On the Calogero–Moser space associated with dihedral groups II. The equal parameter case

We continue the study of Calogero–Moser spaces associated with dihedral groups by investigating in more details the equal parameter case: we obtain explicit equations, some informations about the Poisson bracket, the structure of the Lie algebra associated with the cuspidal point and the action of SL 2 ( C ) .

the author [BoTh].This algorithm was implemented by Thiel [Thi] in his CHAMP package for MAGMA [Mag].Explicit computer computations in small cases (i.e.d ∈ {4, 5, 6, 7}) were necessary to find the general pattern.So, even though this does not appear in this paper, it is fair to say that the above results owe their existence to MAGMA.[Bon1].-Wewill use the notation of the first part [Bon1] and we recall here some of them, the most important ones.We set V = 2 and (x , y ) denotes its canonical basis while (X , Y ) is the dual basis of V * .We identify GL (V ) with GL 2 ( ).We also fix a non-zero natural number d , as well as a primitive d -th root of unity ζ ∈ × .If i ∈ or /d , we set

Recollection of notation from
s = s 0 , t = s 1 and W = 〈s , t 〉: it is the dihedral group of order 2d .The set Ref(W ) of reflections of W is {s i | i ∈ /d }.Finally, let w 0 denote the longest element of W (we have w 0 = t (s t ) (d −1)/2 if d is odd and w 0 = (s t ) d /2 if d is even): this notation was used in the first part [Bon1,Rem. 6.4] but we had forgotten to define it!It will be used here in Section 4. We set q = x y , Q = X Y , r = x d + y d , R = X d + Y d and, if 0 i d , In this second part, we will not use the notation r or R as r = a 0,0 and R = a d ,0 : we prefer this second notation.If i 0, we set and eu 0 = eu (1)  0 .We fix a map c : Ref(W ) → and we set a = c s and b = c t .We denote by H c the rational Cherednik algebra at t = 0, with parameter c , whose presentation is given in [Bon1,(3.2)].Its center is denoted by Z c and we denote by c the affine variety whose algebra of regular functions [ c ] is precisely Z c .
We denote by Trunc c the -linear map Trunc c ( f w ) = f if w = 1, 0 otherwise.It is the map induced by the map Trunc defined in [Bon1,§3.4].Its restriction Trunc c : W is an isomorphism of -graded vector spaces [Bon1,Lem. 3.5].Recall that it is P • -linear, where We add a further notation which will be useful in this second part, namely, we set e = 0 1 0 0 , h = 1 0 0 −1 and f = 0 0 1 0 , so that (e , h , f ) is the standard basis of the Lie algebra sl 2 ( ).
Hypothesis.All along this paper, together with the above notation, we make the additional assumption that a = b .Recall that it is automatically satisfied if d is odd.
The unicity comes from the fact that eu 0 , q and Q are algebraically independent.Note that Ψ 0 = 1 and Ψ 1 = T .Now the sequence (Ψ i ) i 0 is easily determined by the following recursive formula: if i 1, then (1.4) Indeed, this follows from the fact that (x Note also for future reference the following two relations: if i 1, then Proof of (1.5).-We prove only the first identity, the second one being obtained by exchanging the roles of (x , y ) and (X , Y ).Let us consider the two identities obtained by applying ∂ /∂ X and ∂ /∂ Y to (1.3): Multiplying the first equality by y , the second by x , and adding the results yields exactly as expected.

1.B.
Presentation.-We rewrite slightly differently the presentation of Bon1,Theo. 2.1] according to our needs.A straightforward computation shows that, if Using (1.1), this gives This equation can also be obtained by substracting the equation (Z 0 i , j ) to the equation ) (with the notation of [Bon1,§2]).Consequently, the presentation given in [Bon1, Theo.2.1] can be rewritten as follows: Theorem 1.6.-The algebra of invariants [V × V * ] W admits the following presentation: which is completely determined by the following rules: Therefore, a straightforward computation shows that the Poisson bracket between the generators of [V × V * ] W is given by: (1.7) where the last equality only holds if 0 i < j d .In particular, (Q , eu 0 , −q ) is an sl 2 -triple (for the Lie algebra structure on [V × V * ] W induced by the Poisson bracket).Note that -Since W is a Coxeter group, the W -modules V and V * are isomorphic.In our situation, the map is an isomorphism of W -modules.One then gets an action of SL 2 ( ) on V × V * as follows: By construction, this action commutes with the action of W , so induces an action of SL 2 ( ) on the -algebras This induces an action of the Lie algebra sl 2 ( ) by derivations on and ξ ∈ sl 2 ( ), we denote by ξ•ϕ the image of ϕ under the action of − t ξ.
It is easily checked on the generators

Calogero-Moser space at equal parameters
Notation.We denote by q , Q , eu , a 0 , a 1 ,. . ., a d the respective images of q , Q , eu, a 0 , a 1 ,. . ., a d in Z c .

