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17:00
So only when $X$ is a CW-complex. OK. Now that is more believable.
the cantor set is of course not homotopy equivalent to any CW complex, because Whitehead does not hold for it
right. precisely.
@MikeMiller This is interesting. How to prove it?
is there a piece of terminology for the following? suppose i've got a partition of $m$ containing a term $n$. then each partition of $n$ generates a partition of $M$
something along the lines of those partitions being subpartitions of the initial one
@DanielFischer How did you deduce from the fact that $x^{15} = 1$ for all $x\in \mathbb{F}_{16}^{\ast}$, that the order of $b+1$ is $15$?
never mind, i found it. refinement of a partition.
17:05
"How does one prove it?" also, what ate you referring to
I was referring to the statement that $S(X)$ is weak htpy equivalent to $X$. I am not sure if I am being dumb, but I can't see immediately why that is true.
Do you see the map S(X) -> X?
Send simplices of $S(X)$ to $X$ by the corresponding singular simplices, yes.
Why is it an isomorphism on pi_0?
Because it's an isomorphism on $H_0$. There's a bijection on path connected components.
17:10
Why
You literally just said "Cuz it's true"
Just quick question @MikeMiller your doing low dimensional topology right ?
what topology do you use mainly in your studies ?
@L33ter: I'm not sure what that means
I mean do you use algebraic topology, differential topology in your research ? I have zero knowledge about what do you do in low dimensional topology.
Sure, you use every tool you have.
Well, if two points in $S(X)$ can be joined by a path $\gamma$, then such a path runs through the simplices $\Delta_i^n$ of $S(X)$. I can push that path over to $X$ by doing $\sigma_i^n \circ \gamma$ (i.e., push the path simplex-wise), where $\sigma_i^n$ are the singular simplices corresponding to $\Delta_i^n$. This is continuous by the gluing lemma.
Conversely, if there is a path in $X$ joining two points, that's a singular 1-simplex, so corresponds to a 1-simplex in $S(X)$ jointing the preimage points.
17:16
ok
Sorry, I am having dinner right now, so the above argument is a bit garbled.
But I hope the idea is clear.
@Balarka: Why is the map surjectir on pi_1
So, for $A=(1,n)$ and $B=(1,-n)$ for $n\in \mathbb{Z}_{>0}$, we have that $A+B=(0,0)$, right?

$A$ and $B$ have infinite order and $A+B$ has finite order, or not? @DanielFischer
@Evinda From that, I deduce that the order divides $15$. I deduce that it is $15$ from the fact that $b+1 = b^4$ and $\gcd(4,15) = 1$.
@L33ter: To be more specific. The contents of Ch1-3 of hatcher are an absolute baseline for any topologist. The first line of whatever paper I write will be assumptions on the homology of a 3-manifold, and those only suffice because I understand the universal coefficient theorem and Poincare duality. When calculating certain imvariants of certain 3-manifolds I want to know a bunch of what's in chapter 4. I want to use characteristic classes. etc.
There are certainly things I use less than others, but they still see use. Last year I saw someone talking about Massey products as a way of fistinguishing certain spaces associated to 3-manifolds.
A popular trend nowadays is to have the imvariants not be groups, but rather stable homotopy types, and the way you extract info from these is stable homotopy theory. I've seen people talking about how the right way to write down their stuff is using infinity categories. You use whatever you have available.
On the other hand everything I do is predicated on ideas of morse theory and morse homology and I say the word "transverse" every day.
17:24
A loop $\gamma$ in $X$ can be realized as map of two singular 1-simplices, which takes the same value at the endpoints. Then that gives me two edges in $S(X)$ which share the same ends too. Going around one edge and then the other edge gives me a loop in $S(X)$ which pushforward to $\gamma$ by the $\pi_1$-level map, I think.
@DanielFischer You mean the order of b+1 ? Doesn't it hold that 15 divides the order of b+1 ?
@Balarka: Prove pi_1-injectivity then prove pi_n-isomorphism for all n
Hey Mike and Balarka
so in one sentence you want me to reproduce the proof. ok, sure. let me finish dinner first :D
@Evinda Well, yes, since $b$ has order $15$, and $\gcd(4,15) = 1$. So we have $15 \mid \operatorname{ord} (b+1) \mid 15$.
