What do you even mean by similar properties

because if I can sphere separate everything, I can pick it apart into lots of pieces, right?

I am not sure the limit has anything to do with a link

and then if I do that infinitely many times, messing with smaller and smaller pieces of $\Bbb R^3$ each time, I should end up with a tame Cantor set

@Ali I'll find you the link

@AliCaglayan Sure it is. Exiercise: Understand why the Antoine's necklace doesn't have simply connected complement.

sciencedirect.com/science/article/pii/016686418690060X

Its construction is like that of Antoine's necklace but weirder

The point is at each step you're interlinking stuff so that in the limit there is no way to nullhomotopy of the obvious loop which "winds" around the solid torus at the beginning.

huh that is weird

@Ali Look at Figure 3 in that paper, actually. That's all finite stuff, but can you see how to get from the left image to the right?

They're meant to be isotopic to each other

(Those are 1D objects, if you like)

(though looking at Figure 1 would probably help you understand what's going on in that figure better)

@Ali In addition, if you get rid of the component labeled $\overline{S^c}$ in Figure 3, the rest of the pieces become unlinked.

I stand corrected

Thanks @AkivaWeinberger

Figure 3 is only relevant for proving that the embedding is wild (inequivalent to the standard embedding). It's actually pretty simple to see that its complement is simply connected, which I did not expect.

I don't think Akiva said anything much which tells you why the limit has anything to do with the finite step :P

He's just saying there's some weird move going on at each finite step.

The paper established that I think?

The paper establishes a wild cantor set with simply connected complement, which honestly doesn't have much to do with your question.

It just gives an example of a case which tells you finite steps are relevant in understanding the limit.

What am I even thinking anymore

(Proof sketch that the embedding is wild: Call the first solid double torus $H_0$ and the embedding of the Cantor set $W$. They prove that the "waist" of $H_0$'s boundary is not contractible in $H_0\setminus W$.)

Which, like, Antoine does that for you.

Yeah, I was just trying to explain part of the headachiness.

Right.

I don't actually see the move.

my coffee has completely worn off

@BalarkaSen What move, the Figure 3 thing?

Yeah.

It's easier once you realize that the "links" between each genus-2 component are actually symmetric, despite not being drawn that way

@Danu that was not really explicit

I am not sure I understand what you mean by symmetric. What's tripping me up is the middle thing in left picture which links on the "two separate handles"

Without that it seems to become a standard self-unlinked genus 2 surface with occasional links with other genus 2 surface in different places.

Basically the picture on the right with one of the genus 2 surfaces missing, I think

@Krijn Hmm?

@Krijn Do you want me to tell you how to make it?

@DanielSank It's a path integral, so...

@MikeMiller so?

Is it right that I can realize as $SO(n+1)$ as $\{x_1,\dots x_n\in S^n\mid x_i \text{mutually orthogonal}\}$?

not quite, the latter would be $O(n+1)$

also need to write $(x_1,\cdots,x_n)$

Right

@Danu If you have the time

@Krijn Yeah, it's no problem

so you set up the environment with

\begin{tikzpicture} \end{tikzpicture}

So far so good

I also know some draw thingies

Like \draw (0,0) (2,2) and stuff

But that's about it

So you can probably draw the straight (and dotted) lines, right

To make curved lines, typically it's enough to use

\draw (x,y) .. controls (a,b) and (c,d) .. (z,w);

controls?

This will "pull" towards $(a,b)$ and $(c,d)$ in between

Ah!

It makes a Bezier curve, I think

So you want something like

$(0,0)$ .. controls $(-0.5,1)$ and $(0.5,2)$ .. $(0,3)$

for those wiggly lines

Oh, I see, yeah

and more extreme "pulling points" for the horizontal line

How do I add letters to these spaces?

For the little balls, use \fill (0,0) circle (3pt); or something like that

You mean labels for your lines?

