Table Of ContentTHE SHUFFLE ALGEBRA REVISITED
2 ANDREINEGUT
1
0
2
p
Abstract. In this paper we introduce certain new features of the shuffle
e
S algebraof[1]thatwillallowustoobtainexplicitformulasfortheisomorphism
between its Drinfelddoubleand the ellipticHall algebraof [3]. These results
5 arenecessaryforourworkin[2].
1
]
A 1. Introduction
Q In [3], the authors have constructed an isomorphism Υ between the positive
. half of the elliptic Hall algebra E+ and the shuffle algebra A+. This isomorphism
h
t is given by generators and relations, and it extends to the Drinfeld doubles of the
a
algebras in question. Our goal in this paper is to make this isomorphism Υ more
m
explicit, by proving:
[
1
Theorem 1.1. We have:
v
9
4
Υ(u )=P (1.1)
3 k,d k,d
3 where the u are the standard generators of E+ (see Subsection 6.1 for the
k,d
.
9 definition) and the P are the minimal shuffle elements of Subsection 4.9.
k,d
0
2
1 Relation (4.12) gives an explicit formula for P , albeit a non-closed one. It
k,d
: will feature in [2], where we will use it to identify P with certain geometric
v k,d
i operators that act on the K−theory of the moduli spaces of sheaves. Let us say a
X
few things about the structure of this paper:
r
a
• In Section 2, we define the shuffle algebra A+ and introduce the crucial
notion of degree.
• In Section 3, we take a notion of limit from [1] and generalize it.
It will provide us with certain coproducts, in an appropriate sense,
andwewillshowthattheyarecoassociativeandrespectthemultiplication.
• In Section 4, we introduce the minimality property that defines the shuffle
elements P . We also introduce the shuffle elements Q , which will be
k,d k,d
important for the proof of Theorem 1.1.
• In Section 5, we present explicit formulas for the P and Q .
k,d k,d
1
2 ANDREINEGUT
• In Section 6, we define the elliptic hall algebra E+ and prove Theorem1.1.
• In Section 7, we introduce the doubles A and E of the above algebras,and
prove an analogue of Theorem 1.1 in Theorem 7.7.
I would like to thank Boris Feigin, Andrei Okounkov, Aleksander Tsymbaliuk
and Eric Vasserot for their interest and numerous helpful discussions.
2. The Shuffle Algebra
2.1. We will work over the field K = C(q ,q ), and let us write for convenience
1 2
q = q q . Consider an infinite set of variables z ,z ,..., and let us look at the
1 2 1 2
K−vector space:
SymK(z1,...,zk), (2.1)
k≥0
M
bigradedbykandhomogenousdegree. WecanendowitwithaK−algebrastructure
via the shuffle product:
P(z ,...,z )∗Q(z ,...,z )=
1 k 1 l
k k+l
z
i
=Sym P(z ,...,z )Q(z ,...,z ) ω (2.2)
1 k k+1 k+l
z
i=1j=k+1 (cid:18) j(cid:19)
Y Y
where:
(x−1)(x−q)
ω(x)= (2.3)
(x−q )(x−q )
1 2
and Sym denotes the symmetrization operator on rational functions. It is normal-
ized such that relation (2.2) can be rewritten as:
z
a
P(z ,...,z )∗Q(z ,...,z )= P(z )Q(z ) ω
1 k 1 l A B
z
{1,...,kX+l}=A⊔B aY∈AbY∈B (cid:18) b(cid:19) (2.4)
|A|=k,|B|=l
where for A={a ,...,a }, we write P(z )=P(z ,...,z ).
