Semisimple algebras

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

Last update: 9 April 2013

Semisimple algebras

Let $R$ be a integral domain and let $A_R$ be an algebra over $R,$ so that $A_R$ has an $R\text{-basis}$ $\{b_1,\dots ,b_d\},$

$$A_R=R\text{-span}\hspace{0.17em}\{b_1,\dots ,b_d\}\text{and}{b}_i{b}_j=\sum _{k=1}^d{r}_{ij}^k{b}_k,\text{with}\hspace{0.17em}{r}_{ij}^k\in R,$$

making $A_R$ a ring with identity. Let $𝔽$ be the field of fractions of $R,$ let $\overline{𝔽}$ be the algebraic closure of $𝔽,$ and set

$$A=\overline{𝔽}{\otimes }_R{A}_R=\overline{𝔽}\text{-span}\{b_1,\dots ,b_d\},$$

with multiplication determined by the multiplication in $A_R\text{.}$ Then $A$ is an algebra over $\overline{𝔽}\text{.}$

A trace on $A$ is a linear map $\overrightarrow{t}:A\to \overline{𝔽}$ such that

$$\overrightarrow{t}\left(a_1{a}_2\right)=\overrightarrow{t}\left(a_2{a}_1\right),\text{for all}\hspace{0.17em}{a}_1,a_2\in A\text{.}$$

A trace $\overrightarrow{t}$ on $A$ is nondegenerate if for each $b\in A$ there is an $a\in A$ such that $\overrightarrow{t}\left(ba\right)\ne 0\text{.}$

Lemma 5.1. Let $A$ be a finite dimensional algebra over a field $𝔽;$ let $\overrightarrow{t}$ be a trace on $A\text{.}$ 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 $a_1,a_2\in A\text{.}$ Let $B$ be a basis of $A\text{.}$ Let $G={\left(⟨b,b\prime ⟩\right)}_{b,b\prime \in B}$ be the matrix of the form $⟨,⟩$ with respect to $B\text{.}$ The following are equivalent:

1. The trace $\overrightarrow{t}$ is nondegenerate.
2. $\text{det}\hspace{0.17em}G\ne 0\text{.}$
3. The dual basis $B^{*}$ to the basis $B$ with respect to the form $⟨,⟩$ exists.

		Proof.
	(2) $\iff$ (1): The trace $\overrightarrow{t}$ is degenerate if there is an element $a\in A,$ $a\ne 0,$ such that $\overrightarrow{t}\left(ac\right)=0$ for all $c\in B\text{.}$ If $a_b\in \overline{𝔽}$ are such that $$a=\sum _{b\in B}{a}_{b}b,\text{then}0=⟨a,c⟩=\sum _{b\in B}{a}_b⟨b,c⟩$$ for all $c\in B\text{.}$ So $a$ exists if and only if the columns of $G$ are linearly dependent, i.e. if and only if $G$ is not invertible. (3) $\iff$ (2): 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 $B^{*}\text{.}$ Then $$d^{*}=\sum _{b\in B}{P}_{db}b,\text{and}{\delta }_{bc}=⟨b,d^{*}⟩=\sum _{d\in B}{P}_{dc}⟨b,c⟩={\left(G{P}^t\right)}_{b,c}\text{.}$$ So $P^t,$ the transpose of $P,$ is the inverse of the matrix $G\text{.}$ So the dual basis to $B$ exists if and only if $G$ is invertible, i.e. if and only if $\text{det}\hspace{0.17em}G\ne 0\text{.}$ $\square$

Proposition 5.2. Let $A$ be an algebra and let $\overrightarrow{t}$ be a nondegenerate trace on $A\text{.}$ Define a symmetric bilinear form $⟨,⟩:A\times A\to \overline{𝔽}$ on $A$ by $⟨a_1,a_2⟩=\overrightarrow{t}(a_1,a_2),$ for all $a_1,a_2\in A\text{.}$ Let $B$ be a basis of $A$ and let $B^{*}$ be the dual basis to B with respect to $⟨,⟩\text{.}$

