Mathematics - 2020-01-26

12:00 AM

I was looking at your questions. Sobolev spaces, all sorts of measure theory.

@manooooh yes, you want to conclude $X\in\{\emptyset\}$, which is the same as $X=\emptyset$. is this the same as $X\in\mathcal{P}(A)$ and $X=\emptyset$?

Alek Murt

Math is just a hobby for me, so i have studied on my own multiple things, i dont have the formal part

Ted Shifrin

Weird that you've done measure theory and Sobolev spaces and are asking this question.

What exactly are you confused about?

@Thorgott: Do you not agree that what he's trying to prove is wrong?

Alek Murt

Im starting to see some things on sigma-algebras

Ted Shifrin

But when you wrote those semicolons, user76284 appropriately asked what you were writing. I interpret it as three separate questions that should not have been combined in the way you wrote it. Am I correct? @Alek

@manooooh: Oh, I see, you just had containment. So there are lots of subsets on the right-hand side as I described that are not subsets of $A-B$, but you just claimed containment. My mistake.

Alek Murt

12:03 AM

I really dont know how u interpret that, for me, i'm trying to write the inverse image ofsome collection of subsets of a set $Y$

manooooh

@TedShifrin yes: take $U=\{1,2\}$, $A=\{1,2\}$ and $B=\{2\}$. Then $A-B=\{1\}$ so $P(A-B)=\{\emptyset,\{1\}\}$, and $P(A)=\{\emptyset,\{1\},\{2\},\{1,2\}\}$, $P(B)=\{\emptyset,\{2\}\}$. So $P(A)-P(B)=\{\{1\},\{1,2\}\}$, and hence $P(A-B)\subseteq(P(A)-P(B))\cup\{\emptyset\}$

Alek Murt

I dont think that I'm trying to see that as 3 different inverse images

Ted Shifrin

So are you asking three separate questions, as I said, @Alek, or are you taking the inverse image of the union of those subsets?

user76284

@AlekMurt The inverse image is defined for each of those subsets, not a collection of them.

Thorgott

I don't think it's wrong

user76284

12:04 AM

Are you trying to get the collection of inverse images?

Ted Shifrin

No, I was thinking it claimed equality, not containment. I apologized above. @Thorgott, @manooooh.

manooooh

No problem

Ted Shifrin

I guess there's no way to write it other than $\{X\in\mathcal P(A): X\cap B = \emptyset\}$.

Thorgott

ah, ok

user76284

If you love using masochistic notation then $f^\star \circ ([0,1],(1,4],(4,\infty))$ might work. $f^\star$ is the preimage with the pullback notation, and I'm exploiting this notational convenience. Don't try this at home though.

Alek Murt

12:06 AM

No, I'm not talking about the union of those subsets is more of something like @user76284 says

manooooh

@Thorgott showed me that $X\in B'$ implies $X\subseteq (P(B))'$ or $X=\emptyset$, but I can't figure out how to use this to finish the proof here: chat.stackexchange.com/transcript/message/53348033#53348033

Thorgott

you have already used it

Ted Shifrin

I think that never in my 45+ years as a professional mathematician have I encountered this notation/notion. @Alek, @user76284.

Thorgott

the proof is essentially finished

you just need to realize that the last statement is equivalent to the conclusion

manooooh

@Thorgott do you agree we have $(X\in P(A)\wedge X\in(P(B))')\vee(X\in P(A)\wedge X=\emptyset)$?

Ted Shifrin

12:08 AM

What does $X\subseteq (\mathcal P(B))'$ mean?

Alek Murt

So let's enter in context, because for me is also super weird and i would like to know if I'm writing this ok

Ted Shifrin

You mean element, @manooooh? And complement in $\mathcal P(A)$?

manooooh

@TedShifrin $(P(B))'$ is the set not containing elements of the power set of $B$. i.e. $(P(B))'=U-P(B)$, where $U$ is the universe

Ted Shifrin

I'm complaining about $\subseteq$ instead of $\in$.

manooooh

I didn't write subset but $\in$

Alek Murt

12:11 AM

I have a $\sigma ([0,1],(1,4],(4,∞))$ and i have the function that i wrote above, and i would like to know $f^{-1}\sigma ([0,1],(1,4],(4,∞))$

Ted Shifrin

@manooooh: You did write it and that's why I asked what it meant.

Look up 12 "chats."

