Classification of graded Hecke algebras for complex reflection groups

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

Last update: 22 January 2014

Graded Hecke algebras

In this section, we define the graded Hecke algebra following Drinfeld [Dri1986]. Our main result in this section is Theorem 1.9c, which determines exactly how many degrees of freedom one has in defining a graded Hecke algebra.

Let $V$ be an $n$ dimensional vector space over $ℂ$ and let $G$ be a finite subgroup of $G L (V) .$ The group algebra of $G$ is $ℂ G = ℂ -span {t_{g} | g \in G}, with t_{g} t_{h} = t_{g h} .$ Let $a_{g} : V \times V ⟶ ℂ$ be skew symmetric bilinear forms indexed by the elements of $G$ and let $A$ be the associative algebra generated by $V$ and $ℂ G$ with the additional relations $\begin{matrix} t_{h} v t_{h^{- 1}} = h v and [v, w] = \sum_{g \in G} a_{g} (v, w) t_{g}, for h \in G and v, w \in V, & (1.1) \end{matrix}$ where $[v, w] = v w - w v .$ These relations allow every element $a \in A$ to be written in the form $\begin{matrix} a = \sum_{g \in G} p_{g} t_{g}, p_{g} \in S (V), & (1.2) \end{matrix}$ where $S (V)$ is the symmetric algebra of $V .$ More precisely, one must fix a section of the canonical surjection $T (V) \to S (V)$ from the tensor algebra of $V$ to $S (V)$ and take the elements $p_{g}$ to be in the image of this section.

The structure of $A$ depends on the choices of the “parameters” $a_{g} (v, w) \in ℂ .$ Our goal is to determine when the algebra $A$ will be a “semidirect product” of $S (V)$ and $ℂ G .$ This idea motivates the following definition [Dri1986].

The algebra $A$ is a graded Hecke algebra for $G$ if $A ≅ S (V) \otimes ℂ G as a vector space,$ or, equivalently, if the expression in (1.2) is unique for each $a \in A .$ A general element $a \in A$ is a linear combination of products of elements $t_{g}$ and $u_{i},$ where ${u_{1}, u_{2}, \dots, u_{n}}$ is a basis of $V .$ There are two straightening operations needed to put $a$ in the form (1.2): $(a) moving t_{h}'s to the right, and (b) putting u_{i} u_{j} pairs in the correct order.$ These two straightening operations correspond to the two identities in (1.1). Note that the “correct order” of $u_{i} u_{j}$ is determined by the choice of the section of the projection $T (V) \to S (V) .$ Let $v_{1}, v_{2}, v_{3}$ be arbitrary elements of $V$ and let $h \in G .$ Applying the straightening operations to $t_{h} v_{1} v_{2}$ gives $\begin{matrix} t_{h} v_{1} v_{2} & = & t_{h} [v_{1}, v_{2}] + t_{h} v_{2} v_{1} & (rearrange v_{1} and v_{2}) \\ = & t_{h} [v_{1}, v_{2}] + (h v_{2}) (h v_{1}) t_{h} & (move t_{h} to the right), \end{matrix}$ and applying the straightening operations in a different order gives $\begin{matrix} t_{h} v_{1} v_{2} & = & (h v_{1}) (h v_{2}) t_{h} & (move t_{h} to the right) \\ = & [h v_{1}, h v_{2}] t_{h} + (h v_{2}) (h v_{1}) t_{h} . \end{matrix}$ Setting these two expressions equal gives the relation $\begin{matrix} t_{h} [v_{1}, v_{2}] t_{h^{- 1}} = [h v_{1}, h v_{2}], for all h \in G, v_{1}, v_{2} \in V . & (1.3) \end{matrix}$ Similarly, applying the straightening operations to $v_{1} v_{2} v_{3}$ gives $\begin{matrix} v_{1} v_{2} v_{3} & = & [v_{1}, v_{2}] v_{3} + v_{2} v_{1} v_{3} & (moving v_{1} past v_{2}) \\ = & [v_{1}, v_{2}] v_{3} + v_{2} [v_{1}, v_{3}] + v_{2} v_{3} v_{1} & (moving v_{1} past v_{3}) \\ = & [v_{1}, v_{2}] v_{3} + v_{2} [v_{1}, v_{3}] + [v_{2}, v_{3}] v_{1} + v_{3} v_{2} v_{1} & (straightening v_{2} past v_{3}), \end{matrix}$ and applying the straightening operations in a different order gives $\begin{matrix} v_{1} v_{2} v_{3} & = & v_{1} [v_{2}, v_{3}] + v_{1} v_{3} v_{2} & (moving v_{2} past v_{3}) \\ = & v_{1} [v_{2}, v_{3}] + [v_{1}, v_{3}] v_{2} + v_{3} v_{1} v_{2} & (moving v_{1} past v_{3}) \\ = & v_{1} [v_{2}, v_{3}] + [v_{1}, v_{3}] v_{2} + v_{3} [v_{1}, v_{2}] + v_{3} v_{2} v_{1} & (straightening v_{1} past v_{2}). \end{matrix}$ These are equal if $\begin{matrix} [v_{1}, [v_{2}, v_{3}]] + [v_{2}, [v_{3}, v_{1}]] + [v_{3}, [v_{1}, v_{2}]] = 0, for all v_{1}, v_{2}, v_{3} \in V . & (1.4) \end{matrix}$ Conversely, the identities (1.3) and (1.4) are exactly what is needed to guarantee that any order of application of the straightening operations (a) and (b) will produce the same normal form (1.2) for the element $a .$ Thus we have