1. Let $a\in A\text{.}$ Then $$\left[a\right]=\sum _{b\in B}ba{b}^{*}\hspace{0.17em}\text{is an element of the center}\hspace{0.17em}Z\left(A\right)\hspace{0.17em}\text{of}\hspace{0.17em}A$$ and $\left[a\right]$ does not depend on the choice of the basis $B\text{.}$
2. Let $M$ and $N$ be $A\text{-modules}$ and let $\varphi \in {\text{Hom}}_{\overline{𝔽}}(M,N)$ and define $$\left[\varphi \right]=\sum _{b\in B}b\varphi b^{*}\text{.}$$ Then $\left[\varphi \right]\in {\text{Hom}}_A(M,N)$ and $\left[\varphi \right]$ does not depend on the choice of the basis $B\text{.}$

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

Let $A$ be an algebra and let $M$ be an $A-module.$ Define

$${\text{End}}_A\left(M\right)=\{T\in \text{End}\left(M\right)\hspace{0.17em}\mid \hspace{0.17em}Ta=aT\hspace{0.17em}\text{for all}\hspace{0.17em}a\in A\}\text{.}$$

Theorem 5.3 (Schur’s lemma). Let $A$ be a finite dimensional algebra over an algebraically closed field $\overline{𝔽}\text{.}$

1. Let $A^{\lambda }$ be a simple $A\text{-module.}$ Then ${\text{End}}_A\left(A^{\lambda }\right)=\overline{𝔽}·{\text{Id}}_{A^{\lambda }}\text{.}$
2. If $A^{\lambda }$ and $A^{\mu }$ are nonisomorphic simple $A\text{-modules}$ then ${\text{Hom}}_A(A^{\lambda },A^{\mu })=0\text{.}$

		Proof.
	Let $T:A^{\lambda }\to A^{\mu }$ be a nonzero $A\text{-module}$ homomorphism. Since $A^{\lambda }$ is simple, $ker\hspace{0.17em}T=0$ and so $T$ is injective. Since $A^{\mu }$ is simple, $\text{im}\hspace{0.17em}T=A^{\mu }$ and so $T$ is surjective. So $T$ is an isomorphism. Thus we may assume that $T:A^{\lambda }\to A^{\lambda }\text{.}$ Since $\overline{𝔽}$ is algebraically closed $T$ has an eigenvector and a corresponding eigenvalue $\alpha \in \overline{𝔽}\text{.}$ Then $T-\alpha ·Id\in {\text{Hom}}_A(A^{\lambda },A^{\lambda })$ and so $T=\alpha ·Id$ is either 0 or an isomorphism. However, since $\text{det}(T-\alpha ·Id)=0,$ $T=\alpha ·Id$ is not invertible. So $T-\alpha ·Id=0\text{.}$ So $T=\alpha ·Id\text{.}$ So ${\text{End}}_A\left(A^{\lambda }\right)=\overline{𝔽}·Id\text{.}$ $\square$

Theorem 5.4 (The Centralizer Theorem). Let $A$ be a finite dimensional algebra over an algebraically closed field $\overline{𝔽}\text{.}$ Let $M$ be a semisimple $A\text{-module}$ and set $A={\text{End}}_A\left(M\right)\text{.}$ Suppose that

$$M\cong \underset{\lambda \in \hat{M}}{⨁}{\left(A^{\lambda }\right)}^{\oplus m_{\lambda }},$$

where $\hat{M}$ is an index set for the irreducible $A\text{-modules}$ $A^{\lambda }$ which appear in $M$ and the $m_{\lambda }$ are positive integers.

1. $Z\cong {⨁}_{\lambda \in \hat{M}}{M}_{m_{\lambda }}\left(\overline{𝔽}\right)\text{.}$
2. As an $(A,Z)\text{-bimodule,}$ $$M\cong \underset{\lambda \in \hat{M}}{⨁}{A}^{\lambda }\otimes Z^{\lambda },$$ where $Z^{\lambda },\lambda \in \hat{M},$ are the simple $Z\text{-modules.}$

