the rogers-ramanujan identities: from sums, hopefully to products . - - PowerPoint PPT Presentation

the rogers-ramanujan identities: from sums, hopefully to products

.

Shashank Kanade University of Alberta

. introduction

rogers-ramanujan identities .

RR 1 Partitions of n whose adjacent parts differ by at least 2 are equinumerous with partitions of n with each part 1 4 mod 5 RR 2 Partitions of n whose adjacent parts differ by at least 2 and whose smallest part is at least 2 are equinumerous with partitions of n with each part 2 3 mod 5

2

rogers-ramanujan identities .

RR 1 Partitions of n whose adjacent parts differ by at least 2 are equinumerous with partitions of n with each part 1 4 mod 5 RR 2 Partitions of n whose adjacent parts differ by at least 2 and whose smallest part is at least 2 are equinumerous with partitions of n with each part 2 3 mod 5

2

rogers-ramanujan identities .

RR 1 Partitions of n whose adjacent parts differ by at least 2 are equinumerous with partitions of n with each part ≡ 1, 4 (mod 5) RR 2 Partitions of n whose adjacent parts differ by at least 2 and whose smallest part is at least 2 are equinumerous with partitions of n with each part 2 3 mod 5

2

rogers-ramanujan identities .

RR 1 Partitions of n whose adjacent parts differ by at least 2 are equinumerous with partitions of n with each part ≡ 1, 4 (mod 5) RR 2 Partitions of n whose adjacent parts differ by at least 2 and whose smallest part is at least 2 are equinumerous with partitions of n with each part 2 3 mod 5

2

rogers-ramanujan identities .

RR 1 Partitions of n whose adjacent parts differ by at least 2 are equinumerous with partitions of n with each part ≡ 1, 4 (mod 5) RR 2 Partitions of n whose adjacent parts differ by at least 2 and whose smallest part is at least 2 are equinumerous with partitions of n with each part 2 3 mod 5

2

rogers-ramanujan identities .

RR 1 Partitions of n whose adjacent parts differ by at least 2 are equinumerous with partitions of n with each part ≡ 1, 4 (mod 5) RR 2 Partitions of n whose adjacent parts differ by at least 2 and whose smallest part is at least 2 are equinumerous with partitions of n with each part ≡ 2, 3 (mod 5)

2

rogers-ramanujan identities - example .

Rogers-Ramanujan 1 9 = 9 9 = 9 = 8 + 1 = 6 + 1 + 1 + 1 = 7 + 2 = 4 + 4 + 1 = 6 + 3 = 4 + 1 + 1 + 1 + 1 + 1 = 5 + 3 + 1 = 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 Rogers-Ramanujan 2 9 = 9 9 = 7 2 = 7 2 = 3 3 3 = 6 3 = 3 2 2 2

3

rogers-ramanujan identities - example .

Rogers-Ramanujan 1 9 = 9 9 = 9 = 8 + 1 = 6 + 1 + 1 + 1 = 7 + 2 = 4 + 4 + 1 = 6 + 3 = 4 + 1 + 1 + 1 + 1 + 1 = 5 + 3 + 1 = 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 Rogers-Ramanujan 2 9 = 9 9 = 7 + 2 = 7 + 2 = 3 + 3 + 3 = 6 + 3 = 3 + 2 + 2 + 2

3

rogers-ramanujan identities - generating functions .

d(n) Number of partitions of n with adjacent parts differing by at least 2. RR 1

n

d n qn 1 1 q 1 q4 1 q6 1 q9 1 q

5 1 2 1

q

n 1 1

qn Using Jacobi triple product identity

4

rogers-ramanujan identities - generating functions .

d(n) Number of partitions of n with adjacent parts differing by at least 2. RR 1 ∑

n≥0

d(n)qn = 1 (1 − q)(1 − q4)(1 − q6)(1 − q9) · · · 1 q

5 1 2 1

q

n 1 1

qn Using Jacobi triple product identity

4

rogers-ramanujan identities - generating functions .

d(n) Number of partitions of n with adjacent parts differing by at least 2. RR 1 ∑

n≥0

d(n)qn = 1 (1 − q)(1 − q4)(1 − q6)(1 − q9) · · · = ∑

λ≥0

(−1)λ · qλ(5λ−1)/2(1 + qλ) ∏

n≥1(1 − qn)

Using Jacobi triple product identity

4

rogers-ramanujan identities - generating functions .

d(n) Number of partitions of n with adjacent parts differing by at least 2. RR 1 ∑

n≥0

d(n)qn = 1 (1 − q)(1 − q4)(1 − q6)(1 − q9) · · · = ∑

λ≥0

(−1)λ · qλ(5λ−1)/2(1 + qλ) ∏

n≥1(1 − qn)

. . . Using Jacobi triple product identity

4

. sums to products

setup .

A = C[x−1, x−2, . . . ] IΛ0 = Ideal generated by { r−n =

n−1

∑

i=1

x−ix−n+i ; n ≥ 2 } . r

2

x

1x 1

x2

1

r

3

x

1x 2

x

2x 1

2x

1x 2

r

4

x

1x 3

x

2x 2

x

3x 1

2x

1x 3

x2

2

r

5

x

1x 4

x

2x 3

x

3x 2

x

4x 1

2x

1x 4

2x

2x 3

and so on Definition (actually, a Theorem of Cal-L-M): Principal Subspace We call W I

0 a principal subspace.