Yes, or in general anywhere in the picture

You can do something like \draw (x,y) .. controls (a,b) and (c,d) .. (z,w) node[pos=0.5,anchor=north]{$x$};

For general text nodes, something like \draw (x,y) node{$lala$}; will work

(the anchor will "push away" in that direction)

TikZ is easier than I remember it

That should be all you need

Yeah, thanks, that's really helpful!

TikZ is the best

Are any of you guys particularly well-versed in probability?

i'm not, but ask, don't ask to ask

@arctictern So I just need to put in a notion of orientation?

yeah

det = 1

@MikeMiller I didn't have a specific question in mind this time. Probability is just something I'm taking a course in at the moment, I'm wondering who I should look out for

@arctictern But wait, if I just specify $n$ mutually orthogonal, ordered points on the sphere in $\Bbb R^{n+1}$, I can uniquely complete to a positively oriented basis anyways. So.. isn't the orientation thing fine anyways?

the notation you gave for your set didn't say anything about uniquely completing a positively oriented basis

No, that's right, but if I just give $(x_1,\dots x_n)$, the set of all such is still just $SO(n+1)$, correct? Because I can uniquely complete

but the set you wrote chooses to complete it in all ways, subject only to the condition the $x_i$s are orthonormal

for example with n=1, your set contains the matrix diag(1,-1), because its columns are orthonormal

@arctictern he chooses n vectors in his set, and picks the last column of the matrix such that det =1, unit vector, etc

If I write it as $\{(x_1,\dots,x_n)\in (S^n)^n\mid x_i\text{'s mutually orthogonal}\}$, shouldn't that be ok?

if he does that, then sure

when n=1 he's only picking the first column

Exactly

oh

I just leave out the last one, knowing it is uniquely determined

I didn't catch it's missing the last column. then sure.

Oh, now it becomes clear to me how stupid this reformulation is haha

Just $SO(n)$ being the set of pos. orient. bases

derp

@Danu, last question, can I add something to these curves at a certain height?

Decorations? Sure!

Arrows, braces, anything

you'll need a tikz library

Just a small non-filled circle, really

Ah, okay

Add \usetikzlibrary{decorations.markings} in your preamble

Then try something like

...but with a circle

I can check if a circle is part of that library

Where do I add this?

What is Tikz? A WYSI-not-WYG text-based image creator? That sound horrible

like

OH

Found it

\draw[decoration={markings, mark=at position 0.625 with {\arrow{>}}}, postaction={decorate}] (0,0) --(1,1);

@AkivaWeinberger ??

What is Tikz

It's the "native" LaTeX graphics package.

And it's freaking awesome

meh

But you're using text to manipulate pictures??

Yeah.

It's pretty great

Tikz is awesome

I've posted many things I made before

Maybe I should try it before I judge

@Danu Like what

one can make super cool pictures with a pen and paper too

@BalarkaSen These look a lot better though, in scientific documents.

It's also great for making neat diagrams

@Danu people.mpim-bonn.mpg.de/sbehrens/files/Freedman_2013-04-05.pdf

And for making not-so-neat stuff, like

Oh tikz-cd is definitely good for making commutative diagram

@BalarkaSen Yeah, and if I'd TikZ those they'd look 10x better.

@Danu I am honestly skeptic you could tikz all those :P

eg figure 13

@BalarkaSen not cd lol

@BalarkaSen Honestly wouldn't be that hard.

Would take some time for sure

But not technically difficult.

figure 18

Tikz' limits are usually those of the person using it

There's a special knots package @BalarkaSen

Could deal with that just fine

that looks terrible tbh

Figure 28 is hilarious :O

@BalarkaSen Stop kidding yourself :P

i am not dood

Here's a bigger one so you can see it better

Is that elfish from Tolkien?

