I learnt them about 2 years ago and didn't think of that question. But my bro in an engineering uni told me he's learning them now. Assuming they mostly learn applicable math to the real world, I can't think of any applications.

your bro could figure out how much beer fits into a funnel

or a keg, or anything with rotational symmetry, just by knowing a cross-section

Is there such a thing as posting too long of an answer? I'm wondering what my limit should be, so I can improve and provide better answers in the future

@teadawg I'd say there isn't, but it might be appropiate to provide a summation, not going into details, at the top - and below the details and deeper explanation, if you feel it is really long

@cxseven yea that makes sense

Especially the beer example. Anyway good night people.

@Studentmath There comes a point where linking a more detailed answer on Google Docs or something similar becomes more of a hassle than writing in LaTeX

This was VERY challenging to format properly....

Plus my internet was absolutely abysmal at the time of writing it, but I digress

I don't like breaking the silence, it makes me feel like the center of attention...

Seen any interesting questions lately, @Ted?

I guess that's a no.

I wasn't here, @Mike. A few.

Any in particular? I haven't seen too many fun ones lately...

Hmm, I can't seem to connect to the main site at the moment...

A few diff geo types I haven't taken the time to answer, a cool prob one @Mike

Hmm... I'm hunting down a reference for one to the fact that if $X$ is a finite CW-complex, then $\Omega X$ has the homotopy type of a CW complex

Found it... not a very exciting proof.

Hello!!! Is there anybody that knows ALGOL??

@MaryStar What is that?

@TedShifrin Hello, professor.

@PedroTamaroff A programming language...

@MaryStar Then no.

@PedroTamaroff Ok... :/

I also don't

Mike, you don't even.

So clearly you don't.

there's a question at qr.ae/lgats about what portion of a diagonal will pass through the black squares of an MxN checkerboard, and Alon Amit's answer is understandable except for the case where M and N are both odd. What symmetry could he mean should be exploited?

@cxseven Probably the Vick-Eckertt symmetry.

lol

You're calling me professor, @Pedro?

@TedShifrin OK. Hello, Ted. I have good news. Are you on Facebook?

I can be

Close pls

@TedShifrin OK.

Could you?

Hello, can anyone access stats.stackexchange.com? I can access chat rooms but I can't access the site.

I can.

Why can't I? "This webpage is not available"

No idea!

The entire network at SE is experiencing issues right now, they're working on fixing it

They said so in a tweet, so you know it's true :P

But why Mike Miller can?

I'm able to access it as well, idk

It's going to depend on your location. Wait a bit and when they fix stuff you'll be able to see it again.

I'm super frustrated, the site went down right before I was finished writing an answer...

Algebraic Topology question: If I have a nullhomotopic loop in a path-connected space, am I assured there exists a homotopy that is a path-homotopy?

Can you define path-homotopy?

it keeps the basepoint fixed throughout the homotopy

at least in the loop case I mentioned, in the general path case, it keeps the starting and endpoints fixed.

@RobertCardona You can be a cool lad and call it a homotopy relative to a point.

@RobertCardona Yes, and there's more true in general: two elements in the fundamental group are homotopic (i.e., ignoring base point) iff they're conjugate.

It's a good exercise to prove this, so I'll let you do that yourself.

I was trying to earlier, but my approach was creating a product from the existing homotopy, but the product of paths doesn't keep the basepoint fixed, so I started thinking it might not even be true, which is why I asked.

Oh! I see. Well, it definitely is.

Since it's nullhomotopic, there exists a homotopy H, since the space in question is path-connected, there exists a map $\alpha$ from the point $H(s, 1) = c$ to the basepoint of the loop, call it $a$.

Thanks!

I'll keep thinking about it! At least I know it's true now!

Yeah, the key here is to try and draw a picture before you start pushing the symbols around.

For your special case of null-homotopy, the picture definitely helps (and once you have that it should give clues in general).

Thanks! :)

I'm heading to bed. Night, guys

@MikeMiller, I appear to have encountered the same problem. Say I have a loop $f$ with basepoint $a$ that's nullhomotopic. So I have a homotopy $H$ from $f$ to some point $c$ in $X$. For each point on the loop, I can map it to $c$ via $H$ and then map $c$ to $a$ since $X$ is path-connected.