6

setup .

A = C[x−1, x−2, . . . ] IΛ0 = Ideal generated by { r−n =

n−1

∑

i=1

x−ix−n+i ; n ≥ 2 } . r−2 = x−1x−1 = x2

−1

r−3 = x−1x−2 + x−2x−1 = 2x−1x−2 r−4 = x−1x−3 + x−2x−2 + x−3x−1 = 2x−1x−3 + x2

−2

r−5 = x−1x−4 + x−2x−3 + x−3x−2 + x−4x−1 = 2x−1x−4 + 2x−2x−3 and so on . . . Definition (actually, a Theorem of Cal-L-M): Principal Subspace We call W I

0 a principal subspace.

6

setup .

A = C[x−1, x−2, . . . ] IΛ0 = Ideal generated by { r−n =

n−1

∑

i=1

x−ix−n+i ; n ≥ 2 } . r−2 = x−1x−1 = x2

−1

r−3 = x−1x−2 + x−2x−1 = 2x−1x−2 r−4 = x−1x−3 + x−2x−2 + x−3x−1 = 2x−1x−3 + x2

−2

r−5 = x−1x−4 + x−2x−3 + x−3x−2 + x−4x−1 = 2x−1x−4 + 2x−2x−3 and so on . . . Definition (actually, a Theorem of Cal-L-M): Principal Subspace We call WΛ0 = A/IΛ0 a principal subspace.

6

setup .

Recall: r−2 = x−1x−1 = x2

−1

r−3 = x−1x−2 + x−2x−1 = 2x−1x−2 r−4 = x−1x−3 + x−2x−3 + x−3x−1 = x2

−2 + 2x−1x−3

r−5 = x−1x−4 + x−2x−3 + x−3x−2 + x−4x−1 = 2x−2x−3 + 2x−1x−4 and so on . . . . ∙ W I

0 has a basis of monomials satisfying difference-2

conditions. ∙ For a proof using Gröbner bases, See Bruschek-Mourtada-Schepers ‘13. (A slightly different space.) ∙ In this paper, it comes up while calculating Hilbert-Poincaré series

• f arc space of a double point.

∙ Shows up in a lot of different problems — more later.

7

setup .

Recall: r−2 = x−1x−1 = x2

−1

r−3 = x−1x−2 + x−2x−1 = 2x−1x−2 r−4 = x−1x−3 + x−2x−3 + x−3x−1 = x2

−2 + 2x−1x−3

r−5 = x−1x−4 + x−2x−3 + x−3x−2 + x−4x−1 = 2x−2x−3 + 2x−1x−4 and so on . . . . ∙ WΛ0 = A/IΛ0 has a basis of monomials satisfying difference-2 conditions. ∙ For a proof using Gröbner bases, See Bruschek-Mourtada-Schepers ‘13. (A slightly different space.) ∙ In this paper, it comes up while calculating Hilbert-Poincaré series

• f arc space of a double point.

∙ Shows up in a lot of different problems — more later.

7

setup .

Recall: r−2 = x−1x−1 = x2

−1

r−3 = x−1x−2 + x−2x−1 = 2x−1x−2 r−4 = x−1x−3 + x−2x−3 + x−3x−1 = x2

−2 + 2x−1x−3

r−5 = x−1x−4 + x−2x−3 + x−3x−2 + x−4x−1 = 2x−2x−3 + 2x−1x−4 and so on . . . . ∙ WΛ0 = A/IΛ0 has a basis of monomials satisfying difference-2 conditions. ∙ For a proof using Gröbner bases, See Bruschek-Mourtada-Schepers ‘13. (A slightly different space.) ∙ In this paper, it comes up while calculating Hilbert-Poincaré series

• f arc space of a double point.

∙ Shows up in a lot of different problems — more later.

7

setup .

Recall: r−2 = x−1x−1 = x2

−1

r−3 = x−1x−2 + x−2x−1 = 2x−1x−2 r−4 = x−1x−3 + x−2x−3 + x−3x−1 = x2

−2 + 2x−1x−3

r−5 = x−1x−4 + x−2x−3 + x−3x−2 + x−4x−1 = 2x−2x−3 + 2x−1x−4 and so on . . . . ∙ WΛ0 = A/IΛ0 has a basis of monomials satisfying difference-2 conditions. ∙ For a proof using Gröbner bases, See Bruschek-Mourtada-Schepers ‘13. (A slightly different space.) ∙ In this paper, it comes up while calculating Hilbert-Poincaré series

• f arc space of a double point.

∙ Shows up in a lot of different problems — more later.

7

setup .

Recall: r−2 = x−1x−1 = x2

−1

r−3 = x−1x−2 + x−2x−1 = 2x−1x−2 r−4 = x−1x−3 + x−2x−3 + x−3x−1 = x2

−2 + 2x−1x−3

r−5 = x−1x−4 + x−2x−3 + x−3x−2 + x−4x−1 = 2x−2x−3 + 2x−1x−4 and so on . . . . ∙ WΛ0 = A/IΛ0 has a basis of monomials satisfying difference-2 conditions. ∙ For a proof using Gröbner bases, See Bruschek-Mourtada-Schepers ‘13. (A slightly different space.) ∙ In this paper, it comes up while calculating Hilbert-Poincaré series

• f arc space of a double point.

