Mathematics - 2019-02-08 (page 1 of 4)

Mike Miller

00:00

Huh? How would you want to work with characteristic classes?

I guess you could instead define them to be elements of the cohomology of $BU$ or $BO$ but then you need to do some calculations.

Hatcher's vector bundle notes are reasonably geodesic.

anakhro

00:13

Well I typically like to see things developed kind of with motivation.

Milnor likes to just throw it at you and show you cool stuff, which is good if I wanted to immediately work with them.

But I want to grok them

Ted Shifrin

And then there's curvature and Chern's original definition of Chern classes using curvature ...

user193319

If $n = n_1 + ... + n_r$ is some partition of the positive integer $n$, why is it true that $\{1,...,n\} = \{1,...,n_1\} \cup \{n_1 +1,...,n_1 + n_2 \} \cup ... \cup \{n_1 + n_2 + .... n_{r-1} +1,....,n_1 + n_2 + ... + n_r \}$?

Ted Shifrin

You can find this in lots of differential geometry and complex geometry books.

What do you mean, @user193319? You've just split the set up into $r$ (disjoint) subsets.

user193319

Hmm...I can see that they are disjoint...but it isn't clear to me that the two sets are equal...Perhaps there is something wrong with me.

Ted Shifrin

As I always advise in here, write down simple examples.

Mike Miller

00:17

@anakhro The motivation is what you can do with them, though.

Ted Shifrin

@anakhro: In some sense, you can think of them as generalizations of the residue theorem and Gauss-Bonnet.

anakhro

That is one type of motivation.

@TedShifrin do you know a book that goes about it this route? Or at least mentions this and expands on it?

Mike Miller

'geometry of characteristic classes' by Morita does something with this but I don't remember if that book is too fancy.

Ted Shifrin

Lots of them. Complex geometry books like Wells and Griffiths/Harris for starters. Chern. Kobayashi-Nomizu. The 2- or 3-volume book by the Canadians Greub, ....

Mike Miller

Far too fancy. Don't read it.

anakhro

00:21

Far too fancy to Morita?

Mike Miller

@TedShifrin I think the goal is to find something that won't cause a panic attack. At least two on that list would, I think.

yes

Ted Shifrin

LOL ...

Rithaniel

Alright, so, given $L=\langle a,b,c\vert b^{-1}ab=c, ac=ca \rangle$. A homomorphism $f:L\rightarrow\mathbb{Z}$ given by $f(a)=f(c)=1$ and $f(b)=0$. Is this valid? If it is, one thing I think it demonstrates is that, in order for $\langle a\rangle$ to be normal, then $a=c$ must be true.

anakhro

Kobayashi-Nomizu panic attack already occurred when I tried to learn what a manifold was in 3rd year.

Mike Miller

@Rithaniel I feel like there's a disconnect here. You want to show that $a^n \neq c$ for any $n$. So you should choose a homomorphism so that $f(a^n) \neq f(c)$ for any $n$.

But your homomorphisms keep sending these things to the same element!

anakhro

00:24

Thanks for the references though, Ted! I will have a look and find something that won't make my heart flutter too much.

Rithaniel

Ah, I'm going the wrong direction

Mike Miller

@anakhro Did you read Hatcher's notes?

Ted Shifrin

I taught this stuff in various courses, @anakhro. If you're not going to bitch like @MikeM does, I can send you handwritten notes, but they're not full of the comments and details that I would make teaching.

anakhro

No, I will have a look at those, too.

@TedShifrin I wouldn't mind handwritten notes if that isn't too much of a bother to send. Shall I email your email from your website?

anakhro

00:51

Hmmm, actually it would probably be really useful to look into these relatively soon.

I need to write the intro to my thesis

And the linking number and that uses Chern classes

Rithaniel

Okay, so I want to preserve the part of the structure where $b^{-1}ab=c$, but I don't have a good sense of what conjugation translates to in more general groups. I guess it only comes up in non-abelian groups and I don't have enough practice with those.

anakhro

I liked looking at conjugation actions when I first learned about them.

So if you think of the conjugation action on the set of all subgroups of a group, then a neat thing is that "fixed points" are just normal subgroups.

Ted Shifrin

@anakhro: Yeah, email me and I'll send you some stuff.

You have to get good at differential forms, though, for my notes :)

anakhro

Well I guess we will see if I am good enough at forms!

