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7:00 PM
@BalarkaSen $\gamma^2$ not being the identity reminds me of an old question I asked here: Can torsion in the fundamental group happen for something embedded in $\mathbb{R}^3$?
 
@LeakyNun I don't know this enough, so I wouldn't be able to say.
 
yes

also yes it's scary in hong kong
 
@TobiasKildetoft That's a good question. I remember thinking about this for unrelated reasons a few days ago, but can't remember how.
 
@BalarkaSen I asked it on main, and I think I was directed to an MO question about it
 
suppose $\varphi : X \to \Bbb R^3$ is the embedding. clearly this gives an injection $\pi_1(X) \to \pi_1(\Bbb R^3)$, but the latter is zero so the former is zero, so everything embedded in $\Bbb R^3$ has trivial fundamental group, hence no torsion /s
 
7:06 PM
It's certainly false for any decent God fearing subset eg a simplicial subcomplex of R^3. Who cares about the rest imo
 
can you embed Lens space into R^4?
well surely you can embed RP^2 into R^4 so whatever
low dim top is weird
 
You certainly cannot embed RP^3 in R^4
Tangent bundle is not stably 1-trivial.
 
I also asked a related question on MSE a while back math.stackexchange.com/questions/3048957/…
 
@BalarkaSen delet this
an orientable 3-manifold is ______
 
Oh parallelizable. Fuck.
 
7:11 PM
I think the answer is you can't embed RP^n into R^{n+1} for any n but I forget the proof, somehow it involves the Z/2 cup product.
I don't remember about lens spaces in R^4.
 
I only care about the fact that you can't embed $\Bbb P^n_\Bbb R$ into any $\Bbb A^m_\Bbb R$
 
Hm, I can't remember the proof that RP^3 cannot be embedded in R^4. I am sure I worked this out before.
 
@Balarka @MikeMiller do you know a good source on the geometry of the hyperbolic plane?
 
what do you want to know
 
7:16 PM
I need some basics like hyperbolic triangles, hyperbolic volume etc.
 
@MatheinBoulomenos @loch in what sense is $\operatorname{Spec}(A/fA)$ the scheme associated to $V(f) \subseteq \operatorname{Spec}(A)$?
or is it actually not
 
@LeakyNun for a closed subset of a scheme, there are multiple scheme structures
note that $V(f^2)=V(f)$, but the rings are different
 
then how do we translate "open of closed of P^n" into scheme?
 
Yeah I dunno there are so many books that should cover this it almost doesn't matter what you get
 
@TobiasKildetoft Oh I think I remember why I was thinking about this
Say $X$ is a nice subset of $\Bbb R^3$ (let's say compact locally contractible connected). Compactify and consider $S^3 \setminus X$.
This is a $K(G, 1)$ by the Papa sphere theorem. It has zero $\pi_2$ and zero $\pi_3$ and higher because it's an open 3-manifold
 
7:21 PM
21/7 demonstration again...
 
@LeakyNun A standard tactic from the CPC. Of course, many people in HK are too smart for that.
 
Let's see how chaotic Hong Kong will become
Hong Kong is not China yet
 
Hm, so that just means $H_1(S^3 \setminus X) \cong H^1(X)$ by Alexander duality.
So I can only say $H^1(X)$ has no torsion
No
$G/[G, G] = H_1(S^3 \setminus X) = H^1(X)$. It may happen that $G$ has no torsion yet $G/[G, G]$ has. Meh
And now I remember why I thought this was a useless line of approach. Oh well.
 
if X is a scheme, is $\Gamma(X,\mathcal O_X) = \operatorname{Hom}(\operatorname{Spec}(\Bbb Z),X)$?
 
@BalarkaSen I don't understand what you're trying to prove exactly
 
7:25 PM
@LeakyNun no
 
I mean, that $\pi_1(X)$ has no torsion, @MikeMiller. Like Tobias asked.
 
is $\Bbb P^n_\Bbb C(\Bbb C) = \Gamma(\Bbb P^n_\Bbb C, \mathcal O_{\Bbb P^n_\Bbb C})$? ok obviously not
 
Well, maybe something weaker, like $H^1(X)$ doesn't have torsion.
 
Oh sure. But do like the quote you said earlier, and just restrict your hypothesis until it works.
 
