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6:16 AM
@LeakyNun You can use a spectral sequence argument
In fact, I can tell you the idea. Let $X$ be a space and $C^*$ be the presheaf of singular cochains on $X$, which assigns to each open set $U$ the singular cochain group $C^*(U)$.
The inclusion $U \to V$ gives a natural restriction $C^*(V) \to C^*(U)$
This is a "homotopy sheaf" in the sense that there is a homotopy-equalizer diagram $C^*(U \cup V) \to C^*(U) \oplus C^*(V) \Rightarrow C^*(U \cap V)$, in the sense that the pullback of $C^*(U) \to C^*(U \cap V) \leftarrow C^*(V)$ is chain-homotopy equivalent to $C^*(U \cup V)$. This is exactly what Mayer-Vietoris says.
In particular if $\{U_i\}$ is a cover of $X$, there is a big Mayer-Vietoris diagram $$0 \leftarrow C^*(X) \leftarrow \bigoplus C^*(U_i) \leftarrow \bigoplus C^*(U_i \cap U_i) \leftarrow \bigoplus C^*(U_i \cap U_j \cap U_k) \leftarrow \cdots$$
This is actually an exact sequence. The same thing holds in eg de Rham cohomology but arguing this is much simpler there, because the sheaf of differential forms is a fine sheaf. This is sort of like a replacement for that in this weird sheaf of singular cochains!
Now you can append that cochain complex below with the complex of locally constant functions on $X$, and you have a double complex interpolating between the singular complex of $X$ and the Cech complex of $X$, such that the rows are exact.
This is standard spectral sequence argument now. Filter the double complex from top-to-bottom, so that the $E^1$ page consists of the Cech complex at the last row and nothing else, assuming $\{U_i\}$ is a good cover of $X$
$E^2$ page is just the Cech cohomology groups at the last row, and nothing else, and the whole spectral sequence degenerates at $E^2$. Total cohomology is Cech cohomology
Similarly run the filtered specseq for the filtration from right-to-left, and conclude total cohomology is singular cohomology
@RyanUnger Your argument seems good my man
 
6:38 AM
Mornin @Balarka
woke to a wall of text
 
Morning @ÍgjøgnumMeg
lol sorry
 
dw it's beyond me anyway
 
ugh my arrows went all backwards in my big diagram
 
@Mathei @ÍgjøgnumMeg and the other algebraists, do you know a good source to learn about totally ordered groups and the semigroups which can be realized as the positive cone of a totally ordered group?
 
7:20 AM
@AlessandroCodenotti why does that sound like model theory
 
Actually I'm thinking about group and semigroup C*-algebras at the moment, but I can see why it sounds like model theory :P
 
@AlessandroCodenotti have you heard about the "RH independent => RH true" thing?
 
No, but it makes sense that it should be provably false if it is false
 
@AlessandroCodenotti what does "true" mean? like which model are we thinking about?
 
The one that proves the independency? Not sure
 
7:27 AM
then how does it makes sense that false implies provably false
 
Nah that can't be
 
in Logic, Sep 29 '17 at 17:45, by user21820
I 'know' that, in the sense that someone proved (in ZFC) that RH is equivalent to some Π1-sentence.
 
Any model proves the independency if it is actually independent right?
 
so I guess "false" means "false in the standard model of PA"
models don't prove anything
 
Seems reasonable
Sure what I meant is that if ZFC proves "RH is independent" then "RH is independent" holds in every model
So my previous comment about the model in which it holds made no sense
 
7:32 AM
@AlessandroCodenotti that doesn't make any sense
well it can make sense but in a convoluted way
as you need to formalize provability inside the model of ZFC
and you probably doesn't care about models with "nonstandard" natural numbers
 
You can formalize model theory inside the model even
This is routinely done to define $L$ and $HOD$
You just talk about Gödel's code rather than formulas
(yes you might run into funny issues with nonstandard codes)
 
then what's the point
 
The model can prove stuff about what it believes to be the codes of formulas
It has a code for every true formula, but it might have a few extra ones
But again, how do you know that the universe of the metatheory isn't wonky either when it comes to the naturals?
 
I don't
but our natural numbers seem pretty natural
let's say we proved $1 \in 2$, which is certainly true
well we're actually proving it in ZFC, so $ZFC \vdash 1 \in 2$
wait but our metatheory is actually ZFC, so $ZFC \vdash ZFC \vdash 1 \in 2$
where does this chain end?
$\cdots \vdash ZFC \vdash ZFC \vdash ZFC \vdash 1 \in 2$
 
Doesn't it just stop at the metatheory level?
 
