Ricatti cheese is delicious on pasta. I also like going to the regatta boat races. Not sure who the fuck this guy is but he can get it too. You're next.

>Terry is now doing ads for OpenAI
>you were born just in time to watch math become a pay to win esport

>>16991877
It's what the Satanic Reptilian NWO wants, so it's going to happen unfortunately.

>>16991877
wasn't he just complaining that AI is bad for education

>>16991877
It seems like none of these big names have considered how bad it will be that we’re about to have academic research become a pay to win game where you can’t compete unless you have access to the latest models (they have access to internal models btw, you literally cannot even pay to access this unless you’re part of the club)
But I think that’s just the way it seems. I think really what it is is that people like Gowers and Tao know they’re safe so why would they care where this is quickly heading for everyone else? They’re not stupid, they know everyone else is about to get permafucked but they already made it so it’s just tough shit for (You).

File: Diamond.png (99.4 KB)

Is there a closed form answer for the angle alpha? Circles of the same color are same size.

Everything is fucking going AI and ai hate it. I feel like I'm watching IQ points getting slurped out of people's craniums. Future generations may be looking at an idiocracy world.

>>16991899
I was able to learn concepts way faster with AI what the fuck is he talking about.

>>16991907
>we’re about to have academic research become a pay to win game
It literally always has been. Do you know how much journal subscriptions cost? Probably not, because it's the universities that pay for them (and if you're an independent researcher, you can hope it's on Arxiv or get fucked). I imagine it'd be the same for this, if and when there becomes enough demand for it.

>>16988688

>>16991839
How does one start to learn math

>>16992285
No that’s not true, they actually can waive it for people who genuinely can’t pay. Also you know full well that’s a completely different sort of thing than “you must partner with this AI company or you literally cannot compete no matter how smart you are”

>>16992395
Im a retard so dont listen to me but Im going through a series of books called Baldor:
Aritmética Baldor
Then I will through Geometría Baldor
Finally Alegra Baldor
While simultaneity going through The Elements by Euclid.
I do the exercises on an eink tablet so I dont end up with a pile of papers and books.

>>16992417
*Algebra Baldor
*While simultaneously

File: 1000001921.jpg (201.2 KB)

>>16992417
Who decided that you have to start with Algebra, then geometry then algebra 3-4 then either trig, calc or pre calc?

>>16992457
I've also found it weird that geometry comes after algebra since historically it was the other way around. Geometry is also far more general and intuitive and is closer to what math is actually about (namely, abstraction). Algebra is usually taught as tedious computation but I guess is a more natural extension of the arithmetic that's taught in grade school.

Is non-commutative geometry (Connes version) interesting or useful for any other fields of mathematics?
I am asking because the only 2 'applications' I know of are the noncommutative standard model in physics and the Riemann hypothesis, which are not exactly the most reasonable goals for a midwit like me.

It seems very elegant to me because it translates geometry, which I suck at, into functional analysis, which I'm good at.

Revisiting all my shitty highschool math, trying to relearn the majority of it. Failed most of it because I couldn't really understand a bunch. How one thing related, lead into or built on the other. Shit like calculus may as well be some highly alien machine where I couldn't begin to attempt to intuit any sort of interactive pieces.
Currently it's mostly just arithmetic and basic algebra, trying to understand it more thoroughly.

>>16992531
I was forced into this side of things for my doctoral thesis and I found it to be interesting after I gave it a chance. I work with categorification and K-theory of "quantum algebras," and it's kinda neat.

>>16992457
The kitty needs cake.

What is d/dx?
It feels like it should be (dx-xd)\d

>>16992870
If you don't mind me asking, are you more on the representation theory side or just pure category theory?
Would you recommend non commutative algebra to someone or is it just a neat tool for highly specific experts?

>>16994045
NTA
Many (possibly most) things in life are not commutative.
I have been trying to use differential operators in enumerative combinatorics to emulate processes.
Kinda need to see the "matrix code" in generating functions first.

A toy example is for the stirling numbers of the second kind.
[(xD)^n (x-1)^k /k!](x=1)
The xD are the set elements, the (x-1) are the "bins", the k! un-orders the bins. The x=1 kills scenarios where a bin is not used and properly counts the scenarios that use all bins.
It is as obvious to me why this is the answer just as it is obvious why (1+x)^n gives binomial coefficients.