∙ Shows up in a lot of different problems — more later.

7

products? .

Question Where are the products? Idea (J. Lepowsky) First use the Jacobi triple product identity 1 1 q 1 q4 1 q6 1 q9 1 q

5 1 2 1

q

n 1 1

qn alternating sum with each series having non-negative coefficients Could be explained via Euler-Poincaré principle applied to a resolution.

8

products? .

Question Where are the products? Idea (J. Lepowsky) First use the Jacobi triple product identity 1 (1 − q)(1 − q4)(1 − q6)(1 − q9) · · · = ∑

λ≥0

(−1)λ · qλ(5λ−1)/2(1 + qλ) ∏

n≥1(1 − qn)

= alternating sum with each series having non-negative coefficients Could be explained via Euler-Poincaré principle applied to a resolution.

8

products? .

Question Where are the products? Idea (J. Lepowsky) First use the Jacobi triple product identity 1 (1 − q)(1 − q4)(1 − q6)(1 − q9) · · · = ∑

λ≥0

(−1)λ · qλ(5λ−1)/2(1 + qλ) ∏

n≥1(1 − qn)

= alternating sum with each series having non-negative coefficients Could be explained via Euler-Poincaré principle applied to a resolution.

8

koszul complex .

· · · C3 = ⊕

i1,i2≤−2

Aξ i1, i2

∂3

− → C2 = ⊕

i1≤−2

Aξ i1

∂2

− → C1 = A

∂1

− → C0 = WΛ0 ↠ 0 ∙

i j i j j i

. ∙

1 is the projection map

I

0.

∙

k 1 i1 i2 ik k n 1

1 n

1 r in i1 i2 in ik

r

i i

r

i i j

r

i j

r

j i

∙ Hn Ker

n

Im

n 1

H0 H1 0.

9

koszul complex .

· · · C3 = ⊕

i1,i2≤−2

Aξ i1, i2

∂3

− → C2 = ⊕

i1≤−2

Aξ i1

∂2

− → C1 = A

∂1

− → C0 = WΛ0 ↠ 0 ∙ ξ··· ,i,··· , j,··· = · · · ∧ ξi ∧ · · · ∧ ξj ∧ · · · = −ξ··· , j,··· ,i,···. ∙

1 is the projection map

I

0.

∙

k 1 i1 i2 ik k n 1

1 n

1 r in i1 i2 in ik

r

i i

r

i i j

r

i j

r

j i

∙ Hn Ker

n

Im

n 1

H0 H1 0.

9

koszul complex .

· · · C3 = ⊕

i1,i2≤−2

Aξ i1, i2

∂3

− → C2 = ⊕

i1≤−2

Aξ i1

∂2

− → C1 = A

∂1

− → C0 = WΛ0 ↠ 0 ∙ ξ··· ,i,··· , j,··· = · · · ∧ ξi ∧ · · · ∧ ξj ∧ · · · = −ξ··· , j,··· ,i,···. ∙ ∂1 is the projection map A − → A/IΛ0. ∙

k 1 i1 i2 ik k n 1

1 n

1 r in i1 i2 in ik

r

i i

r

i i j

r

i j

r

j i

∙ Hn Ker

n

Im

n 1

H0 H1 0.

9

koszul complex .

· · · C3 = ⊕

i1,i2≤−2

Aξ i1, i2

∂3

− → C2 = ⊕

i1≤−2

Aξ i1

∂2

− → C1 = A

∂1

− → C0 = WΛ0 ↠ 0 ∙ ξ··· ,i,··· , j,··· = · · · ∧ ξi ∧ · · · ∧ ξj ∧ · · · = −ξ··· , j,··· ,i,···. ∙ ∂1 is the projection map A − → A/IΛ0. ∙ ∂k+1(ξ−i1,−i2,··· ,−ik) =

k

∑

n=1

(−1)n−1 · r−in · ξ−i1,−i2,··· ,

−in,··· ,−ik.

∂(r−i) = 0 ∂(ξ−i) = r−i ∂(ξ−i,−j) = r−iξ−j − r−jξ−i ∙ Hn Ker

n

Im

n 1

H0 H1 0.

9

koszul complex .

· · · C3 = ⊕

i1,i2≤−2

Aξ i1, i2

∂3

− → C2 = ⊕

i1≤−2

Aξ i1

∂2

− → C1 = A

∂1

− → C0 = WΛ0 ↠ 0 ∙ ξ··· ,i,··· , j,··· = · · · ∧ ξi ∧ · · · ∧ ξj ∧ · · · = −ξ··· , j,··· ,i,···. ∙ ∂1 is the projection map A − → A/IΛ0. ∙ ∂k+1(ξ−i1,−i2,··· ,−ik) =

k

∑

n=1

(−1)n−1 · r−in · ξ−i1,−i2,··· ,

−in,··· ,−ik.

∂(r−i) = 0 ∂(ξ−i) = r−i ∂(ξ−i,−j) = r−iξ−j − r−jξ−i ∙ Hn = Ker(∂n)/Im(∂n+1) H0 = H1 = 0.

