TemperleyLieb Algebra Idempotence

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Temperley-Lieb AlgebraIdempotence

• Group 3
• Margarita Echavarria
• Julia Plavnik
• Marti Szilagyi

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Review (for those with a 5 second memory)

• Rnm the rectangular disc with m fixed pts at
  the top and n fixed pts at the bottom.
• A diagram in Rnm (m,n) tangle
• Simple tangle diagram has no crossings
• S(Rnm) with finite subset of boundary points is
  the quotient of D(Rnm) by
• i. D a D0 a-1 D8
• ii. D ? 0 ?D (the disjoint union) -(a2a-2)D
  where D? are diagrams
• Theorem 1.2 1 S(Rnm) is spanned by diagrams
  with no crossings and no null-homotopic closed
  curves
• -Proof by induction removes all crossings
  (i) and all closed curves (ii)
• Skein module, Em,n Em,n(a) which is the K
  module generated by all (m,n) tangle diagrams
  quotiened by skein identity.
• S(Rnm) Em,n

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Definition of ? - Skein Cat

• ? ?(a) abelian category with the following
  properties
• i. Class of objects in ? is 0,1,2, N0
• ii. Set of morphisms m ? n Em,n
• iii. ?f,g morphisms ? ? represented by diagrams,
  fg putting the f diagram on top of the g
  diagram and normalizing to ?x0,1 and it extends
  by linearity
• iv. idk k ? k is the diagram with k disjoint
  vertical arcs
• v. id0 0 ? 0 is the empty diagram
• vi. m??n mn for m,n ? ?
• vii. D??D putting the D diagram to the right
  of the D diagram, where D, D are diagrams of
  morphisms in ? and it extends by linearity

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Connecting Skein to TL

• Wiring several surfaces at once induces a map
• S(W) S(F1) x S(F2) x x S(Fk) ? S(F)
• Let F1 Rnm and F2 Rpn
• S(Rnm) x S(Rpn) S(Rpm)
• When nmp ?S(Rnn) is an algebra over K
• n-th Temperley-Lieb Algebra
• TLn

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Temperley-Lieb Algebra

• End?(k) Ek,k
• Denote Ek,k by Ek
• Product xy (x,y ? Ek) is denoted by x?y and
  viewed as a morphism from k ? k. If x,y are
  represented by diagrams, y is placed directly
  below x
• Associative and has unit 1k (k vertical arcs)
• E0 K, E1 K, for n3, Ek is noncommutative
• Ek is generated by 1k, ei i 1, , k-1

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Temperley-Lieb Algebra


ei
i - 1
K - i - 1
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Temperley-Lieb Algebra

• Theorem 1 1k,e1,,ek-1 ? Ek generates Ek as a K
  algebra.
• ? Proof uses the following claim.
• Claim 1 Let D ? ?x0,1 be a simple diagram and
  not 1k. Then it is decomposable as a product of
  e1,,ek-1. The number of decompositions is the
  same as
• Hint To prove, use induction on i(D) i in
  diagram (where i comes from the xi boundary
  point that has the first arc.)
• ? D Dei and i(D)i(D)-1

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Temperley-Lieb Algebra

• Lemma 1 The elements 1k, e1,,ek-1 ? Ek satisfy
  the following
• i. ei2 -(a2a-2)ei
• ii. eiejejei if i - j gt1
• iii. eiei?1eiei
• where i,j 1,,k-1
• If we restrict En to have arcs that run
  monotonically from bottom to top then we get a
  braid, where the composition of two n braids is
  done by placing one on top of the other and two
  braids are isomorphic up to RII and RIII moves
• Proposition 1 There is a multiplicative
  homomorphism Bn? En determined by representing
  ??Bn by a diagram in Rnn and reading the diagram
  as an element of skein En

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Temperley-Lieb Algebra

• The image of Bn spans En since each generator ei
  ? En satisfies
• ? ?i aId a-1ei
• ? ei a?i - a2Id
• Presentation of En can be rewritten in terms of
  ?i (using the relations in Bn) and the following
  relation
• ? (?i-a)(?ia3) 0 or (?ia-3)ei 0
• ?i

i
i1
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Temperley-Lieb Algebra

• Trace For any f ? Ek, let D be the tangle
  diagram of f and D the closure of D (a link).
• We can define the trace of f as follows
• tr(f) ltDgt ? K
• tr(ei) -(a2-a-2)k-1 for i 1,,k-1
• tr(1k) -(a2-a-2)k
• Lemma 2 i. For any f,g ? Ek, tr(fg)tr(gf)
• ii. For every m,n0,1,, f? Em, g? En
• tr(f?g) tr(f)tr(g)
• iii. For the subalgebra
  K(1k,e1,,ek-2) ? Ek,
• tr(fek-1) -(a2-a-2)-1 tr(f)

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Idempotence fk

• Theorem 2 Assume that a4n-1 ? K for n1,,k
  where K is the group of invertible elements of
  K. Then ?! an element fk ? Ek such that
• fk-1k ? K(e1,,ek-1)
• eifkfkei0 for all i 1,..,k-1
• Let f fk, then f2-f (f-1)f 0 which shows
  that fk is IDEMPOTENT
• Since fk commutes with multiplicative generators
  of Ek, then fk ? Z(Ek)
• fk also annihilates all of the additive
  generators of Ek except for 1k

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Idempotence

• Claim 2 For each n 1,,k
• i. f(n)f(n)f(n)
• ii. eif(n) 0 for all i lt n, n 1,,k-1
• iii. (enf(n))2 -(n1/n)enf(n) with
• n(a2n-a-2n)(a2-a-2)-1 ? K
• Proof is by induction and uses
• ?? leads to the bilinear form
• Em?En?Emn
• (f,g) ? f?g
• We can now define fn1 as follows
• (where fk?11 is the image of fk under canonical
  inclusion
• Ek to Ek1. Geometrically adding a vertical arc
  to the right of fk)

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Idempotence

• Graphical calculus for the J-W idempotents

n
1
n


n-1
n1
n1
n
1
n
1
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Idempotence

• Roots of Unity Versus Generic Elements
• The sequence of idempotents f0,f1,f2, (in
  E0(a), E1(a),E2(a),) could be infinite of
  finite.
• For a
• i. generic a4n-1 ? K ?n?1 leads to an
  infinite sequence of idempotents
• ii. primitive an-1 ? K for n1,,k-1 and
  ak1 leads to n idempotents
• ? i.e. a4-1 ? a41 ? f0,f1,f2,f3

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Idempotence

• Jones-Wenzl Idempotents as generators
• Can think of fk as a map k?k in ?
• For a generic, the infinite sequence of
  idempotents, f0,f1, generates all k?k maps in ?
  as a 2-sided ideal.
• For a primitive root, the finite sequence
  generates all k?k maps in ? modulo negligible
  morphisms.

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Idempotence

• Lemma 3 If a?K is generic then any morphism k?l
  in ??(a) may be expressed as a sum,
• where s runs over a finite set of indices,
  is?0,1,, xsis?l, ys k?is are morphisms in ?.
  
• Lemma 4 If a4n-1?K for n1,2,,k then
  tr(fk)(-1)kK1
• Lemma 5 If a?K is a primitive 4r-th root of
  unity, then any morphism k?l in ??(a) may be
  expressed as a sum
• where s runs over a finite set of indices,
  is?0,1,, xsis?l, ys k?is are morphisms in ?
  and zk?l is a negligible morphism in ?.

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• Questions?