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>a circular table top weighs five kilos
>three uniformly random points are selected on the table
>these three points will be the locations where three legs of the table are attached (all legs are the same length and their weight is negligible)
>a fourth random point is selected. This is the location where a flower pot will sit on top of the finished table
>the flower pot weights three kilos

What is the probability that the table is not going to fall over?

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>>16994184

>>16994198
AI slop. Can't trust AI in math

>>16994184
>>16994198
Thank god that is known now. Such possibilities.

is data science a good job

>>16994199
well just fucking read it and check it yourself tard

>>16994199
AI is literally a math machine you low IQ luddite

>>16994418
I am an organic chemistry machine but if you asked me to draw out the exact mechanism of a Corey–Nicolaou macrolactonization I wouldn't know what the fuck to do and you would be wrong to trust any answer I gave

>>16994515
You are a dumb fuck

so far I'm thinking group theory is just dicking around, counting permutations

>>16994541
Frame bundle is where it touches "reality" most obviously. Intro stuff is still useful when you need to utilize discrete symmetry to simplify/solve problems.
Also
https://en.wikipedia.org/wiki/Poincar%C3%A9_group

I am finding algebraic topology hard to self learn. Is it because I’m not witnessing the “practical skills/ experience” that one would witness during an actual class?

>>16994540
No, you, sir. As the product of physics and chemistry we can follow physics and chemistry perfectly without understanding them. In fact, we can't actually do the opposite: we can't NOT follow the laws of the universe. Following isn't understanding though, so we need to build understanding of the machine from within the machine while it's running, without breaking the machine. It's unclear whether a system can perfectly simulate itself, or if our reality has a non mathematical component, so we may be limited in our understanding.

>>16995750
Always found that branch rather impenetrable as well. What book are you using?

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I promised a solution to >>16952247 this thread, so here it is (reposting the image for convenience).
Firstly, it's obvious by conjugation that we can assume [math] \pi_1=\text{id} [/math]. Define [math] \widetilde{M}_\pi=\frac{1}{2}(M_\text{id}+M_\pi). [/math] Then it's easy to check that the entries of [math] \widetilde{M}_\pi [/math] are
[eqn] \widetilde{m}_{ij}=\begin{cases}\text{sgn}(i-j) & \text{if }(i-j)(\pi(i)-\pi(j))>0 \\ 0 & \text{else}\end{cases}. [/eqn]
It suffices to determine [math] \text{det}(\widetilde{M}_\pi) [/math].

(1/9)

>>16995871
Theorem: The following are equivalent constructions of a function [math] F:\displaystyle\bigcup_{n\ge 0}S_n\to\mathbb{Z} [/math].
(a) For [math] \pi\in S_n [/math], [math] F(\pi)=\text{Pf}(\widetilde{M}_\pi) [/math] if [math] n [/math] is even and [math] F(\pi)=\text{Pf}(\widetilde{M}_{\pi\oplus 1}) [/math] if [math] n [/math] is odd.
(b) Let [math] E\subseteq\mathbb{Z} [/math] denote the set of even integers and, for a permutation [math] \pi\in S_n [/math], let [math] \text{Noninv}_\pi\subseteq\mathbb{Z}^2 [/math] be the set of pairs [math] (i, j) [/math] for which [math] i<j [/math] and [math] \pi(i)<\pi(j) [/math]. For a subset [math] S\subseteq \{1, 2, \cdots, n\} [/math], define
[eqn] q_\pi(S)=|S\cap E|+|(S\times S)\cap\text{Noninv}_\pi|. [/eqn]
Then define
[eqn] F(\pi)=2^{-\lceil n/2\rceil}\displaystyle\sum_{S}(-1)^{q_\pi(S)}. [/eqn]
(c) For a permutation [math] \pi\in S_n [/math] and an index [math] 1\le r<n [/math], denote by [math] \pi^{(r)}\in S_n [/math] the permutation obtained by swapping the values of [math] \pi [/math] at [math] r [/math] and [math] r+1 [/math], and denote by [math] \pi\setminus r\in S_{n-2}[/math] the permutation which is order-isomorphic to the function obtained from [math] \pi [/math] by removing [math] r [/math] and [math] r+1 [/math] from its domain. Then define [math] F [/math] axiomatically by
1. [math] F(\text{id}_n)=1 [/math] for every identity element [math] \text{id}_n\in S_n [/math].
2. For every permutation [math] \pi\in S_n [/math] and index [math] 1\le r<n [/math], impose [math] F(\pi\setminus r)=F(\pi)+F(\pi^{(r)}) [/math].