9

koszul complex .

· · · C3 = ⊕

i1,i2≤−2

Aξ i1, i2

∂3

− → C2 = ⊕

i1≤−2

Aξ i1

∂2

− → C1 = A

∂1

− → C0 = WΛ0 ↠ 0 Interpretation

2 i1 2 i1

Ker

2 is precisely the the space of relations amongst the rns. 3 i1 i2 2 i1 i2 i1 2 i1

Im

3 is precisely the the space of trivial relations amongst the rns:

rn rm rm rn 0. H2 Ker

2

Im

3 measures the space of “non-trivial” relations. 10

koszul complex .

· · · C3 = ⊕

i1,i2≤−2

Aξ i1, i2

∂3

− → C2 = ⊕

i1≤−2

Aξ i1

∂2

− → C1 = A

∂1

− → C0 = WΛ0 ↠ 0 Interpretation ∂2 : ⊕

i1≤−2

Aξ i1− → A Ker(∂2) is precisely the the space of relations amongst the rns.

3 i1 i2 2 i1 i2 i1 2 i1

Im

3 is precisely the the space of trivial relations amongst the rns:

rn rm rm rn 0. H2 Ker

2

Im

3 measures the space of “non-trivial” relations. 10

koszul complex .

· · · C3 = ⊕

i1,i2≤−2

Aξ i1, i2

∂3

− → C2 = ⊕

i1≤−2

Aξ i1

∂2

− → C1 = A

∂1

− → C0 = WΛ0 ↠ 0 Interpretation ∂2 : ⊕

i1≤−2

Aξ i1− → A Ker(∂2) is precisely the the space of relations amongst the rns. ∂3 : ⊕

i1,i2≤−2

Aξ i1 i2− → ⊕

i1≤−2

Aξ i1 Im(∂3) is precisely the the space of trivial relations amongst the rns: rn · rm − rm · rn = 0. H2 Ker

2

Im

3 measures the space of “non-trivial” relations. 10

koszul complex .

· · · C3 = ⊕

i1,i2≤−2

Aξ i1, i2

∂3

− → C2 = ⊕

i1≤−2

Aξ i1

∂2

− → C1 = A

∂1

− → C0 = WΛ0 ↠ 0 Interpretation ∂2 : ⊕

i1≤−2

Aξ i1− → A Ker(∂2) is precisely the the space of relations amongst the rns. ∂3 : ⊕

i1,i2≤−2

Aξ i1 i2− → ⊕

i1≤−2

Aξ i1 Im(∂3) is precisely the the space of trivial relations amongst the rns: rn · rm − rm · rn = 0. H2 = Ker(∂2)/Im(∂3) measures the space of “non-trivial” relations.

10

euler-poincaré principle .

· · · C3 = ⊕

i1,i2≤−2

Aξ i1, i2

∂3

− → C2 = ⊕

i1≤−2

Aξ i1

∂2

− → C1 = A

∂1

− → C0 = WΛ0 ↠ 0 Euler-Poincaré principle With being the “dimension” (actually, the x q –character) W

0 x q

n 1

1 n

1

Cn x q Hn x q The Problem Find the precise structure of Hns and calculate Hn x q .

11

euler-poincaré principle .

· · · C3 = ⊕

i1,i2≤−2

Aξ i1, i2

∂3

− → C2 = ⊕

i1≤−2

Aξ i1

∂2

− → C1 = A

∂1

− → C0 = WΛ0 ↠ 0 Euler-Poincaré principle With χ being the “dimension” (actually, the (x, q)–character) χ(WΛ0; x, q) = ∑

n≥1

(−1)n+1 (χ(Cn; x, q) − χ(Hn; x, q)) . The Problem Find the precise structure of Hns and calculate Hn x q .

11

euler-poincaré principle .

· · · C3 = ⊕

i1,i2≤−2

Aξ i1, i2

∂3

− → C2 = ⊕

i1≤−2

Aξ i1

∂2

− → C1 = A

∂1

− → C0 = WΛ0 ↠ 0 Euler-Poincaré principle With χ being the “dimension” (actually, the (x, q)–character) χ(WΛ0; x, q) = ∑

n≥1

(−1)n+1 (χ(Cn; x, q) − χ(Hn; x, q)) . The Problem Find the precise structure of Hns and calculate χ(Hn; x, q).

11

main results .

There is a derivation L−1 of A, that can be extended to the Cj’s: L

1 x j

j x

j 1

L

1 r j

j 1 r

j 1

L

1 i1 ik

i1 1

i1 1 i2 ik

i2 1

i1 i2 1 ik

ik 1

i1 i2 ik 1

L

1 a c

L

1 a c

aL

1 c for all a

c Cj L

1

L

1Ker

Ker L

1Im

Im

12

main results .

There is a derivation L−1 of A, that can be extended to the Cj’s: L−1 · x−j = j · x−j−1 L

1 r j

j 1 r

j 1

L

1 i1 ik

i1 1

i1 1 i2 ik

i2 1

i1 i2 1 ik

ik 1

i1 i2 ik 1

L

1 a c

L

1 a c

aL

1 c for all a

c Cj L

1

L

1Ker

Ker L

1Im

Im

12

main results .

