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19:00
OOOOohhhhhhhhh
I see it!!!!!!
Hey @DonLarynx :D
@Studentmath @Balarka $\mathbb{R}$ is not connected but $\mathbb{R^2}$ is !
Hi @evinda!
@Venus ^^^ see above (and please don't post it on main)
But of course it becomes nontrivial for higher dimensions because if you remove a point, they are both still connected.
@DonLarynx $\Bbb R$ is connected. I guess you mean $\Bbb R \setminus \{x\}$ for some point $x$.
19:02
@robjohn I posted above its closed form. I consider it my POTE(evening).
Yes @Mike
Ah, yeah, you do. Alright.
Oh yes sorry about that
no worries :)
The world is doomed.
19:07
How so, @Pedro? :/
hello jasper
@KhallilBenyattou In a couple of years, the sun will burn out and become a nova or something.
How many years are we talking, @DanielFischer?
@KhallilBenyattou Two or three, give or take $10^{10}$ or so.
How is the unit circle with a point taken away compact?
19:17
We've got plenty of time, @DanielFischer. ^_^
@KhallilBenyattou You'll keep saying that until it's too late.
Touché. That's what I always say before exams.
(Including the one I'll be doing in 2 weeks.) =P
@DanielFischer Could you maybe explain to me this proof?
http://math.stackexchange.com/questions/1076864/why-is-this-function-an-embedding/1076873?noredirect=1#comment2190796_1076873
@KhallilBenyattou This chat is a black hole.
In the sense that once you enter, you can never leave, @Pedro?
19:33
@KhallilBenyattou For one thing.
@MikeMiller I was not even here.
@MikeMiller Oh, I think maybe I get it now.
@user153330 Not funny, lol.
@user153330 Yes, do you know you can click on the left arrow to jump backwards and see what is replied to?
no no, but i know now @JasperLoy
@user153330 Also, you can click on the right arrow to reply to a line, so that we know what is being replied to.
19:40
@JasperLoy thanks for that i did not know
@user153330 I know many other chat tricks, but these two are sufficient for now!
Does anybody know what sets, functions and relations would come under?
(Much like how integration and differentiation come under calculus.)
@user153330 Although I know many many math books, I cannot help you with that question there.
@KhallilBenyattou It all comes under set theory.
@MikeMiller
@KhallilBenyattou naive set theory i guess
19:44
@user153330 Why did you remove your message?
Naïve, @user153330?
@KhallilBenyattou In the sense that you don't need to do axiomatic set theory.
@JasperLoy i dont wont to look like i'm promoting my question :/
@user153330 It's OK to promote your question. This is a math chat for math questions.
@KhallilBenyattou naive because it is very intuitive and can be developed using only intuition for what sets are, and because it suffices for the everyday usage
@JasperLoy ok thanks for that
19:51
Could someone explain me this proof:math.stackexchange.com/questions/1076715/… ?
@evinda It should be 'explain to me'.
@JasperLoy really ?
@LeGrandDODOM Yes.
Is there anyone here who can help me explain the ABC-conjecture? I have a hard time understanding and using this conjecture
19:56
@RubenMeijs The conjecture is stated in terms of three positive integers, a, b and c (hence the name), which have no common factor and satisfy a + b = c. If d denotes the product of the distinct prime factors of abc, the conjecture essentially states that d is usually not much smaller than c. In other words: if a and b are composed from large powers of primes, then c is usually not divisible by large powers of primes.
@user153330 thanks for helping me, but I already understand why people think this conjecture is true and the reasoning behind it. But for my number theory course I need to use this conjecture to show that an equation, such as $x^p-y^q=A$ has finite solutions (with x,y,p,q > 2)
@user153330 I have asked a question about this theorem a few weeks back, but I still don't understand how I should use it.
@JasperLoy I am sorry...
@JasperLoy i was serious
20:26
@evinda what proof? I only see a comment.
@robjohn A comment?
:/
Is your name Zanskar, @N3buchadnezzar?
@evinda You link to a comment...
What do you mean by 'divisible by'? (For instance, is $\frac15$ divisible by $2$ in $\mathbb{Z}_2$? Is $\frac25$?) It's probably easier to think of elements $m\in \mathbb{Z}_p$ as infinitely long vectors $\langle m_i\rangle$ where $0\leq m_i\lt p$ for all $i$ - that is, as their '$p$-adic expansion'. — Steven Stadnicki 3 hours ago
@robjohn I am sorry... Did I send now the proof? :)
How did you quote like that, @robjohn?
20:42
Did I do this right?

