Dror Bar-Natan: Classes: 2000-01: Knot Theory:

Class Notes for April 17th

based on handwritten notes by Romina Guelman

WARNING: These notes are provided "as is", with absolutely no warranty. They can not be assumed to be complete, correct, reliable or relevant. If you don't like them, don't read them.

Creation of this page was assisted by HTMX from the following source files: source.tar.gz.

Reminder. The n'th derivative of a type n invariant defines a function $W:\{\text{$n$-chord diagrams}\}\to F$ , where F is our characteristic 0 ground field, typically $\mathbb Q$ , $\mathbb R$ or $\mathbb C$ . We found a condition for the integrability of a "candidate" n'th derivative, a function on the set of n-chord diagrams, to an invariant of (n-1)-singular knots - the so-called 4T relation. Thus defining

$\calA_n := \Span\{\text{$n$-chord diagrams}\}\left/\{\text{all $4T$ relations}\}\right.$

and $\calA := \bigoplus_{n=0}^\infty \calA_n$ , we see that if W is an n'th derivative, then $W\in\calA_n^\star$ .

The Fundamental Theorem. For every $W\in\calA_n^\star$ there exists an framed-knot invariant V so that $V^{(n)}=W$ .

This theorem is highly non trivial, and so far in this course we are still far from seeing its proof. Notice that at this stage we don't even know yet that the 4T relations are sufficient to guarantee the integrability of W to (n-1)-singular knots, while the theorem asserts that they are sufficient to guarantee integrability all the way up to an invariant of `honest' (non-singular) framed knots.

The computation of $\dim\calA_n$ is a finite problem in linear algebra, as is evident from the handout "Some Dimensions of $\mathcal A$ ". Let us only quote the table at the end of that handout:

n 0 1 2 3 4 5 6 7 8 9 10 11 12

$\dim\calA_n$ 1 1 2 3 6 10 19 33 60 104 184 316 548

The algebra structure of $\mathcal A$ . Given two knots, one can form their "connected sum", as in the picture below:

Connected Sum

(To see that this operation is well defined, shrink one of the components till it's very small indeed, and pull it along the "bridge" connecting it to the second component, until it looks like a tiny freckle on the much larger second component. At this stage you can isotop the second component freely, as if the first didn't exist. Pulling the first component back out (possibly taking a different path for the bridge) and inflating it back to it's original size, we see that the "connected sum" operation is independent of the choice of the bridge and of the specific choice of the embedding of the second component within its isotopy class. Likewise by shrinking the second component into utter freckleness, we see that the end result is independent of the specific choice of the embedding of the first component within its isotopy class.)

The "connected sum" operation on knots suggests that there ought to be a product on $\mathcal A$ - finite type invariants are functionals on knots and they are enumerated by functionals on $\mathcal A$ , and hence the set of knots and the space $\mathcal A$ lie on the same side of the Great Duality Divide, and hence they should behave in similar ways. This also suggests that for the product on $\mathcal A$ we ought to mimick the connected sum product of knots and thus define it by:

Diagram Connected Sum

It is obvious that this operation is well defined modulo 4T relations on either factor. It is less obvious that it is independent of the choice of the placement of the "connection bridge", i.e., of rotations of either factor. Clearly, it is enough to show that (modulo 4T) it is possible to move one end of one chord c from one side of the bridge to the other, without changing the end result:

Is the diagram connected sum well defined?

Laying the left component L flat, this becomes:

The Invariance Statement

This is the first instance of the "general invariance argument" which we shall meet again several times. It is always proven in the same way. Mark two hooks near every vertex of L as shown below,

Adding hooks around the vertices

and then consider the sum S of all ways of connecting the "floating" chord c (coming from the northwest in the above picture) to the hooks, with alternating signs as shown. There are two interesting ways to partition this sum. We can partition the hooks (and hence the terms in the sum) according to the nearest chord to each one. This divides the sum into groups of 4 terms each, with a typical example looking like:

The hooks around a single chord

Clearly each such group of 4 terms forms a 4T relation, and hence modulo 4T, we have that S=0. On the other hand, we can partition the hooks to pairs according to the skeleton interval on which they lie. Each such pair cancells because of the alternating sum, except the left most hook and the right most hook which are not paired with anything. Thus the difference between connecting c to the left most hook and to the right most hook vanishes modulo the 4T relations, and this is what we wanted to prove.

Thus we see that the connected sum operation on chord diagrams is well defined modulo 4T. It is easy to see that with this operation $\mathcal A$ becomes a graded commutative algebra with unit; the commutativity can be seen by rotating the definition of the connected sum operation by 180^o, and the unit is the empty chord diagram, composed of a skeleton circle and no chords.

It is wothwhile to note that the same "invariance argument" as above also proves that the space $\calA=\calA(\circlearrowleft)$ of chord diagrams on a directed circle skeleton (modulo 4T relations) is isomorphic to the space $\calA(\uparrow)$ of chord diagrams on a directed interval skeleton (modulo 4T relations).

