Mathematics - 2012-07-18 (page 1 of 3)

Jonas Teuwen

00:00

It would certainly error there 8-). And then I would have to do the same... but then add it.

robjohn

@JonasTeuwen Now that would not work.

Jonas Teuwen

@robjohn I use a repository (bitbucket) to store my LaTeX files, how do you do this? Bitbucket has free private repositories if you work for an academic institution.

robjohn

@JonasTeuwen I keep them on old fashioned local storage. I am a dinosaur and my feet sink through when I try to use the cloud.

Jonas Teuwen

@robjohn A dinosaur... So cvs?

@robjohn Repositories are not clouds :-).

robjohn

@JonasTeuwen No, just files on a hard disk.

Jonas Teuwen

00:06

@robjohn That is like so retro. I also want to be more retro -> no phone.

robjohn

@JonasTeuwen bitbucket looks pretty cloudlike.

Jonas Teuwen

@robjohn Hmm... A cloud is to me some distributed system. Bitbucket is just a repository server.

So you do stuff like hg commit then you write down what you have changed and then hg push and it stores the new version (plus the differences with the old ones) so you can just delete as you please and still get it back.

Or fork it, make several versions etc. The concept is way older than the modern clouds!

I use the cloud for something else: automatic synchronisation between computers! 8-).

robjohn

@JonasTeuwen but not local. I view the cloud as remote storage that you can access from any location.

Jonas Teuwen

@robjohn You can also have a local repository... hg init :-).

(and it has GUI clients of course, but I don't like that)

robjohn

@JonasTeuwen I use mercurial for work, but for my latex files, I just use a folder.

Jonas Teuwen

00:09

The cool thing about remote repositories is that you automatically have a backup.

Oh right, so you are fossil... at home.

I don't like to make backups, so I do it this way.

I use mercurial for LaTeX and git for larger source code projects.

Not sure why... Historically? I've seen that mercurial is easier to work with and I figured that it would have the "common" features and I work alone usually on my .tex as mathematicians usually don't use repositories. :-).

Peter Tamaroff

@MarianoSuárez-Alvarez Mariano, can I ask you for an hint on an excersice?

I have to prove this:

> Let $(X_i,d_i)$ be metric spaces for $i=1,\dots,n$ and convert $X=\prod_{i=1}^n X_i$ into a metric space $ (X,d)$ by setting $d(x,y)=\max\limits_{1\leq i\leq n}\{d_i(x_i,y_i)\}$. Let $O_i\subset X_i$ be open for each $i$. Prove that $O=\prod_{i=1}^n O_i$ is an open subset of $X$ and that each open subset of $X$ is a union of sets of this form.

It seems to be a generaliation of this one

> Let $(X_i,d_i)$ be metric spaces for $i=1,\dots,n$ and convert $X=\prod_{i=1}^n X_i$ into a metric space $ (X,d)$ by setting $d(x,y)=\max\limits_{1\leq i\leq n}\{d_i(x_i,y_i)\}$. Prove that an open ball in $X$ is the product of open balls from $X_1,\dots,X_n$ respectively

which I already proved

> , and that if $\mathfrak B_{a_i}$ is a basis for the nbhd system at each $a_i\in X_i$ then the collection of all sets of the form $B_1\times\cdots \times B_n$ with each $B_i\in \mathfrak B_{a_i}$ is a basis for the nbhd system at $a=(a_1,\dots,a_n)\in X$.

anon

every open set is a union of balls, so it should suffice to show the elementary-set-theoretic fact that unions distribute through products

Isaac Solomon

Hey, does anyone know of a good resource for r\eading about fibred knots?

anon

for example, $(A\cup B)\times C=(A\times C)\cup (B\times C)$

robjohn

00:29

@anon That's true, but I don't know what you are trying to show.

Peter Tamaroff

@anon But the union might not be over the same indexing set.

anon

$$\large \prod_{i=1}^n O_i=\prod_{i=1}^n \bigcup_{j\in J_{\large i}} B_{i,j}=\bigcup_{(j_{\large i})\in \large\prod\limits_{i=1}^nJ_i}~\underbrace{\prod_{i=1}^n B_{i,j_i}}_{\rm ballz} $$

Peter Tamaroff

It is true that we can take the points of each $O_i$ and each $B(a_i;\delta_{a_i})$ and produce $$O_i=\bigcup_{a_i\in O_i}B(a_i;\delta_{a_i})$$

@anon What is $J_i$?

anon

The indexing set in the union describing $O_i$

You could also work with induction if all those indices appears unseemly.

Peter Tamaroff

@anon Let's consense this notation: The ball around $a$ of radius $r$ is $B(a;r)$

anon

00:37

The centers and radii are totally superfluous so I left them out.

That's how you condense notation ;)

Henry T. Horton

So the tail end of a sequence walks into a $B(a;r)$...