Let $A$ be an algebra defined as in (1.1). Then $A$ is a graded Hecke algebra if and only if the identities (1.3) and (1.4) hold in $A .$

Using (1.1), the relations (1.3) and (1.4) can be rewritten in terms of the bilinear forms $a_{g} : V \times V \to ℂ$ as $\begin{matrix} a_{g} (v_{1}, v_{2}) = a_{h g h^{- 1}} (h v_{1}, h v_{2}) and & (1.6) \\ a_{g} (v_{3}, v_{1}) (g v_{2} - v_{2}) + a_{g} (v_{2}, v_{3}) (g v_{1} - v_{1}) + a_{g} (v_{1}, v_{2}) (g v_{3} - v_{3}) = 0 & (1.7) \end{matrix}$ for $v_{1}, v_{2}, v_{3} \in V$ and $g, h \in G .$

Let $⟨, ⟩ : V \times V \to ℂ$ be a $G -invariant$ nondegenerate Hermitian form on $V .$ For each $g \in G,$ set $\begin{matrix} V^{g} & = & {v \in V | g v = v}, \\ {(V^{g})}^{⊥} & = & {v \in V | ⟨ v, w ⟩ = 0 for all w \in V^{g}}, and \\ ker a_{g} & = & {v \in V | a_{g} (v, w) = 0 for all w \in V} . \end{matrix}$

Let $G$ be a finite subgroup of $G L (V)$ and let $g \in G .$

(a)	${(V^{g})}^{⊥} = {v - g v \| v \in V} .$
(b)	Suppose $g \neq 1 .$ If $codim (V^{g}) = 2$ and $a : V \times V \to ℂ$ is a skew symmetric bilinear form such that $ker a = V^{g},$ then a satisfies (1.7).