1 k A a1 ak
2.2. Theshuffle algebraA+ (see[1])isdefinedasthesubspaceof (2.1)consisting
of rational functions of the form:
(z −z )2·p(z ,...,z )
1≤i<j≤k i j 1 k
P(z ,...,z )= , k ≥1 (2.5)
1 k
(z −q z )(z −q z )
Q1≤i6=j≤k i 1 j i 2 j
where p is a symmetric LauQrent polynomial that satisfies the wheel conditions:
p(z ,q z ,qz ,z ,...,z )=p(z ,q z ,qz ,z ,...,z )=0 (2.6)
1 1 1 1 4 k 1 2 1 1 4 k
THE SHUFFLE ALGEBRA REVISITED 3
Thisconditionisvacuousfork ≤2. WewillcallelementsofA+ shuffle elements.
2.3. The fact that A+ is an algebra follows from:
Proposition 2.4. If P,Q∈A+, then P ∗Q∈A+.
Proof By the definition of multiplication (2.2), for P and Q as in (2.5) we have:
(z −z )2
1≤i<j≤k+l i j
P ∗Q=
ki,Q+j=l1(zi−q1zj)(zi−q2zj)
Q
(z −qz )(z −q z )(z −q z )
i j j 1 i j 2 i
·Sym p(z ,...,z )q(z ,...,z )·
1 k k+1 k+l
z −z
i≤k<j i j (2.7)
Y
The expression on the last line is a rational function with at most simple poles
at z = z . Because it is symmetric, it must necessarily be regular at z = z ,
j i j i
and therefore it is a Laurent polynomial in the z variables. To prove that
P ∗ Q ∈ A+, we need only prove that this expression satisfies the wheel con-
ditions. Infact,wewillshowthateverysummandoftheSymin(2.7)satisfiesthem.
To see this, note that we need to set three of the variablesequalto z,q z,qz (or
1
z,q z,qz, but this case is analogous) and show that the given summand vanishes.
2
If all three of the variables are among {z ,...,z } or {z ,...,z }, then the
1 k k+1 k+l
summand vanishes because p and q satisfy the wheel conditions themselves. If one
of the variables is in {z ,...,z } and two are in {z ,...,z } (or vice-versa)then
1 k k+1 k+l
the product in (2.7) vanishes, and therefore so does the summand.
✷
2.5. In [1] and [3], the shuffle algebra is defined as the subalgebra of (2.1)
generated by SymK(z1). We denote this algebra by A˜+, and Proposition 2.4
implies that A˜+ ⊂A+. In fact, we believe that the two are equal:
Conjecture 2.6.
A˜+ =A+
We will not prove this conjecture, since it is not essential for the results
in this paper. But if we do not accept it, we need to replace the shuffle al-
gebraA+ bythe(apriorismaller)algebraA˜+ inallourdefinitionsandstatements.
2.7. The shuffle algebra is bigraded by the number of variables k and the total
homogenous degree d of shuffle elements:
A+ = A+, A+ = A+
k k k,d
k≥0 d∈Z
M M
Let us consider the linear map ϕ:A+ −→K given by:
k,d
4 ANDREINEGUT
ϕ(P)= P(z ,...,z )· zi−q1zj q−k2+k2d+d+k(q −1)k k q1i−1−q2
1 k z −z 1 1 qi −1
1≤iY6=j≤k i j zi=q−i iY=1 1
1 This map will allow us to normalize our shuffle elements, which will be crucial in
the proof of Theorem 1.1. We will now prove that it behaves nicely with respect
to the shuffle product:
Proposition 2.8. Given P ∈A+ and Q∈A+ , we have:
k,d l,e
ϕ(P ∗Q)=ϕ(P)ϕ(Q)·q(ld−ke)/2
Proof Let us write:
F(k,d):=q−k2+k2d+d+k(q −1)k k q1i−1−q2