Peter Tamaroff

@anon What should $B_{i,j}$ read as, then?

@HenryT.Horton Hahaha.

anon

Each $O_i$ is an open set, and as such a union of balls - we index these balls with $J_i$, so $O_i=\bigcup_{j\in J_i}B_{i,j}$

Peter Tamaroff

@anon OK. Just the notation was messy. =P

Henry T. Horton

That's a lot of balls!

anon

00:42

But again, you might prefer induction. Suppose it's true up to $n-1$. Then we can decompose $$\prod_{i=1}^nO_i=\left(\prod_{i=1}^{n-1}O_i\right)\times O_n=\left(\bigcup_{\ell\in L}\tilde{B}_\ell\right)\times\bigcup_{k\in K}B_k=\bigcup_{(\ell,k)\in L\times K} \tilde{B}_\ell\times B_k$$

Here, since $\prod_{i=1}^{n-1} O_i$ is open (induction hypothesis), it can be written as a union of open balls in $\prod_{i=1}^{n-1}X_i$, which we designate with tildes overhead and index with $L$. In the $n$th component, $O_n$ is a union of open balls in $X_n$, which we denote with just $B$ and index with $K$.

mothafuckin ballz yo

The other direction, that every open set is a union of products of this form is easy: every open set is a union of balls, every ball a product of component balls, and component balls are open in their respective spaces so there we have it.

Peter Tamaroff

@anon Component balls?

anon

balls in the respective component metric spaces $X_i$, $i=1,2,\cdots n$.

Peter Tamaroff

You mean the balls of each $X_i$ right?

Yep.

@anon Question on notation. Why do you change the $B_{i,j}$ to $B_{i,\color{red}{j_i}}$?

Can't we talk bout $j\in K$ where $K=\prod_{i=1}^n J_i$?

@anon Wait, but now $K$ is a cartesian product, so each of its elements is an ordered $n$-tuple. How are we indexing the balls in this case?

anon

01:00

I did say $(j_i)\in\prod\limits_{i=1}^n J_i$. However, it makes sense to write $\prod_{i=1}^n B_{i,j_i}=B_{1,j_1}\times \cdots B_{n,j_n}$, whereas it does not make sense to write $B_{i,(j_1,\cdots,j_n)}$ for fixed $i$. On the LHS I didn't bother indexing the $j$'s with $i$th's because it was unnecessary to get the point across IMO.

Think of $n$ bins, labelled $i=1,2,\cdots,n$, each filled with balls and open sets. We can glue balls together (union) within the bins to form bigger open sets, or we can take open sets from the bins and place them on the floor side-by-side in sequence, and call the result the product of the sets. We can do this multiple times and then glue (union) open sets that exist on the floor (the product of the metric spaces).

Peter Tamaroff

@anon Of course the second doesn't make sense. The notation confuses me, that's all.

@anon Does $(j_i)$ mean something special? (The parenthesis)

anon

Yes, it stands for $(j_1,\cdots,j_n)$.

(Standard shorthand..)

Peter Tamaroff

Bah. I know what to prove, but I need to get things together.

Henry T. Horton

Pull yourself together man

You can do it

anon

The equation I wrote says that gluing balls together in the bins and then combining the open sets $O_i$ formed and putting the result $\prod O_i$ on the floor space, is equivalent to taking every sequence $Y_1,\cdots Y_n$ of balls, where $Y_i$ is some ball from the $i$th bin was used in forming the open set $Y_i$, and combining them, putting the results on the floor space, and then unioning these together.

That's the best I think I can explain it metaphorically atm.

But an easier tack for you might just be the induction route.

Peter Tamaroff

01:10

@anon I understand what you're saying (the last explanation si not necessary) but I need to just get the right words and theorems which I already proved together.

@anon What do you call "floor" space?

anon

see the first part of the metaphor: on the floor (the product of the metric spaces)

Peter Tamaroff

OK. This is my train of thought so far:
$(a)$ Every open ball in $X$ is the product of open balls of the $X_i$.
$(b)$ Every open subset $O_i$ is the union of open balls in $X_i$.
$(c)$ $O$ is the product of the union of open balls in each $X_i$, so $O$ is the union of the product of open balls in each $X_i$, which are open balls in $X$.
$(d)$ Thus, $O$ is the union of open balls of $X$, so it is an open subset of $X$.
$(e)$ Freaking put all of the above in a proof.

anon

Yup. The only part that needs filling in is (c) of course.

Peter Tamaroff

@anon What do you mean by filling?

anon

expanding / elaborating further on

Peter Tamaroff

01:24

@anon On "the product of the union" versus "the union of the product"? What is missing?

anon

You mean within the bullet point (c)? Do you already have available to you that a product of unions is a union of products (of all the underlying sets in their correct arrangements)? If not, you need to prove it. Or pretend it's so trivial to you that you don't need to prove it, see if the grader takes the bait..