There is a derivation L−1 of A, that can be extended to the Cj’s: L−1 · x−j = j · x−j−1 L−1 · r−j = (j − 1) · r−j−1 L−1 · (ξ−i1,··· ,−ik) = (i1 − 1)ξ−i1−1,−i2,··· ,−ik + (i2 − 1)ξ−i1,−i2−1,··· ,−ik + · · · + (ik − 1)ξ−i1,−i2··· ,−ik−1 L

1 a c

L

1 a c

aL

1 c for all a

c Cj L

1

L

1Ker

Ker L

1Im

Im

12

main results .

There is a derivation L−1 of A, that can be extended to the Cj’s: L−1 · x−j = j · x−j−1 L−1 · r−j = (j − 1) · r−j−1 L−1 · (ξ−i1,··· ,−ik) = (i1 − 1)ξ−i1−1,−i2,··· ,−ik + (i2 − 1)ξ−i1,−i2−1,··· ,−ik + · · · + (ik − 1)ξ−i1,−i2··· ,−ik−1 L−1(a · c) = L−1(a)c + aL−1(c) for all a ∈ A, c ∈ Cj L

1

L

1Ker

Ker L

1Im

Im

12

main results .

There is a derivation L−1 of A, that can be extended to the Cj’s: L−1 · x−j = j · x−j−1 L−1 · r−j = (j − 1) · r−j−1 L−1 · (ξ−i1,··· ,−ik) = (i1 − 1)ξ−i1−1,−i2,··· ,−ik + (i2 − 1)ξ−i1,−i2−1,··· ,−ik + · · · + (ik − 1)ξ−i1,−i2··· ,−ik−1 L−1(a · c) = L−1(a)c + aL−1(c) for all a ∈ A, c ∈ Cj [∂, L−1] = 0 L−1Ker(∂) ⊂ Ker(∂) L−1Im(∂) ⊂ Im(∂)

12

main results .

There is an automorphism σ of A, that can be extended to the Cjs: x

1

x

i

x

i 1

r

i

r

i 2

r

2

r

3 i1 ik i1 2 ik 2

a c a c for all a c Cj Ker Ker Im Im

13

main results .

There is an automorphism σ of A, that can be extended to the Cjs: σ(x−1) = 0, σ(x−i) = x−i+1 σ(r−i) = r−i+2, σ(r−2) = 0, σ(r−3) = 0 σ(ξ−i1,··· ,−ik) = ξ−i1+2,··· ,−ik+2 a c a c for all a c Cj Ker Ker Im Im

13

main results .

There is an automorphism σ of A, that can be extended to the Cjs: σ(x−1) = 0, σ(x−i) = x−i+1 σ(r−i) = r−i+2, σ(r−2) = 0, σ(r−3) = 0 σ(ξ−i1,··· ,−ik) = ξ−i1+2,··· ,−ik+2 σ(a · c) = σ(a)σ(c) for all a ∈ A, c ∈ Cj Ker Ker Im Im

13

main results .

There is an automorphism σ of A, that can be extended to the Cjs: σ(x−1) = 0, σ(x−i) = x−i+1 σ(r−i) = r−i+2, σ(r−2) = 0, σ(r−3) = 0 σ(ξ−i1,··· ,−ik) = ξ−i1+2,··· ,−ik+2 σ(a · c) = σ(a)σ(c) for all a ∈ A, c ∈ Cj [∂, σ] = 0 σKer(∂) ⊂ Ker(∂) σIm(∂) ⊂ Im(∂)

13

main results .

Some obvious elements in the Ker(∂2): µ−4 = 2x−2ξ−2 − x−1ξ−3 ∂(µ−4) = 2x−2 · x2

−1 − x−1 · 2x−1x−2 = 0

∂(Ls

−1 · µ−4) = 0 for s ∈ N. 4

2x

2 2

x

1 3 5

4x

3 2

x

2 3

2x

1 4 6

6x

4 2

3x

3 3

3x

1 5 7

8x

5 2

5x

4 3

2x

3 4

x

2 5

4x

1 6 8

10x

6 2

7x

5 3

4x

4 4

x

3 5

2x

2 6

5x

1 7

and so on …

14

main results .

Some obvious elements in the Ker(∂2): µ−4 = 2x−2ξ−2 − x−1ξ−3 ∂(µ−4) = 2x−2 · x2

−1 − x−1 · 2x−1x−2 = 0

∂(Ls

−1 · µ−4) = 0 for s ∈ N.

µ−4 = 2x−2ξ−2 − x−1ξ−3 µ−5 = 4x−3ξ−2 + x−2ξ−3 − 2x−1ξ−4 µ−6 = 6x−4ξ−2 + 3x−3ξ−3 − 3x−1ξ−5 µ−7 = 8x−5ξ−2 + 5x−4ξ−3 + 2x−3ξ−4 − x−2ξ−5 − 4x−1ξ−6 µ−8 = 10x−6ξ−2 + 7x−5ξ−3 + 4x−4ξ−4 + x−3ξ−5 − 2x−2ξ−6 − 5x−1ξ−7 and so on …

14

main results .

Theorem (K.) The second homology H2 is generated by the elements Ls

−1 · µ−4 for

s ∈ N. Remark The proof is very similar to the proof of presentation of W

0 in

Calinescu-Lepowsky-Milas ‘08. Uses the “minimal counter-example” technique.

