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Homework Assignment 6: Deframing

Assigned Thursday October 23; due Thursday October 30 in class.

This document in PDF: HW.pdf

Required reading. Sections 2 and 3 of my paper On the Vassiliev Knot Invariants.

Let $ \Theta:{\mathcal A}\to{\mathcal A}$ be the multiplication operator by the chord diagram $ \theta$, and let $ \partial_\theta=\frac{d}{d\theta}$ be the adjoint of multiplication by $ W_\theta$ on $ {\mathcal A}^\star$, where $ W_\theta$ is the obvious dual of $ \theta$ in $ {\mathcal A}^\star$. Let $ P:{\mathcal A}\to{\mathcal A}$ be defined by

$\displaystyle P=\sum_{n=0}^\infty \frac{(-\Theta)^n}{n!}\partial_\theta^n. $

The following assertions can be verified:

  1. $ \left[\partial_\theta, \Theta\right]=1$, where $ 1:{\mathcal A}\to{\mathcal A}$ is the identity map and where $ [A,B]:=AB-BA$ for any two operators.

  2. $ P$ is a degree 0 operator; that is, $ \deg Pa=\deg a$ for all $ a\in{\mathcal A}$.

  3. $ \partial_\theta$ satisfies Leibnitz' law: $ \partial_\theta(ab)=(\partial_\theta a)b+a(\partial_\theta b)$ for any $ a,b\in{\mathcal A}$.

  4. $ P$ is an algebra morphism: $ P1=1$ and $ P(ab)=(Pa)(Pb)$.

  5. $ \Theta$ satisfies the co-Leibnitz law: $ \Box\circ\Theta=(\Theta\otimes 1+1\otimes\Theta)\circ\Box$ (why does this deserve the name ``the co-Leibnitz law''?).

  6. $ P$ is a co-algebra morphism: $ \eta\circ P=\eta$ (where $ \eta$ is the co-unit of $ {\mathcal A}$) and $ \Box\circ P=(P\otimes P)\circ\Box$.

  7. $ P\theta=0$ and hence $ P\langle\theta\rangle=0$, where $ \langle\theta\rangle$ is the ideal generated by $ \theta$ in the algebra $ {\mathcal A}$.

  8. If $ Q:{\mathcal A}\to{\mathcal A}$ is defined by

    $\displaystyle Q = \sum_{n=0}^\infty \frac{(-\Theta)^n}{(n+1)!}\partial_\theta^{(n+1)} $

    then $ a=\theta Qa+Pa$ for all $ a\in{\mathcal A}$.

  9. $ \ker P=\langle\theta\rangle$.

  10. $ P$ descends to a Hopf algebra morphism $ {\mathcal A}^r\to{\mathcal A}$, and if $ \pi:{\mathcal A}\to{\mathcal A}^r$ is the obvious projection, then $ \pi\circ P$ is the identity of $ {\mathcal A}^r$. (Recall that $ {\mathcal A}^r={\mathcal A}/\langle\theta\rangle$.)

  11. $ P^2=P$.

To be handed in. Verify assertions 4, 5, 7 and 11 above.

Recommended for extra practice. Verify all the other assertions above.

Idea for a good deed. Prepare a beautiful TEX writeup (including the motivation and all the details) of the solution of this assignment for publication on the web. For all I know this information in this form is not available elsewhere.

The generation of this document was assisted by LATEX2HTML.

Dror Bar-Natan 2003-10-22