Tits' Deformation Theorem

## Tits' deformation theorem

(Tits' deformation theorem). Let $R$ be an integral domain, $𝔽$ be the field of fractions of $R$, $\stackrel{‾}{𝔽}$ be the algebraic closure of $𝔽$, and $\stackrel{‾}{R}$ be the integral closure of $R$ in $\stackrel{‾}{𝔽}$. 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 $\stackrel{\to }{A}\left(a\right)$ denote the linear transformation of ${A}_{R}$ induced by left multiplication by $a$. Let t 1 t d be indeterminates and let $p → t 1 … t d x = det x · Id - t 1 A → b 1 + ⋯ + t d A → b d ∈ R t 1 … t d x ,$ so that $\stackrel{\to }{p}$ is the characteristic polynomial of a “generic” element of ${A}_{R}$.

1. Let ${A}_{\stackrel{‾}{𝔽}}=\stackrel{‾}{𝔽}{\otimes }_{R}{A}_{R}$. If $A 𝔽 ‾ ≅ ⨁ λ ∈ A ˆ M d λ 𝔽 ‾ ,$ then the factorisation of $\stackrel{\to }{p}\left({t}_{1},\dots ,{t}_{d};x\right)$ into irreducibles in $\stackrel{‾}{𝔽}\left[{t}_{1},\dots ,{t}_{d},x\right]$ has the form $p → = ∏ λ ∈ A ˆ p → λ d λ , with p ‾ λ ∈ R ‾ t 1 … t d x and d λ = deg p → λ .$ If ${\chi }^{\lambda }\left({t}_{1},\dots ,{t}_{d}\right)\in \stackrel{‾}{R}\left[{t}_{1},\dots ,{t}_{d}\right]$ is given by $p ‾ λ t 1 … t d x = x d λ - χ λ t 1 … t d x d λ - 1 + ⋯ ,$ then $χ A 𝔽 ‾ λ : A 𝔽 ‾ ⟶ 𝔽 ‾ α 1 b 1 + … + α d α d ⟼ χ λ α 1 … α d , λ ∈ A ˆ ,$ are the irreducible characters of ${A}_{\stackrel{‾}{𝔽}}$.
2. Let $𝕂$ be a field and let $\stackrel{‾}{𝕂}$ be the algebraic closure of $𝕂$. Let $\gamma :R\to 𝕂$ be a ring homomorphism and let $\stackrel{‾}{\gamma }:\stackrel{‾}{R}\to \stackrel{‾}{𝕂}$ be the extension of $\gamma$. Let ${\chi }^{\lambda }\left({t}_{1},\dots ,{t}_{d}\right)\in \stackrel{‾}{R}\left[{t}_{1},\dots ,{t}_{d}\right]$ be as in (a). If ${A}_{\stackrel{‾}{𝕂}}=\stackrel{‾}{𝕂}{\oplus }_{R}{A}_{R}$ is semisimple then $A 𝕂 ‾ ≅ ⨁ λ ∈ A ˆ M d λ 𝕂 ‾ ,$ and $χ A 𝕂 ‾ λ : A 𝕂 ‾ ⟶ 𝕂 ‾ α 1 b 1 + … + α d α d ⟼ γ ‾ χ λ α 1 … α d , λ ∈ A ˆ ,$ are the irreducible characters of ${A}_{\stackrel{‾}{𝕂}}$.