		Proof.
	Index the components in the decomposition of $M$ by dummy variables ${\epsilon }_i^{\lambda }$ so that we may write $$M\cong \underset{\lambda \in \hat{M}}{⨁}\underset{i=1}{\overset{m_{\lambda }}{⨁}}{A}^{\lambda }\otimes {\epsilon }_i^{\lambda }\text{.}$$ For each $\lambda \in \hat{M},$ $1\le i,j\le m_{\lambda },$ let ${\varphi }_{ij}^{\lambda }:A^{\lambda }\otimes {\epsilon }_j\to A^{\lambda }\otimes {\epsilon }_i$ be the $A\text{-module}$ isomorphism given by $${\varphi }_{ij}^{\lambda }(m\otimes {\epsilon }_j^{\lambda })=m\otimes {\epsilon }_i^{\lambda },\text{for}\hspace{0.17em}m\in A^{\lambda }\text{.}$$ By Schur's lemma, $$\begin{array}{ccc}{\text{End}}_A\left(M\right)& =& {\text{Hom}}_A(M,M)\\ & \cong & {\text{Hom}}_A(\underset{\lambda }{⨁}\underset{j}{⨁}{A}^{\lambda }\otimes {\epsilon }_j^{\lambda },\underset{\mu }{⨁}\underset{i}{⨁}{A}^{\mu }\otimes {\epsilon }_i^{\mu })\\ & \cong & \underset{\lambda ,\mu }{⨁}\underset{i,j}{⨁}{\delta }_{\lambda \mu }{\text{Hom}}_A(A^{\lambda }\otimes {\epsilon }_j^{\lambda },A^{\mu }\otimes {\epsilon }_i^{\mu })\\ & \cong & \underset{\lambda }{⨁}\underset{i,j=1}{\overset{m_{\lambda }}{⨁}}\overline{𝔽}{\varphi }_{ij}^{\lambda }\text{.}\end{array}$$ Thus each element $z\in {\text{End}}_A\left(M\right)$ can be written as $$z=\sum _{\lambda \in \hat{M}}\sum _{i,j=1}^{m_{\lambda }}{z}_{ij}^{\lambda }{\varphi }_{ij}^{\lambda },\text{for some}\hspace{0.17em}{z}_{ij}^{\lambda }\in \overline{𝔽},$$ and identified with an element of ${⨁}_{\lambda }{M}_{m_{\lambda }}\left(\overline{𝔽}\right)\text{.}$ Since ${\varphi }_{ij}^{\lambda }{\varphi }_{kl}^{\mu }={\delta }_{\lambda \mu }{\delta }_{jk}{\varphi }_{il}^{\lambda }$ it follows that $${\text{End}}_A\left(M\right)\cong \underset{\lambda \in \hat{M}}{⨁}{M}_{m_{\lambda }}\left(\overline{𝔽}\right)\text{.}$$ (b) As a vector space, $Z^{\mu }=\text{span}\{{\epsilon }_i^{\mu }\hspace{0.17em}\mid \hspace{0.17em}1\le i\le m_{\mu }\}$ is isomorphic to the simple ${⨁}_{\lambda }{M}_{m_{\lambda }}\left(\overline{𝔽}\right)$ module of column vectors of length $m_{\mu }\text{.}$ The decomposition of $M$ as $A\otimes Z$ modules follows since $$(a\otimes {\varphi }_{ij}^{\lambda })(m\otimes {\epsilon }_k^{\mu })={\delta }_{\lambda \mu }{\delta }_{jk}(a\otimes {\epsilon }_i^{\mu }),\text{for all}\hspace{0.17em}m\in A^{\mu },a\in A\text{.}$$ $\square$

If $A$ is an algebra then $A^{\text{op}}$ is the algebra $A$ except with the opposite multiplication, i.e.

$$A^{\text{op}}=\{a^{\text{op}}\hspace{0.17em}\mid \hspace{0.17em}a\in A\}\text{with}{a}_1^{\text{op}}{a}_2^{\text{op}}={\left(a_2{a}_1\right)}^{\text{op}},\text{for all}\hspace{0.17em}{a}_1,a_2\in A\text{.}$$

The left regular representation of $A$ is the vector space $A$ with $A$ action given by left multiplication. Here $A$ is serving both as an algebra and as an $A\text{-module.}$ It is often useful to distinguish the two roles of $A$ and use the notation $\overrightarrow{A}$ for the $A\text{-module,}$ i.e. $\overrightarrow{A}$ is the vector space

$$\overrightarrow{A}=\{\overrightarrow{b}\hspace{0.17em}\mid \hspace{0.17em}b\in A\}\text{with}\hspace{0.17em}A\text{-action}a\overrightarrow{b}=\overrightarrow{ab},\text{for all}\hspace{0.17em}a\in A,\overrightarrow{b}\in \overrightarrow{A}\text{.}$$

