Trace of the twisted Heisenberg Category
Can Ozan Oğuz Michael Reeks
















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Department of Mathematics
 University of Southern California
 Los Angeles, CA
coguz@usc.edu



Department of Mathematics
 University of Virginia 
 Charlottesville, VA
mar3nf@virginia.edu


















Trace of the Twisted Heisenberg Category
    Michael Reeks
    December 30, 2023
========================================


	We show that the trace decategorification, or zeroth Hochschild homology, of the twisted Heisenberg category defined by Cautis and Sussan is isomorphic to a quotient of W^-, a subalgebra of W_1+∞ defined by Kac, Wang, and Yan. Our result is a twisted analogue of that by Cautis, Lauda, Licata, and Sussan relating W_1+∞ and the trace decategorification of the Heisenberg category.








§ INTRODUCTION



Categorification is the process of enriching an algebraic object by increasing its categorical dimension by one, e.g. passing from a set to a category or from a 1-category to a 2-category. The original object can be recovered through the inverse process of decategorification. The most commonly used decategorification functor is the split Grothendieck group K_0, but it is natural to ask whether alternative decategorification functors may give additional insight into the categorified object. One such alternative, advocated in <cit.>, is the trace decategorification, which often encodes more information than K_0. 

The trace, or zeroth Hochschild homology, of a -linear additive category 𝒞 is the -vector space given by 

    (𝒞) := (⊕_x∈Ob(𝒞)_𝒞(X))/ span_{fg-gf},

where f and g run through all pairs of morphisms f:x→ y and g:y→ x with x,y ∈Ob(𝒞). If a -linear category 𝒞 carries a monoidal structure, then span{fg-gf} is an ideal, and (𝒞) becomes an algebra where multiplication in the trace is induced from tensor product of 𝒞.  The trace has the advantage that it is, unlike K_0, invariant under passage to the Karoubi envelope, cf. <cit.>. Since passing to the Karoubi envelope often prevents one from working with diagrams, trace seems to be a more suitable option to decategorify diagrammatic categories.  

The traces of several interesting categories have been computed. in <cit.> and <cit.>, the trace of any categorified type ADE quantum group is shown to be isomorphic to a current algebra. In <cit.>, traces of quiver Hecke algebras are studied. In <cit.>, the trace of the Hecke category is shown to be a semidirect product of the Weyl group and a polynomial algebra. A unifying approach to Heisenberg categorifications was given in <cit.> via Frobenius algebras; in <cit.>, the degree zero part of the trace of these categories are computed.  

The trace (𝒞) is closely related to the K_0(𝒞) through the Chern character map 
    h_𝒞: K_0(𝒞) →(𝒞)

which sends the isomorphism class of an object to the class of its identity morphism in the trace.
Interestingly, the map h_𝒞 is usually injective, but is often not surjective. Thus, the trace often contains additional structure which has no analogue in the Grothendieck group.

One interesting example in which h_𝒞 fails to be surjective is given by the Heisenberg category ℋ defined in <cit.>. It is a ℂ-linear additive monoidal category. Therefore (ℋ) carries an algebra structure. There is an injective algebra homomorphism from the Heisenberg algebra 𝔥 to K_0(ℋ) (they are conjecturally isomorphic).  In <cit.>, (ℋ) is shown to be isomorphic to a quotient of W_1+∞, a filtered algebra which is important in conformal field theory. In particular, it properly contains 𝔥 in filtration degree zero. Hence (ℋ) likely contains more information than K_0(ℋ). This fits into a larger framework, studied in <cit.>, involving the elliptic Hall algebra.

 We study a twisted version of Khovanov's Heisenberg category. The twisted Heisenberg algebra 𝔥_tw is a unital associative algebra generated by h_m/2, m∈ 2ℤ+1, subject to the relations 
    [h_n/2, h_m/2] = n/2δ_n,-m.
 In <cit.>, a twisted version of the Heisenberg category, denoted ℋ_tw, is introduced. It is also a ℂ-linear additive monoidal category, with an additional ℤ/2ℤ-grading. It is proved that K_0(ℋ_tw) contains 𝔥_tw (again, they are conjecturally isomorphic).

The goal of this paper is to study the trace , and determine additional structure analogous to that in the untwisted version. We show that the even part of  is isomorphic as an algebra to a quotient of a subalgebra of W_1+∞ that we will denote by W^-. We give explicit descriptions of W_1+∞ and W^- in Section <ref>. This confirms the expectation in <cit.> that there should be a relationship between ℋ_tw and one of two subalgebras of W_1+∞ defined in <cit.>.

  There is an algebra isomorphism 
    _0⟶  W^-/⟨ w_0,0, C-1⟩.
 

Even though the isomorphism between K_0(ℋ_tw) and the twisted Heisenberg algebra 𝔥_tw is still conjectural, we are able to completely characterize .

To prove Theorem <ref>, we first compute sets of algebra generators and relations for both W^- and , adapting arguments used in <cit.> to accommodate the new supercommutative elements arising from the twisting (cf. Section <ref>). We then study actions of each algebra on its canonical level one Fock space representation. These Fock space representations are isomorphic, and so induce a linear map Φ:→ W^-. We prove that Φ is an algebra homomorphism by studying the actions of  both W^- and  on their Heisenberg subalgebras. Finally, we check that the actions of the generators are identified under Φ, and deduce that Φ is an algebra isomorphism.

An important tool in studying the connection between these algebras is the relationship between  and the degenerate affine Hecke-Clifford algebra _n of type A_n-1. The trace of _n as a vector space was computed by the second author in <cit.>. The algebra  admits a triangular decomposition, where (_n) is identified with the upper (respectively lower) half. This identification simplifies some of the computations and the calculation of the graded rank of .

The structure of the paper is as follows. In Section 2, we describe the W-algebra W^- of interest, describe its gradings and a set of generators, and study its Fock space representation. In Section 3, we describe trace decategorification in more detail and present the twisted Heisenberg category studied in <cit.>, as well as its gradings. We also identify a copy of the degenerate affine Hecke-Clifford algebra within the trace. In Section 4, we study a subalgebra of  consisting of circular diagrams called bubbles, and describe how they interact with other elements of the trace. Section 5 contains a number of calculations of diagrammatic relations in the trace that are useful for computing a generating set of . Finally, in Section 6, we describe a triangular decomposition of the trace, and then establish a generating set. This allows us to prove the desired isomorphism by using the action of each algebra on its Fock space.

Acknowledgements
The authors thank Andrea Appel, Victor Kleen, Aaron Lauda,  Joshua Sussan, and Weiqiang Wang for helpful discussions and advice concerning the paper; Maxwell Siegel for his notation suggestions; and Joshua Sussan for suggesting this project.  The first author was partially supported by the NSF grant DMS-1255334. The second author was partially supported by a GRA fellowship in Weiqiang Wang's NSF grant and by a GAANN fellowship.






§ W-ALGEBRA

In this section, we will recall the W-algebra we are interested in, its structure as a ℤ-graded and ℕ-filtered algebra, and one of its subalgebras – the twisted Heisenberg algebra – as well as their Fock space representations.





 §.§ Twisted Heisenberg algebra 𝔥_tw

We recall the definition of the twisted Heisenberg algebra. The twisted Heisenberg algebra 𝔥_tw is a unital associative algebra generated by h_n for n∈ℤ+1/2 subject to the relation that [h_n,h_m]=nδ_n,-m.

  






 §.§ W-algebra W^-


Let 𝒟 denote the Lie algebra of differential operators on the circle. The central extension 𝒟̂  of 𝒟 is described in <cit.>.  It is generated by  by C and by w_k,l=t^kD^l for l∈ℤ and k∈ℤ_≥ 0 where t is a variable over ℂ, and D=td/dt, subject to relations that C and w_0,0 are central, and:

    [t^rf(D),t^sg(D)]=t^r+s(f(D+s)g(D)-f(D)g(D+r))+ψ(t^rf(D),t^sg(D))C,

where

    ψ(t^rf(D),t^sg(D))= ∑_-r≤ j≤ -1f(j)g(j+r)     r=-s≥ 0 
    
          0     r+s≠ 0

for f,g polynomials.

The W-algebra W_1+∞ is the universal envelopping algebra of 𝒟̂. It is shown in <cit.> that trace of Khovanov's Heisenberg category is isomorphic to a quotient of W_1+∞. 



In this paper, we are interested in the universal enveloping algebra of a central extension of a Lie subalgebra of 𝒟 fixed by a degree preserving anti-involution. Define the map:


    σ:   𝒟   ⟶𝒟
       1   ↦σ(1) = -1
       t   ↦σ(t)=-t
       D   ↦σ(D)=-D.


This is a degree preserving anti-involution of 𝒟, and the Lie subalgebra fixed by -σ is 
    𝒟^-:={a∈𝒟|σ(a)=-a}.
 Let 𝒟̂^- be a central extension of 𝒟^- where the 2-cocycle is the restriction of the 2-cocycle ψ given above. Therefore 𝒟̂^- is a Lie subalgebra of 𝒟̂.

  More explicitly, 𝒟̂^- is the Lie algebra over the vector space spanned by {C}∪{ t^2k-1 g(D+(2k-1)/2);  g even}∪{ t^2k f(D+k) ; f odd} where k∈ℤ and even and odd refer to even and odd polynomial functions. Its Lie bracket is given by Equation (<ref>).


Denote by W^- the universal enveloping algebra of 𝒟̂^-. Our main result relates the trace of twisted Heisenberg category to a quotient of W^-.


    [scale=1.1]
    [font=] at (-0.5,1.5) 𝒟;
    [->] (0,1.5) to (3,1.5);
    [font=] at (3.4,1.55) 𝒟;
     at (1.4,1.7) central extension;
    [->] (3.9,1.5) to (6.7,1.5);
    [font=] at (7.5,1.5) W_1+∞;
     at (5.3,1.7) enveloping algebra;
    
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    [text width=1cm] at (-1,.8) fixed by -σ;
    
    [font=] at (-0.4,0.05) 𝒟^-;
    [->] (0,0) to (3,0);
    [font=] at (3.4,0.05) 𝒟^-;
     at (1.4,0.2) central extension;
    [->] (3.9,0) to (6.7,0);
    [font=] at (7.5,0) W^-;
     at (5.3,.2) enveloping algebra;



Note that not all w_k,ℓ are contained in W^-.


When k-ℓ is an even integer, w_k,ℓ∉W^-. Moreover, the difference k-ℓ being odd is not sufficient. For example, t^2D=w_2,1∉W^- since an element starting with t^2 should be followed by f(D+1) where f is an odd polynomial function. Hence t^2D=w_2,1∉W^- but t^2(D+1)=t^2D+t^2=w_2,1+  w_2,0∈ W^- (and, indeed, σ(t^2 (D+1)) = t^2(-D-1) = -  t^2(D+1)).






 §.§ Gradings on W^-



There is a natural ℤ^≥ 0 filtration of W^- called the differential filtration with w_k, ℓ in degree ℓ; denote this filtration by |·|_dot. It is convenient to define an additional filtration: the difference filtration, where w_k, ℓ is in degree ℓ-k, denoted |·|_diff. That this is a filtration follows from the fact that W^- also carries a filtration with w_k, ℓ in degree k.

These filtrations are compatible, so we have a (ℤ×ℤ^≥ 0)-filtration with with an element f=t^j g(D-j/2) ∈ W^- in bidegree ≤(|f|_diff, |f|_dot) = (deg(g)-j, deg(g)), where deg(g) is the polynomial degree of g(w) ∈ℂ[w]. Define the following subalgebras of W^-: 
    W^-,> = ℂ⟨ t^j g(D-j/2) | (g)-j≥ 1⟩;
   
    W^-,< = ℂ⟨ t^j g(D-j/2) | (g)-j≤  1⟩;
  
    W^-,0 = ℂ⟨ g(D) |(g)  odd⟩.

Let W^-,ω[≤ r, ≤ k] denote the set of elements in difference degree ≤ r and differential degree ≤ k, with ω∈{>,<,0}.

Denote by W^- the associated graded object with respect to this filtration. Hence W^- is (ℤ×ℤ^≥ 0)-graded with |w_k,ℓ| = (ℓ-k, ℓ). For ω∈{>,<,0}, define a generating series for the graded dimension of (W^-)^ω by 
    P_ (𝒲^-)^ω(t,q) = ∑_r∈ℤ∑_k∈ℤ, k≥ 0 (W^-)^ω[r,k] t^r q^k.
 

 The graded dimensions of (W^-)^> and (W^-)^< are given by:


    P_^> = ∏_r≥ 0∏_k>01/1-t^2r+1 q^k;


    P_^< = ∏_r≤ 0∏_k>01/1-t^2r-1 q^k.



The algebra W^- is generated by elements of the form t^j g(D-j/2), where (g)-j is odd. Hence ^> is freely generated by elements w_k, ℓ with k-ℓ odd; such elements have bidgegree (k-ℓ, ℓ). The proposition follows. 

Let W^-_r,s denote the rank r, differential filtration s part of W^-. It is easy to see that the differential filtration zero part of W^-, namely ⋃_r∈ℤ W^-_r,0, is spanned as a vector space by {C}∪{t^2n+1}_n∈ℤ. As an algebra, we have that

    [t^2n+1,t^2m+1]=(2n+1)δ_n,-m

  Hence we have an isomorphism between the differential filtration zero part of W^- and the twisted Heisenberg algebra 𝔥_tw given by:
  
    ϕ:   𝔥_tw   ⟶⋃_r∈ℤ W^-_r,0
       h_2n+1/2   ↦1/√(2) t^2n+1

where n∈ℤ.





 §.§ Generators of the algebra W^-


The following lemma describes a generating set for W^- as an algebra.

  The algebra W^-/⟨ w_0,0, C⟩ is generated by w_1,0, w_0,3, and w_± 2,1± w_± 2,0. 
 Let t^k g(D-k/2) be an arbitrary element of W^-. Without loss of generality, we may assume g is a monic monomial of the form g(w) = w^ℓ with ℓ-k odd, since lower terms in g are just monomials of this form with lower degree, and thus can be generated separately. Therefore, we have 
    t^k g(D-k/2) = ∑_i=0^ℓℓi(-1)^ℓ - i (k/2)^ℓ - i t^k D^i.
 The leading term of this element with respect to differential degree is t^k D^ℓ. We will generate the leading term first, and address lower terms afterwards. There are two cases, depending on the parities of k and ℓ.

First, suppose that k=2n is even and ℓ = 2m+1 is odd (recall that k and ℓ must have opposite parity in W^-). Hence, we must generate w_± 2n, 2m+1. The following calculations are easy, using Formula <ref>:

    [w_-2,1 - w_-2,0, w_1,0] = w_-1,0,


    [w_1,0, w_0,3] = -3(w_1,2+ w_1,1)-w_1,0,


    [w_1,2b,w_0,3] = -3w_1, 2b+2 + O(w_1,2b+1),


where O(ω) refers to terms with lower differential degree than ω. Hence, starting with w_1,2 - w_1,1, we can use the Equation (<ref>) above to generate w_1,2b for any b. Now we have: 



    [w_± 2a, 1, w_1,0] = w_± 2a+1,0,


    [w_± 2a+1, 0, w_1,2 - w_1,1] = -(4a+2)w_2a+2,1 - (2a+1)(2a+2)w_2a+2,0.


Thus, starting from w_2,1 + w_2,0, we can generate w_2a,1 for any a. Finally, we have:


    [w_-1,0, w_1,2b] = ∑_i=0^2b-12bi (-1)^2b-i+1 w_0,i = w_0,2b-1 + O(w_0,2b-2),


    [w_± 2a, 1, w_0,2b-1]    = -∑_i=0^2b-22b-1i (± 1)^2b-i(2)^2b-2-i t^2aD^i+1
       = w_2a,2b-1 + O(w_2a,2b-2).


So, we can generate a polynomial with leading term w_± 2n, 2m+1. 
 

Next, suppose that k = 2n+1 is odd and positive and ℓ = 2m is even. Using Formula (<ref>), we have:


    [w_2a+1,0, w_0,2b+1]    =  t^2a+1∑_i=0^2b2b+1i (2a+1)^2b+1 -i D^i 
       = w_2a+1, 2b + O(w_2a+1, 2b-1).


Now Equations (<ref>) and (<ref>) give that we can generate w_2a+1,0 and w_0,2b+1. Hence we can generate a polynomial with leading term w_2a+1,2b.

Finally, assume that k=-(2n+1) is odd and n=2m is even. Using Formula (<ref>), we have:


    [w_-2a,1, w_1,0] = w_1-2a,0.


By Equation (<ref>), we can therefore generate w_-(2a+1),0 for any a. Next, note that:


    [w_-1,0, w_1,2b] = -∑_i=0^2b-12b-1i (-1)^2b-1-i D^i = w_0,2b-1 + O(w_0,2b-2).


By Equation (<ref>), we can generator w_0, 2b+1 for any b. Finally, we have


    [w_-(2a+1),0, w_0,2b-1]    = t^-(2a+1)∑_i=0^2n-22n-1i (-1)^2n-i (2a+1)^2n-1-i D^i
       = w_-(2a+1), 2b-2 + O(w_-(2a+1), 2b-3).


Thus, we can generate a polynomial with leading term w_-(2n+1), 2m.


It remains to adjust the lower terms of these equations so that they match those in Equation (<ref>). But note that each equation used above to generate the leading term results in lower terms which lie in different filtrations of W^-. Therefore we can adjust the coefficients of lower terms by scaling individual equations above. Since there is no dependency between these equations, we can choose constant coefficients for the generators so that our generated polynomial has the correct lower terms.






 §.§ Fock space representation of W^-


The algebra W^- inherits a Fock space representation from W_1+∞. Let W^-,≥ = W^-,0⊕ W^-,>. For parameters c,d ∈ℂ, let ℂ_c,d be a one-dimensional module for W^-,≥ on which each w_k,ℓ with (k,ℓ) ≠ (0,0) acts as zero, C acts as c, and w_0,0 acts as d. Let ℳ_c,d := _W^-,≥^W^-ℂ_c,d. This induced module possesses the following properties:

 <cit.> The W^--module ℳ_c,d has a unique irreducible quotient 𝒱_c,d, which is isomorphic as a vector space to ℂ[w_-1,0, w_-2,0,w_-3,0,… ]. 

 <cit.> The action of W^-/(C-1,w_0,0) is faithful on 𝒱_1,0. 
 This follows immmediately from the argument in <cit.> for W_1+∞ because W^- is a subalgebra. 

Proposition <ref> allows us to the compute the action of the generators on 𝒱_1,0, which we record for convenience below. 

  Let k be a positive integer. The generators of W^- act on 𝒱_1,0 as follows:  

    [w_1,0, w_-k,0]    = δ_1,k,
    
     [w_-2,1-w_-2,0, w_-k,0]    = (k+2) w_-(k+2),0 ,
    
     [w_2,1 + w_2,0, w_-k,0]    = -(k+2)w_2-k,0 ,
    
     [w_0,3, w_-k,0]    = 3k w_-k,2 - 3k^2 w_-k,1 + k^3 w_-k,0.


  






§ TWISTED HEISENBERG CATEGORY


We will now describe the main object of interest in the paper, the twisted Heisenberg category ℋ_tw. After defining the category, we recall the trace decategorification functor and some of its properties. We then describe some filtrations of , identify a copy of the degenerate affine Hecke-Clifford algebra _n, and describe the trace of _n.   Finally, we identify a set of distinguished elements in  which generate the nonzero filtration degrees of the algebra.



 §.§ Definition of ℋ_tw

The twisted Heisenberg category ℋ^t is defined in <cit.> as the Karoubi envelope of a ℂ-linear ℤ/2ℤ-graded additive monoidal category, whose moprhisms are described diagrammatically. There is an injective algebra homomorpshim from 𝔥_tw to the split Grothendieck group of the twisted Heisenberg caterogy K_0(ℋ^t). As in the untwisted case, this map is conjecturally surjective.
   
   The object of our main interest is the trace decategorification or zeroth Hochschild homology of ℋ^t. It is shown in <cit.> that trace of an additive category is isomorphic to the trace of its Karoubi envelope,. Therefore, we can work with the non-idempotent completed version of ℋ^t. We will denote it by ℋ_tw. Focusing our attention to ℋ_tw allows us to work with the diagrammatics introduced in <cit.>. 

