Quandle

Stream: mathlib4

[Kenny Lau]

image.png

I am in the process of catching typos via GPT, and I came across this one, that I have manually reverted due to lack of expertise.

@Kyle Miller Do you know what the correct statements should be?

1. https://ncatlab.org/nlab/show/quandle#examples contains the Alexander quandle with the triangle inverted but also a and b inverted.
2. https://arxiv.org/pdf/2403.14458v1 contains the Lie group quandle but with t in the exponent, i.e. $X \triangleright_t Y = \mathrm{Ad}(e^{tX})Y$

[Kim Morrison]

The thing on the | side of the triangle is acting on the thing on the < side of the triangle, usually. So in the first diff it should be Ad (exp y) x. For the second diff it probably doesn't matter which is which, as long as both a and b appear.

[Kenny Lau]

I think only one option is invertible, because we're in the ring Z[t,t^-1]

[Kenny Lau]

I think it has to be tb + (1-t)a

[Kyle Miller]

The literature is inconsistent about this. For better or worse, Mathlib's has x ◃ y being that x acts on y.

I think the pictured diff is correct @Kenny Lau.

[Kenny Lau]

@Kyle Miller I think it has to be ta + (1-t)b in that case then

[Kyle Miller]

a ◃ b = t * b + (1 - t) * a is correct and is consistent with the geometric meaning of the Alexander quandle (though in some conventions it's more naturally t ^ -1).

It's an "abelianization" of the fundamental quandle, and the idea is that a ^ -1 * b * a as a path read right-to-left when lifted to the infinite cyclic cover and then understood as a 1-chain is t * (-a) + t * b + a; this assumes a and b are both elements of the fundamental quandle, and so both represent the generator in $H_1$ of the knot complement.

The inverse operation is a ◃⁻¹ b = t^-1 * b + (1 - t^-1) * a.

[Kyle Miller]

Section 9 of these notes has some details.

[Kenny Lau]

(19:50) gp > f(a,b)=t*b+(1-t)*a
%1 = (a,b)->t*b+(1-t)*a
(19:50) gp > g(a,b)=t^-1*b+(1-t^-1)*a
%2 = (a,b)->t^-1*b+(1-t^-1)*a
(19:50) gp > f(a,g(a,b))
%3 = b
(19:50) gp > g(a,f(a,b))
%4 = b

[Kenny Lau]

I guess you're correct