Sure - you now have a homotopy from $f$ to the constant loop at $a$.

But it's not a path homotopy, because if I do the same thing for $f(0) = f(1)$, it takes me away from $a$ to $c$ and then back again

Right

Here's my suggestion for the picture. Start with the equator on the sphere; write down a homotopy that contracts it to the north pole. How might you take this homotopy and modify it so that this homotopy ends up giving you a homotopy from the equator circle to the path between the basepoint and the north pole (and back)?

Looking at the equator as a loop?

alright all ya dang analysts help me

$f(s) = (cos 2\pi s, \sin 2 \pi s, 0)$?

suppose i've got a function $f$, and i define $A_n(f)$ to be a sequence of sets $f(i_n)$ for some finite set $\{i_n:1\leq i \leq n\}$

(so i'm taking the image of $f$ on a finite set of points and cranking up the number of points)

Yeah @RobertCardona

Don't do anything explicit, just try to form a picture.

@AlexanderGruber SUP

$A_n(f)$ is a weighted graph, canonically, under the weight function $w_{\{P,Q\}}=d(P,Q)$ where d is the distance function of whatever space $f$ is going into (may as well just take it to be a real-valued function I think)

so, let $B_n$ be the total weight of a minimal spanning tree of $A_n(f)$

what can be said if $B_n\rightarrow \infty$?

@PedroTamaroff hi!

@Alex!

Hi there @TedShifrin!

@AlexanderGruber Hehe, that eludes me.

I have to go now, too.

Bye byes ya'll all alls.

aww :p

Bubye @Pedro

bye pedro

Hi. In what discipline of mathematics would we be interested in problems involving product of two geometrical figures...for e.g. square with a point...?

@deostroll what do you mean by product?

eh, even I don't get it...answer is a cube...

@AlexanderGruber must be a cartesian product...

@deostroll let us know some context, can't help you without much more

@AlexanderGruber It sort of appears in the study of fractals I think...

Check here: en.wikipedia.org/wiki/…

@deostroll i mean, where are you seeing this? is this for a class?

lamer reason: trying to help partner with research...

@deostroll i mean there's lots of stuff where you deal with products of 2d spaces with 1d spaces

@MikeMiller, the homotopy would be $H(s, t) = \frac{(1 - t) ( \cos 2\pi s, \sin 2\pi s, 0) + t(0, 0, 1)}{\|(1 - t) ( \cos 2\pi s, \sin 2\pi s, 0) + t(0, 0, 1)\|}$ and I can visualize the homotopy you mentioned: Map a point on the equator $f(s)$ to the point on the loop $\alpha(s)$, but when I try to define the homotopy, I end up having to pass through $c$, which creates the same problem.

@AlexanderGruber fractals - I can't be more specific than that...

This is the image I was trying to give.

The bottom is the "modified homotopy" that ends up with a path from the basepoint to the north pole. The difference is that at each point, you start with a path from the basepoint to $f_t(0)$ (and you can choose this path quite canonically... by doing $\gamma(t) = f_t(0)$.)

That's what I've been visualizing and writing except the other way around, I go to the contraction point (the north pole) and then to the basepoint.

If I had just thought about it the other way, it would have solved all my problems :/

Thanks a lot!

Oh, I see. Glad to help!

The general picture is not so different.

Is it now-here or no-where: en.wikipedia.org/wiki/Nowhere_dense_set ?

The latter.

Would the homotopy be $F(s, t) = \gamma(t) * H(s, t) * \overline \gamma (t)$?

You need to reparameterize $\gamma$ so that it's defined on $[0,1]$ (the one I did above naturally ends at $t$ instead of $1$), and it should depend on $s$ to determine where it ends... but otherwise, yup.

@MikeMiller How good are we at classifying 3-manifolds up to cobordism.

Smooth, @AlexYoucis?

@MikeMiller Is there anything but?

Not in the world of 3-manifolds, but our cobordisms are 4-dimensional.

@MikeMiller Yeah, but I want smooth everywhere.

Oh, misread. Didn't realize you were being rhetorical.