I still don't know what $\sigma()$ with a bunch of commas means, @Alek.

manooooh

@TedShifrin $X\in P(B)'$ is a relationship like $x\in A=\{1\}$. I write $x$ in capital letter since $P(A)$ contains sets, and sets are commonly denoted with a capital letter

Alek Murt

So i think that i have to look for $\sigma (f^{-1}([0,1],(1,4],(4,∞)))$, that why I ask for that inverse image, where $\sigma ()$ is the sigma algebra generated for those elements inside ()

Ted Shifrin

Commas, semicolons, whatever. Unless $\sigma$ is a function on the set of ordered triples of subsets ...

oh, crazy.

manooooh

If $X$ is an element of $P(A)$ then $X$ is a subset of $A$

Ted Shifrin

12:14 AM

So $f^{-1}$ of the $\sigma$-algebra is again a $\sigma$-algebra, I guess, but you apply $f^{-1}$ to each element of that $\sigma$-algebra, one at a time.

manooooh

and the converse also holds

Ted Shifrin

@manooooh: Yes, I know this stuff. I'm saying you wrote the equivalent of $X\subset P(A)$.

All I said was it should have been $\in$.

You can find what you typed if you count 12 things up from where I put that.

Anyhow, I'm done with this discussion.

Alek Murt

Initially that was my doubt, and my thoughts were that the only thing that makes sense is to apply inverse image for ever subset, but this topic has been really confusing for me, so i wanted to make sure my thoughts were correct

manooooh

Where did I write, with other words, $X\subset P(A)$?

Daminark

Hey there guys!

manooooh

12:16 AM

Hello!

Ted Shifrin

@manooooh: I already told you where a while ago.

Thorgott

Yes, I agree we have that

Ted Shifrin

Demonark!!! Where the hell have you been?

Lukas Heger

hey @Daminark

Ted Shifrin

I guess that since Demonark moved to Wisconsin he became too good for us :P

Daminark

12:17 AM

Lmao, in Chicago fine but Madison???

Nah but yeah thing is last semester I felt I wasn't really getting any math done because the volume of immediate deadlines dropped like a rock

manooooh

Ted here ^^^^^^^^^^? But I didn't write $X\subset P(A)$. I wrote $X\in P(A)$

Daminark

And I was like uh okay I need to pull myself together a bit, gonna try to get offline a bit more and focus a bit

Ted Shifrin

No, not there. Geez.

Is it going better, Demonark?

Daminark

In a way no, I just found other distractions. This semester I'm taking classes that actually all assign psets and I'm gonna try to be a part of some reading groups

At least if I can artificially place some deadlines on my head then maybe that'll help a bit

manooooh

Oh, now I see Ted, yes you are right again. I should have write $X\subset B'$ implies $X\in P(B)'$ or $X=\emptyset$

Daminark

12:22 AM

Or not artificially so much as like, okay with reading groups obviously it's not like the due dates are quite as "over your head" but hopefully this will at least get me to focus a bit

Ted Shifrin

Demonark: In general, faculty expect graduate students to be more grown-up and responsible. Although I knew better and always assigned weekly or bi-weekly problem sets in every graduate course I ever taught. :D

Daminark

That makes sense, yeah. This semester I'm going a bit harder on the algebra though eventually I'll wanna get back to some analysis. I've got homological algebra, elliptic curves, and AG2 (starting with Riemann Roch, then doing some basic scheme theory)

Ted Shifrin

LOL, all algebra.

Daminark

What've you been up to, Ted and Lukas?

Thorgott

are you saying that graduate lectures usually do not have assignments?

manooooh

12:25 AM

@Thorgott oh, yes since we know that $p\wedge q\to p$, so $X=\emptyset$ and $X\in P(A)$ implies $X=\emptyset$, which is the same as $X\in\{\emptyset\}$. And we are done!

Lukas Heger

@Daminark I've been giving seminar talks on group schemes/algebraic groups and rigid-analytic geometry and I'm taking lectures on adic spaces and L functions

Daminark

(On the side I've got a few people and we're gonna read some K-Theory, I'm thinking using Atiyah?)

Lukas Heger

and I'm TAing intro abstract algebra

manooooh

It costs me to understand why if $X$ is the empty set then $X$ belongs to $\{\emptyset\}$

Daminark

Ah dope

Lukas Heger

12:27 AM

oh and I'm in the process of finishing my masters application

Thorgott

are you saying that $p\land q\rightarrow p$ generally?