Peter Tamaroff

@anon Hahaha... wait, grader?

@anon I have that from Halmos I think.

anon

This isn't homework? Then what are you getting all worked up for?

I work out like 1/10 of the exercises I do on actual paper.

Peter Tamaroff

@anon I'm studying on my own. Surprise: I take it seriously! =D

I know you also do .

anon

Then again it's not really conducive towards studying and review..

Peter Tamaroff

01:27

@anon Could you paraphrase that?

anon

Working out very few exercises out explicitly does not help much when one needs to study it some time afterward and review the material.

Peter Tamaroff

@anon Oh, I try to work out most of the exercises. What are you referring to about now?

BenjaLim

@PeterTamaroff What are you trying to prove?

anon

Myself.

BenjaLim

Interesting question: math.stackexchange.com/questions/172172/…

Peter Tamaroff

01:28

@BenjaLim I'm not trying, I'm proving it. =D

BenjaLim

what is it

anon

refer back to chat.stackexchange.com/transcript/message/5404894#5404894

Peter Tamaroff

1 hour ago, by Peter Tamaroff

> Let $(X_i,d_i)$ be metric spaces for $i=1,\dots,n$ and convert $X=\prod_{i=1}^n X_i$ into a metric space $ (X,d)$ by setting $d(x,y)=\max\limits_{1\leq i\leq n}\{d_i(x_i,y_i)\}$. Let $O_i\subset X_i$ be open for each $i$. Prove that $O=\prod_{i=1}^n O_i$ is an open subset of $X$ and that each open subset of $X$ is a union of sets of this form.

I got the decent amount of 175 rep today. Previous days had been pretty "dry".

BenjaLim

By definition of the product topology you have a basis consisting of sets of the form $\prod O_i$ where each $O_i$ is open in $X_i$

Peter Tamaroff

@BenjaLim REMINDER: This is metric spaces!

BenjaLim

01:31

@PeterTamaroff The last part of the problem follows by showing that the collection of all such $O$'s is a basis

anon

his definition apparently has open sets in the product topology formed from unions of balls in the product topology, and a previous exercise showed that balls in the product topology are products of balls in the direct factors

BenjaLim

@anon?

Peter Tamaroff

Mendelson adds: "A collection of sets of a metric space is usually called a basis for the open sets if each open set is a union of set in this collection. For example, the open balls in a metric form a basis for the open sets"

anon

A question does not a single question mark make.

Henry T. Horton

@PeterTamaroff It's dry season for me... I got a bounty awarded today though

Peter Tamaroff

01:33

@BenjaLim If each of the $X_i$ are metric spaces set $X=\prod X_i$ with $d=\max d_i$

BenjaLim

@PeterTamaroff what mendelson is saying is exactly what I'm saying

Peter Tamaroff

Then an open ball in $X$ is the product of open balls in each $X_i$.

BenjaLim

@anon Now my head is getting mixed up between direct product and disjoint union......

@PeterTamaroff Ok take $x \in \prod O_i$

can write $x = (x_1,\ldots,x_n)$ with each $x_i \in O_i$

@PeterTamaroff I don't understand have you proven this yet?

@anon

Peter Tamaroff

@BenjaLim Yes, I have proven that already.

anon

what is a regular orbit of a group action?

Peter Tamaroff

01:41

I have to prove what I quoted there.

BenjaLim

Why not like this:

Peter Tamaroff

In White.

BenjaLim

did it go something like this:

anon

An orbit that the action acts regularly on I guess, nevermind.

Peter Tamaroff

@anon Question.

BenjaLim

01:42

Take $x \in \prod O_i$, then we can write $x = (x_1,\ldots x_n)$ with $x_i \in O_i$. By definition of each $O_i$ being open in $X_i$ there exists $\epsilon_i > 0$ such that $B^{d_i}_{\epsilon_i}(x_i) \subseteq O_i$

anon

Answer.

BenjaLim

now take $\epsilon = \max \epsilon_i$ and you have $B^d_\epsilon(x) \subseteq \prod O_i$ @PeterTamaroff

Peter Tamaroff

@anon Let each set $A_{i,j}$ be indexed by $I_i$, and set $M_i=\bigcup_{j \in I_j}A_{i,j}$

anon

okay

BenjaLim

@PeterTamaroff did you do what I did above?

Peter Tamaroff

01:44

Then how did you express $$\prod_{i=1}^n M_i$$?

Typo in the previous one, it should read $j\in I_i$.

anon

Change the letter M to O, and A to B, and then refer back to the first equation I gave!

Peter Tamaroff

@BenjaLim Let me check.

@anon I mean, where can I find reference of that.

BenjaLim

@PeterTamaroff: Let $X$ be a topological space. Suppose that $\mathcal{C}$ is a collection of open sets of $X$ such that for each open set $U$ of $X$ and each $x \in U$, there is $C \in \mathcal{C}$ such that $x \in C \subset U$. Then $\mathcal{C}$ is a basis for the topology of $X$.