Or did I just randomly insult some existing language?

funny, some figures from that pdf are missing which were in the original

He guys. Say we have a group with 3 elements $(=e,a,b)$. Is the following possible; $aa=b$?

where $e$ is the neutral element

Is anything else possible?

yes

(no)

$aa=e=bb=ab=ba$

that would not be a group

ok that is good news to me

can't happen

finite groups of prime order are unique

$bb=ab$ implies $b=a$

aa = ab means a = b in a group

ah of course

good:) thanks

@Danu And I thought trying to draw the complement of the Hopf link was bad

For beautiful pictures: arxiv.org/pdf/1602.06450.pdf

ah, here's the original : math.uqam.ca/~powell/Freedman2013.pdf

huh, i can still see missing stuff

Maybe there's an even neweer version

figure 73, 77

good luck constructing those!

it seems he didn't sketch them tho

Do you really think those are difficult?

Those will be very easy, honestly

If you check the thing I just showed you it has many more complicated links and stuff

i do, until you construct it for me

what you gave me doesn't really look like a link :P

Check out the paper I just linked

you underestimate the power of the dark side TikZ

Knots/links are just simple with that package

It's just a question of time

they are good but none of them really has a 3d feel to it

(oh, also, hell, figure 1)

It's really easy to depict a knot in 2D I suppose, once you know the crossing information. just project it generically so that the only singularities are double points and draw over/undercrossings accordingly.

but knot-moves should def be harder to draw

3D effects are obtained by deforming circles to ellipses

Thanks Danu, my image is wonderful :)

Is 'Mathematologist' a word?

@Dragneel This guy seems to think so, and indeed believes he is one: mathematicalmysterytour.blogspot.co.uk/2014/10/…

This may be quite a novice question. However, I am curious. What is the difference between let x = 0 and sub x = 0

@jserv Very interesting article. He does have a point, however. Those who use math in their fields such as professors, statisticians, and accountants, are not mathematicians.

@IPAddress what does "sub x = 0" mean?

As he said, "Mathematics itself is really too broad a field for any one person to be an expert in all of it."

i meant substitute x = 0

I'm guessing Sub stands for Substitute.

usually people say "substitute x=0" when talking about x inside of a formula, but "let x=0" could also be used in the context of x appearing in a claim

not sure why you think there's a meaningful difference

Hello. Let Zm denote the congruence classes of integers mod m; i.e., the quotient group Z/mZ. I am trying to show that the function f : Zm ---> Zkm defined by f([x]) = [kx] is injective. If f([x]) = f([y]) then [kx]=[ky] or k(x - y) = pn, where p is some integer. I having trouble deducing that x=y. This is a standard problem; I did it a long time ago, but the solution is being especially elusive. I tried searching through google but nothing came up. Any hints, links, etc. would be appreciated.

@arctictern thanks :)

[Random] Some proof in the h bar, which I turned into a flow chart. One can see how in genral compact sets are oftenmisleading to draw in pictures i.sstatic.net/dvLG6.png

@user193319 km|(kx-ky) iff m|(x-y)

@arctictern But I don't have km|(kx-ky); I have n|(kx-ky).

what in the world is n?

@arctictern Oh, sorry. I mixed up m and n; it should read m|(kx-ky).

[kx] and [ky] are elements of Z_km. their equality means kx-ky is a multiple of km.

Could I have help with a issue I found with a problem using the remainder theorem

maybe

@arctictern Ah shoot! I was thinking we were still in Zm, rather than Zkm.

Can someone help exlpain this: 100000 = n log2 n. We cannot solve this symbolically but the solution is n ≈ 7740.96

I get that n is close to infinit

what is there to explain?

makes sense to me. except for the "close to infinity" part, dunno what you mean by that.

starting with the identity implied from the remainder theorem $2x^{2}+4x+5 = A(x^{2}-1) + B$

@arctictern Thank you for your help!

yep, np

Our next step would be to find $B$

yeah, you get A=2 by comparing leading coefficients, then subtract to get B

so $let x = 1$

um, what?

I get that 100.000 * 2^100.000 = 100.000 * ∞

@FelixRosén 2^100 is most certainly not infinity.

2^100000

whatever, still not infinity

I know, the calculator tells me its ∞. I know its a very large number but not ∞. I still dont get how they got n ≈ 7740.96

The book I am learning from tells me you should always make the divisor zero to work out the arbitrary value (in this case $B$)

that sentence makes no sense. try this sentence: "both sides need to be equal no matter what x is."

