Mathematics

12:15 AM

@Disc So the commutator subgroup is the group of elements $ABA^{-1}B^{-1}$ where both $A$ and $B$ are in the group?

Or where $A$ and $B$ are in the commutator subgroup? Or where one is in the commutator subgroup and the other is in the group?

Disciple of Barney

@Algebra No, it is the group generated by those though.

@Algebra $ABA^{-1}B^{-1}CDC^{-1}D^{-1}$ doesn't have to be a commutator, so you consider the group generated by all commutators

Algebra

I don't get it sorry

$ABA^{-1}B^{-1}$ is generated into the commutator subgroup

But where do $A$ and $B$ live?

(Iteratively speaking^)

$A$ and $B$ live in $G$ it seems

and we take every possible $ABA^{-1}B^{-1}$ and put all of these into the commutator subgroup

Disciple of Barney

You have a group $G$ and a set $C=\{aba^{-1} b^{-1} \mid a,b \in G \}$, then the commutator subgroup is the group generated by the commutator elements, so it is equal to $\langle C \rangle $ and $a,b$ just live in $G$, they do not have to be in the commuator subgroup

@Algebra Did you get the 3UTC thing to work? (I am sure it does, just curious)

Algebra

Yep, I have tested it twice with success

Disciple of Barney

Did what I write clear things up?

Algebra

12:32 AM

Yep, thank you!

Algebra

12:43 AM

Now I am trying to show that for elementary matrices that:

$E_{i,j}^{ab} = [E_{i,k}^a , E_{k,j}^b]$

Where these are the elementary matrices, e.g. the identity matrix with ( for $E_{i,j}^a$) the entry $a$ at $i,j$

Where $i\ne j$

Will Hunting

@DiscipleofBarney I also know that serial voting is reversed at 3 UTC.

Algebra

Hey JL

Will Hunting

12:58 AM

@Algebra Have we talked before?

Algebra

@WillHunting I am AC - CTAC (on an alternative account because I am at uni

Will Hunting

@Algebra Oh AC.

@Algebra Haha, I only use one account at one time, you are so secretive.

Hi @WillHunting

@ᴇʏᴇs Hi BP.

What is AC

1:00 AM

His initials.

@Algebra Have you finished your exams and assignments?

I think he is still working on his final assignment...

We will talk later!

Pedro Tamaroff

1:15 AM

@ᴇʏᴇs Aneurisma cerebral.

ᴇʏᴇs

lol

Algebra

1:38 AM

@WillHunting Final assignment due in a few hours

Karim Mansour

1:57 AM

hey guys

@Algebra mine too I have assignment due on thursday and a project

then I am done for this semester

anonymous

How do you find the integral of x/2 ?

Karim Mansour

@Rememberme sorry I wasn't here

what was your question again ?

@Algebra we'll survive !!!!

Disciple of Barney

2:29 AM

$\DeclareMathOperator{\im}{im}$
I am pretty sure this works. Let $f:F_g \to F_h$ and $a_g,a_h$ are the abelinization homomorphisms, and $f^{ab}: F^{ab}_g \to F^{ab}_h$ is the obvious homomorphism, and we will assume it is injective. So $a_hf = f^{ab}a_g$, and note that if we restrict the codomain of $f$ to the image of $f$ we get the same commutative diagram and $a_h|\im f$. Note that$$\ker a_h |\im f=\ker a_g = \ker f^{ab}a_g$$ they all have isomorphic images, but that means that $\im [a_h |\im f] \cong F_g^{ab}$, since $\im f$ is free that means that $\im f \cong F_g$. Note that (f.g.) fr…

@MikeMiller You might be able to use the same idea for surface groups, (I don't know much about surface groups) but surface groups are also Hopfian (they are residually finite) so that might be able to be exploited.

( f.g. residually finite implies hopfian)

Disciple of Barney

2:52 AM

Actually, without thinking about it too much, pretty sure essentially the same proof goes through without any changes for surface groups since they are Hopfian, I might give it some more thought later...

Mike Miller

3:28 AM

@DiscipleofBarney: I haven't looked at this yet, but it looks exciting. Thanks for sending this to me.

ᴇʏᴇs

Best image

Disciple of Barney

@ᴇʏᴇs Just testing to see iff pdfs could be uploaded

user1667423

Can a trefoil knot be untied in dimensions higher than 3?

Is it chiral in dimensions higher than 3?

Disciple of Barney

Yes

user1667423

3:46 AM

What is the simplest non-trivial knot in four dimensions?