Define $x = \frac{4^k - 1}{3}$. So far I have:

$$k_1 \to 1 \Longrightarrow x_1 \to 1
\\
k_2 \to 2 \Longrightarrow x_2 \to 5
\\
k_3 \to 3 \Longrightarrow x_3 \to 21
\\
k_4 \to 4 \Longrightarrow x_4 \to 85$$

But then it's evident that

$$4^{k_n} = x_{n+1} - x_n$$

Also equivalent to

$$4^{k_n} = \frac{4^{k_{n+1}}}{3} - \frac{4^{k_n}}{3}$$

So we multiply by 3 on each side to get

$$3*4^{k_n} = 4^{k_{n+1}} - 4^{k_n}$$

Take $\mod 3$ on this equation to get

$0 = 0$
Thus I have proved $\frac{4^{k+1} - 4^k}{3} = 4^k$ for all $k \in \Bbb{N}$
I mean, my issue is that I could have taken $k_{n+1}$ to be $k_{n+1000}$ instead...
and the result would still hold true.
I don't see why you've done that, @DonLarynx.
$4^{k+1} - 4^{k} = 4^{k} (4-1) = 3(4^{k})$
Isn't that obvious? Is there a reason why you're working in mod $3$?
@DonLarynx Where did you deduce that?
Recall that $(1+x+x^2+\cdots+x^{k-1})(x-1)=x^k-1$. Let $x=4$. Then $4^k-1=3\times (1+4+\cdots+4^{k-1})$.
@TedShifrin Hello Ted.
Do you like "Fear and Loathing in Las Vegas"?
rehi @Pedro ... dunno ... never saw it
@TedShifrin Oh, noes.
Add it to your "TO DO" list.
didn't know you were into druggie culture, @Pedro :P
20:55
@TedShifrin There's a long run from watching a movie to doing what the movie shows, eh!
if you say so :D
I've seen delToro in several movies ...
@TedShifrin He's good. I like Johnny better, though.
@evinda yes. that is better.
20:57
BTW, I've put a bounty here.
@KhallilBenyattou I just pasted the link
@robjohn Did you take a look at it?
@Pedro: It should just say every analytic vector bundle on $D$ is trivial.
Ok Thanks! @Khallil @Pedro
@TedShifrin Aha?
Tries to remember the definition of vector bundle from a past life.
21:02
@evinda I am not terribly familiar with $p$-adic numbers
@TedShifrin Are you maybe familiar with $p$-adic numbers?
Definitely not, @evinda.
@TedShifrin Ok :)
@r9m did you see the last question I posted?
r9m
r9m
@Chris'ssis oo !! okay :)
I just woke up ..
21:09
@r9m btw, I also posted here my solution to the Basel problem.
r9m
r9m
@Chris'ssis Where ? :D link please
@Chris'ssis got it .. :)
@Chris'ssis okay ! Sweet solution !!! :D
@r9m :-) Also check the last integral, it's very cute.
@robjohn I think I have a very cute solution to the double integral above, it seems unreal. I'm trying to come up with a second solution.
Hey @Ted @Pedro \o
@Chris's may I ask why you delete these after you load to the chat?
@Studentmath Some of the stuff I show here I might add to my book.
21:18
@Studentmath I'd like to add that proof to my book.
21:28
@PedroTamaroff It's just a fiber bundle with some voodoo conditions imposed.
voodoo?
@KhallilBenyattou No, why?
may someone check out my question? math.stackexchange.com/questions/1077002/…
@Pedro just stick a line at each point in the base space, so that it locally looks like $U$ cross fiber $F$.
boo
21:36
@Victor what is it you are looking for?
@skullpatrol - A New York times style website that is specifically for math without any math technical detail. For example, math equation.
@Victor Your question looks good. I'd totally ask her out.
@PedroTamaroff Ah, so you now believe in rebirth?
@BalarkaSen What's "just a fibre bundle?"
@Victor That's a tough one.
21:39
@PedroTamaroff if you're prepared to listen, i am prepared to give a rigorous definition, with examples ;)
I cannot listen.
I can read.
I'm going to show you a more advanced version.
@Pedro that'd do
consider a space E and a space B, with a map p : E \to B
@Victor The newest changes in math are not accessible to the average reader.