C. BONNAFÉ
Note the following formulas: (2.1) Note also the following formula, which follows from [Gor,§3.6 (2.3) An important feature of the equal parameter case is that the elements a j have a reasonably simple form: Proposition 2.4.-If 0 j d , then -For future use of the above formula, we set Proof.-Let b j ∈ H c denote the right-hand side of the equation of the proposition.Since First, an easy computation shows that b j commutes with s and t .Now, by (2.2), we have Now, the first two lines of this last equation compensate each other and it remains we can rewrite the above formula as follows: where This formula implies that Θ j ,k is a linear combination of (non-commutative) monomials of the form But j l + j d − 1 and 0 m j − 1, so l + j ≡ m mod d .This implies in particular that i ∈ /d ζ i (l + j −m ) = 0, and so θ j ,k ,l ,m = 0.This shows that [x ,b j ] = 0.
A similar computation shows that [X ,b j ] = 0 and so b j commutes with s , t , x , X , This completes the proof of the proposition.
This has the following consequence, that will be used later for obtaining a presentation of the algebra Z c .
Proof.-Assume first that 0 i j d .Since a j is central, we get Then the above equality might rewritten Now, if 1 i j d − 1, applying the above formula by replacing i by i − 1 and j by j + 1 yields So, the coefficient of and as expected.

2.B. Poisson bracket. -
We determine here part of the Poisson bracket between the generators: with the convention that a −1 = a d +1 = 0.
The first three equalities of the proposition are standard and hold for any Coxeter group (see [Dez,§4] or [BEG,§3]) and can easily be checked in this case by a little computation.
Similarly, the fact that {eu , a j } = (2 j − d )a j follows from the general fact that, if h ∈ H c is homogeneous of degree k , then {eu , h } = k h (see for instance [BoRo,Prop. 3.3.3]).We now prove that {q , a j } = j a j +1 , the last equality being proved similarly.From the formula given for a j in Proposition 2.4, we get In order to prove the proposition, it is sufficient to check that Trunc c ({q , a j }) = j a j −1 .But, from the above formula and from (2.2), one gets So it remains to prove that Let us compute the big fraction in the above formula.First, Therefore, Simplifying and using the change of variable k → j − 2 − k , one gets We deduce that Since the term corresponding to k = j − 2 vanishes, this implies that Therefore, the left-hand side of the formula (?) is a linear combination of monomials of the form x d − j −1−l y l X j −3−m Y m , where 0 l d − j −1 and 0 m j −3, and the coefficient of this monomial is But j l + j d − 1 and 1 m + 1 j − 2, so this coefficient is 0 and the equality (?) is proved.

2.C. Presentation. -
The main result of this paper is the following: Theorem 2.9.
-If a = b , then the algebra Z c admits the following presentation: • Generators: q , Q , eu , a 0 , a 1 , . . ., a d .
-By [BoTh], a presentation of Z c is obtained by deforming the generators of Z 0 = [V × V * ] W and deforming the relations.Therefore, in order to prove the theorem, it is sufficient to check that the relations given in the statement are satisfied.So let Let us first prove that For this, it is sufficient to prove that Trunc c (eu a i ) = Trunc c (q a i +1 + Q a i −1 ).But the map Trunc c is P • -linear so it is sufficient to prove that Since a i is central, it follows from (2.3) and Proposition 2.4 that W is a linear combination of monomials of the form x d −i −1−l y l X i −1−m Y m where 0 l d −i −1 and 0 m i −1, and the coefficient of this monomial is equal to But i i + l d − 1 and 0 m i − 1, so i + l − m ≡ 0 mod d .This shows that the above sum is zero, and this completes the proof of (?).
Let us now prove that This will be proved by induction on j − i .So let us first consider the case where j − i = 0, i.e.where j = i .Again, it is sufficient to prove the equality after applying the map Trunc c .We deduce from Corollary 2.5 that Since Trunc c is P • -linear and or, equivalently, that It follows directly that as desired.
Assume now that j − i 1 and that Then, by the induction hypothesis, we have Applying {q , −} to this equality, and using Proposition 2.6 and Corollary 2.8, one gets: But, by (1.5), Since the induction hypothesis implies that the result follows.