Proposition 5.5 Let $A$ be an algebra and let $\overrightarrow{A}$ be the regular representation of $A\text{.}$ Then ${\text{End}}_A\left(\overrightarrow{A}\right)\cong A^{\text{op}}\text{.}$ More precisely,

$${\text{End}}_A\left(\overrightarrow{A}\right)=\{{\varphi }_b\hspace{0.17em}\mid \hspace{0.17em}b\in A\},$$

where ${\varphi }_b$ is given by

$${\varphi }_b\left(\overrightarrow{a}\right)=a\overrightarrow{b},\text{for all}\hspace{0.17em}\overrightarrow{a}\in \overrightarrow{A}\text{.}$$

		Proof.
	Let $\varphi \in {\text{End}}_A\left(\overrightarrow{A}\right)$ and let $b\in A$ be such that $\varphi \left(\overrightarrow{1}\right)=\overrightarrow{b}\text{.}$ For all $\overrightarrow{a}\in \overrightarrow{A},$ $$\varphi \left(\overrightarrow{a}\right)=\varphi (a·\stackrel{e_{\alpha }}{1})=a\varphi \left(\overrightarrow{1}\right)=a\overrightarrow{b},$$ and so $\varphi ={\varphi }_b\text{.}$ Then ${\text{End}}_A\left(\overrightarrow{A}\right)\cong A^{\text{op}}$ since $$({\varphi }_{b_1}\circ {\varphi }_{b_2})\left(\overrightarrow{a}\right)=a\overrightarrow{b_2}{b}_1={\varphi }_{b_2{b}_1}\left(\overrightarrow{a}\right),$$ for all $b_1,b_2\in A$ and $\overrightarrow{a}\in \overrightarrow{A}\text{.}$ $\square$

Theorem 5.6. Suppose that $A$ is a finite dimensional algebra over an algebraically closed field $𝔽$ such that the regular representation $\overrightarrow{A}$ of $A$ is completely decomposable. Then $A$ is i somorphic to a direct sum of matrix algebras, i.e.

$$a\cong \underset{\lambda \in \hat{A}}{⨁}{M}_{d_{\lambda }}\left(\overline{𝔽}\right),$$

for some set $\hat{A}$ and some positive integers $d_{\lambda },$ indexed by the elements of $\hat{A}\text{.}$

		Proof.
	If $\overrightarrow{A}$ is completely decomposable then, by Theorem 5.4, ${\text{End}}_A\left(\overrightarrow{A}\right)$ is isomorphic to a direct sum of matrix algebras. By Proposition 5.5, $$A^{\text{op}}\cong \underset{\lambda \in \hat{A}}{⨁}{M}_{d_{\lambda }}\left(\overline{𝔽}\right),$$ for some set $\hat{A}$ and some positive integers $d_{\lambda },$ indexed by the elements of $\hat{A}\text{.}$ The map $$\begin{array}{ccc}{\left(\underset{\lambda \in \hat{A}}{⨁}{M}_{d_{\lambda }}\left(\overline{𝔽}\right)\right)}^{\text{op}}& ⟶& \underset{\lambda \in \hat{A}}{⨁}{M}_{d_{\lambda }}\left(\overline{𝔽}\right)\\ a& ⟼& a^t,\end{array}$$ where $a^t$ is the transpose of the matrix $a,$ is an algebra isomorphism. So $A$ is isomorphic to a direct sum of matrix algebras. $\square$

If $A$ is an algebra then the trace tr of the regular representation is the trace on $A$ given by

$$\text{tr}\left(a\right)=\text{Tr}\left(\overrightarrow{A}\left(a\right)\right),\text{for}\hspace{0.17em}a\in A,$$

where $\overrightarrow{A}\left(a\right)$ is the linear transformation of $A$ induced by the action of $a$ on $A$ by left multiplication.

