07-1352/Class Notes for January 23: Difference between revisions
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==A Numerology Problem== |
==A Numerology Problem== |
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We briefly mentioned a numerology problem from {{Dror}}'s [[paperlet]] [[The HOMFLY Braidor Algebra]]. More about it will come later. |
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'''Question.''' Can you find nice formulas for the functions <math>f_{12}</math> and <math>f_{21}</math> of the variables <math>t_1</math>, <math>t_2</math> and <math>x</math>, whose Taylor expansions begin with |
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<math>f_{12}=x+\frac{x t_2}{3}-\frac{x t_1}{3}</math> |
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:<math>-\frac{1}{5} t_1 x^3+\frac{t_2 x^3}{5}+\frac{t_1^3 x}{45}-\frac{t_2^3 |
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x}{45}+\frac{1}{15} t_1 t_2^2 x-\frac{1}{15} t_1^2 t_2 x</math> |
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:<math>-\frac{1}{7} t_1 x^5+\frac{t_2 x^5}{7}+\frac{11}{315} t_1^3 |
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x^3-\frac{11}{315} t_2^3 x^3+\frac{11}{105} t_1 t_2^2 x^3-\frac{11}{105} t_1^2 t_2 x^3</math> |
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::<math>-\frac{2 t_1^5 x}{945}+\frac{2 t_2^5 |
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x}{945}-\frac{2}{189} t_1 t_2^4 x+\frac{4}{189} t_1^2 t_2^3 x-\frac{4}{189} t_1^3 t_2^2 x+\frac{2}{189} t_1^4 t_2 |
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x</math> |
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:<math>-\frac{1}{9} t_1 x^7+\frac{t_2 x^7}{9}+\frac{598 t_1^3 x^5}{14175}-\frac{598 t_2^3 x^5}{14175}+\frac{1619 t_1 t_2^2 |
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x^5}{14175}-\frac{1619 t_1^2 t_2 x^5}{14175}</math> |
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::<math>-\frac{74 t_1^5 x^3}{14175}+\frac{74 t_2^5 x^3}{14175}-\frac{74 t_1 t_2^4 |
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x^3}{2835}+\frac{148 t_1^2 t_2^3 x^3}{2835}-\frac{148 t_1^3 t_2^2 x^3}{2835}+\frac{74 t_1^4 t_2 x^3}{2835}</math> |
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::<math>+\frac{t_1^7 |
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x}{4725}-\frac{t_2^7 x}{4725}+\frac{1}{675} t_1 t_2^6 x-\frac{1}{225} t_1^2 t_2^5 x+\frac{1}{135} t_1^3 t_2^4 x-\frac{1}{135} |
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t_1^4 t_2^3 x+\frac{1}{225} t_1^5 t_2^2 x-\frac{1}{675} t_1^6 t_2 x</math> |
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:<math>-\frac{1}{11} t_1 x^9+\frac{t_2 |
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x^9}{11}+\frac{2414 t_1^3 x^7}{51975}-\frac{2414 t_2^3 x^7}{51975}+\frac{53243 t_1 t_2^2 |
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x^7}{467775}-\frac{53243 t_1^2 t_2 x^7}{467775}</math> |
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::<math>-\frac{4058 t_1^5 x^5}{467775}+\frac{4058 t_2^5 |
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x^5}{467775}-\frac{3904 t_1 t_2^4 x^5}{93555}+\frac{782 t_1^2 t_2^3 x^5}{10395}-\frac{782 t_1^3 t_2^2 |
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x^5}{10395}+\frac{3904 t_1^4 t_2 x^5}{93555}</math> |
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::<math>+\frac{331 t_1^7 x^3}{467775}-\frac{331 t_2^7 |
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x^3}{467775}+\frac{331 t_1 t_2^6 x^3}{66825}-\frac{331 t_1^2 t_2^5 x^3}{22275}+\frac{331 t_1^3 t_2^4 |
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x^3}{13365}-\frac{331 t_1^4 t_2^3 x^3}{13365}+\frac{331 t_1^5 t_2^2 x^3}{22275}-\frac{331 t_1^6 t_2 |
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x^3}{66825}</math> |
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::<math>-\frac{2 t_1^9 x}{93555}+\frac{2 t_2^9 x}{93555}-\frac{2 t_1 t_2^8 x}{10395}+\frac{8 t_1^2 |
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t_2^7 x}{10395}-\frac{8 t_1^3 t_2^6 x}{4455}+\frac{4 t_1^4 t_2^5 x}{1485}-\frac{4 t_1^5 t_2^4 |
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x}{1485}+\frac{8 t_1^6 t_2^3 x}{4455}-\frac{8 t_1^7 t_2^2 x}{10395}+\frac{2 t_1^8 t_2 |
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x}{10395}</math> |
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and |
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<math>f_{21}=1+\frac{1}{9} x^2 t_1 t_2-\frac{1}{9} x^2 t_1^2 -\frac{13}{135} t_1^2 x^4+\frac{13}{135} t_1 t_2 x^4+\frac{2}{135} |
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t_1^4 x^2+\frac{2}{45} t_1^2 t_2^2 x^2-\frac{8}{135} t_1^3 t_2 x^2</math> |
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:<math>-\frac{1147 t_1^2 x^6}{14175}+\frac{1147 t_1 t_2 |
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x^6}{14175}+\frac{13}{525} t_1^4 x^4+\frac{878 t_1^2 t_2^2 x^4}{14175}-\frac{1229 t_1^3 t_2 x^4}{14175}-\frac{1}{525} t_1^6 |
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x^2+\frac{2}{105} t_1^3 t_2^3 x^2-\frac{1}{35} t_1^4 t_2^2 x^2+\frac{2}{175} t_1^5 t_2 x^2</math> |
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:<math>-\frac{2939 t_1^2 |
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x^8}{42525}+\frac{2939 t_1 t_2 x^8}{42525}+\frac{1327 t_1^4 x^6}{42525}+\frac{2896 t_1^2 t_2^2 |
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x^6}{42525}-\frac{4223 t_1^3 t_2 x^6}{42525}</math> |
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::<math>-\frac{199 t_1^6 x^4}{42525}+\frac{20}{567} t_1^3 t_2^3 |
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x^4-\frac{97 t_1^4 t_2^2 x^4}{1701}+\frac{1124 t_1^5 t_2 x^4}{42525}+\frac{2 t_1^8 |
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x^2}{8505}+\frac{2}{243} t_1^4 t_2^4 x^2-\frac{16 t_1^5 t_2^3 x^2}{1215}+\frac{8 t_1^6 t_2^2 |
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x^2}{1215}-\frac{16 t_1^7 t_2 x^2}{8505}</math>? |