@MikeMiller :)

I don't know off the top of my head. Recall that two manifolds are smoothly cobordant iff their Whitney numbers are the same; in our case this means checking that the following things are the same: $w_1^3, w_1w_2, w_3$.

@MikeMiller What are there Whitney numbers?

The products of Whitney classes so that you get something of top degree. The above is the only possible in dim 3.

@MikeMiller Oh, I see.

@MikeMiller Silly question--what if they are non-orientable?

So the question comes down to computing Whitney numbers, and how good we are at that for 3-manifolds.

Nothing changes... Why would it?

Remember that every manifold is $\Bbb Z/2\Bbb Z$-orientable.

In particular, since there are 3 Whitney numbers, there are at most 8 cobordisms classes. I don't know if this number is realized.

Other things worth noting, @AlexYoucis: the cobordism group (in any dimension) is an abelian 2-group; and all orientable 3-manifolds are null-cobordant, since they're parallelizable.

@MikeMiller How does 3 beget 8?

@AlexYoucis $2^3$ possibilities.

@MikeMiller My god...it checks out.

@AlexYoucis I did a little googling. We're very good at classifying 3-manifolds up to cobordism: they're all the same.

@MikeMiller Also, did you just say that all 3-manifolds are parallelizable

@MikeMiller hooray.

No, I said that all orientable 3-manifolds are parallelizable.

@MikeMiller ...is that true...

apparently antilog is a thing

rather a term

Yes.

To Alex.

@MikeMiller Proof?

I had no idea this was true

@MikeMiller, sorry to bug you again, how do we know $H(s, t)$ is a loop if $t$ is fixed?, I know it's continuous since $H$ is continuous, but how do I know $H(0, t) = H(1, t)$ if all we know is that $H$ is a homotopy? For the equator-sphere case, we explicitly defined it, so it worked out, but what about this general case?

@AlexYoucis I'll just link you to the proof I found a while back, let me find it.

Perfect, it's the first google result here.

Why does it need to be a loop for fixed $t$, @RobertCardona? We should be demanding that it's a loop for fixed $s$.

Oh, you've got your variables in a different order than I do.

@MikeMiller Qiaochu says the proof is wrong.

@AlexYoucis I know. Read Qiaochu's corrected proof.

His proof (without Qiaochu's addendum) does indeed correctly show that 3-manifolds are null-cobordant, though, given the result I told you before...

MY bad, I got used to the notation Munkres used and haven't gotten accustomed to the style you mentioned which is the style Hatcher uses.

Intuitively it makes sense, if they weren't equal, then at some point $H$ would not be continuous because the loop breaks. But I'm not seeing it algebraically.

@RobertCardona So let me write it this way for my sanity. Let $\gamma_s(t)$ be the path from $x_0$ to $f_s(0)$ given by reparameterizing $\gamma(t) = f_t(0)$ to end at $1$ instead of $s$. Given a homotopy $f_s(t)$ from $f$ to a constant map, let $f'_s(t) = (\gamma_s * f_s * \overline{\gamma_s})(t)$.

Where I guess $f*g$ here means do $f$ then $g$.

Yes, but to do that in this case, we need to know $f_s$ is a loop, how do we know that?

By definition of the homotopy!

The original homotopy is a null-homotopy of a loop... each $f_s$ still has to be a loop. We just didn't demand that the basepoint stay fixed.

The definition of homotopy I have: $F : I \times I \to X$, $F(s, 0) = f(s)$, $F(s, 1) = c$, for our case. I know $f$ is a loop so $F(0, 0) = F(1, 0) = a$ where $a$ is the base point. But for arbitrary fixed $t$, I want to show $F(0, t) = F(1, t)$, correct?

To show that $F(s, t)$ is a loop for fixed $t$.

Your definition of homotopy is wrong if you're homotoping loops.

A homotopy between two maps $X \to Y$ is a map $X \times I \to Y$ that restricts (on $X \times 0$ and $X \times 1$) to the two maps.

In our case, $X = S^1$.

Yes.

That's what I wrote, isn't it? the first map is $f(s)$ and the second map is constant at $c$ since it's a nullhomotopy.

You write $I \times I$. That I wrote $S^1 \times I$ is a big difference! It means each of the $f_t$s are loops.