     The category ℋ_tw is the ℂ-linear, ℤ/2ℤ-graded monoidal additive category whose objects are generated by P and Q. A generic object is a sequence of P's and Q's. The morphisms of ℋ_tw are generated by oriented planar diagrams up to boundary fixing isotopies, with generators 

    [->](-2,-0.25) to (-2,0.75);
     (-1.99,0.25) circle [radius=2pt];
    [<-](-1,-0.25) to (-1,0.75);
     (-1,0.25) circle [radius=2pt];
    [->] (0,-0.25) to (1,0.75);
    [->] (1,-0.25) to (0,0.75);
    [->] (1.5,0.5) arc (180:360:5mm);
    [<-] (3,0.5) arc (180:360:5mm);
    [->] (4.5,0) arc (180:0:5mm);
    [<-] (6,0) arc (180:0:5mm);

where the first diagram corresponds to a map P→ P{1} and the second diagram corresponds to a map Q→ Q{1}, where {1} denotes the ℤ/2ℤ-grading shift. The first two diagrams above have degree one, and the last five have degree zero. The identity morphisms of P and Q are indicated by an undecorated upward and downward pointing arrow, respectively. These generators satisfy the following relations:


    [baseline=(current bounding box.center),scale=0.75]
     [->](0,0) to [out=45,in=-45] (0,2);
     [->](0.5,0) to [out=135,in=-135] (0.5,2);
    
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    [baseline=(current bounding box.center),scale=0.75]
    [->] (1.5,0) to (1.5,2);
    [->] (2,0) to (2,2);
    [baseline=(current bounding box.center),scale=0.75]
     [->](0,0) to [out=45,in=-45] (0,2);
     [<-](0.5,0) to [out=135,in=-135] (0.5,2);
    
    =
    [baseline=(current bounding box.center),scale=0.75]
    [->] (1.5,0) to (1.5,2);
    [<-] (2,0) to (2,2);
    [baseline=(current bounding box.center),scale=0.75]
     [->](0,0) to (1.5,2);
     [->](1.5,0) to (0,2);
    [->](0.75,0) to [out=45,in=-45] (0.75,2);
    
    =
    [baseline=(current bounding box.center),scale=0.75]
     [->](0,0) to (1.5,2);
     [->](1.5,0) to (0,2);
    [->](0.75,0) to [out=135,in=225] (0.75,2);




    [baseline=(current bounding box.center),scale=0.75]
     [<-](0,0) to [out=45,in=-45] (0,2);
     [->](0.5,0) to [out=135,in=-135] (0.5,2);
    
    =
    [baseline=(current bounding box.center),scale=0.75]
    [<-] (1.5,0) to (1.5,2);
    [->] (2,0) to (2,2);
    
    -
    [baseline=(current bounding box.center),scale=0.75]
    [->] (3,2) arc (180:360:5mm);
    [<-] (3,0) arc (180:0:5mm);
    
    -
    [baseline=(current bounding box.center),scale=0.75]
    [->] (3,2) arc (180:360:5mm);
    [<-] (3,0) arc (180:0:5mm);
     (3.05,1.8) circle [radius=3pt];
     (3.95,0.2) circle [radius=3pt];




    [baseline=(current bounding box.center),scale=0.75]
    [->] (3,2) arc (-180:180:5mm);
    
    =1
    [baseline=(current bounding box.center),scale=1.5]
     (1,0) to [out=90, in=-75](0.95,0.5);
     (0.95,0.5) arc (5:355:2mm);
    [->] (0.95,0.46) to [out=75, in=270] (1,1);
    
    =0



    [baseline=(current bounding box.center),scale=0.75]
        [->] (0,0) to (1,2);
        [->] (1,0) to (0,2);
         (0.25,0.5) circle [radius=3pt];
        
        =
        [baseline=(current bounding box.center),scale=0.75]
        [->] (0,0) to (1,2);
        [->] (1,0) to (0,2);
         (0.75,1.5) circle [radius=3pt];
        [baseline=(current bounding box.center),scale=0.75]
        [->] (0,0) to (1,2);
        [->] (1,0) to (0,2);
         (0.75,0.5) circle [radius=3pt];
        
        =
        [baseline=(current bounding box.center),scale=0.75]
        [->] (0,0) to (1,2);
        [->] (1,0) to (0,2);
         (0.25,1.5) circle [radius=3pt];



    [baseline=(current bounding box.center)]
        [<-] (3,2) arc (180:0:5mm);
         (3.05,2.2) circle [radius=2.5pt];
        
        =
        -[baseline=(current bounding box.center)]
        [<-] (3,2) arc (180:0:5mm);
         (3.95,2.2) circle [radius=2.5pt];
        [baseline=(current bounding box.center)]
        [->] (3,2) arc (180:0:5mm);
         (3.05,2.2) circle [radius=2.5pt];
        
        =
        [baseline=(current bounding box.center)]
        [->] (3,2) arc (180:0:5mm);
         (3.95,2.2) circle [radius=2.5pt];



    [baseline=(current bounding box.center)]
        [->] (3,2) arc (180:360:5mm);
         (3.05,1.8) circle [radius=2.5pt];
        
        =
        [baseline=(current bounding box.center)]
        [->] (3,2) arc (180:360:5mm);
         (3.95,1.8) circle [radius=2.5pt];
        [baseline=(current bounding box.center)]
        [<-] (3,2) arc (180:360:5mm);
         (3.05,1.8) circle [radius=2.5pt];
        
        =
        -[baseline=(current bounding box.center)]
        [<-] (3,2) arc (180:360:5mm);
         (3.95,1.8) circle [radius=2.5pt];



    [baseline=(current bounding box.center),scale=0.75]
        [->] (0,0) to (0,2);
        
        =
        [baseline=(current bounding box.center),scale=0.75]
        [->] (0,0) to (0,2);
         (0,0.6) circle [radius=3pt];
         (0,1.2) circle [radius=3pt];
        [baseline=(current bounding box.center),scale=0.75]
        [<-] (0,0) to (0,2);
        
        =
        -[baseline=(current bounding box.center),scale=0.75]
        [<-] (0,0) to (0,2);
         (0,0.6) circle [radius=3pt];
         (0,1.2) circle [radius=3pt];
        [baseline=(current bounding box.center),scale=0.75]
        [->] (3,2) arc (-180:180:5mm);
         (3.95,2.2) circle [radius=3pt];
        
        =0



    [baseline=(current bounding box.center),scale=0.75]
        [->] (0,0) to (0,2);
         (0,1.6) circle [radius=3pt];
        [fill=] (0.2,1) circle [radius=0.3pt];
        [fill=] (0.4,1) circle [radius=0.3pt];
        [fill=] (0.6,1) circle [radius=0.3pt];
        [->] (0.8,0) to (0.8,2);
         (0.8,0.3) circle [radius=3pt];
        
        =-[baseline=(current bounding box.center),scale=0.75]
        [->] (0,0) to (0,2);
         (0,0.3) circle [radius=3pt];
        [fill=] (0.2,1) circle [radius=0.3pt];
        [fill=] (0.4,1) circle [radius=0.3pt];
        [fill=] (0.6,1) circle [radius=0.3pt];
        [->] (0.8,0) to (0.8,2);
         (0.8,1.6) circle [radius=3pt];
        . [baseline=(current bounding box.center),scale=0.75]
        [<-] (0,0) to (0,2);
         (0,1.6) circle [radius=3pt];
        [fill=] (0.2,1) circle [radius=0.3pt];
        [fill=] (0.4,1) circle [radius=0.3pt];
        [fill=] (0.6,1) circle [radius=0.3pt];
        [<-] (0.8,0) to (0.8,2);
         (0.8,0.3) circle [radius=3pt];
        
        =-[baseline=(current bounding box.center),scale=0.75]
        [<-] (0,0) to (0,2);
         (0,0.3) circle [radius=3pt];
        [fill=] (0.2,1) circle [radius=0.3pt];
        [fill=] (0.4,1) circle [radius=0.3pt];
        [fill=] (0.6,1) circle [radius=0.3pt];
        [<-] (0.8,0) to (0.8,2);
         (0.8,1.6) circle [radius=3pt];
        .

Also, if we let 

    [baseline=(current bounding box.center),scale=1.5]
     (1,0) to [out=90, in=85](1.05,0.5);
     (1.05,0.5) arc (-175:175:2mm);
    [->] (1.05,0.46) to [out=95, in=270] (1,1);
    
        :=
        [baseline=(current bounding box.center),scale=0.75]
        [->](0,0) to (0,2);
        [fill](0,1) circle[radius=3pt];

we get the following relations:


    [baseline=(current bounding box.center),scale=0.75]
        [->](0,0) to (0,2);
        [fill](0,0.6) circle[radius=3pt];
        (0,1.2) circle[radius=3pt];
        
        =
        -[baseline=(current bounding box.center),scale=0.75]
        [->](0,0) to (0,2);
        (0,0.6) circle[radius=3pt];
        [fill](0,1.2) circle[radius=3pt];



    [baseline=(current bounding box.center),scale=0.75]
        [->] (0,0) to (0,2);
        [fill] (0,1.6) circle [radius=3pt];
        [fill] (0.2,1) circle [radius=0.3pt];
        [fill] (0.4,1) circle [radius=0.3pt];
        [fill] (0.6,1) circle [radius=0.3pt];
        [->] (0.8,0) to (0.8,2);
        [fill] (0.8,0.3) circle [radius=3pt];
        
        =
        [baseline=(current bounding box.center),scale=0.75]
        [->] (0,0) to (0,2);
        [fill] (0,0.3) circle [radius=3pt];
        [fill] (0.2,1) circle [radius=0.3pt];
        [fill] (0.4,1) circle [radius=0.3pt];
        [fill] (0.6,1) circle [radius=0.3pt];
        [->] (0.8,0) to (0.8,2);
        [fill] (0.8,1.6) circle [radius=3pt];
        [baseline=(current bounding box.center),scale=0.75]
        [->] (0,0) to (0,2);
         (0,1.6) circle [radius=3pt];
        [fill] (0.2,1) circle [radius=0.3pt];
        [fill] (0.4,1) circle [radius=0.3pt];
        [fill] (0.6,1) circle [radius=0.3pt];
        [->] (0.8,0) to (0.8,2);
        [fill] (0.8,0.3) circle [radius=3pt];
        
        =
        [baseline=(current bounding box.center),scale=0.75]
        [->] (0,0) to (0,2);
         (0,0.3) circle [radius=3pt];
        [fill] (0.2,1) circle [radius=0.3pt];
        [fill] (0.4,1) circle [radius=0.3pt];
        [fill] (0.6,1) circle [radius=0.3pt];
        [->] (0.8,0) to (0.8,2);
        [fill] (0.8,1.6) circle [radius=3pt];



    [baseline=(current bounding box.center),scale=0.75]
        [->](0,0) to (1,2);
        [->](1,0) to (0,2);
        [fill](0.25,0.5) circle[radius=3pt];
    
    =
    [baseline=(current bounding box.center),scale=0.75]
        [->](0,0) to (1,2);
        [->](1,0) to (0,2);
        [fill](0.75,1.5) circle[radius=3pt];
    
    +[baseline=(current bounding box.center),scale=0.75]
        [->](0,0) to (0,2);
        [->](0.5,0) to (0.5,2);
    +[baseline=(current bounding box.center),scale=0.75]
        [->](0,0) to (0,2);
        [->](0.5,0) to (0.5,2);
        (0,1.2) circle[radius=3pt];
        (0.5,0.6) circle[radius=3pt];




    [baseline=(current bounding box.center),scale=0.75]
        [->](0,0) to (1,2);
        [->](1,0) to (0,2);
        [fill](0.25,1.5) circle[radius=3pt];
    
    =
    [baseline=(current bounding box.center),scale=0.75]
        [->](0,0) to (1,2);
        [->](1,0) to (0,2);
        [fill](0.75,0.5) circle[radius=3pt];
    
    +
    [baseline=(current bounding box.center),scale=0.75]
        [->](0,0) to (0,2);
        [->](0.5,0) to (0.5,2);
    
    -
    [baseline=(current bounding box.center),scale=0.75]
        [->](0,0) to (0,2);
        [->](0.5,0) to (0.5,2);
        (0,1.2) circle[radius=3pt];
        (0.5,0.6) circle[radius=3pt];





    [baseline=(current bounding box.center),scale=0.75]
        [<-](0,0) to (1,2);
        [<-](1,0) to (0,2);
        [fill](0.25,0.5) circle[radius=3pt];
    
    =
    [baseline=(current bounding box.center),scale=0.75]
        [<-](0,0) to (1,2);
        [<-](1,0) to (0,2);
        [fill](0.75,1.5) circle[radius=3pt];
    
    +[baseline=(current bounding box.center),scale=0.75]
        [<-](0,0) to (0,2);
        [<-](0.5,0) to (0.5,2);
    -[baseline=(current bounding box.center),scale=0.75]
        [<-](0,0) to (0,2);
        [<-](0.5,0) to (0.5,2);
        (0,1.2) circle[radius=3pt];
        (0.5,0.6) circle[radius=3pt];



    [baseline=(current bounding box.center),scale=0.75]
        [<-](0,0) to (1,2);
        [<-](1,0) to (0,2);
        [fill](0.25,1.5) circle[radius=3pt];
    
    =
    [baseline=(current bounding box.center),scale=0.75]
        [<-](0,0) to (1,2);
        [<-](1,0) to (0,2);
        [fill](0.75,0.5) circle[radius=3pt];
    
    +[baseline=(current bounding box.center),scale=0.75]
        [<-](0,0) to (0,2);
        [<-](0.5,0) to (0.5,2);
    +[baseline=(current bounding box.center),scale=0.75]
        [<-](0,0) to (0,2);
        [<-](0.5,0) to (0.5,2);
        (0,1.2) circle[radius=3pt];
        (0.5,0.6) circle[radius=3pt];
.



If x and y are morphisms, the diagram corresponding to x⊗ y is obtained by placing the diagram of y to the right of the diagram of x. Since the relative positions of the hollow dots are important, we will work with the convention that the hollow dots in the diagram of y will be placed below the height of hollow dots in the diagram of x.




 §.§ Trace decategorification

In <cit.>, the trace or zeroth Hochschild homology of a -linear additive category 𝒞 is proposed as an alternative decategorification functor. Here we will recall its definition, and point out one subtlety occuring in our case due to the supercommutative nature of hollow dots and solid dots.

Let 𝒞 be a -linear additive category. Then its trace decategorification, denoted (𝒞), is defined as follows:


    (𝒞)≃( ⊕_x∈𝒪b(𝒞)(x) ) /ℐ,

where ℐ is the ideal generated by span_{fg-gf} for all f:x→ y and g:y→ x , x,y∈𝒪b(𝒞). Note that here we quotient out by an ideal, so (𝒞) has an algebra structure.

Trace decategorification has a nice diagrammatic interpretation, in which we consider our string diagrams to be drawn on an annulus instead of a plane. The annulus recaptures the trace relation fg=gf diagrammatically since we can slide f or g around the annulus to change their composition order.


    [baseline=(current bounding box.center),rounded corners]
    	[blue,dashed,fill=blue!5!white] (0.5,1) circle [radius=40pt];
    	[fill,blue,fill=white] (0.5,1) circle [radius=4pt];
    	 (0,0.3) to (0,0.5);
    	 (-0.25,0.5) rectangle (0.25,0.9);
    	 (0,0.9) to (0,1.2);
    	 (-0.25,1.2) rectangle (0.25,1.6);
    	[->] (0,1.6) to (0,1.7);
    	 (0,1.7) arc(180:0:5mm);
    	 (1,1.7) to (1,0.3);
    	 (1,0.3) arc(360:180:5mm);
    	 at (0,0.7)f;
    	 at (0,1.4)g;
    	
    	=
    	[baseline=(current bounding box.center),rounded corners]
    	[blue,dashed,fill=blue!5!white] (0.5,1) circle [radius=40pt];
    	[fill,blue,fill=white] (0.5,1) circle [radius=4pt];
    	 (0,0.3) to (0,0.5);
    	 (-0.25,0.5) rectangle (0.25,0.9);
    	 (0,0.9) to (0,1.2);
    	 (-0.25,1.2) rectangle (0.25,1.6);
    	[->] (0,1.6) to (0,1.7);
    	 (0,1.7) arc(180:0:5mm);
    	 (1,1.7) to (1,0.3);
    	 (1,0.3) arc(360:180:5mm);
    	 at (0,0.7)g;
    	 at (0,1.4)f;


As described in Section <ref>,  has a ℤ/2ℤ-grading where [baseline=(current bounding box.center),scale=0.75]
[->](-2,-0.25) to (-2,0.75);
(-2,0.25) circle [radius=3pt];
 and 
[baseline=(current bounding box.center),scale=0.75]
[<-](-1,-0.25) to (-1,0.75);
(-1,0.25) circle [radius=3pt];
 have degree one, and other generating diagrams have degree zero. We also have supercommutativity relations (<ref>) and (<ref>) and supercyclicity relations (<ref>) and (<ref>). These relations have several interesting diagrammatic consequences.


	Working with relation (<ref>), we have the following compuation:
	
    [baseline=(current bounding box.center),rounded corners]
    		[blue,dashed,fill=blue!5!white] (0.5,1) circle [radius=40pt];
    		[fill,blue,fill=white] (0.5,1) circle [radius=4pt];
    		[->] (0,0.3) to (0,1.7);
    		 (0,1.7) arc(180:0:5mm);
    		 (1,1.7) to (1,0.3);
    		 (1,0.3) arc(360:180:5mm);
    		 (0,0.7) circle [radius=2pt];
    		 (0,1.4) circle [radius=2pt];
    		
    		=
    		[baseline=(current bounding box.center),rounded corners]
    		[blue,dashed,fill=blue!5!white] (0.5,1) circle [radius=40pt];
    		[fill,blue,fill=white] (0.5,1) circle [radius=4pt];
    		[->] (0,0.3) to (0,1.7);
    		 (0,1.7) arc(180:0:5mm);
    		 (1,1.7) to (1,0.3);
    		 (1,0.3) arc(360:180:5mm);
    		 (0,0.7) circle [radius=2pt];
    		 (0,1.4) circle [radius=2pt];
    		
    		=
    		-[baseline=(current bounding box.center),rounded corners]
    		[blue,dashed,fill=blue!5!white] (0.5,1) circle [radius=40pt];
    		[fill,blue,fill=white] (0.5,1) circle [radius=4pt];
    		[->] (0,0.3) to (0,1.7);
    		 (0,1.7) arc(180:0:5mm);
    		 (1,1.7) to (1,0.3);
    		 (1,0.3) arc(360:180:5mm);
    		 (0,0.7) circle [radius=2pt];
    		 (0,1.4) circle [radius=2pt];
    		
    		=0.

	Here the first equality is obtained by sending the solid dot around the annulus using trace relation, and the second equality is a consequence of relation (<ref>). Therefore the above diagram is equal to zero in the trace.



	To demonstrate the subtlety with supercyclicity relations (<ref>) and (<ref>), consider the following situation:
	
    [baseline=(current bounding box.center),rounded corners]
    		[blue,dashed,fill=blue!5!white] (0.5,1) circle [radius=40pt];
    		[fill,blue,fill=white] (0.5,1) circle [radius=4pt];
    		[->] (0,0.3) to (0,1.7);
    		 (0,1.7) arc(180:0:5mm);
    		 (1,1.7) to (1,0.3);
    		 (1,0.3) arc(360:180:5mm);
    		 (0,1.4) circle [radius=2pt];
    		
    		=-[baseline=(current bounding box.center),rounded corners]
    		[blue,dashed,fill=blue!5!white] (0.5,1) circle [radius=40pt];
    		[fill,blue,fill=white] (0.5,1) circle [radius=4pt];
    		[->] (0,0.3) to (0,1.7);
    		 (0,1.7) arc(180:0:5mm);
    		 (1,1.7) to (1,0.3);
    		 (1,0.3) arc(360:180:5mm);
    		 (0,0.7) circle [radius=2pt];

	If we denote [baseline=(current bounding box.center),scale=0.75]
	[->](-2,-0.25) to (-2,0.75);
	(-2,0.25) circle [radius=3pt];
	 by f, with the usual trace relation, we would get f∘ id=id∘ f. However, in this case, we gain an extra negative sign from the supercyclicity relations. So, we must replace the usual trace relation fg=gf with the supertrace relation fg=(-1)^|f||g|gf in the ideal ℐ, where |f|,|g| are the degrees of f and g with respect to the ℤ/2ℤ grading.  This example can be generalized to show that composition of an an odd morphism with a cycle of odd length is zero in the supertrace, since it will be equal to its negative when a hollow dot travels around the annulus and arrives to its original position.


We wish to restrict our study to the following subalgebra of the trace.

 The even trace of  is defined by
	
    ≃( ⊕_x∈𝒪b()_0(x) ) /ℐ_0

	where _0(x) consists of even degree endomorphisms and ℐ_0 is its ideal generated by span_ℂ{fg-gf} for all f:x→ y and g:y→ x , x,y∈𝒪b().


This is the restriction of the trace to only the even part (with respect to the ℤ/2ℤ grading induced by (c_i) = 1). The odd part of the trace is not zero (it contains, e.g., [baseline=(current bounding box.center),scale=0.75]
	[->](-2,-0.25) to (-2,0.75);
	 (-2,0.25) circle [radius=3pt];), but is not interesting from a representation theoretic viewpoint as explained above. The example of trace functions on the finite Hecke-Clifford algebra in <cit.> demonstrates the importance of the even trace.

Wan and Wang study the space of trace functions on the finite Hecke-Clifford algebra ℋ_n: linear functions ϕ: ℋ_n →ℂ such that ϕ([h,h'])=0 for all h,h'∈ℋ_n, and ϕ(h)=0 for all h∈ (ℋ_n)_1. This latter requirement encodes the information that odd elements act with zero trace on any ℤ_2-graded ℋ_n-module (because multiplication by an odd element results in a shift in degree). The space of such trace functions is clearly canonically isomorphic to the dual of the even cocenter, rather than of the full cocenter. The same observation holds for the trace of the affine Hecke-Clifford algebra, as studied in <cit.>.

We will see in Section 4 that the structure of  is largely controlled by the even trace of the degenerate affine Hecke-Clifford algebra in type A; we therefore do not lose interesting representation-theoretic information by restricting to , and greatly simplify our calculations by doing so. For instance, the ambiguity in the supercyclicity relations identified in Example 3.<ref> does not interfere with calculations in .

Since ℐ_0 is an ideal of ⊕_x∈𝒪b()_0(x), the compositions fg and gf must be even morphisms, even though individually f and g may be odd morphisms. This situation is analogous to the even cocenter of the degenerate affine Hecke-Clifford algebra studied in <cit.>, where Clifford generators c_i do no appear individually (as they are odd generators), but still have an impact on the cocenter via the relation c_i^2=-1.

Diagrammatically, the above definition means that we will have an even number of hollow dots on our diagrams. In a diagram with 2n hollow dots, sliding one around the annulus from top to the bottom will multiply the diagram by (-1)^2n-1(-1)=1 where (-1)^2n-1 is a result of changing relative height with the remaining 2n-1 hollow dots using relation (<ref>) and (-1) is the result of sliding it through a clockwise cup using relation (<ref>).



	For the sake of clarity, when working with diagrams in the even trace we will not draw them on an annulus, but will instead draw them inside square brackets, e.g.  
	[ [baseline=(current bounding box.center),scale=0.5]
		[->] (0,0) to (1,2);
		[->] (1,0) to (0,2);
		  ]. This notation refers to the equivalence class of the diagram in .


Our main theorem will relate  and W^-. In particular, we will establish that the correspondence in Table 1 gives an isomorphism between  and W^-. Recall that w_k,ℓ = t^ℓ D^k ∈ W^-. 



 
 






 §.§ Degenerate affine Hecke-Clifford algebra


 
 We recall the definition of the degenerate affine Hecke-Clifford algebra of type A_n-1, denoted _n, which was first studied in <cit.>. Let 𝒞ℓ_n be the Clifford algebra with generators c_1, …, c_n, subject to the relations: 
    c_i^2 = -1     for  1≤ i ≤ n,
 
    c_i c_j = - c_j c_i     if  i≠j.
 The symmetric group S_n has a natural action on 𝒞ℓ_n by permuting the generators.

Define the Sergeev algebra, or finite Hecke-Clifford algebra of type A_n-1, to be the semidirect product 
    𝕊 := 𝒞ℓ_n ⋊ℂ S_n
 corresponding to this action.

The degenerate affine Hecke algebra, _n, is isomorphic as a vector space to 𝕊⊗ℂ[x_1, …, x_n]. It is an associative unital algebra over ℂ[u], where u is a formal parameter usually set to 1, generated by s_1,s_2,...,s_n-1, x_1,x_2,...,x_n, and c_1,c_2,...,c_n subject to relations making ℂ[x_1, …, x_n], Cℓ_V, and ℂ S_n subalgebras, along with the additional relations: 
  

    x_i c_i    = -c_i x_i,    x_i c_j = c_j x_i    (i≠ j) ,
    σ c_i     = c_σ(i)σ   (1≤ i ≤ n, σ∈ S_n),
    
    x_i+1s_i-s_ix_i     =u(1-c_i+1c_i)  , 
    
    x_js_i     =s_ix_j   (j≠ i,i+1) .


It also has a ℤ/2ℤ grading via (s_i)=(x_i)=0 and (c_i)=1. This algebra is also called the affine Sergeev algebra, and later on we will see that it controls the endomorphisms of up strands in ℋ_tw.



 §.§ Trace of the degenerate affine Hecke-Clifford algebra


The second author computes the trace (or zeroth Hochschild homology) of the even part of _n as a vector space in <cit.>, where he gives an explicit description of a vector space basis for types A, B and D. Here we recall the result for type A.

Let I be the standard root system of type A_n-1, and let W= S_n be the Weyl group. For a partition λ = (λ_1, λ_2, …, λ_k) ⊢ n, let J_λ be the unique minimal subset of I (up to conjugation by W) such that W_J_λ contains an element of cycle type λ. Let w_λ∈ W be the element (1,…, λ_1)(λ_1 +1, …, λ_1 + λ_2)… (n-λ_n +1, …, n). Then w_λ∈ W_J_λ. 

Let V be the standard representation of 𝔰𝔩_n, with basis {x_1, …, x_n}. Denote by V^2 the vector space with basis {x_1^2, …, x_n^2}.

Finally, fix a basis {f_J_λ;i} of the vector space S((V^2)^W_J_λ)^N_W(W_J_λ), where S(U) denotes the symmetric algebra of the vector space U, and N_W(W_J_λ) denotes the normalizer of the parabolic subgroup W_J_λ in W. We have the following description of a basis for (_n)_0 in type A.

<cit.>
 The set {w_λf_J_λ;i}_λ∈𝒪𝒫_n is a basis of (_n)_0, where 𝒪𝒫_n is the set of partitions of n with all odd parts.



Let n=3. Then we have 𝒪𝒫_3={(1,1,1),(3)}. 