Let $A$ be a graded Hecke algebra for $G$ defined by skew symmetric bilinear forms $a_{g} : V \times V \to ℂ .$

(c)	If $g \neq 1$ then $ker a_{g} \supseteq V^{g} .$
(d)	If $g \neq 1$ and $a_{g} \neq 0$ then $ker a_{g} = V^{g}$ and $codim (V^{g}) = 2 .$
(e)	If $g \neq 1$ and $a_{g} \neq 0$ then, for all $h \in G,$ $a_{h^{- 1} g h} (b_{1}, b_{2}) = det (h^{⊥}) a_{g} (b_{1}, b_{2}),$ where ${b_{1}, b_{2}}$ is a basis of ${(V^{g})}^{⊥}$ and $h^{⊥} : {(V^{g})}^{⊥} \to {(V^{g})}^{⊥}$ is the composition of $h$ restricted to ${(V^{g})}^{⊥}$ with the canonical projection $V \to V / V^{g} .$

Proof.

(a) Consider the map $ϕ : V \to V$ given by $ϕ (v) = v - g v .$ Then $ker ϕ = V^{g}$ and $im ϕ \subseteq {(V^{g})}^{⊥}$ since, if $v \in V,$ $w \in V^{g},$ then $⟨ v - g v, w ⟩ = ⟨ v, w ⟩ - ⟨ g v, w ⟩ = ⟨ v, w ⟩ - ⟨ g v, g w ⟩ = ⟨ v, w ⟩ - ⟨ v, w ⟩ .$ Since $dim (im ϕ) = codim (ker ϕ) = codim (V^{g})$ it follows that $im ϕ = {(V^{g})}^{⊥} .$

(b) Let $v_{1}, v_{2}, v_{3} \in V .$ If any $v_{i} \in V^{g},$ then (1.7) holds trivially for the skew symmetric form $a .$ So assume each $v_{i} \notin V^{g}$ and write each $v_{i}$ as $v_{i}^{+} + v_{i}^{-}$ where $v_{i}^{+} \in V^{g}$ and $v_{i}^{-} \in {(V^{g})}^{⊥} .$ Then $a (v_{i}, v_{j}) = a (v_{i}^{-}, v_{j}^{-}) and v_{i} - g v_{i} = v_{i}^{-} - g v_{i}^{-} .$ Since $dim {(V^{g})}^{⊥} = 2,$ at least one of the $v_{i}^{-}$ is a linear combination of the other two. Say $v_{i}^{-} = c_{2} v_{2}^{-} + c_{3} v_{3}^{-}$ with $c_{2}, c_{3} \in ℂ .$ Substituting $v_{i} - g v_{i} = v_{i}^{-} - g v_{i}^{-}$ and $v_{1}^{-} = c_{2} v_{2}^{-} + c_{3} v_{3}^{-}$ then yields $\begin{matrix} a (v_{3}, v_{1}) (g v_{2} - v_{2}) + a (v_{2}, v_{3}) (g v_{1} - v_{1}) + a (v_{1}, v_{2}) (g v_{3} - v_{3}) \\ = a (v_{3}^{-}, v_{1}^{-}) (g v_{2}^{-} - v_{2}^{-}) + a (v_{2}^{-}, v_{3}^{-}) (g v_{1}^{-} - v_{1}^{-}) + a (v_{1}^{-}, v_{2}^{-}) (g v_{3}^{-} - v_{3}^{-}) = 0, \end{matrix}$ and so (1.7) holds.

(c) Let $v_{3} \in V^{g}$ and $v_{2} \in V .$
If $v_{2} \in V^{g},$ then $a_{g} (v_{2}, v_{3}) (g v_{1} - v_{1}) = 0$ for all $v_{1} \in V$ by (1.7). Since $V^{g} \neq V,$ there is some $v_{1}$ such that $g v_{1} \neq v_{1}$ and so $a_{g} (v_{2}, v_{3}) = 0 .$
If $v_{2} \notin V^{g},$ let $v_{1} = \sum_{k = 1}^{r - 1} g^{k} v_{2},$ where $r$ is the order of $g .$ By (1.6), $a_{g} (v_{3}, g^{k} v_{2}) = a_{g} (g^{- k} v_{3}, v_{2}) = a_{g} (v_{3}, v_{2}),$ for any $k,$ and so $\begin{matrix} 0 & = & a_{g} (v_{3}, v_{1}) (g v_{2} - v_{2}) + a_{g} (v_{2}, v_{3}) (g v_{1} - v_{1}) \\ = & (r - 1) a_{g} (v_{3}, v_{2}) (g v_{2} - v_{2}) + a_{g} (v_{3}, v_{2}) (g v_{2} - v_{2}) = r a_{g} (v_{3}, v_{2}) (g v_{2} - v_{2}) . \end{matrix}$ Thus $a_{g} (v_{3}, v_{2}) = 0 .$
Hence, for all $v_{2} \in V,$ $a_{g} (v_{3}, v_{2}) = 0$ and so $V^{g} \subseteq ker a_{g} .$

