Algebras

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

Department of Mathematics
University of Wisconsin, Madison
Madison, WI 53706 USA
ram@math.wisc.edu

Last updates: 22 January 2010

Algebras

Let $R$ be a integral domain and let $A_R$ be an algebra over $R$, so that $A_R$ has an $R$-basis $\left\{b_1,\dots ,b_d\right\}$, $$A_R=R\text{-span}\left\{b_1,\dots ,b_d\right\}\text{and}{b}_i{b}_j=\sum _{k=1}^d{r}_{ij}^k{b}_k\text{,}\text{with}{r}_{ij}^k\in R\text{,}$$ making $A_R$ a ring with identity. Let $𝔽$ be the field of fractions over $R$, let $\overline{𝔽}$ be the algebraic closure of $𝔽$, and set $$A=\overline{𝔽}{\otimes }_R{A}_R=\overline{𝔽}\text{-span}\left\{b_1,\dots ,b_d\right\}$$ with multiplication determined by the multiplication in $A_R$. Then $A$ is an algebra over $\overline{𝔽}$.

A trace on $A$ is a linear map $\overrightarrow{t}:A\to \overline{𝔽}$ such that $$\overrightarrow{t}\left(a_1{a}_2\right)=\left(a_2{a}_1\right)\text{,}\text{for all}{a}_1,a_2\in A\text{.}$$ A trace $\overrightarrow{t}$ on $A$ is nondegenerate if for each $b\in A$ there is an $a\in A$ such that $\overrightarrow{t}\left(ba\right)\ne 0$.

Let $A$ be a finite dimensional algebra over a field $𝔽$; let $\overrightarrow{t}$ be a trace over $A$. Define a symmetric bilinear form $⟨,⟩:A\times A\to 𝔽$ on $A$ by $⟨a_1,a_2⟩=\overrightarrow{t}\left(a_1{a}_2\right)$, for all $a_1,a_2\in A$. Let $B$ be a basis of $A$. Let $G={\left(⟨b,b\prime ⟩\right)}_{b,b\prime ,\in ,B}$ be the matrix of the form $⟨,⟩$ with respect to $B$. The following are then equivalent.

1. The trace $\overrightarrow{t}$ is nondegenerate.
2. $\det G\ne 0$.
3. The dual basis $B^{*}$ to the basis $B$ with respect to the form $⟨,⟩$ exists.

		Proof.
	(b) $\iff$ (a): The trace $\overrightarrow{t}$ is degenerate if there is an element $a\in A$, $a\ne 0$, such that $\overrightarrow{t}\left(a,c\right)=0$ for all $c\in B$. If $a_b\in \overline{𝔽}$ are such that $$a=\sum _{b\in B}{a}_b\text{,}\text{then}0=⟨a,c⟩=\sum _{b\in B}{a}_b⟨b,c⟩$$ for all $c\in B$. So $a$ exists if and only if the columns of $G$ are linearly dependent, i.e. if and only if $G$ is not invertible. (c) $\iff$ (b): Let $B^{*}=\left\{b^{*}\right\}$ be the dual basis to $\left\{b\right\}$ with respect to $⟨,⟩$ and let $P$ be the change of basis matrix from $B$ to $B^{*}$. Then $$d^{*}=\sum _{b\in B}{P}_{db}b\text{,}\text{and}{\delta }_{bc}=⟨b,d^{*}⟩=\sum _{b\in B}{P}_{dc}⟨b,c⟩={\left(G{P}^t\right)}_{b,c}\text{.}$$ So $P^t$, the transpose of $P$, is the inverse of the matrix $G$. So the dual basis to $B$ exists if and only if $G$ is invertible, i.e. if and only if $\det G\ne 0$. $\square$

Let $A$ be an algebra and let $\overrightarrow{t}$ be a nondegenerate trace on $A$. Define a symmetric bilinear form $⟨,⟩:A\times A\to \overline{𝔽}$ on $A$ by $⟨a_1,a_2⟩=\overrightarrow{t}\left(a_1,a_2\right)$, for all $a_1,a_2\in A$. Let $B$ be a basis of $A$ and let $B^{*}$ be the dual basis to $B$ with respect to with respect to $⟨,⟩$.

1. Let $a\in A$. Then $$\left[a\right]=\sum _{b\in B}ba{b}^{*}\text{is an element in the center}Z\left(A\right)\text{of}A$$ and $\left[a\right]$ does not depend on the choice of basis $B$.
2. Let $M$ and $N$ be $A$-modules and let $\phi \in {\mathrm{Hom}}_{\overline{𝔽}}\left(M,N\right)$ and define $$\left[\phi \right]=\sum _{b\in B}b\phi b^{*}\text{.}$$ Then $\left[\phi \right]\in {\mathrm{Hom}}_A\left(M,N\right)$ and $\left[\phi \right]$ does not depend on the choice of basis $B$.

		Proof.
	(a): Let $c\in A$. Then $$\begin{array}{rcl}c\left[a\right]& =& \sum _{b\in B}cba{b}^{*}\\ & =& \sum _{b\in B}\sum _{d\in B}⟨cb,d^{*}⟩da{d}^{*}\\ & =& \sum _{d\in B}da\sum _{b\in B}⟨d^{*}c,b⟩b^{*}\\ & =& \sum _{d\in B}da{d}^{*}c\\ & =& \left[a\right]c\text{,}\end{array}$$ since $⟨cd,d^{*}⟩=\overrightarrow{t}\left(cb{d}^{*}\right)=\overrightarrow{t}\left(d^{*}cb\right)=⟨d^{*}c,b⟩$. So $\left[a\right]\in Z\left(A\right)$. Let $D$ be another basis for $A$ and let $D^{*}$ be the dual basis to $D$ with respect to $⟨,⟩$. Let $P=P_{db}$ be the transition matrix from $D$ to $B$ and let $P^{-1}$ be the inverse of $P$. Then $$d=\sum _{b\in B}{P}_{db}b\text{and}{d}^{*}={\left(P^{-1}\right)}_{\tilde{b}d}{\tilde{b}}^{*}\text{,}$$ since $$\begin{array}{rcl}⟨d,{\tilde{d}}^{*}⟩& =& ⟨\sum _{b\in B}{P}_{db},\sum _{\tilde{b}\in B}{\left(P^{-1}\right)}_{\tilde{b}\tilde{d}}{\tilde{b}}^{*}⟩\\ & =& \sum _{b,\tilde{b}\in B}{P}_{db}{\left(P^{-1}\right)}_{\tilde{b}\tilde{d}}{\delta }_{b\tilde{b}}\\ & =& {\delta }_{d\tilde{d}}\text{.}\end{array}$$ So $$\begin{array}{rcl}\sum _{d\in D}da{d}^{*}& =& \sum _{d\in D}\sum _{b\in B}{P}_{db}ba\sum _{\tilde{b}\in B}{\left(P^{-1}\right)}_{\tilde{b}d}{\tilde{b}}^{*}\\ & =& \sum _{b,\tilde{b}}ba{\tilde{t}}^{*}{\delta }_{b\tilde{b}}\\ & =& \sum _{b\in B}ba{b}^{*}\text{.}\end{array}$$ So $\left[a\right]$ does not depend on the choice of the basis $B$. (b): The proof of part (b) is the same as the proof for part (b) except $a$ is replaced by $\phi$. $\square$

Reference

[HA] T. Halverson and A. Ram, Partition algebras, European Journal of Combinatorics 26, (2005), 869-921; arXiv:math/040131v2.

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