For λ=(1,1,1), we have w_λ = 1 and J_λ = ∅, since  W_J_λ = {1}. Thus N_W(W_J_λ) = S_3. So, we choose a basis {f_J_λ;i} of the vector space S((V^2)^W_J)^N_W(W_J) = ℂ[x_1^2 , x_2^2, x_3^2]^S_3, i.e. the symmetric polynomials in 3 variables. We can take {f_J_λ;i} = {s_ν}, the Schur polynomials in 3 variables.

For λ=(3), we have w_λ=(123), a 3-cycle in S_3. Thus J_λ = I, W_J_λ=W and N_W(W_J_λ)=W. Therefore f_J_λ,i is a basis of ℂ[x_1^2+x_2^2+x_3^2], polynomials in the variable (x_1^2+x_2^2+x_3^2) (in this case, the N_W(W_J_λ)-invariance is superfluous).

 
Therefore a basis of Tr(_3)_0 is given by { s_ν}∪{(123)(x_1^2+ x_2^2+ x_3^2)^n}_n∈ℕ where {s_ν} are the Schur polynomials in 3 variables.

Note that this bases does not contain any classes indexed by partitions with even parts. Correspondingly, we will see that degree zero diagrams in  containing even cycles are zero.




 §.§ Distinguished elements h_n

Define the elements: 

    n(x_1^j_1⋯ x_n^j_n)  (c_1^ϵ_1⋯ c_n^ϵ_n)   := [  [scale=0.8]
     
      [->] (3.2,0) .. controls (3.2,1.25) and (0,.25) .. (0,2)
         node[pos=0.85, shape=coordinate](X);
      [->] (0,0) .. controls (0,1) and (.8,.8) .. (.8,2);
      [->] (.8,0) .. controls (.8,1) and (1.6,.8) .. (1.6,2);
      [->] (2.4,0) .. controls (2.4,1) and (3.2,.8) .. (3.2,2);
       at (1.6,.35) …;
       at (2.4,1.65) …;
      
        (.05,1.6) circle (2pt);
        (.78,1.6) circle (2pt);  
        (1.58,1.6) circle (2pt);
        (3.18,1.6) circle (2pt);
    
       (.35,1.2) circle (2.5pt);
        (.62,1.2) circle (2.5pt); 
        (1.425,1.2) circle (2.5pt);
        (3.025,1.2) circle (2.5pt);
      
       at (-0.18,1.7) j_1;
       at (.55,1.7) j_2;
       at (1.35,1.7) j_3;
        at (3.45,1.7) j_n;
    
      at  (.051,1.25) ϵ_1;
       at  (.99,1.25) ϵ_2; 
       at  (1.76,1.25) ϵ_3;
       at  (3.37,1.25) ϵ_n;
     ],
    
    -n(x_1^j_1⋯ x_n^j_n)  (c_1^ϵ_1⋯ c_n^ϵ_n)   := [  [scale=0.8]
     
      [<-] (3.2,0) .. controls (3.2,1.25) and (0,.25) .. (0,2)
         node[pos=0.85, shape=coordinate](X);
      [<-] (0,0) .. controls (0,1) and (.8,.8) .. (.8,2);
      [<-] (.8,0) .. controls (.8,1) and (1.6,.8) .. (1.6,2);
      [<-] (2.4,0) .. controls (2.4,1) and (3.2,.8) .. (3.2,2);
       at (1.6,.35) …;
       at (2.4,1.65) …;
      
        (.05,1.6) circle (2pt);
        (.78,1.6) circle (2pt);  
        (1.58,1.6) circle (2pt);
        (3.18,1.6) circle (2pt);
    
       (.35,1.2) circle (2.5pt);
        (.62,1.2) circle (2.5pt); 
        (1.425,1.2) circle (2.5pt);
        (3.025,1.2) circle (2.5pt);
      
       at (-0.18,1.7) j_1;
       at (.55,1.7) j_2;
       at (1.35,1.7) j_3;
        at (3.45,1.7) j_n;
    
      at  (.051,1.25) ϵ_1;
       at  (.99,1.25) ϵ_2; 
       at  (1.76,1.25) ϵ_3;
       at  (3.37,1.25) ϵ_n;
     ],


where ϵ_i ∈{0,1}. In both of these elements, we consider the hollow dots to be descending in height from left to right, so that the dot labeled ϵ_1 is the highest.



These elements are analogues to those denoted h_± n⊗ (x_1^j_1⋯ x_n^j_n) in <cit.>. 


Additionally, set 
    n∑ x_i^j_i = ∑nx_i^j_i.


  For n≥ 1 and 1≤ i ≤ n-1 we have  
  * ± nx_i  =± nx_i+1±± i±  (n-i). 

  * ± n x_i   c_j = - ± nx_i+1c_j+1. 
 
 Part (1) is just <cit.>, except our solid dot sliding relation through crossing involves an extra term with hollow bubbles. But cycles with single hollow dot are zero since sending the hollow dot around the annulus gives us the same diagram with a negative sign. For the above calculations, our n-cycles split into smaller cycles with single hollow dot at least on one of them. The proof of part 2 depends on the relative position of i and j, but is a straightforward computation. 

Let w∈ S_n, and define the elements:

    f_w; j_1, …, j_n; ϵ_1, …, ϵ_n= 
    [very thick][->] (-.55,0) – (-.55,1.5);
        [very thick][->] (.55,0) – (.55,1.5);
        [fill=white!20,] (-.8,.4) rectangle (.8,.8);
         () at (0,.55) w;
         () at (0,1.25) ⋯;
         () at (0,.25) ⋯;
      (-.55,1.25) circle (2pt);
      (.55,1.25) circle (2pt);
      (-.55,.9) circle (2pt);
      (.55,.9) circle (2pt);
     (-.75,.9) node ϵ_1;
     (.8,.9) node ϵ_n;
     (-.75,1.25) node j_1;
     (.8,1.25) node j_n;

and

    f_w; j_1, …, j_n; ϵ_1, …, ϵ_n= 
    [very thick][<-] (-.55,0) – (-.55,1.5);
        [very thick][<-] (.55,0) – (.55,1.5);
        [fill=white!20,] (-.8,.4) rectangle (.8,.8);
         () at (0,.55) w;
         () at (0,1.25) ⋯;
         () at (0,.25) ⋯;
      (-.55,1.25) circle (2pt);
      (.55,1.25) circle (2pt);
      (-.55,.9) circle (2pt);
      (.55,.9) circle (2pt);
     (-.75,.9) node ϵ_1;
     (.8,.9) node ϵ_n;
     (-.75,1.25) node j_1;
     (.8,1.25) node j_n;
    .



 Let w∈ S_n and (n_1, …, n_r) be a composition of n. Then 
    [f_± w; j_1, …, j_n; ϵ_1, …, ϵ_n] = ∑ d_n_1, …, n_rn_1  p_n_1 c_n_1…n_r p_n_r  c_n_r
 for constants d_n_1, …, n_r∈ℂ, polynomials p_n_i in i variables, and elements c_n_i consisting of at most i Clifford generators (e.g. c_n_3 = { c_1^ϵ_1 c_2^ϵ_2 c_3^ϵ_3 | ϵ_i ∈{0,1}}). 

We proceed by induction on ∑ϵ_i. The base case is ∑ϵ_i = 0; then 
    [f_± w; j_1, …, j_n; ϵ_1, …, ϵ_n] = [f_± w; j_1, …, j_n]
 and we apply <cit.>. 

Now assume the statement is true for ∑ϵ_i = k for all k<m≤ n. Take (ϵ_1, …, ϵ_n) so that ∑ϵ_i = m. Choose g∈ S_n such that gwg^-1 = w_λ, where λ is the cycle type of w (so gwg^-1 = (s_1 … s_n_1 -1)… (s_n_1 + … + n_r-1… s_n_1 +… +n_r -1)).  Let p= x_1^j_1… x_n^j_n and c= c_1^ϵ_1… c_n^ϵ_n.  Then we have 
    f_± w; j_1, …, j_n; ϵ_1, …, ϵ_n = pcw = (-1)^ϵ cpw
 where 
    ϵ = ∑_ϵ_i = 1 j_i.
 

Thus conjugating by g gives that 
    gpcwg^-1    = (-1)^ϵ gcpwg^-1
       =(-1)^ϵ (g.c) gpwg^-1
       = (-1)^ϵ[ (g.c)(g.p)gwg^-1 + (g.c) p_L wg^-1],
 where p_L is a polynomial of degree less than j_1 + … + j_n. Note that gwg^-1 is a product of cycles, so the first term in the above expression has the correct form. In the second term, we have {i | ϵ_g(i) =1}≤ m (strict inequality can occur if g has fixed points). If {i | ϵ_g(i) =1} < m, we are done by induction, so assume that we have equality.

Now repeat the process on the second term, choosing a g' ∈ S_n such that g'(wg^-1) (g')^-1 is a product of cycles, and conjugating (g.c) p_L wg^-1. Each application of this process results in one term in which the symmetric group element is a product of cycles (which has the desired form), and one term with the degree of the polynomial part strictly lesser and the degree of the Clifford part weakly lesser.

 If the degree of the Clifford part ever strictly decreases, we are done. If not, the conjugation will eventually reduce the degree of the polynomial part to 0, so we have an element of the form c' σ, c' ∈ Cℓ_n and σ∈ S_n. Choose a g”∈ S_n such that g”σ (g”)^-1 is a product of cycles; then 
    g” c σ (g”)^-1 =  (g”c) g”σ (g”)^-1.
 This now has the desired form.


  Let w∈ S_n and (n_1, …, n_r) a composition of n. Then 
    [f_± w; j_1, …, j_n; ϵ_1, …, ϵ_n] = ∑ d_n_1, …, n_r± n_1x_1^ℓ_1   c_1^k_1…± n_rx_1^ℓ_r c_1^k_r
 where d_n_1, …, n_r∈ℂ and ℓ_1, …, ℓ_r, k_1, …, k_r ∈ℕ. 
 This follows immediately from the preceding lemmas.  

Proposition <ref> allows us to write any element in  or  as a linear combination of the elements n. We will therefore direct our attention to these elements in future computations.



 §.§ Gradings in 


The next lemma follows from diagrammatic computations in the next section. We record it here for convenience of terminology.
 The algebra  is ℤ-filtered where (nx_1^2a) ≤ n for any a≥ 0. 


This is called the rank filtration. Denote by  (resp. ) the subalgebra of  generated by nx_1^2a, n≥ 1  (resp. n≤ 1).

 The algebra  is ℤ^≥ 0-filtered where (n x_1^2a)≤ a for any a ≥ 0. 
 Dots can slide through crossings modulo a correction term containing fewer dots. 

This is called the dot filtration, and corresponds to the differential filtration (given by (w_ℓ, k) = k) in W^-. 

These filtrations are compatible, so  is (ℤ×ℤ^≥ 0)-filtered with n x_1^2a in bidegree (n, a). For ω∈{>,<,0} denote the associated graded object by . Define a generating series for the graded dimension of  by 
    P_(t,q) = ∑_r∈ℤ∑_k∈ℤ, k≥ 0 [r,k] t^r q^k.
 


The following is an easy calculation using Proposition <ref> and Proposition <ref>. They are not used in the proof of the main result, but we record them here for convenience.
  The graded dimensions of  and  are given by: 
	
    P_^> = ∏_r≥ 0∏_k>01/1-t^2r+1 q^k;

	
    P_^< = ∏_r≤ 0∏_k>01/1-t^2r-1 q^k.




Note that the rank grading and dot gradings are shifted by 1 for clockwise bubbles (so d_2 is in bidegree (1,2) and d_4 is in bidegree (1,3)). This is a consequence of the decomposition formula in Lemma <ref>. 





§ BUBBLES


We investigate the endomorphisms of 1 in , known as bubbles. We prove that all bubbles can be written in terms of clockwise bubbles, and deduce formulas for moving bubbles past strands in the trace.



 §.§ Definition and basic properties

 
 Elements of _ℋ_tw(1) are ℂ-linear combinations of possibly intersecting or nested closed diagrams, which may have dots. We can always separate the nested pieces, and resolve any crossing that occur between different closed diagrams using the defining relations and end up with non intersecting, not nested closed oriented diagrams. Each one can be deformed into an oriented circle, possibly with dots, via an isotopy. A single closed, oriented, non self intersecting diagram is called a bubble. They are the building blocks of  endomorphisms of the identity object in ℋ_tw.
 
We define 

    d̅_k,l:=[baseline=(current bounding box.center)]
        [->] (3,2) arc (-180:180:5mm);
     (3.95,2.2) circle [radius=2pt];
     (3.95,1.8) circle [radius=2pt];
     at (4.2,1.8) l;
     at (4.2,2.2) k;
        and
        d_k,l:=[baseline=(current bounding box.center)]
        [<-] (3,2) arc (-180:180:5mm);
     (3.95,2.2) circle [radius=2pt];
     (3.95,1.8) circle [radius=2pt];
     at (4.2,1.8) l;
     at (4.2,2.2) k;
        for k,l∈ℤ_≥ 0.


   Given any closed diagram with any configuration of dots, it is possible to collect the hollow dots and the solid dots together, possibly after multiplying the diagram by -1, by using relation (<ref>). Solid dots move freely along caps and cups, and hollow dots may capture a negative sign while moving along caps or cups, depending on the orientation. After regrouping, we may assume that the dots are placed on the right middle side of the diagram as above.
   
   Moreover, using the left two equations in relation (<ref>), we can erase a pair of hollow dots, possibly by changing the sign of the diagram.
   
   Therefore the set {d_k,l,d̅_k,l|k∈ℤ_≥0, l∈{0,1}} is a spanning set for _ℋ_tw(1).
   

In our defining relations, we have that 


    d̅_0,0=[baseline=(current bounding box.center)]
    [->] (3,2) arc (-180:180:5mm);
    
    
    
    =1
    andd̅_0,1=[baseline=(current bounding box.center)]
    [->] (3,2) arc (-180:180:5mm);
     (4,2) circle [radius=2pt];
    
    
    
    =0.

Further, we have the following. 

We have that d̅_k,1=0 and d_k,1=0 for all non-negative integers k.



An example computation shows that

    d̅_1,1=[baseline=(current bounding box.center)]
        [->] (3,2) arc (-180:180:5mm);
     (3.95,2.2) circle [radius=2pt];
     (3.95,1.8) circle [radius=2pt];
        
        =
        -[baseline=(current bounding box.center)]
        [->] (3,2) arc (-180:180:5mm);
     (3.95,2.2) circle [radius=2pt];
     (3.95,1.8) circle [radius=2pt];
    
       =
       -[baseline=(current bounding box.center)]
       [->] (3,2) arc (-180:180:5mm);
     (3.75,1.6) circle [radius=2pt];
     (3.95,1.8) circle [radius=2pt];
       
       =
       -d̅_1,1=0,

where in the second equality, negative sign comes from relation (11), and the third equality comes from sliding the solid dot around.

More generally, if we have k solid dots where k is an even integer, then sliding the hollow dot around the circle and passing it through k solid dots multiplies the diagram by (-1)^k+1, so the diagram is zero.
If k is an odd number, sliding a solid dot around the circle and passing it through a hollow dot catches a minus sign, so these diagrams are zero as well.

These arguments do not depend on the orientation of the bubble, hence the result follows.


From now on, we will assume that the second index in d̅_k,l and d_k,l is always zero. We will omit it from our notation and write d_k instead of d_k,0, and  d̅_k instead of d̅_k,0.


We have that d_2n+1=d̅_2n+1=0 for all non-negative integers n.




Note that

    d̅_1=[baseline=(current bounding box.center)]
    [->] (3,2) arc (-180:180:5mm);
     (4,2) circle [radius=2pt];
    
    
    
    =
    [baseline=(current bounding box.center)]
    [->] (3,2) arc (-180:180:5mm);
     (4,2) circle [radius=2pt];
     (3.95,2.2) circle [radius=2pt];
     (3.95,1.8) circle [radius=2pt];
    
    
    
    =
    -[baseline=(current bounding box.center)]
    [->] (3,2) arc (-180:180:5mm);
     (4,2) circle [radius=2pt];
     (3.05,2.2) circle [radius=2pt];
     (3.95,1.8) circle [radius=2pt];
    
    
    
    =
    [baseline=(current bounding box.center)]
    [->] (3,2) arc (-180:180:5mm);
     (4,2) circle [radius=2pt];
     (3.95,2.2) circle [radius=2pt];
     (3.05,1.8) circle [radius=2pt];
    
    
    
    =
    [baseline=(current bounding box.center)]
    [->] (3,2) arc (-180:180:5mm);
     (4,2) circle [radius=2pt];
     (3.95,2.2) circle [radius=2pt];
     (3.95,1.8) circle [radius=2pt];
 
    =
    -[baseline=(current bounding box.center)]
    [->] (3,2) arc (-180:180:5mm);
     (4,2) circle [radius=2pt];
     (3.95,2.2) circle [radius=2pt];
     (3.95,1.8) circle [radius=2pt];
    
    
    
    =
    -[baseline=(current bounding box.center)]
    [->] (3,2) arc (-180:180:5mm);
     (4,2) circle [radius=2pt];
    
    
    
    =0.

The same arguments works for any odd number of solid dots and works for clockwise oriented bubbles.




 
We have that 
    d̅_2n=∑_2a+2b=2n-2[baseline=(current bounding box.center)]
        [->] (0,0) arc (0:180:5mm);
         (0,0) arc (360:180:5mm);
        [<-] (0.1,0) arc (-180:0:5mm);
         (0.1,0) arc (180:0:5mm);
         (-0.2,0.4) circle[radius=2pt];
         (0.9,0.4) circle[radius=2pt];
         at (-0.04,0.6) 2a;
         at (1.05,0.6) 2b;
        
        =∑_2a+2b=2n-2d̅_2ad_2b

    for any integers a,b and n≥1.



For the n=1 case, we have the following computation:


    d̅_2=[baseline=(current bounding box.center)]
            [->] (3,2) arc (-180:180:5mm);
     (3.95,2.2) circle [radius=2pt];
     (3.95,1.8) circle [radius=2pt];
           =
        [baseline=(current bounding box.center)]
        [->] (0,0) arc (10:180:5mm);
         (0,-0.18) arc (350:180:5mm);
         (0.1,0) arc (170:-170:5mm);
         (-0.02,-0.2) to (0.1,0.01);
         (-0.01,0.02) to (0.11,-0.2);
         (-0.1,0.23) circle[radius=2pt];
        
       =
        [baseline=(current bounding box.center)]
        [->] (0,0) arc (10:180:5mm);
         (-1,-0.08) arc (180:350:5mm);
         (0.1,0) arc (170:-170:5mm);
         (-0.02,-0.2) to (0.1,0.01);
         (-0.01,0.02) to (0.11,-0.2);
         (0.2,0.23) circle[radius=2pt];
        
        +[baseline=(current bounding box.center)]
        [->] (0,0) arc (0:180:5mm);
         (0,0) arc (360:180:5mm);
        [<-] (0.1,0) arc (-180:0:5mm);
         (0.1,0) arc (180:0:5mm);
        
        +[baseline=(current bounding box.center)]
        [->] (0,0) arc (0:180:5mm);
         (0,0) arc (360:180:5mm);
        [<-] (0.1,0) arc (-180:0:5mm);
         (0.1,0) arc (180:0:5mm);
         (-0.04,-0.15) circle[radius=2pt];
         (0.13,0.1) circle[radius=2pt];
        
        =d_0

where the first diagram on right hand side is zero since it contains a left curl, the second term is d̅_0d_0=d_0 and the last term is zero by Lemma <ref>.

 For general n, if you replace one of the solid dots with a right-twist curl, and slide the remaining 2n-1 dots through the crossings using relations <ref> and <ref> repeatedly, we will get many resolution terms, consisting of a sum of product of counterclockwise and clockwise bubbles, some with only solid dots, some with hollow dots as well. The terms with hollow dots are zero, and so are the terms with an odd number of solid dots. Also, the figure eight shape contains a left twist curl, so it is zero as well, which proves the statement. 






 §.§ Algebraic independence of bubbles



 A categorified Fock space action for ℋ_tw is described in <cit.>.  ℋ_tw acts on the category 𝔖, whose objects are induction and restriction functors between ℤ/2ℤ-graded finite dimensional 𝕊_n-modules, for all n≥1. Morphisms of 𝔖 are natural transformations between the induction and restriction functors. 

Following Khovanov's approach from <cit.>, let 𝔖_n be the subcategory of 𝔖, whose objects start with induction or restriction functors from ℤ/2ℤ-graded finite dimensional 𝕊_n-modules. For every n∈ℤ_≥1, we have a functor F_n:ℋ_tw→𝔖_n sending P to _n^n+1 and sending Q to _n^n-1.

Note that F_n sends _ℋ_tw(1) to the center of 𝕊_n, which is same as the center of ℂ [S_n]. 

Explicit descriptions of the actions of a crossing, a cup and a cap are provided in <cit.>. We would like to study the action of clockwise bubbles to show their algebraic independence. Note that d_2k  is obtained as composition of a cup, k copies of 1x_1 and a cap.

    [baseline=(current bounding box.center)]
    [->] (3,2) arc (-180:180:5mm);
     (3.95,2.2) circle [radius=2pt];
     at (4.2,2.2) k;
    
    =
    [baseline=(current bounding box.center)]
    [->] (1,-3.5) arc (0:180:5mm);
     (1,-3.8) circle[radius=2pt];
    [->] (1,-4) to (1,-3.5);
    [->] (0,-3.5) to (0,-4);
     at (0,-4.15) ⋮;
     at (1,-4.15) ⋮;
     (1,-4.8) circle[radius=2pt];
    [->] (1,-5) to (1,-4.5);
    [->] (0,-4.5) to (0,-5);
    [->] (0,-5) arc (-180:0:5mm);
     [decorate,decoration=brace,amplitude=2pt,xshift=-4pt,yshift=0pt] (1.3,-3.5) – (1.3,-5) node [black,midway,xshift=15pt] k dots;


Therefore to study the action of d_2k, we need to know the action of 1x_1 in addition to actions of cups and caps. Now 1x_1 is defined as a combination of caps, cups and crossings:

    [baseline=(current bounding box.center),scale=1.15]
    [->] (0,0) to (0,1);
      (0,.5) circle[radius=2pt];
    
    =
    [baseline=(current bounding box.center),scale=0.75]
    [->] (0,0) to (0,.5);
    [<-] (0.5,0.5) arc (-180:0:3mm);
    [->] (0,.5) to (.5,1);
    [->] (.5,.5) to (0,1);
    [->] (1.1,1) to (1.1,.5);
    [->] (0,1) to (0,1.5);
    [<-] (1.1,1) arc (0:180:3mm);


Using the explicit description of Fock space representation of ℋ in <cit.>, we compute the required actions. These computations give that -1x_1 acts by sending 
    1 ↦ J_n+1=∑_i=1^n(1-c_n+1c_i)(i,n+1).
 This is the (n+1)-st even Jucys-Murphy element. Therefore 1x_1^2 acts by multiplication by J_n+1^2. This is analogous to the untwisted case where the same element acts as multiplication by a Jucys-Murphy element.

 Finally, the action of the bubble d_2k is given by multiplication by 
    ∑_i=1^n(i↔ n+1)J_n+1^2k(n+1↔ i)-c_n(i↔ n+1)J_n+1^2k(n+1↔ i)c_1,
 
 where (i↔ n) denotes the n-cycle s_is_i+1....s_n-1.

Here we can apply the filtration argument on the number of disturbances of permutations as done in <cit.> to obtain the following.


The elements {d_2k}_k≥0 are algebraically independent, i.e. there is an isomorphism 
    _ℋ_tw(1)≅ k[d_0,d_2,d_4,...].


Therefore the bubbles are algebraically independent, and they form of a copy of a polynomial ring in infinitely many variables.