Your topology now on the product space is already completely determined by the metric $d$ so we just need to use the lemma

Peter Tamaroff

@BenjaLim I don't know what a freaking topological space is, pal.

anon

good. let's move on to locally compact abelian topological groups.

BenjaLim

01:47

@anon no we need to start talking about moduli stacks and etale cohomology

@PeterTamaroff Let's kill your problem now

I claim we can do this problem entirely without a metric

you know what a basis is yes?

Peter Tamaroff

@BenjaLim A basis for the system of neighborhoods at $a\in X$, in a metric space $(X,d)$

BenjaLim

ok.

maybe not then...

hmmmm

Peter Tamaroff

@anon I just down want to copy what you wrote there. I need to understand.

BenjaLim

@PeterTamaroff What are you and @anon going on about?

anon

the big latex equations I gave

Peter Tamaroff

01:50

@BenjaLim Some rule to write $\prod \bigcup$ in terms of $\bigcup \prod$

BenjaLim

ok

but your problem on proving it is a basis

have you done it?

I think I have it

anon

Do you want an intuitive understanding of it first, or do you feel you already have that? Try to distribute $$(A_1\cup A_2 \cup A_2)\times (B_1\times B_2)\times (C_{cat}\times C_{dog}) $$ as an exercise..

BenjaLim

@anon why not?

Peter Tamaroff

2 hours ago, by Peter Tamaroff

> Let $(X_i,d_i)$ be metric spaces for $i=1,\dots,n$ and convert $X=\prod_{i=1}^n X_i$ into a metric space $ (X,d)$ by setting $d(x,y)=\max\limits_{1\leq i\leq n}\{d_i(x_i,y_i)\}$. Let $O_i\subset X_i$ be open for each $i$. Prove that $O=\prod_{i=1}^n O_i$ is an open subset of $X$ and that each open subset of $X$ is a union of sets of this form.

A collection of sets of a metric space is usually called a basis for the open sets if each open set is a union of set in this collection. For example, the open balls in a metric form a basis for the open sets

BenjaLim

@PeterTamaroff Here's the proof:

Let $V$ be open in the product space.

Henry T. Horton

01:53

That was 2 hours ago!?

BenjaLim

take $x \in V$

Henry T. Horton

I've spent a long time staring at the wall

anon

Is it a beautiful wall?

BenjaLim

Then you know that given any $x = (x_i) \in V$, there exists $\epsilon > 0$ such that $B_\epsilon^d (x) \subset V$

Henry T. Horton

No

BenjaLim

01:55

@PeterTamaroff Now consider $\prod_{i=1}^n B_{\epsilon}^{d_i} (x_i)$

I claim that $\prod_{i=1}^n B_\epsilon^{d_i}(x_i) \subset B_\epsilon(x)$ @PeterTamaroff

Peter Tamaroff

@BenjaLim That is basically what I proved before.

The a ball in $X$ is the product of balls in $X_i$.

BenjaLim

not exactly.

My proof is much easier it's just showing containment and then invoking a lemma :D

completing the problem :D

Peter Tamaroff

@BenjaLim OK, move on.

BenjaLim

12 mins ago, by BenjaLim

@PeterTamaroff: Let $X$ be a topological space. Suppose that $\mathcal{C}$ is a collection of open sets of $X$ such that for each open set $U$ of $X$ and each $x \in U$, there is $C \in \mathcal{C}$ such that $x \in C \subset U$. Then $\mathcal{C}$ is a basis for the topology of $X$.

Peter Tamaroff

@BenjaLim I don't know about topological spaces. Why do you insist?

J. M.

01:59

"Because I don't have a latex editor. I'm an amateur. Mathematics is just my hobby." - that's his excuse? Now I'm pissed...

BenjaLim

@PeterTamaroff let me tell you something

you don't need to know what a topological space is

just think about it for now as a space with open sets

Peter Tamaroff

@J.M. At whom?

BenjaLim

A topology on $X$ pretty much means just all the open sets

@PeterTamaroff ARE YOU PAYING ATTENTION IN CLASS?

Peter Tamaroff

@BenjaLim Yes, yes.

anon

well this is awkward

Peter Tamaroff

02:00

@anon LOL

J. M.

@PeterTamaroff That guy who edits a lot. That was the excuse he gave for his behavior. The hell?

BenjaLim

@PeterTamaroff Now explain to me why $\prod B^{d_i}_\epsilon(x_i) \subset B_\epsilon^d(x)$

Peter Tamaroff

@BenjaLim Because $d=\max d_i$ =D

J. M.

@anon It's risky being your namesake... :)

BenjaLim

Ok and in the guy on the left we just have one $\epsilon$

the epsilon on the left is the same as that on the right

@PeterTamaroff Let me put it

correctly

you want to show given any $y$ on the left that it is on the right too

showing it is in the right is the same thing as saying that $d(y,x) < \epsilon$, or that

Peter Tamaroff

02:02

That $d <\epsilon$ mean that $\max d_i <\epsilon$, so that each $d_i$ is less than $\epsilon$ too.

