B}\hspace{0.17em}\text{be an index set for irreducible}\hspace{0.17em}B\text{-modules}\hspace{0.17em}{B}^{\lambda }\text{, and let}\\ {\hat{A}}^{\mu }=\{P\to \mu \}\hspace{0.17em}\text{be an index set for a basis of the simple}\hspace{0.17em}A\text{-module}\hspace{0.17em}{A}^{\mu },\end{array}$$

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

$$\begin{array}{cc}\begin{array}{c}\Gamma \hspace{0.17em}\text{with vertices on level A:}\hspace{0.17em}\hat{A}\text{, vertices on level B:}\hspace{0.17em}\hat{B},\text{and}\\ m_{\mu }^{\lambda }\hspace{0.17em}\text{edges}\hspace{0.17em}\mu \to \lambda \hspace{0.17em}\text{if}\hspace{0.17em}{A}^{\mu }\hspace{0.17em}\text{appears with multiplicity}\hspace{0.17em}{m}_{\mu }^{\lambda }\hspace{0.17em}\text{in}\hspace{0.17em}{\text{Res}}_a^B\left(B^{\lambda }\right)\text{.}\end{array}& \text{(4.21)}\end{array}$$

For example, the graph $\Gamma$ for the symmetric group algebras $A=ℂS_3$ and $B=ℂS_4$ is

$$A^: B^:$$

If $\lambda \in \hat{B}$ then

$$\begin{array}{cc}{\hat{B}}^{\lambda }=\{P\to \mu \to \lambda \mid \mu \in \hat{A},P\to \mu \in {\hat{A}}^{\mu }\hspace{0.17em}\text{and}\hspace{0.17em}\mu \in \lambda \hspace{0.17em}\text{is an edge in}\hspace{0.17em}\Gamma \}& \text{(4.22)}\end{array}$$

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

$$\begin{array}{cc}\{a_{\underset{\mu }{PQ}}\mid \mu \in \hat{A},P\to \mu ,Q\to \mu \in {\hat{A}}^{\mu }\}\text{and}\{b_{\underset{\underset{\lambda }{\mu \nu }}{PQ}}\mid \lambda \in \hat{B},P\to \mu \to \lambda ,Q\to \nu \to \lambda \in {\hat{B}}^{\lambda }\},& \text{(4.23)}\end{array}$$

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

$$\begin{array}{cc}{a}_{\underset{\mu }{PQ}}{a}_{\underset{\nu }{ST}}={\delta }_{\mu \nu }{\delta }_{QS}{a}_{\underset{\mu }{PT}}\text{and}{b}_{\underset{\underset{\lambda }{\mu \gamma }}{PQ}}{b}_{\underset{\underset{\sigma }{\tau \nu }}{ST}}={\delta }_{\lambda \sigma }{\delta }_{QS}{\delta }_{\gamma \tau }{b}_{\underset{\underset{\lambda }{\mu \nu }}{PT}},& \text{(4.24)}\end{array}$$

and such that, for all $\mu \in \hat{A},P,Q\in {\hat{A}}^{\mu },$

$$\begin{array}{cc}{a}_{\underset{\mu }{PQ}}^{\mu }=\sum _{\mu \to \lambda }{b}_{\underset{\underset{\lambda }{\mu \mu }}{PQ}}^{\lambda }& \text{(4.25)}\end{array}$$

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 $\hat{C}=\hat{A},$ and extend the graph $\Gamma$ to a graph $\hat{\Gamma }$ 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,

$$\begin{array}{cc}\hat{\Gamma }\hspace{0.17em}\text{has vertices on level C:}\hspace{0.17em}\hat{C},\text{and}\text{an edge}\hspace{0.17em}\lambda \to \mu ,\lambda \in \hat{B},\mu \in \hat{C},\hspace{0.17em}\text{for each edge}\hspace{0.17em}\mu \to \lambda ,\mu \in \hat{A},\lambda \in \hat{B}\text{.}& \text{(4.26)}\end{array}$$