1 1 qi −1
i=1 1
Y
Then the LHS equals:
i≤k
z −q z (z −z )(z −qz )
i 1 j i j i j
Sym P(z ,...,z )Q(z ,...,z ) ·
1 k k+1 k+l
z −z (z −q z )(z −q z )
i j i 1 j i 2 j
1≤i6=Yj≤k+l j>Yk+1 zi=q1−i
z −q z z −q z
i 1 j i 1 j
F(k+l,d+e)=Sym P(z ,...,z )Q(z ,...,z )
1 k k+1 k+l
z −z z −z
i j i j
1≤i6=j≤k k+1≤i6=j≤k+l
Y Y
i≤k
(z −qz )(z −q z )
i j j 1 i
·F(k+l,d+e)
(z −q z )(z −z )
i 2 j j i
j>Yk+1 zi=q1−i
The last factor in the numerator means that the only terms which do not vanish
in the aboveSym are those which put the variables{z ,...,z } before the variables
1 k
{z ,...,z }, leaving the above equal to:
k+1 k+l
z −q z z −q z
i 1 j i 1 j
Sym P(z ,...,z ) Sym Q(z ,...,z )
1 k k+1 k+l
z −z z −z
i j i j
1≤iY6=j≤k zi=q1−i k+1≤Yi6=j≤k+l zi=q1−i
1≤i≤k (q−i−qq−j)(q−j −q1−i)
1 1 1 1 ·F(k+l,d+e)=
(q−i−q q−j)(q−j −q−i)
k+1≤j≤k+l 1 2 1 1 1
Y
1≤i≤k (qj−i−1−q )(qj−i+1−1) F(k+l,d+e)
=ϕ(P)ϕ(Q)q−ek+kl 1 2 1 ·
1 (qj−i−q )(qj−i−1) F(k,d)F(l,e)
k+1≤j≤k+l 1 2 1
Y
1Wecouldhavejustaswellchosenq2 insteadofq1,bothmapsworkequallywell
THE SHUFFLE ALGEBRA REVISITED 5
The products on the last line cancel out, leaving the desired RHS.
✷
3. Degrees and Coproducts
3.1. Given a shuffle element P(z ,...,z )∈A+ and a vector of integers:
1 k k,d
s=(0=s ,s ,...,s ,s =d)∈Zk+1,
0 1 k−1 k
we can consider the limit:
P(ξz ,...,ξz ,z ,...,z )
1 i i+1 k
lim (3.1)
ξ−→∞ ξsi
If this limit exists for all i ∈ {0,...,k}, then we say that the degree of P
is≤s. WewilldenotebyBs ⊂A+ thesubspaceofshuffleelementsofdegree≤s. 2
k,d
3.2. More explicitly, recall that any shuffle element can be written as:
(z −z )2 c m (z ,...,z )
1≤a<b≤k a b λ λ λ 1 k
P(z ,...,z )= (3.2)
1 k
(z −q z )(z −q z )
Q 1≤a6=b≤k a P1 b a 2 b
where mλ is the monomial symmQetric function corresponding to the integer parti-
tion with k parts λ = (λ ≥ ... ≥ λ ), and c ∈ K are constants. It is easy to see
1 k λ
that the limit (3.1) exists if and only if:
c 6=0=⇒λ≤λs, where λs =2(n−i)+s −s
λ i i i−1
where the dominance ordering is defined for both degrees and partitions:
s≤s′ ⇔ s =s′ and s ≤s′ ∀j
k k j j
λ≤λ′ ⇔ |λ|=|λ′| and λ +...+λ ≤λ′ +...+λ′ ∀j
1 j 1 j
(3.3)
3.3. With the above notations, if the limit (3.1) exists, then it equals:
1≤a<b≤i(za−zb)2 i<a<b≤k(za−zb)2 ′λcλmλ+i(z1,...,zi)mλ−i(zi+1,...,zk)
qi(k−i) (z −q z )(z −q z ) (z −q z )(z −q z )
Q 1≤a6=b≤iQa 1 b a 2 Pb i<a6=b≤k a 1 b a 2 b (3.4)
where ′ gQoesoverthosepartitionsλoftheQoriginalsumwhichsatisfyλ +...+λ =
λ 1 i
λs+...+λs, and λ±i denote the truncated partitions:
1 P i
λ+i =(λ ≥...≥λ ), λ−i =(λ ≥...≥λ ) (3.5)
1 i i+1 k
2Thegradingsk anddareimplicitlycontained inthevector s
6 ANDREINEGUT
3.4. Less explicitly, (3.4) implies that for any P ∈Bs we can write:
P(ξz ,...,ξz ,z ,...,z )
1 i i+1 k
lim = Q (z ,...,z )R (z ,...,z )
ξ−→∞ ξsi a,i 1 i a,i i+1 k
a
X
forcertainrationalfunctionsQ , R ,withtheindexagoingoversomefiniteset.