Lukas Heger

I've written CV, motivation letter, asked for recommendation letters and all that

Daminark

Nice, still going with the plan you mentioned that one time? ALGANT?

Lukas Heger

yeah

manooooh

@Thorgott yes of course. It is one law from Logic

Lukas Heger

12:28 AM

but I probably can't go to LA :/

Ted Shifrin

I was wondering what happened with that, Lukas.

Lukas Heger

due to illness, I won't finish my bachelor as early as I planned

Thorgott

$X=\emptyset\Leftrightarrow X\in\{\emptyset\}$ holds qua definition

Ted Shifrin

I certainly don't like writing $\{A\}$ when $A$ is a subset.

But I give up on this stuff.

manooooh

@Thorgott oh, I see, it is pretty obvious

Daminark

12:30 AM

But yeah on my end I'm TAing Calc 2 again

Thorgott

i agree with Ted in that this is too heavy on symbols

Ted Shifrin

When I took point-set topology in college from Munkres, he literally would not allow us to write any symbols other than $\in$, $\implies$, $\subset$, $\cup$, $\cap$ (and I'm sure I've forgotten a few) and the abbreviation s.t. for "such that."

manooooh

Let me clear it for you and me: \begin{align*}\text{PROVE: $P(A-B)\subseteq (P(A)-P(B))\cup\{\emptyset\}$}\\\hline X\in P(A-B)=P(A\cap B')&\to X\subseteq A\wedge X\subseteq B'\\&\to X\in P(A)\wedge(X\in(P(B))'\vee X=\emptyset)\\&\to(X\in P(A)\wedge X\in(P(B))')\vee(X\in P(A)\wedge X=\emptyset)\\&\to(X\in P(A)\cap(P(B))')\vee(X=\emptyset)\\&\to X\in P(A)-P(B)\vee X\in\{\emptyset\}\\&\to X\in(P(A)-P(B))\cup\{\emptyset\}\end{align*}

Thanks @Thorgott and @TedShifrin for your huge help!

Thorgott

@TedS I see that it left a stain on you

Daminark

I buy that, at least for psets

On boards I tend to be a bit more symbolsy

$\forall$ and $\exists$ in particular

manooooh

12:40 AM

One can use symbols or can't use symbols. Is up to you. I prefer to use symbols in this particular situation because does not include quantifiers and the expressions are not long enough to get lost on the road

Daminark

So, it's officially up to you in the sense of, we all have free will. But if you want other people to read your stuff then that's something you wanna take into consideration

manooooh

Erm... Is up to you to read my question or not...

Daminark

I mean sure, just like, if there over comes a time where you ever do need others to read what you're doing, and it will likely come, be ready to make some concessions

Ted Shifrin

1:02 AM

@Thorgott Not at all. I found that every mathematician from whom I took courses or with whom I studied used hardly any symbols at all. If you're going to work in formal logic, then have at it. But if you ever are going to write a paper for publication, mathematics (not logic) journals want readable mathematics, i.e., words and not symbols.

And it's important for pedagogy to instill an understanding of concepts, not of symbols.

In the meantime, I'll avoid reading what manoooh chooses to write.

anakhro

Let's do like they used to do and use words for operations too.

No more equals sign.

NO LONGER WILL WE BE BOUND TO MATHEMATICAL NOTATION.

LANGUAGE WILL TAKE BACK WHAT IS RIGHTFULLY THEIRS.

Mathphile

3:06 AM

Can the product of a transcendental number and an algebraic irrational number be algebraic?

Mike Miller

products of algebraic numbers are algebraic and reciprocals kf algebraic numbers are algebraic so no

Leaky Nun

showing that the sum of an algebraic number and a transcendental number is transcendental is actually an instructive exercise in logic

Lucas Henrique

3:35 AM

@LukasHeger thank you! Also, @BalarkaSen, @Thorgott and @Ted: thank you all :)

Mathphile

also can an integer to the power an irrational be transcendental?

Lucas Henrique

41

Q: irrationality of $\sqrt{2}^{\sqrt{2}}$.

The fact that there exists irrational number $a,b$ such that $a^b$ is rational is proved by the law of excluded middle, but I read somewhere that irrationality of $\sqrt{2}^{\sqrt{2}}$ is proved constructively. Do you know the proof?