BenjaLim

$\max d_i(x_i,y_i) < \epsilon$

J. M.

@PeterTamaroff See here.

BenjaLim

but this is clear because if $y$ is a priori on the left

we have that $d_i(y_i,x_i) < \epsilon $ for all $i$

and so the maximum is less than epsilon.

Peter Tamaroff

@J.M. Just "his hobby"? Why the hell is he proving theorems by Weil?

BenjaLim

@PeterTamaroff and that concludes your problem.

Peter Tamaroff

02:03

@BenjaLim Yep.

Henry T. Horton

He does have a LaTeX editor, math.SE is his LaTeX editor

2

Peter Tamaroff

I knew that already. But I don't know what you proved.

anon

it's like he's never heard of paper

BenjaLim

@PeterTamaroff huh?

J. M.

@PeterTamaroff That's not the point. I'm a hobbyist myself, but I don't seek out to be disruptive...

Peter Tamaroff

02:04

@BenjaLim I can't use your fancy nancy lemma.

BenjaLim

@PeterTamaroff I am just saying there is no need to talk about how to express this as some union and that.

Peter Tamaroff

@BenjaLim See this

Henry T. Horton

I use IE as a browser. – Makoto Kato 1 hour ago

anon

@HenryT.Horton I had no feeling about the matter until this.

BenjaLim

@PeterTamaroff see what?

Peter Tamaroff

02:05

43 mins ago, by Peter Tamaroff

OK. This is my train of thought so far:
$(a)$ Every open ball in $X$ is the product of open balls of the $X_i$.
$(b)$ Every open subset $O_i$ is the union of open balls in $X_i$.
$(c)$ $O$ is the product of the union of open balls in each $X_i$, so $O$ is the union of the product of open balls in each $X_i$, which are open balls in $X$.
$(d)$ Thus, $O$ is the union of open balls of $X$, so it is an open subset of $X$.
$(e)$ Freaking put all of the above in a proof.

BenjaLim

@HenryT.Horton that guy should be on the front page of troll science comics

@PeterTamaroff my god it's so complicated....

Peter Tamaroff

@BenjaLim What is complicated?

BenjaLim

@PeterTamaroff I am saying that the problem is not so bad. Sometimes there is the over tendency to split too many hairs.

Henry T. Horton

@BenjaLim Peter is bald

Peter Tamaroff

@BenjaLim I know is not complicated, in fact I'm happy with that train of thought, so I just want to put that in a good proof.

BenjaLim

02:08

@PeterTamaroff yes. But remember being succinct is also good.

Peter Tamaroff

@BenjaLim Well, is that train of thought there succint enough?

BenjaLim

huhuuhhuh?

Peter Tamaroff

@HenryT.Horton HAHAHAHHA. Is it too trivial each subset of the discrete metric is open?

BenjaLim

@PeterTamaroff how do you know (a)?

Peter Tamaroff

@BenjaLim I proved it before

BenjaLim

02:10

where

Peter Tamaroff

Basically by a very similar argument of yours.

Well, not that similar.

BenjaLim

I told you mine was a lot shorter I just a bunch of containments

Peter Tamaroff

@BenjaLim Let me show you:

Let $B_i=B(a_i;\delta_i)$ be open balls about $a_i\in X_i$.

Then we have that

BenjaLim

type it up I'm going for a shower bbl

Peter Tamaroff

$$B=\prod_{i=1}^n B_i=\left\{ x\in X:\begin{cases} d_1(a_1,x_1)<\delta_1 \cr \cdots\cr d_1(a_n,x_n)<\delta_n \end{cases}\right\}$$

@BenjaLim Wait

Come on, it is not that long.

Henry T. Horton

02:14

He's dead, bro

Peter Tamaroff

Conversely, a ball $B(a;\delta)$ about $a\in X$ is $=\{x\in X:\max \{d_i(x_i,a_i)\}<\delta\}$

Choosing each $\delta_n=\delta$ we see that $B=B(a;\delta)$. So we can write $B(a;\delta)$ as a product of open balls in each $X_i$.

I'll go too. BBL.

BenjaLim

02:31

god

user19161

@HenryT.Horton I don't think so, bro.

Henry T. Horton

Hello

user19161

WHy do I get so much spam these days?

user19161

@HenryT.Horton Hope your eye is OK now.

Peter Tamaroff

02:56

@BenjaLim Vishnu

Henry T. Horton

Does someone here know a thing or two about Lie algebras?

