http://knotebook.org ITEP, Moscow September 21, 2015 ttp://knotebook.org (ITEP) September 21, 2015 1 / 73
H K R (q A) A=q N = Tr R P exp A K SU(N) September 21, 2015 2 / 73
S = κ 4π d 3 x Tr (AdA + 2 ) 3 A3 q = exp 2πi κ + N September 21, 2015 3 / 73
Can be lifted to extended polynomials H B R{q p k } which depend on infinitely many time-variables {p k } and on particular braid representation B of the oriented knot K. September 21, 2015 4 / 73
H K R (q A) H B R{q p k} p k = Ak A k q k q k At this topological locus the r.h.s. does not depend on the choice of B for a given K. September 21, 2015 5 / 73
Character expansion separates dependencies on the braid and the time variables: HR{q p B k } = crq(q)s B Q {p k } Q m R Sum is over Young diagrams Q of the size Q = m R September 21, 2015 6 / 73
More artful character expansions (especially in the Hall-Littlewood or MacDonald polynomials) can be even more useful for particular applications September 21, 2015 7 / 73
An m-strand braid is parameterized by a sequence of integers a (j) i with i = 1, 2,..., m 1 and j = 1,..., n a (1) 1 = 2, a (1) 2 = 2, a (2) 1 = 1, a (2) 2 = 3 (the closure is the knot 8 10 ) September 21, 2015 8 / 73
crq B (q) are represented as traces in the spaces of intertwining operators of combinations of simple matrices, c B RQ(q) = Tr Q j ( m 1 i=1 R a(j) i i,i+1 ) September 21, 2015 9 / 73
Representation is parameterized by the Young diagram: an ordered set of l(r) positive integers R = {r 1 r 2... r l(r) > 0} = [r 1, r 2,..., r l(r) ] September 21, 2015 10
Find and investigate the equations, which the HOMFLY and extended HOMFLY polynomials satisfy as functions of a (j) i and r k. September 21, 2015 11
Definition of knot polynomials Evaluation of knot polynomials Properties of knot polynomials September 21, 2015 12
Definitions/evaluation: Perturbative expansion in 1/k leads to Vassiliev invariants and to Vogel s algebra of triple-ended diagrams (related to Konnes-Kreimer formalism) September 21, 2015 13
Gauge choices quadratic actions (free field representations): A z = 0: A0 A propagator 1 z Kontsevich integral A 0 = 0: ɛab A a Ȧ b sign(t)δ (2) (z) 2d knot diagrams September 21, 2015 14
Knot diagrams: Integrable lattice theory on arbitrary graph Reidemeister invariance = YB relation R-matrix calculus Conformal blocks Hypercube method (good also for virtual knots and for superpolynomials) September 21, 2015 15
R-matrix approach: c B RQ(q) = Tr Q j ( m 1 i=1 Skein relations Û i ˆRa ij Û i ) September 21, 2015 16
R-matrix, acting in the channel [1] [1] = [2] [11] has two eigenvalues, q and 1/q, associated with the two irreducible representations [2] and [11] respectively, and thus satisfies the Hecke algebra constraint ( ) R q )(R + q 1 = 0 September 21, 2015 17
Skein relations in fund.rep.: H (... a ij+1...) H (... a ij 1...) = = ( q q 1) H (... a ij...) September 21, 2015 18
For bigger representations R, the product R R = S S and R-matrix in this channel has many different eigenvalues: as many as there are irreducible representations S in this expansion. September 21, 2015 19
These eigenvalues are equal to ɛ S q κ S, where κ S is an eigenvalue of the cut-and-join operator: Ŵ [2], associated with its eigenfunction character S S {p k }, and ɛ S = ±1, depending on R and S. Therefore, this R-matrix satisfies ( ) R ɛ S q κ S = 0 S September 21, 2015 20
S ( ) e / a ij ɛ S q κ S H (...a ij...) R = 0 for any variable a ij. September 21, 2015 21
For the first symmetric representation )( ) (R q )(R 6 + q 2 R 1 = 0 H (... a ij+3...) [2] (1+q 2 +q 6 )H (... a ij+2...) [2] + +q 2 (1+q 4 +q 6 )H (... a ij+1...) [2] q 8 H (... a ij...) [2] = September 21, 2015 22