$$\int _0^{\frac{\pi }{2}}\int _0^{\frac{\pi }{2}}\int _0^{\frac{\pi }{2}}\frac{\sin (y) (\sin (z)-\sin (x)) \log (\sin (y)+1)+\sin (x) \log (\sin (x)+1) (\sin (y)-\sin (z))+\sin (z) (\sin (x)-\sin (y)) \log (\sin (z)+1)}{(\sin (x)-\sin (y)) (\sin (x)-\sin (z)) (\sin (y)-\sin (z))}dzdydx$$
21:41
take a point x of E
@Chris'ssis: No thanks
@BalarkaSen Taken.
@skullpatrol - I want the pure math website to be written in English word for my recreational purpose similar to that of science news
E is called an F-bundle over B if there is an nbhd U around p(x) in B such that there is a homeo f : p^{-1}(U) \to U \times F such that f \circ p = h, where g is the projection U \times F \to U and
@robjohn @r9m just look at the beautiful closed form of this one $$\frac{3}{4} \pi ^2 \log (2)-\frac{21 \zeta (3)}{8}$$
21:44
@Victor The problem is that there's not much you can say without being at least a little technical.
@Victor perhaps some "recreational math" web sites?
@MikeMiller - The technical detail should be the math equations
@Victor Equations are usually not the technicalities.
@Pedro E = B \times F is trivially a bundle, for example.
@BalarkaSen What about the Jothilengheim space?
21:47
no idea what that is
:P
$$\Large{\text{WAIT!!!}}$$
Here is the closed form of the integral in 4 dimensions
$$\frac{1}{128} \left(\pi \left(64 \left(24+\pi ^2\right) G-32 \pi ^2-\psi ^{(3)}\left(\frac{1}{4}\right)+\psi ^{(3)}\left(\frac{3}{4}\right)\right)+96 (31 \zeta (5)-28 \zeta (3))\right)$$
@r9m ^^^
@Chris'ssis "Closed".
@BalarkaSen no, that is the definition of the beauty.
21:49
@PedroTamaroff - But i want to know about math current event.
@PedroTamaroff Maybe you give it a try and "open" it. :-)
i still haven't been able to figure out how S^3 is an S^1-bundle over S^2
looks rather false to me
@Victor Current events in math are very technical.
@Chris'ssis no it's ugly
ugly ugly ugly
@Chris'ssis I'd rather kill a baby panda.
10
21:50
LEL @Pedro
@PedroTamaroff lollllllll
@Chris'ssis Seriously, doing that would be extremely painful.
:'( poor baby panda
@PedroTamaroff What if I told you there is absolutely no pain with that?
21:51
@Chris'ssis Really?
@PedroTamaroff Let's not make fun of baby pandas.
@JasperLoy I'm not making fun of baby pandas.
@PedroTamaroff Yeap.
@Chris'ssis OK.
@BalarkaSen You could try to find an action of $S^1$ on $S^3$ such that the orbit space is homeomorphic to $S^2$.
21:52
oh?
ponders
he wait no way
A message about killing baby pandas gets 6 stars? This chat is sick.
if S^3 is an S^1-bundle over S^2, then homotopy long exact sequence tells us \pi_3(S^2) = Z
that's obviously false
@JasperLoy "Everyone's so stingy around here. Everyone's so stingy everywhere!"
@BalarkaSen Why?
@Jasper It's clear that Pedro has no desire to kill a baby panda.
21:54
@BalarkaSen Stop.
hi @Mike @Pedro @DanielF...
@DanielFischer no idea
Hi @Ted.
but it sure is false
@MikeMiller stop defending the panda killer
21:55
In the mathematical field of algebraic topology, the homotopy groups of spheres describe how spheres of various dimensions can wrap around each other. They are examples of topological invariants, which reflect, in algebraic terms, the structure of spheres viewed as topological spaces, forgetting about their precise geometry. Unlike homology groups, which are also topological invariants, the homotopy groups are surprisingly complex and difficult to compute. The n-dimensional unit sphere — called the n-sphere for brevity, and denoted as Sn — generalizes the familiar circle (S1) and the ordinary sphere...