Take a look at the accepted answer

Gelfond–Schneider theorem

In mathematics, the Gelfond–Schneider theorem establishes the transcendence of a large class of numbers. == History == It was originally proved independently in 1934 by Aleksandr Gelfond and Theodor Schneider. == Statement == If a and b are algebraic numbers with a ≠ 0, 1, and b irrational, then any value of ab is a transcendental number. === Comments === The values of a and b are not restricted to real numbers; complex numbers are allowed (they are never rational when they have an imaginary part not equal to 0, even if both the real and imaginary parts are rational).In general, ab ...

Mathphile

4:25 AM

If $a$ is algebraic, $b$ is irrational and $a^b$ trancendental, then is $b$ always algebraic?

Mathphile

4:54 AM

does the Gelfond–Schneider theorem imply the above?

Lucas Henrique

5:05 AM

nope, not trivially if true

New Student

5:15 AM

which book has set operations defined in terms of logical operations, For example union defined as $\or$

Lucas Henrique

any formal set theory book

Paul Halmos "Naive Set Theory", probably

Leaky Nun

5:58 AM

@AkivaWeinberger what does ELO as in ELO rating system stand for?

(don't google!)

Akiva Weinberger

6:52 AM

@LeakyNun It's Elo, it's a guy's name

Leaky Nun

7:08 AM

@AkivaWeinberger nice

Jacksoja

9:44 AM

@LeakyNun Hi

Jacksoja

10:23 AM

@AlessandroCodenotti Hi

@AlessandroCodenotti I have a question

Alessandro Codenotti

11:05 AM

You should just ask your question instead of pinging everybody, if somebody who knows an answer reads it, it will be answered

7

Anush

11:39 AM

Sample 3 binary strings uniformly of length 30. What is the exact expected minimum distance between the closest pair?

maths student

12:05 PM

0

Q: Bayes theorem and application

A pregnancy test is accurate 97% of the time when someone is pregnant and 98% accurate when someone is not. Assuming that 60% of people who take the test are pregnant, and that someone tests positive twice in a row, what is the probability they actually are pregnant? I think this is application ...

probability probability-theory bayes-theorem

user

12:21 PM

Is it true that $(A/B) \cup C= A \cup (B \C)$?

sorry

that should be

$(A/B) \cap C = A \cap (B/C)$

or better $(A/B) \cap C = A \cap (C/B)$

user

12:56 PM

So I ask is it true that $(A/B) \cap C = A \cap (C/B)$

Thorgott

1:25 PM

draw a diagram

user

1:38 PM

I think it is true

by basic properties of sets

I am just sanity checking, because it makes a question rather too trivial

@Thorgott

Thorgott

2:04 PM

it is

Lucas Henrique

hi @user, @Thorgott

Thorgott

2:26 PM

hi Lucas

Edward Evans

Yo chat

Jacksoja

5:57 PM

@LukasHeger hey

fomin

math.stackexchange.com/questions/3523121/… has been closed but for no reason that I can see. Is it possible to have it reopened?

Ted Shifrin

No one in here can help with that. You need a moderator. The question seems clear. It would have been nice if you'd shown some of your attempts or an approach other than just having a numerical hunch. But I personally do not even understand the question, so I can't help.

Jacksoja

@TedShifrin Hey Ted

Ted Shifrin

Howdy @Jacksoja.

Jacksoja

@TedShifrin need some hint Ted ! Been stuck for days

Ted Shifrin

6:13 PM

Days, not weeks?

Jacksoja

Want to show order of any element in a finite cyclic group has to divide order of the element of highest order.

No, going crazy as it is

Ted Shifrin

Oh, there's a version of that question that's much harder, but this isn't hard.

Jacksoja

What version

Ted Shifrin

Leave out cyclic, just say abelian.

Jacksoja

I tried alot of ways

Ted Shifrin

6:15 PM

So a cyclic group has a generator. What is the order of the element of highest order?

Jacksoja

Oh yes

That is my question haha

Gonna prov it is cyclic

Ted Shifrin

Huh????

Jacksoja

@TedShifrin it is p-1 but am not alowed to use that

Ted Shifrin

You told me the group was cyclic, and now you're saying you're gonna prove that.

What is the correct statement?

Jacksoja

My question is the hard version

Ted Shifrin

6:19 PM

So you have a finite abelian group.

Nothing more.

Jacksoja

Yes

Ted Shifrin

What tools do you have? Do you know that such a group can be decomposed as a direct sum of cyclic groups?