N. Maneesh

1

Q: If $\text{Boundary}(A)\subseteq A.$, then $\overline A=A$?

Definition 1:- Let $(X,d)$ be a metric space. $A$ be a subset of $X$. Then $\text{Boundary}(A)=\{x\in X:$open ball centered at $x$ intersects both $A$ and $A^c\}$ Definition 2:- $\overline A=A\cup A'$, $A'$ is the set of all limit points of $A$. Using these two definitions My aim is to prove ...

In Jose Carlos sir's answer. He uses method of contradiction. But where did he get contradiction?

anakhro

01:09

$x\in bdry(A)\subseteq A\Longrightarrow x\in A$

N. Maneesh

He assumes the proposition and $x\notin A$, then He arrives at $x\in A$. But $x\in A$ is not come in the proposition. It is the conclusion. right?

anakhro

You have A is a subset of its closure

So it remains to show that the closure is a subset of A.

Mike Miller

@Rithaniel Okay, but $\Bbb Z$ has no conjugation action. If $a, b \in \Bbb Z$ then "$bab^{-1}$" (or better, $b + a - b$) is the same thing as $a$.

anakhro

So you need to show that $x\in\bar{A}\Longrightarrow x\in A$.

N. Maneesh

@anakhro In the method of contradiction the proof must contradict to our proposition. right?

anakhro

01:11

So Jose is saying "so suppose that x is not in A".

And then he arrives at "x is in A". Hence x must have been in A to begin with.

Proof by contradiction is where you prove something of the form "if P, then Q" by supposing "not Q" and then showing that this reduces to a contradiction, hence "Q" must be true.

Leaky Nun

the homotopy classes of paths from $x$ to $y$ form a $\pi_1(X,x)$-$\pi_1(X,y)$-bi-torsor

N. Maneesh

okay. I got it. It is not I think the method of contradiction. Case1 $x\in A$ and case 2$x\in A^C$. Both cases $x\in A$

@anakhro right?

Hence the proof

anakhro

Sure, why not. But really he's just doing a plain proof by contradiction.

Rithaniel

So, I want to build a homomorphism $L\rightarrow G$ where $f(a^n)\neq f(c)$ for any $n$, but I want to specifically be $G=\mathbb{Z}$ at your recommendation, but that's impossible because then, by the requirement $b^{-1}ab=c$ means that $a=c$ for any given $b$.

Mike Miller

Oh god, have I been sending you on a wild goose chase this whole time by misreading your question? One second.

It seems that I have.

I apologize. Let me come up with a hint that works.

You are correct that all you can conclude so far is that if $c = a^n$ for some $n$, then $c = a$.

Rithaniel

01:19

Ah, shoot, well, I've learned a little bit about constructing homomorphisms, at least.

Mike Miller

OK. Your group is $\langle a, b, c \mid b^{-1} a b = c, ca = ac\rangle$. The first relation says that we don't even need to include $c$ as a generator! This is the same as the group $\langle a, b \mid b^{-1} aba = ab^{-1} ab\rangle$.

Rithaniel

Ah, that's clever.

Mike Miller

Your observation is correct: if we want to show that $\langle a\rangle$ is not normal by showing that $bab^{-1} \neq a$, we'll need to construct a homomorphism to a nonabelian group.

Rithaniel

Alright, the first that comes to mind are 2x2 matrices with nonzero determinant. Perhaps a subgroup thereof?

Ultradark

@TedShifrin if one takes two geodesics on $\Bbb S^2$ and they are mapped to $\Bbb R^3$ and linked, this is the Hopf link right?

anakhro

01:26

Any two (maximal length) geodesics on S^2 intersect...

Rithaniel

(Typing up matrices in latex is annoying)

Mike Miller

@Rithaniel I need to think for a moment - a bit distracted on my end so I may be slow.

Some of my favorite groups are the quaternions, and dihedral groups. So I might think of those.

Rithaniel

Take your time, I think I have a potential idea.

Mike Miller

It's a pretty nasty relation we have here.

Rithaniel

Yeah, I'm specifically trying to make it so that $\langle a,b^{1}ab\rangle$ is a normal subgroup, but $\langle a\rangle$ isn't.