15

main results .

Theorem (K.) The second homology H2 is generated by the elements Ls

−1 · µ−4 for

s ∈ N. Remark The proof is very similar to the proof of presentation of WΛ0 in Calinescu-Lepowsky-Milas ‘08. Uses the “minimal counter-example” technique.

15

. context

• 1. principal subspaces

.

g = sl2 = C{xα, α, x−α} ⟨a, b⟩ = Tr(ab) x t t

1

c x tn y tm x y tm

n

x y n m

n 0c

c A 1

1

t t

1

t

1

t

1 17

• 1. principal subspaces

.

g = sl2 = C{xα, α, x−α} ⟨a, b⟩ = Tr(ab) n = C xα t t

1

c x tn y tm x y tm

n

x y n m

n 0c

c A 1

1

t t

1

t

1

t

1 17

• 1. principal subspaces

.

g = sl2 = C{xα, α, x−α} ⟨a, b⟩ = Tr(ab) n = C xα

• g = g ⊗ C[t, t−1] ⊕ Cc

[x ⊗ tn, y ⊗ tm] = [x, y] ⊗ tm+n + ⟨x, y⟩nδm+n,0c [c, g] = 0

• g ∼

= A(1)

1

t t

1

t

1

t

1 17

• 1. principal subspaces

.

g = sl2 = C{xα, α, x−α} ⟨a, b⟩ = Tr(ab) n = C xα

• g = g ⊗ C[t, t−1] ⊕ Cc

[x ⊗ tn, y ⊗ tm] = [x, y] ⊗ tm+n + ⟨x, y⟩nδm+n,0c [c, g] = 0

• g ∼

= A(1)

1

¯ n = n ⊗ C[t, t−1] ⊂ g ¯ n− = n ⊗ t−1C[t−1] ⊂ g [¯ n−, ¯ n−] = 0.

17

• 1. principal subspaces

.

L(Λ) : Irreducible, integrable g − module generated by highest wt. vector vΛ WΛ := U(¯ n) · vΛ ∼ = U(¯ n−) · vΛ x

1 x 2

f W 1 v Theorem (Calinescu-Lepowsky-Milas ‘08) Ker f I

0 and Ker f 1

I x

1. 18

• 1. principal subspaces

.

L(Λ) : Irreducible, integrable g − module generated by highest wt. vector vΛ WΛ := U(¯ n) · vΛ ∼ = U(¯ n−) · vΛ U(¯ n−) ∼ = C[x−1, x−2, . . . ] = A f W 1 v Theorem (Calinescu-Lepowsky-Milas ‘08) Ker f I

0 and Ker f 1

I x

1. 18

• 1. principal subspaces

.

L(Λ) : Irreducible, integrable g − module generated by highest wt. vector vΛ WΛ := U(¯ n) · vΛ ∼ = U(¯ n−) · vΛ U(¯ n−) ∼ = C[x−1, x−2, . . . ] = A fΛ : U(¯ n−) − → WΛ 1 → vΛ Theorem (Calinescu-Lepowsky-Milas ‘08) Ker f I

0 and Ker f 1

I x

1. 18

• 1. principal subspaces

.

L(Λ) : Irreducible, integrable g − module generated by highest wt. vector vΛ WΛ := U(¯ n) · vΛ ∼ = U(¯ n−) · vΛ U(¯ n−) ∼ = C[x−1, x−2, . . . ] = A fΛ : U(¯ n−) − → WΛ 1 → vΛ Theorem (Calinescu-Lepowsky-Milas ‘08) Ker(fΛ0) = IΛ0 and Ker(fΛ1) = IΛ0 + Ax−1.

18

• 2. garland-lepowsky resolution

.

. . . E2 = ⟨vr0r1·Λ0⟩ . E1 = ⟨vr0·Λ0⟩ . . E0 = ⟨vΛ0⟩ . . LΛ0 . . ⊕

i,j≤−2

Aξi,j . ⊕

i≤−2

Aξi . . A . . WΛ0 vr0 x2 1 v x2

1v

vr0

2

vr0r1 2x 2 vr0 x 1 L 1 vr0 2x

2 2

x

1L 1 2

2x

2 2

x

1 3 4 19

• 2. garland-lepowsky resolution

.

. . . E2 = ⟨vr0r1·Λ0⟩ . E1 = ⟨vr0·Λ0⟩ . . E0 = ⟨vΛ0⟩ . . LΛ0 . . ⊕

i,j≤−2

Aξi,j . ⊕

i≤−2

Aξi . . A . . WΛ0 vr0Λ0 → x2

α(−1)vΛ0 = x2 −1vΛ0

vr0

2

vr0r1 2x 2 vr0 x 1 L 1 vr0 2x

2 2

x

1L 1 2

2x

2 2

x

1 3 4 19

• 2. garland-lepowsky resolution

.

. . . E2 = ⟨vr0r1·Λ0⟩ . E1 = ⟨vr0·Λ0⟩ . . E0 = ⟨vΛ0⟩ . . LΛ0 . . ⊕

i,j≤−2

Aξi,j . ⊕

i≤−2

Aξi . . A . . WΛ0 vr0Λ0 → x2

α(−1)vΛ0 = x2 −1vΛ0

vr0Λ0 ∼ ξ−2 vr0r1 2x 2 vr0 x 1 L 1 vr0 2x

2 2

x

1L 1 2

2x

2 2

x

1 3 4 19

• 2. garland-lepowsky resolution

.