 §.§ Counter-clockwise bubble slide moves


In order to describe  as a vector space, it would be convenient to have a standard form for our diagrams in the trace. In particular, we want to collect all the bubbles appearing in a diagram on the rightmost part of the diagram. In order to do so, we must describe how bubbles slide through upward and downward strands. Note that since we can work with local relations, the bubbles don't have to interact with solid dots or crossings, they can simply slide through under a crossing or under a solid dot.

All calculations in this section take place in the trace, though we omit the brackets in some situations for readability.


We have that [d̅_2n,1]=2∑_k=1^n [ [baseline=(current bounding box.center),scale=1.75]
 (1,0) to [out=90, in=-75](0.95,0.5);
 (0.95,0.5) arc (5:355:2mm);
[->] (0.95,0.46) to [out=75, in=270] (1,1);
 (0.8,0.67) circle (1pt);
 at (0.8,0.8)  2k-1;
 ] in  for any positive integer n.



The proof is a direct computation, given below:

    [baseline=(current bounding box.center),scale=0.75]
    [->] (2.5,1) arc (-180:180:5mm);
    [->] (4,0) to (4,2);
     (3,1.5) circle (2pt);
     at (3,1.8) 2n;
    
     =
     [baseline=(current bounding box.center),scale=0.75]
    [->] (2.5,1) arc (-180:180:5mm);
    [->] (3.3,0) to (3.3,2);
     (3,1.5) circle (2pt);
     at (2.5,1.8) 2n;
       =
    [baseline=(current bounding box.center),scale=0.75]
    [->] (2.5,1) arc (-180:180:5mm);
    [->] (3.3,0) to (3.3,2);
     (3,1.5) circle (2pt);
     (3.5,1) circle (2pt);
     at (2.7,1.8) 2n-1;
     at (3.6,1) ;
    
    +
    [baseline=(current bounding box.center),scale=1.60]
     (1,0) to [out=90, in=-75](0.95,0.5);
     (0.95,0.5) arc (5:355:2mm);
    [->] (0.95,0.46) to [out=75, in=270] (1,1);
     (0.8,0.67) circle (1pt);
     at (0.75,0.8) 2n-1;
    
    -
    [baseline=(current bounding box.center),scale=1.60]
     (1,0) to [out=90, in=-75](0.95,0.5);
     (0.95,0.5) arc (5:355:2mm);
    [->] (0.95,0.46) to [out=75, in=270] (1,1);
     (0.7,0.67) circle (1pt);
     (0.85,0.65) circle [radius=1pt];
     (0.98,0.57) circle (1pt);
     at (0.65,0.8) 2n-1;
    
       =
    [baseline=(current bounding box.center),scale=0.75]
    [->] (2.5,1) arc (-180:180:5mm);
    [->] (3.3,0) to (3.3,2);
     (3,1.5) circle (2pt);
     (3.5,1) circle (2pt);
     at (2.7,1.8) 2n-1;
     at (3.6,1) ;
    
    +
    2
    [baseline=(current bounding box.center),scale=1.60]
     (1,0) to [out=90, in=-75](0.95,0.5);
     (0.95,0.5) arc (5:355:2mm);
    [->] (0.95,0.46) to [out=75, in=270] (1,1);
     (0.8,0.67) circle (1pt);
     at (0.75,0.8) 2n-1;
    
       =
    [baseline=(current bounding box.center),scale=0.75]
    [->] (2.5,1) arc (-180:180:5mm);
    [->] (3.3,0) to (3.3,2);
     (3,1.5) circle (2pt);
     (3.5,1) circle (2pt);
     at (2.7,1.8) 2n-2;
     at (3.7,1) 2;
    
    +
    [baseline=(current bounding box.center),scale=1.60]
     (1,0) to [out=90, in=-75](0.95,0.5);
     (0.95,0.5) arc (5:355:2mm);
    [->] (0.95,0.46) to [out=75, in=270] (1,1);
     (0.8,0.67) circle (1pt);
     at (0.65,0.8) 2n-2;
     (1,0.75) circle (1pt);
     at (1.1,0.75) ;
    
    -
    [baseline=(current bounding box.center),scale=1.60]
     (1,0) to [out=90, in=-75](0.95,0.5);
     (0.95,0.5) arc (5:355:2mm);
    [->] (0.95,0.46) to [out=75, in=270] (1,1);
     (0.7,0.67) circle (1pt);
     (0.85,0.65) circle (1pt);
     (0.98,0.57) circle (1pt);
     at (0.65,0.8) 2n-2;
     (1,0.75) circle (1pt);
     at (1.1,0.75) ;
    
    +
    2
    [baseline=(current bounding box.center),scale=1.60]
     (1,0) to [out=90, in=-75](0.95,0.5);
     (0.95,0.5) arc (5:355:2mm);
    [->] (0.95,0.46) to [out=75, in=270] (1,1);
     (0.8,0.67) circle (1pt);
     at (0.75,0.8) 2n-1;
    
       =
    [baseline=(current bounding box.center),scale=0.75]
    [->] (2.5,1) arc (-180:180:5mm);
    [->] (3.3,0) to (3.3,2);
     (3,1.5) circle (2pt);
     (3.5,1) circle (2pt);
     at (2.7,1.8) 2n-2;
     at (3.7,1) 2;
    
    +
    2
    [baseline=(current bounding box.center),scale=1.60]
     (1,0) to [out=90, in=-75](0.95,0.5);
     (0.95,0.5) arc (5:355:2mm);
    [->] (0.95,0.46) to [out=75, in=270] (1,1);
     (0.8,0.67) circle (1pt);
     at (0.75,0.8) 2n-1;
    
       =
    [baseline=(current bounding box.center),scale=0.75]
    [->] (2.5,1) arc (-180:180:5mm);
    [->] (3.3,0) to (3.3,2);
     (3,1.5) circle (2pt);
     (3.5,1) circle (2pt);
     at (2.7,1.8) 2n-4;
     at (3.7,1) 4;
    
    +
    2
    [baseline=(current bounding box.center),scale=1.60]
     (1,0) to [out=90, in=-75](0.95,0.5);
     (0.95,0.5) arc (5:355:2mm);
    [->] (0.95,0.46) to [out=75, in=270] (1,1);
     (0.8,0.67) circle (1pt);
     at (0.75,0.8) 2n-3;
    
    +
    2
    [baseline=(current bounding box.center),scale=1.60]
     (1,0) to [out=90, in=-75](0.95,0.5);
     (0.95,0.5) arc (5:355:2mm);
    [->] (0.95,0.46) to [out=75, in=270] (1,1);
     (0.8,0.67) circle (1pt);
     at (0.75,0.8) 2n-1;

Continuing to slide dots in the first term in this way, we obtain:


    [baseline=(current bounding box.center),scale=0.75]
    [->] (2.5,1) arc (-180:180:5mm);
    [->] (4,0) to (4,2);
     (3,1.5) circle (2pt);
     at (3,1.8) 2n;
       =
    [baseline=(current bounding box.center),scale=0.75]
    [->] (2.5,1) arc (-180:180:5mm);
    [->] (3.3,0) to (3.3,2);
     (3.5,1) circle (2pt);
     at (3.8,1) 2n;
    
    +
    2∑_k=1^n[baseline=(current bounding box.center),scale=1.75]
     (1,0) to [out=90, in=-75](0.95,0.5);
     (0.95,0.5) arc (5:355:2mm);
    [->] (0.95,0.46) to [out=75, in=270] (1,1);
     (0.8,0.67) circle (1pt);
     at (0.8,0.8) 2k-1;
    .



 
We have that 
    [ [baseline=(current bounding box.center),scale=1.75]
     (1,0) to [out=90, in=-75](0.95,0.5);
     (0.95,0.5) arc (5:355:2mm);
    [->] (0.95,0.46) to [out=75, in=270] (1,1);
     (0.8,0.67) circle (1pt);
     at (0.75,0.8) 2n+1;
     ]
    =
    ∑_a+b=n[ [baseline=(current bounding box.center),scale=0.75]
    [->] (2.5,1) arc (-180:180:5mm);
    [->] (4,0) to (4,2);
     (3,1.5) circle (2pt);
     (4,1) circle (2pt);
     at (3,1.8) 2a;
     at (4.3,1) 2b;
     ]

in  for any non-negative integer n.



This is an easy computation using induction on n. The base case is 

    [baseline=(current bounding box.center),scale=1.75]
     (1,0) to [out=90, in=-75](0.95,0.5);
     (0.95,0.5) arc (5:355:2mm);
    [->] (0.95,0.46) to [out=75, in=270] (1,1);
     (0.8,0.67) circle (1pt);
    
    =
    [baseline=(current bounding box.center),scale=1.75]
     (1,0) to [out=90, in=-75](0.95,0.5);
     (0.95,0.5) arc (5:355:2mm);
    [->] (0.95,0.46) to [out=75, in=270] (1,1);
     (1,0.27) circle (1pt);
    
    +
    [baseline=(current bounding box.center),scale=0.75]
    [->] (2.5,1) arc (-180:180:5mm);
    [->] (4,0) to (4,2);
    
    -
    [baseline=(current bounding box.center),scale=0.75]
    [->] (2.5,1) arc (-180:180:5mm);
    [->] (4,0) to (4,2);
     (3.5,1) circle (2pt);
     (4,0.8) circle (2pt);
    
    =
    [baseline=(current bounding box.center),scale=0.75]
    [->] (2.5,1) arc (-180:180:5mm);
    [->] (4,0) to (4,2);

where the first term after the first equality contains a left twist curl, and the last term is zero since a bubble with a hollow dot is zero.

For the induction step, suppose the statement holds for n≥ 1. Then

    [baseline=(current bounding box.center),scale=1.75]
     (1,0) to [out=90, in=-75](0.95,0.5);
     (0.95,0.5) arc (5:355:2mm);
    [->] (0.95,0.46) to [out=75, in=270] (1,1);
     (0.8,0.67) circle (1pt);
     at (0.7,0.78) 2n+3;
       =
    [baseline=(current bounding box.center),scale=1.75]
     (1,0) to [out=90, in=-75](0.95,0.5);
     (0.95,0.5) arc (5:355:2mm);
    [->] (0.95,0.46) to [out=75, in=270] (1,1);
     (0.8,0.67) circle (1pt);
     at (0.7,0.78) 2n+2;
     (1,0.27) circle (1pt);
     at (1.1,0.27) ;
    
    +
    [baseline=(current bounding box.center),scale=0.75]
    [->] (2.5,1) arc (-180:180:5mm);
    [->] (4,0) to (4,2);
     (3,1.5) circle (2pt);
    
     at (3,1.8) 2n+2;
     at (4.3,1) ;
    
    -
    [baseline=(current bounding box.center),scale=0.75]
    [->] (2.5,1) arc (-180:180:5mm);
    [->] (4,0) to (4,2);
     (3,1.5) circle (2pt);
    
     (3.5,1) circle (2pt);
     (4,0.8) circle (2pt);
     at (3,1.8) 2n+2;
     at (4.3,1) ;
    
    =
    [baseline=(current bounding box.center),scale=1.75]
     (1,0) to [out=90, in=-75](0.95,0.5);
     (0.95,0.5) arc (5:355:2mm);
    [->] (0.95,0.46) to [out=75, in=270] (1,1);
     (0.8,0.67) circle (1pt);
     at (0.7,0.78) 2n+2;
     (1,0.27) circle (1pt);
     at (1.1,0.27) ;
    
    +
    [baseline=(current bounding box.center),scale=0.75]
    [->] (2.5,1) arc (-180:180:5mm);
    [->] (4,0) to (4,2);
     (3,1.5) circle (2pt);
    
     at (3,1.8) 2n+2;
     at (4.3,1) ;
    
       =
    [baseline=(current bounding box.center),scale=1.75]
     (1,0) to [out=90, in=-75](0.95,0.5);
     (0.95,0.5) arc (5:355:2mm);
    [->] (0.95,0.46) to [out=75, in=270] (1,1);
     (0.8,0.67) circle (1pt);
     at (0.7,0.78) 2n+1;
     (1,0.27) circle (1pt);
     at (1.1,0.27) 2;
    
    +
    [baseline=(current bounding box.center),scale=0.75]
    [->] (2.5,1) arc (-180:180:5mm);
    [->] (4,0) to (4,2);
     (3,1.5) circle (2pt);
     (4,1) circle (2pt);
     at (3,1.8) 2n+1;
     at (4.3,1) ;
    
    -
    [baseline=(current bounding box.center),scale=0.75]
    [->] (2.5,1) arc (-180:180:5mm);
    [->] (4,0) to (4,2);
     (3,1.5) circle (2pt);
     (4,1) circle (2pt);
     (3.5,1) circle (2pt);
     (4,0.8) circle (2pt);
     at (3,1.8) 2n+1;
     at (4.3,1) ;
    
    +
    [baseline=(current bounding box.center),scale=0.75]
    [->] (2.5,1) arc (-180:180:5mm);
    [->] (4,0) to (4,2);
     (3,1.5) circle (2pt);
    
     at (3,1.8) 2n+2;
     at (4.3,1) ;
    
       =
    [baseline=(current bounding box.center),scale=1.75]
     (1,0) to [out=90, in=-75](0.95,0.5);
     (0.95,0.5) arc (5:355:2mm);
    [->] (0.95,0.46) to [out=75, in=270] (1,1);
     (0.8,0.67) circle (1pt);
     at (0.7,0.78) 2n+1;
     (1,0.27) circle (1pt);
     at (1.1,0.27) 2;
    
    +
    [baseline=(current bounding box.center),scale=0.75]
    [->] (2.5,1) arc (-180:180:5mm);
    [->] (4,0) to (4,2);
     (3,1.5) circle (2pt);
    
     at (3,1.8) 2n+2;
     at (4.3,1) ;
    ,

where on the second line, we know that counter-clockwise bubbles with odd number of hollow dots are zero by Proposition <ref>, and the terms with hollow dots are zero by Proposition <ref>.

Now we can apply our induction hypothesis to the upper part of [baseline=(current bounding box.center),scale=1.75]
 (1,0) to [out=90, in=-75](0.95,0.5);
 (0.95,0.5) arc (5:355:2mm);
[->] (0.95,0.46) to [out=75, in=270] (1,1);
 (0.8,0.67) circle (1pt);
 at (0.7,0.78) 2n+1;
 (1,0.27) circle (1pt);
 at (1.1,0.27) 2; to get that


    [baseline=(current bounding box.center),scale=1.75]
     (1,0) to [out=90, in=-75](0.95,0.5);
     (0.95,0.5) arc (5:355:2mm);
    [->] (0.95,0.46) to [out=75, in=270] (1,1);
     (0.8,0.67) circle (1pt);
     at (0.7,0.78) 2n+3;
    
    =
    ∑_a+b=n[baseline=(current bounding box.center),scale=0.75]
    [->] (2.5,1) arc (-180:180:5mm);
    [->] (4,0) to (4,2);
     (3,1.5) circle (2pt);
     (4,1) circle (2pt);
     at (3,1.8) 2a;
     at (4.3,1) 2b;
     (4,0.5) circle (2pt);
     at (4.2,0.5) 2;
    
    +
    [baseline=(current bounding box.center),scale=0.75]
    [->] (2.5,1) arc (-180:180:5mm);
    [->] (4,0) to (4,2);
     (3,1.5) circle (2pt);
    
     at (3,1.8) 2n+2;
     at (4.3,1) ;
    
    =
    ∑_a+b=n+1[baseline=(current bounding box.center),scale=0.75]
    [->] (2.5,1) arc (-180:180:5mm);
    [->] (4,0) to (4,2);
     (3,1.5) circle (2pt);
     (4,1) circle (2pt);
     at (3,1.8) 2a;
     at (4.3,1) 2b;
    ,

 as desired.


Obtaining an explicit formula for sliding counter-clockwise bubbles is difficult since we express their commutators in terms of left twist curls with some dots on the curl, whose resolution terms still leave us with counter-clockwise bubbles on the left side of 1x_i^a. However, the situation is better with clockwise oriented bubbles.




 §.§ Clockwise bubble slide moves


We can compute an explicit formula for clockwise bubble slides.


We have 
    [d_2n,1]=2[ [baseline=(current bounding box.center),scale=1.5]
    [->] (0,0) to (0,1);
     (0,0.75) circle (1pt);
     at (0.2,0.75) 2n;
     ]
    +2∑_a+b=2n-1[ [baseline=(current bounding box.center),scale=1.5]
     (1,0) to [out=90, in=85](1.05,0.5);
     (1.05,0.5) arc (-175:175:2mm);
    [->] (1.05,0.46) to [out=95, in=270] (1,1);
     (1.02,0.75) circle (1pt);
     at (0.8,0.75) a;
     (1.45,0.5) circle (1pt);
     at (1.55,0.5) b;
     ]

in  for all n≥ 0.



This is a direct computation, given below:

    [baseline=(current bounding box.center),scale=0.75]
    [<-] (2.5,1) arc (-180:180:5mm);
    [->] (4,0) to (4,2);
     (3,1.5) circle (2pt);
     at (3,1.8) 2n;
       =
    [baseline=(current bounding box.center),scale=0.75]
    [<-] (2.5,1) arc (-180:180:5mm);
    [->] (3.3,0) to (3.3,2);
     (3,1.5) circle (2pt);
     at (2.5,1.8) 2n;
    
    +2[baseline=(current bounding box.center),scale=0.75]
    [->] (0,0) to (0,2);
     (0,1.6) circle (2pt);
     at (0.34,1.6) 2n;
    
       =
    [baseline=(current bounding box.center),scale=0.75]
    [<-] (2.5,1) arc (-180:180:5mm);
    [->] (3.3,0) to (3.3,2);
     (3,1.5) circle (2pt);
     at (2.7,1.8) 2n-1;
     (3.5, 1) circle (2pt);
     at (3.8,1) ;
    
    +
    [baseline=(current bounding box.center),scale=1.5]
     (1,0) to [out=90, in=85](1.05,0.5);
     (1.05,0.5) arc (-175:175:2mm);
    [->] (1.05,0.46) to [out=95, in=270] (1,1);
     (1.01,0.8) circle (1pt);
     at (0.7,0.8) 2n-1;
    
    +
    [baseline=(current bounding box.center),scale=1.5]
     (1,0) to [out=90, in=85](1.05,0.5);
     (1.05,0.5) arc (-175:175:2mm);
    [->] (1.05,0.46) to [out=95, in=270] (1,1);
     (1.01,0.8) circle (1pt);
     at (0.7,0.8) 2n-1;
     (1,0.9) circle (1pt);
     (1.12,0.67) circle (1pt);
    
    +2[baseline=(current bounding box.center),scale=0.75]
    [->] (0,0) to (0,2);
     (0,1.6) circle (2pt);
     at (0.34,1.6) 2n;
    
       =
    [baseline=(current bounding box.center),scale=0.75]
    [<-] (2.5,1) arc (-180:180:5mm);
    [->] (3.3,0) to (3.3,2);
     (3,1.5) circle (2pt);
     at (2.7,1.8) 2n-1;
     (3.5, 1) circle (2pt);
     at (3.8,1) ;
    
    +2
    [baseline=(current bounding box.center),scale=1.5]
     (1,0) to [out=90, in=85](1.05,0.5);
     (1.05,0.5) arc (-175:175:2mm);
    [->] (1.05,0.46) to [out=95, in=270] (1,1);
     (1.01,0.8) circle (1pt);
     at (0.7,0.8) 2n-1;
    
    +2[baseline=(current bounding box.center),scale=0.75]
    [->] (0,0) to (0,2);
     (0,1.6) circle (2pt);
     at (0.34,1.6) 2n;
    
       =
    [baseline=(current bounding box.center),scale=0.75]
    [<-] (2.5,1) arc (-180:180:5mm);
    [->] (3.3,0) to (3.3,2);
     (3,1.5) circle (2pt);
     at (2.7,1.8) 2n-2;
     (3.5, 1) circle (2pt);
     at (3.8,1) 2;
    
    +
    [baseline=(current bounding box.center),scale=1.5]
     (1,0) to [out=90, in=85](1.05,0.5);
     (1.05,0.5) arc (-175:175:2mm);
    [->] (1.05,0.46) to [out=95, in=270] (1,1);
     (1.01,0.8) circle (1pt);
     at (0.7,0.8) 2n-2;
     (1.24,0.72) circle (1pt);
    
    +
    [baseline=(current bounding box.center),scale=1.5]
     (1,0) to [out=90, in=85](1.05,0.5);
     (1.05,0.5) arc (-175:175:2mm);
    [->] (1.05,0.46) to [out=95, in=270] (1,1);
     (1.01,0.8) circle (1pt);
     at (0.7,0.8) 2n-2;
     (1.24,0.72) circle (1pt);
     (1,0.9) circle (1pt);
     (1.12,0.67) circle (1pt);
    
    +
    2
    [baseline=(current bounding box.center),scale=1.5]
     (1,0) to [out=90, in=85](1.05,0.5);
     (1.05,0.5) arc (-175:175:2mm);
    [->] (1.05,0.46) to [out=95, in=270] (1,1);
     (1.01,0.8) circle (1pt);
     at (0.7,0.8) 2n-1;
    
    +2[baseline=(current bounding box.center),scale=0.75]
    [->] (0,0) to (0,2);
     (0,1.6) circle (2pt);
     at (0.34,1.6) 2n;
    
       =
    [baseline=(current bounding box.center),scale=0.75]
    [<-] (2.5,1) arc (-180:180:5mm);
    [->] (3.3,0) to (3.3,2);
     (3,1.5) circle (2pt);
     at (2.7,1.8) 2n-2;
     (3.5, 1) circle (2pt);
     at (3.8,1) 2;
    
    +
    2[baseline=(current bounding box.center),scale=1.5]
     (1,0) to [out=90, in=85](1.05,0.5);
     (1.05,0.5) arc (-175:175:2mm);
    [->] (1.05,0.46) to [out=95, in=270] (1,1);
     (1.01,0.8) circle (1pt);
     at (0.7,0.8) 2n-2;
     (1.24,0.72) circle (1pt);
    
    +
    2
    [baseline=(current bounding box.center),scale=1.5]
     (1,0) to [out=90, in=85](1.05,0.5);
     (1.05,0.5) arc (-175:175:2mm);
    [->] (1.05,0.46) to [out=95, in=270] (1,1);
     (1.01,0.8) circle (1pt);
     at (0.7,0.8) 2n-1;
    
    +
    2[baseline=(current bounding box.center),scale=0.75]
    [->] (0,0) to (0,2);
     (0,1.6) circle (2pt);
     at (0.34,1.6) 2n;

Continuing to slide dots in the first term in this way, we obtain:

    [baseline=(current bounding box.center),scale=0.75]
    [<-] (2.5,1) arc (-180:180:5mm);
    [->] (4,0) to (4,2);
     (3,1.5) circle (2pt);
     at (3,1.8) 2n;
    
    =
    [baseline=(current bounding box.center),scale=0.75]
    [<-] (2.5,1) arc (-180:180:5mm);
    [->] (2,0) to (2,2);
     (3,1.5) circle (2pt);
     at (3,1.8) 2n;
    
    +2[baseline=(current bounding box.center),scale=0.75]
    [->] (0,0) to (0,2);
     (0,1.6) circle (2pt);
     at (0.34,1.6) 2n;
    
    +
    2∑_a+b=2n-1[baseline=(current bounding box.center),scale=1.5]
     (1,0) to [out=90, in=85](1.05,0.5);
     (1.05,0.5) arc (-175:175:2mm);
    [->] (1.05,0.46) to [out=95, in=270] (1,1);
     (1.02,0.75) circle (1pt);
     at (0.8,0.75) a;
     (1.45,0.5) circle (1pt);
     at (1.55,0.5) b;
    .



In particular, we can refine this statement to obtain the following recursive formula for computing [d_2n,1].