Ultradark

01:30

@anakhro then I'll just use two different spheres with one geodesic each, and map each geodesic to $\Bbb R^3$ and then link them together

Mike Miller

I guess the moral is to not make up your own group presentations! :P

Rithaniel

Okay, so would $b=\begin{bmatrix}
2 & 3 \\
1 & 2
\end{bmatrix}$ and $a=\begin{bmatrix}
1 & 1 \\
1 & 2
\end{bmatrix}$ work?

Also, yeah, it seems to be a headache

Mike Miller

Do they satisfy the relation? (I'd plug it into wolfram alpha)

Rithaniel

They do, at a glance. $b^{-1}ab=\begin{bmatrix}
2 & 3 \\
-1 & -1
\end{bmatrix}$

Well, actually, didn't check thoroughly enough.

Mike Miller

I'm plugging it in now.

Rithaniel

01:34

That's just the first one.

Nope, they don't

Mike Miller

Unfortunately I do not think it satisfies the big relation.

Yeah

It seems hard to cook something up where that's true. I don't see a way to substantially simplify it.

anakhro

@Ultradark I don't understand what you are trying to do. Of course if you take any two embeddings of a circle and "link" them (whatever that means), you will get a Hopf link.

But whatever this "link" operation is, you would have to define.

N. Maneesh

math.stackexchange.com/questions/3104032/… @anakhro

Mike Miller

@Rithaniel Here's a reformulation, I dunno if it's useful. Your relation is $(ab) a (ab)^{-1} = bab^{-1}$.

anakhro

@N.Maneesh what?

Mike Miller

01:38

You're trying to show, then, that $(ab) a (ab)^{-1}$ isn't $a$.

N. Maneesh

@anakhro Now am I correct? I have answered with the help of Jose sir's answer.

Rithaniel

Does that imply $ab=b$

Mike Miller

Nah.

Life's too hard for that.

Rithaniel

Darn

Leaky Nun

1 hour ago, by Rithaniel

Alright, so, given $L=\langle a,b,c\vert b^{-1}ab=c, ac=ca \rangle$. A homomorphism $f:L\rightarrow\mathbb{Z}$ given by $f(a)=f(c)=1$ and $f(b)=0$. Is this valid? If it is, one thing I think it demonstrates is that, in order for $\langle a\rangle$ to be normal, then $a=c$ must be true.

anakhro

01:40

@N.Maneesh what's "$A'$"?

Leaky Nun

Is this the question?

N. Maneesh

@anakhro Set of all limit points of $A$

Ted Shifrin

@Ultradark: No, of course not. Two great circles on the 2-sphere always intersect (twice).

anakhro

@N.Maneesh It's the same proof.

N. Maneesh

@anakhro He uses the method of contradiction right?

Rithaniel

01:42

Nah, the question is (in current form) Given $\langle a,b\vert (ab)a(ab)^{−1}=bab^{−1}\rangle$ is it true that $\langle a,b^{-1}ab\rangle$ is normal, but $\langle a\rangle$ is not? @LeakyNun

Leaky Nun

word problem is undecidable

Mike Miller

That's not helpful, and also seems like you don't know quite what that sentence means. The word problem is undecidable in arbitrary presentations, and there are specific presentations in which the word problem is undecidable.

I have no reason to believe the word problem is undecidable for this particular presentation.

anakhro

@N.Maneesh you do as well.

You just hid it with a curtain and didn't say anything about it.

Mike Miller

Actually, I just looked it up, and the word problem is decidable for 1-relator groups.

Leaky Nun

interesting

N. Maneesh

01:44

@anakhro I use the method of direct proof.

Mike Miller

Of course I would rather not find and run the algorithm here, I'd rather have a more elementary argument.

Rithaniel

(Hmmm, wanna look up 1-relator groups now)

Ultradark

@TedShifrin okay, so if those two great circles intersect at the north pole and south pole, then it's a singular knot after you map them to $\Bbb R^3$?

Mike Miller

@Rithaniel Here is the document I found.

anakhro

I would philosophically object to that being called "direct".

Mike Miller

01:45

Andy writes very well.

Ted Shifrin

It's not a knot.

Mike Miller

@anakhro I sent you an email not long ago.

Ultradark

*link

I meant to say link

Mike Miller

It's also not a link. "Singular link" and "singular knot" are things you will have to define.

Rithaniel

Yeah, gets right into it with what a one-relator group is. I don't immediately understand the explanation, but I definitely appreciate a direct approach.

N. Maneesh

01:46

@anakhro Why Here I don't assume the conclusion is wrong. I started with the proposition then reached the conclusion.