. . . E2 = ⟨vr0r1·Λ0⟩ . E1 = ⟨vr0·Λ0⟩ . . E0 = ⟨vΛ0⟩ . . LΛ0 . . ⊕

i,j≤−2

Aξi,j . ⊕

i≤−2

Aξi . . A . . WΛ0 vr0Λ0 → x2

α(−1)vΛ0 = x2 −1vΛ0

vr0Λ0 ∼ ξ−2 vr0r1·Λ0 → 2xα(−2)vr0·Λ0 − xα(−1)L(−1)vr0·Λ0 2x

2 2

x

1L 1 2

2x

2 2

x

1 3 4 19

• 2. garland-lepowsky resolution

.

. . . E2 = ⟨vr0r1·Λ0⟩ . E1 = ⟨vr0·Λ0⟩ . . E0 = ⟨vΛ0⟩ . . LΛ0 . . ⊕

i,j≤−2

Aξi,j . ⊕

i≤−2

Aξi . . A . . WΛ0 vr0Λ0 → x2

α(−1)vΛ0 = x2 −1vΛ0

vr0Λ0 ∼ ξ−2 vr0r1·Λ0 → 2xα(−2)vr0·Λ0 − xα(−1)L(−1)vr0·Λ0 ∼ 2x−2ξ−2 − x−1L−1ξ−2 2x

2 2

x

1 3 4 19

• 2. garland-lepowsky resolution

.

. . . E2 = ⟨vr0r1·Λ0⟩ . E1 = ⟨vr0·Λ0⟩ . . E0 = ⟨vΛ0⟩ . . LΛ0 . . ⊕

i,j≤−2

Aξi,j . ⊕

i≤−2

Aξi . . A . . WΛ0 vr0Λ0 → x2

α(−1)vΛ0 = x2 −1vΛ0

vr0Λ0 ∼ ξ−2 vr0r1·Λ0 → 2xα(−2)vr0·Λ0 − xα(−1)L(−1)vr0·Λ0 ∼ 2x−2ξ−2 − x−1L−1ξ−2 = 2x−2ξ−2 − x−1ξ−3

4 19

• 2. garland-lepowsky resolution

.

. . . E2 = ⟨vr0r1·Λ0⟩ . E1 = ⟨vr0·Λ0⟩ . . E0 = ⟨vΛ0⟩ . . LΛ0 . . ⊕

i,j≤−2

Aξi,j . ⊕

i≤−2

Aξi . . A . . WΛ0 vr0Λ0 → x2

α(−1)vΛ0 = x2 −1vΛ0

vr0Λ0 ∼ ξ−2 vr0r1·Λ0 → 2xα(−2)vr0·Λ0 − xα(−1)L(−1)vr0·Λ0 ∼ 2x−2ξ−2 − x−1L−1ξ−2 = 2x−2ξ−2 − x−1ξ−3 = µ−4

19

• 3. relations to khovanov homology of torus knots

.

Stable unreduced Khovanov homology. Kh T n lim

m

q

n 1 m 1 1Kh T n m .

This limit exists (Stošić). Conjecture (Gorsky-Oblomkov-Rasmussen ‘12) Kh T n is dual to the homology of the Koszul complex determined by the elements r

2

r

n

• 1. (Note: the gradings are

different than ours.)

20

• 3. relations to khovanov homology of torus knots

.

Stable unreduced Khovanov homology. Kh(T(n, ∞)) = lim

m→∞ q−(n−1)(m−1)+1Kh(T(n, m)).

This limit exists (Stošić). Conjecture (Gorsky-Oblomkov-Rasmussen ‘12) Kh T n is dual to the homology of the Koszul complex determined by the elements r

2

r

n

• 1. (Note: the gradings are

different than ours.)

20

• 3. relations to khovanov homology of torus knots

.

Stable unreduced Khovanov homology. Kh(T(n, ∞)) = lim

m→∞ q−(n−1)(m−1)+1Kh(T(n, m)).

This limit exists (Stošić). Conjecture (Gorsky-Oblomkov-Rasmussen ‘12) Kh(T(n, ∞)) is dual to the homology of the Koszul complex determined by the elements r−2, . . . , r−n−1. (Note: the gradings are different than ours.)

20

relations to khovanov homology of torus knots .

Every Koszul complex is a dg-algebra: ξ−i1,...,−ij = ξ−i1 ∧ · · · ∧ ξ−ij. Z

n

Ker

n is a sub-algebra with

B

n

Im

n , a two-sided ideal.

H

n

Hn is a graded algebra.

21

relations to khovanov homology of torus knots .

Every Koszul complex is a dg-algebra: ξ−i1,...,−ij = ξ−i1 ∧ · · · ∧ ξ−ij. Z = ⊕

n≥0

Ker(∂n) is a sub-algebra with B = ⊕

n≥0

Im(∂n), a two-sided ideal. H

n

Hn is a graded algebra.

21

relations to khovanov homology of torus knots .