Proposition 5.7. Let $A={⨁}_{\lambda \in \hat{A}}{M}_{d_{\lambda }}\left(\overline{𝔽}\right)\text{.}$ The trace of the regular representation is nondegenerate if and only if the integers $d_{\lambda }$ are all nonzero in $\overline{𝔽}\text{.}$ In characteristic $p$ they could be $0\text{.}$

		Proof.
	As $A\text{-modules,}$ the regular representation $$\overrightarrow{A}\cong \underset{\lambda \in \hat{A}}{⨁}{\left(A^{\lambda }\right)}^{\oplus d_{\lambda }},$$ where $A^{\lambda }$ is the irreducible $A\text{-module}$ consisting of column vectors of length $d_{\lambda }\text{.}$ For $a\in A$ let $A^{\lambda }\left(a\right)$ be the linear transformation of $A^{\lambda }$ induced by the action of $a\text{.}$ Then the trace tr of the regular representation is given by $$\text{tr}=\sum _{\lambda \in \hat{A}}{d}_{\lambda }{\chi }^{\lambda },\text{where}\begin{array}{cccc}{\chi }_A^{\lambda }:& A& ⟶& \overline{𝔽}\\ & a& ⟼& \text{Tr}\left(A^{\lambda }\left(a\right)\right),\end{array}$$ where ${\chi }_A^{\lambda }$ are the irreducible characters of $A\text{.}$ Since the $d_{\lambda }$ are all nonzero the trace tr is nondegenerate. $\square$

Theorem 5.8 (Maschke’s Theorem). Let $A$ be a finite dimensional algebra over a field $𝔽$ such that the trace $\text{tr}$ of the regular representation of $A$ is nondegenerate. Then every representation of $A$ is completely decomposable.

		Proof.
	Let $B$ be a basis of $A$ and let $B^{*}$ be the dual basis of $A$ with respect to the form $⟨,⟩:A\times A\to \overline{𝔽}$ defined by $$⟨a_1,a_2⟩=\text{tr}\left(a_1{a}_2\right),\text{for all}\hspace{0.17em}{a}_1,a_2\in A\text{.}$$ The dual basis $B^{*}$ exists because the trace tr is nondegenerate. Let $M$ be an $A\text{-module.}$ If $M$ is irreducible then the result is vacuously true, so we may assume that $M$ has a proper submodule $N\text{.}$ Let $p\in \text{End}\left(M\right)$ be a projection onto $N,$ i.e. $pM=N$ and $p^2=p\text{.}$ Let $$\left[p\right]=\sum _{b\in B}bp{b}^{*},\text{and}e=\sum _{b\in B}b{b}^{*}\text{.}$$ For all $a\in A,$ $$\text{tr}\left(ea\right)=\sum _{b\in B}\text{tr}\left(b{b}^{*}a\right)=\sum _{b\in B}⟨ab,b^{*}⟩=\sum _{b\in B}ab{\mid }_b=\text{tr}\left(a\right)\text{.}$$ So $\text{tr}\left((e-1)a\right)=0,$ for all $a\in A\text{.}$ Thus, since tr is nondegenerate, $e=1\text{.}$ Let $m\in M\text{.}$ Then $p{b}^{*}m\in N$ for all $b\in B,$ and so $\left[p\right]m\in N\text{.}$ So $\left[p\right]M\subseteq N\text{.}$ Let $n\in N\text{.}$ Then $p{b}^{*}n=b^{*}n$ for all $b\in B,$ and so $\left[p\right]n=en=1·n=n\text{.}$ So $\left[p\right]M=N$ and ${\left[p\right]}^2=\left[p\right],$ as elements of $\text{End}\left(M\right)\text{.}$ Note that $[1-p]=\left[1\right]-\left[p\right]=e-\left[p\right]=1-\left[p\right]\text{.}$ So $$M=\left[p\right]M\oplus (1-\left[p\right])M=N\oplus [1-p]M,$$ and, by Proposition 5.2(b), $[1-p]M$ is an $A\text{-module.}$ So $[1-p]M$ is an $A\text{-submodule}$ of $M$ which is complementary to $M\text{.}$ By induction on the dimension of $M,N$ and $[1-p]M$ are completely decomposable, and therefore $M$ is completely decomposable. $\square$

Together, Theorem 5.6, 5.8 and Proposition 5.7 yield the following theorem.