Mike Miller

@Rithaniel An element of the free group just means a word.

Ultradark

A singular knot is a smooth map $f:\Bbb S^1\to\Bbb R^3$ whose image has singularities.

Mike Miller

lol

Rithaniel

Okay, so this is groups which have a single "rule" for it's generators.

I suppose the term would be "a single relation"

anakhro

@N.Maneesh your proof is almost literally Jose's proof with "$x\in A$" appended to the beginning.

It's still as non-constructive.

Mike Miller

01:49

@Rithaniel I do like that you play with specific examples. This one seems like it's getting a bit out of hand. :)

anakhro

You just fail to mention the obvious contradiction.

Rumplestillskin

Hello one and all! Do we have any people interested in nonlinear DE's in attendance? Would love some advice!

Rithaniel

Indeed. I should probably try and state the specifics of the group I want to construct.

Ultradark

It feels good to define something

N. Maneesh

@anakhro What shall I add in the statement? so that proof is correct. Please help me.

Rithaniel

01:54

I want a group $L$ with with a subgroup $L'\unlhd L$ which itself has a subgroup $L''\unlhd L$, but where $L''$ is not normal in $L$.

anakhro

@N.Maneesh It's not that it's incorrect. It's that it's the same.

Mike Miller

@Rithaniel You mean $L''$ is normal in $L'$, right? That's a lot easier. :)

N. Maneesh

@anakhro Okay.

Mike Miller

For instance, what if $L'$ is abelian, but has nontrivial subgroups?

You'd have to rig it so those subgroups are not normal in the big group.

The dihedral group is your friend.

Leaky Nun

If $\langle a \rangle$ is normal then $b^{-1} a^n b = a$, then $a^n = bab^{-1}$, so $a^{n+2} = a^n$, so $a^2=1$

doesn't seem useful

Rithaniel

02:01

Yes, exactly, I was trying to ensure that specifically with that tangled group I invented. Though, hmmmm, dihedral group you say? Do I want odd or even rotations?
Actually, Leaky, if that's true, I could just require that $a^3=1$ and force $\langle a\rangle$ to not be normal. Though, I'm not sure. Shouldn't it be $b^{-1}a^nb=a^m$? It doesn't have to go to the generator after conjugation, necessarily, does it?

Ah, wait, I see, something has to go to the generator.

Mike Miller

The example I have in mind only works with $n$ even. It works for $D_4$.

Or $D_8$ if the subscript to you denotes the number of elements.

Rithaniel

I'll probably describe it as "dihedral group of the square" to avoid ambiguity.

anakhro

I typically just wipe the tears off my exam instead of asking which one the professor meant.

Mike Miller

"Let $D_8$ be the dihedral group so that the rest of this problem makes sense."

Leaky Nun

oh and if $n$ is even then $a=1$ (so the group becomes $\Bbb Z$); if $n$ is odd then $ab=ba$ (so the group becomes $C_2 \times \Bbb Z$)

so it suffices to surject the group to another group that cannot be a quotient of $C_2 \times \Bbb Z$

Rithaniel

02:07

So, $\langle r\rangle$ for the first level subgroup and $\langle r^2\rangle$ for the second? But isn't $r^{-1}=r^2$, thereby making $\langle r^2\rangle$ normal in $D_{4\text{ or maybe }8}$?

Mike Miller

That's not the pair I have in mind.

Leaky Nun

did you guys surject it to $D_{4,8}$?

anakhro

Heh, love that subscript, Rithaniel.

Mike Miller

@LeakyNun We moved on to a totally different question which motivated the original one.

That presentation is too hard for me.

Leaky Nun

14 mins ago, by Rithaniel

I want a group $L$ with with a subgroup $L'\unlhd L$ which itself has a subgroup $L''\unlhd L$, but where $L''$ is not normal in $L$.

this?

Rithaniel

02:09

Yeah, that's the one. (How do you quote messages?)

Mike Miller

Yea

Rithaniel

Also, danke @anakhro

Well, $\langle r^3\rangle=\langle r\rangle$ so those are normal, so it has to be a different first level.

$\langle r, sr\rangle$ doesn't work because then you just have the dihedral group again.

Maybe $\langle r^2, sr\rangle$?

Mike Miller

Why $sr$ instead of $s$?

Just curious.

Rithaniel

Testing out ideas, it's the first one I came to. I was actually running on the idea of dihedral group being the same as one generated by two elements of order two.