Proof. It's easy to see that the axioms of (c) uniquely define [math] F [/math], and to check that (a) and (b) both satisfy these axioms.
We'll refer to these respectively as the matrix, Gauss sum, and axiomatic definitions of [math] F [/math].

(2/9)

>>16995875
Lemma 1: (a) [math] F [/math] is invariant under reverse-complement.
(b) For direct sums, [math] F [/math] is multiplicative; that is,
[eqn] F(\pi_1\oplus\pi_2)=F(\pi_1)F(\pi_2) [/eqn].
For skew sums,
[eqn] F(\pi_1\ominus\pi_2)=\begin{cases}F(\pi_1)F(\pi_2) & \text{if }|\pi_1||\pi_2|\text{ even} \\ 0 & \text{else}\end{cases}. [/eqn]
Here [math] |\pi| [/math] is the degree of [math] \pi [/math].
(c) For odd [math] n [/math], let [math] \sigma_n=(1\,2\,\cdots\,n) [/math] in cycle notation. Then [math] F(\pi\sigma_n)=F(\sigma_n\pi)=F(\pi) [/math] for all [math] \pi\in S_n [/math].

Proof: (a) is immediate from the Gauss sum definition, since the reverse-complement preserves non-inversion pairs.
(b) follows straightforwardly from the matrix definition.
(c) follows from the uniqueness of [math] F [/math] in axiomatic definition; if we define [math] G(\pi)=F(\pi\sigma_n) [/math] or [math] G(\pi)=F(\sigma_n\pi) [/math], then [math] G [/math] satisfies the same axioms as [math] F [/math].

Theorem: [math] F [/math] takes values only in [math] \{-1, 0, 1\} [/math].

For each [math] \pi\in S_n [/math], define the quadratic form [math] Q_\pi [/math] on [math] \mathbb{F}_2^n [/math] by
[eqn] Q_\pi(x)=\displaystyle\sum_{i\text{ even}}x_i+\displaystyle\sum_{\substack{i<j\\\pi(i)<\pi(j)}}x_ix_j. [/eqn]
Then [math] 2^{\lceil n/2\rceil}F(\pi)=\displaystyle\sum_{x\in\mathbb{F}_2^n}(-1)^{Q_\pi(x)} [/math]. If we define
[eqn] B_\pi(x, y)=Q_\pi(x+y)+Q_\pi(x)+Q_\pi(y), [/eqn]
so that [math] B_\pi(x, y)=\displaystyle\sum_{\substack{i<j\\\pi(i)<\pi(j)}}(x_iy_j+x_jy_i) [/math], and we define the radical of this alternating bilinear form
[eqn] R_\pi=\text{rad}(B_\pi)=\{x\in\mathbb{F}_2^n\,|\,B_\pi(x, \cdot)\equiv 0\}, [/eqn]
then a standard argument gives
[eqn] 2^{2\lceil n/2\rceil}F(\pi)^2=2^n\displaystyle\sum_{x\in R_\pi}(-1)^{Q_\pi(x)}=\begin{cases}0 & \text{if }Q_\pi\big|_{R_\pi}\not\equiv 0 \\ 2^{n+\dim R_\pi} & \text{if }Q_\pi\big|_{R_\pi}\equiv 0\end{cases}. [/eqn].

(3/9)

>>16995877
Lemma 2: Define the permutation [math] \tau_\pi [/math] on [math] \{0, 1, \cdots, n\} [/math] by [math] \tau_\pi=\pi^{-1}\sigma\pi\sigma [/math], where [math] \sigma=(0\,1\,\cdots\,n) [/math] in cycle notation and we extend [math] \pi [/math] by taking [math] \pi(0)=0 [/math]. Then [math] \dim R_\pi=c(\tau_\pi)-1 [/math], where [math] c(\tau_\pi) [/math] is the number of cycles in [math] \tau_\pi [/math].