We have
    [d_2n,1]=[d_2n-2,1]∘  x_1^2 + 4[ [baseline=(current bounding box.center),scale=1.5]
     (1,0) to [out=90, in=85](1.05,0.5);
     (1.05,0.5) arc (-175:175:2mm);
    [->] (1.05,0.46) to [out=95, in=270] (1,1);
     (1.01,0.85) circle (1pt);
     at (1.1,0.85) 2;
     (1.45,0.5) circle (1pt);
     at (1.7,0.5) 2n-3;
     ]
    -2[ [baseline=(current bounding box.center),scale=0.75]
    [<-] (0.5,1) arc (-180:180:5mm);
    [->] (0,0) to (0,2);
     (1.5,1) circle (2pt);
     at (1.8,1) 2n-2;
     ]
 in  for all n≥ 0.



This lemma follows from the observation that 

    [baseline=(current bounding box.center),scale=1.5]
     (1,0) to [out=90, in=85](1.05,0.5);
     (1.05,0.5) arc (-175:175:2mm);
    [->] (1.05,0.46) to [out=95, in=270] (1,1);
     (1.02,0.75) circle (1pt);
     at (0.8,0.75) a;
     (1.45,0.5) circle (1pt);
     at (1.65,0.5) 2k;
    
    =
    [baseline=(current bounding box.center),scale=1.5]
     (1,0) to [out=90, in=85](1.05,0.5);
     (1.05,0.5) arc (-175:175:2mm);
    [->] (1.05,0.46) to [out=95, in=270] (1,1);
     (1.02,0.75) circle (1pt);
     at (0.8,0.75) a+1;
     (1.45,0.5) circle (1pt);
     at (1.70,0.5) 2k-1;
    
    -
    [baseline=(current bounding box.center),scale=0.75]
    [<-] (2.5,1) arc (-180:180:5mm);
    [->] (2,0) to (2,2);
     (3,1.5) circle (2pt);
     at (3,1.8) 2k-1;
     (2,1.4) circle (2pt);
     at (1.8,1.4) a;
    
    +
    [baseline=(current bounding box.center),scale=0.75]
    [<-] (2.5,1) arc (-180:180:5mm);
    [->] (2,0) to (2,2);
     (3,1.5) circle (2pt);
     at (3,1.8) 2k-1;
     (2,1.4) circle (2pt);
     at (1.8,1.4) a;
     (2,1.2) circle (2pt);
     (2.5,1.1) circle (2pt);
    
    =
    [baseline=(current bounding box.center),scale=1.5]
     (1,0) to [out=90, in=85](1.05,0.5);
     (1.05,0.5) arc (-175:175:2mm);
    [->] (1.05,0.46) to [out=95, in=270] (1,1);
     (1.02,0.75) circle (1pt);
     at (0.8,0.75) a+1;
     (1.45,0.5) circle (1pt);
     at (1.70,0.5) 2k-1;
    ,

where the second term after the first equality is zero by Lemma <ref>, and the third term is zero by Lemma <ref>. Applying this result to the summands in the statement of Lemma <ref> yields the result.


Finally, we obtain an explicit formula for computing [d_2n,1].

We have 
    [d_2n,1]=(2+4n)[ [baseline=(current bounding box.center),scale=1.5]
    [->] (0,0) to (0,1);
     (0,0.75) circle (1pt);
     at (0.2,0.75) 2n;
     ]
    -∑_a+b=n-1(2+4a)
    [ [baseline=(current bounding box.center),scale=0.75]
    [<-] (0.5,1) arc (-180:180:5mm);
    [->] (0,0) to (0,2);
     (1.5,1) circle (2pt);
     (0,1.5) circle (2pt);
     at (0.3,1.5) 2a;
     at (1.8,1) 2b;
     ]
 in  for all n≥ 0.



We claim that 
    [baseline=(current bounding box.center),scale=1.5]
     (1,0) to [out=90, in=85](1.05,0.5);
     (1.05,0.5) arc (-175:175:2mm);
    [->] (1.05,0.46) to [out=95, in=270] (1,1);
     (1.01,0.85) circle (1pt);
     at (1.1,0.85) 2;
     (1.45,0.5) circle (1pt);
     at (1.7,0.5) 2n-3;
    
    =
    [baseline=(current bounding box.center),scale=1.5]
    [->] (0,0) to (0,1);
     (0,0.75) circle (1pt);
     at (0.2,0.75) 2n;
    
    -
    ∑_a+b=n-1
    a≠0[baseline=(current bounding box.center),scale=0.75]
    [<-] (2.5,1) arc (-180:180:5mm);
    [->] (2,0) to (2,2);
     (3,1.5) circle (2pt);
     at (3,1.8) 2b;
     (2,1.4) circle (2pt);
     at (1.7,1.4) 2a;
 for n≥2.  We proceed via induction on n. The base case n=2 is a direct computation. Now suppose the formula holds for some n≥ 2. Then 

    [baseline=(current bounding box.center),scale=1.5]
     (1,0) to [out=90, in=85](1.05,0.5);
     (1.05,0.5) arc (-175:175:2mm);
    [->] (1.05,0.46) to [out=95, in=270] (1,1);
     (1.01,0.85) circle (1pt);
     at (1.1,0.85) 2;
     (1.45,0.5) circle (1pt);
     at (1.8,0.5) 2n-3;
    
    =
    [baseline=(current bounding box.center),scale=1.5]
     (1,0) to [out=90, in=85](1.05,0.5);
     (1.05,0.5) arc (-175:175:2mm);
    [->] (1.05,0.46) to [out=95, in=270] (1,1);
     (1.01,0.85) circle (1pt);
     at (1.1,0.85) 3;
     (1.45,0.5) circle (1pt);
     at (1.8,0.5) 2n-4;
    
    -
    [baseline=(current bounding box.center),scale=0.75]
    [<-] (2.5,1) arc (-180:180:5mm);
    [->] (2,0) to (2,2);
     (3,1.5) circle (2pt);
     at (3,1.8) 2n-4;
     (2,1.4) circle (2pt);
     at (1.7,1.4) 2;
       =
    [baseline=(current bounding box.center),scale=1.5]
     (1,0) to [out=90, in=85](1.05,0.5);
     (1.05,0.5) arc (-175:175:2mm);
    [->] (1.05,0.46) to [out=95, in=270] (1,1);
     (1.01,0.85) circle (1pt);
     at (1.1,0.85) 4;
     (1.45,0.5) circle (1pt);
     at (1.8,0.5) 2n-5;
    
    -
    [baseline=(current bounding box.center),scale=0.75]
    [<-] (2.5,1) arc (-180:180:5mm);
    [->] (2,0) to (2,2);
     (3,1.5) circle (2pt);
     at (3,1.8) 2n-4;
     (2,1.4) circle (2pt);
     at (1.7,1.4) 2;
    
       =
    [baseline=(current bounding box.center),scale=1.5]
     (1,0) to [out=90, in=85](1.05,0.5);
     (1.05,0.5) arc (-175:175:2mm);
    [->] (1.05,0.46) to [out=95, in=270] (1,1);
     (1.02,0.75) circle (1pt);
     at (1.12,0.75) 2;
     (1,0.92) circle (1pt);
     at (1.1,0.92) 2;
     (1.45,0.5) circle (1pt);
     at (1.8,0.5) 2n-5;
    
    -
    [baseline=(current bounding box.center),scale=0.75]
    [<-] (2.5,1) arc (-180:180:5mm);
    [->] (2,0) to (2,2);
     (3,1.5) circle (2pt);
     at (3,1.8) 2n-4;
     (2,1.4) circle (2pt);
     at (1.7,1.4) 2;
    .

 Now we can apply the induction hypothesis to the lower part of the first term in the last expression. This gives us:


    [baseline=(current bounding box.center),scale=1.5]
     (1,0) to [out=90, in=85](1.05,0.5);
     (1.05,0.5) arc (-175:175:2mm);
    [->] (1.05,0.46) to [out=95, in=270] (1,1);
     (1.01,0.85) circle (1pt);
     at (1.1,0.85) 2;
     (1.45,0.5) circle (1pt);
     at (1.8,0.5) 2n-3;
       =
    ( [baseline=(current bounding box.center),scale=1.5]
    [->] (0,0) to (0,1);
     (0,0.65) circle (1pt);
     at (0.3,0.65) 2n-2;
     (0,0.85) circle (1pt);
     at (0.15,0.85) 2;
    
    -
    ∑_a+b=n-2
    a≠0[baseline=(current bounding box.center),scale=0.75]
    [<-] (2.5,1) arc (-180:180:5mm);
    [->] (2,0) to (2,2);
     (3,1.5) circle (2pt);
     at (3,1.8) 2b;
     (2,1.4) circle (2pt);
     at (1.7,1.4) 2a;
     (2,1.7) circle (2pt);
     at (1.7,1.7) 2;
    )-
    [baseline=(current bounding box.center),scale=0.75]
    [<-] (2.5,1) arc (-180:180:5mm);
    [->] (2,0) to (2,2);
     (3,1.5) circle (2pt);
     at (3,1.8) 2n-4;
     (2,1.4) circle (2pt);
     at (1.7,1.4) 2;
    
       =
    ( [baseline=(current bounding box.center),scale=1.5]
    [->] (0,0) to (0,1);
     (0,0.65) circle (1pt);
     at (0.2,0.65) 2n;
    
    -
    ∑_a+b=n-2
    a≠0[baseline=(current bounding box.center),scale=0.75]
    [<-] (2.5,1) arc (-180:180:5mm);
    [->] (2,0) to (2,2);
     (3,1.5) circle (2pt);
     at (3,1.8) 2b;
     (2,1.4) circle (2pt);
     at (1.4,1.4) 2a+2;
    )-
    [baseline=(current bounding box.center),scale=0.75]
    [<-] (2.5,1) arc (-180:180:5mm);
    [->] (2,0) to (2,2);
     (3,1.5) circle (2pt);
     at (3,1.8) 2n-4;
     (2,1.4) circle (2pt);
     at (1.7,1.4) 2;
    
       =
    [baseline=(current bounding box.center),scale=1.5]
    [->] (0,0) to (0,1);
     (0,0.65) circle (1pt);
     at (0.2,0.65) 2n;
    
    -
    ∑_a+b=n-1
    a≠0[baseline=(current bounding box.center),scale=0.75]
    [<-] (2.5,1) arc (-180:180:5mm);
    [->] (2,0) to (2,2);
     (3,1.5) circle (2pt);
     at (3,1.8) 2b;
     (2,1.4) circle (2pt);
     at (1.4,1.4) 2a;


Applying this result to the recursive formula in Lemma <ref> proves the statement.



Commutators of bubbles with downward strands are similar to those of bubbles with upward strands. 

 We have

    [d_2n,-1]=-2[ [baseline=(current bounding box.center),scale=1.5]
    [<-] (0,0) to (0,1);
     (0,0.75) circle (1pt);
     at (0.2,0.75) 2n;
     ]
    -2∑_a+b=2n-1[ [baseline=(current bounding box.center),scale=1.5]
    [<-] (1,0) to [out=90, in=-75](0.95,0.5);
     (0.95,0.5) arc (5:355:2mm);
     (0.95,0.46) to [out=75, in=270] (1,1);
     (1.00,0.75) circle (1pt);
     at (0.85,0.75) a;
     (0.55,0.5) circle (1pt);
     at (0.40,0.5) b;
     ]

in  for all n≥ 0.

 This follows from a computation similar to those in the proofs of Lemmas <ref> and <ref>. 

Finally we have an explicit formula for commutators of clockwise oriented bubbles and a single downward strand.
  We have

    [d_2n,-1]=-(2+4n)[ [baseline=(current bounding box.center),scale=1.5]
    [<-] (0,0) to (0,1);
     (0,0.75) circle (1pt);
     at (0.2,0.75) 2n;
     ]
    +∑_a+b=n-1(2+4a)[ [baseline=(current bounding box.center),scale=0.75]
    [<-] (0.5,1) arc (-180:180:5mm);
    [<-] (2.3,0) to (2.3,2);
     (1.5,1) circle (2pt);
     (2.3,1.5) circle (2pt);
     at (2.6,1.5) 2a;
     at (1.8,1) 2b;
     ]
 in  for n≥0.

 This follows from Lemma <ref>, using a similar argument as in the proof of Proposition <ref>.

Note that in this formula, we are still left with clockwise bubbles on the left side of a downward strand, but with fewer dots on it. Hence the formula may be applied inductively in order to move all the bubbles to the rightmost part of the diagram.



§ DIAGRAMMATIC LEMMAS


This section contains some technical computations to derive relations between diagrams consisting of up and down strands. These relations allow us to find a generating set of  in Section 6.


 §.§ Differential degree zero part of 


The differential degree zero part of  consists of elements {n}_n∈ℤ. First, we have the following basic fact.

 <cit.>  We have 
    2n≅ 0
 for any n∈ℤ.   By Proposition <ref>, the proof in the Hecke-Clifford algebra applies here, as well.  

The elements of  satisfy the following relations. 
 The following commutators are zero for all non-negative integers n,m:

    
  * [n,m]=0,
    
  * [-n,-m]=0,
    
  * [2n,-2n]=0.




Parts (1) and (2) follow from the fact that similarly oriented strands can be split apart when they cross twice.
Part (3) follows immediately from Proposition <ref>.


To obtain a copy of the twisted Heisenberg algebra in the , we need to look at commutators between elements with odd numbers of oppositely oriented strands.


We have, for any n,m ∈ℤ^≥ 0, 
    [2n+1,-2m+1]=(δ_n,-m)(-2(2n+1)).



First note that  <cit.> and <cit.> holds in our twisted case with a small modification, since all the arguments in their proofs use the fact that the resolution terms contain left twist curls, hence are zero. There are extra resolution terms with hollow dots due to relation (<ref>), but two hollow dots on a diagram containing a left twist curl still gives zero. The only modification comes in the case m=n where we get two copies of counter-clockwise bubbles instead of one, since a two hollow dots on a counter-clockwise bubble end up canceling each other without changing the sign of the diagram. We immediately get that when m≠ n, our commutator is zero since we have no solid dots. Therefore we have 


    h_2n+1h_-2m+1   =[  [baseline=(current bounding box.center),scale=0.8]
           [->] (3.2,0) .. controls (3.2,1.25) and (0,.25) .. (0,2)
         node[pos=0.85, shape=coordinate](X);
      [->] (0,0) .. controls (0,1) and (.8,.8) .. (.8,2);
      [->] (.8,0) .. controls (.8,1) and (1.6,.8) .. (1.6,2);
      [->] (2.4,0) .. controls (2.4,1) and (3.2,.8) .. (3.2,2);
       at (1.6,.35) …;
       at (2.4,1.65) …;
    [baseline=(current bounding box.center),scale=0.8]
     [<-] (3.2,0) .. controls (3.2,1.25) and (0,.25) .. (0,2)
         node[pos=0.85, shape=coordinate](X);
      [<-] (0,0) .. controls (0,1) and (.8,.8) .. (.8,2);
      [<-] (.8,0) .. controls (.8,1) and (1.6,.8) .. (1.6,2);
      [<-] (2.4,0) .. controls (2.4,1) and (3.2,.8) .. (3.2,2);
       at (1.6,.35) …;
       at (2.4,1.65) …;
     ]
    
       =
        [  [baseline=(current bounding box.center),scale=0.8]
     
     
       (3,0) .. controls ++(0,1.25) and ++(0,-1.75) .. (0,1.5);
       (0,0) .. controls ++(0,1) and ++(0,-.7) .. (.6,1.5);
       (.6,0) .. controls ++(0,1) and ++(0,-.7) .. (1.2,1.5);
       (1.8,0) .. controls ++(0,1) and ++(0,-.7) .. (2.4,1.5);
       (2.4,0) .. controls ++(0,1) and ++(0,-.7) .. (3,1.5);
       at (1.2,.35) …;
       at (1.8,1.15) …;
      
      
       (6,0) .. controls ++(0,1.25) and ++(0,-1.75) .. (3.6,1.5) ;
       (3.6,0) .. controls ++(0,1) and ++(0,-.7) .. (4.2,1.5);
       (4.2,0) .. controls ++(0,1) and ++(0,-.7) .. (4.8,1.5);
       (5.4,0) .. controls ++(0,1) and ++(0,-.7) .. (6,1.5);
       at (4.8,-.65) …;
       at (5.4,1.85) …;
      
      
        [blue, dotted] (-0.4,0) – (6.4,0);
        [blue, dotted] (-0.4,1.5) – (6.4,1.5);
      
       [->]  (3,1.5) .. controls ++(0,0.5) and ++(0,-.5) .. (3.6,2.5);
        (3.6,1.5) .. controls ++(0,1) and ++(0,-1) .. (0,2.5);
       [->]  (2.4,1.5) .. controls ++(0,0.5) and ++(0,-.5) .. (3,2.5);
       [->]  (1.2,1.5) .. controls ++(0,0.5) and ++(0,-.5) .. (1.8,2.5);
       [->]  (.6,1.5) .. controls ++(0,0.5) and ++(0,-.5) .. (1.2,2.5);
       [->]  (0,1.5) .. controls ++(0,0.5) and ++(0,-.5) .. (.6,2.5);
        (4.2,1.5) – (4.2,2.5);
        (4.8,1.5) – (4.8,2.5);
        (6,1.5) – (6,2.5);
      
         (2.4,0) .. controls ++(0,-.5) and ++(0,.5) .. (3,-1);
         (1.8,0) .. controls ++(0,-.5) and ++(0,.5) .. (2.4,-1);
         (.6,0) .. controls ++(0,-.5) and ++(0,.5) .. (1.2,-1);
         (0,0) .. controls ++(0,-.5) and ++(0,.5) .. (.6,-1);
         (3,0) .. controls ++(0,-.5) and ++(0,.5) .. (3.6,-1);
      [,->] (3.6,0) .. controls ++(0,-1) and ++(0,+1) .. (0,-1);
       [->] (4.2,0) – (4.2,-1);
       [->] (5.4,0) – (5.4,-1);
       [->] (6,0) – (6,-1);
     ]
    
       =h_-(2m+1)h_2n+1(-2d̅_0(2n+1)).


Hence [(2n+1),-(2m+1)]=δ_n,-m(-2(2n+1)).
 

Therefore the subset A={(2n+1)}_n∈ℤ of the filtration degree zero part of  is isomorphic to the twisted Heisenberg algebra via


    ϕ:   𝔥_tw   ∼⟶ A 
       2n+1/2   ↦1/2-(2n+1).


In the W-algebra W^-, we have an isomophic copy of the twisted Heisenberg algebra as well, given by B={ω_2n+1,0}_n∈ℤ, with the isomorphism given by


    ψ:   𝔥_tw   ∼⟶ B 
       2n+1/2   ↦1/√(2) ω_2n+1,0.


Therefore we have an isomorphism between the degree zero part of  and the degree zero part of W^-:


    ψ∘ϕ^-1:   A   ∼⟶ B 
       -(2n+1)   ↦√(2) w_2n+1,0.




 §.§ Nonzero differential degree part of 


We have the following basic facts about diagrams in ^>, which we may copy from the corresponding facts in the trace of the affine Hecke-Clifford algebra because of the triangular decomposition of  described in Proposition <ref>.

 <cit.> In Tr() for any m,n ∈ℤ, we have 
    2n+1x_1^2m+1 = 0,


    2nx_1^2m =0.
  

Hence any diagram containing an odd cycle with an odd number of dots or an even cycle with an even number of dots is zero. Therefore, the difference of the number of strands and number of solid dots must be odd. This agrees with the fact that in the W-algebra W^-, l-k has to be an odd number for w_l,k.

The generators of ^> satisfy the following relations.


    [[baseline=(current bounding box.center),scale=0.75]
        [->](0,0) to (1,2);
        [->](1,0) to (0,2);
        [fill](0.25,1.5) circle[radius=3pt];
    
    ,[baseline=(current bounding box.center),scale=0.75]
    [<-](0,0) to (0,2);
    ]
    =-4
    [baseline=(current bounding box.center),scale=0.75]
    [->](0,0) to (0,2);
    and[[baseline=(current bounding box.center),scale=0.75]
        [->](0,0) to (1,2);
        [->](1,0) to (0,2);
        [fill](0.25,1.5) circle[radius=3pt];
    
    ,[baseline=(current bounding box.center),scale=0.75]
    [->](0,0) to (0,2);
    ]
    =2
    [baseline=(current bounding box.center),scale=0.75]
    [->] (2,0) .. controls (2,1.25) and (0,.25) .. (0,2);
    [->] (0,0) .. controls (0,1) and (.8,.8) .. (1,2);
    [->] (1,0) .. controls (1,1) and (1.8,.8) .. (2,2);



    [[baseline=(current bounding box.center),scale=0.75]
        [->](0,0) to (1,2);
        [->](1,0) to (0,2);
        [fill](0.75,0.5) circle[radius=3pt];
    
    ,[baseline=(current bounding box.center),scale=0.75]
    [<-](0,0) to (0,2);
    ]
    =0
    and[[baseline=(current bounding box.center),scale=0.75]
        [->](0,0) to (1,2);
        [->](1,0) to (0,2);
        [fill](0.25,1.5) circle[radius=3pt];
    
    ,[baseline=(current bounding box.center),scale=0.75]
    [->](0,0) to (0,2);
    ]
    =2
    [baseline=(current bounding box.center),scale=0.75]
    [->] (2,0) .. controls (2,1.25) and (0,.25) .. (0,2);
    [->] (0,0) .. controls (0,1) and (.8,.8) .. (1,2);
    [->] (1,0) .. controls (1,1) and (1.8,.8) .. (2,2);




    [[baseline=(current bounding box.center),scale=0.75]
        [->](0,0) to (0,2);
        [->](.5,0) to (.5,2);
        
    ,[baseline=(current bounding box.center),scale=0.75]
    [<-](0,0) to (0,2);
    ]
    =-4
    [baseline=(current bounding box.center),scale=0.75]
    [->](0,0) to (0,2);
    and[[baseline=(current bounding box.center),scale=0.75]
       [->](0,0) to (0,2);
        [->](.5,0) to (.5,2);
    
    ,[baseline=(current bounding box.center),scale=0.75]
    [->](0,0) to (0,2);
    ]
    =0




 For m,n∈ℤ with mn>0, we have  
  * [2m x_1, 2n x_1] = 2(n-m) 2n+2mx_1.

  * [mc_1, nc_1] = -2nc_1.


 Part (1) is a slight modification of <cit.>. By Proposition <ref>, if at least one of the indices inside the commutator is odd, the commutator will be zero. Hence we will work with the case where both indices are even numbers. The modification we need in <cit.> is a result of us having two resolution terms in our relations (<ref>) and (<ref>). As a consequence of having even number of strands in both of our elements, canceling the two empty dots in our resolution terms give rise to the same sign as the other resolution term, hence we have a coefficient of two in front of our result.

Part (2) follows easily the proof of <cit.> since moving an empty dot through a crossing is for free in , and we get a negative sign from changing relative heights of hollow dots.


  For n≥0, we have 
    [± 2n(x_1 + … + x_2n),  1] = ± 4n ±(2n+1).