Disciple of Barney

@user1667423 There are none (there might be a higher dimentional "knot theory", but I am not discussing that)

math.harvard.edu/archive/21a_spring_06/exhibits/unknotting/…

Mike Miller

@user1667423 Every smooth embedding $S^n \hookrightarrow S^m$ is isotopic to the standard embedding if $n<m-2$; e.g., $S^1 \hookrightarrow S^4$. Indeed, we also have that every smooth embedding $S^{n-1} \hookrightarrow S^n$ is isotopic to the standard one. The only interesting knot theory in higher dimensions is codimension 2: $S^{n-2} \hookrightarrow S^n$.

If you want to find a nontrivial such knot, pick a knotted arc in the half-space $H^3$ whose endpoints lie on the boundary, and spin this around in a circle to get a knotted sphere $S^2$ in $\Bbb R^4$. This is known as Artin spinning. You can show that the complement of this sphere has a different fundamental group than the complement of the standard emedding.

An identical idea lets you construct knotted emeddings $S^{n-2} \hookrightarrow S^n$ for all $n \geq 3$.

Mike Miller

4:07 AM

@DiscipleofBarney: It was essential in a few places that the groups be free, because otherwise we wouldn't be able to show that the image of $f$ was free and there's no way to show that the image of $f$ is isomorphic to $\pi_1(\Sigma_g)$ or anything. But for free groups your argument works great. I can modify your thing a bit to get it to do what I want for surfaces. Thanks so much.

Disciple of Barney

@MikeMiller Subgroups of free groups are free, so the image of $f$ will be free regardless. There are probably some problem with looking at $\pi_1( \sigma_g)$, I will look at it in a bit

($\Sigma_g$)

Mike Miller

Capital sigma

Right, I was saying that the codomain being free was essential. Actually, I can get that to work anyway - we know the image has to be $F_{2g}$; quotient out by the evident relation to get a map to itself

Disciple of Barney

So if it is true then the only surface groups that have this property are free

Mike Miller

which there are none of

Disciple of Barney

Ah, okay

Glad to help @MikeMiller

Jane Smith

4:18 AM

Hello, Sorry for interrupting, I recently posted a question:

2

Q: Are there general guidelines to make "assumptions" when proving limits?

I am studying the definition of the limit using Paul's Online Notes When proving the following limit (Example 3) $\lim\limits_{x \to 2} x^2+x-11 = 9$ At one point he assumes: $|x+5| < K$ He also asumes: $|x-4| < 1$ Because one is a nice number to work with. I have read multiple explanatio...

And I cant still understand why I have to replace |x+5| by some constant K Can I get some help here?

Disciple of Barney

@JaneSmith The short answer is that in these sort of arguments you only care about local things, so you can assume you are in any small neighbourhood around a point (it doesn't matter how small). The long answer is that you more carefully go through the existing answer. (JMorivitz basically said the same thing I did, with more examples and discussion of the details.)

Jane Smith

Yes I know, and completely understand the answer

But I still dont get why

|x+5| needs to be replaced with K. Why not |x-4|?

I currently have a formulaic view of the issue.

Will Hunting

Did you read the comments?

Jane Smith

4:35 AM

yes and multiple times, not only that I have been studying that issue for three days in a row so far. I'm just stuck, its frustating

Will Hunting

Oh dear...

Jane Smith

Im thinking in just skipping this whole delta epsilon madness and jump directly into derivatives, Im wasting to much time

Will Hunting

That might be a good idea for the time being...

@JaneSmith In short, you are not replacing |x+5| by K, but rather restricting the size of |x+5| to some degree.

Jane Smith

@WillHunting Is |x+5| also an interval?

Will Hunting

@JaneSmith |x+5| is not an interval. It is a number.

@JaneSmith By saying that |x+5|<K, of course, you are restricting x to lie in some interval.

Jane Smith

4:46 AM

@WillHunting So I'm guessing the number right?

Will Hunting

@JaneSmith In the notes, he just wants to show that |x+5|<K holds for some K. Then the proof works.

Jane Smith

This is what I understand so far:

|x-4| works because is in the range of delta

Will Hunting

It then turns out that K=10 does work.

I think that Paul's notes are very detailed.

Jane Smith

So thats way I have to replace |x+5| with a number in order to continue with the proof

@WillHunting and very easy

Will Hunting

You want to make |x+5||x-4| small, so you try to make each one small.

That's the idea.

@JaneSmith Your effort is commendable. It is not your fault.

Jane Smith

4:54 AM

Paul also uses the min function to decide between the assumptions. Why?

Will Hunting

When a number is smaller than the min of two, it is smaller than each of them. That's why.

You want to make the number small enough for things to work. That's what limits are all about.

