07-1352/Class Notes for January 23

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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

Question. Can you find nice formulas for the functions [math]\displaystyle{ f_{12} }[/math] and [math]\displaystyle{ f_{21} }[/math] of the variables [math]\displaystyle{ t_1 }[/math], [math]\displaystyle{ t_2 }[/math] and [math]\displaystyle{ x }[/math], whose Taylor expansions are

[math]\displaystyle{ f_{12}=x+\frac{x t_2}{3}-\frac{x t_1}{3} }[/math]

[math]\displaystyle{ -\frac{1}{5} t_1 x^3+\frac{t_2 x^3}{5}+\frac{t_1^3 x}{45}-\frac{t_2^3 x}{45}+\frac{1}{15} t_1 t_2^2 x-\frac{1}{15} t_1^2 t_2 x }[/math]

[math]\displaystyle{ -\frac{1}{7} t_1 x^5+\frac{t_2 x^5}{7}+\frac{11}{315} t_1^3 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]

[math]\displaystyle{ -\frac{2 t_1^5 x}{945}+\frac{2 t_2^5 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 x }[/math]

[math]\displaystyle{ -\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 x^5}{14175}-\frac{1619 t_1^2 t_2 x^5}{14175} }[/math]

[math]\displaystyle{ -\frac{74 t_1^5 x^3}{14175}+\frac{74 t_2^5 x^3}{14175}-\frac{74 t_1 t_2^4 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]

[math]\displaystyle{ +\frac{t_1^7 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} 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]

[math]\displaystyle{ -\frac{1}{11} t_1 x^9+\frac{t_2 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 x^7}{467775}-\frac{53243 t_1^2 t_2 x^7}{467775} }[/math]

[math]\displaystyle{ -\frac{4058 t_1^5 x^5}{467775}+\frac{4058 t_2^5 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 x^5}{10395}+\frac{3904 t_1^4 t_2 x^5}{93555} }[/math]

[math]\displaystyle{ +\frac{331 t_1^7 x^3}{467775}-\frac{331 t_2^7 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 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 x^3}{66825} }[/math]

[math]\displaystyle{ -\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 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 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 x}{10395} }[/math]

and

[math]\displaystyle{ 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} 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]

[math]\displaystyle{ -\frac{1147 t_1^2 x^6}{14175}+\frac{1147 t_1 t_2 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 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]

[math]\displaystyle{ -\frac{2939 t_1^2 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 x^6}{42525}-\frac{4223 t_1^3 t_2 x^6}{42525} }[/math]

[math]\displaystyle{ -\frac{199 t_1^6 x^4}{42525}+\frac{20}{567} t_1^3 t_2^3 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 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 x^2}{1215}-\frac{16 t_1^7 t_2 x^2}{8505} }[/math]?

A HOMFLY Braidor

The Algebra

Let [math]\displaystyle{ 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]\displaystyle{ S_n }[/math] on [math]\displaystyle{ \{1,\ldots,n\} }[/math] and by formal variables [math]\displaystyle{ x }[/math] and [math]\displaystyle{ t_1\ldots t_n }[/math], and let [math]\displaystyle{ A^1_n }[/math] be the quotient of [math]\displaystyle{ A^0_n }[/math] by the following "HOMFLY" relations:

1. [math]\displaystyle{ x }[/math] commutes with everything else.
2. The product of permutations is as in the symmetric group [math]\displaystyle{ S_n }[/math].
3. If [math]\displaystyle{ \sigma }[/math] is a permutation then [math]\displaystyle{ t_i\sigma=\sigma t_{\sigma i} }[/math].
4. [math]\displaystyle{ [t_i,t_j]=x\sigma_{ij}(t_i-t_j) }[/math], where [math]\displaystyle{ \sigma_{ij} }[/math] is the transposition of [math]\displaystyle{ i }[/math] and [math]\displaystyle{ j }[/math].

Finally, declare that [math]\displaystyle{ \deg x=\deg t_i=1 }[/math] while [math]\displaystyle{ \deg\sigma=0 }[/math] for every [math]\displaystyle{ 1\leq i\leq n }[/math] and every [math]\displaystyle{ \sigma\in S_n }[/math], and let [math]\displaystyle{ A_n }[/math] be the graded completion of [math]\displaystyle{ A^1_n }[/math].