(d) By (c), $codim (V^{g}) \geq codim (ker a_{g}) .$ Since $a_{g} \neq 0,$ there exist $v, w \in V$ with $a_{g} (v, w) \neq 0$ and so $codim (ker a_{g}) \geq 2 .$ Let $v_{1} - g v_{1}$ and $v_{2} - g v_{2}$ be linearly independent elements of ${(V^{g})}^{⊥} .$ Then (1.7) implies that any element $v_{3} - g v_{3} \in {(V^{g})}^{⊥}$ is a linear combination of $v_{1} - g v_{1}$ and $v_{2} - g v_{2},$ and so $2 \geq dim ({(V^{g})}^{⊥}) = codim (V^{g}) \geq codim (ker a_{g}) \geq 2 .$ Thus $V^{g} = ker a_{g}$ and $codim (V^{g}) = 2 .$

(e) Write $h b_{1} = h_{11} b_{1} + h_{21} b_{2} + {(h b_{1})}^{g}$ and $h b_{2} = h_{12} b_{1} + h_{22} b_{2} + {(h b_{2})}^{g}$ with $h_{i j} \in ℂ$ and ${(h b_{i})}^{g} \in V^{g} .$ Then $\begin{matrix} a_{h^{- 1} g h} (b_{1}, b_{2}) & = & a_{g} (h b_{1}, h b_{2}) = a_{g} (h_{11} b_{1} + h_{21} b_{2} + {(h b_{1})}^{g}, h_{12} b_{1} + h_{22} b_{2} + {(h b_{2})}^{g}) \\ = & (h_{11} h_{22} - h_{21} h_{12}) a_{g} (b_{1}, b_{2}) = det (h^{⊥}) a_{g} (b_{1}, b_{2}) \end{matrix}$ since $a_{g}$ is skew symmetric and $V^{g} \subseteq ker a_{g} .$

$□$

The following theorem is a slightly strengthened version of statements (given without proof) in [Dri1986].

Let $G$ be a finite subgroup of $G L (V)$ and let $Z_{G} (g) = {h \in G | h g = g h}$ denote the centralizer of an element $g$ in $G .$

(a)	If $A$ is a graded Hecke algebra for $G,$ then the values of $a_{h^{- 1} g h}$ are determined by the values of $a_{g}$ via the equation $a_{h^{- 1} g h} (v_{1}, v_{2}) = a_{g} (h v_{1}, h v_{2}), for all g, h \in G, v_{1}, v_{2} \in V .$
(b)	For $g \neq 1,$ there is a graded Hecke algebra $A$ with $a_{g} \neq 0$ if and only if $ker a_{g} = V^{g}, codim (V^{g}) = 2, and det (h^{⊥}) = 1, for all h \in Z_{G} (g),$ where $h^{⊥}$ is $h$ restricted to the space ${(V^{g})}^{⊥} .$ In this case, $a_{g}$ is determined by its value $a_{g} (b_{1}, b_{2})$ on a basis ${b_{1}, b_{2}}$ of ${(V^{g})}^{⊥} .$
(c)	Let $d$ be the number of conjugacy classes of $g \in G$ such that $codim (V^{g}) = 2$ and $det (h^{⊥}) = 1$ for all $h \in Z_{G} (g),$ where $h^{⊥}$ is $h$ restricted to the space ${(V^{g})}^{⊥} .$ The sets ${a_{g}}_{g \in G}$ corresponding to graded Hecke algebras $A$ form a vector space of dimension $d + dim ({(⋀^{2} V)}^{G}) .$

Proof.