Mike Miller

Well, $s$ and $sr$ more or less do the same thing, so I'm not worried.

Rithaniel

02:17

Would be easier to work with $s$.

Okay, so the pair is $\langle r^2,s\rangle$ and $\langle s\rangle$?

Yeah, I believe that works.

Daminark

waves hello

Rithaniel

Heya Daminark

Mike Miller

@Rithaniel Yup! :)

In fact, that works in any $D_{2n}$ for $n > 2$ even.

Rithaniel

Excellent, thank you for all the help, Mike

Mike Miller

Sure thing, Nathaniel.

Daminark

02:28

Lmao

Rithaniel

Actually, it's Adam. Nathaniel is the middle name, which works, I suppose.

Oh wait, I said Mark

Oops

*Mike

Daminark

l m f a o

Alessandro Codenotti

lol that was great though

Daminark

Whoa Alessandro you changed

Alessandro Codenotti

Yeah yesterday. After almost 5 years it was time to

Fargle

02:31

Are you saying that one day Daminark won't be a thonk?

Daminark

:0

Alessandro Codenotti

As unlikely as it sounds

Mike Miller

@Rithaniel I'm astonished I got your middle name right.

Rithaniel

Well, I got the name Rithaniel when I was a kid from my middle name. You just reverse engineered it.

Skyler

02:59

is there a reason chatjax wont render this $$ \overline{x^2} = \frac{1}{n}\sum_{i=1}^n {xi}^2 \geq \frac{1}{n^2}(\sum{i=1}^n x_i)^2 = \overline{x}^2 $$

Dair

@Skyler It renders fine for me.

Skyler

huh, weird, it does it in this window but I have a markdown styled doc where it's bricked

see inertia7.com/projects/196

Mike Miller

Given a manifold with boundary $M$, the book I'm reading uses the notation $\sqsupset$ for the subset of $I \times M$ given by $I \times \partial M \cup \{1\} \times M$.

This is amazing notation.

Fargle

Holy crap, that's great.

Dair

@Skyler None of the latex on that page renders for me.

Skyler

03:01

I have a Note at the top for running the chatjax script, click the "link"

it works almost completely fine but there's one spot where it seems like it bricks

Daminark

Oh it's like a cup, that's clever

Dair

btw it should be $x_i^2$

Skyler

I already stripped out all the formatting in case there was something unsupported there...

@Dair thanks

ok, for sum reason it was the sum_{i=1} part causing most of the problems...

Dair

@Skyler Idk, I need to go, but can you not import the mathjax library and use that instead of asking the reader to use ChatJax? That way you could get errors in the webconsole.

And it would also be more convenient for the reader.

Skyler

@Dair They would still need to click the link right?

Dair

03:17

@Skyler No, there are ways to embed MathJax in the page so that no link is needed.

Take for instance math.stackexchange.com

Skyler

As in I can make some javascript which auto-executes MathJax without user interaction? I am not the site-admin btw

Dair

oh nvm, idk what inertia7 allows, I thought you could embed directly.

Skyler

I might be able to, though I'm not sure how I would embed MathJax in the first place

Skyler

03:34

one last test: $$ = \sum_{i=1}^n ( (x_i - \overline{x})^2 + 2 \overline{x} ( x_i - \overline{x) + \overline{x^2} ) $$

BAYMAX

Hi chat!!

Any method to solve for the system -

$F(x_{0},y_{0}) = 0$

and

$G(x_{0},y_{0}) = 0$

where F and G are non-linear

user616128

03:57

What are F and G?

Skyler

I think he's trying for the general case where all we know is the initial conditions are static state

actually idk

BAYMAX

04:31

F and G are continuous functions / nonlinear involving $x_{0},y_{0}$

Zee

05:08

No

Akiva Weinberger

06:37

@Rudi_Birnbaum

A more abstract but crazy beautiful one: here's a Penrose tiling and its Fourier transform.

Tweeted by aatishb on September 10, 2014 at 5:47 PM

Akiva Weinberger

06:48

(Also: a neat poster)

Sir Roger Penrose standing on his tiling

Rudi_Birnbaum

@AkivaWeinberger One then can ask why do we have ten-fold symmetry in the diffraction pattern while the tiling (apparently) is only 5-fold.