Cabling procedure to evaluate colored polynomials: H K R m = H K m R where K m denotes m-cabling of the knot K September 21, 2015 23
HOMFLY for the l.h.s. can be presented as a sum over the irreducible representations, H K R m = S H K S To get HS K, one calculates HK m R in rep.r and then project the result onto irrep Q. September 21, 2015 24
( ) R q )(R + q 1 = 0 = P [2] = 1 + qr 1 + q 2 P [11] = q2 qr 1 + q 2 September 21, 2015 25
[1] [1] [1] = [3] + 2[21] + [111] R 3 = R 12 R 13 )( ) (R 3 q )(R 2 3 q 2 R 2 3+R 3 +1 = 0 September 21, 2015 26
P [3] = (1 q2 R 3 )(R 2 3 + R 3 + 1) (1 q 4 )(1 + q 2 + q 4 ) P [111] = q 6(R 3 q 2 )(R 2 3 + R 3 + 1) (1 q 4 )(1 + q 2 + q 4 ) September 21, 2015 27
P ± [21] = ±e±πi/6 (q 2 R 3 1)(R 3 q 2 ) 3 (R 3 e ±2πi/3 ) 1 + q 2 + q 4 September 21, 2015 28
Mixing matrixes U R 3 = R 12 R 13 = RURU R i,i+1 = sum over paths in rep.graph September 21, 2015 29
Representation graph [1] [2] [11] [3] [21] [111] [4] [31] [22] [211] [1111]... September 21, 2015 30
The multiplicity M Q of the representation Q in Q is obviously equal to the number of directed paths in the representation graph, connecting and Q. More generally, M RQ is equal to the number of directed paths between R and Q. September 21, 2015 31
The matrices ˆR i,i+1, i = 1,..., m 1 can be represented in the basis of paths between Q and and they have extremely simple form in this basis. September 21, 2015 32
First of all, with each index i of the matrix ˆR i,i+1 one associates a level i in the graph. A given path P is passing through exactly one vertex P i at level i and through some two adjacent vertices P i 1 and P i+1 at levels i 1 and i + 1. September 21, 2015 33
The structure of the representation graph is such that these P i 1 and P i+1 are connected either by a single two-segment path (singlet) (then it is a fragment of our P) or by two such paths (doublet), the segments of our path P and another path P. September 21, 2015 34
In the former case (singlet) our path P provides a diagonal element in ˆR (i) and it is equal to either q or 1/q. In the language of Young diagrams the singlet appears when the two boxes added to the diagram P i 1 in order to form P i+1 lie either in the same row, then we put q at the diagonal of ˆRi ; or in the same column, then we put 1/q. September 21, 2015 35
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In the latter case (the doublet) the two boxes are neither in the same row nor in the same column, and the two paths P and P form a 2 2 block in ˆR (i). This block is described as follows. September 21, 2015 38
First, that of the paths P and P which lies to the left of the other, corresponds to the left column and to the first row of the 2 2 block. September 21, 2015 39
Second, the Young diagrams P i+1 is obtained by adding two boxes to the diagram P i 1, and the two paths correspond to doing this in two different orders, thus providing at the intermediate stage the two adjacent vertices P i and P i. September 21, 2015 40
The two added boxes are connected by a hook in the Young diagram, which has length n (measured between the centers of the two boxes). September 21, 2015 41
q n c n s n s n q n c n c n = 1 = q q 1 [n] q q n q n s n = 1 c 2 n = [n 1]q [n + 1] q [n] q September 21, 2015 42
7 n = 7 [5521] [6521] [5522] [6522] September 21, 2015 43
ALGEBRAIC PROPERTIES September 21, 2015 44
Special polynomials: H K R (q = 1 A) = ( ) R σ (A) K representations R September 21, 2015 45
exp ( H K R (q A) = ( σ K (A)) R z +l( ) 2 S ( A z 2 ) ϕ R ( ) ϕ R ( ) = symmetric-group characters Hurwitz τ-function ) September 21, 2015 46
