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7 The general Case

Recall that our goal is to obtain the relations among the generators {D(r)a;i,j, D0a;i,j0(r)}, {Ea;i,j(r) }, and {Fa;i,j(r) } associated to a composition µ of (M|N). To that end, we divide them into 3 disjoint parts as following:

A : for all admissible indices i, j, r.

If we choose two elements from Part A, then their bracket is obtained by Theorem 2.

If we choose two elements from Part B, then they are the images of some elements from the Part A in Y(glN|M) under the swap map ζN|M, and the bracket is obtained by Theorem 2 as well.

Now suppose one of them is from Part A and the other is from Part B. Note that every element in Part A is in the north-westernM ×M corner of T(u) and hence is in the subalgebra Y(glM) of Y(glM|N) (see Section 4). On the other hand, every element in Part B is in the south-eastern N ×N corner ofT(u) and hence is in the subalgebra ψM Y(gl0|N)

of Y(glM|N). Thus, their bracket is zero by Lemma 4.3.

Therefore, we only have to focus on the cross section where the odd blocks and even blocks are “close”, and this is done in Proposition 5.1, Lemma 6.1 and Lemma 6.2.

Moreover, there are some non-trivial ternary brackets relations in the non-super case, and the corresponding ternary relations in the super case are found in Lemma 6.3.

The following proposition summarizes the results we have obtained up to now.

Proposition 7.1. For all admissible a, b, f, g, h, i, j, k, we have the following equalities in the super Yangian Yµ((u−1, v−1, w−1)).

Fa;i,j(u),[Fa;h,k(v), Fb;f,g(w)]

+

Fa;i,j(v),[Fa;h,k(u), Fb;f,g(w)]

= 0, |a−b| ≥1, where a:= 0 if 1≤a≤m anda:= 1if m+ 1≤a≤m+n.

Proof. This is the consequence of Theorem 2, Proposition 5.1, Lemmas 6.1−6.3, together with the mapsψk and ζM|N.

The next lemma is a block generalization of [Go, Lemma 5] and the proof is essentially the same, except that we are using block decompositions. The relations are purely super phenomenons.

Lemma 7.2. Associated toµ= (µ1, µ2, . . . µmm+1, . . . , µm+n)withm >1andn >1, we have the following identities in Yµ.

[Em−1;i,j(r) , Em;h,k(1) ],[Em;h(1)

0,k0, Em+1;f,g(s) ]

= 0, (7.1)

[Fm−1;i,j(r) , Fm;h,k(1) ],[Fm;h(1)

0,k0, Fm+1;f,g(s) ]

= 0, (7.2)

for all admissible f, g, h, i, j, k, h0, k0, r, s.

Proof. By using the mapsζM|N andψ, it is enough to show (7.1) in the casem=n= 2 only. Therefore, we want to show (7.1) inY1234), i.e.,

[E1;i,j(r) , E(1)2;h,k],[E2;h(1)

0,k0, E3;f,g(s) ]

= 0. (7.3)

We first claim that for all admissiblei, j, h, k,

[E1,3;i,j(u), E2;h,q(v)E3;q,k(v)−E2,4;h,k(v) ] = 0, (7.4) where the index q is summed over 1,2, . . . , µ3. To prove the claim, we use (4.11) and (4.13) associated to the composition (µ1, µ23, µ4) to derive the following identities.

E1,3;i,j(u) = D1;i,p0 (u)tp,µ12+j(u), E2;h,q(v)E3;q,k(v)−E2,4;h,k(v) = t0µ1+h,µ123+r(v)D4;r,k(v),

for all 1 ≤ i ≤ µ1, 1≤ j ≤ µ3, 1 ≤ h ≤ µ2, 1≤ k ≤ µ4, and the indices p, q, r are summed over µ1, µ34, respectively. Substituting these identities into the bracket in (7.4) and setting a notation na:=µ12+. . .+µa for short, we have

[E1,3;i,j(u), E2;h,q(v)E3;q,k(v)−E2,4;h,k(v)]

= [D1;i,p0 (u)tp,n2+j(u), t0µ1+h,n3+r(v)D4;r,k(v)]

=D01;i,p(u)tp,n2+j(u)t0µ1+h,n3+r(v)D4;r,k(v) +t0µ1+h,n3+r(v)D4;r,k(v)D01;i,p(u)tp,n2+j(u)

=D01;i,p(u)tp,n2+j(u)t0µ1+h,n3+r(v)D4;r,k(v) +t0µ1+h,n3+r(v)D01;i,p(u)D4;r,k(v)tp,n2+j(u)

=D01;i,p(u)tp,n2+j(u)t0µ1+h,n3+r(v)D4;r,k(v) +D1;i,p0 (u)t0µ1+h,n3+r(v)tp,n2+j(u)D4;r,k(v)

=D01;i,p(u)[tp,n2+j(u), t0µ1+h,n3+r(v)]D4;r,k(v) = 0, and the claim follows.