Oh god, I think I figured it out

Peter Tamaroff

@anon $$\left( {{A_1} \times {B_1} \times {B_2} \times {C_1} \times {C_2}} \right) \cup \left( {{A_2} \times {B_1} \times {B_2} \times {C_1} \times {C_2}} \right) \cup \left( {{A_3} \times {B_1} \times {B_2} \times {C_1} \times {C_2}} \right) = $$

anon

no

Peter Tamaroff

@anon FUCK

anon

Try $(A_1\cup A_2)\times(B_1\cup B_2)$ first.

Henry T. Horton

03:04

@PeterTamaroff Reported for language.

Peter Tamaroff

I just used $(A\cup B)\times C=(A\times C)\cup (B\times C)$

anon

@PeterTamaroff Unless you mean $$(A_1\times (B_1\cup B_2)\times(C_1\cup C_2))\times(A_2\times (B_1\cup B_2)\times(C_1\cup C_2))\times(A_3\times (B_1\cup B_2)\times(C_1\cup C_2)),$$ in which case you've only just started!

Perhaps this would be easier: what's $(a_1+a_2)(b_1+b_2)$ ;)

Henry T. Horton

$\cup \to \subset \to \cap \to \supset$

anon

"barrel" shall henceforth refer to all containments, unions and intersections

(barrel roll)

Peter Tamaroff

@anon OK, I get

$$ = \left( {{A_1} \times {B_1}} \right) \cup \left( {{A_1} \times {B_2}} \right) \cup \left( {{A_2} \times {B_1}} \right) \cup \left( {{A_2} \times {B_2}} \right)$$

anon

03:10

yup. notice the index pairings: (1,1), (1,2), (2,1), (2,2)

you don't have to do the bigger one, too much writing

Peter Tamaroff

@anon OK.

@HenryT.Horton ¿¿¿¿¿?????

Henry T. Horton

$\cup^\ast \leftarrow \subset^\ast \leftarrow \cap^\ast \leftarrow \supset^\ast$

anon

$\vee\leftrightarrow<\leftrightarrow\wedge\leftrightarrow>$

Peter Tamaroff

@anon SO $$\bigcup\limits_{i = 1}^2 {{A_i}} \times \bigcup\limits_{i = 1}^2 {{B_i}} = \bigcup\limits_{\left( {i,j} \right) \in \left\{ {1,2} \right\} \times \left\{ {1,2} \right\}} {{A_i} \times {B_j}} $$

anon

mmhmm

Peter Tamaroff

03:16

@anon But how do I make it clear the sequence should be $1,1$-$1,2$-$2,1$-$2,2$?

anon

sequence?

order doesn't matter..

CLarue

The point is every possibility is covered

anon

no, $U\cup V= V\cup U$

Peter Tamaroff

Oh, right.

anon

you can rearrange the pairs as (2,1),(1,1),(1,2),(2,2) or whatever you want, every pair will be in the cartesian product of index sets

CLarue

03:18

In fact, I may surmise you can permute within the tuples in a consistent manner.

Peter Tamaroff

So I can grow a pair and say $$\bigcup\limits_{i \in I} {{A_i}} \times \bigcup\limits_{j \in J} {{B_j}} = \bigcup\limits_{\left( {i,j} \right) \in I \times J} {{A_i} \times {B_j}} $$

anon

yes

Peter Tamaroff

@anon OK.

@anon See, little by little.

CLarue

Given index sets $I_0$ through $I_\alpha$, and sets $A_{\beta,i}$ for $i\in I_\beta$, $$\prod_{i\leq\alpha}\bigcup_{k\in I_k}A_{i,k}=\bigcup_{p\in\prod I_\beta}\prod_{i\leq\alpha}A_{i,p(i)}$$

Peter Tamaroff

@anon So now $$\bigcup\limits_{{i_1} \in {I_1}} {{A_{1,{i_1}}}} \times \bigcup\limits_{{i_2} \in {I_2}} {{A_{2,{i_2}}}} \times \cdots \times \bigcup\limits_{{i_n} \in {I_n}} {{A_{n,{i_n}}}} = \bigcup\limits_{\left( {i,j} \right) \in \prod\limits_{m = 1}^n {{I_m}} } {\left( {{A_{1,{i_1}}} \times {A_{2,{i_2}}} \times \cdots \times {A_{n,{i_n}}}} \right)} $$

or shortly

anon

03:26

@CLarue already covered

Peter Tamaroff

$$\prod\limits_{m = 1}^n {\bigcup\limits_{i \in {I_n}} {{A_{n,i}}} } = \bigcup\limits_{\left( i \right) \in \prod\limits_{m = 1}^n {{I_m}} } {\prod\limits_{m = 1}^n {{A_{n,i}}} } $$

CLarue

for infinite $\alpha$?