jaws drop
I'm guessing Balarka isn't quite ready to be unignored yet ...
the hell
@TedShifrin Not at all.
Only a fool would unignore him.
he is on the campaign trail he is suppose to be kissing baby pandas
21:56
And you're no fool Ted.
everyone dislikes me :(
it's painful
i'm hurt
not me
Are you sure, @Pedro?
@BalarkaSen There, there
@BalarkaSen I do not dislike you.
21:57
hi @Studentmath
@DanielFischer i always used to think \pi_n(S^m) for m < n is trivial :O
@BalarkaSen You have to remember your first messages in this chat.
do not vote for the panda killer^
@BalarkaSen Tsk tsk tsk.
@Ted $O(2)$ sits inside $S^3$ (operator norm). Is it linked?
21:58
@PedroTamaroff which messages?
Hello!!! Can someone describe the computation model RAM (Random Access Machine) ?? What is this?? What does it do??
@BalarkaSen I don't dare look at them.
i dunno what you're talking about
You were a Bart when you first talked here.
I've never contemplated that, @Mike. Feels very un-Hopf linkish, so I'm going to say NO. But I have to ponder.
21:58
@BalarkaSen No, homotopy groups are weird.
if you're talking about super-ignorant and prat-like messages, yes, i do remember them
Sard's Theorem @DanielF
@PedroTamaroff i agree i was a prat then
A friend of mine had a proof attempt that doesn't work: show it's contained in a torus, as no nontrivial link sits inside a torus. Unfortunately his torus didn't live inside $S^3$.
but i have changed. i do know how little i know.
22:00
@BalarkaSen I think you know much more than me.
@Mike: There's an obvious way to fill in $SO(2)$ with a disk in $\Bbb R^3$. We just need to see how $-SO(2)$ intersects it.
superficially, yes. serious knowledge/understanding, maybe not so much
@BalarkaSen Looks like I have fooled everyone then, lol.
@Ted Question motivated by the fact that Pin(2) is linked.
Ah, cool, @Mike. Haven't pondered Pin since grad school.
22:02
Hello @DanielFischer @janos Could you maybe describe the computation model RAM (Random Access Machine) ?? What is this?? What does it do??
what the hell is Pin?
@BalarkaSen It's like Spin, but with no "S".
oh must be some rotation group
hmm. and that's all i want to know for the moment :P
22:04
@Ted It's interesting that $O(2)$ and $Pin(2)$ are topologically the same but that their canonical embeddings are quite different... So I hope $O(2)$ isn't linked :)
So $SO(2)$ is the Hopf fibre over the identity coset, @Mike. We just need to see an obvious way to interpret the other component.
Huh? Are you interpreting it as sitting in $SU(2)$?
Yes, I am. I realize you didn't want that.
In $SU(2)$, what is the double covering of $SO(2)\subset SO(3)$?
So when you said operator norm, that gives a different embedding from the obvious $\ell^2$ norm.
I've never thought about what all these different embeddings look like. throws up hands in despair and goes to fix a martini
wonder what a Mnyflds is.
must be some very complicated Poincare orbifold :P
@BalarkaSen That person clearly needs to read books before asking these questions.
I am having one of the darkest periods in my life now. I hope this means the sun is about to rise.
22:12
@DanielFischer I've been meaning to compute $\pi_1$ of knot complements and link complements. Do I need advanced tech to do this for, say, the Hopf link or the Treefoil knot?
If not, I'll try it.
@JasperLoy why, what happened?
@BalarkaSen I don't know. Never did such stuff.
@BalarkaSen Just more disturbing thoughts than usual. Nothing new happened recently.
@Jasper: You know that we all wish you nothing but the very best.