@MatheinBoulomenos but $\Gamma(\Bbb P^n_\Bbb C, \mathcal O_{\Bbb P^n_\Bbb C}) = \Bbb C = \operatorname{Hom}(\Bbb P^n_\Bbb C, \Bbb A^1)$ right
 
7:27 PM
If X is a simplicial subcomplex then it has a mapping cylinder neighborhood so you just need to prove it for open 3-manifolds in R^3.
 
@LeakyNun yeah
@LeakyNun $\Gamma(X,\mathcal{O}_X)=\mathrm{Hom}(X,\mathrm{Spec}(\Bbb Z[x]))$, though
 
yeah I was thinking about that
really?
 
@MikeMiller Ah alright then you do sphere theorem on that open manifold to prove it's a K(G, 1).
 
because it's obviously true for affine $X$
 
Then if pi_2 is nonzero there's an embedded sphere representing it and your open manifold splits as a connected sum of things which embed in R^3. Since the free product of torsion-free groups is torsion-free (is this true?) you reduce to the case pi_2 = 0. Then X is a K(G,1) by the usual argument and is equivalent to a 2-complex so G can't have torsion.
 
7:29 PM
@LeakyNun use the adjunction of spec and global sections
 
S^3 \ S^0 is not a a K(G,1) so you need to be careful
 
did I just rediscover it lol
 
Yeah you should chuck out connected subsets of S^3
compact/connected
 
I'm not taking complements
 
I know, I'm just saying which things have good complements
Unrelated
Still reading your argument
 
7:31 PM
global section is left adjoint to Spec... i.e. $\operatorname{Hom}(\Gamma (X,\mathcal O_X), A) = \operatorname{Hom}(X, \operatorname{Spec}(A))$? this feels wrong because something is contravariant @MatheinBoulomenos
 
I implicitly use the Alexander theorem that a 2-sphere bounds a ball on both sides
 
and for the thing we said to be true, the left hom should be reversed
 
@LeakyNun yes, the left hom goes the other direction
 
so $\operatorname{Hom}(A,\Gamma(X,\mathcal O_X)) = \operatorname{Hom}(X,\operatorname{Spec}(A))$?
 
7:32 PM
what is the convention?
 
for locally ringed spaces, even
 
we're taking CRing^op as the category?
 
@MikeMiller OK so you reduce to the case pi_2 = 0 on each of those pieces. Got it. Nice.
 
there is no convention, I think you're confused about the definition of a morphism of schemes
global sections is contravariant for locally ringed spaces
 
so... what on earth is an adjunction
@BalarkaSen what is $\pi_1(\Bbb Q \times \Bbb Q \cup (\Bbb R \setminus \Bbb Q) \times (\Bbb R \setminus \Bbb Q))$?
 
7:34 PM
Is the thing about free products true? Maybe by Grigorchuk?
@LeakyNun lol fuck you
 
@LeakyNun idfk man
 
lol why
 
I mean Kurosh, not Grigorchuk
 
Yes, something like that.
My brain is a mush
My brain has been Kuroshed
 
winnie the pooh: prove that it is path-connected
winnie the pooh in a tuxedo: find its fundamental group
 
7:36 PM
@Leaky $\mathrm{Spec}$ and global sections (of ringed spaces, not of sheaves on a fixed base!) are both contravariant. Adjunctions for contravariant functors are defined as you would expect
 
oh right I proved the adjunction in Lean lmfao
therefore it is true
 
@BalarkaSen Here is the question I asked back then math.stackexchange.com/questions/631177/…
 
@MatheinBoulomenos what can be said about the "forgetful functor" Sch=>Top?
 
@LeakyNun it commutes with coproducts
 
@MikeMiller In this case you don't need to use Kurosh, I guess. You break up the submanifold of R^3 into connected sum of submanifolds with pi_2 = 0 in R^3 and each of those are finite-dimensional models of K(G_i, 1). The big submanifold has pi_1 = G_1 * ... * G_n and K(G_1 * ... * G_n, 1) = bigvee K(G_i, 1) which is a finite-dimensional model, so G_1 * ... * G_n is torsion-free.
 