Every Koszul complex is a dg-algebra: ξ−i1,...,−ij = ξ−i1 ∧ · · · ∧ ξ−ij. Z = ⊕

n≥0

Ker(∂n) is a sub-algebra with B = ⊕

n≥0

Im(∂n), a two-sided ideal. H = ⊕

n≥0

Hn is a graded algebra.

21

• 3. relations to khovanov homology of torus knots

.

Conjecture (Gorsky-Oblomkov-Rasmussen ‘12) For T(n, ∞), H is generated as by the elements µ−4, . . . , µ−n−2, with the defining relations being x(z)2 = 0 x(z)µ(z) = 0 x′′(z)µ(z) − x′(z)µ′(z) = 0 µ(z)µ′(z) = 0, recall that µ(z) = 2x′(z)ξ(z) − x(z)ξ′(z). : unreduced reduced.

22

• 3. relations to khovanov homology of torus knots

.

Conjecture (Gorsky-Oblomkov-Rasmussen ‘12) For T(n, ∞), H is generated as by the elements µ−4, . . . , µ−n−2, with the defining relations being x(z)2 = 0 x(z)µ(z) = 0 x′′(z)µ(z) − x′(z)µ′(z) = 0 µ(z)µ′(z) = 0, recall that µ(z) = 2x′(z)ξ(z) − x(z)ξ′(z). σ: unreduced − → reduced.

22

• 4. relations to meurman-primc’s work

.

E0 − → L(Λ0) R = U(g)x2

−11 ⊂ E0

¯ R = Coeffs of r(x) = Y(r, x), r ∈ R R1 E0 E0 u v u

1v

Y u x Y v x Y u x Y v x u R1 Meurman-Primc ‘99, Primc ‘02 The kernel of the map above is generated, in some sense, by: d dx x x 2 x x 2 x z 2 d dxx z L 1 x2

11

x

11

2x2

11

L 1 x

11

L

1 2 x 1

2

2 L 1x 1 3 x 1

2

2 x 2 4 23

• 4. relations to meurman-primc’s work

.

E0 − → L(Λ0) R = U(g)x2

−11 ⊂ E0

¯ R = Coeffs of r(x) = Y(r, x), r ∈ R ¯ R1 ⊗ E0 − → E0 u ⊗ v − → u−1v Y(u, x) ⊗ Y(v, x) − → •

• Y(u, x)Y(v, x)•
• , u ∈ ¯

R1 Meurman-Primc ‘99, Primc ‘02 The kernel of the map above is generated, in some sense, by: d dx x x 2 x x 2 x z 2 d dxx z L 1 x2

11

x

11

2x2

11

L 1 x

11

L

1 2 x 1

2

2 L 1x 1 3 x 1

2

2 x 2 4 23

• 4. relations to meurman-primc’s work

.

E0 − → L(Λ0) R = U(g)x2

−11 ⊂ E0

¯ R = Coeffs of r(x) = Y(r, x), r ∈ R ¯ R1 ⊗ E0 − → E0 u ⊗ v − → u−1v Y(u, x) ⊗ Y(v, x) − → •

• Y(u, x)Y(v, x)•
• , u ∈ ¯

R1 Meurman-Primc ‘99, Primc ‘02 The kernel of the map above is generated, in some sense, by: d dx ( xθ(x)2) ⊗ xθ(x) − 2(xθ(z)2) ⊗ d dxxθ(z) L 1 x2

11

x

11

2x2

11

L 1 x

11

L

1 2 x 1

2

2 L 1x 1 3 x 1

2

2 x 2 4 23

• 4. relations to meurman-primc’s work

.

E0 − → L(Λ0) R = U(g)x2

−11 ⊂ E0

¯ R = Coeffs of r(x) = Y(r, x), r ∈ R ¯ R1 ⊗ E0 − → E0 u ⊗ v − → u−1v Y(u, x) ⊗ Y(v, x) − → •

• Y(u, x)Y(v, x)•
• , u ∈ ¯

R1 Meurman-Primc ‘99, Primc ‘02 The kernel of the map above is generated, in some sense, by: d dx ( xθ(x)2) ⊗ xθ(x) − 2(xθ(z)2) ⊗ d dxxθ(z) L(−1)x2

−11 ⊗ x−11 − 2x2 −11 ⊗ L(−1)x−11

L

1 2 x 1

2

2 L 1x 1 3 x 1

2

2 x 2 4 23

• 4. relations to meurman-primc’s work

.

E0 − → L(Λ0) R = U(g)x2

−11 ⊂ E0

¯ R = Coeffs of r(x) = Y(r, x), r ∈ R ¯ R1 ⊗ E0 − → E0 u ⊗ v − → u−1v Y(u, x) ⊗ Y(v, x) − → •

• Y(u, x)Y(v, x)•
• , u ∈ ¯

R1 Meurman-Primc ‘99, Primc ‘02 The kernel of the map above is generated, in some sense, by: d dx ( xθ(x)2) ⊗ xθ(x) − 2(xθ(z)2) ⊗ d dxxθ(z) L(−1)x2

−11 ⊗ x−11 − 2x2 −11 ⊗ L(−1)x−11

∼ L−1ξ−2 · x−1 − 2ξ−2 · L−1x−1 = ξ−3 · x−1 − 2ξ−2 · x−2 = µ−4.

23

Questions?

24