Theorem 5.9 (Artin–Wedderburn Theorem). Let $A$ be a finite dimensional algebra over an algebraically closed field $\overline{𝔽}\text{.}$ Let $\{b_1,\dots ,b_d\}$ be a basis of $A$ and let $\text{tr}$ be the trace of the regular representation of $A\text{.}$ The following are equivalent:

1. Every representation of $A$ is completely decomposable.
2. The regular representation of $A$ is completely decomposable.
3. $A\cong {⨁}_{\lambda \in \hat{A}}{M}_{d_{\lambda }}\left(\overline{𝔽}\right)$ for some finite index set $\hat{A},$ and some $d_{\lambda }\in {\mathbb{Z}}_{>0}\text{.}$
4. The trace of the regular representation of $A$ is nondegenerate.
5. $\text{det}\left(\text{tr}\left(b_i{b}_j\right)\right)\ne 0\text{.}$

Remark. Let $R$ be an integral domain, and let $A_R$ be an algebra over $R$ with basis $\{b_1,\dots ,b_d\}\text{.}$ Then $\text{det}\left(\text{tr}\left(b_i{b}_j\right)\right)$ is an element of $R$ and $\text{det}\left(\text{tr}\left(b_i{b}_j\right)\right)\ne 0\text{.}$ in $\overline{𝔽}$ if and only if $\text{det}\left(\text{tr}\left(b_i{b}_j\right)\right)\ne 0\text{.}$ in $R\text{.}$ In particular, if $R=ℂ\left[x\right],$ then $\text{det}\left(\text{tr}\left(b_i{b}_j\right)\right)$ is a polynomial. Since a polynomial has only a finite number of roots, $\text{det}\left(\text{tr}\left(b_i{b}_j\right)\right)\left(n\right)=0$ for only a finite number of values $n\in ℂ\text{.}$

Theorem 5.10 (The Tits Deformation Theorem). Let $R$ be an integral domain, $𝔽,$ the field of fractions of $R,$ $\overline{𝔽}$ the algebraic closure of $𝔽,$ and $\bar{R},$ the integral closure of $R$ in $\overline{𝔽}\text{.}$ Let $A_R$ be an $R\text{-algebra}$ and let $\{b_1,\dots ,b_d\}$ be a basis of $A_R\text{.}$ For $a\in A_R$ let $\overrightarrow{A}\left(a\right)$ denote the linear transformation of $A_R$ induced by left multiplication by $a\text{.}$ Let $t_1,\dots ,t_d$ be indeterminates and let

$$\overrightarrow{p}(t_1,\dots ,t_d;x)=\text{det}(x·\text{Id}-(t_1\overrightarrow{A}\left(b_1\right)+\dots t_d\overrightarrow{A}\left(b_d\right)))\in R[t_1,\dots ,t_d]\left[x\right],$$

so that $\overrightarrow{p}$ is the characteristic polynomial of a “generic” element of $A_R\text{.}$