17:30
@BalarkaSen Then go do calculus.
That's what I have been doing all day.
Good! So you must be excited to do more.
Sure, I definitely am.
I find calculus more exciting than topology now.
What
Balarka, ironic or sincere?
17:32
I am serious.
@DanielFischer I still haven't understood how you deduced that the order divides 15.. :/
How on earth... What book are you reading?
@Evinda Because the order of every element of $\mathbb{F}_{16}^{\ast}$ divides $15$.
I wonder if it's possible to not feel more excited about something you don't know and which you're actively learning than the things you aren't actively thinking about.
@AndrewThompson T. Shifrin, "Multivariable Mathematics".
Out of curiosity, what grade are you in, Balarka?
17:35
@BalarkaSen Ah, the book might save it.
I base that solely on the author.
(I don't know how the Indian schooling system works but I'm guessing it's more-or-less the same as here)
@AndrewThompson The book is very good.
I haven't read any of his books, but I bet they're very good :)
@AkivaWeinberger Depends on the graded ring structure you want to impose on me.
(Don't tell him I said that!)
@BalarkaSen Meaning, you don't want to tell me?
17:37
@AkivaWeinberger Balarka enters elementary school this summer
@DanielFischer So do we deduce from the fact that the order of b+1 divides 15 and 15 divides the order of b+1, that the order of b+1 is equal to 15?
@Akiva Nah, was joking. I am on 10th grade.
17:39
Where are you from, Akiva?
Oh, cool
If you can make it there you can make it anywhere
@BalarkaSen Are you still doing math in school or do you only study independently at this point?
@Evinda Yes.
@AndrewThompson School's off.
17:40
I think the question was whether you're still asked to take algebra
Yes, it was.
I am not sure what that means. If you mean whether I am still in school, yes, I am. Until 12th grade, I will be there.
I'm not sure why you're not sure what it means. In school, do you still have to take the standard math curricula?
@DanielFischer Ok... This is only true since we ar in $\mathbb{F}_2[x]$, right?
17:43
Yes, I do. Isn't that tautological because taking standard math curricula is a part of studying in high school?
Many schools in many countries offer students in your situation (i.e., being in the 10th grade and conversant in elementary algebraic topology) to skip grades in certain subjects.
"Standard math curricula" doesn't include topology, does it?
@AndrewThompson Ah, I was not familiar with that. No, our country has no such facilities.
Or at least, any that I know of.
I'm guessing you learn calc in school and topology on your own time?
17:46
10th grade math does not include calculus.
So, both on your own time.
10th grade in India does include Hatcher chapters 0-3
Well, chapters 0 and 1 are done in grade 9
17:47
For some reason I don't believe you.
We will do stable homotopy theory in 11th grade, in fact.
You will not be provided anything more than basic differentiation and limits in 11th grade
You're all just messing with me.
@AkivaWeinberger Well why not?
Except for Albas
17:48
India's output of people who do PDE has always been low, partially due to the overemphasis of algebraic topology in the high school curriculum
Not in India, Albas. Didn't you know that the Bourbaki group are actually Indian high school students, the texts being compiled from their exam-answers?
OK, I get it, I'm gullible.
@AkivaWeinberger No, he means in terms of calculus. The point being that the 11th grade curriculum is focused on topology as opposed to calculus.
"Nikhil Bourbakichandram"
That was their real name.
Nicolas Bourbaki is just a French version.
Really?? Now you are messing with me
17:50
I don't know if you've seen it online, but there are a growing number of people studying algebraic topology and category theory instead of more geometric or analytic disciplines. I'm confident that the strange choice of schooling is why.
Hatcher has made a fortune from that. A large percentage of Indian high school students owns his book.
No. The Americans decided to attribute Bourbaki to the French, as they feared Indian dominance in mathematics after the growing popularity of the subject after Ramanujan, @Albas.
a Stewart-level fortune
I do not own hatcher
He put it up for free online
17:52
Must've been good karma from the fact that he even made it available for free on his website
@BalarkaSen: Now answer my question or get back to work. :)
Back to work everyone.