Hope you recover quickly, @JasperLoy
22:14
I will go eat some breakfast now, later.
Bon appétit
OK, @Mike, so I have no idea what your circle(s) look(s) like in $S^3$.
@Mike: OK, I'm confuzled. $\text{Pin}(2)$ is a double-cover of $O(2)$, not a double-cover of $SO(2)$.
I can't visualize the fibration @DanielFischer. Any such bundle would have trivialization U cross S^1 (homeo to R^2 cross S^1, I guess) around any point, no idea how on a sudden we jump up a dimension.
Indeed @BalarkaSen. I ended up just deleting my answer because it felt weird.
22:29
@BalarkaSen If it were easy to see, it probably wouldn't have lasted unknown until Hopf.
i saw that, i was even meaning to delete mine but the guy accepted it @Kaj
Besides, I didn't want to break my streak of like 8 accepted answers in a row :)
Oh wait, not quite. A single one of those a newbie didn't accept, but I got 4 upvotes :/
i'm still grumph at having 2 upvotes on my explicitly written survey on modern interest in absolute galois theory
22:32
@Balarka: So is life in academia
And likewise, @Balarka, I got one of those "nice answer" badges yesterday for minimal effort.
@Ted Can't hold a conversation with you when there's no service in this whole city
i think i'm gonna go sleep soon
Why, if $n$ is a composite, then so is $2^n -1$?
r9m
r9m
22:39
@DonLarynx if $n = ab$, $2^{a} - 1|2^{ab} - 1$
$2^{r \mod 2} - 1 = s \mod 2 = 0$
We have 3 pins on the wall, no wonder we cannot see so many starred messages.
@r9m isn't it the other way around? the right term is larger,
r9m
r9m
@DonLarynx $a|b$ is a notation .. it means $a$ divides $b$, ie $b \equiv 0\pmod{a}$
ugh. Armstrong has weird exercises.
22:45
@BalarkaSen Why skip between so many books?
to look for exercises
There aren't enough in one book?
munkres has less exercises than theory
I would think that doing all the exercises in Munkres is more than enough.
one funny exercises in armstrong has a sketch of a cool surface made out of deleting 5 huge discs out of them and pasting a moebius strip at each of the discs.
and he's asking to find fundamental group of that space.
22:47
Are you studying point set or algebraic now?
algebraic
Hmm, OK. Munkres has 3 books on topology.
One point set, one algebraic, one differential.
i will do diff geo/diff topo if and only if Ted unignores me
That's stupid. What has Ted ignoring got to do with diff geo?
math has nothing to do with who is or is not paying attention to you
22:52
hi @skull!
i'd rather Ted not ignore me
@DonLarynx hi pal
Ted's ignoring me too, in fact I think he's ignoring everyone in this room.
you too?
nah
Well, if you don't like him ignoring you, just ignore him.
22:54
true^
i want him to unignore me, not ignore him.
Well, I have already helped you, so I won't ping him again.
i am not asking you to do it.
why does his opinion mean so much to you?
22:56
he is a great mathematician @skull
@BalarkaSen so are you
@BalarkaSen How old are you now?
i don't know anything
@Ted I know Pin(2) is a double cover of O(2), but it's still homeomorphic to it (just as Spin(2) is homeo to SO(2)). The description I'm thinking of is Lawson's - Pin(V,q) is the subgroup of the invertible elts of Cl(V,q) generated by the $v \in V$ with $q(v)=\pm 1$. By Pin(2) I mean $\text{Pin}(\Bbb R^2,\|\cdot \|^2)$.
@BalarkaSen then act like you don't know anything
22:57
sure, i do so
not really
they don't call you your "highness" for nothing pal :-)
that's just those stupid guys. Sawarnik started doing that for no reason.
there was a reason

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