Mathematics - 2019-12-09 (page 3 of 3)

Alessandro Codenotti

8:04 PM

@TedShifrin I pinged both then, because I first wrote Ted and then edited in TedS :P

Do you know if the thing I was wondering about holds? (I can write it again if you didn't see it)

Rithaniel

I am currently learning that even accurately modeling a trefoil is tricky

FuzzyPixelz

If $f,g: E \to F$ are two bounded functions, is it true that $\| fg \|_{\infty} \le \| f \|_{\infty} \| g \|_{\infty}$ when $F$ isn't $\mathbb R$ nor $\mathbb C$?

Ted Shifrin

Yes, if you use the Riemannian metric to define your metric balls (in the definition of Hausdorff), then $dV = d\mathscr H_n$.

Alessandro Codenotti

I see, thanks

Ted Shifrin

You should check in usual Euclidean space.

Rather than Federer, look at Morgan's gentle GMT book.

Alessandro Codenotti

8:17 PM

So the point of defining the integral of forms and of compactly supported smooth functions is that you can actually compute stuff? Because by using the Hausdorff measure I can integrate many more things

Ted Shifrin

But you can't actually perform any calculations.

Alessandro Codenotti

@TedShifrin I was joking when I said I said I was going to check Federer! But thanks for suggesting a (hopefully) more reasonable alternative

@TedShifrin with the Hausdorff measure you mean?

Ted Shifrin

The point of Hausdorff measure, to me, is to get a canonical measure on lower-dimensional creatures.

Yes, that's what I meant.

Alessandro Codenotti

Makes sense

Ted Shifrin

And, of course, to talk about non-integral Hausdorff dimensions.

Alessandro Codenotti

8:19 PM

But we saw how to get a canonical measure on embedded hypersurfaces, can this be done for lower dimensional submanifolds?

Ted Shifrin

You mean in the smooth, oriented case?

Alessandro Codenotti

Yes

Ted Shifrin

Hausdorff measure, of course, makes sense on far more general objects.

Alessandro Codenotti

Sure

Ted Shifrin

Of course, if you have a submanifold of any dimension, there's an induced metric, and you can write down the associated volume (in appropriate dimension) form.

It's a density in the case that the submanifold fails to be orientable.

Alessandro Codenotti

8:23 PM

A density?

@TedShifrin I see, thanks

user193319

Let $G$ be a locally compact topological group. If $V$ is a symmetric, relatively compact open neighborhood of $1$, twhy is it true that $\overline{V}^n \subseteq V \cdot V^n$?

Ted Shifrin

A density in this case is basically the absolute value of a top-degree form.

Physicists might call it a pseudo-volume form.

Alessandro Codenotti

If $M$ is orientable is an embedded hypersurface necessarily orientable?

Ted Shifrin

NO. And you know a counterexample.

Alessandro Codenotti

I do

Moebius strip in $\Bbb R^3$

Ted Shifrin

8:29 PM

LOL. Is there a ? or a ! ?

The right question is a closed (compact) hypersurface.

Alessandro Codenotti

And what's the right answer to the right question?

Ted Shifrin

It's still NO.

Alessandro Codenotti

anticlimactic

Ted Shifrin

LOL ... no, more interesting. Here's the cool theorem. An embedded (closed) hypersurface in a simply connected manifold is orientable.

Muhammed Ç. TUFAN

Hello!

Alessandro Codenotti

8:31 PM

Do I still know a counterexample?

Muhammed Ç. TUFAN

May I ask you question?

Ted Shifrin

Yes, you definitely know a counterexample. Give me the simplest compact non-orientable manifold.

You just did, @Muhammed :P

Alessandro Codenotti

$\Bbb R\Bbb P^2$ but I don't know how to embed it into a 3-dimensional thing

Muhammed Ç. TUFAN

Wow, I really did hahah :D

Ted Shifrin

Sure you do, @Alessandro :)

Do you care to ask a second question, @Muhammed? :)

Muhammed Ç. TUFAN

8:32 PM

So, may I ask you my next question? :P

you did

@TedShifrin Hm

Hi all! btw

This is kind of a infinite question loop. I should better ask directly hahah

Ted Shifrin

Indeed, @Muhammed :)

hi, @Rudi

Muhammed Ç. TUFAN

8:33 PM

I'm working on something and my brain is about to emergency shut off...