7:46 AM
metametatheory...
"what are the rules of mathematics?" "the theory"
"what are the rules of the theory?" "the metatheory"
what are the rules of the metatheory? the metametatheory?
 
This stuff is too close to logic for me, why can't you do some honest set theory lol
 
no
 
You should ask in the logic room, user82... surely knows more than me about those issues
 
8:36 AM
@LeakyNun "at God" - Aristotle
 
God doesn't exist
 
Aristotle's argument literally was God exists because infinite regress isn't possible lol
 
oh
well the bible says that through god all things are possible, so from the contrapositive we infer that "God doesnt exist because infinite regress isnt possible"
there you go
a nonsense response to a nonsense argument
 
yup lol
well Aristotle was fairly reasonable about his cosmological argument, it's more applicable to people like Aquinas who tried to incorporate it into Christian theology
 
how on earth can a triangle be exact at all three corners
any argument that points to God isn't reasonable
 
8:42 AM
think of wrapping up an exact sequence on a triangle
if you have an exact sequence $C_\bullet$ just take $A_i = \bigoplus_{n \equiv i \pmod{3}} C_i$ for $i = 0, 1, 2$ and then you have an exact triangle $A_0 \to A_1 \to A_2$
 
oh
cool
 
ya its just a convenient language after the confusion is cleared
@LeakyNun arguable but yeah
 
is it always how you make an exact triangle
 
not necessarily, but in making spectral sequences out of filtered complexes, this is how you do it, yeah
the exact triangle "unwraps" if the abelian objects at the vertices are graded abelian objects and the maps do something to do the grades
you can make that rigorous maybe
 
depression rates rose from 1.3% to 9.1% during the period of protests against the extradition bill
this isn't good
 
8:48 AM
is there any indication from the government that they may back down
 
@BalarkaSen arguments are wordplays, if you wanna prove god you gotta do it scientifically
 
or is it similarly tense
 
@BalarkaSen no
 
ugh
@LeakyNun well the idea is people like to axiomatize god in a blackbox that cannot be accessed by scientific methodology
 
the next protest is on 21st
and my prediction is that it will escalate into police-protester conflict (just like it did in the last N protests), and then the govt will condemn the "violence" (just like they did in the last N protests) without doing anything (just like the last N protests)
 
8:50 AM
which is, like, a consistent thing to do. but they you can neither prove or disprove the existence of god in any """"useful"""" way
 
right, so it's unfalsifiable
 
@LeakyNun Ugh, that sucks. Is the police explicitly told to be violently suppressive?
What's the root of the police-protester conflicts?
 
police abusing their power
power corrupts
the police hate the protesters
the protesters hate the police
the police who abuse their powers never seem to get punished
 
Do you buy that, though? Why would the police hate the protesters to the extreme that they'd be as violent that the government has to condemn them afterwards?
 
the police think that we're causing chaos
 
8:53 AM
Isn't it more likely it's a string-pulling by the government itself
 
and breaking the law
well the government is pulling the strings by not doing anything
but the police themselves, I believe, are violent people
or at least violent towards the protesters
 
I see. That's worse, because then this internal conflict will pull apart the protests
whereas the administrators can just sit and watch
 
we've been demanding an independent investigation of the police since june 16
police (at least those policing the protests) behave sometimes like gang members
this is ridiculous
 
@LeakyNun You might be amused by Bishop Berkeley's argument for God, which is that events occur only because they are observed by someone - as long as no one is observing an event it isn't happening (if it's happening, you know it did, or at least you know someone who knows it did, so on). So occurrence is a causation of collective conscious, but then if this has to be consistent in a way independent of the intersubjective space of consciousness and objective in any way, there must be God
which is the entity which perceives all that happens :P
(because otherwise they wouldn't be happening)
 
ugh
 
9:03 AM
its kinda cute u gotta give it to em
 
Einstein: i'm about to end this man's whole career
 
lol
he actually made quite a big of fun relativism arguments but then went off in a tangent to this god business
I recommend reading Borges, "A New Refutation of Time"
its very short and very enjoyable
 
9:20 AM
@Alessandro alas, I have no idea what those are
lol
 
Like a group, but ordered :P
 
9:38 AM
@AlessandroCodenotti I know almost nothing about that. I only know the fun fact that an abelian group can be totally ordered iff it is torsion-free
 
Oh that's cool
One direction is trivial, the other is magic to me
 
If $A$ is torsion-free, then $A$ embeds into $\Bbb Q \otimes_{\Bbb Z} A$. That's just a $\Bbb Q$-vector space, you can choose a basis and order lexicographically
 
Ah well it's clear for f.g. groups by the structure theorem and extending to non f.g. seems reasonable
@MatheinBoulomenos That's a neat construction
 
Orderable groups occur quite often in geometry and topology.
 