For each $\nu \in \hat{C}$ define

$$\begin{array}{cc}{\hat{C}}^{\nu }=\{P\to \mu \to \lambda \to \nu \hspace{0.17em}|\hspace{0.17em}\genfrac{}{}{0ex}{}{\mu \in \hat{A},\lambda \in \hat{B},\nu \in \hat{C},P\to \mu \in {\hat{A}}^{\mu }\hspace{0.17em}\text{and}}{\mu \to \lambda \hspace{0.17em}\text{and}\hspace{0.17em}\lambda \to \nu \hspace{0.17em}\text{are edges in}\hspace{0.17em}\hat{\Gamma }}\},& \text{(4.27)}\end{array}$$

so that ${\hat{C}}^{\nu }$ is the set of “paths to $\nu \text{"}$ in the graph $\hat{\Gamma }\text{.}$ Continuing with our previous example, $\hat{\Gamma }$ 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 $B{\otimes }_{A}B$ is given by $$b_{\underset{\underset{\lambda }{\mu \gamma }}{PT}}\otimes b_{\underset{\underset{\sigma }{\gamma \nu }}{TQ}}$$ do not depend on the choice of $T\to \gamma \in {\hat{A}}^{\gamma }$ 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 $$\begin{array}{cc}{\epsilon }_1\left(b_{\underset{\underset{\lambda }{\mu \mu }}{PP}}\right)={\epsilon }_{\mu }^{\lambda }{a}_{\underset{\mu }{PP}}\text{.}& \text{(4.29)}\end{array}$$ Then ${\epsilon }_{\mu }^{\lambda }$ is independent of the choice of $P\to \mu \in {\hat{A}}^{\mu }$ and $$\begin{array}{c}(b_{\underset{\underset{\lambda }{\mu \gamma }}{PT}}\otimes b_{\underset{\underset{\sigma }{\gamma \nu }}{TQ}})(b_{\underset{\underset{\rho }{\tau \pi }}{RX}}\otimes b_{\underset{\underset{\eta }{\pi \xi }}{XS}})={\delta }_{\gamma \pi }{\delta }_{QR}{\delta }_{\nu \tau }{\delta }_{\sigma \rho }{\epsilon }_{\gamma }^{\sigma }(b_{\underset{\underset{\gamma }{\pi \mu }}{PT}}\otimes b_{\underset{\underset{\eta }{\gamma \xi }}{TS}})\text{.}\\ \text{Rad}\left(B{\otimes }_{A}B\right)\text{has basis}\{b_{\underset{\underset{\lambda }{\mu \gamma }}{PT}}\otimes b_{\underset{\underset{\sigma }{\gamma \nu }}{TQ}}\hspace{0.17em}|\hspace{0.17em}{\epsilon }_{\mu }^{\lambda }=0\hspace{0.17em}\text{or}\hspace{0.17em}{\epsilon }_{\nu }^{\sigma }=0\},\end{array}$$ and the images of the elements $$e_{\underset{\underset{\underset{\gamma }{\lambda \sigma }}{\mu \nu }}{PQ}}=\left(\frac{1}{{\epsilon }_{\gamma }^{\sigma }}\right)(b_{\underset{\underset{\lambda }{\mu \gamma }}{PT}}\otimes b_{\underset{\underset{\sigma }{\gamma \nu }}{TQ}}),\text{such that}{\epsilon }_{\mu }^{\lambda }\ne 0\hspace{0.17em}\text{and}\hspace{0.17em}{\epsilon }_{\nu }^{\sigma }\ne 0,$$ 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 $$\begin{array}{cc}{\text{tr}}_A\left({\epsilon }_1\left(b\right)\right)={\text{tr}}_B\left(b\right),\text{for all}\hspace{0.17em}b\in B\text{.}& \text{(4.30)}\end{array}$$ Let ${\chi }_A^{\mu },\mu \in \hat{A},$ and ${\chi }_B^{\lambda },\lambda \in \hat{B},$ be the irreducible characters of the algebras $A$ and $B,$ respectively. Define constants ${\text{tr}}_A^{\mu },\mu \in \hat{A},$ and ${\text{tr}}_B^{\lambda },\lambda \in \hat{B},$ by the equations $$\begin{array}{cc}{\text{tr}}_A=\sum _{\mu \in \hat{A}}{\text{tr}}_A^{\mu }{\chi }_A^{\mu }\text{and}{\text{tr}}_B=\sum _{\lambda \in \hat{B}}{\text{tr}}_B^{\lambda }{\chi }_B^{\lambda },& \text{(4.31)}\end{array}$$ respectively. Then the constants ${\epsilon }_{\mu }^{\lambda }$ defined in (4.29) satisfy $${\text{tr}}_B^{\lambda }={\epsilon }_{\mu }^{\lambda }{\text{tr}}_A^{\mu }\text{.}$$
4. In the algebra $B{\otimes }_{A}B,$ $$1\otimes 1=\sum _{ P ↓ μ ↓ ↓ λ γ }{b}_{\underset{\underset{\lambda }{\mu \mu }}{PP}}\otimes b_{\underset{\underset{\gamma }{\mu \mu }}{PP}}$$
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 $$\iota \left(b_{\underset{\underset{\pi }{\tau \beta }}{RS}}\right)=\sum _{\pi \to \gamma }{e}_{\underset{\underset{\underset{\gamma }{\pi \pi }}{\tau \beta }}{RS}}\text{.}$$