Yours is the same as mine as long as you demand that $F(0,t) = F(1,t)$ for all $t$.

If you want to talk about homotopy classes of loops, you've gotta demand that.

I was looking at the equator as a loop $\eta : I \to S^2$ defined by $\eta(s) = (\cos 2pi s, \sin 2\pi s, 0)$. In this case it would be $I \times I$?

Sure, that's fine. Then just demand that $F(0,t) = F(1,t)$ for all $t$ (i.e., that $f_t$ is a loop).

Here's the thing: I have $f$ a loop that's nullhomotopic, so I know there exists a homotopy $H : I \times I \to X$ where $H(s, 0) = f(s)$ and $H(s, 1) = c$ for some $c \in X$. How am I able to add restriction to it, given that all I know is that it exists?

No, the definition of null-homotopic includes that restriction. If it didn't, every loop would be null-homotopic. It's a stupid definition without that restriction.

So if I say a path $f$ is nullhomotopic, by definition that nullhomotopy is a path homotopy?

Well, every path $f$ is null-homotopic (by a homotopy of paths, i.e., a map $I \times I \to X$ that restricts to $f$ at $I \times 0$). I don't know that I would define it to fix a basepoint, but it doesn't really matter.

I feel like I've not been very helpful; sorry. I need to go. I suggest just checking the definitions involved again.

Thanks a lot for your help!

in general a homotopy between two maps doesn't require that either of them is a loop

but chances are you're working with a case where it is required

@cxseven, I'm starting with a loop, so it means all of them must be loops, correct?

I feel like it needs some showing though

probably, and if that's the case, a good text would state that explicitly somewhere

wait, do you mean required by definition or required as a consequence of one of the maps being a loop?

as a consequence

in the second case it depends on what you mean by "loop", since many people mean a continuous map from $S^1$ into the space

$\alpha : I \to X$ where $\alpha(0) = \alpha(1)$.

I have a homotopy, $H$ between a loop $f$ and the constant map at $c$. My question was, does it mean that $H(0, t) = H(1, t)$ throughout the homotopy?

for a homotopy $H$, the condition $H(0,0) = H(1,0)$ does not necessarily imply that $H(0,1)=H(1,1)$

chances are that your text required that $H(0,t)=H(1,t)$ for all $t\in[0,1]$

in other words that it was not a consequence of anything but rather assumed

of course, if instead of an interval $I$, you use $S^1$, then the topology of the domain would imply that $H(s,t)$ is a loop for every $t$

So here's what I'm trying to prove: Let $f$ be a loop in $X$. If $f$ is nullhomotopic, then there exists a homotopy which is a path homotopy.

a path homotopy between what and what?

between $f$ and a constant map, in this case, I'm saying it has to be the basepoint.

I probably should have said "..then there exists a nullhomotopy which is a path homotopy"

this question seems like it's too simple and so i must be missing something with regard to your book's terminology

This isn't from a book; and maybe it is too simple.

My question was as follows: If I have a loop which is nullhomotopic, do I know the homotopy is a path homotopy?

Not necessarily, an example would be taking the equator on the sphere to the north pole. It's not a path homotopy because the basepoint doesn't remain fixed.

from what I understand, a path homotopy keeps the endpoints fixed

right, so the homotopy demonstrating null-homotopy is not necessarily a path homotopy

yes! but all I know is that the loop is nullhomotopic, how do I know there exists a nullhomotopy that is also a path homotopy, that is, keeps the endpoints fixed, which in this case is the constant map at the basepoint

yes

but I want to show that one necessarily exists

yes, one must exist

why?

you can construct a homotopy that takes the loop to a point

now you can concatenate that with a homotopy that takes that point back to one of the original endpoints

That was the first thing I tried

but there is a problem

at the base point, it goes to the other point and then back again

hence it's not fixed there

unless they happen to be the same point already, in which case we're done

if it's not fixed there, it's not a path homotopy.

true, so you will need a "thread"

The problem there is that I need to know $H(s, t)$ is a loop for fixed $t$

which takes me back to the question I had a few minutes ago

If I know $f$ is a loop, is $H(s, t)$ a loop for fixed $t$?

If i know that, I can do the following concatenation $\alpha * H(s, t) * \overline \alpha$.