a,i a,i
Then we may define the map:
∆s(P)= Q ⊗R (3.6)
i a,i a,i
a
X
Let us define the truncated degree vectors by analogy with (3.5):
s+i =(0,s ,...,s ,s ) s−i =(0,s −s ,...,s −s ,s −s )
1 i−1 i i+1 i k−1 i k i
(3.7)
From the explicit description of ∆s in (3.4), we can claim that:
i
∆s :Bs −→Bsi+ ⊗Bsi− (3.8)
i
This entails the fact that Q and R are both shuffle elements and that they
a,i a,i
have the correct degree, which is easy to check.
3.5. In the next section, the maps ∆s will be shown to induce coproducts on
i
certain commutative subalgebras of A+. For the time being, it is easy to see that
they satisfy “coassociativity”,in the sense that:
(∆sj+ ⊗1)◦∆s =(1⊗∆si−)◦∆s, ∀i<j (3.9)
i j j−i i
Indeed, this follows by applying formula (3.4) twice. Moreover, the following
Lemma shows that the maps ∆s “respect the multiplication”:
i
Lemma 3.6. Consider any degree vectors s and s′, and shuffle elements P ∈ Bs
1
and P ∈Bs′. Then P ∗P ∈Bs+s′, where:
2 1 2
(s+s′)i =jm+ja′=xisj +s′j′ (3.10)
Moreover, we have:
∆is+s′(P1∗P2)= ∆sj(P1)∗∆sj′′(P2) (3.11)
i=j+j′
X
(s+s′)i=sj+s′j′
Proof Let us assume P ∈A+ and write k =k +k . The LHS of (3.11) equals:
t kt 1 2
(P ∗P )(ξz ,...,ξz ,z ,...,z )
1 2 1 i i+1 k
lim =
ξ−→∞ ξ(s+s′)i
THE SHUFFLE ALGEBRA REVISITED 7
= lim |A1|=k1,|A2|=k2 P1(ξ∗zA1)P2(ξ∗zA2) x∈A1 y∈A2ω zzxy
ξ−→∞ ξ(s+Qs′)i Q (cid:16) (cid:17)
{1,...,kX}=A1⊔A2
where the exponent ∗ is 1 or 0, depending on whether or not the corresponding
subscript of z lies in {1,...,i} or not. The above quantity equals:
|B1|=j,|B2|=j′
lim |C1|=k1−j,|C2|=k2−j′ P1(ξzB1,zC1)P2(ξzB2,zC2) xy∈∈BB21∪∪CC21ω zzxy
ξ−→∞ ξ(s+s′)i Q (cid:16) (cid:17)
i=Xj+j′ {1,...,iX}=B1⊔B2
{i+1,...,k}=C1⊔C2
By using the fact that ω(0) = ω(∞) = 1, the above limit is finite, and P ∗P
1 2
therefore lies in Bs+s′. Moreover,we can write the above as:
|B1|=j,|B2|=j′ P (ξz ,z )P (ξz ,z )x∈B1ω zx x∈C2ω zx
lim |C1|=k1−j,|C2|=k2−j′ 1 B1 C1 2 B2 C2 yY∈B2 (cid:18)zy(cid:19)yY∈C2 (cid:18)zy(cid:19)
ξ−→∞i=Xj+j′ {1,...,iX}=B1⊔B2 ξmaxi=j+j′sj+s′j′
{i+1,...,k}=C1⊔C2
which is easily seen to be the RHS of the desired relation.