Note that in the above computation we have used the facts that D1;i,j(u) =tij(u) and D4;i,j0 (u) =t0n3+i,n3+j(u), therefore [D1;i,j(u), t0µ

1+h,n3+k(v)] = 0 and [D04;i,j(u), th,n2+k(v)] = 0 by (2.4).

It suffices to prove (7.3) when h=j and k0 =f, by Lemma 6.1(b). Computing the following bracket by Lemma 6.1(b), we have

(u−v)(w−z)

Recall the fact stated in Theorem 1 that Y(glM|N) is generated as an algebra by the set

Da;i,j(r) , D0(r)a;i,j, Ea;i,j(r) , Fa;i,j(r) . The following theorem describes the relations among these generators.

Theorem 3. The following relations hold inY(glM|N)for all admissible indicesa, b, f, g, h, i, j, k, l, r, s, h0, k0:

[D(r)a;i,j, Fb;h,k(s) ] =

Proof. (7.5)−(7.7) follow from Proposition 4.5, while the others come from Proposi-tion 7.1, Lemma 7.2 and the identity

S(v)−S(u)

u−v = X

r,s≥1

S(r+s−1)u−rv−s, for any formal seriesS(u) =P

r≥0S(r)u−r.

Remark 7.1. In the special case where all µi = 1, the right hand side of (7.10) and (7.11) degenerate to zero when a=m. See [Go, Theorem 3].

In fact, the relations in Theorem 3 are enough as defining relations of the super Yangian Y(glM|N).

Theorem 4. The super Yangian Y(glM|N) is generated by the elements {Da;i,j(r) , D0(r)a;i,j|1≤a≤m+n,1≤i, j≤µa, r≥0}, {Ea;i,j(r) |1≤a < m+n,1≤i≤µa,1≤j≤µa+1, r≥1}, {Fa;i,j(r) |1≤a < m+n,1≤i≤µa+1,1≤j ≤µa, r≥1}, subject to the relations (7.5)−(7.20).

Proof. Recall the notationYµ:=Y(glM|N) defined in Section 6. Let Ybµ denote the ab-stract algebra generated by the elements and relations as in the statement of Theorem 4.

We may further define all the otherEa,b;i,j(r) and Fb,a;i,j(r) inYbµ by the relations (3.9), and it is not hard to show that this definition is independent of the choices ofk[BK1, p.22].

Let Γ be the map

Γ :Ybµ−→Yµ

sending every element inYbµinto the element in Yµ with the same name. By Theorem 1 and Theorem 3, the map Γ is a surjective algebra homomorphism. Therefore, it remains to prove that Γ is also injective. The injectivity will be proved in Section 8.

8 Injectivity of Γ

Our strategy of proving the injectivity of Γ is as follows: we find a spanning set for Ybµ (see Proposition 8.1) and show that the images of the spanning set for Ybµ under Γ is linearly independent inYµ (see Proposition 8.4).

Proposition 8.1. Ybµ is spanned as a vector space by the monomials in the elements {D(r)a;i,j, Ea,b;i,j(r) , Fb,a;i,j(r) } taken in certain fixed order.

Proof. LetYbµ0 (resp. Ybµ+, Ybµ) denote the subalgebras of Ybµ generated by the elements {D(r)a;i,j} (resp. {Ea,b;i,j(r) }, {Fb,a;i,j(r) }). By the relations in Theorem 3, Ybµ is spanned by the monomials where allF’s come before allD’s and allD’s come before allE’s.

Define a filtration on Ybµ by setting

deg(D(r)a;i,j) = deg(Ea,b;i,j(r) ) = deg(Fb,a;i,j(r) ) =r−1, for all r≥1,

and denote the associated graded algebra by grLYbµ. The above argument implies that the multiplication map is surjective,

grLYbµ⊗grLYbµ0⊗grLYbµ+grLYbµ.

Moreover, grLYbµ0 is commutative by Proposition 4.5. It follows that Ybµ0 is spanned by the monomials in{Da;i,j(r) }in certain fixed order. Hence it is enough to show thatgrLYbµ+ is spanned by the monomials in E’s in certain order, and the swap mapζN|M will show thatgrLYbµ is spanned by the monomials inF’s in certain order.

We denote the image of Ea,b;i,j(r) in the graded algebra grLr−1Ybµ+ by E(r)a,b;i,j. We have the following.