Proof: Take a fixed [math] x\in R_\pi [/math]. Define the sums [math] r_k=x_1+\cdots+x_k [/math] and [math] s_k=\displaystyle\sum_{\pi(j)<k}x_j [/math], with [math] r_0=s_0=0 [/math]. Let [math] m=r_n=s_n [/math]. Compute straightforwardly that [math] B_\pi(x, e_i)=m+r_{i-1}+s_{\pi(i)} [/math], from which we get that [math] x\in R_\pi [/math] if and only if [math] s_{\pi(i)}=m+r_{i-1} [/math] for all [math] 1\le i\le n [/math].
Let [math] p_k=\pi^{-1}(k) [/math] for [math] 1\le k\le n [/math]. We can rewrite the above radical condition as [math] s_k=m+r_{p_k-1} [/math]. Noticing that [math] s_k+s_{k-1}=x_{p_k}=r_{p_k}+r_{p_k-1} [/math], the radical condition becomes [math] r_{p_k}=r_{p_{k-1}-1} [/math] for [math] 2\le k\le n [/math], with edge cases [math] r_{p_1}=m [/math] and [math] r_{p_n-1}=0 [/math].
For [math] \tau_\pi [/math] as defined in the lemma statement, [math] \tau_\pi(p_k-1)=p_{k+1} [/math] for [math] 1\le k<n [/math], [math] \tau_\pi(p_n-1)=0 [/math], and [math] \tau_\pi(n)=p_1 [/math]. This says exactly that the [math] r_i [/math] are constant for indices in the same cycle of [math] \tau_\pi [/math], with that constant being [math] 0 [/math] for the cycle containing [math] 0 [/math].
Therefore, there's one degree of freedom in [math] R_\pi [/math] for each cycle of [math] \tau_\pi [/math] not containing [math] 0 [/math]. Conversely, one can run this construction in reverse and see that any consistent choice of the [math] r_i [/math] with respect to this condition will give a valid element [math] x\in R_\pi [/math].

(4/9)

>>16995880
Lemma 3: Let [math] C [/math] be a cycle of [math] \tau_\pi [/math] with [math] 0, n\not\in C [/math]. Define [math] r_i=1_{i\in C} [/math] for [math] 0\le i\le n [/math] and [math] x_i=r_i+r_{i-1} [/math] for [math] 1\le i\le n [/math]. Then [math] x\in R_\pi [/math] and [math] Q_\pi(x)=1 [/math].

Proof: [math] x\in R_\pi [/math] follows by the previous lemma.
We compute [math] Q_\pi(x) [/math]. Write [math] C=\{c_1<\cdots<c_l\} [/math]. For the linear term,
[eqn] \displaystyle\sum_{i\text{ even}}x_i=\displaystyle\sum_{i=1}^{n}r_i=l\;(\text{mod }{2}), [/eqn]
since [math] r_n=0 [/math]. Since [math] x\in R_\pi [/math],
[eqn] 0=\displaystyle\sum_{i=1}^{n}r_iB_\pi(x, e_i) [/eqn] [eqn] =\displaystyle\sum_{\substack{i<j\\\pi(i)<\pi(j)}}(r_jx_i+r_ix_j) [/eqn] [eqn]=\displaystyle\sum_{\substack{i<j\\\pi(i)<\pi(j)}}(r_jr_{i-1}+r_ir_{j-1}). [/eqn]
Let [math] P=\displaystyle\sum_{\substack{i<j\\\pi(i)<\pi(j)}}x_ix_j [/math] be the quadratic part of [math] Q_\pi(x) [/math]. Expanding and cancelling gives
[eqn] P=\displaystyle\sum_{\substack{i<j\\\pi(i)<\pi(j)}}(r_ir_j+r_{i-1}r_{j-1}). [/eqn]
Since [math] r_i=0 [/math] for [math] i\not\in C [/math], we can rewrite this as
[eqn] P=\displaystyle\sum_{u<v}(1_{\pi(c_u)<\pi(c_v)}+1_{\pi(c_u+1)<\pi(c_v+1)}). [/eqn]
Define [math] \rho\in S_l [/math] by [math] \tau_\pi(c_i)=c_{\rho(i)} [/math]; [math] \rho [/math] is necessarily an [math] l [/math]-cycle. Now,
[eqn] \pi(c_{\rho(i)})=\pi(\tau_\pi(c_i))=\sigma(\pi(\sigma(c_i)))=\pi(c_i+1)+1. [/eqn]
This gives that [math] \pi(c_u+1)<\pi(c_v+1) [/math] if and only if [math] \pi(c_{\rho(u)})<\pi(c_{\rho(v)}) [/math]. Denoting [math] a_i=\pi(c_i) [/math],
[eqn] P=\displaystyle\sum_{u<v}(1_{a_u<a_v}+1_{a_{\rho(u)}<a_{\rho(v)}}). [/eqn]
Thus, on general principles, [math] P=\text{inv}(\rho)\;(\text{mod }{2}) [/math], and since [math] \rho [/math] is a [math] l [/math]-cycle, [math] P=l-1\;(\text{mod }{2}) [/math].
Therefore, [math] Q_\pi(x)=l+(l-1)=1\in\mathbb{F}_2 [/math].