Jacksoja

No

I tried using

Ted Shifrin

Do you know something about the order of $ab$ if the orders of $a$ and $b$ are relatively prime?

Jacksoja

Euclid algorithm ,sorry typing from phone

Bit slow

Ted Shifrin

6:21 PM

Yes, that might be relevant.

Jacksoja

I tried that

Ted Shifrin

So my hint (this exercise is actually in the algebra book I wrote, so I remember it) is to answer the question I wrote 5 lines up.

Jacksoja

Product of their order

Ok thanks

Ted Shifrin

OK, find a way to use that for your question. You have to be slightly tricky.

Jacksoja

Btw your book

On diff geo

What are prereq ?

@TedShifrin can i use your Youtube lectures ? And give it a go at Reading?

Ted Shifrin

6:24 PM

Multivariable calc and some linear algebra (like linear maps and eigenvalues/eigenvectors in 2D).

The YouTube lectures are for the most part much more advanced than the diff geo material.

You don't need analysis for the diff geo.

Thorgott

So the task is to show that the order of an element in a finite abelian group divides the order of the element of highest order?

Ted Shifrin

Don't give it away, @Thorgott or DogAteMy!

Jacksoja

Ok thanks so much !

Yes let me try the hint

Thorgott

6:38 PM

Ah, I see it now. Yeah, that's a nice exercise.

Lucas Henrique

@Jacksoja oh, lol. I came here to ask the same

Jacksoja

@LucasHenrique haha good luck

Tell me if you solve it

You taking algebra as well ?

Lucas Henrique

Kind of. I'm studying Galois theory

Ted Shifrin

7:15 PM

The same @Lucas?!!

Lucas Henrique

Ok. Let $a$ be one of the elements with maximum order and let $b \not \in <\!a\!>$ (otherwise it's trivial)

(I can't typeset this properly)

Ted Shifrin

Oh, because of finite fields this could show up in Galois theory (although you have more ways around it, I think).

@Jacksoja didn't ask for more hint, but I would suggest proceeding by contradiction ... using the beautiful fact I asked him above.

Lucas Henrique

I'm trying to show that if $ \operatorname{ord}b \not \mid \operatorname{ord}a$, we can choose $r,s$ such that $\operatorname{ord}(a^r b^s) > \operatorname{ord} a$

Which looks like Bezout's lemma

Ted Shifrin

Yes, that's correct, but it's a bit sneaky, because you need relatively prime orders for our lemma above.

Lucas Henrique

@Ted, I thought about that direct sum of cyclic groups thing. Is that done by finding cyclic subgroups of maximal order, taking quotients (since each subgroup is normal cuz commutativity) and using induction?

Ted Shifrin

7:21 PM

Whoa. I haven't thought about it in a while, but I think it's easier than that.

Lucas Henrique

@TedShifrin oh. :(. I'm not sure of how I'd get relatively prime orders.

Ted Shifrin

That's why it's a bit sneaky ;P

Lucas Henrique

Could you give me a hint, please?

Ted Shifrin

Well, first of all, what is your set-up if you're doing a contradiction argument?

Lucas Henrique

I'm not sure if I understand your question, but I'd try to show that... my hypotheses are contradictory. :P Specifically, I'd try to show that there's an element with strictly bigger order than the one we supposed to have maximum order

Ted Shifrin

7:27 PM

Yes, but with what starting hypothesis?

Lucas Henrique

There's an element with order not dividing the maximum order

Ted Shifrin

So can you phrase that as a gcd statement?

Lucas Henrique

I was literally writing that down haha

Ted Shifrin

That's gonna be my hint, then.

Lucas Henrique

If $m$ is the maximum and $n$ is the other one, $\operatorname{gcd}(m,n) < n$

Ted Shifrin

7:33 PM

Yup, so you want to play with that gcd to come up with your $r,s$.

Lucas Henrique

Ohh

So that $\operatorname{gcd}(m,n) \geq n$

Ted Shifrin

I'm not saying any more. You're more clever in algebra stuff than I am, anyhow :P

Lucas Henrique

Thank you @Ted. I'll play with these symbols for a while

Ted Shifrin

That last sentence is too confusing, because we already had an $m$.

BTW, do you think that lemma generalizes to lcm?

Lucas Henrique

Sure

Ted Shifrin

7:35 PM

Nope. Find a counterexample.