1. Let $A_{\overline{𝔽}}=\overline{𝔽}{\otimes }_R{A}_R\text{.}$ If $$A_{\overline{𝔽}}\cong \underset{\lambda \in \hat{A}}{⨁}{M}_{d_{\lambda }}\left(\overline{𝔽}\right),$$ then the factorization of $\overrightarrow{p}(t_1,\dots ,t_d,x)$ into irreducibles in $\overline{𝔽}[t_1,\dots ,t_d,x]$ has the form $$\overrightarrow{p}=\prod _{\lambda \in A^}{\left({\overrightarrow{p}}^{\lambda }\right)}^{d_{\lambda }},\text{with}{\overrightarrow{p}}^{\lambda }\in \bar{R}[t_1,\dots ,t_d,x]\text{and}{d}_{\lambda }=\text{deg}\left({\overrightarrow{p}}^{\lambda }\right)\text{.}$$ If ${\chi }^{\lambda }(t_1,\dots ,t_d)\in \bar{R}[t_1,\dots ,t_d]$ is given by $${\overrightarrow{p}}^{\lambda }(t_1,\dots ,t_d,x)=x^{d_{\lambda }}-{\chi }^{\lambda }(t_1,\dots ,t_d)x^{d_{\lambda }-1}+\dots ,$$ then $$\begin{array}{cccc}{\chi }_{A_{\overline{𝔽}}}^{\lambda }:& A_{\overline{𝔽}}& ⟼& \overline{𝔽}\\ & {\alpha }_1{b}_1+\dots +{\alpha }_d{b}_d& ⟼& {\chi }^{\lambda }({\alpha }_1,\dots ,{\alpha }_d),\end{array}\lambda \in \hat{A},$$ are the irreducible characters of $A_{\overline{𝔽}}\text{.}$
2. Let $𝕂$ be a field and let $\overline{𝕂}$ be the algebraic closure of $𝕂\text{.}$ Let $\gamma :R\to 𝕂$ be a ring homomorphism and let $\bar{\gamma }:\bar{R}\to \overline{𝕂}$ be the extension of $\gamma \text{.}$ Let ${\chi }^{\lambda }(t_1,\dots ,t_d)\in \bar{R}[t_1,\dots ,t_d]$ be as in (a). If $A_{\overline{𝕂}}=\overline{𝕂}{\otimes }_R{A}_R$ is semisimple then $$A_{\overline{𝕂}}\cong \underset{\lambda \in \hat{A}}{⨁}{M}_{d_{\lambda }}\left(\overline{𝕂}\right),$$ and $$\begin{array}{cccc}{\chi }_{A_{\overline{𝕂}}}^{\lambda }:& A_{\overline{𝕂}}& ⟼& \overline{𝕂}\\ & {\alpha }_1{b}_1+\dots +{\alpha }_d{b}_d& ⟼& \left(\bar{\gamma }{\chi }^{\lambda }\right)({\alpha }_1,\dots ,{\alpha }_d),\end{array}$$ for $\lambda \in \hat{A},$ are the irreducible characters of $A_{\overline{𝕂}}\text{.}$

		Proof.
	First note that if $\{b_1^{\prime },\dots ,b_d^{\prime }\}$ is another basis of $A_R$ and the change of basis matrix $P=\left(P_{ij}\right)$ is given by $$b_i^{\prime }=\sum _j{P}_{ij}{b}_j\text{then the transformation}{t}_i^{\prime }=\sum _j{P}_{ij}{t}_j,$$ defines an isomorphism of polynomial rings $R[t_1,\dots ,t_d]\cong R[t_1^{\prime },\dots ,t_d^{\prime }]\text{.}$ Thus it follows that if the statements are true for one basis of $A_R$ (or $A_{\overline{𝔽}}\text{)}$ then they are true for every basis of $A_R$ (resp. $A_{\overline{𝔽}}\text{).}$ (a) Using the decomposition of $A_{\overline{𝔽}}$ let $\{e_{ij}^{\mu },\mu \in \hat{A},1\le i,j\le d_{\lambda }\}$ be a basis of matrix units in $A_{\overline{𝔽}}$ and let $t_{ij}^{\mu }$ be corresponding variables. Then the decomposition of $A_{\overline{𝔽}}$ induces a factorization $$\begin{array}{cc}\overrightarrow{p}(t_{ij}^{\mu },x)=\prod _{\lambda \in \hat{A}}{\left({\overrightarrow{p}}^{\lambda }\right)}^{d_{\lambda }},\text{where}{\overrightarrow{p}}^{\lambda }(t_{ij}^{\mu };x)=\text{det}(x-\sum _{\mu ,i,j}{t}_{ij}^{\mu }{A}^{\lambda }\left(e_{ij}\right))\text{.}& \text{(5.11)}\end{array}$$ The polynomial ${\overrightarrow{p}}^{\lambda }(t_{ij}^{\mu };x)$ is irreducible since specializing the variables gives $$\begin{array}{cc}{\overrightarrow{p}}^{\lambda }(t_{j+1,j}^{\lambda }=1,t_{1,n}^{\lambda }=t,t_{i,j}^{\mu }=0\hspace{0.17em}\text{otherwise};x)=x^{d_{\lambda }}-t,& \text{(5.12)}\end{array}$$ which is irreducible in $\bar{R}[t;x]\text{.}$ This provides the factorization of $\overrightarrow{p}$ and establishes that $\text{deg}\left({\overrightarrow{p}}^{\lambda }\right)=d_{\lambda }\text{.}$ By (5.11) $${\overrightarrow{p}}^{\lambda }(t_{ij}^{\mu };x)=x^{d_{\lambda }}-\text{Tr}\left(A^{\lambda }\left(\sum _{\mu ,i,j}{t}_{ij}^{\mu }{e}_{ij}^{\mu }\right)\right)x^{d_{\lambda }-1}+\dots ,$$ which establishes the last statement. Any root of $\overrightarrow{p}(t_1,\dots ,t_d,x)$ is an element of $\overline{R[t_1,\dots ,t_d]}=\bar{R}[t_1,\dots ,t_d]\text{.}$ So any root of ${\overrightarrow{p}}^{\lambda }(t_1,\dots ,t_d,x)$ is an element of $\bar{R}[t_1,\dots ,t_d]$ and therefore the coefficients of ${\overrightarrow{p}}^{\lambda }(t_1,\dots ,t_d,x)$ (symmetric functions in the roots of ${\overrightarrow{p}}^{\lambda }\text{)}$ are elements of $\bar{R}[t_1,\dots ,t_d]\text{.}$ (b) Taking the image of the Eq. (5.11), give a factorization of $\gamma \left(\overrightarrow{p}\right),$ $$\gamma \left(\overrightarrow{p}\right)=\prod _{\lambda \in \hat{A}}\gamma {\left({\overrightarrow{p}}^{\lambda }\right)}^{d_{\lambda }},\text{in}\hspace{0.17em}\overline{𝕂}[t_1,\dots ,t_d,x]\text{.}$$ For the same reason as in (5.12) the factors $\gamma \left({\overrightarrow{p}}^{\lambda }\right)$ are irreducible polynomials in $\overline{𝕂}[t_1,\dots ,t_d,x]\text{.}$ On the other hand, as in the proof of (a), the decomposition of $A_{\overline{𝕂}}$ induces a factorization of $\gamma \left(\overrightarrow{p}\right)$ into irreducibles in $\overline{𝕂}[t_1,\dots ,t_d,x]\text{.}$ These two factorizations must coincide, whence the result. $\square$