Algebras and co-algebras. We all know algebras. One standard definition is as follows: A (graded) algebra is a triple $(A,m:A\otimes A\to A,\epsilon:\bbF\to A)$ where A is a (graded) vector space over some ground field $\mathbb F$ and $m:A\otimes A\to A$ and $\epsilon:\bbF\to A$ are (graded) linear maps satisfying the further diagram-commutativity conditions:

$\xymatrix{ A\otimes A\otimes A\ar[rr]^{m\otimes 1}\ar[d]_{1\otimes m} & & A\otimes A\ar[d]^m \\ A\otimes A\ar[rr]_m & & A }$

"The product m is associative" $\xymatrix{ \bbF\otimes A\ar[rr]^{\epsilon\otimes I}\ar@{ = }[rd] & & A\otimes A\ar[ld]^m \\ & A & }$

" epsilon is a left unit"

$\xymatrix{ A\otimes\bbF\ar[rr]^{I\otimes\epsilon}\ar@{ = }[rd] & & A\otimes A\ar[ld]^m \\ & A & }$

" epsilon is a right unit" $\xymatrix{ A\otimes A\ar@{<->}[rr]^\sigma\ar[rd]_m & & A\otimes A\ar[ld]^m \\ & A & }$

"The product m is commutative"
(not always required)
( sigma is the transposition of the two factors)

A (graded) co-algebra is the dual notion to the notion "algebra". This only means that all the arrows in the definition of an algebra are reveresed. Thus a (graded) co-algebra is a triple $(A,\Delta:A\to A\times A,\eta:A\to\bbF)$ where A is a (graded) vector space over some ground field $\mathbb F$ and $\Delta:A\to A\times A$ and $\eta:A\to\bbF$ are (graded) linear maps satisfying the following diagram-commutativity conditions:

$\xymatrix{ A\ar[rr]^\Delta\ar[d]_\Delta & & A\otimes A\ar[d]^{I\otimes\Delta} \\ A\otimes A\ar[rr]_{\Delta\otimes A} & & A\otimes A\otimes A }$

"The co-product Delta is co-associative" $\xymatrix{ & A\ar[ld]_\Delta\ar@{ = }[rd] & \\ A\otimes A\ar[rr]^{\eta\otimes I} & & \bbF\otimes A }$

" eta is a left co-unit"

$\xymatrix{ & A\ar[ld]_\Delta\ar@{ = }[rd] & \\ A\otimes A\ar[rr]^{I\otimes\eta} & & A\otimes\bbF }$

" eta is a right co-unit" $\xymatrix{ & A\ar[ld]_\Delta\ar[rd]^\Delta & \\ A\otimes A\ar@{<->}[rr]_\sigma & & A\otimes A }$

"The co-product Delta is co-commutative"
(not always required)

The primary example of a co-algebra is the dual space of a finite-dimensional algebra; that is, if $(A,m:A\otimes A\to A,\epsilon:\bbF\to A)$ is a finite-dimensional algebra, than the dual space with the adjoint maps $(A^\star,\Delta=m^\star:A^\star\to A^\star\otimes A^\star,\eta=\epsilon^\star:A^\star\to\bbF)$ is a co-algebra.

Proposition. $\mathcal A$ is also a co-commutative co-algebra.

Exercise. Show that if V₁ and V₂ are finite type invariants of types n₁ and n₂ respectively, then their product V₁V₂ if of type n₁+n₂. Show that there is a unique map $\Delta:\calA\to\calA\otimes\calA$ so that on the level of weight systems, $W_{V_1V_2}=(W_{V_1}\otimes W_{V_2})\circ\Delta$ . Finally, if $\eta:\calA\to\bbF$ denotes the weight system of the type 0 constant invariant 1, show that the triple $(\calA,\Delta,\eta)$ is a co-commutative co-algebra.

$\xymatrix{ A\otimes A\otimes A\ar[rr]^{m\otimes 1}\ar[d]_{1\otimes m} & & A\otimes A\ar[d]^m \\ A\otimes A\ar[rr]_m & & A }$ "The product m is associative"	$\xymatrix{ \bbF\otimes A\ar[rr]^{\epsilon\otimes I}\ar@{ = }[rd] & & A\otimes A\ar[ld]^m \\ & A & }$ " is a left unit"
$\xymatrix{ A\otimes\bbF\ar[rr]^{I\otimes\epsilon}\ar@{ = }[rd] & & A\otimes A\ar[ld]^m \\ & A & }$ " is a right unit"	$\xymatrix{ A\otimes A\ar@{<->}[rr]^\sigma\ar[rd]_m & & A\otimes A\ar[ld]^m \\ & A & }$ "The product m is commutative" (not always required) ( is the transposition of the two factors)

$\xymatrix{ A\ar[rr]^\Delta\ar[d]_\Delta & & A\otimes A\ar[d]^{I\otimes\Delta} \\ A\otimes A\ar[rr]_{\Delta\otimes A} & & A\otimes A\otimes A }$ "The co-product is co-associative"	$\xymatrix{ & A\ar[ld]_\Delta\ar@{ = }[rd] & \\ A\otimes A\ar[rr]^{\eta\otimes I} & & \bbF\otimes A }$ " is a left co-unit"
$\xymatrix{ & A\ar[ld]_\Delta\ar@{ = }[rd] & \\ A\otimes A\ar[rr]^{I\otimes\eta} & & A\otimes\bbF }$ " is a right co-unit"	$\xymatrix{ & A\ar[ld]_\Delta\ar[rd]^\Delta & \\ A\otimes A\ar@{<->}[rr]_\sigma & & A\otimes A }$ "The co-product is co-commutative" (not always required)