Lucas Henrique

:(

Ted Shifrin

smirks

I said the same thing (when I was writing my algebra book) and then realized it was wrong.

Thorgott

That question seemed too innocent for the answer to be positive

Ted Shifrin

LOL @Thorgott ... particularly if one knows Ted.

Lucas Henrique

$2$ and $6$ over $\mathbb Z_{12}$?

$\operatorname{ord}(2+6) < \operatorname{lcm}(\operatorname{ord}2, \operatorname{ord}6)$

Thorgott

7:41 PM

Also, if you are willing to use the structure theorem, the question is trivial, because you can explicitly classify the orders of all elements, but that's not necessary here

Ted Shifrin

OK. I had a slightly different version of the same example.

Lucas Henrique

But couldn't you have $a^rb^s$ with order being the lcm?

Thorgott

Yes, in fact, that's always possible

Lucas Henrique

In this case, $1*2 + 3*8$

@Thorgott that's what I'll be trying then

(I think I should, at least)

Thorgott

I think you can do it with even less work, but that definitely suffices

Lucas Henrique

7:44 PM

OH. So that's the subtlety

Ted Shifrin

losted

Lucas Henrique

$a^{\operatorname{gcd}(m,n)}b$

the first one's order is $\frac m{\operatorname{gcd}(m,n)}$ which is relatively prime to $n$

Ted Shifrin

Are you sure about that?

I think I made a mistake like that, too.

Lucas Henrique

it looks I shouldn't be hahaha

Thorgott

I just noticed I made a mistake like that too lol

But it's easily fixed

Ted Shifrin

7:49 PM

@Thorgott: Many years ago it took me a while to figure out how to fix it.

Lucas Henrique

that's wrong

oh hell. for $4, 6 \in \mathbb Z_{12}$, $4^1+6 = 10$ has order $6$ as expected

idk what I'm doing wrong

but these orders are relatively prime, duh

Ted Shifrin

nods

Lucas Henrique

Is it about the chosen orders?

Ted Shifrin

Come back to your false claim. If $m/d$ and $n$ aren't necessarily relatively prime, how do you fix that? Forget the group theory for a while.

Lucas Henrique

@TedShifrin okay, found a counterexample. I'd take $m/d$ and $n/d$ but it I don't think this choice will work

Ted Shifrin

8:02 PM

Now comes the tricky stuff, then.

Heya, a @Balarka.

Balarka Sen

Hi

Alessandro Codenotti

Hi @Balarka @Ted

Ted Shifrin

@Balarka, with regard to a question on main on which I'm commenting ... there's no elementary way to see that the commutators of the generators of $\pi_1(\Bbb R^2-p-q)$ are nontrivial, right?

hi, demonic @Alessandro (will I get flagged yet again?).

3

vesii

Hi guys, I'm trying to solve the limit $\lim_{x\to 0} \frac{\sqrt{1+x}-\frac{1}{2}\sin(x)-\cos(x)}{\arctan(x^2)\sin(\frac{\pi}{3}\cos(x))}$. I can't seem to find the trick in order to solve it. Any suggestions please?

Ted Shifrin

I always use Taylor polynomials, @vesii.

Alessandro Codenotti

8:07 PM

@TedShifrin there's only one way to find out

vesii

Unfortunately, I did not learn taylor yet :(

Ted Shifrin

The $\sin(\frac{\pi}3\cos x)$ term is sort of a red herring (i.e., irrelevant to the issue). Can you do it without that?

There are some super-zealous people voting to close everything. I've seen a few where they're over-doing it.

And I haven't heard of most of the people so voting, either.

Too many people have vote-to-close privileges now.

@vesii: So you have no choice but to use L'Hôpital, but I truly hate that.

Lucas Henrique

@TedShifrin $m$, $n/d$?

I tested it a few times, it worked and I don't know why, haha

Ted Shifrin

You can make counterexamples for that one, too, just like for $m/d,n$, right?

Lucas Henrique

Not sure if the restriction $m > n$ implies anything.

Ted Shifrin

8:12 PM

Oh, true, it isn't totally symmetric.

What about $m=12$, $n=9$?

vesii

@TedShifrin, I'll try, thanks :)

Lucas Henrique

I guess I'll have to change both orders simultaneously

Ted Shifrin

@vesii: Factor out the $\sin(\pi/3)$ limit of that one term. It's a mess otherwise.

@LucasHenrique Bingo.