Rudi_Birnbaum

then better sleep

Ted Shifrin

@Alessandro: In your topology course, how did you learn about projective spaces?

Rudi_Birnbaum

hi @Ted :-)

Ted E

I read "This is kind of a infinite question loop" as 'This is a question about infinite loop spaces'

Muhammed Ç. TUFAN

$\cos\theta=-\frac{M_2}{M_1}$ so what is $\sin\theta$?

Ted Shifrin

8:34 PM

smacks other Ted

Ted E

:-)

Ted Shifrin

It could have one of two possible values, @Muhammed.

Ted E

Hi @Rudi_Birnbaum

Alessandro Codenotti

Either as a quotient of $S^2$ or by gluing together a square

Rudi_Birnbaum

or even more ...

Ted Shifrin

8:35 PM

Unless you know what quadrant $\theta$ is in, $\sin\theta$ could be either positive or negative.

Rudi_Birnbaum

Hi other@TedE

Muhammed Ç. TUFAN

You guys really funny, I can sit and talk with you guys all of my life lol

Leaky Nun

and it can't be any other value by the factor theorem

Rudi_Birnbaum

we didn't have the pleasure, yet, did we?

Ted Shifrin

@Alessandro: Did you not do higher-dimensional ones?

Alessandro Codenotti

8:36 PM

Sure, quotients of bigger spheres

Or starting from a vector space as well

Ted Shifrin

Cell structure?

Alessandro Codenotti

Oh right, I forgot about that

Muhammed Ç. TUFAN

Hm, @TedShifrin can you check my question on math stackexchange please?

Ted Shifrin

@Muhammed: So don't call it $-M_2/M_1$. Just call it $w$. If $\cos\theta = w$, what can $\sin\theta$ be?

Well, you need to give a link :P

Muhammed Ç. TUFAN

I think it would be a bad decision to ask little questions about my main question and finally connect them together..

0

Q: What is the domain of the function $\tan\theta_{1}(\theta)=\frac{\sin\theta}{\cos\theta+1}$ and what is $\theta_1^{\max}$?

I have some questions about the lecture that I took today on Physics. Consider that the cosine is defined as below, $\cos\theta=-\frac{M_2}{M_1}$ $M_1$: Mass of the first object. $M_2$: Mass of the second object. (Sorry for the physical terms, I was undecided between open this question here ...

trigonometry

Leaky Nun

8:37 PM

@AlessandroCodenotti play?

Muhammed Ç. TUFAN

Here it is, thank you by the way.

Alessandro Codenotti

@LeakyNun I already played waaaaay too much today

Leaky Nun

ok sure

Ted Shifrin

@Muhammed: What is this $\theta_1(\theta)$ nonsense?

Rudi_Birnbaum

@MuhammedÇ.TUFAN that's actually a wise statement (that however may not be appreaciated by mathematicians ..)

Muhammed Ç. TUFAN

8:38 PM

Every second you spend to talk to me and help to me is a piece gold to me..

Ted Shifrin

@Alessandro: So thinking about cell structure gives you an idea?

Alessandro Codenotti

I won a game against a player slightly above 2100 earlier though (it was on time and he was berserking but I'm proud of it nonetheless lol)

Leaky Nun

nice

Ultradark

nice!

Ted Shifrin

Oh, I see, we want $\tan \tau$ to be the given fraction and we're saying this gives $\tau$ as a function of $\theta$.

Leaky Nun

8:39 PM

@user193319 what on earth is $\overline{V}$?

oh closure

Muhammed Ç. TUFAN

@TedShifrin I don't really know, my teacher did it, I should better to ask him what he is up to.

Ted Shifrin

No, I get it. But there is a mistake. If $M_1=M_2$, then the denominator should be $\cos\theta-1$ not $\cos\theta+1$.

Muhammed Ç. TUFAN

But, the $\tan\theta_1$ is a function of $\theta$.