 First note that we have: 


    [[baseline=(current bounding box.center),scale=0.75]
        [->](0,0) to (1,2);
        [->](1,0) to (0,2);
        [fill](0.25,1.5) circle[radius=3pt];
         [->](1.5,0) to (1.5,2);
    ]   =
    [[baseline=(current bounding box.center),scale=0.75]
     [->](0,0) to (1.5,2);
     [->](1.5,0) to (0,2);
     [fill](.25,1.63) circle[radius=3pt];
    [->](0.75,0) to [out=135,in=225] (0.75,2);
    ] 
       = 
    [[baseline=(current bounding box.center),scale=0.75]
     [->](0,0) to (1.5,2);
     [->](1.5,0) to (0,2);
     [fill](.65,1.2) circle[radius=3pt];
    [->](0.75,0) to [out=135,in=225] (0.75,2);
    ]
     +
     [[baseline=(current bounding box.center),scale=0.75]
    [->] (2,0) .. controls (2,1.25) and (0,.25) .. (0,2);
    [->] (0,0) .. controls (0,1) and (.8,.8) .. (1,2);
    [->] (1,0) .. controls (1,1) and (1.8,.8) .. (2,2);
    ]
    -
    [[baseline=(current bounding box.center),scale=0.75]
    [->] (2,0) .. controls (2,1.25) and (0,.25) .. (0,2);
    [->] (0,0) .. controls (0,1) and (.8,.8) .. (1,2);
    [->] (1,0) .. controls (1,1) and (1.8,.8) .. (2,2);
     (.08,1.5) circle[radius=3pt];
     (.67, 1.2) circle[radius=3pt];
    ]
       =
    [[baseline=(current bounding box.center),scale=0.75]
        [->](0,0) to (1,2);
        [->](1,0) to (0,2);
        [fill](0.25,1.5) circle[radius=3pt];
         [->](-.5,0) to (-.5,2);
    ]
    +
    2
    [[baseline=(current bounding box.center),scale=0.75]
    [->] (2,0) .. controls (2,1.25) and (0,.25) .. (0,2);
    [->] (0,0) .. controls (0,1) and (.8,.8) .. (1,2);
    [->] (1,0) .. controls (1,1) and (1.8,.8) .. (2,2);
    ].


Hence [2x_1, 1] = 2 3. 

Next, moving the solid dot in 2x_2 around to the bottom of the crossing using the trace relation gives:


    [[baseline=(current bounding box.center),scale=0.75]
        [->](0,0) to (1,2);
        [->](1,0) to (0,2);
        [fill](.75,.5) circle[radius=3pt];
         [->](1.5,0) to (1.5,2);
    ]   =
    [[baseline=(current bounding box.center),scale=0.75]
     [->](0,0) to (1.5,2);
     [->](1.5,0) to (0,2);
    [->](0.75,0) to [out=45,in=-45] (0.75,2);
     [fill](.9,.75) circle[radius=3pt];
    ] 
       = 
    [[baseline=(current bounding box.center),scale=0.75]
     [->](0,0) to (1.5,2);
     [->](1.5,0) to (0,2);
    [->](0.75,0) to [out=45,in=-45] (0.75,2);
     [fill](1.3,.25) circle[radius=3pt];
    ]
     +
     [[baseline=(current bounding box.center),scale=0.75]
    [->] (2,0) .. controls (2,1.25) and (0,.25) .. (0,2);
    [->] (0,0) .. controls (0,1) and (.8,.8) .. (1,2);
    [->] (1,0) .. controls (1,1) and (1.8,.8) .. (2,2);
    ]
    -
    [[baseline=(current bounding box.center),scale=0.75]
    [->] (2,0) .. controls (2,1.25) and (0,.25) .. (0,2);
    [->] (0,0) .. controls (0,1) and (.8,.8) .. (1,2);
    [->] (1,0) .. controls (1,1) and (1.8,.8) .. (2,2);
     (1.77,.5) circle[radius=3pt];
     (1.05, .3) circle[radius=3pt];
    ]
       =
    [[baseline=(current bounding box.center),scale=0.75]
        [->](0,0) to (1,2);
        [->](1,0) to (0,2);
        [fill](.75,.5) circle[radius=3pt];
         [->](-.5,0) to (-.5,2);
    ]
    + 
    2[[baseline=(current bounding box.center),scale=0.75]
    [->] (2,0) .. controls (2,1.25) and (0,.25) .. (0,2);
    [->] (0,0) .. controls (0,1) and (.8,.8) .. (1,2);
    [->] (1,0) .. controls (1,1) and (1.8,.8) .. (2,2);
    ].

So, [2(x_1 + x_2), 1 ] = 43. 

Next, we claim that [2nx_2n, 1] = 2 2n+1 for any n. Indeed, we have:

    [[baseline=(current bounding box.center),scale=0.75]
      [->] (3.2,0) .. controls (3.2,1.25) and (0,.25) .. (0,2)
         node[pos=0.85, shape=coordinate](X);
      [->] (0,0) .. controls (0,1) and (.8,.8) .. (.8,2);
      [->] (.8,0) .. controls (.8,1) and (1.6,.8) .. (1.6,2);
      [->] (2.4,0) .. controls (2.4,1) and (3.2,.8) .. (3.2,2);
       at (1.6,.35) …;
       at (2.4,1.65) …;
        (3.18,.2) circle (3pt);
    [->](3.5,0) to (3.5,2);
    ]   =
    [[baseline=(current bounding box.center),scale=0.75]
      [->] (3.2,0) .. controls (3.2,1.25) and (0,.25) .. (0,2)
         node[pos=0.85, shape=coordinate](X);
      [->] (0,0) .. controls (0,1) and (.8,.8) .. (.8,2);
      [->] (.8,0) .. controls (.8,1) and (1.6,.8) .. (1.6,2);
      [->] (2.4,0) .. controls (2.4,1) and (3.2,.8) .. (3.2,2);
       at (1.6,.35) …;
       at (2.4,1.65) …;
        (2.8,.55) circle (3pt);
    [->](3,0) ..controls (3.2,1) and (3,.8)..(2.8,2);
    ] 
       =
    [[baseline=(current bounding box.center),scale=0.75]
      [->] (3.2,0) .. controls (3.2,1.25) and (0,.25) .. (0,2)
         node[pos=0.85, shape=coordinate](X);
      [->] (0,0) .. controls (0,1) and (.8,.8) .. (.8,2);
      [->] (.8,0) .. controls (.8,1) and (1.6,.8) .. (1.6,2);
      [->] (2.4,0) .. controls (2.4,1) and (3.2,.8) .. (3.2,2);
       at (1.6,.35) …;
       at (2.4,1.65) …;
        (3.18,.2) circle (3pt);
    [->](2.8,0) ..controls (3,1) and (2.8,.8)..(2.6,2);
    ]
    +
    [[baseline=(current bounding box.center),scale=0.75]
      [->] (3,0) .. controls (3,1.25) and (0,.25) .. (0,2)
         node[pos=0.85, shape=coordinate](X);
      [->] (0,0) .. controls (0,1) and (.8,.8) .. (.8,2);
      [->] (.8,0) .. controls (.8,1) and (1.6,.8) .. (1.6,2);
      [->] (2.4,0) .. controls (2.4,1) and (3.2,.8) .. (3.2,2);
       at (1.6,.35) …;
       at (2.4,1.65) …;
      
    [->](3.4,0) ..controls (3.6,1) and (3,.8)..(2.8,2);
    ]
       -
    [[baseline=(current bounding box.center),scale=0.75]
      [->] (3,0) .. controls (3,1.25) and (0,.25) .. (0,2)
         node[pos=0.85, shape=coordinate](X);
      [->] (0,0) .. controls (0,1) and (.8,.8) .. (.8,2);
      [->] (.8,0) .. controls (.8,1) and (1.6,.8) .. (1.6,2);
      [->] (2.4,0) .. controls (2.4,1) and (3.2,.8) .. (3.2,2);
       at (1.6,.35) …;
       at (2.4,1.65) …;
       (2.85,.4) circle (3pt);
     (3.43, .2) circle (3pt);
    [->](3.4,0) ..controls (3.6,1) and (3,.8)..(2.8,2);
    ]
       =
    [[baseline=(current bounding box.center),scale=0.75]
      [->] (3.2,0) .. controls (3.2,1.25) and (0,.25) .. (0,2)
         node[pos=0.85, shape=coordinate](X);
      [->] (0,0) .. controls (0,1) and (.8,.8) .. (.8,2);
      [->] (.8,0) .. controls (.8,1) and (1.6,.8) .. (1.6,2);
      [->] (2.4,0) .. controls (2.4,1) and (3.2,.8) .. (3.2,2);
       at (1.6,.35) …;
       at (2.4,1.65) …;
        (3.18,.2) circle (3pt);
    [->](-.3,0) to (-.3,2);
    ]
    +
    2
    [[baseline=(current bounding box.center),scale=0.75]
      [->] (3.2,0) .. controls (3.2,1.25) and (0,.25) .. (0,2)
         node[pos=0.85, shape=coordinate](X);
      [->] (0,0) .. controls (0,1) and (.8,.8) .. (.8,2);
      [->] (.8,0) .. controls (.8,1) and (1.6,.8) .. (1.6,2);
      [->] (2.4,0) .. controls (2.4,1) and (3.2,.8) .. (3.2,2);
       at (1.6,.35) …;
       at (2.4,1.65) …;
      
      [->] (2.8,0) .. controls (2.8,1) and (3.6,.8) .. (3.6,2);
    
    ],

where the last equality is obtained by pushing the crossings at the bottom of the diagrams without dots to the top. Indeed, diagrammatic calculations similar to the above give that 
    [2n x_a, 1] = 22n+1
 for any 1 < a ≤ 2n.

Finally, note that 
    2nx_11 = [[baseline=(current bounding box.center),scale=0.75]
      [->] (3.2,0) .. controls (3.2,1.25) and (0,.25) .. (0,2)
         node[pos=0.85, shape=coordinate](X);
      [->] (0,0) .. controls (0,1) and (.8,.8) .. (.8,2);
      [->] (.8,0) .. controls (.8,1) and (1.6,.8) .. (1.6,2);
      [->] (2.4,0) .. controls (2.4,1) and (3.2,.8) .. (3.2,2);
       at (1.6,.35) …;
       at (2.4,1.65) …;
        (.1,1.55) circle (3pt);
    [->](.4,0) ..controls (.4,1) and (.2,.8)..(.4,2);
    ].

The dot will slide over the top-leftmost crossing in the same manner as in Equation (<ref>), meaning the correction terms will cancel out. Hence, we have the desired result.
 

 Let m be an odd integer. We have 
    [ 2(x_1+ x_2), m] = 4mm+2.
 


We compute directly: 


    m2x_1   = [    [scale=0.8]
     
     
      [->] (3,0) .. controls ++(0,1.25) and ++(0,-1.75) .. (0,2);
      [->] (0,0) .. controls ++(0,1) and ++(0,-1.2) .. (.6,2);
      [->] (.6,0) .. controls ++(0,1) and ++(0,-1.2) .. (1.2,2);
      [->] (1.8,0) .. controls ++(0,1) and ++(0,-1.2) .. (2.4,2);
      [->] (2.4,0) .. controls ++(0,1) and ++(0,-1.2) .. (3,2);
       at (1.2,.35) …;
       at (1.8,1.65) …;
      
      
      [->] (3.4,0)–(4.5,2);
      [->] (4.5,0)–(3.4,2);
         (3.7,1.4) circle (2pt);
       ]
       =
    [    [scale=0.8]
     
     
       (3,0) .. controls (3,1.25) and (0,.25) .. (0,1.5)
         node[pos=0.85, shape=coordinate](X);
       (0,0) .. controls (0,1) and (.6,.8) .. (.6,1.5);
       (.6,0) .. controls (.6,1) and (1.2,.8) .. (1.2,1.5);
       (1.8,0) .. controls (1.8,1) and (2.4,.8) .. (2.4,1.5);
       (2.4,0) .. controls (2.4,1) and (3,.8) .. (3,1.5);
       at (1.2,.35) …;
       at (1.8,1.15) …;
      
      
       (4.2,0) .. controls (4.2,1.25) and (3.6,.25) .. (3.6,1.5)
         node[pos=0.85, shape=coordinate](X);
       (3.6,0) .. controls (3.6,1) and (4.2,.8) .. (4.2,1.5);
      
      
        [blue, dotted] (-0.4,0) – (4.4,0);
        [blue, dotted] (-0.4,1.5) – (4.4,1.5);
      
       [->]  (3,1.5) .. controls (3,2) and (3.6,2) .. (3.6,2.5);
       [->] (3.6,1.5) .. controls (3.6,2.5) and (0,1.5) .. (0,2.5);
       [->]  (2.4,1.5) .. controls (2.4,2) and (3,2) .. (3,2.5);
       [->]  (1.2,1.5) .. controls (1.2,2) and (1.8,2) .. (1.8,2.5);
       [->]  (.6,1.5) .. controls (.6,2) and (1.2,2) .. (1.2,2.5);
       [->]  (0,1.5) .. controls (0,2) and (.6,2) .. (.6,2.5);
       [->] (4.2,1.5) – (4.2,2.5);
           (3.6,1.5) circle (2pt);
      
         (2.4,0) .. controls (2.4,-.5) and (3,-.5) .. (3,-1);
         (1.8,0) .. controls (1.8,-.5) and (2.4,-.5) .. (2.4,-1);
         (.6,0) .. controls (.6,-.5) and (1.2,-.5) .. (1.2,-1);
         (0,0) .. controls (0,-.5) and (.6,-.5) .. (.6,-1);
         (3,0) .. controls (3,-.5) and (3.6,-.5) .. (3.6,-1);
       (3.6,0) .. controls (3.6,-1) and (0,0) .. (0,-1);
        (4.2,0) – (4.2,-1);
    ]
    
       =
    [    [scale=0.8]
     
     
       (3,0) .. controls (3,1.25) and (0,.25) .. (0,1.5)
         node[pos=0.85, shape=coordinate](X);
       (0,0) .. controls (0,1) and (.6,.8) .. (.6,1.5);
       (.6,0) .. controls (.6,1) and (1.2,.8) .. (1.2,1.5);
       (1.8,0) .. controls (1.8,1) and (2.4,.8) .. (2.4,1.5);
       (2.4,0) .. controls (2.4,1) and (3,.8) .. (3,1.5);
       at (1.2,.35) …;
       at (1.8,1.15) …;
      
      
       (4.2,0) .. controls (4.2,1.25) and (3.6,.25) .. (3.6,1.5)
         node[pos=0.85, shape=coordinate](X);
       (3.6,0) .. controls (3.6,1) and (4.2,.8) .. (4.2,1.5);
      
      
        [blue, dotted] (-0.4,0) – (4.4,0);
        [blue, dotted] (-0.4,1.5) – (4.4,1.5);
      
       [->]  (3,1.5) .. controls (3,2) and (3.6,2) .. (3.6,2.5);
       [->] (3.6,1.5) .. controls (3.6,2.5) and (0,1.5) .. (0,2.5);
       [->]  (2.4,1.5) .. controls (2.4,2) and (3,2) .. (3,2.5);
       [->]  (1.2,1.5) .. controls (1.2,2) and (1.8,2) .. (1.8,2.5);
       [->]  (.6,1.5) .. controls (.6,2) and (1.2,2) .. (1.2,2.5);
       [->]  (0,1.5) .. controls (0,2) and (.6,2) .. (.6,2.5);
       [->] (4.2,1.5) – (4.2,2.5);
           (3.0,1.95) circle (2pt);
      
         (2.4,0) .. controls (2.4,-.5) and (3,-.5) .. (3,-1);
         (1.8,0) .. controls (1.8,-.5) and (2.4,-.5) .. (2.4,-1);
         (.6,0) .. controls (.6,-.5) and (1.2,-.5) .. (1.2,-1);
         (0,0) .. controls (0,-.5) and (.6,-.5) .. (.6,-1);
         (3,0) .. controls (3,-.5) and (3.6,-.5) .. (3.6,-1);
       (3.6,0) .. controls (3.6,-1) and (0,0) .. (0,-1);
        (4.2,0) – (4.2,-1);
    ]
     
    -
    [    [scale=0.8]
     
     
       (3,0) .. controls (3,1.25) and (0,.25) .. (0,1.5)
         node[pos=0.85, shape=coordinate](X);
       (0,0) .. controls (0,1) and (.6,.8) .. (.6,1.5);
       (.6,0) .. controls (.6,1) and (1.2,.8) .. (1.2,1.5);
       (1.8,0) .. controls (1.8,1) and (2.4,.8) .. (2.4,1.5);
       (2.4,0) .. controls (2.4,1) and (3,.8) .. (3,1.5);
       at (1.2,.35) …;
       at (1.8,1.15) …;
      
      
       (4.2,0) .. controls (4.2,1.25) and (3.6,.25) .. (3.6,1.5)
         node[pos=0.85, shape=coordinate](X);
       (3.6,0) .. controls (3.6,1) and (4.2,.8) .. (4.2,1.5);
      
      
        [blue, dotted] (-0.4,0) – (4.4,0);
        [blue, dotted] (-0.4,1.5) – (4.4,1.5);
      
       [->]  (3,1.5) .. controls (3,2) and (0,2) .. (0,2.5);
       [->] (3.6,1.5)–(3.6,2.5);
       [->]  (2.4,1.5) .. controls (2.4,2) and (3,2) .. (3,2.5);
       [->]  (1.2,1.5) .. controls (1.2,2) and (1.8,2) .. (1.8,2.5);
       [->]  (.6,1.5) .. controls (.6,2) and (1.2,2) .. (1.2,2.5);
       [->]  (0,1.5) .. controls (0,2) and (.6,2) .. (.6,2.5);
       [->] (4.2,1.5) – (4.2,2.5);
        
      
         (2.4,0) .. controls (2.4,-.5) and (3,-.5) .. (3,-1);
         (1.8,0) .. controls (1.8,-.5) and (2.4,-.5) .. (2.4,-1);
         (.6,0) .. controls (.6,-.5) and (1.2,-.5) .. (1.2,-1);
         (0,0) .. controls (0,-.5) and (.6,-.5) .. (.6,-1);
         (3,0) .. controls (3,-.5) and (3.6,-.5) .. (3.6,-1);
       (3.6,0) .. controls (3.6,-1) and (0,0) .. (0,-1);
        (4.2,0) – (4.2,-1);
    ]
       +
     [    [scale=0.8]
     
     
       (3,0) .. controls (3,1.25) and (0,.25) .. (0,1.5)
         node[pos=0.85, shape=coordinate](X);
       (0,0) .. controls (0,1) and (.6,.8) .. (.6,1.5);
       (.6,0) .. controls (.6,1) and (1.2,.8) .. (1.2,1.5);
       (1.8,0) .. controls (1.8,1) and (2.4,.8) .. (2.4,1.5);
       (2.4,0) .. controls (2.4,1) and (3,.8) .. (3,1.5);
       at (1.2,.35) …;
       at (1.8,1.15) …;
      
      
       (4.2,0) .. controls (4.2,1.25) and (3.6,.25) .. (3.6,1.5)
         node[pos=0.85, shape=coordinate](X);
       (3.6,0) .. controls (3.6,1) and (4.2,.8) .. (4.2,1.5);
     (3.6, 1.2) circle (3pt);
      
      
        [blue, dotted] (-0.4,0) – (4.4,0);
        [blue, dotted] (-0.4,1.5) – (4.4,1.5);
      
       [->]  (3,1.5) .. controls (3,2) and (0,2) .. (0,2.5);
       [->] (3.6,1.5)–(3.6,2.5);
       [->]  (2.4,1.5) .. controls (2.4,2) and (3,2) .. (3,2.5);
       [->]  (1.2,1.5) .. controls (1.2,2) and (1.8,2) .. (1.8,2.5);
       [->]  (.6,1.5) .. controls (.6,2) and (1.2,2) .. (1.2,2.5);
       [->]  (0,1.5) .. controls (0,2) and (.6,2) .. (.6,2.5);
       [->] (4.2,1.5) – (4.2,2.5);
     (2.8,1.7) circle (3pt);
        
      
         (2.4,0) .. controls (2.4,-.5) and (3,-.5) .. (3,-1);
         (1.8,0) .. controls (1.8,-.5) and (2.4,-.5) .. (2.4,-1);
         (.6,0) .. controls (.6,-.5) and (1.2,-.5) .. (1.2,-1);
         (0,0) .. controls (0,-.5) and (.6,-.5) .. (.6,-1);
         (3,0) .. controls (3,-.5) and (3.6,-.5) .. (3.6,-1);
       (3.6,0) .. controls (3.6,-1) and (0,0) .. (0,-1);
        (4.2,0) – (4.2,-1);
    ].


Cancelling the empty dots in the last term results in a change in sign, and both of the latter diagrams are (m+2)-cycles. Hence we have:

    =[    [scale=0.8]
     
     
       (3,0) .. controls (3,1.25) and (0,.25) .. (0,1.5)
         node[pos=0.85, shape=coordinate](X);
       (0,0) .. controls (0,1) and (.6,.8) .. (.6,1.5);
       (.6,0) .. controls (.6,1) and (1.2,.8) .. (1.2,1.5);
       (1.8,0) .. controls (1.8,1) and (2.4,.8) .. (2.4,1.5);
       (2.4,0) .. controls (2.4,1) and (3,.8) .. (3,1.5);
       at (1.2,.35) …;
       at (1.8,1.15) …;
      
      
       (4.2,0) .. controls (4.2,1.25) and (3.6,.25) .. (3.6,1.5)
         node[pos=0.85, shape=coordinate](X);
       (3.6,0) .. controls (3.6,1) and (4.2,.8) .. (4.2,1.5);
      
      
        [blue, dotted] (-0.4,0) – (4.4,0);
        [blue, dotted] (-0.4,1.5) – (4.4,1.5);
      
       [->]  (3,1.5) .. controls (3,2) and (3.6,2) .. (3.6,2.5);
       [->] (3.6,1.5) .. controls (3.6,2.5) and (0,1.5) .. (0,2.5);
       [->]  (2.4,1.5) .. controls (2.4,2) and (3,2) .. (3,2.5);
       [->]  (1.2,1.5) .. controls (1.2,2) and (1.8,2) .. (1.8,2.5);
       [->]  (.6,1.5) .. controls (.6,2) and (1.2,2) .. (1.2,2.5);
       [->]  (0,1.5) .. controls (0,2) and (.6,2) .. (.6,2.5);
       [->] (4.2,1.5) – (4.2,2.5);
           (3.0,1.95) circle (2pt);
      
         (2.4,0) .. controls (2.4,-.5) and (3,-.5) .. (3,-1);
         (1.8,0) .. controls (1.8,-.5) and (2.4,-.5) .. (2.4,-1);
         (.6,0) .. controls (.6,-.5) and (1.2,-.5) .. (1.2,-1);
         (0,0) .. controls (0,-.5) and (.6,-.5) .. (.6,-1);
         (3,0) .. controls (3,-.5) and (3.6,-.5) .. (3.6,-1);
       (3.6,0) .. controls (3.6,-1) and (0,0) .. (0,-1);
        (4.2,0) – (4.2,-1);
    ]
     - 2m+2


Sliding the solid dot in the first diagram all the way to the left results in m total crossing resolutions, each of which yieds a term of -2 m+2. So,

    =[    [scale=0.8]
     
     
       (3,0) .. controls (3,1.25) and (0,.25) .. (0,1.5)
         node[pos=0.85, shape=coordinate](X);
       (0,0) .. controls (0,1) and (.6,.8) .. (.6,1.5);
       (.6,0) .. controls (.6,1) and (1.2,.8) .. (1.2,1.5);
       (1.8,0) .. controls (1.8,1) and (2.4,.8) .. (2.4,1.5);
       (2.4,0) .. controls (2.4,1) and (3,.8) .. (3,1.5);
       at (1.2,.35) …;
       at (1.8,1.15) …;
      
      
       (4.2,0) .. controls (4.2,1.25) and (3.6,.25) .. (3.6,1.5)
         node[pos=0.85, shape=coordinate](X);
       (3.6,0) .. controls (3.6,1) and (4.2,.8) .. (4.2,1.5);
      
      
        [blue, dotted] (-0.4,0) – (4.4,0);
        [blue, dotted] (-0.4,1.5) – (4.4,1.5);
      
       [->]  (3,1.5) .. controls (3,2) and (3.6,2) .. (3.6,2.5);
       [->] (3.6,1.5) .. controls (3.6,2.5) and (0,1.5) .. (0,2.5);
       [->]  (2.4,1.5) .. controls (2.4,2) and (3,2) .. (3,2.5);
       [->]  (1.2,1.5) .. controls (1.2,2) and (1.8,2) .. (1.8,2.5);
       [->]  (.6,1.5) .. controls (.6,2) and (1.2,2) .. (1.2,2.5);
       [->]  (0,1.5) .. controls (0,2) and (.6,2) .. (.6,2.5);
       [->] (4.2,1.5) – (4.2,2.5);
           (.15,2.2) circle (2pt);
      
         (2.4,0) .. controls (2.4,-.5) and (3,-.5) .. (3,-1);
         (1.8,0) .. controls (1.8,-.5) and (2.4,-.5) .. (2.4,-1);
         (.6,0) .. controls (.6,-.5) and (1.2,-.5) .. (1.2,-1);
         (0,0) .. controls (0,-.5) and (.6,-.5) .. (.6,-1);
         (3,0) .. controls (3,-.5) and (3.6,-.5) .. (3.6,-1);
       (3.6,0) .. controls (3.6,-1) and (0,0) .. (0,-1);
        (4.2,0) – (4.2,-1);
    ]
         - 2mm+2
        
    =  [    [baseline=(current bounding box.center),scale=.8]
      
      
      [->] (0,0)–(1.2,2);
      [->] (1.2,0)–(0,2);
         (.35,1.4) circle (2pt);
       [baseline=(current bounding box.center),scale=0.8]
     
     
      [->] (3,0) .. controls ++(0,1.25) and ++(0,-1.75) .. (0,2);
      [->] (0,0) .. controls ++(0,1) and ++(0,-1.2) .. (.6,2);
      [->] (.6,0) .. controls ++(0,1) and ++(0,-1.2) .. (1.2,2);
      [->] (1.8,0) .. controls ++(0,1) and ++(0,-1.2) .. (2.4,2);
      [->] (2.4,0) .. controls ++(0,1) and ++(0,-1.2) .. (3,2);
       at (1.2,.35) …;
       at (1.8,1.65) …;
       ]
        - 2mm+2

Hence we have 
    [2x_1, m] = 2mm+2.
 A similar computation gives that 
    [2x_2, m] = 2mm+2,
 giving the desired result.