DIFFERENTIAL EXPANSION: H K r (A, q 2 ) = 1+ {A/q} r s=1 [r]! [s]![r s]! G K s (A, q) s 1 {Aq r+j } j=0 {x} = x x 1 and [n] = {q n }/{q} September 21, 2015 47
For generic knot Gs K is a non-factorizable Laurent polynomial of A and q, but for some knots it can be further factorized. September 21, 2015 48
What is important, if Gs K is divisible by some differential {Aq k }, the same is true for all other Gs K with s > s. This property allows one to introduce defect functions νs K and µ K s = s 1 νs K : September 21, 2015 49
G K s (A, q) = F K s (A, q) ν K s 1 j=0 {Aq j } = = F K s (A, q) s 2 µ K s j=0 {Aq j } September 21, 2015 50
which are both(!) monotonically increasing function of s, ν K s ν K s, µ K s µ K s for s < s i.e. both grow but not faster than s. September 21, 2015 51
For A = q N with any fixed N, positive or negative, F s (q N, q) {q} µk s (1) i.e. at fixed N the s-the term of differential expansion is actually of the order {q} 2s. September 21, 2015 52
It turns out that νs K as a function of s has a very special shape, fully parameterized by a single integer δ K 1, which we call the defect of differential expansion: defect δ K = 1 = µ K s = s 2, ν K s = September 21, 2015 53 K K K
7 6 5 4 3 2 1 s 2 3 4 5 6 7 8 91011121314 : µ K s s 2 6 5 4 3 2 1 s 3 4 5 6 7 8 91011121314 September 21, 2015 54
10 9 8 7 6 5 4 3 2 1 s 2 3 4 5 6 7 8 9101112131415 : µ K s 2s 3 s 4 5 6 7 8 91011121314 4 3 2 1 September 21, 2015 55
GENERALIZATIONS: H K r+m = H K r H K m q 2m =1 September 21, 2015 56
H K R+M? = H K m H K R q 2m =1 connecte provided both R and R + M are Young diagrams. The following picture is an explanation of what we mean by R + M: September 21, 2015 57
connected M of unit width discon September 21, 2015 58
Kashaev polynomial: K K R (A) = H K R ( ) q 2 = e 2πi/ R, A is the value of colored HOMFLY at a primitive root of unity q 2 R = 1. September 21, 2015 59
For all single hook diagrams R Kashaev polynomial is easily expressed through the special polynomial: K K R (A) = K K [1] (A R ) = H K [1] R = [r, 1 k ] (q 2 = 1, A R ) September 21, 2015 60
Dual to a property of colored Alexander polynomials: H K r (A = 1, q) = H K 1 (A = 1, q r ) R = [r, 1 k ] September 21, 2015 61
When A moves away from 1 concentric circles deform and collide with Alexander circles (we plot zeros of H 10 ): September 21, 2015 62
A = 1 A = 1 10 10 2 2 2 1 1 1-2 -1 1 2-2 -1 1 2-2 -1 1 2-1 -1-1 -2-2 -2 September 21, 2015 63
A = 1 10 6 A = 1 10 4 2 2 2 1 1 1-2 -1 1 2-2 -1 1 2-2 -1 1 2-1 -1-1 -2-2 -2 September 21, 2015 64
We can also consider perturbation in purely imaginary direction A = 1 A = 1 i 10 10 2 2 2 1 1 1-2 -1 1 2-2 -1 1 2-2 -1 1 2-1 -1-1 -2-2 September 21, 2015 65-2
A = 1 i 10 6 A = 1 i 10 4 2 2 2 1 1 1-2 -1 1 2-2 -1 1 2-2 -1 1 2-1 -1-1 -2-2 -2 September 21, 2015 66
For the trefoil all the roots of Alexander polynomial are unimodular. A = 1 A = 1 10 10 1.5 2 1.0 1 0.5-1.5-1.0-0.5 0.5 1.0 1.5-2 -1 1 2-0.5 September 21, 2015 67-1
A = 1 10 4 A = 1 10 2 2 2 2 1 1 1-2 -1 1 2-2 -1 1 2-2 -1 1 2-1 -1-1 -2-2 -2 September 21, 2015 68
and A = 1 A = 1 i 10 10 1.5 2 2 1.0 1 1 0.5-1.5-1.0-0.5 0.5 1.0 1.5-2 -1 1 2-2 -1 1 2-0.5-1 -1-1.0-1.5-2 -2 September 21, 2015 69
A = 1 i 10 2 A = 1 i 10 1 2 2 2 1 1 1-2 -1 1 2-2 -1 1 2-2 -1 1 2-1 -1-1 -2-2 -2 September 21, 2015 70
Zeroes of resultant q 2(H 4 1 1, H4 1 k ) for the figure eight knot K = 4 1 : 1.5 1.0 0.5-1.5-1.0-0.5 0.5 1.0 1.5-0.5-1.0-1.5 September 21, 2015 71
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THANKS FOR YOUR KIND ATTENTION! September 21, 2015 74