Proof.
1. 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 $b ′ i = ∑ j P i j b j$ then the transformation $t ′ i = ∑ j P i j t j ,$ 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. ${A}_{\stackrel{‾}{𝔽}}$).
2. (a): Using the decomposition of ${A}_{\stackrel{‾}{𝔽}}$ let $\left\{{e}_{ij}^{\mu }\phantom{\rule{.5em}{0ex}}|\phantom{\rule{.5em}{0ex}}\mu \in \stackrel{ˆ}{A}\text{,}\phantom{\rule{.5em}{0ex}}1\le i,j\le {d}_{\lambda }\right\}$ be a basis of matrix units in ${A}_{\stackrel{‾}{𝔽}}$ and let ${t}_{ij}^{\mu }$ be the corresponding variables. Then the decomposition of ${A}_{𝔽‾}$ induces a factorisation
 $p → t i j μ x = ∏ λ ∈ A ˆ p → λ d λ , where p → λ t i j μ x = det x - ∑ μ i j t i j μ A λ e i j .$
The polynomial ${\stackrel{\to }{p}}^{\lambda }\left({t}_{ij}^{\mu };x\right)$ is irreducible since specialising the variables gives
 $p → λ t j + 1 j λ = 1 t 1 n λ = t t i j μ = 0 otherwise ; x = d d λ - t ,$
which is irreducible in $\stackrel{‾}{R}\left[t;x\right]$. This provides the factorisation of $\stackrel{\to }{p}$ and establishes that $\mathrm{deg}\left({\stackrel{\to }{p}}^{\lambda }\right)={d}_{\lambda }$. By (1.1) $p → λ t i j μ x = x d λ - Tr A λ ∑ μ i j t i j μ e i j μ x d λ - 1 + ⋯ ,$ which establishes the last statement.
3. Any root of $\stackrel{\to }{p}\left({t}_{1},\dots ,{t}_{d},x\right)$ is an element of $R\left[{t}_{1},\dots ,{t}_{d}\right]=\stackrel{‾}{R}\left[{t}_{1},\dots ,{t}_{d}\right]$. So any root of ${\stackrel{\to }{p}}^{\lambda }\left({t}_{1},\dots ,{t}_{d},x\right)$ is an element of $\stackrel{‾}{R}\left[{t}_{1},\dots ,{t}_{d}\right]$ and therefore the coefficients of ${\stackrel{\to }{p}}^{\lambda }\left({t}_{1},\dots ,{t}_{d},x\right)$ (symmetric functions in the roots of ${\stackrel{\to }{p}}^{\lambda }$) are elements of $\stackrel{‾}{R}\left[{t}_{1},\dots ,{t}_{d}\right]$.
4. (b): Taking the image of equation (1.1), give a factorisation of $\gamma \left(\stackrel{\to }{p}\right)$, $γ p → = ∏ λ ∈ A ˆ γ p → λ d λ , in 𝕂 ‾ t 1 … t d x .$ for the same reason as in (1.2) the factors $\gamma \left({\stackrel{\to }{p}}^{\lambda }\right)$ are irreducible polynomials in $\stackrel{‾}{𝕂}\left[{t}_{1},\dots ,{t}_{d},x\right]$.
5. On the other hand, as in the proof of (a), the decomposition of ${A}_{\stackrel{‾}{𝕂}}$ induces a factorisation of $\gamma \left(\stackrel{\to }{p}\right)$ into irreducibles in $\stackrel{‾}{𝕂}\left[{t}_{1},\dots ,{t}_{d},x\right]$. These two factorisations must coincide, whence the result.
6. $\square$

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 $\stackrel{ˆ}{A}$ 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 $\stackrel{ˆ}{A}$ is an index set for the simple $ℂA\left(n\right)$-modules ${A}^{\lambda }\left(n\right)$ and $\mathrm{dim}\left({A}^{\lambda }\left(n\right)\right)=\mathrm{dim}\left({A}^{\lambda }\left(\alpha \right)\right)$ for each $\lambda \in \stackrel{ˆ}{A}$.
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 ${\chi }^{\lambda }\left({b}_{1},\dots ,{b}_{d}\right)\in ℂ\left[{b}_{1},\dots ,{b}_{d},x\right]$, $\lambda \in \stackrel{ˆ}{A}$, such that for every $n\in ℂ$ such that $ℂA\left(n\right)$ is semisimple, $χ A n λ : ℂ A n ⟶ ℂ α 1 b 1 + ⋯ + α d b d ⟼ χ λ α 1 … α d n , λ ∈ A ˆ ,$ 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.