 We have 
    [2n(x_1+x_2 +…+ x_2n), -(2m+1)] = {[ -4(2m+1) 2n-2m-1       if  n>m≥ 1;                0       if  n=m≥ 1; -2(2m+1) 2n-2m-1    if  1≤ n < m. ].

 We follow the methods of <cit.>, substituting our new relations as necessary.

As in that case, let β_n = 2nx_1 and α_m = 2m+1x_1, and proceed by induction on m. When m=1, we can compute directly:

    [    [scale=0.8]
      [->] (3,0) .. controls ++(0,1.25) and (0,.25) .. (0,2);
      [->] (0.0,0) .. controls ++(0,1) and ++(0,-1.2) .. (0.6,2);
      [->] (0.6,0) .. controls ++(0,1) and ++(0,-1.2) .. (1.2,2);
      [->] (1.2,0) .. controls ++(0,1) and ++(0,-1.2) .. (1.8,2);
      [->] (2.4,0) .. controls ++(0,1) and ++(0,-1.2) .. (3,2);
       at (1.8,.35) …;
       at (2.4,1.65) …;
      [<-] (-0.6,0) – (-0.6,2);
        (.05,1.6) circle (2pt);
      ]
         (<ref>)  [    [scale=0.8]
       (3.0,0) .. controls ++(0,1.25) and ++(0,-1.1) .. (0,1.5);
       (0.0,0) .. controls ++(0,.5) and ++(0,-.6) .. (0.6,1.5);
       (0.6,0) .. controls ++(0,.5) and ++(0,-.6) .. (1.2,1.5);
       (1.2,0) .. controls ++(0,.5) and ++(0,-.6) .. (1.8,1.5);
       (2.4,0) .. controls ++(0,.5) and ++(0,-.6) .. (3,1.5);
       at (1.8,.35) …;
       at (2.4,1.35) …;
      [<-] (-0.6,0) – (-0.6,1.5);
        (.1,1.2) circle (2pt);
        
      
        [blue, dotted] (-1,0) – (3.4,0);
        [blue, dotted] (-1,1.5) – (3.4,1.5);
      
       [->] (0.0,1.5) .. controls ++(0,.45) and ++(0,-.6) .. (-0.6,2.5);
        (-0.6,1.5) .. controls ++(0,.45) and ++(0,-.6) .. (0.0,2.5);
       [->] (0.6,1.5) –  (0.6,2.5);
       [->] (1.2,1.5) –  (1.2,2.5);
       [->] (1.8,1.5) –  (1.8,2.5);
       [->] (3,1.5) –  (3,2.5);
       
       [<-] (0.0,0) .. controls ++(0,-.45) and ++(0,.6) .. (-0.6,-1);
       [->] (-0.6,0) .. controls ++(0,-.45) and ++(0,.6) .. (0.0,-1);
       [<-] (0.6,0) –  (0.6,-1);
       [<-] (1.2,0) –  (1.2,-1);
       [<-] (2.4,0) –  (2.4,-1);
       [<-] (3.0,0) –  (3,-1);
      ]
    
       +2   [    [scale=0.8]
       (3.0,0) .. controls ++(0,1.25) and ++(0,-1.1) .. (0,1.5);
       (0.0,0) .. controls ++(0,.5) and ++(0,-.6) .. (0.6,1.5);
       (0.6,0) .. controls ++(0,.5) and ++(0,-.6) .. (1.2,1.5);
       (1.2,0) .. controls ++(0,.5) and ++(0,-.6) .. (1.8,1.5);
       (2.4,0) .. controls ++(0,.5) and ++(0,-.6) .. (3,1.5);
       at (1.8,.35) …;
       at (2.4,1.35) …;
       (-0.6,0) – (-0.6,1.5);
        (.1,1.2) circle (2pt);
        
      
        [blue, dotted] (-1,0) – (3.4,0);
        [blue, dotted] (-1,1.5) – (3.4,1.5);
      
      
        (-0.6,1.5) .. controls ++(0,.35) and ++(0,.35) .. (0.0,1.5);
       [->] (0.6,1.5) –  (0.6,2.5);
       [->] (1.2,1.5) –  (1.2,2.5);
       [->] (1.8,1.5) –  (1.8,2.5);
       [->] (3,1.5) –  (3,2.5);
       
       
       [->] (-0.6,0) .. controls ++(0,-.35) and ++(0,-.35) .. (0.0,0);
       [<-] (0.6,0) –  (0.6,-1);
       [<-] (1.2,0) –  (1.2,-1);
       [<-] (2.4,0) –  (2.4,-1);
       [<-] (3.0,0) –  (3,-1);
      ]

where the trailing terms arising from relation (<ref>) have the same sign after cancelling the empty dots, and thus add together. We claim that the diagram in the second term is 2n-1. Indeed, sliding the dot gives:

    [    [scale=0.8]
       (3.0,0) .. controls ++(0,1.25) and ++(0,-1.1) .. (0,1.5);
       (0.0,0) .. controls ++(0,.5) and ++(0,-.6) .. (0.6,1.5);
       (0.6,0) .. controls ++(0,.5) and ++(0,-.6) .. (1.2,1.5);
       (1.2,0) .. controls ++(0,.5) and ++(0,-.6) .. (1.8,1.5);
       (2.4,0) .. controls ++(0,.5) and ++(0,-.6) .. (3,1.5);
       at (1.8,.35) …;
       at (2.4,1.35) …;
       (-0.6,0) – (-0.6,1.5);
        (.3,.65) circle (2pt);
        
      
        [blue, dotted] (-1,0) – (3.4,0);
        [blue, dotted] (-1,1.5) – (3.4,1.5);
      
      
       (-0.6,1.5) .. controls ++(0,.35) and ++(0,.35) .. (0.0,1.5);
       [->] (0.6,1.5) –  (0.6,2.5);
       [->] (1.2,1.5) –  (1.2,2.5);
       [->] (1.8,1.5) –  (1.8,2.5);
       [->] (3,1.5) –  (3,2.5);
       
       
       [->] (-0.6,0) .. controls ++(0,-.35) and ++(0,-.35) .. (0.0,0);
       [<-] (0.6,0) –  (0.6,-1);
       [<-] (1.2,0) –  (1.2,-1);
       [<-] (2.4,0) –  (2.4,-1);
       [<-] (3.0,0) –  (3,-1);
      ]      (<ref>)  d_0,02n-1 + d_0,12n-1
    = 2n-1

by relations (<ref>) and (<ref>).

 Now, sliding the solid dot over the crossing on the right hand side of Equation (<ref>) gives:


    (<ref>)  [    [scale=0.8]
       (3.0,0) .. controls ++(0,1.25) and ++(0,-1.1) .. (0,1.5);
       (0.0,0) .. controls ++(0,.5) and ++(0,-.6) .. (0.6,1.5);
       (0.6,0) .. controls ++(0,.5) and ++(0,-.6) .. (1.2,1.5);
       (1.2,0) .. controls ++(0,.5) and ++(0,-.6) .. (1.8,1.5);
       (2.4,0) .. controls ++(0,.5) and ++(0,-.6) .. (3,1.5);
       at (1.8,.35) …;
       at (2.4,1.35) …;
      [<-] (-0.6,0) – (-0.6,1.5);
        (-0.55,2.2) circle (2pt);
        
      
        [blue, dotted] (-1,0) – (3.4,0);
        [blue, dotted] (-1,1.5) – (3.4,1.5);
      
       [->] (0.0,1.5) .. controls ++(0,.45) and ++(0,-.6) .. (-0.6,2.5);
        (-0.6,1.5) .. controls ++(0,.45) and ++(0,-.6) .. (0.0,2.5);
       [->] (0.6,1.5) –  (0.6,2.5);
       [->] (1.2,1.5) –  (1.2,2.5);
       [->] (1.8,1.5) –  (1.8,2.5);
       [->] (3,1.5) –  (3,2.5);
       
       [<-] (0.0,0) .. controls ++(0,-.45) and ++(0,.6) .. (-0.6,-1);
       [->] (-0.6,0) .. controls ++(0,-.45) and ++(0,.6) .. (0.0,-1);
       [<-] (0.6,0) –  (0.6,-1);
       [<-] (1.2,0) –  (1.2,-1);
       [<-] (2.4,0) –  (2.4,-1);
       [<-] (3.0,0) –  (3,-1);
      ]
    +2   [    [scale=0.8]
       (3.0,0) .. controls ++(0,1.25) and ++(0,-1.1) .. (0,1.5);
       (0.0,0) .. controls ++(0,.5) and ++(0,-.6) .. (0.6,1.5);
       (0.6,0) .. controls ++(0,.5) and ++(0,-.6) .. (1.2,1.5);
       (1.2,0) .. controls ++(0,.5) and ++(0,-.6) .. (1.8,1.5);
       (2.4,0) .. controls ++(0,.5) and ++(0,-.6) .. (3,1.5);
       at (1.8,.35) …;
       at (2.4,1.35) …;
      [<-] (-0.6,0) – (-0.6,1.5);
        
      
        [blue, dotted] (-1,0) – (3.4,0);
        [blue, dotted] (-1,1.5) – (3.4,1.5);
      
    
    
       [->] (0.0,2.5) arc (360:180:3mm);
       [->] (0.0,1.5) arc (0:180:3mm);
       [->] (0.6,1.5) –  (0.6,2.5);
       [->] (1.2,1.5) –  (1.2,2.5);
       [->] (1.8,1.5) –  (1.8,2.5);
       [->] (3,1.5) –  (3,2.5);
       
       [<-] (0.0,0) .. controls ++(0,-.45) and ++(0,.6) .. (-0.6,-1);
       [->] (-0.6,0) .. controls ++(0,-.45) and ++(0,.6) .. (0.0,-1);
       [<-] (0.6,0) –  (0.6,-1);
       [<-] (1.2,0) –  (1.2,-1);
       [<-] (2.4,0) –  (2.4,-1);
       [<-] (3.0,0) –  (3,-1);
      ]

where the trailing terms arising from relation (<ref>) have the same sign after canceling the empty dots, and thus add together.  We can use the trace relation to slide the top cup in the second term to the bottom; after simplication, this term is therefore equal to n-1.  The first term is equal to β_nα_-1 as in <cit.>. Thus, 



[    [scale=0.8]
  [->] (3,0) .. controls ++(0,1.25) and (0,.25) .. (0,2);
  [->] (0.0,0) .. controls ++(0,1) and ++(0,-1.2) .. (0.6,2);
  [->] (0.6,0) .. controls ++(0,1) and ++(0,-1.2) .. (1.2,2);
  [->] (1.2,0) .. controls ++(0,1) and ++(0,-1.2) .. (1.8,2);
  [->] (2.4,0) .. controls ++(0,1) and ++(0,-1.2) .. (3,2);
   at (1.8,.35) …;
   at (2.4,1.65) …;
  [<-] (-0.6,0) – (-0.6,2);
    (.05,1.6) circle (2pt);
  ]
  
=
   [    [scale=0.8]
  [->] (3,0) .. controls ++(0,1.25) and (0,.25) .. (0,2);
  [->] (0.0,0) .. controls ++(0,1) and ++(0,-1.2) .. (0.6,2);
  [->] (0.6,0) .. controls ++(0,1) and ++(0,-1.2) .. (1.2,2);
  [->] (1.2,0) .. controls ++(0,1) and ++(0,-1.2) .. (1.8,2);
  [->] (2.4,0) .. controls ++(0,1) and ++(0,-1.2) .. (3,2);
   at (1.8,.35) …;
   at (2.4,1.65) …;
  [<-] (3.6,0) – (3.6,2);
    (.05,1.6) circle (2pt);
   ] +4 2n-1


as desired. The base case of the induction is proved. The induction step follows from examination of the Jacobi identity, exactly as in <cit.>, using our Lemma <ref> in place of <cit.>.



 Let n ∈ℤ. We have 
    [1x_1^2, 2n-1] = 22nx_1 + … + x_2n +22nx_2+…+x_2n-1.

 This is a straightforward diagrammatic calculation similar to Lemmas <ref> and <ref>. We have 

    [[baseline=(current bounding box.center),scale=0.75]
      [->] (3.2,0) .. controls (3.2,1.25) and (0,.25) .. (0,2)
         node[pos=0.85, shape=coordinate](X);
      [->] (0,0) .. controls (0,1) and (.8,.8) .. (.8,2);
      [->] (.8,0) .. controls (.8,1) and (1.6,.8) .. (1.6,2);
      [->] (2.4,0) .. controls (2.4,1) and (3.2,.8) .. (3.2,2);
       at (1.6,.35) …;
       at (2.4,1.65) …;
        (-.3,1.5) circle (3pt);
     at (-.7,1.5) 2n;
    [->](-.3,0) to (-.3,2);
    ] 
    
    =
    [    [scale=0.8]
       (2.4,0) .. controls (2.4,1.25) and (0,.25) .. (0,1.5)
         node[pos=0.85, shape=coordinate](X);
       (0,0) .. controls (0,1) and (0.6,.8) .. (0.6,1.5);
       (0.6,0) .. controls (0.6,1) and (1.2,.8) .. (1.2,1.5);
       (1.8,0) .. controls (1.8,1) and (2.4,.8) .. (2.4,1.5);
       at (1.2,.35) …;
       at (1.8,1.65) …;
       (3,0) – (3,1.5);
      
        (0.05,2.25) circle (2pt);
     at (-.4,2.25) 2n;
     
        [blue, dotted] (-0.4,0) – (3.4,0);
        [blue, dotted] (-0.4,1.5) – (3.4,1.5);
        
       [->]  (3,1.5) .. controls ++(0,0.95) and ++(0,-1.15) .. (0,2.5);
       [->]  (2.4,1.5) .. controls ++(0,0.5) and ++(0,-0.5) .. (3,2.5);
       [->]  (0,1.5) .. controls ++(0,0.5) and ++(0,-0.5) .. (0.6,2.5);
       [->]  (0.6,1.5) .. controls ++(0,0.5) and ++(0,-0.5) .. (1.2,2.5);
       [->]  (1.2,1.5) .. controls ++(0,0.5) and ++(0,-0.5) .. (1.8,2.5);
       
         (3,0) .. controls ++(0,-0.95) and ++(0,+0.95) .. (0,-1);
         (0,0) .. controls ++(0,-0.5) and ++(0,+0.5) .. (0.6,-1);
         (0.6,0) .. controls ++(0,-0.5) and ++(0,+0.5) .. (1.2,-1);
         (1.8,0) .. controls ++(0,-0.5) and ++(0,+0.5) .. (2.4,-1);
        (2.4,0) .. controls (2.4,-.5) and (3,-.5) .. (3,-1);
       ]


Sliding the dots all the way to the right side of the diagram results in 2(2n-1) resolution terms. Each of these resolution terms contains a 2n-cycle and a single solid dot - there are 2 resolution terms containing a solid dot on the first strand and 2 containing a solid dot on the last strand, and 4 resolution terms with a dot on each other strand. All empty dots cancel in such a way that no resolution terms cancel with each other. The result follows.






The following lemmas will allow us to generate bubbles with arbitrary numbers of dots using just ± 1x_1^2.

We have 
    ∑_a+b=2n-1[baseline=(current bounding box.center),rotate=90]
    3mm[->] (-0.6,0) .. controls ++(0,.4) and ++(0,-.5) .. (-0.0,1);
      [<-] (0.0,0) .. controls ++(0,.4) and ++(0,-.5) .. (-0.6,1);
       [<-] (-0.6,1) .. controls ++(0,.4) and ++(0, .4) .. (-0.0,1);
      [->] (0.0,0) .. controls ++(0,-.5) and ++(0,-.5) .. (-0.6,0);
        (-0.05,1.2) circle (2pt);
         (-0.1,-.3) circle (2pt);
       at (0.1,1.3) a;
       at (0.1,-.4) b;
    =
    ∑_i+j=n-1(1+2j)[baseline=(current bounding box.center)]
        [->] (3,2) arc (-180:180:5mm);
     (3.95,2.2) circle [radius=2pt];
     at (4.2,2.2) 2i;
        [baseline=(current bounding box.center)]
        [<-] (3,2) arc (-180:180:5mm);
     (3.95,2.2) circle [radius=2pt];
     at (4.2,2.2) 2j;




We compute:

    ∑_a+b=2n-1[baseline=(current bounding box.center),rotate=90]
    3mm[->] (-0.6,0) .. controls ++(0,.4) and ++(0,-.5) .. (-0.0,1);
      [<-] (0.0,0) .. controls ++(0,.4) and ++(0,-.5) .. (-0.6,1);
       [<-] (-0.6,1) .. controls ++(0,.4) and ++(0, .4) .. (-0.0,1);
      [->] (0.0,0) .. controls ++(0,-.5) and ++(0,-.5) .. (-0.6,0);
        (-0.05,1.2) circle (2pt);
         (-0.1,-.3) circle (2pt);
       at (0.1,1.3) a;
       at (0.1,-.4) b;
 
    =
    [baseline=(current bounding box.center),rotate=90]
    3mm[->] (-0.6,0) .. controls ++(0,.4) and ++(0,-.5) .. (-0.0,1);
      [<-] (0.0,0) .. controls ++(0,.4) and ++(0,-.5) .. (-0.6,1);
       [<-] (-0.6,1) .. controls ++(0,.4) and ++(0, .4) .. (-0.0,1);
      [->] (0.0,0) .. controls ++(0,-.5) and ++(0,-.5) .. (-0.6,0);
        (-0.05,1.2) circle (2pt);
       at (0.2,1.3) 2n-1;
    +
    [baseline=(current bounding box.center),rotate=90]
    3mm[->] (-0.6,0) .. controls ++(0,.4) and ++(0,-.5) .. (-0.0,1);
      [<-] (0.0,0) .. controls ++(0,.4) and ++(0,-.5) .. (-0.6,1);
       [<-] (-0.6,1) .. controls ++(0,.4) and ++(0, .4) .. (-0.0,1);
      [->] (0.0,0) .. controls ++(0,-.5) and ++(0,-.5) .. (-0.6,0);
        (-0.05,1.2) circle (2pt);
         (-0.1,-.3) circle (2pt);
       at (0.2,1.3) 2n-2;
    +
    [baseline=(current bounding box.center),rotate=90]
    3mm[->] (-0.6,0) .. controls ++(0,.4) and ++(0,-.5) .. (-0.0,1);
      [<-] (0.0,0) .. controls ++(0,.4) and ++(0,-.5) .. (-0.6,1);
       [<-] (-0.6,1) .. controls ++(0,.4) and ++(0, .4) .. (-0.0,1);
      [->] (0.0,0) .. controls ++(0,-.5) and ++(0,-.5) .. (-0.6,0);
        (-0.05,1.2) circle (2pt);
         (-0.1,-.3) circle (2pt);
       at (0.2,1.3) 2n-3;
       at (0.2,-.4) 2;
    +
    …
    +
    [baseline=(current bounding box.center),rotate=90]
    3mm[->] (-0.6,0) .. controls ++(0,.4) and ++(0,-.5) .. (-0.0,1);
      [<-] (0.0,0) .. controls ++(0,.4) and ++(0,-.5) .. (-0.6,1);
       [<-] (-0.6,1) .. controls ++(0,.4) and ++(0, .4) .. (-0.0,1);
      [->] (0.0,0) .. controls ++(0,-.5) and ++(0,-.5) .. (-0.6,0);
        (-0.05,1.2) circle (2pt);
         (-0.1,-.3) circle (2pt);
       at (0.2,-.4) 2n-2;
 
    =
    [baseline=(current bounding box.center),rotate=90]
    3mm[->] (-0.6,0) .. controls ++(0,.4) and ++(0,-.5) .. (-0.0,1);
      [<-] (0.0,0) .. controls ++(0,.4) and ++(0,-.5) .. (-0.6,1);
       [<-] (-0.6,1) .. controls ++(0,.4) and ++(0, .4) .. (-0.0,1);
      [->] (0.0,0) .. controls ++(0,-.5) and ++(0,-.5) .. (-0.6,0);
        (-0.05,1.2) circle (2pt);
       at (0.2,1.3) 2n-1;
    +
    2
    [baseline=(current bounding box.center),rotate=90]
    3mm[->] (-0.6,0) .. controls ++(0,.4) and ++(0,-.5) .. (-0.0,1);
      [<-] (0.0,0) .. controls ++(0,.4) and ++(0,-.5) .. (-0.6,1);
       [<-] (-0.6,1) .. controls ++(0,.4) and ++(0, .4) .. (-0.0,1);
      [->] (0.0,0) .. controls ++(0,-.5) and ++(0,-.5) .. (-0.6,0);
        (-0.05,1.2) circle (2pt);
         (-0.1,-.3) circle (2pt);
       at (0.2,1.3) 2n-3;
       at (0.2,-.4) 2;
    +
    2
    [baseline=(current bounding box.center),rotate=90]
    3mm[->] (-0.6,0) .. controls ++(0,.4) and ++(0,-.5) .. (-0.0,1);
      [<-] (0.0,0) .. controls ++(0,.4) and ++(0,-.5) .. (-0.6,1);
       [<-] (-0.6,1) .. controls ++(0,.4) and ++(0, .4) .. (-0.0,1);
      [->] (0.0,0) .. controls ++(0,-.5) and ++(0,-.5) .. (-0.6,0);
        (-0.05,1.2) circle (2pt);
         (-0.1,-.3) circle (2pt);
       at (0.2,1.3) 2n-5;
       at (0.2,-.4) 4;
    +
    …
    +
    2
    [baseline=(current bounding box.center),rotate=90]
    3mm[->] (-0.6,0) .. controls ++(0,.4) and ++(0,-.5) .. (-0.0,1);
      [<-] (0.0,0) .. controls ++(0,.4) and ++(0,-.5) .. (-0.6,1);
       [<-] (-0.6,1) .. controls ++(0,.4) and ++(0, .4) .. (-0.0,1);
      [->] (0.0,0) .. controls ++(0,-.5) and ++(0,-.5) .. (-0.6,0);
        (-0.05,1.2) circle (2pt);
         (-0.1,-.3) circle (2pt);
       at (0.2,-.4) 2n-2;

because we have 
    [baseline=(current bounding box.center),rotate=90]
    3mm[->] (-0.6,0) .. controls ++(0,.4) and ++(0,-.5) .. (-0.0,1);
      [<-] (0.0,0) .. controls ++(0,.4) and ++(0,-.5) .. (-0.6,1);
       [<-] (-0.6,1) .. controls ++(0,.4) and ++(0, .4) .. (-0.0,1);
      [->] (0.0,0) .. controls ++(0,-.5) and ++(0,-.5) .. (-0.6,0);
        (-0.05,1.2) circle (2pt);
         (-0.1,-.3) circle (2pt);
       at (0.2,1.3) 2n-2i;
       at (0.2,-.4) 2i-1;
    =
    [baseline=(current bounding box.center),rotate=90]
    3mm[->] (-0.6,0) .. controls ++(0,.4) and ++(0,-.5) .. (-0.0,1);
      [<-] (0.0,0) .. controls ++(0,.4) and ++(0,-.5) .. (-0.6,1);
       [<-] (-0.6,1) .. controls ++(0,.4) and ++(0, .4) .. (-0.0,1);
      [->] (0.0,0) .. controls ++(0,-.5) and ++(0,-.5) .. (-0.6,0);
        (-0.05,1.2) circle (2pt);
         (-0.1,-.3) circle (2pt);
       at (0.2,1.3) 2n-2i-1;
       at (0.2,-.4) 2i;.