I just finished dinner.
Its evening here, I'm just sitting in my bed being lazy.
I'm watching a video of Wildberger explain homology
17:53
Oh, I'm giving a talk tomorrow. I forgot.
starts taking his pens out
Give me a few minutes, I'll answer the question.
But I'm doing that instead of studying for real stuff
I'm also just chillin here
@BalarkaSen Did you ever give thought to that question about closed sets on spheres I gave you?
17:53
@AkivaWeinberger Those videos are terrible
Bye and night everyone
isn't that the guy who doesn't believe in calculus
@AkivaWeinberger I am betting he does homology with $\Bbb Q$ coefficients and does them with rational simplices.
Huy
Huy
I don't believe in it either.
17:54
Nah, he's doing it fine
@AndrewThompson Whatchu gonna talk about
Huy
Huy
@MikeMiller: did you read the blog post about the guy who did a PhD in math and five years later doesn't understand his thesis anymore and what this means for maths?
I'm giving a student lecture in homological algebra. In 90 minutes I go from "This is a module." to "This is how you describe Ext in terms of syzygies."
@AkivaWeinberger Not the second one, no.
17:55
@AndrewThompson Is that an actual word
Wow, it is
Huy
Huy
he did it in functional analysis and says "therefore it has no practical use" or soemthing like that
would he prefer to build a better bomb
@DanielFischer Also I want to express all the roots of $g=x^4+x+1 \in \mathbb{F}_2[x]$ as also three generators of the multiplicative group $\mathbb{F}_{16}^{\ast}$ as for the basis $\{ 1, b, b^2, b^3 \}$.


In order to express all the roots do we use the fact that if b is a root of an irreducible polynomial then a^q is also a root?
17:57
A-bombs have no practical use, either :P
Well, all but two
@AkivaWeinberger Hiroshima and Nagasaki?
Yeah, of course
Question: Do $(v_0~v_1~v_2~v_3)$ and $(v_1~v_2~v_0~v_3)$ define the same orientation of the simplex?
18:00
Like, any even permutation fixes the orientation?
Thanks a lot @MikeMiller for those information.
What is taniyama shimura conjecture?
that's a nice summary of Japan's history
modularity theorem?
18:04
no I meant the link that Andrew posted a couple minutes ago
@MikeMiller Injection on $\pi_1$: Take two loops $\sigma_1, \sigma_2 : [0, 1] \to S(X)$. If $p \circ \sigma_1$ and $p \circ \sigma_2$ are homotopic ($p : S(X) \to X$) then I can cut open the square of the homotopy into two simplices. This in turn gives me two simplices between the images of $\sigma_i$, which match up to a homotopy, I think.
I am a theoretical physicist and i don't understand mathematics
@BalarkaSen: I'm going to leave now. I want to hear the words "Simplicial approximation". When you're mapping from S(X) to X, you have NO IDEA what this map does except on the simplices.
Hmm, true.
Huy
Huy
@HariPrasad try learning it
18:10
@AndrewThompson To quote the comments section: It's like CrashCourse but high
Haha, yes.
is there any Structure in the Prime Numbers?
Big, ill-posed question
My 0.3 mechanical pencil is broken :( 0.5, I'll have to settle for you.
Lots of work done with regards to that question, lots of work to be done
@HariPrasad
I guess the answer is, "At least some. Probably a bit more, but we're not sure. Also, there's a bit of structure that it looks like they have but we can't actually prove that they have yet but they probably do have it."
And the primes are connected with the complex numbers (like $\pi+\sqrt 2i$) for some reason.
18:14
@AkivaWeinberger Can we use Higher-order Fourier analysis to find Structure in the Prime Numbers?
(I shouldn't try to answer questions after watching a video that may or may not have been made while high.)
Um, maybe? I'm not the person to ask for that
Huy
Huy
that's like asking if you can use your eyes to discover a new physical phenomenon
Heya, @user276387. You should get a username.