Applying the Tits deformation theorem to the case where $R=ℂ\left[x\right]$ (so that $𝔽=ℂ\left(x\right)\text{)}$ gives the following theorem. The statement in (a) is a consequence of Theorem 5.6 and the remark which follows Theorem 5.9.

Theorem 5.13. Let $ℂA\left(n\right)$ be a family of algebras defined by generators and relations such that the coefficients of the relations are polynomials in $n\text{.}$ Assume that there is an $\alpha \in ℂ$ such that $ℂA\left(\alpha \right)$ is semisimple. Let $\hat{A}$ be an index set for the irreducible $ℂA\left(\alpha \right)\text{-modules}$ $A^{\lambda }\left(\alpha \right)\text{.}$ Then

1. $ℂA\left(n\right)$ is semisimple for all but a finite number of $n\in ℂ\text{.}$
2. If $n\in ℂ$ is such that $ℂA\left(n\right)$ is semisimple then $\hat{A}$ is an index set for the simple $ℂA\left(n\right)\text{-modules}$ $A^{\lambda }\left(n\right)$ and $\text{dim}\left(A^{\lambda }\left(n\right)\right)=\text{dim}\left(A^{\lambda }\left(\alpha \right)\right)$ for each $\lambda \in \hat{A}\text{.}$
3. Let $x$ be an indeterminate and let $\{b_1,\dots ,b_d\}$ be a basis of $ℂ\left[x\right]A\left(x\right)\text{.}$ Then there are polynomials ${\chi }^{\lambda }(t_1,\dots ,t_d)\in ℂ[t_1,\dots ,t_d,x],\lambda \in \hat{A},$ such that for every $n\in ℂ$ such that $ℂA\left(n\right)$ is semisimple, $$\begin{array}{cccc}{\chi }_{A\left(n\right)}^{\lambda }:& ℂA\left(n\right)& ⟶& ℂ\\ & {\alpha }_1{b}_1+\dots +{\alpha }_d{b}_d& ⟼& {\chi }^{\lambda }({\alpha }_1,\dots ,{\alpha }_d,n),\end{array}\lambda \in \hat{A},$$ are the irreducible characters of $ℂA\left(n\right)\text{.}$

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).

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