Leaky Nun

24 mins ago, by FuzzyPixelz

If $f,g: E \to F$ are two bounded functions, is it true that $\| fg \|_{\infty} \le \| f \|_{\infty} \| g \|_{\infty}$ when $F$ isn't $\mathbb R$ nor $\mathbb C$?

Alessandro Codenotti

@TedShifrin I guess I'm supposed to attach a 3-cell to make it orientable. I'm still confused though, but I'm also very tired so I'll blame it on that and think about this tomorrow

Leaky Nun

8:40 PM

@FuzzyPixelz so what is $F$?

16 mins ago, by user193319

Let $G$ be a locally compact topological group. If $V$ is a symmetric, relatively compact open neighborhood of $1$, twhy is it true that $\overline{V}^n \subseteq V \cdot V^n$?

what should "relatively compact" mean

FuzzyPixelz

A normed vector space, sorry.

Leaky Nun

over $\Bbb R$ or $\Bbb C$?

Ted Shifrin

Think about the cell decomposition of $\Bbb RP^3$, @Alessandro. You're building $\Bbb RP^n$ by attaching a disk to $\Bbb RP^{n-1}$.

Leaky Nun

oh I guess it means its closure is compact

Ted Shifrin

@Muhammed. You are messing up your arccos stuff pretty bad there.

Alessandro Codenotti

8:42 PM

WAIT

Are you suggesting that $\Bbb R\Bbb P^3$ is orientable?

FuzzyPixelz

Either @LeakyNun

Muhammed Ç. TUFAN

@TedShifrin the tangent is $\tan\theta_{1}(\theta)=\frac{\sin\theta}{\cos\theta+\frac{M_1}{M_2}}$. So the denominator is $\cos\theta+1$>

Leaky Nun

@FuzzyPixelz for any $x \in E$, $|(fg)(x)| = |f(x)| |g(x)| \le \|f\|_\infty \|g\|_\infty$ since $|f(x)| \le \|f\|_\infty$ and $|g(x)| \le \|g\|_\infty$ for all $x$; taking the supremum gives $\|fg\|_\infty \le \|f\|_\infty \|g\|_\infty$ as required

you only need the two defining properties of $\sup$

which some might call universal properties

(and get smacked by Ted)

Ted Shifrin

Oh, sorry, I looked too quickly @Muhammed. OK, so the denominator is $\cos\theta+1$, which approaches $0$ (but always with positive numbers) as $\theta\to\pi$.

FuzzyPixelz

Umm, my whole problem is with the first equality..

Leaky Nun

8:44 PM

@AlessandroCodenotti well its homology is $\Bbb Z, \Bbb Z/2\Bbb Z, 0, \Bbb Z$...

the final $\Bbb Z$ implies that it is orientable

or not

FuzzyPixelz

(may Ted have mercy on me)

Leaky Nun

the final $\Bbb Z$ must have some meaning

Muhammed Ç. TUFAN

It's OK, I understand @TedShifrin.

Alessandro Codenotti

I don't know @Leaky

Ted Shifrin

But we have to be more careful, because the numerator also goes to $0$, @Muhammed.

Do you want me to write an answer to your question? You know basic calculus stuff, right?

Muhammed Ç. TUFAN

8:46 PM

But I didn't understand what you mean at your last message :(

Ted Shifrin

What is the limit of $\dfrac{\sin\theta}{\cos\theta+1}$ as $\theta\to\pi$?

Muhammed Ç. TUFAN

I know but I don't understand how denominator and nominator goes to $0$ :/

Ted Shifrin

$\sin\pi = 0$ and $\cos\pi+1=0$.

Muhammed Ç. TUFAN

I'm a little bad at chained limits..

Ted Shifrin

Chained?

Muhammed Ç. TUFAN

8:47 PM

When something goes to something, something goes to something thing you know

I mean ummm

Ted Shifrin

Note that $\dfrac{\sin\theta}{\cos\theta+1} = \dfrac{\sin\theta(1-\cos\theta)}{(1+\cos\theta)(1-\cos\theta)} = \dfrac{\sin\theta}{\sin^2\theta} \cdot (1-\cos\theta)$.