In showing limits, you want to make something small enough so that something else is small enough.

And you try different ways to do that, there is no single method to do all questions.

Maybe put it aside first and come back to it later. This is hard for a beginner.

Jane Smith

so because he restricted |x+5| you also need to restrict delta, to avoid epsiolon from taking values outside the range

@WillHunting Im trying to seld study all this by myself

Will Hunting

@JaneSmith Delta is what you need to find, given epsilon.

And to find delta, he tries to restrict |x+5|.

@JaneSmith To take AP test?

Jane Smith

no epsilon then, you avoid delta from taking values outside the range

@WillHunting to learn Partial Differential Equations :)

crazy isnt?

Will Hunting

@JaneSmith Have you already taken a calculus course in college?

Jane Smith

5:02 AM

nope

Will Hunting

@JaneSmith Then why are you trying to learn PDE?

@JaneSmith Yes, crazy, like me.

Jane Smith

because I like them, I would like to implement the wave equation in 2D some day

Will Hunting

@DiscipleofBarney You seem very knowledgeable in knot theory. =)

@JaneSmith I see. If you like PDE, you should study analysis properly first and understand epsilons and deltas.

Jane Smith

that's why I don't want to skip it, they look "important"

again: no epsilon then, you avoid delta from taking values outside the range

Will Hunting

@JaneSmith What do you mean by implement some day?

@JaneSmith The idea is that you fix a positive epsilon from the start and you try to find a positive delta corresponding to this epsilon.

Jane Smith

5:06 AM

10

Q: How to Self-Study Mathematical Methods?

Edit: Ok, user Chinny84 made comment that truly helps narrow the focus of my question. Basically, I'm asking for a self-study course of Mathematical Methods. Thanks to his recommendation I identified a book that is very close to the answer: Mathematical Methods in the Physical Sciences by Mary...

self-learning education book-recommendation

Will Hunting

@JaneSmith Is it possible for you to go back to college to do math?

@JaneSmith To go into PDE, you will need to know a lot of math, one after another topic...

Jane Smith

I had to drop for personal reason, Im currently 21 I could start again

Will Hunting

@JaneSmith No problem, even if you are 91.

@JaneSmith You have to ask yourself what your intention is first.

Jane Smith

@WillHunting I choose Paul for self studying everything is well organized

Will Hunting

@JaneSmith Your intention cannot be just to implement that thing, it has to be part of something bigger right?

Ask yourself what that bigger thing is, and then decide what you need to do.

Jane Smith

5:12 AM

No, just for fun. Computer Graphics more exactly

Will Hunting

Just for fun. Hmm...

Well, fun sometimes requires lots of effort...

Jane Smith

Calculus 1 2 3 and Linear Algebra, so yes a lot of effort

Will Hunting

And to do hardcore PDE, much much more...

So it is impossible to know exactly what you need to learn if it is just for fun.

Jane Smith

well in order to learn PDE I have to start with the basics

Will Hunting

And depending on what you learn, you may not be able to implement what you want, whatever that means.

I do not know anything about simulations to tell you exactly what you need to learn.

You should not even assume that PDE is what you need to implement that thing.

Jane Smith

5:18 AM

wave equation is not PDE?

Will Hunting

It is. But PDE is such a huge topic, and there are other topics.

So I don't know exactly what you need if it is just for fun to implement one small thing.

But I can tell you that epsilon and delta is not easy to understand for most people on earth.

Anyway, why I am talking so much is because I am rather free right now. =)

Jane Smith

Its not just for fun, I would also learn other topics like derivatives, integrals and series, they will be helpful for other things

Will Hunting

Then I really suggest you do a whole degree in college.

Jane Smith

@WillHunting I see you are in Europe right?

Will Hunting

@JaneSmith I am not. And how do you see that?

There is no way you can see that, LOL.

Jane Smith

5:24 AM

Its 1:24 AM

Will Hunting

Well, you don't know anything about me. =) It could be anytime here.

Basically, I could be anywhere in the world.

@JaneSmith So what you should do is ask yourself what you really want to do.

@JaneSmith Mathematical Methods can refer to many many things and can be very very advanced.

Jane Smith

OK, thanks for your help, very grateful for the advises and taking your time to discuss this delta epsilon madness. Bye Bye :)

See you

Will Hunting

Hello @KajHansen!

Disciple of Barney

5:44 AM

Huhh?! Will Hunting talking about math, what is the world coming too?

@WillHunting I use to know two things about knots, and the thing I stated is the only thing I know.

Kaj Hansen

5:58 AM

Hey @WillHunting. Just relaxing with some video games right now

Transcript for

$Mathematics$ Mathematics