18.12.23 (The monoidal groupoid of a Coxeter group)
Hi everyone. Today a hint how to make your life easier or at least more visual. Imagine you are in this real life situation: Someone gives you two words in the generators of a group and asks you: Are they the same? The general answer is: No idea, this is some very hard problem. However if you work with Coxeter groups you can replace some of the work, by thinking of words in the generators of your Coxeter group as element of a monoidal category. Every generator of the Coxeter group is it's own color, say for S_3 you have s (red), t (blue) and you have generating morphisms of the category corresponing to the Coxeter relations, for instance the relation sts=tst, you replace by a pair of inverse isos sts->tst and tst->sts, which you think of as a trivalent vertex connecting sts with tst in alternating colors. The nontrivial relations between these isos come from the Zamolodhikov equation coming from every rank 3 parabolic subgroup. Basically you take the reduced expression graph, which is up to trivial loops one big circle and you demand that going around the circle is trivial. You should try it, it's similar to working with string diagrams for S_n (Indeed: S_n is really the groupoid describing the different ways to write to longest element in (A_1)^n, i.e. s_1s_2...s_n :)
01.12.23 (sl2 tensor product numbers as defect in dihedral Hecke algebra)
Hello everyone. So currently I am reading Elias and Williamson's people.mpim-bonn.mpg.de/geordie/aarhus/GR4allSB.pdf and one really cool thing they say and which I wanted to share with the world is that the Clebsch-Gordan multiplicities have a nice other way how they pop up:
Take the Hecke algebra of a dihedral group I(m) (generators s, t and relation (st)^m=e). Write down the two possible words for the longest element in I(m) (For type I(2)=A_1 x A_1 this would be st=ts, for I(3)=A_2=S_3 this would be sts=tst, for type I(4)=B_2 this would be stst=tsts, for type I(5) this would be ststs=tstst). Then compute the product of the KL generators B_s and B_t and take their product: (say B_sB_tB_s for I(3), B_sB_tB_sB_t for I(4) and B_sB_tB_sB_tB_s for I(5), write these elements in the standard basis and compute the coefficient of 1 (This coefficient is the standard trace tr(B(w_0)) of the element B(w_0)=B_sB_tB_s... . If we do this for the above dihedral groups this trace will become by Deodhar's defect formula (by counting with Bruhat strolls): For I(2) we get q^2, for I(3) we get q^3+q, for I(4) we get q^4+2q^2, for I(5) we get q^5+3q^3+2q^1. Now these coefficients appearing are exactly counting multiplicites in tensor products for sl2: L(1)^{\otimes 1}=L(1), which explains the coefficient 1 in front of the first q^2 (Here the exponent 2 is the dimension of L(1) ), L(1)^{\otimes 2}=L(2)\oplus L(1) which explains the coefficients of q^3+q, etc. etc. Beautiful stuff!!! Apparently this has to do with the fact that the Temperley--Lieb algebra at a root of unity is connected with the endomorphism algebra of the corresponding standard bimodule in the categorified picture!
09.11.23 (How symmetries are connected with Fibonacci-numbers, part 2)
Welcome back. Let's continue a little bit with the consequences of the connection between Fibonacci numbers and the Hecke algebra of S_2. Just as computing the powers of A gives Fibonacci-numbers, computing powers of H_s for general q gives H_s^n=(F_(n+1) & F_n \\ F_n & F_(n-1)) where F_n is the n-th quantum Fibonacci number iteratively defined as:
F_0=0, F_1=1, F_n=(q^{-1}-q)F_(n-1)+F_(n-2).
One obtains the following generalizations of Formulas for Fibonacci numbers:
By looking at the determinant of H_s one gets F_(n+1)F_(n-1)-F_n^2=(-1)^n:
By diagonalizing one gets a version of the Binot formula: F_n = ( (q^(-1))^n- (-q)^n) / (q+q^(-1)), which becomes for actual Fibonacci numbers F_n=(((1+sqrt(5))/2)^n + ((1 - sqrt(5))/2)^n ) / sqrt(5).
Conclusion: Deformed symmetries are everywhere!
09.11.23 (How symmetries are connected with Fibonacci-numbers, Part 1)
Hello everyone,
Today something really cool, I noticed yesterday, which you might think is deep or not. I am pretty unsure. Before I tell you how the symmetric group S_2 is connected to Fibonacci-numbers, let me recall quantization:
So a typical idea of good Lie theoretic symmetries is quantazation of these symmetries, meaning putting some extra parameter q such that when q approches 1 one recovers the original symmetries. Since things like finite groups G are discrete, one usually doesn't deform them but rather linearizations of them, think kG the group algebra or k[G] the ring of regular functions of G.