So you have $\dfrac{1-\cos\theta}{\sin\theta}$. What happens to that when $\theta\to \pi$?

From the physics do you know that $\theta$ has to be in certain intervals (quadrants)?

The point is that this fraction blows up, but it can go to either $+\infty$ or $-\infty$ depending on whether $\theta<\pi$ or $\theta>\pi$.

@AlessandroCodenotti I'm not suggesting it. It's a fact (that you should know) that $\Bbb RP^n$ is orientable iff $n$ is odd.

One way you can convince yourself (without homology) is that the antipodal map on $S^n$ is orientation-preserving precisely when $n$ is odd.

Alessandro Codenotti

Uhm I'm not sure if I knew this and forgot or not

Leaky Nun

@AlessandroCodenotti orientability is famously equivalent to the existence of a non-vanishing top form. One such form for $S^n$ is $\omega_{(x_0, x_1, \cdots, x_n)} = \sum x_i \ \mathrm dx_0 \land \cdots \widehat{\mathrm dx_i} \cdots \land \mathrm dx_n$ corresponding to the normal vector field $v_r = r$. Induce this via the quotient map to $\Bbb RP^n$ and you get $\omega_{[x_0 : x_1 : \cdots : x_n]} = (1 + (-1)^{n+1}) \sum x_i \ \mathrm dx_0 \land \cdots \widehat{\mathrm dx_i} \cdots \land \mathrm dx_n$

@TedShifrin am I doing the induction correctly?

Muhammed Ç. TUFAN

@TedShifrin it goes to $0$. I hope I'm not wrong. By the way Google Chrome Mobile is not converting MathJax commands into rendered formulas, do you know why it is not?

Leaky Nun

you need a desktop

Ted Shifrin

8:57 PM

No, slow down. First you're pulling back the form by the antipodal map and getting a factor of $(-1)^{n+1}$. This means that the form is invariant precisely when $n$ is odd.

Muhammed Ç. TUFAN

Oh, okay. Thanks for the information @LeakyNun.

Ted Shifrin

See the LaTeX in chat link on the right, but I'm not sure you see that other than on a desktop.

@Leaky: So $A^*\omega=\omega$ when $n$ is odd. This means that $\omega$ descends to a well-defined form on the quotient.

Leaky Nun

yeah but there must be some general notion of induction for a covering map between manifolds

Ted Shifrin

When $n$ is even, you can average and get $0$, but there's no well-defined form other than $0$ that descends.

There's something called the trace (analogous to the Galois theory trace).

Induction is a confusing word to use. You really need a form invariant under the $G$-action.

Leaky Nun

and it's what I wrote down?

Ted Shifrin

9:00 PM

You have to do local pullbacks by local inverses and then average, but in this case that gives you $0$.

Leaky Nun

oh no

why would I get 0?

Ted Shifrin

$0$ when $n$ is even, OK when $n$ is odd, but you're making it too difficult. An invariant form descends to the quotient.

Leaky Nun

I suppose so

Ted Shifrin

@MuhammedÇ.TUFAN Sorry, too much going on. What goes to $0$? The fraction does not. The numerator goes to $2$ and the denominator goes to $0$, so the quotient goes to $\pm\infty$.

Leaky Nun

@AlessandroCodenotti can you concretely prove that there are exactly two vertices of degree 1?

I like my concrete approach

label the $n^{\text{th}}$ graph as $0 - 1 - 2 - \cdots - n$

so an element of the ultraproduct is $(a_n)_{n \in \Bbb N}$ where $0 \le a_n \le n$

Muhammed Ç. TUFAN

9:03 PM

Oh, the numerator goes to $2$ true. But when we think that denominator goes to $0$ the fraction must go to only $+\infty$, is it wrong @TedShifrin?

Ted Shifrin

This is wrong. Because the denominator can be negative, approaching $0$. This is what I was talking about above. If $\theta>\pi$, then $\sin\theta$ will be negative. That's why I asked if you had some limitations on where $\theta$ can live.