"can we use topological conformal field theories of (infty, 2) type to prove the hochschild conjectures?"
@Huy if one haven't seen it yet
18:15
Hi, @AkivaWeinberger
I can't decide what to make it. No imagination, you see.
Maybe we could give you suggestions.
Hey, guys, what should @user276387's username be?
Grotesque Gorillas of Guinea.
@AkivaWeinberger Eisenstein primes
lol Balarka.
Eisenstein primes, eh? That guy had some beautiful ideas. I remember trying to follow his proof of quadratic reciprocity, even though I had no business in doing that.
@HariPrasad I mean, weirder complex numbers. $\frac12+14.1347251417\dots i$ is a fairly important one.
18:21
@AkivaWeinberger yup
Hello world!
I changed my username but it's still the same on here.
Random question but im confused how this works - how do the stats not add up to 100%
@user276387 just reload the chatroom
like how are they getting 61% from that graph
it clearly doesn't add up to 61%
@GGG @user276387 Maybe leave the chat and come back
18:27
@tabchas came from h bar?
@Evinda What is only true because we started from $\mathbb{F}_2[X]$? It's always true that an element of $\mathbb{F}_q^{\ast}$ has an order dividing $q-1$, and if $x$ has order $q-1$ and $y = x^m$ with $\gcd(m,q-1) = 1$ then $y$ also has order $q-1$.
@HariPrasad haha yep. Was tryin to see if someone could help me out :). Would you happen to have any ideas?
@tabchas well, let me try
@HariPrasad thank you!
Left and came back. Never mind, it will fix itself.
user276387 my name forever more - in the voice of Vincent Price reading an Edgar Alan Poe poem
Maybe logging in and out of your account?
Have you tried turning the universe off and on again?
@tabchas Got from "Nielsen Global Consumer Exercise Trends Survey 2014" slideshare? I think its a mistake and its just for visualization
@DanielFischer What do we consider as y in this case?
Yeah, done the logging in and out. Maybe clearing my cache.
I'm not ready to put the plug off the universe just yet lol.
18:42
Try threatening Atwood and Spolsky.
@HariPrasad yep. What do you mean its just for visualization (like its just relative numbers with no real meaning)?
I got my monthly burst of inspiration last night and developed an algorithm for constructing something which I know exists but that I'm having trouble characterizing
What novel trick am I missing in the space between recurrence relations and finite differences?
@tabchas yes that's what it seems to be and by the design of the slides its clear that its not well drafted (it seems like took from different sources and put together)
@HariPrasad ah ok. makes sense. i feel like a lot of these stats pages gives these relative data points
@tabchas maybe
18:47
i guess its serves its purpose to give you a relative estimation of the market
thanks tho @HariPrasad
@tabchas you are welcome
If I write "@GGG" does it ping you?
@user276387
@GGG
Nope. Only when you write @user... I cleared history, logged in and out. Maybe I'm logged in to chat with one of my other computers.
@mickLH what do you have in mind?
@user276387 Maybe something on this page is useful?
Wildberger just wrote $\frac00=0$. (Don't worry, it's the quotient of groups. $0$ is the additive group with one element.)
Just thought it looked weird.
(He changed it to $“0/0”$ to emphasize the non-fraction-ness)
GGG
GGG
18:58
Thanks @AkivaWeinberger that worked!
@AkivaWeinberger Not sure it looks that much less like a fraction that way
Yay!
@TobiasKildetoft True.
@Semiclassical I have a function which interpolates equidistant points smoothly, the definition of it is recursive though, a fractal even
But what can you do.
@Evinda $x = b$ and $y = b+1$.
18:59
sounds like a spline?
Yes, it's meant to be a spline
@AkivaWeinberger Not much I suppose. $0$ is certainly a fine thing to call the trivial group (at least when working in the category of abelian groups)
I'm having trouble formulating a solution for a real-valued point on the curve though
I can obtain any arbitrary rational point
@TobiasKildetoft Compare this to Euler, who wrote $\dfrac{p^0-1}0=\ln p$ !
@DanielFischer Ah I see... And we prove it as we did above, right?
19:01
(Remember, he lived before limits.)