Do you have explicit examples of situations in which they naturally occur?
 
9:54 AM
I think there's some conjectures around the fact that if $M$ is a $3$-manifold which admits a taut foliation then $\pi_1 M$ is left-orderable.
If $G$ is left-orderable that's the same thing as saying $G$ acts on $\Bbb R$ by orientation-preserving homeomorphisms
If $M$ admits a taut foliation then you pass to the universal cover. You'd expect the closed transversal to become a copy of $\Bbb R$ upstairs (it may not be; it can be many copies of $\Bbb R$), then $\pi_1 M$ will act on $\Bbb R$ by deck transformations
This will give the purported left-order on $\pi_1 M$
But I think this is wide open
Ah OK here is the conjecture
Let $Y$ be a rational homology 3-sphere. The following are equivalent: (1) $\widehat{HF}(Y)$ is non-minimal, which apparently means $|\widehat{HF}(Y)| > |H_1(F; \Bbb Z)|$, where $\widehat{HF}(Y)$ is some flavour of Heegaard-Floer homology which always has cardinality $\geq |H_1(F; \Bbb Z)|$
(2) $\pi_1 Y$ is left-orderable
(3) $Y$ admits a co-orientable taut foliation (a locally nice decomposition of $Y$ into $2$-manifolds, the leaves of the foliation, such that there's a closed loop transverse to all those leaves).
@MikeMiller should explain (1)
@Alessandro See Theorem 6.8 in Ghys for a proof that $G$ is countable left-orderable iff $G$ injects into $\text{Homeo}^+(\Bbb R)$.
 
10:11 AM
@BalarkaSen Sounds interesting, thanks
 
 
3 hours later…
12:43 PM
@BalarkaSen no
There is no explicit conjectural construction of the ordering on pi_1
That's how it works for certain hyperbolic foliations and the idea is called the "circle at infinity" or something and of course due to Thurston
 
Oof
Hyperbolic foliation = there's a hyperbolic metric which makes the leaves totally geodesic?
(therefore also taut)
 
@BalarkaSen Hell if I know.
arxiv.org/abs/1705.08817 is a nice paper too that isn't about any of those legs of the conjecture but gets geometric results
 
I didn't mean totally geodesic, I meant minimal surfaces
@MikeMiller Wow, extremely short paper
 
@BalarkaSen yeah I was way overcomplicating that argument at first
my argument does not say if $\dim_B(A)=k$ if $A$ is merely $k$-rectifiable and $H^k(A)<\infty$
maybe add compact for good measure
 
1:02 PM
@BalarkaSen That's how you know it's good
 
 
2 hours later…
2:35 PM
What is up
 
I'm up
 
Just moved 90 miles
 
For good?
 
Yeah at least for the next 2 yrs
 
You like your new area?
 
2:40 PM
I do
But am not so keen on my living situation lol
 
How come? Because you have share/flatmates, I'm assuming lol.
If that's the case, I can tell you what I've done to mine to get them to move out.
 
2:55 PM
First start moving everything slightly than they had left. Just a few cms. Tilting things at a slight angle etc. Unpairing their socks. Hiding things then "finding" it for them at an obvious place. Nothing too obvious. Install tiny cams preemptively because they inevitably will, then mess with those too.
If they become superstitious, you hit gold. Tell them you spend your portion of the rent to exorcise the apartment every month. Pay someone to recite some nonsense at the start of every month and spray holy water at the place and then cut it all out -- until at the very end of the month.
 
@RyanUnger lol
 
@J.Doe yeah I share the place with other flatmates. I'm probably going to look for an upgrade as soon as i can lol
 
3:12 PM
Hello chat
 
Hi @Albas
 
Hey @BalarkaSen. How are things going at TIFR?
 
's okay
 
Cool. Did you face the infamous Mumbai rain?
 