2.D. Back to Poisson bracket. -
In Proposition 2.6, we did not determine the Poisson brackets {a i , a j }.This was only determined for a = 0 in (1.7): it is proven that there exists a polynomial Π i , j ∈ [T , T ′ , T ′′ ], which is homogeneous of degree d − 1, such that {a i ,0 , a j ,0 } = Π i , j (eu 0 , q ,Q ).
This will be deformed to the unequal parameter case as follows: Proposition 2.10.-If 0 i < j d , there exists a polynomial Φ i , j ∈ [T, T ′ , T ′′ ], homogeneous of degree d − 3, such that Proof.-We will prove that there exist polynomials Φ • i , j , Φ i , j ∈ [T, T ′ , T ′′ ], homogeneous of degree d − 1 and d − 3 respectively, such that This is sufficient because, by specializing a to 0, one gets that Φ • i , j = Π i , j .Let us first assume that i = 0. To make an induction argument on j work, we will prove a slightly stronger result, namely that where ϕ j , θ j ∈ [T , T ′ , T ′′ ] are homogeneous of degree j − 1 and j − 3 respectively.For this, let us apply {a 0 , −} to the following two relations given by Theorem 2.9 Using Proposition 2.6, this gives Thanks to the first equality, we can replace the term q {a 0 , a 2 } in the second equation by eu {a 0 , a 1 }, and this yields (a 0 eu − 2q a 1 ){a 0 , a 1 } = 2d q d −1 (a 0 eu − 2q a 1 ).
Since a 0 eu − 2q a 1 = 0 (by computing its image by Trunc c ) and since Z c is an integral domain, we get {a 0 , a 1 } = 2d q d −1 , which proves (℘ + 0,1 ).We also deduce that Now, assume that j 3 and that (℘ + 0, j ′ ) holds for j ′ < j .Applying {a 0 , −} to by Theorem 2.9 and {a 0 , by the induction hypothesis.This gives which proves that (℘ + 0, j ) holds.We will now prove that (℘ i , j ) holds by induction on i .The case i = 0 has just been treated, so assume that i 1 and that (℘ i −1, j ′ ) holds for all j ′ .Then By the Jacobi identity, we get So the result follows from the induction hypothesis because, if Θ ∈ [T, T ′ , T ′′ ] is an homogeneous polynomial of degree k , then is of the form Θ # (eu , q ,Q ) where Θ # is homogeneous of degree k .The proof of the proposition is complete.

2.E.
Lie algebra structure at the cuspidal point.-By Theorem 2.9, the affine variety c might be described as If d = 3 and a = 0, then c is smooth.So assume from now on that d 4 and a = 0. Then the homogeneous component of minimal degree of all the above equations is equal to 2, so the point 0 = (0, ..., 0) ∈ c is singular and the tangent space of c at 0 has dimension d +4.It is the only singular point and it is a cuspidal point in the sense of [Bel] (see [Bon1,§5.2]).This means that the corresponding maximal ideal m 0 of Z c is a Poisson ideal (since m 0 = 〈q ,Q , eu , a 0 , a 1 , . . . a d 〉, this can also be checked thanks to Proposition 2.10).This implies that the cotangent space m 0 /m 2 0 of c at 0 inherits a Lie algebra structure from the Poisson bracket: we denote by Lie 0 ( c ) the vector space m 0 /m 2 0 endowed with its Lie algebra structure.It has been proved in [Bon1,Prop. 8.4] that Proof.-If m ∈ m 0 , we denote by ṁ its image in Lie 0 ( c ).Then ( q , Q , ė u , ȧ 0 , ȧ 1 , . .., ȧ d ) is a basis of Lie 0 ( c ).We set It follows from Proposition 2.6 that g is a Lie subalgebra of Lie 0 ( c ) isomorphic to sl 2 ( ), and that S d is normalized by g and is isomorphic to Sym d ( 2 ) as an sl 2 ( )-module.
Since d 5 (and so d − 3 2), we get from Proposition 2.10 that {a i , a j } ∈ m 2 0 and so [ ȧ i , ȧ j ] = 0.This completes the proof of the proposition.