7:38 PM
@BalarkaSen yes Kurosh is sufficient: if A * B has torsion, it has a finite subgroup, which is the free product of a free group, a subgroup of A, and a subgroup of B. This can only happen if the free product has only one nontrivial factor
 
@MatheinBoulomenos that is not good
we care about products not coproducts right
what is the coproduct even
 
disjoint union
 
oh...
stupid construction
 
@BalarkaSen Smart. You still need to know that B(H * G) = BH vee BG, but I guess this follows from clever universal cover trickery
 
@MatheinBoulomenos is there left adjoint to any of the functor in the chain Sch => LRS => RS => AS => S => Top?
locally ringed spaces, ringed spaces, abelian-grouped spaces (?), sheaves, (I forgot presheaves)
 
7:41 PM
@MikeMiller No, that's a good point. B(H * G) = BH v BG is basically the same circle of ideas as Kurosh.
 
oh Shf => PShf is sheafification
 
@LeakyNun that chain doesn't even make sense, you need "sheaved" spaces, not sheaves
 
AS => S is free abelian group? does that work?
oh right, sheaved spaces
 
Lol what the fuck is Leaky even doing
"sheaved spaces"
JESUS
 
The forgetful functor Sch=>Top has really bad properties. you can have a product of two one-point schemes that is uncountable
 
7:42 PM
what
 
in fact the product of two copies of $\mathrm{Spec}(\overline{\Bbb Q})$ is in bijection to $\mathrm{Gal}(\overline{\Bbb Q}/\Bbb Q)$
 
The forgetful functor from schemes to topological spaces essentially forgets everything we care about (so it is indeed very forgetful)
 
yeah
it's better to work with $S$-valued points than with the forgetful functor
 
basechange whoosh fwoosh
 
@LeakyNun there's no canonical way to do that for the reason mathei mentioned that you can have multiple scheme structures on a closed set - so usually you should be clear what scheme structure you're taking (most of the time its the reduced one)
 
7:44 PM
@BalarkaSen It feels strictly easier to me
 
his name is mathemi now lmao
 
taking open subschemes is fine though
 
@MikeMiller Yeah, you can just draw BH v BG as a graph of spaces. I'm sulking because I should have seen Kurosh lol
 
@MatheinBoulomenos that makes sense... somehow
 
I know how to write down the universal cover, and by cutting it into pieces I understand and how they fit together I get either an inductive MV description or an MV specseq depending on how you want to think about it
 
7:45 PM
field theoretically
but not topologically
 
the topologies agree though
 
what is S-valued point?
 
morphism S -> X
 
oh
but that doesn't have a topology?
 
yeah, but who cares
 
7:47 PM
so you're basically saying, it's better to work with the functor to set
 
It has... the Grothendieck topology!!! how exciting
 
@LeakyNun A scheme is a functor to the category of sets
 
lol ^
 
a matrix is a linear map
 
not the functor to set. It's better to work with the Yoneda embedding, basically
 
7:48 PM
sry im really sleepy and i shouldnt dunk on algebraic geometers
 
@BalarkaSen Yes you should
 
@MatheinBoulomenos that is the functor to set
how does X(S)=Hom(S,X) carry the Grothendieck topology?
it isn't even a category...?
 
p sure he's messing w/ you
 
:c
 
8:29 PM
-_-
 
8:42 PM
Question ------> Does a homeomorphism exist between the sphere and the twice pointed sphere?
 
What do you think?
 
what on earth is the twice pointed sphere?
 
yes because they have genus 1
but
 
Oh I didn't even noticed, my brain replaced it with punctured automatically
 
I don't know how to deal with the points that are "pinched"
 
8:52 PM
What does twice pointed mean then?
 
t's a sphere with two cusps
on the opposite poles
picture forthcoming
that second one looks like a bloody raindrop
 
9:40 PM
can someone make a better or more elegant fit of this?
 
I'm confused by this equality. If $p$ is a prime and $x \in \mathbb{R}^{+}$

$\displaystyle \prod_{p \le x}\left(1-\frac{1}{p}\right)^{-1} = \sum_{n: p|n \Rightarrow p \le x} \frac{1}{n}$
I'm guessing they first wrote it as $\displaystyle \prod_{p \le x} \sum_{n \ge 0}\frac{1}{p^n}$
But how does that weird index follow?
I'm only really interested in showing that the LHS > log(x).
There's a technique on the wiki page on zeta function but the method requires Re(s) > 1. This one diverges. And when we get to the harmonic series we easily get that this is > log(x).
 
 
2 hours later…
11:23 PM
@BalarkaSen meme paper arxiv.org/pdf/1907.05004.pdf
 

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