Claim*: For all admissiblea, b, c, d, i, j, h, k, r, s, we have

[E(r)a,b;i,j, E(s)c,d;h,k] = (−1)bδb,cδh,jE(r+s−1)a,d;i,k −(−1)a b+a c+b cδa,dδi,kE(r+s−1)c,b;h,j . (8.1) Assuming the claim, we have that the graded algebra grLYbµ+ is spanned by the monomials in {E(r)a,b;i,j} in certain order and hence Ybµ+ is spanned by the monomials in {Ea,b;i,j(r) }in certain order as well and therefore Proposition 8.1 is established.

To establish the claim*, we first prove some special cases.

Lemma 8.2. The following identities hold in grLYbµ+: (a)

[E(r)a,a+1;i,j, E(s)b,b+1;h,k] = 0, if|a−b| 6= 1, (8.2) (b)

[E(r)a,a+1;i,j, E(s)b,b+1;h,k] = [E(r−1)a,a+1;i,j, E(s+1)b,b+1;h,k], if|a−b|= 1, (8.3) (c)

E(r)a,a+1;i,j,[E(s)a,a+1;h,k, E(t)b,b+1;f,g]

=−

E(s)a,a+1;i,j,[E(r)a,a+1;h,k, E(t)b,b+1;f,g]

, (8.4) if |a−b|= 1,

(d)

E(r)a,b;i,j = (−1)b−1[E(r)a,b−1;i,h, E(1)b−1,b;h,j] = (−1)a+1[E(1)a,a+1;i,k, E(r)a+1,b;k,j], (8.5) for allb > a+ 1 and any 1≤h≤µb−1, 1≤k≤µa+1.

Proof. (8.2) and (8.3) follow from (7.15) and (7.13). (8.4) follows from (7.17) and (8.5) follows from (3.9).

Lemma 8.3. The following identities hold in grLYbµ+: (a)

[E(r)a,a+2;i,j, E(s)a+1,a+2;h,k] = 0, for all 1≤a≤m+n−2, (8.6) (b)

[E(r)a,a+1;i,j, E(s)a,a+2;h,k] = 0, for all 1≤a≤m+n−2, (8.7) (c)

[E(r)a,a+2;i,j, E(s)a+1,a+3;h,k] = 0, for all 1≤a≤m+n−3, (8.8) (d)

[E(r)a,b;i,j, E(s)c,c+1;h,k] = 0, for all 1≤a < c < b≤m+n. (8.9) Proof. (a) By (8.5) and (8.4), we have

(−1)a+1[E(r)a,a+2;i,j, E(s)a+1,a+2;h,k] =

[E(r)a,a+1;i,f, E(1)a+1,a+2;f,j], E(s)a+1,a+2;h,k

=−

[E(r)a,a+1;i,f, E(s)a+1,a+2;f,j], E(1)a+1,a+2;h,k

=−

[E(r+s−1)a,a+1;i,f, E(1)a+1,a+2;f,j], E(1)a+1,a+2;h,k

and the last term is zero by (8.4).

(b) The same method in (a) works, except that we apply (8.5) on the termE(s)a,a+2;h,k. (c) It takes some effort in this case due to the Z2-grading. First assume that a 6= m−1. We apply (8.5) on the left hand side of (8.8) and use the super-Jacobi identity:

[E(r)a,a+2;i,j, E(s)a+1,a+3;h,k] = (−1)a+1+a+2

[E(r)a,a+1;i,h, E(1)a+1,a+2;h,j],[E(1)a+1,a+2;h,j, E(s)a+2,a+3;j,k]

= (−1)a+1+a+2h

[E(r)a,a+1;i,h, E(1)a+1,a+2;h,j], E(1)a+1,a+2;h,j

, E(s)a+2,a+3;j,k

i

+ε(−1)a+1+a+2h

E(1)a+1,a+2;h,j,

[E(r)a,a+1;i,h, E(1)a+1,a+2;h,j], E(s)a+2,a+3;j,k

i , where ε is (−1)αβ, α is the degree of [E(r)a,a+1;i,h, E(1)a+1,a+2;h,j] and β is the degree of E(1)a+1,a+2;h,j. By (8.4), the first term is zero. Moreover, by our assumption thata6=m−1, the elementsE(1)a+1,a+2;h,j is even and henceεis 1. Keep using the super-Jacobi identity and Lemma 8.2, we may deduce that the above equals to