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==A HOMFLY Braidor== |
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===The Algebra=== |
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Let <math>A^0_n=\langle S_n, x, t_1,\ldots t_n\rangle</math> be the free associative (but non-commutative) algebra generated by the elements of the symmetric group <math>S_n</math> on <math>\{1,\ldots,n\}</math> and by formal variables <math>x</math> and <math>t_1\ldots t_n</math>, and let <math>A^1_n</math> be the quotient of <math>A^0_n</math> by the following "HOMFLY" relations: |
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# <math>x</math> commutes with everything else. |
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# The product of permutations is as in the symmetric group <math>S_n</math>. |
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# If <math>\sigma</math> is a permutation then <math>t_i\sigma=\sigma t_{\sigma i}</math>. |
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# <math>[t_i,t_j]=x\sigma_{ij}(t_i-t_j)</math>, where <math>\sigma_{ij}</math> is the transposition of <math>i</math> and <math>j</math>. |
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Finally, declare that <math>\deg x=\deg t_i=1</math> while <math>\deg\sigma=0</math> for every <math>1\leq i\leq n</math> and every <math>\sigma\in S_n</math>, and let <math>A_n</math> be the graded completion of <math>A^1_n</math>. |
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We say that an element of <math>A_n</math> is "sorted" if it is written in the form <math>x^k\cdot\sigma t_1^{k_1}t_2^{k_2}\cdots t_n^{k_n}</math> where <math>\sigma</math> is a permutation and <math>k</math> and the <math>k_i</math>'s are all non-negative integer. The HOMFLY relations imply that every element of <math>A_n</math> is a linear combinations of sorted elements. Thus as a vector space, <math>A_n</math> can be identified with the ring <math>B_n</math> of power series in the variables <math>x,t_1,\ldots,t_n</math> tensored with the group ring of <math>S_n</math>. The product of <math>A_n</math> is of course very different than that of <math>B_n</math>. |
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'''Examples.''' |
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# The general element of <math>A_1</math> is <math>(1)f(x,t_1)</math> where <math>(1)</math> denotes the identity permutation and <math>f(x,t_1)</math> is a power series in two variables <math>x</math> and <math>t_1</math>. <math>A_1</math> is commutative. |
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# The general element of <math>A_2</math> is <math>(12)f(x,t_1,t_2)+(21)g(x,t_1,t_2)</math> where <math>f</math> and <math>g</math> are power series in three variables and <math>(12)</math> and <math>(21)</math> are the two elements of <math>S_2</math>. <math>A_2</math> is not commutative and its product is non-trivial to describe. |
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# The general element of <math>A_3</math> is described using <math>3!=6</math> power series in 4 variables. The general element of <math>A_n</math> is described using n! power series in <math>n+1</math> variables. |
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The algebra <math>A_n</math> embeds in <math>A_{n+1}</math> in a trivial way by regarding <math>\{1,\ldots,n\}</math> as a subset of <math>\{1,\ldots,n+1\}</math> in the obvious manner; thus when given an element of <math>A_n</math> we are free to think of it also as an element of <math>A_{n+1}</math>. There is also a non-trivial map <math>\Delta:A_n\to A_{n+1}</math> defined as follows: |
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# <math>\Delta(x)=x</math>. |
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# <math>\Delta(t_i)=t_{i+1}+x\sigma_{1,i+1}</math>. |
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# <math>\Delta</math> acts on permutations by "shifting them one unit to the right", i.e., by identifying <math>\{1,\ldots,n\}</math> with <math>\{2,\ldots,n+1\}\subset\{1,\ldots,n+1\}</math>. |
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===The Equations=== |
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We seek to find a "braidor"; an element <math>B</math> of <math>A_2</math> satisfying: |
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* <math>B=(21)+x(12)+</math>(higher order terms). |
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* <math>B(\Delta B)B=(\Delta B)B(\Delta B)</math> in <math>A_3</math>. |
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With the vector space identification of <math>A_n</math> with <math>B_n</math> in mind, we are seeking two power series of three variables each, whose low order behaviour is specified and which are required to satisfy 6 functional equations written in terms of 4 variables. |
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===The Equations in Functional Form=== |
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===A Solution=== |
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The first few terms of a solution can be computed using a computer, as shown above and below. But a true solution, written in a functional form, is still missing. |
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===Computer Games=== |
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A primitive mathematica program to play with these objects is [http://katlas.math.toronto.edu/svn/06-1350/ComputingTheJonesPolynomial.nb here]. |
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Latest revision as of 12:32, 26 January 2007
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A Second Look
We'll take a second look at 06-1350/Class Notes for Tuesday October 24.
A Numerology Problem
We briefly mentioned a numerology problem from Dror's paperlet The HOMFLY Braidor Algebra. More about it will come later.