✷
4. Minimal shuffle elements
4.1. It is quite difficult to express the wheel conditions (2.6) explicitly. But the
notion of degree of the previous section allows us to assert the following:
Lemma 4.2. Consider the degree vector:
id
s(k,d) = −1+δk, i∈{1,...,k} (4.1)
i k i
(cid:24) (cid:25)
Then dimBs(k,d) =1.
Lemma 4.3. Consider the degree vector:
id
S(k,d) = , i∈{1,...,k} (4.2)
i
k
(cid:22) (cid:23)
Then dimBS(k,d) = p(gcd(k,d)), where p(n) denotes the number of partitions of
the number n.
Remark 4.4. The above numbers have a geometric interpretation. If we consider
(k,d) to be a lattice point, then s(k,d) (respectively, S(k,d) ) is the highest
i i
y−coordinate of a lattice point on the vertical line x = i which is strictly below
(respectively, on or below) the line (0,0),(k,d).
8 ANDREINEGUT
✻
q q q q q q q
q q q q q q q
q q q q q q q
✻ ✑
✑
q q q q q ✑q q
✑
✻
q q q ✑q ✑q qs(6,4q)5=S(6,4)5
4
✑
✻
✑
q q ✑q q q q q
✑
✻
❄q✑✑q S(6q,4)3❄q❄s(6,4q)3 q❄ q ✲
✛ ✲
6
In particular, when gcd(k,d)=1 the two lemmas assert the exact same thing.
Remark 4.5. Theaboveinterpretationiscloselyconnectedtothefollowinggeomet-
ric setup, which will be very important to us in Section 6. Consider triangles with
vertices (0,0),(i,j),(k,d), for some i∈{1,...,k−1} and j =s(k,d) :
i
✻
q q q q q q q
q q q q q q q
q q q q q q q
✑✚
✑✚
q q q q q ✑✚q q
✑
✚
✑
q q q ✑q ✚q q q
✑✚
✑✚
q q ✑✚q q q q q
✑✟✟
✟
✑✟
q✑✟ q q q q q q ✲
We will only look at triangles with no inside lattice points. Such a triangle is
calledminimal if it has no lattice points on the edges, except the vertices. Such a
triangleis pictured above. If it has lattice points on onlyone edge, it willbe called
semi-minimal. We will never consider triangles with lattice points on ≥2 edges.
Note that for any (k,d) with k > 1, there exists such a minimal triangle for some
i, simply by taking one of minimal area.