3.C. Hermite's reciprocity law
Theorem 2.9 shows that the natural morphism of algebras σ : Sym(E ) −→ Z c is surjective and it describes its kernel.For avoiding the confusion between multiplication in Z c and multiplication in Sym(E ), we will denote by ⋆ the multiplication in Sym(E ).For instance, 1 is an element of Sym(E ) whereas a 0 a 2 − a 2 1 is an element of Z c , which is equal to σ(a 0 ⋆ a 2 − a ⋆2 1 ).Similarly, if e 1 ,. . ., e n are elements of E and if Ψ ∈ [T 1 , . .., T n ] is a polynomial in n indeterminates, we denote by Ψ ⋆ (e 1 , . .., e n ) the evaluation of Ψ at (e 1 , . . . e n ) inside the algebra Sym(E ) whereas Ψ(e 1 , . .., e n ) denotes the evaluation of Ψ inside the algebra Z c : they satisfy the equality σ(Φ ⋆ (e 1 , . .., e n )) = Ψ(e 1 , . .., e n ).
Proposition 2.6 and (3.1) imply that E is an SL 2 ( )-stable subspace of Z c , so that σ is SL 2 ( )-equivariant.Let us denote by V 2 ≃ 2 another copy of 2 viewed as the standard representation of SL 2 ( ) (or sl 2 ( )), and we denote by (t , u) its canonical basis.We then have two isomorphisms of vector spaces Proposition 2.6 and (3.1) imply that σ ♯ and σ d are SL 2 ( )-equivariant and we will identify Let us first interprete the equations and that we have a natural morphism given by multiplication.We denote by Der(Sym( So the family of equations (Z i ) 1 i d −1 can be summarized by (3.2) Ker(D (2) ) is contained in Ker(σ).
The interpretation of the equations (Z i , j ) 1 i j d −1 is somewhat more subtle and is related with Hermite's reciprocity law (see the upcoming Remark 3.7).First, evaluation induces a surjective morphism of SL 2 ( )-modules In the special case where m = 2 and n = d , then: In fact, the family (a i −1 ⋆ a j +1 − a i ⋆ a j ) 1 i j d −1 generates the ideal equal to the kernel of the natural morphism ǫ •,d : Sym(Sym d (V 2 )) → Sym(V 2 ).On the other hand, it follows from (1.2) that: Lemmas 3.3 and 3.4 allow to define a linear map It is an isomorphism of vector spaces but an important fact is the following: Proof.-This is more or less the computation done in the end of the proof of Theorem 2.10.It is sufficient to prove that it is an isomorphism of sl 2 ( )-modules.By (3.1) Proposition 2.6, we have Therefore, ρ d ( f • (a i −1 ⋆ a j +1 − a i ⋆ a j ) = (i − 1)q ⋆d − j −1 ⋆ Q ⋆i −2 ⋆ Ψ ⋆ j −i +1 (eu , q ,Q ) + j q ⋆d − j ⋆ Q ⋆i −1 ⋆ Ψ ⋆ j −i −1 (eu , q ,Q ).
and so one gets where the last equality follows from (1.4).Applying now (1.5) yields Putting things together and using again (3.1) and Proposition 2.6 yields as desired.The fact that follows from a similar computation and this completes the proof of the Lemma.
Using the isomorphism of SL 2 ( )-modules ρ d , the family of equations (Z i , j ) can be rewritten as follows: commutative.In particular, h m ,n induces an isomorphism, still denoted by h m ,n , between Ker(ǫ m ,n ) and Ker(ǫ n ,m ).
This shows that P is an SL 2 ( )-stable subvariety of d +4 ≃ E * .
form of the generators.-The elements eu , a 0 , a 1 ,. . ., a d are characterized by the fact that Trunc c (eu ) = eu 0 and Trunc c (a j ) = a j ,0 .Recall from[BoRo,   §3.3 and  §4.1] that Proof.-First, note that the Poisson bracket on Z c is in fact the restriction of a Poisson bracket {, } : H c × Z c −→ H c .This Poisson bracket satisfies the following property: if If d = 4, then Lie 0 ( c ) ≃ sl 3 ( ).We now determine Lie 0 ( c ) in the remaining cases: Proposition 2.12.-If d 5, then Lie 0 ( c ) = sl 2 ( ) ⊕ S d , where S d is a commutative ideal of Lie 0 ( c ) of dimension d + 1 on which sl 2 ( ) acts irreducibly (i.e. S d ≃ Sym d ( 2 ) as an sl 2 ( )-module).