Akiva Weinberger

07:03

Because, despite having only one axis of 5-fold symmetry and no axis of 10-fold symmetry, if you rotate any finite patch by a tenth of a turn (36 deg) you can find the rotated version infinitely many times in the original

So, if you are given a finite patch, you can only know the orientation of the tiling modulo a tenth-turn (instead of the fifth-turn you'd expect from the axis of symmetry).

Rudi_Birnbaum

Right!

:-)

Akiva Weinberger

Oh you already knew the answer

Rudi_Birnbaum

No, but I just saw it.

Akiva Weinberger

Ah

Ah, yes, the Penrose tiling's natural habitat: math/science buildings in universities

Akiva Weinberger

07:39

@Rudi_Birnbaum In other words:

$user image$

Even if you had a map of the whole thing, your local surroundings wouldn't give you a clue as to where you are, and they wouldn't even tell you which way you're facing except mod 36.

(If that were a regular tiling, those bees wouldn't know where they were facing except mod 60)

ÍgjøgnumMeg

08:18

Morning!

Secret

Trying to think what nontrivial tiling can give all the same mod 1

Akiva Weinberger

09:36

@Secret You could use the idea used in this applet (the cut-and-project method)

gregegan.net/APPLETS/12/12.html

I think if you want 360-fold symmetry, you'd need 90 different types of rhombus

Every rhombus whose angles are an integer number of degrees

(with that algorithm, anyway)

If you want to see an actual lecture on aperiodic tilings (which is what got me looking at this stuff in the first place): https://youtu.be/a0wo_yAh0Ps

54min15sec

Eran

Let X be a set, if F is a sigma algebra generated by some collection of subsets of X, E, I know that countable union and complementation of sets from E is in the sigma algebra, BUT is it correct that every set in F is countable union/complementation/intersection of sets in E?

Secret

fail

Secret

09:57

@Eran I think arbitrary intersections of a family of sets that are arbitrary union of E can fall outside E though they are elements in F

Akiva Weinberger

10:10

@Eran Let X be the set of real numbers, and E be the set of open intervals. Show that ℚ, the set of rationals, is in F.

Eran

each $a\in \mathbb{Q} $ , $a = \bigcap_{n=1}^{\infty} (a-\frac{1}{n},a+\frac{1}{n})$, $\mathbb{Q}$ is countable so $ \mathbb{Q} = \bigcup_{a\in \mathbb{Q}} \bigcap_{n=1}^{\infty} (a-\frac{1}{n},a+\frac{1}{n} )$

Secret

That is why $F_{\sigma}$ and $G_{\delta}$ sets are such classics

Eran

Countable union of countable intersection intervals is in F

@AkivaWeinberger, am I correct?

Secret

I wonder how to prove the general case where the set $F,X,E$ are not specified...

somehow that means at least countable intersections must have triggered something that caused the falling out from $E$ in the first place...

It is easy to see why in the context of topology, but what if $F,X,E$ are not topologies?

Akiva Weinberger

10:32

@Eran Yes

So you do need two levels. You can't do it with just intersections or just unions.

Secret

how does intersection and union trigger the falling out in the general case?

Akiva Weinberger

Neat pics

Eran

10:49

@AkivaWeinberger Ok, assuming I can do countable number of levels, is it correct that every set in F can be written as countable union/intersection/complement countable times of any operation you want, of sets in E?

Secret

Akiva, do you know why combination of intersection and union can cause the resulting set to fall outside a given set?

Akiva Weinberger

@Eran I'm not actually sure… if you had a subset of $[0,1]$ that required one level, a subset of $[1,2]$, that required two levels, a subset of $[2,3]$ that required three levels, etc., and union'd them together, you'd get something that required infinitely many levels

and then if you intersected a bunch of things of that type, you could get a level higher

and go up the ordinal hierarchy

Eran

דם איק

oops

Akiva Weinberger

@Secret I dunno beyond "infinity is weird"

@Eran "Blood eek"?

Secret

hmm... ok

Eran

10:58

So what you're saying is that the sets in the sigma-algebra can't be defined by the sets of the collection generating this sigma-algebra

I assume that you can read hebrew (also by your name)

Not sure if you're israeli or not

Akiva Weinberger

Lo, ani amerikái, aval ani baáretz lashana

Ani lomed beyeshiva

Eran

Beyerushalaym?

Akiva Weinberger

Ken

Eran

I study Math and cs here in huji

Akiva Weinberger

Katamon

Eran

11:00

haha nice

Akiva Weinberger

Ah, I was there over the summer in Ulpan

Eran

In Givat Ram?