Muhammed Ç. TUFAN

It would be nice to work on limits hard, I guess. Our teacher did yell at the class about that :/

Oh, OK. I get it now.

I wish I know the limitations, the teacher didn't told anything about it.

Leaky Nun

I find it instructive to look at a graph

Muhammed Ç. TUFAN

Hmm, now if the denominator goes to $\pm\infty$. What would happend, let me think please.

@TedShifrin thanks for your all help, I'm grateful to you so much.

Ted Shifrin

It means that when we're near $\theta=\pi$, $\tan\theta_1$ is going crazy, going to $+\infty$ when $\theta<,\pi$ and to $-\infty$ when $\theta>\pi$. So $\theta_1$ is going crazy, going to $+\pi/2$ sometimes and $-\pi/2$ other times. I think you need to ask your teacher more questions.

Muhammed Ç. TUFAN

9:09 PM

I will spend some more hours on that thing myself I think :/

Ted Shifrin

If you know that $0<\theta<\pi$, for example, then you can say what he said, that $\theta_1$ tends to $\pi/2$.

Muhammed Ç. TUFAN

Hmm, I got.

Ted Shifrin

Let me know what happens :)

And yes, you need to get good at basic calculus and limits :P

Muhammed Ç. TUFAN

I want to ask you a question shyly.

Leaky Nun

have we lost @AlessandroCodenotti?

Ted Shifrin

9:11 PM

@Leaky: I think he was offended that $\Bbb RP^3$ is orientable.

Muhammed Ç. TUFAN

I want to tell you all the background of this question.

I tried to simplify the question. You might understand the event.

Do you know a little about physics?

Ted Shifrin

Hello @D.ZackGarza ... I see you're at my former home of 34 years.

@Muhammed Yes, I know some physics.

Muhammed Ç. TUFAN

Okay, than here I go

Ted Shifrin

We even have a professional physicist, @Semiclassical, who is often here.

Muhammed Ç. TUFAN

Imagine a collision event, there are only 2 objects.

Oh, thats really good to hear. I will see him often in the future I think :P

Ted Shifrin

9:14 PM

Or you could go to Physics.SE.

Muhammed Ç. TUFAN

The first object has $\vec{V}_{1}$ velocity.

Ted Shifrin

and mass $m_1$, no doubt.

Thorgott

lol

Muhammed Ç. TUFAN

Yeah, it would be more accurate choice.

D. Zack Garza

@TedShifrin Oh how cool! And I think I've seen some of your linear algebra videos. A hearty hello from Boyd :)

Muhammed Ç. TUFAN

9:15 PM

Yep absolutely

Ted Shifrin

A hearty hello back. Say hi to my old office (444). :)

Muhammed Ç. TUFAN

And the second object has $\vec{V}_{2}=\vec{0}$ velocity. So it is inertial.

Ted Shifrin

Well, until it gets hit.

Is this going to be an elastic collision?

Muhammed Ç. TUFAN

Yeah everything will be conserved.

Ted Shifrin

OK.

Leaky Nun

9:17 PM

@user193319 here's a high-powered way to kill it using what I call the half-lemma: by the half-lemma, there's a neighbourhood $W$ of $1$ such that $W^n \subseteq V$. Then, for $x \in \overline{V}$, we have $xW \cap V \ne \varnothing$, so $xw \in V$ for some $w \in W$, so $x \in W^{-1} V$. So $\overline{V} \subseteq W^{-1} V$. So $\overline{V}^n \subseteq W^{-n} \cdot V^n \subseteq V \cdot V^n$

the proof of the existence of $W$ is to note that the multiplication map $G^n \to G$ is continuous, so the preimage of $V$ is a neighbourhood of $1$, and by the product topology it must contain $W \times W \times \cdots \times W$ for some neighbourhood $W$ of $1 \in G$

so "$G$ locally-comapct" and "$V$ relatively compact" are unnecessary conditions

Ted Shifrin

@Leaky: That sounds tube lemma-ish.

So don't you need compactness?

Leaky Nun

I don't think I used compactness

Ted Shifrin

As I said, it sounds tube lemma-ish. How do you get a neighborhood like that?