Somewhat in the first week and a half, yeah
 
3:16 PM
I had to cancel two flights because of that shit.
 
rip
 
What math have you been doing?
 
a bit of this, a bit of that. primarily thinking about singular spaces
 
Oh, don't know what they are. I am predominantly doing the second chapter of Atiyah McDonald to prep up for my upcoming algebraic geometry course.
 
spaces with singularities :P think of manifold away from a subspace which is also a manifold
A&M is good
 
3:20 PM
Hmm okay.
Yeah, I kinda disliked algebra before reading A&M but the exercises kind of make me interested in reading more
I learned about Spec for the first time. Its cool
$\operatorname{Spec}(\Bbb Z)$ is one of my favorite things right now.
 
I like stuff. Especially things.
 
Spec is a good construction. '
@Albas Why so?
 
will i learn about spec in abstract?
 
3:47 PM
Can $S^2$ be expressed as a 3-dim. projection of a 4-dim. shape?
 
how can i define the notion of tangent objects without saying infinitely close .
 
@Ultradark what
 
@RyanUnger Is that not a well formulated question? I wasn't sure...lol
 
@Ultradark Abstract?
 
@J.Doe abstract algebra
 
3:51 PM
@Ultradark I took two abstract algebra courses and have no idea what Spec(Z) is.
 
@Ultradark well I have no idea what the question means
 
I think it's only if you get to commutative algebra.
 
@J.Doe it's just the set of all primes of a ring
 
@ÍgjøgnumMeg I didn't know that! I googled earlier and got the wiki page on schemes.
 
4:08 PM
Okay so the definition is pretty simple but it gets complicated pretty quickly lol
 
@BalarkaSen @ÉricoMeloSilva very interesting read web.math.princeton.edu/~seri/homepage/papers/telaviv.pdf
 
ive read this it’s good
 
@RyanUnger This looks very nice
 
is it possible to re-construct a curved surface, based on quadrilaterals that tile a portion of the plane
 
@BalarkaSen I like the topology, I like how it is basically the cofinite topology. The same topology is given in the case of the affine space I think. I guess thats the relation between algebraic geometry and abstract algebra. Though this is Spec in general but Spec(Z) was the first example where how the topology was cofinite was easy to see.
 
4:22 PM
the main thing being that the quadrilaterals are not regular
they are "stretched"
I'm just thinking of the picture of triangles on different types of curved surfaces
 
@RyanUnger This looks so cool. I have a PDE course next semester and it goes into the Einstein field equations and Yang Mills (the physics aspect). I hope I can learn this.
 
@Albas Spec(Z) \ 0 is cofinite, rather.
But sure. I just don't know why it's an inherently interesting object, I guess :P
The Zariski topology is everywhere, on the other hand
 
Balarka, Spec(Z) corresponds to knots
 
boom
 
Are you saying knots re not interesting
 
4:26 PM
lol
 
this is the picture I'm thinking of
 
The problem with 20 year old articles like this (which are outside of my area) is I have no idea if the problems are still open
 
@RyanUnger I was in a discussion today and an algebraic geometer had to clarify at some point in the conversation that by finite group he meant a finite group scheme
 
That's pretty normal.
 
4:32 PM
@Balarka arxiv.org/abs/1511.05088 a classmate suggested me a cool thing on ordered groups
 
I couldn't decide if I should laugh or cry
@Alessandro Oh damn, Rolfsen
 
I don't know him
 
Doesn't Milne say that the word "scheme" is omitted everywhere but is understood in every construction?
 
Author of the classic knot theory book
@RyanUnger no clue man lol
 
Oh ok I think I actually know which book you mean
 
4:36 PM
I should read this paper as well actually
I'll save it away in my ever increasing reading list
 
If one projects evenly spaced geodesics of the $S^2$ onto the unit disk lol
 
I'm going to read at least the first sections, we can do that together
 
I'll have to start anything new after a week, when I go back to college
We can read bits and pieces togather then
 
and then projects another copy of evenly spaced geodesics of the $S^2$ onto the unit disk (but this one is rotated 90 degrees) such that the geodesics all create a mesh
then
My question is: Does the mesh support a hyperbolic space
 
@BalarkaSen sounds good to me, I have plenty of stuff to read until then!
 
4:59 PM
can anyone point me to some good resources that help explain godel's theorems in an easy to understand way
i don't really understand all these incompleteness/completeness theorems or what they are or how we prove them and what system these proofs assume
 
You might start by looking up youtube videos on them. People tend to try to phrase problems/proofs in easy to understand ways over there.
(There is also the issue that the more easily understood a statement is, the more likely it is leaving out minutia about the subject it concerns. So keep that in mind.)
 