@arctictern let's go with that

so 2x^2+4x+5 = 2(x^2-1) + (4x+7).

A=2, B=4x+7

@arctictern I understand what you're saying. Also B can only have x where the degree is 0

B can only have x where the degree is 0?

where what's degree is 0?

@arctictern B can not be 4x+7

Whoah, look at Figure 4.4 (page 66) on this

@IPAddress it can and is

It's a link

@arctictern oh my bad... I looked at an online solution... completely wrong

Hi chat

Hi @Astyx

Hi cat h

Hi @Alessandro. Anything interesting?

Hi @Ted

Salut @Ted

We finished the classification of compact surfaces today in class

How was it done?

Oh, cool, @Alessandro, so your fingers are all full of paste.

Hi, MikeM.

Salut, @Astyx :)

I'm wet and smell.

Probably by triangulation and combinatorial arguments

really like the proof on Mike's blog, on the other hand

How was maths @Balarka ?

Not bad at all, @Astyx.

I messed up a coordinate geometry problem in part (but I did the other part right) but everything was good otherwise

I expected nothing less :p

@MikeMiller we showed that every compact surface is a connected sum of tori and projective planes a couple of months ago, today we computed their fundamental groups to decide that they're actually distinct spaces

the combinatorics was super-easy, fortunately

@Alessandro How did you show every compact surface is a connected sum of tori/proj planes?

Now Balarka gave me the fundamental group of the complement of the Hopf link to think about, I haven't made many progresses, but at least I'm getting good at drawing

Grumph @Balarka :)

@BalarkaSen in a painful way cutting and pasting polygons with identified sides

We probably skipped a few details here and there

@Alessandro: My goal is to turn every person in here into a good drawer. :)

I can only do the mental pictures.

@TedShifrin Is the grumph for super-easy combinatorics or the messed up coordinate geometry? :P

The latter, of course.

Vectors, man, vectors.

@Balarka I do find my argument a bit more natural, but I should find time to clear it up a bit and make it simpler.

Oh well. I did good on the conics part. But I messed up while computing the two sides of a equilateral triangle with given side and a vertex (I computed one side right, but the other side wrong). It was something silly, but the line was horrendous to manipulate with and was pressed with time. :(

@TedShifrin True

If I remember right dealing with the sphere is a nontrivial additional prknown.

@TedShifrin Whoops.

at which address is that blog?

@Balarka: No big deal that you're incompetent at easy stuff :P

Hi @Fargle

@TedShifrin who needs 4+ dimensions!

Hiya, how goes it?

Well, thinking about a complex manifold of dimension > 1, I do. @Alessandro

@AlessandroCodenotti Did you prove the existence of a triangulation?

Doing well, Fargle, and you?

No, we just assumed there is one

@MikeMiller That a homology sphere is a sphere in dimension 2, right? Hm.

Quite well. Just finished a take-home midterm for real analysis.

Hey everyone, I had a quick question. I know that multiplication in time domain is convolution in frequency domain and vice versa. However, when does the factor of (1/2pi) get tagged on?

Anything interesting, Fargle? I wonder how many students got someone on MSE to do their questions :P

There're only 6 of us >_>

IMO if you're going to take something for granted, you should take the (brief!!) time to learn what a handlebody diagram is and use those.

@Christian: Different people have different conventions with the Fourier transform. So you'd better check your text/notes.

The classification theorem is completely trivial with the simple-for-surfaces Kirby calculus.

@MikeM: I would take it for granted, as well.

But yeah, there were some neat questions. It was all true-false with proof/counterexample. Some simple things like "no infinite set is compact", but then some less trivial things like "every sequence in $\Bbb R$ has a monotone subsequence"

I've forgotten whether Munkres proves it.

@AlessandroCodenotti assorteddetails.wordpress.com

Right, but I don't think is just a matter of convention.

@Fargle: That's one of my all-time favorite proofs in Spivak. He uses peak points.