Moreover, we can decompose these figure eights into a linear combination of products of two bubbles using dot slide relations <ref> and <ref> as follows:

    [baseline=(current bounding box.center),rotate=90]
    3mm[->] (-0.6,0) .. controls ++(0,.4) and ++(0,-.5) .. (-0.0,1);
      [<-] (0.0,0) .. controls ++(0,.4) and ++(0,-.5) .. (-0.6,1);
       [<-] (-0.6,1) .. controls ++(0,.4) and ++(0, .4) .. (-0.0,1);
      [->] (0.0,0) .. controls ++(0,-.5) and ++(0,-.5) .. (-0.6,0);
        (-0.05,1.2) circle (2pt);
         (-0.1,-.3) circle (2pt);
       at (0.2,1.3) 2n-2a-1;
       at (0.2,-.4) 2a;
    =
    ∑_i+j=n-1
    j≥ a[baseline=(current bounding box.center)]
        [->] (3,2) arc (-180:180:5mm);
     (3.95,2.2) circle [radius=2pt];
     at (4.2,2.2) 2i;
        [baseline=(current bounding box.center)]
        [<-] (3,2) arc (-180:180:5mm);
     (3.95,2.2) circle [radius=2pt];
     at (4.2,2.2) 2j;
        .

Combining these results, we get that 

    ∑_a+b=2n-1[baseline=(current bounding box.center),rotate=90]
    3mm[->] (-0.6,0) .. controls ++(0,.4) and ++(0,-.5) .. (-0.0,1);
      [<-] (0.0,0) .. controls ++(0,.4) and ++(0,-.5) .. (-0.6,1);
       [<-] (-0.6,1) .. controls ++(0,.4) and ++(0, .4) .. (-0.0,1);
      [->] (0.0,0) .. controls ++(0,-.5) and ++(0,-.5) .. (-0.6,0);
        (-0.05,1.2) circle (2pt);
         (-0.1,-.3) circle (2pt);
     at (0.1,1.3) a;
       at (0.1,-.4) b;
    =
    ∑_i+j=n-1(1+2j)[baseline=(current bounding box.center)]
        [->] (3,2) arc (-180:180:5mm);
     (3.95,2.2) circle [radius=2pt];
     at (4.2,2.2) 2i;
        [baseline=(current bounding box.center)]
        [<-] (3,2) arc (-180:180:5mm);
     (3.95,2.2) circle [radius=2pt];
     at (4.2,2.2) 2j;
        .




 We have 
    [1x_1^2a,-1x_1^2b]=-2d̅_2(a+b)-∑_i+j=2(a+b)-1(2+4j)d̅_2id_2j
 for a,b∈ℤ_≥0.



We compute: 
    [baseline=(current bounding box.center),scale=0.75]
    [->] (0,0) to (0,2);
    [<-] (0.75,0) to (0.75,2);
     (0,1.2) circle (2pt);
     (0.75,1.2) circle (2pt);
     at (-0.3,1.3) 2a;
     at (1.1,1.3) 2b;
       =
    [baseline=(current bounding box.center),scale=0.75]
    [->] (-0.6,0) .. controls ++(0,.4) and ++(0,-.5) .. (-0.0,1);
    [<-] (0.0,0) .. controls ++(0,.4) and ++(0,-.5) .. (-0.6,1);
    [<-] (-0.6,1) .. controls ++(0,.4) and ++(0,-.5) .. (-0.0,2);
    [->] (0.0,1) .. controls ++(0,.4) and ++(0,-.5) .. (-0.6,2);
      (-0.55,1.7) circle (2pt);
      (-0.1,1.7) circle (2pt);
     at (-.95,1.8) 2a;
     at (0.35,1.8) 2b;
    
    =[baseline=(current bounding box.center),scale=0.75]
     [->] (-0.6,0) .. controls ++(0,.4) and ++(0,-.5) .. (-0.0,1);
      [<-] (0.0,0) .. controls ++(0,.4) and ++(0,-.5) .. (-0.6,1);
       [<-] (-0.6,1) .. controls ++(0,.4) and ++(0,-.5) .. (-0.0,2);
      [->] (0.0,1) .. controls ++(0,.4) and ++(0,-.5) .. (-0.6,2);
        (-0.05,1.2) circle (2pt);
         (-0.1,1.7) circle (2pt);
       at (0.3,1.3) 2a;
       at (0.35,1.8) 2b;
    
    -2
    ∑_j=0^2a-1[baseline=(current bounding box.center),scale=0.75]
     [->] (-0.6,0) .. controls ++(0,.4) and ++(0,-.5) .. (-0.0,1);
      [<-] (0.0,0) .. controls ++(0,.4) and ++(0,-.5) .. (-0.6,1);
       [<-] (-0.6,1) .. controls ++(0,.4) and ++(0, .4) .. (-0.0,1);
      [->] (0.0,2) .. controls ++(0,-.5) and ++(0,-.5) .. (-0.6,2);
        (-0.05,1.2) circle (2pt);
         (0,1.8) circle (2pt);
       at (0.2,1.2) j;
       at (1.2,1.8) 2(a+b)-1-j;
    
       =
    [baseline=(current bounding box.center),scale=0.75]
     [->] (-0.6,0) .. controls ++(0,.4) and ++(0,-.5) .. (-0.0,1);
      [<-] (0.0,0) .. controls ++(0,.4) and ++(0,-.5) .. (-0.6,1);
       [<-] (-0.6,1) .. controls ++(0,.4) and ++(0,-.5) .. (-0.0,2);
      [->] (0.0,1) .. controls ++(0,.4) and ++(0,-.5) .. (-0.6,2);
        (-0.05,1.2) circle (2pt);
         (-0.55,1.2) circle (2pt);
       at (0.3,1.3) 2a;
       at (-0.9,1.3) 2b;
      
    -
    2
    ∑_i=0^2b-1[baseline=(current bounding box.center),scale=0.75]
     [->] (-0.6,0) .. controls ++(0,.4) and ++(0,-.5) .. (-0.0,1);
      [<-] (0.0,0) .. controls ++(0,.4) and ++(0,-.5) .. (-0.6,1);
       [<-] (-0.6,1) .. controls ++(0,.4) and ++(0, .4) .. (-0.0,1);
      [->] (0.0,2) .. controls ++(0,-.5) and ++(0,-.5) .. (-0.6,2);
        (-0.05,1.2) circle (2pt);
         (0,1.8) circle (2pt);
       at (0.66,1.2) 2a+i;
       at (0.66,1.8) 2b-1-i;
    
    -
    2
    ∑_j=0^2a-1[baseline=(current bounding box.center),scale=0.75]
     [->] (-0.6,0) .. controls ++(0,.4) and ++(0,-.5) .. (-0.0,1);
      [<-] (0.0,0) .. controls ++(0,.4) and ++(0,-.5) .. (-0.6,1);
       [<-] (-0.6,1) .. controls ++(0,.4) and ++(0, .4) .. (-0.0,1);
      [->] (0.0,2) .. controls ++(0,-.5) and ++(0,-.5) .. (-0.6,2);
        (-0.05,1.2) circle (2pt);
         (0,1.8) circle (2pt);
       at (0.2,1.2) j;
       at (1.2,1.8) 2(a+b)-1-j;
    
       =
    [baseline=(current bounding box.center),scale=0.75]
    [<-] (-0.6,0) .. controls ++(0,.4) and ++(0,-.5) .. (-0.0,1);
    [->] (0.0,0) .. controls ++(0,.4) and ++(0,-.5) .. (-0.6,1);
    [->] (-0.6,1) .. controls ++(0,.4) and ++(0,-.5) .. (-0.0,2);
    [<-] (0.0,1) .. controls ++(0,.4) and ++(0,-.5) .. (-0.6,2);
      (-0.55,1.7) circle (2pt);
      (-0.1,1.7) circle (2pt);
     at (-.95,1.8) 2b;
     at (0.35,1.8) 2a;
    
    -
    2
    ∑_j=0^2(a+b)-1[baseline=(current bounding box.center),scale=0.75]
     [->] (-0.6,0) .. controls ++(0,.4) and ++(0,-.5) .. (-0.0,1);
      [<-] (0.0,0) .. controls ++(0,.4) and ++(0,-.5) .. (-0.6,1);
       [<-] (-0.6,1) .. controls ++(0,.4) and ++(0, .4) .. (-0.0,1);
      [->] (0.0,2) .. controls ++(0,-.5) and ++(0,-.5) .. (-0.6,2);
        (-0.05,1.2) circle (2pt);
         (0,1.8) circle (2pt);
       at (0.2,1.2) j;
       at (1.2,1.8) 2(a+b)-1-j;
      
       =
    [baseline=(current bounding box.center),scale=0.75]
    [<-] (-0.6,0) .. controls ++(0,.4) and ++(0,-.5) .. (-0.0,1);
    [->] (0.0,0) .. controls ++(0,.4) and ++(0,-.5) .. (-0.6,1);
    [->] (-0.6,1) .. controls ++(0,.4) and ++(0,-.5) .. (-0.0,2);
    [<-] (0.0,1) .. controls ++(0,.4) and ++(0,-.5) .. (-0.6,2);
      (-0.55,1.7) circle (2pt);
      (-0.1,1.7) circle (2pt);
     at (-.95,1.8) 2b;
     at (0.35,1.8) 2a;
    
    -
    2
    ∑_j=0^2(a+b)-1[baseline=(current bounding box.center),scale=0.75]
     [->] (-0.6,0) .. controls ++(0,.4) and ++(0,-.5) .. (-0.0,1);
      [<-] (0.0,0) .. controls ++(0,.4) and ++(0,-.5) .. (-0.6,1);
       [<-] (-0.6,1) .. controls ++(0,.4) and ++(0, .4) .. (-0.0,1);
      [->] (0.0,0) .. controls ++(0,-.5) and ++(0,-.5) .. (-0.6,0);
        (0,1.2) circle (2pt);
         (-0.1,-.3) circle (2pt);
       at (0.3,1.3) j;
       at (1.2,-.2) 2(a+b)-1-j;
    
       =
    [baseline=(current bounding box.center),scale=0.75]
    [<-] (0,0) to (0,2);
    [->] (0.75,0) to (0.75,2);
     (0,1.2) circle (2pt);
     (0.75,1.2) circle (2pt);
     at (-0.3,1.3) 2b;
     at (1.1,1.3) 2a;
    
    -
    2[baseline=(current bounding box.center)]
        [->] (3,2) arc (-180:180:5mm);
     (3.95,2.2) circle [radius=2pt];
     at (4.5,2.2) 2(a+b);
        
    -
    2
    ∑_j=0^2(a+b)-1[baseline=(current bounding box.center),scale=0.75]
     [->] (-0.6,0) .. controls ++(0,.4) and ++(0,-.5) .. (-0.0,1);
      [<-] (0.0,0) .. controls ++(0,.4) and ++(0,-.5) .. (-0.6,1);
       [<-] (-0.6,1) .. controls ++(0,.4) and ++(0, .4) .. (-0.0,1);
      [->] (0.0,0) .. controls ++(0,-.5) and ++(0,-.5) .. (-0.6,0);
        (0,1.2) circle (2pt);
         (-0.1,-.3) circle (2pt);
       at (0.3,1.3) j;
       at (1.2,-.2) 2(a+b)-1-j;
    .


Therefore [1x_1^2a,-1x_1^2b]=-2d̅_2(a+b)-∑_i+j=2(a+b)-1(2+4j)d̅_2id_2j.






§ ALGEBRA ISOMORPHISM

In this section, we will study the structure of , first as a vector space and then as an algebra. We show that  has a triangular decomposition into two copies of the trace of _n and a polynomial algebra. We then describe a generating set for , which allows us to define the algebra homomorphism to W^-. Finally, we prove that this homomorphism is an isomorphism.



 §.§ Trace of ℋ_tw as a vector space

Let m,n ≥ 0 and define J_m,n to be the 2-sided ideal in _ℋ_tw(P^mQ^n) generated by diagrams which contain at least one arc connecting a pair of upper points. 

There exists a split short exact sequence 

    0 → J_m,n→_ℋ_tw(P^mQ^n) → ()^op_m ⊗_n ⊗ℂ[d_0,d_2,d_4....] → 0.




In _ℋ_tw(P^mQ^n), due to the middle diagram in relation (<ref>), we can assume our diagrams have no crossing between opposite oriented strands. Taking the quotient _ℋ_tw(P^mQ^n)/J_m,n kills diagrams with cups connecting two upper points, and those with caps connecting two lower points. Therefore we are left with diagrams, possibly with bubbles, which have no caps or cups and have crossings only among like-oriented strands.
   Note that in the quotient _ℋ_tw(P^mQ^n)/J_m,n, the diagram in relation (<ref>) simplifies to 

    [baseline=(current bounding box.center),scale=0.75]
     [<-](0,0) to [out=45,in=-45] (0,2);
     [->](0.5,0) to [out=135,in=-135] (0.5,2);
    
    =
    [baseline=(current bounding box.center),scale=0.75]
    [<-] (1.5,0) to (1.5,2);
    [->] (2,0) to (2,2);

and therefore we can move the bubbles to the rightmost part of our diagrams for free. This gives us a short exact sequence 

    0 → J_m,n→_ℋ_tw(P^mQ^n) →_ℋ_tw(P^m) ⊗_ℋ_tw(Q^n) ⊗_ℋ_tw(1) → 0.

By <cit.>, we have that _ℋ_tw(P^m) is isomorphic to ()^op_m and that _ℋ_tw(Q^n) is isomoprhic to _n. By Proposition <ref>, it follows that _ℋ_tw(1) is isomorphic to ℂ[d_0,d_2,d_4....]. Hence the result follows.



If f,g∈_n such that fg=1,  then f,g∈⋊ℂ[S_n]⊂_n.


There is an ℕ-filtration on _n given by (x_i)=1 for i∈{1,...,n} and other generators have degree zero. Under this filtration, the degree zero part of _n is the semidirect product Cl_n⋊ℂ[S_n]. Therefore, in the associated graded object, we see that if fg=1, (gr(f)gr(g))=(gr(f))+(gr(g))=(1)=0, hence gr(f),gr(g) are in degree zero part. Therefore f,g∈⋊ℂ[S_n].



The indecomposable objects of ℋ_tw are of the form P^mQ^n for m,n∈ℤ_≥ 0.


First, note that if QP appears in an object, that object can be decomposed into more components using the diagram in relation (<ref>). Hence all indecomposable objects must be of the form P^mQ^n. 

On the other hand, to see that every sequence of the form P^mQ^n is an indecomposable object, we will show that any idempotent in (P^mQ^n) has to be the identity.

Let f,g be two maps as mentioned in Lemma <ref>. Note that gf is an idempotent since (gf)(gf)=g(fg)f=gf. Since we had the splitting short exact sequence 0 → J_m,n→_ℋ_tw(P^mQ^n) →(P^m) ⊗ (Q^n) ⊗(id) → 0 in Lemma <ref>, we know that the maps f and g will decompose into (f_1,f_2) and (g_1,g_2) where f_1,g_1:P^m→ P^m and (f_2,g_2):Q^n→ Q^n. Now g_1f_1 is the identity map in (P^m), and by the above lemma g_1,f_1∈𝒞ℓ_n⋊ℂ[S_n]. Similarly, f_2, g_2∈𝒞ℓ_n⋊ℂ[S_n].

 But in 𝒞ℓ_n⋊ℂ[S_n], g_1f_1=1 implies that f_1g_1=1 as well. To see this, consider the diagrams corresponding to g_1 and f_1 which consist of a permutation and some hollow dots on top. After composing these diagrams, we can collect all the hollow dots on the top since hollow dots can pass through crossing for free, possibly gaining a sign. Furthermore, each strand has an even number of hollow dots, since this composition is the identity map. So, the hollow dots cancel with each other. This shows that the corresponding permutations of f_1 and g_1 are inverses of each other, and in particular they commute. Therefore f_1g_1=1. Similarly, f_2g_2=1. Thus we have that fg=1.





We have the triangular decomposition of :

    ≅⊕_m,n∈ℤ_≥0((_m)^op⊗_n ⊗ℂ[d_0,d_2,d_4....]).




As shown in <cit.>, to find , it is enough to consider the direct sum over indecomposable objects of endomorphism spaces of objects of H_tw. Let I=span_ℂ{fg-gf} where f:x→ y and g:y→ x for x,y objects of a ℂ-linear category. Therefore by Lemma <ref> we have

    ≅(⊕_m,n∈ℤ_≥0_(P^mQ^n))/I.


By Lemma <ref>, this gives us 

    ≅(⊕_m,n∈ℤ_≥0 ((_m)^op⊗_n ⊗ℂ[d_0,d_2,d_4....])⊕ J_m,n)/I.

Recall that the ideal J_m,n is generated by diagrams containing at least one cup connecting two upper points. Therefore, the diagrams in J_m,n must also contain caps, since they are dealing with endomorphisms. Using the trace relation and the relations in ℋ_tw, we can express the elements of J_m,n as direct sum of endomorphisms of P^m'Q^n' for m'≤ m and n'≤ n. Hence we have 

    ≅⊕_m,n∈ℤ_≥0((_m)^op⊗_n ⊗ℂ[d_0,d_2,d_4....])
        
       ≅(⊕_m,n∈ℤ_≥0((_m)^op⊗_n)) ⊗ℂ[d_0,d_2,d_4....].








 §.§ Generators of the algebra 


The following gives a generating set for  as an algebra.

  The algebra  is generated by -1, ± 2(x_1 + x_2), and d_0 + d_2. 
 

First, Proposition <ref> implies that 1 and (d_0+d_2) allow us generate a differential degree two element 1x_1^2; since all relations in ℋ_tw are local, we can evaluate the commutator [1x_1^2, (d_0+d_2)] by moving the dot to the bottom of the upward strand and sliding the bubbles over the upper portion. We can therefore apply Lemma <ref> repeatedly to show that ad(d_0+d_2)^n1 has a leading term of 1x_1^2n.

By Lemma <ref>, the elements -1 and 2x_1+x_2 are sufficient to generate 2m+1 for all integers m>0. Then we can generate 2nx_1+…+x_n from 1x_1^2 and 2m+1 by using Lemma <ref>.  Lemma <ref>, -1 and 2nx_1+x_2+… + x_n allow us to generate 2r+1 for all integers r.

 Proposition <ref> implies that all elements with nonzero rank degree can be written as a sum of elements of the form ± nx_1^ℓ c_1^k. By Propositions <ref> and <ref>, all elements of this form except for the ones generated in the preceding paragraphs are 0 in , so we have generated all of  and .

Finally, Lemma <ref> allows us to generate d_2n, applying Lemma <ref> to split up the d_2n terms. 



 §.§ The isomorphism


There is an obvious isomorphism of vector spaces between the Fock space representations of  and W^-: 


    ϕ: V= ℂ[1, 2, …] →ℂ[w_-1,0, w_-2,0, …] = 𝒱_1,0.


Recall that each algebra acts faithfully on its Fock space representation.

 The map ϕ in Equation (<ref>) commutes with the action of the twisted Heisenberg subalgebras in V and 𝒱_1,0, i.e.:

    ϕ(rv) =√(2)w_-r,0ϕ(v).
 

The vector space realizations of V and 𝒱_1,0 in Equation (<ref>) imply that the action of r on V is simply the adjoint action of r on the subalgebra , and the action of w_-r,0 on ϕ(v) is the adjoint action of w_-r,0 on (W^-)^-. The Lemma follows from our computation of these twisted Heisenberg relations in Propositions <ref> and <ref>.


 For any v∈ V we have ϕ((d_0+d_2)v) = -2w_0,3ϕ(v). 
 Propositions <ref> and <ref> give that w_0,3 maps w_-1,0 to an element with leading term w_-1,2, and (d_0+d_2) maps 1 to an element with leading term 1x_1^2. Comparision of the actions of these terms on the twisted Heisenberg subalgebras on either side gives that that their images in the endomorphisms of the Fock space are identical. 

  For any v ∈ V we have ϕ(± 2(x_1 + x_2) v) = 2√(2)(w_∓ 2, 1 + w_∓ 2, 0) ϕ(v). 
 This follows from comparision of Lemma <ref> and Proposition <ref>. 

Now extend ϕ to a map 
    Φ: ⟶ W^-/⟨ w_0,0, C-1 ⟩
 by mapping 
    1↦√(2)w_-1,0    ± 2(x_1 + x_2)→ 2√(2)w_∓ 2, 1 + w_∓ 2, 0     d_2+d_0 ↦ -2w_0,3
 and extending algebraically, i.e. 
    Φ(a_1 … a_k) = Φ(a_1) …Φ(a_k)
 for generators a_1, …, a_k of . 

 The map Φ above is well defined. 

 Suppose  A∈ has two representations in terms of generators, A= a_i_1… a_i_k = a_j_1… a_j_ℓ. Then a_i_1… a_i_k. V = a_j_1… a_j_ℓ.V, so applying Φ gives Φ(a_i_1… a_i_k).𝒱_1,0=Φ(a_j_1… a_j_ℓ).𝒱_1,0. Hence Φ(a_i_1… a_i_k) = Φ(a_j_1… a_j_ℓ) by the faithfulness of the Fock space representation for W^-.

  The map Φ is an isomorphism of algebras. 

 We immediately have that Φ is surjective, because it maps generators to generators. Thus, it remains to show that Φ is injective. Let A := a_i_1… a_i_k∈ and assume that Φ(A).𝒱_1,0 = 0. Then Φ(A) = 0 by the faithfulness of the representation. But then Φ(a_i_1)…Φ(a_i_k) .𝒱_1,0=0. Then, by Lemmas <ref>, <ref>, and <ref>, we have Φ(a_i_1)…Φ(a_i_k ). 𝒱_1,0 = ϕ(a_i_1… a_i_k. V) = ϕ(A.v)=0. But ϕ is an isomorphism, so this implies that A.V=0. Hence A = 0 by the faithfulness of the Fock space representation of . 


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