(−1)a+1+a+2h

E(1)a+1,a+2;h,j,

[E(r)a,a+1;i,h, E(1)a+1,a+2;h,j], E(s)a+2,a+3;j,k

i

=(−1)a+1+a+2h

E(1)a+1,a+2;h,j,

E(r)a,a+1;i,h,[E(1)a+1,a+2;h,j, E(s)a+2,a+3;j,k]i + 0

=(−1)a+1+a+2

[E(1)a+1,a+2;h,j, E(r)a,a+1;i,h],[E(1)a+1,a+2;h,j, E(s)a+2,a+3;j,k] + 0

=−(−1)a+1+a+2

[E(r)a,a+1;i,h, E(1)a+1,a+2;h,j],[E(1)a+1,a+2;h,j, E(s)a+2,a+3;j,k]

=−[E(r)a,a+2;i,j, E(s)a+1,a+3;h,k].

Therefore, (8.8) is true for all a6=m−1.

Now let a=m−1, same method shows that [E(r)m−1,m+1;i,j, E(s)m,m+2;h,k]

=

(−1)m[E(r)m−1,m;i,f, E(1)m,m+1;f,j],(−1)m+1[E(1)m,m+1;h,g, E(s)m+1,m+2;g,k]

[E(r)m−1,m;i,f, E(1)m,m+1;f,j],[E(1)m,m+1;h,g, E(s)m+1,m+2;g,k] , which is zero by (7.1) and hence (8.8) is true when a=m−1 as well.

(d) By super-Jacobi identity and (8.5), it is enough to show the following 2 cases:

[E(r)a,c+1;i,j, E(s)c,c+1;h,k] = 0, for alla < c, (8.10) and

[E(r)a,c+1;i,j, E(s)c,c+2;h,k] = 0, for alla < c. (8.11) They can be proved by using (8.2)−(8.8) and induction on c−a. We show (8.10) in detail here. Whenc=a+ 1, it follows directly from (8.6). Now assume c > a+ 1. By (8.5) and super-Jacobi identity, we have

[E(r)a,c+1;i,j, E(s)c,c+1;h,k] =

(−1)a+1[E(1)a,a+1;i,f, E(r)a+1,c+1;f,j], E(s)c,c+1;h,k

= (−1)a+1

E(1)a,a+1;i,f,[E(r)a+1,c+1;f,j, E(s)c,c+1;h,k]

±

E(r)a+1,c+1;f,j,[E(1)a,a+1;i,f, E(s)c,c+1;h,k] . The first term is zero by induction hypothesis and the second term is also zero by (8.2).

Proof of claim*. Without loss of generality, we may assume that a ≤ c. The proof is split into 7 cases and we prove them one by one.

Case 1. a < b < c < d:

It follows directly from (8.2) and (8.5) that the bracket in (8.1) is zero.

Case 2. a < b=c < d:

By (8.3) and (8.5), we have

[E(r+1)b−1,b;i1,j, E(s+1)b,b+1;h,k1] = [E(r+s+1)b−1,b;i1,j, E(1)b,b+1;h,k1] =δh,j(−1)bE(r+s+1)b−1,b+1;i1,k1. (8.12) Note that whenh6=j, the bracket is zero by (7.13) and hence theδh,j comes out.

Taking the bracket on both sides of the equation (8.12) with the elements E(1)b+1,b+2;k1,k2, E(1)b+2,b+3;k2,k3,· · ·, E(1)d−1,d;kd−1,k

from the right and using the super-Jacobi identity, (8.2) and (8.5), we have [E(r+1)b−1,b;i1,j, E(s+1)b,d;h,k] =δh,j(−1)bE(r+s+1)b−1,d;i1,k. (8.13) Taking brackets on both sides of (8.13) with the elements

E(1)b−2,b−1;i2,i1, E(1)b−3,b−2;i3,i2,· · ·, E(1)a,a+1;i,ib−a−1 from the left and using exactly the same method as above, we have

[E(r)a,b;i,j, E(s)b,d;h,k] =δh,j(−1)bE(r+s−1)a,d;i,k , as desired.

Case 3. a < c < b=d:

Using the super-Jacobi identity, (8.5) and (8.9), we have [E(r)a,b;i,j, E(s)c,b;h,k] =

Using the same method as in Case 3, we have [E(r)a,b;i,j, E(s)c,d;h,k] = Now the bracket in the first term is zero by Case 3, and we may rewrite the whole second term as±[E(r)a,b+1;i,k, E(s)c,b;h,j], which is zero by Case 4. Assume thatd−b >1, The bracket in the first term is zero by induction hypothesis, while the bracket in the second term is zero as well by Case 1.