		Proof.
	By (4.11) and (4.25), $$\begin{array}{cc}\begin{array}{ccc}B& \stackrel{\sim }{⟶}& \underset{\mu \in \hat{A}}{⨁}{\overrightarrow{A}}^{\mu }\otimes L^{\mu }\\ b_{\underset{\underset{\lambda }{\mu \nu }}{PQ}}& ⟼& {\overrightarrow{a}}_{\underset{\mu }{P}}\otimes {\ell }_{\underset{\underset{\lambda }{\mu \nu }}{Q}}^{\mu }\end{array}\text{and}\begin{array}{ccc}B& \stackrel{\sim }{⟶}& \underset{\nu \in \hat{A}}{⨁}{R}^{\nu }\otimes {\overleftarrow{A}}^{\nu }\\ b_{\underset{\underset{\lambda }{\mu \nu }}{PQ}}& ⟼& r_{\underset{\mu }{P}}^{\nu }\otimes {\overleftarrow{a}}_{\underset{\nu }{Q}}\end{array}& \text{(4.32)}\end{array}$$ 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 $$\begin{array}{cc}{\stackrel{\sim }{m}}_{\underset{\underset{\underset{\gamma }{\lambda \sigma }}{\mu \nu }}{PQ}}=r_{\underset{\underset{\lambda }{\mu \gamma }}{P}}^{\gamma }\otimes {\overleftarrow{a}}_{\underset{\gamma }{T}}\otimes {\overrightarrow{a}}_{\underset{\gamma }{T}}\otimes {\ell }_{\underset{\underset{\sigma }{\gamma \nu }}{Q}}^{\gamma }=b_{\underset{\underset{\lambda }{\mu \gamma }}{PT}}\otimes b_{\underset{\underset{\sigma }{\gamma \nu }}{TQ}}& \text{(4.33)}\end{array}$$ do not depend on $T\to \gamma \in {\hat{A}}^{\gamma }$ 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 $$\begin{array}{cc}{\epsilon }_{\underset{\underset{\underset{\mu }{\lambda \sigma }}{\gamma \tau }}{TQ}}^{\mu }\in 𝔽\text{given by}{\epsilon }_{\underset{\underset{\underset{\mu }{\lambda \sigma }}{\gamma \tau }}{TQ}}^{\mu }{a}_{\underset{\mu }{PP}}=\epsilon ({\overrightarrow{a}}_{\underset{\mu }{P}}\otimes {\ell }_{\underset{\underset{\lambda }{\mu \gamma }}{T}}^{\mu }\otimes r_{\underset{\underset{\sigma }{\tau \mu }}{Q}}^{\mu }\otimes {\overleftarrow{a}}_{\underset{\mu }{P}}),& \text{(4.34)}\end{array}$$ since $$\begin{array}{ccccc}{\epsilon }_{\underset{\underset{\underset{\mu }{\lambda \sigma }}{\gamma \tau }}{TQ}}^{\mu }{a}_{\underset{\mu }{PP}}& =& \epsilon (b_{\underset{\underset{\lambda }{\mu \gamma }}{PT}}\otimes b_{\underset{\underset{\sigma }{\tau \mu }}{QP}})& =& {\epsilon }_1(b_{\underset{\underset{\lambda }{\mu \gamma }}{PT}}\otimes b_{\underset{\underset{\sigma }{\tau \mu }}{QP}})\\ & =& {\delta }_{\underset{\underset{\lambda \sigma }{\gamma \tau }}{TQ}}{\epsilon }_1\left(b_{\underset{\underset{\lambda }{\mu \mu }}{PP}}\right)& =& {\delta }_{\underset{\underset{\lambda \sigma }{\gamma \tau }}{TQ}}{\epsilon }_1\left(b_{\underset{\underset{\lambda }{\mu \mu }}{PP}}{b}_{\underset{\underset{\lambda }{\mu \mu }}{PP}}\right)& =& {\delta }_{\underset{\underset{\lambda \sigma }{\gamma \tau }}{TQ}}{\epsilon }_{\underset{\underset{\lambda \lambda }{\mu \mu \mu }}{PP}}^{\mu }{a}_{\underset{\mu }{PP}}\text{.}\end{array}$$ 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 $P\to \mu \in {\hat{A}}^{\mu }\text{.