(5/9)

>>16995882
Proof of theorem: We've shown before the lemmas that the possible values of [math] F(\pi)^2 [/math] are [math] 0 [/math] if [math] Q_\pi [/math] is non-trivial on [math] R_\pi [/math], or otherwise, [math] 2^{\dim R_\pi} [/math] if [math] n [/math] is even and [math] 2^{\dim R_\pi-1} [/math] if [math] n [/math] is odd. Furthermore, Lemma 2 shows that these dimensions are one less than the cycle count of [math] \tau_\pi=\pi^{-1}\sigma\pi\sigma [/math].
By Lemma 3, if this cycle count is at least [math] 3 [/math], then [math] Q_\pi [/math] can't restrict trivially to [math] R_\pi [/math], so [math] F(\pi)\not=0 [/math] implies [math] c(\tau_\pi)\le 2 [/math], or [math] \dim R_\pi \le 1 [/math]. But there's a further parity restriction: since [math] B_\pi [/math] is alternating, [math] \dim R_\pi=n\;(\text{mod }2) [/math], which restricts [math] \dim R_\pi [/math] fully to [math] 1 [/math] if [math] n [/math] is odd and [math] 0 [/math] if [math] n [/math] is even, presuming [math] F(\pi)\not=0 [/math].
Therefore the possible values of [math] F(\pi)^2 [/math] are [math] 0 [/math] or [math] 1 [/math] in all cases.

This solves Kurisu's problem: the only possible values of the determinant are [math] 0 [/math] and [math] 2^n [/math]. If [math] n [/math] is odd, clearly only [math] 0 [/math] is achievable, since the matrix is skew-symmetric. If [math] n [/math] is even, it's easy to check both are achievable.

(6/9)

>>16995884
Corollary: Let [math] \pi\in S_n [/math]. If [math] n [/math] is even, [math] F(\pi)\not=0 [/math] if and only if [math] \tau_\pi [/math] is a [math] (n+1) [/math]-cycle. If [math] n [/math] is odd, [math] F(\pi)\not=0 [/math] if and only if [math] c(\tau_\pi)=2 [/math] and [math] 0 [/math] and [math] n [/math] aren't in the same cycle.

Proof: If [math] n [/math] is even and [math] c(\tau_\pi)=1 [/math], then [math] Q_\pi\big|_{R_\pi}=0 [/math] trivially, so [math] F(\pi)\not=0 [/math].
If [math] n [/math] is odd and [math] c(\tau_\pi)=2 [/math] but [math] 0 [/math] and [math] n [/math] are same-cycle, we can go through the proof of Lemma 3 but with [math] C [/math] as the other cycle to get [math] Q_\pi\big|_{R_\pi}\not\equiv0 [/math], so [math] F(\pi)\not=0[/math]. Or, if [math] 0 [/math] and [math] n [/math] are in different cycles, we can let [math] C [/math] be the cycle containing [math] n [/math] to get that [math] Q_\pi(x)=0 [/math]. Since [math] \dim R_\pi=1 [/math] in this case, this [math] x [/math] generates [math] R_\pi [/math], and so we've shown [math] Q_\pi\big|_{R_\pi}\equiv0 [/math], whence [math] F(\pi)\not=0 [/math].
We've already shown [math] c(\tau_\pi)\ge 3 [/math] implies [math] F(\pi)=0 [/math].

Corollary: If [math] F(\pi)\not=0 [/math] so that [math] Q_\pi\big|_{R_\pi}\equiv0 [/math], then [math] F(\pi)=(-1)^{\text{Arf}(\overline{Q}_\pi)} [/math], where [math] \text{Arf}(\overline{Q}_\pi) [/math] is the [math] \mathbb{F}_2 [/math]-valued Arf invariant of the quadratic form induced from [math] Q_\pi [/math] on the quotient [math] \mathbb{F}_2^n/R_\pi [/math].