@AlessandroCodenotti, @BalarkaSen: The book is published already by AMS.
 
5:17 PM
@RyanUnger idk if ur on fb but incoming class ppl are making a group chat
 
5:49 PM
why cannot a square number end with more than three 4s?
 
Hi @Ted
 
Hi, a @Balarka!
@Mathphile: I've never heard of that. Do you know a square that ends with 3 of 'em?
 
@TedShifrin yes
213444
289444
925444
1077444
2137444
2365444
3849444
4153444
6061444
6441444
8773444
9229444
 
Wow, I guess I don't know squares that high.
So I guess asking about ending in 4 4's is asking why you can't have $n\equiv 4444\pmod 10,000$. We should be able to get something out of factoring.
 
$X^2 \equiv 4444 \bmod{10000}$ and then use legendre symbols and QR
I guess
 
5:55 PM
This is asking which 4*(10^k-1)/9 number appear as quadratic residues
This seems like a job for number theorists... like @ÍgjøgnumMeg
 
Certainly not a job for me.
 
clumsy number quack*
 
I'm enjoying unemployment more and more. :)
 
@ÍgjøgnumMeg any ideas?
 
Note that $k \equiv 4444 \pmod{10000}$ implies $k\equiv 12 \pmod{16}$, but squares are congruent to $0,1,4,9$ mod $16$
 
6:04 PM
ah
 
or that
 
Yes, that was the sort of argument I was thinking of, but was too lazy to do the factoring.
And hi, @Mathein!
 
Hi @Ted
 
Hey @MatheinBoulomenos
 
Hi @ÍgjøgnumMeg
 
6:06 PM
can we extend this to $n$ number of 4's, where $n\gt 3$ ?
 
We used to have a huge French contingent in here. It seems that expat German grad students (plus one native) will be the largest contingent :P
 
hahaha
 
@Mathphile: What do you mean?
If $n>4$, then the number will have to end in 4 4's.
 
@TedShifrin sorry i do not know what i was thinking lol
 
Well, you confuzled me.
 
6:10 PM
can we prove or disprove that we cannot prove RH using Zermelo-Frenkel axioms?
 
@TedShifrin working do be suck
 
LOL, @Eric, you're awfully young to be so cynical.
 
Hi everyone
 
Hi demonic @Alessandro
 
Hi @Alessandro
 
6:18 PM
I finally finished my exams today
 
Hiya @Alessandro, congrats
 
Ausgezeichnet! I presume you did swimmingly, as always.
 
Congrats, @Alessandro!
 
Yep, they went well
I'm at the airport now, waiting for my flight back to Italy
 
Well, try not to be overly demonic!
 
6:23 PM
@TedShifrin im a latin american prole tbf
@AlessandroCodenotti grats dude
 
congrats @Alessandro
I met with a grad student today to talk about my thesis (he's not my advisor or anything, just interested)
 
That's always exciting, @Mathein.
Did it go well?
 
yes, it went well
 
Every time I read math at the airport or on a flight I think of this Italian economist
 
LOL
I remember that one
 
6:33 PM
@AlessandroCodenotti lmfao
 
Luckily I'm not an algebraic geometer and I don't blow up planes
 
A fine observation
 
@ÉricoMeloSilva I do have an FB but I don't really use it
is anything interesting happening
 
6:50 PM
@Alessandro: I have only blown up more interesting varieties.
 
@RyanUnger no just gathering the ppl
do u want me to add u
 
sure...maybe I should update my fb
 
new gossip zone for ryan
 
probably says I'm a sophomore in hs
@BalarkaSen I don't gossip
 
yeah right
 
6:56 PM
@RyanUnger add me cuz i can’t find u
no one else has my name so it should be ez to find me
 
done
 
@ÉricoMeloSilva Actually, my real name is Erico Melo Silva
 
Eric do you have 4 names
 
I also added you
 
@Balarka sick welcome to the club
 
6:58 PM
quick sanity check
 
@RyanUnger i have a first name a middle name and two last names yes
 
I am FB friends with very few chatters ...
 
The standard n-simplex is the convex hull of the points (1,0,...), (0,1,..,) etc. in RR^(n+1)
 
i only do gchat
 
y isn’t Ted friends w me
 
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