}$ By Theorem 4.18(a), $${\bar{m}}_{\underset{\underset{\underset{\gamma }{\lambda \sigma }}{\mu \nu }}{PQ}}{\bar{m}}_{\underset{\underset{\underset{\pi }{\rho \eta }}{\tau \xi }}{RS}}={\delta }_{\gamma \pi }{\epsilon }_{\underset{\underset{\underset{\gamma }{\sigma \rho }}{\nu \tau }}{QR}}{\bar{m}}_{\underset{\underset{\underset{\gamma }{\lambda \eta }}{\mu \xi }}{PS}}={\delta }_{\gamma \pi }{\delta }_{\underset{\underset{\sigma \rho }{\nu \tau }}{QR}}{\epsilon }_{\gamma }^{\sigma }{\bar{m}}_{\underset{\underset{\underset{\gamma }{\lambda \eta }}{\mu \xi }}{PS}}$$ 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 $$\begin{array}{cc}{\text{tr}}_B^{\lambda }={\text{tr}}_B\left(b_{\underset{\underset{\lambda }{\mu \mu }}{PP}}\right)={\text{tr}}_A\left({\epsilon }_1\left(b_{\underset{\underset{\lambda }{\mu \mu }}{PP}}\right)\right)={\epsilon }_{\mu }^{\lambda }{\text{tr}}_A\left(a_{\underset{\mu }{PP}}\right)={\epsilon }_{\mu }^{\lambda }{\text{tr}}_A^{\mu }\text{.}& \text{(4.35)}\end{array}$$ (d) Since $$1=\sum _{P\to \mu \to \lambda }{b}_{\underset{\underset{\lambda }{\mu \mu }}{PP}}\text{in the algebra}\hspace{0.17em}B,$$ it follows from part (b) and (4.16) that $$\begin{array}{ccc}1\otimes 1& =& \left(\sum _{P\to \mu \to \lambda }{b}_{\underset{\underset{\lambda }{\mu \mu }}{PP}}\right)\otimes \left(\sum _{Q\to \nu \to \gamma }{b}_{\underset{\underset{\gamma }{\nu \nu }}{QQ}}\right)\\ & =& \sum _{\underset{Q\to \nu \to \gamma }{P\to \mu \to \lambda }}{\delta }_{PQ}{\delta }_{\mu \nu }(b_{\underset{\underset{\lambda }{\mu \mu }}{PP}}\otimes b_{\underset{\underset{\gamma }{\nu \nu }}{QQ}})=\sum _{ P ↓ μ ↓ ↓ λ γ }{\bar{m}}_{\underset{\underset{\underset{\mu }{\lambda \gamma }}{\mu \mu }}{PP}},\end{array}$$ 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 $$b_{\underset{\underset{\pi }{\tau \beta }}{RS}}{e}_{\underset{\underset{\underset{\gamma }{\lambda \sigma }}{\mu \nu }}{PQ}}=\left(\frac{1}{{\epsilon }_{\gamma }^{\sigma }}\right)b_{\underset{\underset{\pi }{\tau \beta }}{RS}}(b_{\underset{\underset{\lambda }{\mu \gamma }}{PT}}\otimes b_{\underset{\underset{\sigma }{\gamma \nu }}{TQ}})=\left(\frac{1}{{\epsilon }_{\gamma }^{\sigma }}\right){\delta }_{\underset{\underset{\pi \lambda }{\beta \mu }}{SP}}(b_{\underset{\underset{\lambda }{\tau \gamma }}{RT}}\otimes b_{\underset{\underset{\sigma }{\gamma \nu }}{TQ}})={\delta }_{\underset{\underset{\pi \lambda }{\beta \mu }}{SP}}{e}_{\underset{\underset{\underset{\gamma }{\pi \sigma }}{\tau \nu }}{RQ}}\text{.}$$ 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 $$\begin{array}{ccccc}\iota \left(b_{\underset{\underset{\pi }{\tau \beta }}{RS}}\right)& =& \iota \left(b_{\underset{\underset{\pi }{\tau \beta }}{RS}}\right)·1& =& b_{\underset{\underset{\pi }{\tau \beta }}{RS}}\sum _{P\to \mu \to \lambda \to \gamma }{e}_{\underset{\underset{\underset{\gamma }{\lambda \lambda }}{\mu \mu }}{PP}}\\ & & & =& \sum _{P\to \mu \to \lambda \to \gamma }{\delta }_{\underset{\underset{\pi \lambda }{\beta \mu }}{SP}}{e}_{\underset{\underset{\underset{\gamma }{\pi \lambda }}{\tau \mu }}{RP}}& =& \sum _{\pi \to \gamma }{e}_{\underset{\underset{\underset{\gamma }{\pi \pi }}{\tau \beta }}{RS}}\text{.}\end{array}$$ $\square$

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).

page history