Proof:
[eqn] 2^{\lceil n/2\rceil}F(\pi)=\displaystyle\sum_{x\in\mathbb{F}_2^n}(-1)^{Q_\pi(x)}=2^{\dim R_\pi}\displaystyle\sum_{\overline{x}\in\mathbb{F}_2^n/R_\pi}(-1)^{\overline{Q}_\pi(\overline{x})}=2^{\dim R_\pi+(\dim(\mathbb{F}_2^n/R_\pi))/2}(-1)^{\text{Arf}(\overline{Q}_\pi)}. [/eqn]
As [math] F(\pi)=\pm 1 [/math], all powers of [math] 2 [/math] cancel.

(7/9)

>>16995886
Theorem: The values of [math] F [/math] distribute as
[eqn] \frac{1}{n!}|F^{-1}(0)\cap S_n|=\frac{\lfloor n/2\rfloor}{\lfloor n/2\rfloor +1} [/eqn]
[eqn] \frac{1}{n!}|F^{-1}(1)\cap S_n|=\frac{1}{2}\left(\frac{1}{\lfloor n/2\rfloor+1}+\frac{1}{2^{\lfloor n/2\rfloor}}\right) [/eqn]
[eqn] \frac{1}{n!}|F^{-1}(-1)\cap S_n|=\frac{1}{2}\left(\frac{1}{\lfloor n/2\rfloor+1}-\frac{1}{2^{\lfloor n/2\rfloor}}\right). [/eqn]

Proof: There's a uniformly [math] n [/math]-to-[math] 1 [/math] map [math] f: S_n\to S_{n-1} [/math] defined by [math] f(\pi)\oplus 1=\pi\sigma_n^k [/math] for some integer [math] k [/math] dependent on [math] \pi [/math]. Assuming [math] n [/math] is odd, by Lemma 1, [math] F(\pi)=F(f(\pi)) [/math], so the normalized distribution of [math] F [/math] is the same on [math] S_n [/math] and [math] S_{n-1} [/math].
Thus assume [math] n=2m [/math] is even. By a previous corollary, [math] F(\pi)\not=0 [/math] if and only if [math] \pi^{-1}\sigma\pi\sigma [/math] is a [math] (n+1) [/math]-cycle. Let [math] \alpha_\pi=\pi^{-1}\sigma\pi [/math], so the map [math] \pi\mapsto\alpha_\pi [/math] is a bijection from [math] S_n [/math] to [math] (n+1) [/math]-cycles in [math] S_{n+1} [/math]. It's a known result (Zagier-Stanley) that the product of two random [math] (n+1) [/math]-cycles is also one with probability [math] \frac{2}{n+2} [/math]. Since letting [math] \pi [/math] range over [math] S_n [/math] is equivalent to letting [math] \alpha_\pi [/math] range over [math] (n+1) [/math]-cycles,
[eqn] \displaystyle\sum_{\pi\in S_n}F(\pi)^2=\frac{2}{n+2}n!=\frac{1}{m+1}(2m)!. [/eqn]
Finding the first moment of [math] F [/math] is easier; just fix an [math] r [/math] and sum the recursion in the Gauss sum definition of [math] F [/math] to inductively get
[eqn] \displaystyle\sum_{\pi\in S_n}F(\pi)=\frac{1}{2^m}(2m)!. [/eqn]
Finally, knowing that [math] F(\pi)\in\{-1, 0, 1\} [/math], the first two moments of [math] F [/math] fully determine its distribution.

(8/9)