Case 6. a=c < b < d:

Note that [E(r)a,b;i,j, E(s)a+1,d;f,k] = 0 by Case 5. Hence it is enough to show that [E(r)a,b;i,j, E(1)a,a+1;h,f] = 0, for all b > a. (8.14) We prove (8.14) by induction onb−a≥1. When b−a= 1, it follows from (8.2).

Now assumeb−a >1. By (8.5), we have [E(r)a,b;i,j, E(1)a,a+1;h,f] =

(−1)b−1[E(r)a,b−1;i,g, E(1)b−1,b;g,j], E(1)a,a+1;h,f

= (−1)b−1

E(r)a,b−1;i,g,[E(1)b−1,b;g,j, E(1)a,a+1;h,f]

±(−1)b−1

E(1)b−1,b;g,j,[E(r)a,b−1;i,g, E(1)a,a+1;h,f] . Note that [E(r)a,b−1;i,g, E(1)a,a+1;h,f] = 0 by induction hypothesis. Also by (8.2), [E(1)b−1,b;g,j, E(1)a,a+1;h,f] = 0 unless b−1 = a+ 1. When b −1 = a+ 1, (8.14) becomes [E(r)a,a+2;i,j, E(1)a,a+1;h,f], which is zero by (8.7).

Case 7. a=c < b=d:

We claim that

[E(r)a,b;i,j, E(s)a,b;h,k] = 0. (8.15) Ifb=a+ 1, it follows directly from (8.2). If b > a+ 1, we may expand one term in the bracket of (8.15) by (8.5) as follow.

[E(r)a,b;i,j, E(s)a,b;h,k] =

(−1)b−1[E(r)a,b−1;i,f, E(1)b−1,b;f,j], E(s)a,b;h,k

= (−1)b−1

E(r)a,b−1;i,f,[E(1)b−1,b;f,j, E(s)a,b;h,k]

±(−1)b−1

E(1)b−1,b;f,j,[E(r)a,b−1;i,f, E(s)a,b;h,k] .

Note that [E(1)b−1,b;f,j, E(s)a,b;h,k] = 0 by Case 3 and [E(r)a,b−1;i,f, E(s)a,b;h,k] = 0 by Case 6.

Therefore, we have proved (8.15).

This completes the proof of claim*.

Proposition 8.4. The images of the monomials in Proposition 8.1 underΓare linearly independent.

Proof. By Corollary 2.2, we may identifygrLY(glM|N) =grLYµ with the loop superal-gebraU(glM|N[t]) via

grr−1L t(r)ij 7−→(−1)iEijtr−1. We consider the following composition

grLYbµ⊗grLYbµ0⊗grLYbµ+grLYbµ

−→Γ grLYµ∼=U(glM|N[t]).

Let na := µ12 +. . .+ µa for short. By Proposition 3.1, the image of E(r)a,b;i,j (resp. D(r)a;i,j, F(r)b,a;i,j) under the above composition map is (−1)na+iEna+i,nb+jtr−1

(resp. (−1)na+iEna+i,na+jtr−1, (−1)nb+iEnb+i,na+jtr−1 ). By the PBW theorem for U(glM|N[t]), the set of all monomials in

grr−1L D(r)a;i,j|1≤a≤m+n, 1≤i, j≤µa, r≥1

grLr−1Ea,b;i,j(r) |1≤a < b≤m+n, 1≤i≤µa,1≤j≤µb, r≥1

grLr−1Fb,a;i,j(r) |1≤a < b≤m+n, 1≤i≤µb,1≤j≤µa, r≥1

taken in certain fixed order forms a basis forgrLYµand hence Proposition 8.4 follows.

Let Yµ0,Yµ+ and Yµ denote the subalgebras of Yµ generated by all theD’s,E’s and F’s, respectively. Along the proofs of Proposition 8.1 and Proposition 8.4, we have found the PBW bases for each of these algebras.

Corollary 8.5. (1) The set of monomials in {Da;i,j(r) }1≤a≤m+n,1≤i,j≤µa,r≥1 taken in certain fixed order forms a basis forYµ0.

(2) The set of monomials in {Ea,b;i,j(r) }1≤a<b≤m+n,1≤i≤µa,1≤j≤µb,r≥1 taken in certain fixed order forms a basis forYµ+.

(3) The set of monomials in{Fb,a;i,j(r) }1≤a<b≤m+n,1≤i≤µb,1≤i≤µa,r≥1 taken in certain fixed order forms a basis forYµ.

(4) The set of monomials in the union of the elements listed in (1), (2) and (3) taken in certain fixed order forms a basis forYµ.

Acknowledgements

Thanks to my PhD supervisor Weiqiang Wang for his patient guidance and helpful comments improving this article.

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