Drorbn at 01:18, 2 November 2006

2006-11-02T01:18:16Z

← Older revision		Revision as of 21:18, 1 November 2006
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	With a tiny bit of further algebra and quoting an old theorem of Milnor and Moore, it follows that <math>{\mathcal A}(\bigcirc)</math> is a commutative graded polynomial algebra with finitely many generators at each degree. The dimension of the spaces of generators at degrees up two 12 are known and are denoted <math>\dim{\mathcal P}_m</math> in the table below, which is reproduced from [[06-1350/Class Notes for Thursday October 12]]:		With a tiny bit of further algebra and quoting an old theorem of Milnor and Moore, it follows that <math>{\mathcal A}(\bigcirc)</math> is a commutative graded polynomial algebra with finitely many generators at each degree. The dimension of the spaces of generators at degrees up two 12 are known and are denoted <math>\dim{\mathcal P}_m</math> in the table below, which is reproduced from [[06-1350/Class Notes for Thursday October 12]]:


	{\|align=center border=1 cellspacing=0 cellpadding=4		{\|align=center border=1 cellspacing=0 cellpadding=4

Drorbn at 01:15, 2 November 2006

2006-11-02T01:15:12Z

New page

{{06-1350/Navigation}}

'''Solve the following problem''' and submit your solution in class by November 16, 2006:

'''Problem.''' The product of two polynomials is again a polynomial; there must be an analog for that in the world of "polynomial" invariants of knots.
# Prove that the product of two finite type invariants (of, say, knotted <math>\Gamma</math>'s) is again a finite type invariant. Of what type will it be, as a function of the types of the two factors of the product?
# In what way does the product of finite type invariant induces a map <math>\Box:{\mathcal A}(\Gamma)\to{\mathcal A}(\Gamma)\otimes{\mathcal A}(\Gamma)</math>?
# Describe the map <math>\Box</math> of above in explicit terms. First use the "chords and 4T" description of <math>{\mathcal A}(\Gamma)</math>, and then the "trivalent diagrams and AS, IHX and STU" description of the same object.
# Learn somewhere about coalgebras and show that <math>\Box</math> is always coassociative and cocommutative.
# Learn somewhere about bialgebras (Hopf algebras without an antipode) and show that <math>{\mathcal A}(\bigcirc)</math> becomes a commutative associative cocommutative and coassociative bialgebra, if taken with the "connected sum" product and with <math>\Box</math> as a coproduct.

With a tiny bit of further algebra and quoting an old theorem of Milnor and Moore, it follows that <math>{\mathcal A}(\bigcirc)</math> is a commutative graded polynomial algebra with finitely many generators at each degree. The dimension of the spaces of generators at degrees up two 12 are known and are denoted <math>\dim{\mathcal P}_m</math> in the table below, which is reproduced from [[06-1350/Class Notes for Thursday October 12]]:

{|align=center border=1 cellspacing=0 cellpadding=4
|- align=right
|align=center|m
|0
|1
|2
|3
|4
|5
|6
|7
|8
|9
|10
|11
|12
|- align=right
|align=center|<math>\dim{\mathcal A}_m^r</math>
|1
|0
|1
|1
|3
|4
|9
|14
|27
|44
|80
|132
|232
|- align=right
|align=center|<math>\dim{\mathcal A}_m</math>
|1
|1
|2
|3
|6
|10
|19
|33
|60
|104
|184
|316
|548
|- align=right
|align=center|<math>\dim{\mathcal P}_m</math>
|0
|1
|1
|1
|2
|3
|5
|8
|12
|18
|27
|39
|55
|}

06-1350/Homework Assignment 3 - Revision history

Drorbn at 01:18, 2 November 2006

Drorbn at 01:15, 2 November 2006