>>16995888
Finally, an interesting further extension for the more geometrically-inclined:
Given a permutation [math] \pi\in S_n [/math], write the numbers [math] 1, 2, \cdots, 2n [/math] around a circle. Start with the closed disk representing circle, and attach handles to its boundary from [math] i [/math] to [math] \pi(i)+n [/math], giving that handle a full twist for even [math] i [/math]. There's a natural embedding [math] \Sigma_\pi\hookrightarrow S^3 [/math], and also a natural basis [math] \{e_1, \cdots, e_n\} [/math] of [math] H_1(\Sigma_\pi;\,\mathbb{F}_2) [/math] given by the ribbons.
Exercises for the reader:
a) The intersection form [math] B_\pi [/math] on [math] H_1(\Sigma_\pi;\,\mathbb{F}_2) [/math] is exactly the [math] B_\pi [/math] previously defined algebraically.
b) The quadratic refinement [math] Q_\pi [/math] of [math] B_\pi [/math] associated with the spin structure on [math] \Sigma_\pi [/math] inherited from the embedding [math] \Sigma_\pi\hookrightarrow S^3 [/math] is exactly the [math] Q_\pi [/math] previously defined algebraically.
c) [math] F(\pi)\not=0 [/math] if and only if every connected component of [math] \partial\Sigma_\pi [/math] inherits the bounding spin structure.
For any boundary component curve [math] \gamma [/math] with a bounding spin structure, we can "fill it in" with a disk that matches the spin structure. If we can do this with all boundary components, label the resulting closed spin surface [math] \hat{\Sigma}_\pi [/math].
d) The value of [math] F(\pi) [/math] can be fully characterized geometrically as follows.
-If [math] \hat{\Sigma}_\pi [/math] doesn't exist, then [math] F(\pi)=0 [/math].
-If [math] \hat{\Sigma}_\pi [/math] is null-cobordant in [math] \Omega^{Spin}_2\cong\mathbb{Z}_2 [/math], then [math] F(\pi)=1 [/math].
-If [math] \hat{\Sigma}_\pi [/math] represents the non-trivial class in [math] \Omega^{Spin}_2\cong\mathbb{Z}_2 [/math], then [math] F(\pi)=-1 [/math].

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I am mathematical laity who only barely passes highschool math. I wish to attain a solid graduate level apprehension of semigroups and group theory at large of course

>>16995871
Thanks anon, I actually have a different solution to this problem, I just forgot to post it last thread
You'll have to excuse my laziness with the lack of formatting
If M~ is singular, we're done, so assume M~ is not, we need to show det M~ =1
It's sufficient to prove that M~^-1 is an integer matrix, since M is an integer matrix, and skew symmetry of M implies Det M~ >= 0
Then, let x be the unique vector such that M~x=ej, so if x is an integer for all j, M~^-1 is an integer matrix
Let u(k) = sum i>k x(i), v(k) = sum π(i)<k x(i), v(1)=u(n)=0,v(n+1)=u(0) (note this implies the map v->x is an injection)
x(k) = u(k-1) - u(k) = v(π(k)+1) - v(π(k)) (1)
It can be shown (Mid x)_k = u(0) - u(k-1) - u(k), (Mπ x)_k = v(π(k)) - v(n+1) + v(π(k)+1)
By our definition of x, ((Mid+Mπ)x)_k = 2δ(k,j)
So, v(π(k)) + v(π(k)+1) - u(k-1) - u(k) = 2δ(k,j) (2)
and, v(π(k-1)) + v(π(k-1)+1) - u(k-1) - u(k-2) = 2δ(k-1,j)
The difference is v(π(k))+ v(π(k)+1)-v(π(k-1))-v(π(k-1)+1) + u(k-2) - u(k-1) + u(k-1) - u(k) (3)
From (1) u(k-2) - u(k-1) + u(k-1) - u(k) = v(π(k)+1) - v(π(k)) + v(π(k-1)+1) - v(π(k-1)) (4)
So subbing (4) into (3) and halving: v(π(k)+1) - v(π(k-1)) = δ(k,j) - δ(k-1,j) (k=2..n) (5)
Let k=n, u(k)=0, u(k-1)=u(k) + v(π(k)+1) - v(π(k)) from (1), subbing into (2) and halving simplifies to v(π(n)) = δ(n,j) (6)
Collectively, v(1)=0, (5) and (6) are a system of n+1 equations in n+1 variables that can be written as Dv=a
D has at most one 1 and one -1 in each row, so is totally unimodular by a known result of Poincaré
D is invertible, since otherwise we'd have distinct solution v, implying distinct x, contradicting uniqueness
a is an integer vector, so then v is too, because D^-1 is integer, and x is too for any j
So M~^-1 is an integer matrix, and det M~=1

Therefore, in any case, det M~= 0 or 1

>>16995995
Interesting. I spent a while looking for a "direct" proof like this but couldn't find one before I found the quadratic form business. I wonder whether your proof is just a computational distillation of mine or more fundamentally different, since my proof isn't much more than dressed-up elementary linear algebra anyway.

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