I'm afraid that it may not even exist at the real values "in-between" the rational values
How would you interpolate, say, $f(0)=1$, $f(1)=2$, and $f(2)=4$ with your thing?
@micklh i don't know much about interpolation myself. but you might check out Wilf's short discussion on the corresponding 3-term boundary value problem in Generatingfunctionology
page 10 is the start of the section, and page 11 mentions cubic splines specifically
I remember that section
he solves it in terms of generating functions---as you'd expect from the title of the book---but the statement of the problem shouldn't require that
19:04
It's not essential to the rest of the book
sure. but 1) it's what i remember, 2) IMO generating functions makes recurrence relations + finite differences a lot clearer
@AkivaWeinberger Simply put: I can define the midpoints given those values
So, $f(0.5)=1.5$, $f(1.5)=3$?
fyi, he actually discusses the example of interpolating $f(n)=2^n$
page 13
Wouldn't that just be linear interpolation?
19:08
No I meant I have a formula in terms of the given points, which can give the $y$ values for the $x$ midpoints
Here's a render of that case you specified
the green values are the spline?
Yes
Here's a runge function to show that it's not just a polynomial through $n$ points
Interesting
Sounds like fun
@DanielFischer Did you also see my other question?
19:13
Does anyone happen to know of any work involving rational manifold-like structures?
@Evinda No, sorry.
Or even just rationality-preserving functions.
@DanielFischer Also I want to express all the roots of $g=x^4+x+1 \in \mathbb{F}_2[x]$ as also three generators of the multiplicative group $\mathbb{F}_{16}^{\ast}$ as for the basis $\{ 1, b, b^2, b^3 \}$.


In order to express all the roots do we use the fact that if b is a root of an irreducible polynomial then a^q is also a root?
Is there a subset $A\subset\Bbb Q^2$ such that if $d(x,y)=1$, we have $x\in A\iff y\notin A$?
Where $d$ is the distance function
@Evinda Where $q$ is what?
19:23
@Evinda Sorry, I don't understand what you want to do. Find all roots of $g$, probably. By assumption, $b$ is one, it's easy to check that $b+1$ is also one. You can then find the remaining two. But then what?
@TobiasKildetoft Yes, I meant g.
@DanielFischer How can we find the remaining two?
@Evinda That does not make any sense
@TobiasKildetoft You mean the proposition that I stated?
@Evinda I mean your assertion about $a^q$ being a root
@TobiasKildetoft Ok... Knowing that b and b+1 are roots , how can we find the other ones?
20:22
I just came up with a good question while I was solving some topology problems @BalarkaSen
Suppose $\{\tau_{\alpha}\}_{\alpha \in I}$ can you give me a condition for when $\bigcup \tau_{\alpha}$ is a topology ?
Anyone want to translate some German for me
@PVAL if it is short, sure.
it is good I am dedicating like 1 hr a day solving top problems that is good
@Tobias If you have access to Springer, its lemma 4 here link.springer.com/article/10.1007/BF01403388
The proof of mainly
@PVAL not from home, sorry
ahh, got it
@PVAL it gets a bit long to start typing out
20:34
@L33ter That doesn't sound much like a good question, to be fair. What's wrong with "closed under union and finite intersection"?
I guess what I was looking for is that union will be a topology when the subbasis generated by the union is the same as the union
@PVAL But if you have some specific thing you want to know, it seems fairly straightforward
I have to go, working on something right now.
when r u coming bk ?
Hi! I have a seemingly simple question about Gaussian random variables, but would like to know if there is a very simple argument (hopefully a one-liner or few-words-reference) for it
Namely, if u,v are two fixed, orthogonal vectors in $\mathbb{R]^n$
and X is a spherical Gaussian r.v. in $\mathbb{R}^n$; then how to show that <u,X> and <v,X> are independent?
20:51
@BalarkaSen I want to re-assure you, I am still working on that software! Nearly all the progress has been in graph algorithms for representing and manipulating parametric manipulations to and representations of ...algorithms. Though visible progress has also been made: You can now communicate a small subset of formal concepts to it using LaTeX directly!

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