4.6. In [1], the authors have proved the case d = 0 of Lemma 4.3. We will use a
slightly modified version of their argument in order to prove half of Lemmas 4.2
and 4.3, namely the fact that the required dimensions are ≤ than what we claim
they are. For any degree vector s =(0=s ,s ,...,s ,s =d) and any partition
0 1 k−1 k
µ=(µ ≥...≥µ ) of k, consider the evaluation map:
1 t
ϕ :Bs −→K[y±1,...,y±1],
µ 1 t
THE SHUFFLE ALGEBRA REVISITED 9
ϕµ(P)=p(z1,...,zk)|zµ1+...+µs−1+x=ysq1x, ∀s∀1≤x≤µs
where p is the Laurent polynomial of (2.5). Note that ϕ is, up to a constant,
(k)
simply the map of Section (2.7). This construction gives rise to the subspaces:
Bs ⊃Bs = kerϕ (4.3)
µ ν
ν>µ
\
where > refers to the dominance ordering of (3.3). It is easy to see that these
subspaces form a filtration of Bs (if we set Bs =Bs), since:
(k)
µ≤µ′ =⇒Bµs ⊂Bµs′
Proof of the ≤ half of Lemma 4.2 and Lemma 4.3: We will now prove the
inequalities:
dimBs ≤1, dimBS ≤p(gcd(k,d))
where s = s(k,d) and S = S(k,d) are the degree vectors of (4.1) and (4.2). By
using the filtration (4.3), it is easy to see that these inequalities are equivalent to:
k ·ρ
dimϕ (Bs)≤δ(k), dimϕ (BS)≤ δgcd(k,d) , ∀|µ|=k
µ µ µ µ µ µ
|ρ|=Xgcd(k,d) (4.4)
where for a partition ρ = (ρ ≥ ρ ≥ ...) and a positive integer n, we write
1 2
n·ρ = (nρ ≥ nρ ≥ ...). Let σ ∈ {s,S}, take a shuffle element P ∈ Bσ and let
1 2 µ
r = ϕ (P). Because P satisfies the wheel condition (2.6), the Laurent polynomial
µ
r vanishes for:
y =q qa−by , a∈{0,1,...,µ −2}, b∈{0,...,µ −1} (4.5)
j 2 i i j
y =q qa−by , a∈{0,1,...,µ −2}, b∈{0,...,µ −1} (4.6)
j 1 i i j
for i < j, with the correct multiplicities. Because P lies in Bσ = kerϕ , we
µ ν>µ ν
see that r also vanishes for:
T
y =qµi−by and y =q−by , b∈{0,1,...,µ −1} (4.7)
j 1 i j 1 i j
for i<j. Therefore, the Laurent polynomial r is divisible by:
µj−1
r (y ,...,y )= (y −qµi−by )(y −q−ky )
0 1 t j 1 i j 1 i
"
1≤i<j≤k b=0
Y Y
µj−1µi−2
(y −q qa−by )(y −q qa−by ) (4.8)
j 2 i j 1 i
#
b=0 a=0
Y Y
This polynomial has total degree:
10 ANDREINEGUT
deg(r )= 2µ µ =k2− µ2
0 i j i
i<j i
X X
and degree at most:
deg (r )= 2µ µ =2kµ −µ2
yi 0 i j i i
i6=j
X
in each variable y . As for r, it’s easy to see that it has total degree:
i
deg(r)=k(k−1)+d
Because the degree of P is ≤σ, then it has degree at most:
deg (r)=2kµ −µ (µ +1)+σ
yi i i i µi
in each variable y . So the quotient r/r is a Laurent polynomial of total degree:
i 0
µ (µ −1)+d
i i
i
X
and degree at most:
µ (µ −1)+σ
i i µi
in each variable y . If t σ <d, then r/r =0 and therefore dimϕ (Bσ)=0.
i i=1 µi 0 µ µ
If t σ = d, then r/r = const·yσµ1...yσµt and therefore dimϕ (Bσ) ≤ 1. So
i=1 µi P 0 1 t µ µ
looking at the definition of s in (4.1), the first inequality of (4.4) follows from the
P
relation:
t t
µ d
µ =k =⇒ i −1+δk ≤d
i k µi
i=1 i=1(cid:24) (cid:25)
X X
with equality if and only if µ = (k). In the case of S from (4.2), the second
inequality of (4.4) follows from:
t t
µ d
i
µ =k=⇒ ≤d
i
k
i=1 i=1(cid:22) (cid:23)
X X
with equality if and only if k/gcd(k,d) divides µ for all i. Both these inequalities
i
are elementary.
✷
4.7. Take a pair of integers (a,b) with gcd(a,b)=1 and look at:
C(a,b) = BS(na,nb) ⊂ A+
na,nb
n≥0 n≥0
M M
where recall that S(na,nb) = ia . We will often abbreviate this subspace by C
i b
(when a,b are clear from context), and write C for its graded parts.
n
(cid:4) (cid:5)