I looked up Babai after you mentioned he's still teaching yesterday (I thought he was older) @Ted, turns out he had a phd student called Codenotti, I wonder if he his a distant relative

I also like the proof of classification of surfaces by Riemann mapping

Oh, that's cool, @Alessandro. Go find your (pseudo-) relative.

modulo that every top. manifold admits a complex structure.

@Ted Topologicallt you need Schoenflies, after which it is not too hard. Smoothly you may as well use Morse theory and handlebodies.

@TedShifrin I guess that it's similar to what I did, but I felt a little shaky.

(which I guess is more or less trivial if you're smoothable so that's what's needed)

I wish I had the exact text of my answer in front of me.

@Balarka: Then you need smoothness and isothermal parameters. Pretty big sledgehammer.

Because often times, I see multiplication being simply replaced with the convolution sign. However, I'm learning about nyquist rates

@MikeM: An introductory alg top course won't have smooth manifolds and Morse theory.

@BalarkaSen Did you ever read the Hatcher note where he smooths surfaces?

and here they are multiplying by (1/2pi) when going frmo multi in time domain to conv in freq domain.

@TedShifrin Right, I don't know how to prove topological manifolds admit smooth structures yet. Mike sent me something but I didn't read it yet.

@MikeMiller I started reading it a little, but the exams came butting in. Definitely plan to read it after 25th

A lot of the questions were pretty obviously false, or "obviously" if you were a careful reader. For example, "For $x,y,z \in \Bbb R$, if $x < y$, then $xz < yz$"

Yeah, Munkres stipulates the existence of a triangulation.

He gives references.

@Fargle: Sounds like the course isn't getting to what I consider the interesting stuff. Actual analysis. :P

@Christian: This is too technical for me at this point in my life. You'd better ask your professor.

@Balarka The isothermal parameters are the hard part. And after that you need to prove uniformization and understand the uniformizing subgroup of SL(2,R).

@TedShifrin What would you consider the interesting stuff? We have gotten to Lebesgue measure (though with a lot of handwaving as to what sets are actually measurable).

Oh ... Stuff with derivatives and convergence of functions, etc.

we finished the alg top part of the course today however and we'll start complex analysis next time

@Balarka: Maybe you can help with this.

@TedShifrin Yeah, we won't get to derivatives this semester, but you know me, I take joy in scattered autodidactism.

I just don't see the point of measure theory before people learn the basics.

@Fargle According to my analysis professor there's "a shitload of measurable sets" (which means $2^{2^{\aleph_0}}$ many sets)

@TedShifrin Hmm, wait, I know this but I am forgetting how to split the fiber short exact sequence.

Well, the hypothesis should help, @Balarka :P

I guess I could try going through Rudin again. Or I have a copy of Spivak, if I want to start more easily.

Oh, I think I do know how to do it.

Or ... my book. :D

@AlessandroCodenotti That sounds about right, I'd expect there to be as many as $|\mathcal{P}(\Bbb R)|$ measurable sets.

The fibers are nullhomotopic. Any map $S^{n-1} \to F$ bounds a ball.

(Even though there are supposedly many that aren't measurable. Damn you, AC!)

Project that ball down to the base.

That sounds reasonable, @Balarka.

@TedShifrin True, I do have that as well. And I could always use more linear algebra.

Whole boundary gets squished to a point because it's in the fiber.

So that's a map $\pi_{n-1}(F) \to \pi_n(B)$

That's probably the section

I just feel a much greater attraction to algebra than to analysis.

Not I, @Fargle, but I tolerate algebra.

I did love teaching certain parts of it, though.

In the best of all possible worlds, I'd like to be really, really good at both.

That's an excellent goal.

There's just a lot of intuition left for me to build, in both branches. Hnnng.

(In all branches, really. Though my arithmetic is pretty sharp.)

My intuition is much better for analysis/geometry/topology, but writing an algebra book made me a lot better at explaining/intuiting undergraduate-level algebra.

@Fargle yup, to see this you need to show that measure 0 sets are measurable and then that the Cantor set is an uncountable measurable set with 0 measure