Oh, no. I'm wrong.

Leaky Nun

the tube lemma seems to be the "reverse" direction

Muhammed Ç. TUFAN

Ted Shifrin

9:22 PM

It's a neighborhood of $1\times\dots\times 1$.

Leaky Nun

right

Ted Shifrin

Yeah, this is the standard sort of problem, @Muhammed. So what's your question?

Muhammed Ç. TUFAN

On the top, you see the before collision and the bottom you see after the collision.

Ted Shifrin

So you just take the intersection of the finite number of neighborhoods @Leaky.

Muhammed Ç. TUFAN

I'm going into it now.

Leaky Nun

9:22 PM

yup

Muhammed Ç. TUFAN

Do you know about the center of mass frame?

That we stay at the center of the mass of the system and watch everything from there.

Ted Shifrin

OK. Let's not take an hour with this. Get to the question.

Muhammed Ç. TUFAN

Okay, the aim of this example is finding the $\theta_1^{\max}$ by examining the collision with center of mass frame.

The center of mass position is moving as the first object moves to second object.

Ted Shifrin

In other words, when $M_2$ is essentially infinite.

Muhammed Ç. TUFAN

Try to imagine it.

Mmm, sorry I would not answer you no, but not that way :/

Let me take a shot of the collision from the perspective of the center of mass frame.

That image shows the collision from the perspective of center of mass frame.

We sit at the center of mass position of those objects and watch everything happening there.

Ted Shifrin

9:33 PM

OK

Muhammed Ç. TUFAN

If the CM has a velocity (in this example it has, as the $M_1$ closes to $M_2$ CM closes to $M_2$) we have a velocity of that velocity of CM

Now I'm going to write the conservation of kinetic energy in the non CM frame.

$\frac{1}{2}{M_1}{V_1}^2+\frac{1}{2}{M_2}{V_2}^2=\frac{1}{2}{M_1}{V_1^{'}}^2+\frac{1}{2}{M_2}{V_2^{`}}^2$

The ${V_1^{}}$ and ${V_2^{}}$ are the velocities of the objects after the collision.

Now I will write the conservation of momentum

{M_1}{\vec{V}_1}+{M_2}{\vec{V}_2}={M_1}{\vec{V}_1^{'}}+{M_2}{\vec{V}_2^{'}}

Is everything OK?

I'm sorry If im boring you, if you are boring you can be honest to me. It would make me happier

Ted Shifrin

I asked you 20 minutes ago to make it fast. This isn't fast.

And this really isn't a physics question site. Although I think this is nice stuff, you're turning it into more than an hour.

Muhammed Ç. TUFAN

Yeah, It's not. I'm sorry. I'm trying to explain it from my mobile phone :/

Yeah, I agree with you.

Ted Shifrin

It's almost impossible to do complicated math/physics on a phone.

Muhammed Ç. TUFAN

I should really better to stop it.

Ted Shifrin

9:45 PM

And it takes forever.

When you have something specific and brief, by all means ask. But if you're going to do long things, it's better to make a question on the main site and take your time putting it up.

Muhammed Ç. TUFAN

Yeah, you are right. I will do that for the better results.

I'm thankful to you for everything.

You helped me with your all, I believe it.

Have a good day, thanks again.

Ted Shifrin

I will give you a physics puzzle in exchange next time.

You're welcome.

Muhammed Ç. TUFAN

I would be happy if you do that :P

Good bye!

Ted Shifrin

Take care!

user12692

10:26 PM

Dec 6 at 15:34, by Jack

I may miss something, but this answer seems too simple to be true:

user12692

Dec 6 at 15:34, by Jack

0

A: Dirichlet kernel inequality

All you need to know is that $3 < \pi < 4$ and $e > 2$. As $n \ge 1$ you have $1 + \dfrac 1{2n} \le \dfrac 32$ so that $\dfrac 2\pi \left( 1 + \dfrac 1{2n} \le \dfrac 32 \right) \le \dfrac 3 \pi < 1$, and $\ln \pi < \log_2 4 = 2$.

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$Mathematics$ Mathematics