Now I'm done!

Lunchtime for Ted.

Thorgott

bon appétit

Balarka Sen

@TedShifrin Doubtful. I at least need to draw the covering space and invoke the lifting results.

That's the easiest proof I know

Lucas Henrique

8:21 PM

Thank you, @Ted :). Bon appétit!

Ted Shifrin

9:20 PM

@BalarkaSen Yup, me too.

Balarka Sen

@TedShifrin I suppose one way to argue could be to show that $T^2$ is not homotopy equivalent to $S^1 \vee S^1 \vee S^2$. This you can conclude via ring structure in cohomology.

This proves the commutator is not nullhomotopic, because that's what the $2$-cell in $T^2$ is attached to the 1-skeleton by.

Lucas Henrique

$a^r b^s$, if $m/r$ and $n/s$ are relatively prime, has order $\frac{mn}{rs}$ which we want to be $>m \implies n > rs$

Balarka Sen

I wouldn't call this easy by any means, just interesting. You could potentially write a purely de Rham theoretic proof maybe.

Replacing cell structure by handlebodies to make everything a manifold

Fun fact: $\Sigma T^2$ is nonetheless homotopy equivalent to $\Sigma(S^1 \vee S^1 \vee S^2) = S^2 \vee S^2 \vee S^3$

Suspending completely "untangles" the cell structure, thanks to higher homotopy groups being abelian (commutator becomes nullhomotopic after suspending)

Ted Shifrin

9:47 PM

@Balarka: This was an exercise (?) in a complex variables book, however :P

@LucasHenrique You'd better specify something about $r$ and $s$ ...

Lucas Henrique

$0 \le r < m,\ 0 \le s < n$?

I feel dumb compared to you guys, lol

Edward Evans

10:08 PM

@Balarka speaking of de Rham, there's a swiss theoretical physicist in London called de Rham who funnily enough works in cosmology

Ted Shifrin

@LucasHenrique I don't think that'll do it.

Don't you need something like $rs|n$ (and $rs<n$)?

Lukas Heger

@Edward in Germany we usually say De Shane

Edward Evans

w h a t

I assume this is a joke but I don't get it

Lukas Heger

it's not good anyway

Edward Evans

From my experience in Germany thus far I didn't expect it to be

lol

Ted Shifrin

10:19 PM

@Edward, you have my permission to smack @Lukas.

Lukas Heger

lol

copper.hat

I was thinking of proposing a mathunderflow site to handle the periodic surge of homework questions :-)

Ted Shifrin

Periodic? It seems periodic with period 0.

But I have to run off to a concert. Hi/bye @copper.hat :)

copper.hat

Bye :-)

Lukas Heger

@edward do you still not get it?

Edward Evans

10:21 PM

Indeed I do not

hahaha

ah okey

ich habs

duh

Mathphile

can anyone help me to prove this integral?

$\int_0^{\pi/2} (\tan x)^{1/3}dx = \frac{\pi}{\sqrt{3}}$

Lucas Henrique

@TedShifrin why do I need $rs \mid n$?

I mean, $s\mid n$ is sufficient (I think)

Maybe I'll have to appeal to prime factorization

Thorgott

Just write down your proof completely and then we'll see whether it works

Lucas Henrique

I don't have one and this problem looks too dumb to waste hours

I'm sad to have spent so much time on it, to be honest

Lucas Henrique

11:31 PM

Jeez, finally

Let $\ell = \operatorname{lcm}(m,n)$ with $m = \prod_{i=1}^k p_i^{\alpha_i}$ and $n = \prod_{i=1}^k p_i^{\beta_i}$. Break $\ell$ down in two relatively prime products $m' = \prod_{\alpha_i \geq \beta_i} p_i^{\alpha_i}$ and $n' = \prod_{\beta_i > \alpha_i} p_i^{\beta_i}$

$a^{m/m'}b^{n/n'}$ is our element

Could anyone check this out, please?

Thorgott

11:47 PM

Yup, that works

And also gives you the general procedure to find an element whose order is the lcm (granted the elements commute)

Here's what I did: Let $d=\mathrm{gcd}(m,n)$. Then $a^d$ has order $m/d$ and $b^d$ has order $n/d$, which are coprime, hence the order of $a^db^d$ is $mn/d^2=l/d$. But, by commutativity, this is $(ab)^d$ and that implies the order of $ab$ is $l$.

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