(a) is simply a restatement of (1.6).

(b) $⟹ :$ If $A$ is a graded Hecke algebra and $a_{g} \neq 0$ then by Lemma 1.8d, $codim (V^{g}) = 2$ and $ker a_{g} = V^{g} .$ So $a_{g}$ is determined by its value $a_{g} (b_{1}, b_{2})$ on a basis $b_{1}, b_{2}$ of ${(V^{g})}^{⊥} .$ Suppose $h \in Z_{G} (g) .$ Then, by Lemma 1.8e, $a_{g} (b_{1}, b_{2}) = a_{h g h^{- 1}} (h b_{1}, h b_{2}) = a_{g} (h b_{1}, h b_{2}) = det (h^{⊥}) a_{g} (b_{1}, b_{2}),$ and so $det (h^{⊥}) = 1 .$ Note that $h (V^{g}) = V^{g}$ and $h {(V^{g})}^{⊥} = {(V^{g})}^{⊥}$ since, for each $v \in V^{g},$ $h (v) = h g (v) = g h (v) .$

$⟸ :$ If $codim (V^{g}) = 2$ then, up to constant multiples, there is a unique skew symmetric form on $V$ which is nondegenerate on ${(V^{g})}^{⊥}$ and which has $ker a_{g} = V^{g} .$ Fix such a form and then define forms $a_{h},$ $h \in G,$ by $\begin{matrix} a_{h} (v_{1}, v_{2}) = {\begin{matrix} a_{g} (k v_{1}, k v_{2}) & if h = k^{- 1} g k, \\ 0 & otherwise, \end{matrix} & (1.10) \end{matrix}$ for $v_{1}, v_{2} \in V .$ Let $a_{1}$ be any $G -invariant$ skew symmetric form on $V .$ Then this collection ${a_{g}}_{g \in G}$ of skew symmetric bilinear forms satisfies (1.6) by definition and (1.7) by Lemma 1.8b. Thus (by Lemma 1.5), it determines a graded Hecke algebra $A$ via (1.1).

(c) From (a) and (b) it follows that the sets ${a_{g}}_{g \in G},$ running over all graded Hecke algebras $A$ for $G,$ form a vector space. Since each of the collections ${a_{g}}_{g \neq 1}$ constructed by (1.10) has its support on a single conjugacy class, these collections form a basis of the vector space of sets ${a_{g}}_{g \neq 1} .$ The only condition on the form $a_{1}$ is that it satisfies (1.6), which means that it is a $G -invariant$ element of ${(⋀^{2} V)}^{*} .$

$□$

The following consequence of Theorem 1.9 will be useful for completing the classification of graded Hecke algebras for complex reflection groups.

Assume that $G$ contains $h = ξ \cdot 1$ for some $ξ \in ℂ \ {\pm 1} .$ If $A$ is a graded Hecke algebra for $G,$ then $a_{g} = 0$ for all $g \neq 1 .$

Proof.

If $h = ξ \cdot 1 \in G,$ then $h \in Z_{G} (g)$ for every $g \in G$ and $det (h^{⊥}) = ξ^{2}$ if $codim (V^{g}) = 2 .$ The statement then follows from Theorem 1.9b.

$□$

Notes and references

This is a typed version of Classification of graded Hecke algebras for complex reflection groups by Arun Ram and Anne V. Shepler.

Research of the first author supported in part by the National Security Agency and by EPSRC Grant GR K99015 at the Newton Institute for Mathematical Sciences. Research of the second author supported in part by National Science Foundation grant DMS-9971099.

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