Peter Tamaroff

@CLarue $p(i)$ is what? The $n$th coordinate of $p$?

anon

same diff

CLarue

;)

anon

03:27

@PeterTamaroff yes

CLarue

Arbitrary products are sets of choice functions

Peter Tamaroff

@CLarue Yes, I know.

CLarue

so p is a choice function

anon

@PeterTamaroff and your indexing is a bit weird; you want $A_{m,i_m}$

Peter Tamaroff

@CLarue Though I don't really grasp it yet.

@anon Yes, that is something I need to get used to.

CLarue

03:28

It just returns some element in the set it should like in coordinate system

$p(i)\in I_i$

But what is it you think you don't understand about choice functions?

Alex Becker

@anon If you're not too busy, I have a quick topology question.

Peter Tamaroff

@CLarue I know given a arbitrary family of indexed sets $X_i$, then the arbitrary product $\prod_{i\in I}X_i$ is the set of all function $x:I\to X_i$ such that $x(i)=x_i \in X_i$.

@CLarue In general, I know little about set theory. I never took a course on set theory whatsever.

CLarue

$x:I\rightarrow\bigcup X_i$

anon

@AlexBecker doubt I'd be of help. what's up?

Peter Tamaroff

@CLarue Yes, of course, sorry.

CLarue

03:33

It's a function that, when given a direction, picks a point in space along that direction.

Alex Becker

@anon Is the orientable surface of genus g with a single puncture homotopic to a figure eight?

Zhen Lin

@AlexBecker That seems unlikely if $g \ne 1$.

anon

no idea

Peter Tamaroff

@CLarue Mendelson reccommends Halmos Naive Set Theory from which I read through sectinons 1-8 and I'm a little stuck on section 9 which is families.

Alex Becker

@ZhenLin My reasoning is that you can consider a fundamental polygon with 4g sides, remove the center, and identify the opposite sides to get something homeomorphic to the genus g surface with a single puncture.

Zhen Lin

03:35

Yes.

It's not quite opposite sides.

Alex Becker

@ZhenLin That might be the issue

@ZhenLin Let me fire up wikipedia.

Zhen Lin

It should be $A_1 B_1 {A_1}^{-1} {B_1}^{-1} \cdots$

Henry T. Horton

A torus with $n$ punctures is homotopy equivalent to a wedge of $n + 1$ circles

Not sure about higher genus

Alex Becker

@ZhenLin Wait, but doesn't the octagon with opposite sides identified give the orientable surface of genus 2?

Henry T. Horton

Yes, puncture it and retract onto the edges and count the circles

Zhen Lin

03:41

I dunno. The standard construction uses the identification I mention.

anon

"give you the general case of a mathematical technique when you only need it for a specific usage"

6

CLarue

@PeterTamaroff From the little that I know, a family of sets is simply a way to notate something that would otherwise confuse. Of course, I assume deeper matters are covered in that section than defining families.

Peter Tamaroff

@anon HAHAHAHHAHA

Henry T. Horton

@AlexBecker I can't use my math hand currently to verify, but it seems that $\Sigma_g \setminus \{p\}$ will have the homotopy type of a wedge of $2g$ circles

Zhen Lin

That's what I guess too.

Alex Becker

03:47

@HenryT.Horton That makes sense to me. I think my problem was that the surfaces I'm looking at are not actually $\Sigma_g$ for $g> 2$, and I simply prematurely extrapolated from the cases of $g=1,2$ which I actually drew.

Henry T. Horton

I have a severe problem with premature extrapolation myself.

3

Alex Becker

Yeah. It wouldn't be so bad if piecing together higher-genus surfaces from fundamental polygons weren't such a pain.

Alright, after some thinking the surface I'm really interested in (identifying opposite sides of a $2n$-gon with punctures at the center and vertices) seems to be homotopy equivalent to a circle with $n$ loops attached to it in a certain pattern (different from wedge sum).

Henry T. Horton

I really need to start working for my advisor meeting tomorrow

The vertices are punctured?

Alex Becker

@HenryT.Horton Yeah, I decided that version is more useful to me. I'm working with polygonal billiards.

Henry T. Horton

Wouldn't it end up being contractible then?

The vertices are identified with the wedge point

So it's like you've removed the wedge point

Alex Becker

03:59

@HenryT.Horton But the center is also removed, so it can't be contractible.

Zhen Lin

Isn't that homotopy equivalent to a circle with $n$ circles attached?

That should in turn be homotopy equivalent to a wedge of $(n + 1)$ circles...

Alex Becker

@ZhenLin I don't think so. Since I'm identifying opposite sides, each pair of sides looks like it would turn into a line segment with each end attached to the same circle, but at different points, and in such a way that between the two ends is always an end of a different such segment.

Let me see if I can draw this

Zhen Lin

Sure, I didn't say they were homeomorphic.

Alex Becker

@ZhenLin Good point. I guess I can pass through the points.

