AKT-09/HW1

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Solve the following problems and submit them in class by October 13, 2009:

Problem 1. If f \in {\mathcal V}_n and g \in {\mathcal V}_m then f \cdot g \in {\mathcal V}_{n+m} (as what one would expect by looking at degrees of polynomials) and W_{f \cdot g} = m_\mathbb{Q} \circ (W_f \otimes W_g) \circ \Box where (W_f \otimes W_g) \circ \Box: {\mathcal A} \rightarrow \mathbb{Q} \otimes \mathbb{Q} and m_\mathbb{Q} is the multiplication of rationals. (See 090924-2, minute 36:01).

Problem 2. Let \Theta:{\mathcal A}\to{\mathcal A} be the multiplication operator by the 1-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

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

Verify the following assertions, but submit only your work on assertions 4,5,7,11:

  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
    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.

Idea for a good deed. Later than October 13, 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.

Mandatory but unenforced. Find yourself in the class photo and identify yourself as explained in the photo page.

AKT-09-ClassPhoto.jpg