Tits' Deformation Theorem

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: 20 January 2010

Tits' deformation theorem

(Tits' deformation theorem). Let $R$ be an integral domain, $𝔽$ be the field of fractions of $R$, $\overline{𝔽}$ be the algebraic closure of $𝔽$, and $\bar{R}$ be the integral closure of $R$ in $\overline{𝔽}$. Let $A_R$ be an $R$-algebra and let $\left\{b_1,\dots ,b_d\right\}$ be a basis of $A_R$. For $a\in A_R$ let $\overrightarrow{A}\left(a\right)$ denote the linear transformation of $A_R$ induced by left multiplication by $a$. Let t 1 … t d be indeterminates and let $$\overrightarrow{p}\left(t_1,\dots ,t_d;x\right)=\det \left(x·\mathrm{Id}-\left(t_1\overrightarrow{A}\left(b_1\right)+\cdots +t_d\overrightarrow{A}\left(b_d\right)\right)\right)\in R\left[t_1,\dots ,t_d\right]\left[x\right]\text{,}$$ so that $\overrightarrow{p}$ is the characteristic polynomial of a “generic” element of $A_R$.

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

		Proof.
	First note that if $\left\{{b\prime }_1,\dots ,{b\prime }_d\right\}$ is another basis of $A_R$ and the change of basis matrix $P=\left(P_{ij}\right)$ is given by $${b\prime }_i=\sum _j{P}_{ij}{b}_j$$ then the transformation $${t\prime }_i=\sum _j{P}_{ij}{t}_j\text{,}$$ defines an isomorphism of polynomial rings $R\left[t_1,\dots ,t_d\right]\cong R\left[{t\prime }_1,\dots ,{t\prime }_d\right]$. Thus it follows that if the statements are true for one basis of $A_R$ then they are true for every basis of $A_R$ ( resp. $A_{\overline{𝔽}}$). (a): Using the decomposition of $A_{\overline{𝔽}}$ let $\left\{e_{ij}^{\mu }|\mu \in \hat{A}\text{,}1\le i,j\le d_{\lambda }\right\}$ be a basis of matrix units in $A_{\overline{𝔽}}$ and let $t_{ij}^{\mu }$ be the corresponding variables. Then the decomposition of $A_{𝔽‾}$ induces a factorisation $$\overrightarrow{p}\left(t_{ij}^{\mu },x\right)=\prod _{\lambda \in \hat{A}}{\left({\overrightarrow{p}}^{\lambda }\right)}^{d_{\lambda }}\text{,}\text{where}{\overrightarrow{p}}^{\lambda }\left(t_{ij}^{\mu };x\right)=\det \left(x-\sum _{\mu ,i,j}{t}_{ij}^{\mu }{A}^{\lambda }\left(e_{ij}\right)\right)\text{.}$$ The polynomial ${\overrightarrow{p}}^{\lambda }\left(t_{ij}^{\mu };x\right)$ is irreducible since specialising the variables gives $${\overrightarrow{p}}^{\lambda }\left(t_{j+1,j}^{\lambda }=1,t_{1,n}^{\lambda }=t,t_{i,j}^{\mu }=0\text{otherwise}\text{;}x\right)=d^{d_{\lambda }}-t\text{,}$$ which is irreducible in $\bar{R}\left[t;x\right]$. This provides the factorisation of $\overrightarrow{p}$ and establishes that $\mathrm{deg}\left({\overrightarrow{p}}^{\lambda }\right)=d_{\lambda }$. By (1.1) $${\overrightarrow{p}}^{\lambda }\left(t_{ij}^{\mu };x\right)=x^{d_{\lambda }}-\mathrm{Tr}\left(A^{\lambda }\left(\sum _{\mu ,i,j}{t}_{ij}^{\mu }{e}_{ij}^{\mu }\right)\right)x^{d_{\lambda }-1}+\cdots \text{,}$$ which establishes the last statement. Any root of $\overrightarrow{p}\left(t_1,\dots ,t_d,x\right)$ is an element of $R\left[t_1,\dots ,t_d\right]=\bar{R}\left[t_1,\dots ,t_d\right]$. So any root of ${\overrightarrow{p}}^{\lambda }\left(t_1,\dots ,t_d,x\right)$ is an element of $\bar{R}\left[t_1,\dots ,t_d\right]$ and therefore the coefficients of ${\overrightarrow{p}}^{\lambda }\left(t_1,\dots ,t_d,x\right)$ (symmetric functions in the roots of ${\overrightarrow{p}}^{\lambda }$) are elements of $\bar{R}\left[t_1,\dots ,t_d\right]$. (b): Taking the image of equation (1.1), give a factorisation of $\gamma \left(\overrightarrow{p}\right)$, $$\gamma \left(\overrightarrow{p}\right)=\prod _{\lambda \in \hat{A}}\gamma {\left({\overrightarrow{p}}^{\lambda }\right)}^{d_{\lambda }}\text{,}\text{in}\overline{𝕂}\left[t_1,\dots ,t_d,x\right]\text{.}$$ for the same reason as in (1.2) the factors $\gamma \left({\overrightarrow{p}}^{\lambda }\right)$ are irreducible polynomials in $\overline{𝕂}\left[t_1,\dots ,t_d,x\right]$. On the other hand, as in the proof of (a), the decomposition of $A_{\overline{𝕂}}$ induces a factorisation of $\gamma \left(\overrightarrow{p}\right)$ into irreducibles in $\overline{𝕂}\left[t_1,\dots ,t_d,x\right]$. These two factorisations must coincide, whence the result. $\square$

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 $n$. 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 $ℂA\left(\alpha \right)$-modules $A^{\lambda }\left(\alpha \right)$. Then

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

		Proof.
	Applying the Tits' deformation theorem to the case where $R=ℂ\left[x\right]$ (so that $𝔽=ℂ\left(x\right)$) gives the following theorem. The statement in (a) is a consequence of theorem 1.2, Regular Representations and the remark which follows theorem 1.2, Complete Reducibility. $\square$

Reference

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

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