The easiest non-trivial group of symmetries Z/2Z, also known as symmetric group S_2 generated by one element s, satifying s^2=1. The action of s on the regular representation k^2=kS_2 can be written as the matrix s=( 0 & 1 // 1 & 0 ). It has two eigen-values 1 and -1, corresponding to the trivial representation of S_2 and sign representation of S_2. The projection onto the 1 and -1 eigenspace can be written nicely as 1/2 (id + s) and 1/2 (id - s) respectively.
The non-trivial quantization of this matrix is H_s=(q^{-1}-q & 1 \\ 1 & 0 ) (this matrix is the representing matrix of the generator of the Iwahori-Hecke algebra of S_2 on the algebra itself for experts). Note that if q -> 1 the matrix H_s becomes the matrix s, since one has 1 -1 =0.
The interesting thing is that the matrix H_s behaves suprisingly like the original one we started with: It has two eigenvalues q^{-1} and -q. The projections onto these eigenspaces can be expressed as 1/(q+q^{-1}) (q * id + H_s) and 1/(q+q^{-1}) (q^{-1} - H_s) respectively. The factor q+q^{-1} appearing in the denominator is an example of a quantum integer, here 'quantum two' often denoted [2].
What did we gain here? We got some bigger family of matrices which all behave like the generator in the symmetric group. Now you might think, you have never seen this deformation but this is not true! An example of H_s is the matrix A=(1 & 1 \\ 1 & 0 )=H_s (Here q= (-1+\sqrt(5))/2, so q^{-1} becomes (1+sqrt(5))/2, also known as the golden ratio!). A is famous, since powers of it are nice: A^n = (F_(n+1) & F_n \\ F_n & F_{n-1}), where F_n is the n-th Fibonacci number, defined by
F_0=0, F_1=1, For n>1: F_n=F_(n-1)+F_(n-2).
so computing powers of A is the same as computing Fibonacci numbers. What is pretty famous is that A has two eigenvalues: The golden ratio (1+sqrt(5))/2=q^{-1} and (1- sqrt(5))/2=-q. What is quantum two in case: Well it's [2]=q+q^{-1}=(-1+\sqrt(5))/2 + (1+sqrt(5))/2 = sqrt(5). In particular it is not suprising that in formulas for Fibonacci numbers sqrt(5) has to appear, since Fibonacci numbers are basically an example of the group algebra of S_2 deformed and 2 appears as the cardinality of S_2 :) !!! To be continued!
05.10.23 (Higher brackets gice multiplication on the symmetric coalgebra)
Welcome to the autumn 2023 section, the moment where I start writing more regulary cool math stuff here.
Here is something from my current project, which is related to things like the Cartier-Milnor-Moore Theorem (CMM).
If you don't know the statement of CMM: It states that in characteristic zero, the universal enveloping algebra U(g) of a Lie algebra g is isomorphic as a coalgebra to the symmetric coalgebra S^c(g), i.e. as a vector space the subspace of T(g) consisting of all symmetric tensors (i.e. in degree n (g^{\otimes n})^{S_n}), but equipped with the deconcatenation coproduct. As a special case: In characteristic 0 the symmetric algebra of a vector space V called Sym(V) (which is a Hopf algebra, by making all elements of V primitive), is isomorphic to the symmetric coalgebra as coalgebra. They are even isomorphic as Hopf algebras, when you equipp S^c(V) with the shuffle product.
The isomorphism is given by symmetrization, i.e. applying the sum of all permuations to your given element of Sym(V). So for instance the polynomial x^2 in Sym(V) would be mapped to the symmetric 2x\otimes x in the symmetric coalgebra, while the polynomial xy would be mapped to x\otimes y + y\otimes x.
Now using this coalgebra isomorphism between U(g) and S^c(g), allows to translate the multiplication in U(g) to the multiplication in the symmetric coalgebra (which is nuts):
Here some examples: The product of two degree 1 elements x and y is x*y=[x,y]/2 + x \otimes y + y\otimes x
The product of a degree 2 symmetric tensor xy+yx and a degree 1 element z is (x\otimes y+y\otimes x)*z=([x,[y,z,]]+[y[x,z]])/12 + ( [x,z]\otimes y+y\otimes [x,z] + [y,z]\otimes x + x \otimes [y,z])/2 + x\otimes y \otimes z + y\otimes x \otimes z + x\otimes z \otimes y + y \otimes z \otimes x + z \otimes x \otimes y + z\otimes y \otimes x
Note that the fact that U(g) is filtered can be seen here as the product of two symmetric tensors with respect to the U(g) product, becomes also filtered and if the Lie algebra would be abelian, i.e. [-,-]=0, this would make all the lower terms disappear.
What's the conclusion: Life is weird, when you don't allow for quotients in your life, but doable at least in characteristic 0.