@ZhenLin This raises a new question for me though. The fundamental groups of the surface I'm considering and the wedge of $n+1$ circles are isomorphic, but is the isomorphism canonical? I suppose if we had a homeomorphism then we would have different isomorphisms corresponding to elements of the mapping class groups, but I have no idea where the isomorphisms between fundamental groups come from in the case of homotopy equivalence.

Although really I suppose I could get away with some reasonably uniform way of defining isomorphisms of the homology groups.

Zhen Lin

04:14

It's induced by the homotopy equivalence, of course.

A homotopy equivalence, after all, is a fortiori a weak equivalence, and a weak equivalence is by definition a map which induces isomorphisms in homotopy groups.

Alex Becker

@ZhenLin So if we have a map $f:X\to Y$ and $g:Y\to X$ defining a homotopy equivalence, then the map induced by $f$ is an isomorphism between $\pi_1(X)$ and $\pi_1(Y)$?

Zhen Lin

Yes, with inverse induced by $g$.

Alex Becker

Ah. Is there any nice classification of the isomorphisms thus induced?

Zhen Lin

The point, after all, is that there is a category whose objects are pointed spaces and morphisms are homotopy equivalence classes of continuous maps, and $\pi_1$ is a (representable!) functor from this category to $\textbf{Grp}$.

I don't know about classifying the isomorphisms. But it's easy enough to describe the homomorphisms in terms of generators: $\pi_1 f$ maps a class represented by a loop $\gamma$ to the class of the loop $f \circ \gamma$.

Alex Becker

@ZhenLin Right. Is there a standard name for $Hom(X,Y)$ in this category?

Zhen Lin

04:22

I think the standard notation is $[X, Y]$.

Dylan Moreland

@anon It's like we're on 4chan.

Zhen Lin

Oh, wait, that's for the un-pointed version.

Alex Becker

@DylanMoreland That is perhaps one of the largest indictments of any site, ever.

2

Dylan Moreland

@ZhenLin Hatcher uses $\langle X, Y\rangle$ when he keeps track of basepoints.

But I've never seen that anywhere else.

Zhen Lin

There are too many brackets flying around anyway!

I'd just write something like $\textbf{Ho}_* (X, Y)$.

Although that might be confused with the actual homotopy category...

shrug

Alex Becker

04:26

My issue is that I want to compare curves in these spaces for different $n$.

If I can get nice isomorphisms between the fundamental groups of the surfaces I'm working with and the wedge of $n+1$ circles then I could simply compare paths in the wedge of $n+1$ circles, which would be pretty nice because wedge is a coproduct.

Dylan Moreland

Have the Makoto threads on meta gotten anywhere or did nothing happen again because that's how meta works?

Alex Becker

But if I can't figure out a "uniform" way of selecting isomorphisms between the fundamental groups of the surfaces and the wedge sums, the scheme falls apart.

@DylanMoreland Joriki gave a nice answer which seems fairly widely accepted.

Zhen Lin

Well, there's an obvious continuous map from your space to $\bigvee^{n+1} S^1$ and it seems reasonable to believe this one is a homotopy equivalence.

Henry T. Horton

@ZhenLin When we did model categories in class $\mathrm{Ho}(X,Y)$ was for $\hom$ in the homotopy category and $\pi^r(X,Y)$ was used for (right) homotopy classes of maps in a model category

But for based spaces I think I just see people use $[X,Y]_\ast$

leo

some help is welcome

if you all are no busy

Alex Becker

04:33

@ZhenLin guess I forgot to mention that I have particular curves in mind that I want to show are "equivalent" in some sense (I know for dynamical reasons they should be). It might be that there is another, less-obvious but still somehow "uniform" way of defining isomorphisms between these fundamental groups which does what I want. I suppose I should probably try to check if this isomorphism does what I need.

Henry T. Horton

Or $[X,Y]_0$

Alex Becker

@HenryT.Horton I like that last notation, since I prefer $x_0$ for my base point.

Zhen Lin

The idea I have in mind seems uniform enough. Roughly speaking, we contract all but one of the edges of the fundamental polygon to a point and map to the circles.

Dylan Moreland

@AlexBecker This is nice, I agree.

Alex Becker

@ZhenLin All but one? The way I have in mind is contracting the surface to a circle with $2n$ line segments coming out, and identifying all the base points of the segments and the endpoints of the opposite ones.

Zhen Lin

04:39

Hmmm... that's more complicated. The one I'm thinking of sends circles to circles, more or less, so it is definitely at least a weak homotopy equivalence.

Henry T. Horton

We need a blackboard in here

Zhen Lin

aye...

leo

@robjohn are you around?

robjohn

@leo yes

leo

:-)

@robjohn can you help me? I'm a bit confused

Alex Becker

04:50

@ZhenLin Wait, do you mean all but one of the edges get contracted to a single point, or each of the edges gets contracted to a point except for one?

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