Essentially go from the basepoint, to a loop in the homotopy and then back to the basepoint.

But I get to pass through the original homotopy that I started with.

But you said that $H(s, t)$ isn't necessarily a loop for fixed $t$?

yes for each t you can have $s\in[0,1/3)$ be a continuous map from the original endpoint ($H(0,0)$) to the new endpoint, $s\in[1/3,2/3]$ be a rescaled continuous map from the original homotopy giving you a loop starting and ending at that endpoint, $s\in(2/3,1]$ be a continuous map from that endpoint back to the original endpoint

I can be explicit

So I'll need two distinct paths, one from the basepoint $a$ to $H(0, t)$ and the other from $H(1, t)$ back to the basepoint?

no, it's okay! I can work it out on my own.

yes

Thanks!

I was trying to use the same path both ways, but to do that I needed $H(s, t)$ to be a loop for fixed $t$, and wasn't sure if that was true or not

Thanks.

np

where's everyone?

hi @anon

just learned that the artist behind the Injustice comics is named Mike Miller

I just learned that there are two Mike Miller

there are a lot of us

only 2 on MSE, though

Oops.. $3$

there's a third?

Greetings

$$\int_0^1 \frac{\log(1-x)\log(x)\log(1+x)}{1-x} \ dx$$

$$\int_0^1 \frac{\log(1-x)\log(x)\log(1+x)}{1+x} \ dx$$

It would be interesting to see the way Terence Tao would approach some of these integrals I use to post here (the more advanced ones).

(or other worldwide known mathematicians)

oh noes. am i seriously on ignore?

@Chris'ssis Terence Tao doesn't do integrals.

He is a harmonic analyst.

@BalarkaSen I highly doubt he doesn't do integrals at all when needed (or maybe for fun?).

I am sure he doesn't do ad hoc type integrals.

Very small proportion of mathematicians research on just integrals.

@BalarkaSen Did you talk to him on this topic?

@Chris'ssis You can look at his papers.

None of them are on say finding closed form of some very large and complicated integral.

@BalarkaSen This doesn't mean he doesn't attend integrals at all. Maybe some arise in his work and need to be treated.

Maybe, but we're just speculating at this point.

@BalarkaSen As you did above: "Terence Tao doesn't do integrals."

I didn't.

@BalarkaSen Well, you did it.

He doesn't. He never wrote a research paper on integrals.

@BalarkaSen This leads to the conclusion that "Terence Tao doesn't do integrals."?

Yes.

@BalarkaSen This is absurd to me.

@BalarkaSen Anyway.

Let's just agree that we both like to contradict each other and drop this discussion.

thinks to himself 'I should come here more often, good conversation occurs'

@DanZimm ^ this is good conversation to you? :P

made me chuckle at least

confus. why should it make someone chuckle?

it's a highly philosophical super speculative contradictory discussion.

it should make people run away :P

O.o

"highly philosophical" I would say is debatable

other than that surely

Let me see what other books on integrals, series and limits appeared in the last period of time ...

Anything which is not mathematics is philosophy, @DanZimm

:P

If I may, I'm going to quote that

a lot

LOL

@UserX Waste of time, I should say :P

I plan to study the metrization theorem at some point.

It looks interesting.

@BalarkaSen agreed.

@DanZimm what branch of mathematics do you study?

Currently in school so I tend to study everything (not doing any sort of dissertation yet), but my interests are in modern PDE theory

currently working on a dual problem to the p-laplacian, if you're familair

ugh.

ughh.

$$ \begin{cases} \operatorname{curl}\left( \lvert b \rvert^{q-2} b \right) = 0 \\ \operatorname{div}(b) = 0 \end{cases} $$ where $b : \mathbb{R}^3 \to \mathbb{R}^3$ and $q \in (1, \infty)$ - I'm interested in the limiting problem when $q \to \infty$

OK, I have no intuition for that stuff.

Also why ugh?

It's fairly technical at this point

There's no hope to find classical solutions (or solutions which are smooth and you can write out the formula for them)

so why are you interested in finding a solution?

i mean, what's the motivation?

