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11:00 PM
It sounds like a wiggle of the standard degeneracy locus interpretation in G/H and my lecture notes (proved with Schubert cycles in the universal case).
 
Sounds reasonable.
 
Some year Balarka will read some off the junk I’ve sent him.
 
Yup
 
it's probably a good exercise to work out for someone who understands this stuff
sadly I don't
 
Someone should write down what this means for Pontryagin classes
Nobody understands those
Except MacPherson
 
11:04 PM
Pontryagin understood Pontryagin classes
should read his original paper
 
Seems doubtful. Oh, Chern understood them
 
I mean, depends on what understanding means
 
True.
 
there's a very concrete geometric definition of Pontryagin classes that was known to Pontryagin already
 
Yeah, similar kind of section zero locus definition
I find that too mysterious though
 
11:06 PM
it's about when sections span a something-dimensional subspace, I think
but I forgot
 
Apparently MacPherson understood $p_1$ of a 4-manifold in terms of singularities of a generic map $M \to \Bbb R^4$
And then went on to write a purely combinatorial definition of Pontryagin classes
Which was supposed to be a hard problem.
 
idk, I think zero locus stuff is understandable
combinatorics is incomprehensible
 
I was sick for a few days post-vaccination and kept thinking about random things. Someone asked me about quandle homology and I went into a quandle rabbithole.
@Thorgott Want to know what a rack is?
 
I feel compelled to say no
 
I will take that as a yes, because silence is consent.
A rack is a set with a binary operation such that multiplication is an automorphism
 
11:13 PM
not even associative?
 
No associativity, it's distributive: $a * (b * c) = (a * b) * (a * c)$. This is the same as saying multiplication is an automorphism.
I forgot one other condition, though, which is right-invertibility: for any $a, b$ there is a unique $c$ with $a * c = b$.
 
My exercise is: Construct a sequence of continuous functions $f_n : [0, 1] → [−1, 1]$ such that $\int_0^1 f_n(x) dx = 0$ for all $n$ and $f_n(x)$ does not converge for any $x\in [0, 1]$. Now I'm asking myself if $f_n(x)=\sin(\pi \cdot n)$ is a valid solution? (sorry to interrupt here)
 
ok, that's at least something
cause otherwise that would just be a set with a map $X\rightarrow\mathrm{Sym}(X)$, which would seem like a pretty dumb notion
 
I said multiplication is an automorphism.
That's not the same as saying it's just a set with a self-action.
 
oh right
I get you now
now convince me this is interesting
@vitamind no, that's just zero
 
11:19 PM
A natural example of a rack is a group with binary operation being $a * b = aba^{-1}$. Then $a * (b * c) = a(bcb^{-1})a^{-1} = aba^{-1} aca^{-1} ab^{-1} a^{-1} ac^{-1}a^{-1} = (a * b) * (a * c)$.
 
Can't you just take three different functions and alternate them?
$f_1(x) = x-1/2, f_2(x) = 1/2-x$
and $f_3(x)$ a function that isn't $0$ at $1/2$
 
@Thorgott Do you remember, given a knot, how to compute a presentation of the fundamental group of its complement, in $S^3$ let's say?
 
I would not a remember a thing I've never learned
I don't know anything about knots lol
 
Ah, alas. Well, long story short, given any link you can do the following: Whenever there's a crossing you can draw 3 loops, one which winds around the overcrossing, one which winds around one half of the undercrossing, and one which winds around the other half of the undercrossing
 
@TedShifrin: I found my post. What I had proven was that the Binomial Transform of $-H_n$ is $\frac1n[n>0]$, but since the Binomial Transform is its own inverse, I had what I wanted (that the Binomial Transform of $-\frac1n[n>0]$ is $H_n$).
 
11:25 PM
Let's call them $a, b, c$. These things, for each crossing, generate the link group. The relators are $aba^{-1} = c$, and that's it.
This is called the Wirtinger presentation. Good exercise in Siefert-van Kampen theorem
 
I think I did the Schubert cycle proof for Pontryagin in some class. Chern has that in his old Princeton notes. @Balarka
Never heard of binomial transform, @robjohn … Is this a well-known thing?
 
I wonder if Leslie will let me actually create Leslie Coin given he's giving it free publicity
 
The link group is a really dumb invariant of the link. For a knot, the knot group is typically good but because of chance: There is a famous theorem of Gordon-Luecke which says knots are isotopic iff the knot complements are homeomorphic. There is also a famous theorem of Papakyriakapoulos which says knot complements are $K(G, 1)$, so determined by their fundamental groups, upto homotopy.
And there is also a famous theorem of Thurston which says knot complements are mostly hyperbolic, and finally hyperbolization theorem says hyperbolic manifolds are often determined by fundamental group and some extra structure at the end...
 
@TedShifrin Sort of. It is often useful for dealing with series that involve binomial coefficients.
 
However, the above chain of reasoning has a screeching hault at the very beginning for links. Gordon-Luecke theorem fails. It's really easy to come up with non-isotopic links with homeomorphic complements.
So the link group is going to lose a lot of info
Some genius had the following idea: Instead of creating this link group with $a, b, c$, $aba^{-1} = c$, why not create a rack generated by $a, b, c$, $a*b = c$?
 
11:30 PM
The $n^\text{th}$ term of the transformed sequence is the $n^\text{th}$ forward difference of the first $n+1$ terms.
 
That's called the fundamental rack of the link. This is a complete invariant of the link, aka provably totally determines the link.
 
Forward differences are alternating sums?
 
Almost by construct, you lose no information. Unfortunately, racks are insane new notions and studying them would mean you have to create foundations for racks like people for the past few centuries have been creating foundations for groups
So that's the story
I'm actually being tongue and cheek. It's an insane idea but you can extract very good link invariants from the fundamental rack. This is by Rourke and Sanderson.
It's 5 AM now though so I am going to sleep
G'night
 
night
 
I woke up at 1 am
 
11:33 PM
5 am is a good time to wake up
I've been wakin' up at 7
 
11:44 PM
Night !
 
11:54 PM
@Yorch What country?
 
Does anyone happen to know if there's a way of computing $\int_0^1\frac{\arctan(x)}{x+1}dx$ with complex analysis?
 
7 am in my country
which is Mexico
well, I don't own it yet
I meant the country I live in
 
wait, its 7pm in MX now
MX City I mean
 
that happened to me too in lockdown, sometimes I didn't know whether it was day or night outside :-)
 
I don't like being inside.
 
11:57 PM
yeah
I was referring to what balarka said
I woke up 12 hours ago I guess
I don't like being inside either
I want to move to the countryside
and grow my own yams
 
I want to be able to learn another language. Unfortunately I am a monoglot.
Romance languages are beyond my capabilities.
 
me to some place which has lot of greenery and rivers and lakes and hills also :-)
 
what language would you like to learn?
 
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