Akiva Weinberger

Mount Scopus campus

Wikipedia has a construction involving the ordinal hierarchy, but the simplest discription is probably just "the smallest sigma-algebra containing these things"

"Take all the sigma-algebras that contain these sets nd intersect them"

Borel set

In mathematics, a Borel set is any set in a topological space that can be formed from open sets (or, equivalently, from closed sets) through the operations of countable union, countable intersection, and relative complement. Borel sets are named after Émile Borel. For a topological space X, the collection of all Borel sets on X forms a σ-algebra, known as the Borel algebra or Borel σ-algebra. The Borel algebra on X is the smallest σ-algebra containing all open sets (or, equivalently, all closed sets). Borel sets are important in measure theory, since any measure defined on the open sets of a...

Eran

yup I know that's the definition

Akiva Weinberger

The ordinal hierarchy, for the record, is essentially "What if we decided to count past infinity"

So it starts $0$, $1$, $2$, etc

$0,1,2,3,\dots,\omega,\omega+1,\omega+2,\dots,\omega+\omega$

${}=\omega\cdot2,\omega\cdot2+1,\dots, \omega\cdot3,\dots$

Eran

11:05

Okay, thanks

Akiva Weinberger

But yeah in the "Generating the Borel algebra" section of that Wikipedia article it mentions those

By the way

Fibonacci

$user image$

The red tile has size $1$, the purple(?) tile has size $\varphi$, the next one has size $1+\varphi=\varphi^2$, and inductively you can continue that as far as you want

and the number of tiles in each thing trace the Fibonacci numbers

Rudi_Birnbaum

11:31

@AkivaWeinberger lol!

@ÍgjøgnumMeg God morgon!

@AkivaWeinberger: Do you know any program that could fourier-transform jpg images?

ÍgjøgnumMeg

@Rudi hej rudi :)

Perturbative

Hey everyone!

Hey @ÍgjøgnumMeg :)

ÍgjøgnumMeg

Hullo

Perturbative

Hope things are going well on your side!

Alessandro Codenotti

The first $\omega_1$ levels of the Borel hierarchy are all distinct for Polish spaces, I don't know in general

ÍgjøgnumMeg

11:40

Living the dream

Perturbative

Silly question, the statement of the dual universal coeffcients theorem is the following above, and if we have $n = 0$, then $H_{n-1}(X) = \emptyset$ (if I recall correctly) which isn't a group so $\operatorname{Ext}(H_{n-1}(X), A)$ isn't even defined

Ohh wait maybe $H_{-1}(X) = 0$

Or maybe $H_{-1}(X)$ is undefined

Tobias Kildetoft

@Perturbative chanin complexes are always infinite on both sides. We just fill them with $0$'s to make them appear finite, so all (co)homology is always defined.

Alessandro Codenotti

Assume that there are zero modules in you chain complex in negative dimensions

Damn I got sniped

Perturbative

Ahh okay that makes sense

I don't ever think about the negative ones so I must've forgot that :p

Alessandro Codenotti

You can have homology theories with negative homology modules as well sometimes

Perturbative

11:56

Hmm, none of the topological ones I've seen so far (singular, simplicial, cellular) have had non-zero negative homology modules (though I think there's a theorem that states they are all the same homology theory essentially)

Alessandro Codenotti

For CW-complexes

Perturbative

Oh derp, well I need to look at cellular again

Tobias Kildetoft

I assume spectra also have those (seeing as they are essentially like CW complexes which allow things in negative degrees).

Akiva Weinberger

12:21

@Rudi_Birnbaum Not that I know of

Secret

12:56

yesterday, by Akiva Weinberger

Even the identity matrix doesn't work normally

I acutally knew what $$\frac{d ♥}{dt}$$ is

the answer lies in $\mathcal{C}_0$

Rithaniel

13:20

I never have been able to get mathjax to work on this phone.

Nick

13:57

@Rithaniel if your phone supports javascript, all you actually need to run mathjax is a link to the code that will execute on click.

Akiva Weinberger

14:07

4 hours ago, by Akiva Weinberger

If you want to see an actual lecture on aperiodic tilings (which is what got me looking at this stuff in the first place): https://youtu.be/a0wo_yAh0Ps

Transcript for

$Mathematics$ Mathematics