We say that an element of [math]\displaystyle{ A_n }[/math] is "sorted" if it is written in the form [math]\displaystyle{ x^k\cdot\sigma t_1^{k_1}t_2^{k_2}\cdots t_n^{k_n} }[/math] where [math]\displaystyle{ \sigma }[/math] is a permutation and [math]\displaystyle{ k }[/math] and the [math]\displaystyle{ k_i }[/math]'s are all non-negative integer. The HOMFLY relations imply that every element of [math]\displaystyle{ A_n }[/math] is a linear combinations of sorted elements. Thus as a vector space, [math]\displaystyle{ A_n }[/math] can be identified with the ring [math]\displaystyle{ B_n }[/math] of power series in the variables [math]\displaystyle{ x,t_1,\ldots,t_n }[/math] tensored with the group ring of [math]\displaystyle{ S_n }[/math]. The product of [math]\displaystyle{ A_n }[/math] is of course very different than that of [math]\displaystyle{ B_n }[/math].

Examples.

1. The general element of [math]\displaystyle{ A_1 }[/math] is [math]\displaystyle{ (1)f(x,t_1) }[/math] where [math]\displaystyle{ (1) }[/math] denotes the identity permutation and [math]\displaystyle{ f(x,t_1) }[/math] is a power series in two variables [math]\displaystyle{ x }[/math] and [math]\displaystyle{ t_1 }[/math]. [math]\displaystyle{ A_1 }[/math] is commutative.
2. The general element of [math]\displaystyle{ A_2 }[/math] is [math]\displaystyle{ (12)f(x,t_1,t_2)+(21)g(x,t_1,t_2) }[/math] where [math]\displaystyle{ f }[/math] and [math]\displaystyle{ g }[/math] are power series in three variables and [math]\displaystyle{ (12) }[/math] and [math]\displaystyle{ (21) }[/math] are the two elements of [math]\displaystyle{ S_2 }[/math]. [math]\displaystyle{ A_2 }[/math] is not commutative and its product is non-trivial to describe.
3. The general element of [math]\displaystyle{ A_3 }[/math] is described using [math]\displaystyle{ 3!=6 }[/math] power series in 4 variables. The general element of [math]\displaystyle{ A_n }[/math] is described using n! power series in [math]\displaystyle{ n+1 }[/math] variables.

The algebra [math]\displaystyle{ A_n }[/math] embeds in [math]\displaystyle{ A_{n+1} }[/math] in a trivial way by regarding [math]\displaystyle{ \{1,\ldots,n\} }[/math] as a subset of [math]\displaystyle{ \{1,\ldots,n+1\} }[/math] in the obvious manner; thus when given an element of [math]\displaystyle{ A_n }[/math] we are free to think of it also as an element of [math]\displaystyle{ A_{n+1} }[/math]. There is also a non-trivial map [math]\displaystyle{ \Delta:A_n\to A_{n+1} }[/math] defined as follows:

1. [math]\displaystyle{ \Delta(x)=x }[/math].
2. [math]\displaystyle{ \Delta(t_i)=t_{i+1}+x\sigma_{1,i+1} }[/math].
3. [math]\displaystyle{ \Delta }[/math] acts on permutations by "shifting them one unit to the right", i.e., by identifying [math]\displaystyle{ \{1,\ldots,n\} }[/math] with [math]\displaystyle{ \{2,\ldots,n+1\}\subset\{1,\ldots,n+1\} }[/math].

The Equations

We seek to find a "braidor"; an element [math]\displaystyle{ B }[/math] of [math]\displaystyle{ A_2 }[/math] satisfying:

• [math]\displaystyle{ B=(21)+x(12)+ }[/math](higher order terms).
• [math]\displaystyle{ B(\Delta B)B=(\Delta B)B(\Delta B) }[/math] in [math]\displaystyle{ A_3 }[/math].

With the vector space identification of [math]\displaystyle{ A_n }[/math] with [math]\displaystyle{ 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.

A Solution

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.

Computer Games

A primitive mathematica program to play with these objects is here.