These PDEs arise from physical problems

so since nature abides by the PDEs we should probably be able to come up with some sort of solution

If you wanna go really applied you can start to work with engineers and try to find properties of the solutions

e.g. convex a.e. on this domain

(although a property like that may be fairly difficult to find)

is still not a mathematician so he can't really speak much on these things

I never really liked this PDE stuff

to each his own

I'm generally interested in analysis also, so optimal control + calculus of variations tempts me too

optimal control I really know just about nothing though, but calculus of variations I know a fair bit

i like point-set topology though. what about you?

Very fun! Unfortunately my topology teacher skipped my favorite part (sequences/nets) in order to cover an intro to algebraic topology

algebraic topology is a bit technical and to some extent boring.

not sure, as i am only a beginner

The worst to me is how my professor insists on the class working in groups the majority of the time

so we have been struggling through the basics instead of being lectured about them

I much prefer to struggle on my own doing homework ;P

if you don't struggle, you'll never really learn it

My point was I don't like our classes being only group work

I'm fine with it here and there but I enjoy lectures generally

well cool ideas always seem cool

but to get used to it, you need to nail it up.

I suppose

@BalarkaSen what do you study?

or like to study rather

algebra and number theory mostly.

Ah fun!

thinks all math is fun

except straight calculations

Pretty merh about calculations

i am actually pretty interested in galois theory

@DanZimm calculations are important :)

to some extent yes - but the calculations I'm talking about are physics calculations - after being a physics major, I'm much more interested in a more rigorous mathematical approach

you can't just do mathematics with the click of a mouse.

@DanZimm ah i see

i believe mathematics is a place to bake your ideas :)

e.g. "here's a formula, here are what the variables are except for "x", find "x""

from my experience that's what a ton of it is

even if the formula you have to derive from several other formulas

formulas are meh if you don't have a motivation

yea

hahaha, an answer posted to one of my questions made my day! :-)

@robjohn see above the last answer posted, and particularly the last integral.

Can we find a fast, easy way to evaluate $$\int_0^1 \frac{\log(2)-\log(1+\sqrt{1-x^2})}{x} \ dx$$?

@BalarkaSen ^^^

what @Chris'ssis?

@BalarkaSen ^^^ (the integral above, yeah)

you mean the integral?

@BalarkaSen It's exceptionally nice.

i am not in the least interested integrals, sorry.

OK

chat.stackexchange.com/rooms/19076/…

@Sawarnik what do you mean waste of time?

@UserX Bathing.

Everyday.

@Sawarnik I don't feel okay the whole day if I haven't bathed the morning. I guess that's my OCD part.

Aww.

Like Balarka's OCD part is he doesn't feel okay the whole day if he hasn't done math that day.

Well I got that too...

Not really..

Well, I do

Nah.

Kay

Not in a mood to argue

I would like to do math each day too, but I won't get really itchy if I don't get to it once a year....

@UserX How is your smoking habits these days? Any improvements?

I don't understand the question here.

@Sawarnik You don't really do any math any day.

@BalarkaSen I do.

You don't get to know.

Nah

@BalarkaSen you owe me a problem

@UserX ?

@BalarkaSen you said you ll give me a problem but you never did

oh that one. have you studied lagrange's theorem?

Only the mean value one

^

then forget about it, @UserX. you need lagrange to do the problem i was going to propose.

@Balarka Do you feel that bathing everyday is a waste of time too? :D

waste

:D :D

everything nonmathematical is waste

waste waste waste

That's such a stupid statement @BalarkaSen.

Surely you're joking Mr. Balarka!

@UserX Agreeed.

@UserX i am aware that i am being silly. go study lagrange rather than critisizing me.

:P

Well why not

Gotta eat chicken nuggets first

@BalarkaSen And you better improve your spelling ...

Spelling*^

LOL

?

@UserX Didn't your girlfriend get irritated when you did math all day?

@UserX OK, Lagrange's theorem states that if H is a subgroup of G then the order of H divides the order of G.

Eww. Boring stuff starts.

Can you use this to determine how many groups of order 4 are there, upto isomorphism? Can you enlist them?

@Sawarnik You should start studying it.

@BalarkaSen Currently I studying much geometry :D, as RMO is one week away. Please forgive me :O :O.

/bye.