Mathematics - 2018-09-15 (page 3 of 4)

Poline Sandra

17:00

the exercise say let $(E,d)$ a metric connected space with an unbounded distance

mercio

why do you think it is the discrete distance

what number is $|3-1|_{\Bbb Q}$ ?

Poline Sandra

I don't know, for the topological space yes, but for the distance I don't know

mercio

but you can't do the exercise if you don't have a good idea of what $|\cdot |_{\Bbb Q}$ is

Poline Sandra

that is why i ask

mercio

maybe you could ask your instructor ?

or pore over the book to see if they gave a definition of that distance ?

but I really really think they meant that it is the usual absolute value

they really is no reason to think otherwise

Poline Sandra

17:08

$|3-1|_{\mathbb{Q}}=2$

mercio

okay

then it is the normal distance and not the discrete distance

and in particular it is unbounded

Poline Sandra

do you know how we find the induced distance in general

mercio

you pick the distance on the bigger set and you restrict it to the smaller set

Poline Sandra

???

mercio

if you have a metric space (X,d)

and a subset Y of X

the restriction of d on Y is a metric on Y called the metric induced by d

Poline Sandra

17:16

yes

Secret

mercio

and usually if someone talks about an induced distance, you should have a pretty good idea of who X and Y are

Secret

got a really weird paradox here:

The three triangles should all be similar to each other due to all having the hypotenuse of slope 2

Using these, we have:

Akiva Weinberger

So I wrapped my head around why $\pi_3(S^2)=\Bbb Z$ and why $\pi_4(S^3)=\Bbb Z_2$ but I keep on trying to visualize $\pi_4(S^2)=\Bbb Z_2$ and failing

Curse you, higher dimensions

Secret

$\frac{e}{b-a\pi} = \frac{e}{ae} \implies b = a(\pi +e)$

But using another triangle, I get instead:

Akiva Weinberger

17:19

and your wibbly wobbly spheres

Ted Shifrin

Howdy, DogAteMy

Secret

$\frac{\pi e - (\pi +e)}{a\pi e + b} = \frac{e}{ae}\implies \pi e - (\pi +e) = \pi e + \frac{b}{a} \implies -a(\pi + e) = b$

So combining the two results together, we got $b=-b \implies b= 0$ which is absurd because the y intercept as shown clearly equals to something

Ted Shifrin

DogAteMy: How did you wrap your head? Do you know about the Thom-Pontryagin result relating this to framed cobordism?

Akiva Weinberger

I don't

All I have are pictures

Ted Shifrin

Ah.

Akiva Weinberger

17:23

$\pi_3(S^2)$ is this:

Ted Shifrin

I guess I'd dig up a fiber bundle and play with the long exact sequence. That's not exactly going to be a picture.

Akiva Weinberger

So those are each meant to be solids of revolution around the $y$-axis

so like the second picture is two sphere bits connected by a cylinder

Ted Shifrin

I don't get it.

The Hopf map generates, and I don't see a picture like that.

mercio

I think it is $S^4$ that you should wrap around your head, not the other way around

Akiva Weinberger

This is a movie of a sequence of slices of a map from $S^3$ to $\Bbb R^4\setminus\{\text{line}\}$

Leaky Nun

17:25

what is the significance of that map?

Ted Shifrin

Even if I understood the pictures, how does that prove that you get $\Bbb Z$?

Akiva Weinberger

aka an element of $\pi_3(\Bbb R^4\setminus\{{\rm line}\})$

Eric Silva

sup my dudes

Ted Shifrin

There's @Eric !!

I inquired of Demonark the other day if he'd seen you.

Eric Silva

i have returned to the western hemisphere

Ted Shifrin

17:26

Western?

Eric Silva

i was in taipei and then tokyo for the past few eeeks

Ted Shifrin

Oh wow.

Akiva Weinberger

In the second frame, that's two sphere parts connected by a cylinder. In the next frame, it transforms to a one-sheeted hyperboloid which transforms into a double cone, 'cause you're rotating the inner sphere

Joe Shmo

I'm trying to show the inexistence of a holomorphic function f = u + iv, who's modulus = K/coshx. After differentiating the squared modulus of the function w.r.t to x, and y, and applying Cauchy-Riemann, I am arriving at (1/v)(du/dy) = tanhx, but I don't see how that's a contradiction.

Eric Silva

eatin the good foooooood

Akiva Weinberger

17:27

In the frame after, you continue rotating the inner sphere so it goes back to a cylinder

Ted Shifrin

I guess Demonark is taking the GRE today.

Sounds exciting, Eric. I hate you.

Eric Silva

should be done by now

Ted Shifrin

Aren't the subject tests usually in the PM?

Eric Silva

lol i would hate me too Ted

Akiva Weinberger

I technically don't have a proof that it's not nullhomotopic or that it generates everything, but that is the Hopf map

Eric Silva

17:28

i think both the ones in the chicagoland area were 8am

Ted Shifrin

DogAteMy: I realized years ago that, although I'm quite geometric, a lot of the pictures in topology aren't intuitive to me.

@JoeShmo: Graph the function $K/\cosh x$ and think.

Akiva Weinberger

@TedShifrin Here's an interesting picture. Imagine a cylinder

Ted Shifrin

OK, I have a cylinder. ... Oh, btw, the Gauss-Bonnet thing is finally settled.

Akiva Weinberger

So that's two circles on top of each other, and each point on the top circle corresponds to a point in the bottom circle and they're connected by a straight line segment

and those line segments together make up the side of the cylinder

Ted Shifrin

Yup, a simple developable ruled surface.

Akiva Weinberger

17:31

So if you rotate the bottom circle by a small amount, but keep those lines straight,

Ted Shifrin

Yes, I know this. I've even done it in class.

Akiva Weinberger

the side of the cylinder becomes a one-sheeted hyperboloid. Yeah?

Ted Shifrin

Yup.

More amazingly, it's doubly ruled.

Akiva Weinberger

And as it continues to rotate, the hyperboloid gets sharper and sharper until - for an instant - it intersects itself and looks like a double cone

Ted Shifrin

Challenge: Classify all the doubly-ruled surfaces.

Yup.

Akiva Weinberger

17:32

and then it continues to rotate until it's back to where it started.

OK?

Ted Shifrin

I think once it tangles, it stays tangled (at least with physical string).

Akiva Weinberger

Not if the strings can pass through themselves

Ted Shifrin

Well, they can't with real strings.

Akiva Weinberger

This is true.

Ted Shifrin

Anyhow ...

Akiva Weinberger

17:33

But yeah, you get the picture?

Ted Shifrin

Yes, sure.

Akiva Weinberger

Now

Look at the second frame

If you rotate that around the y-axis, you get two sphere bits connected by a cylinder, yeah?

I mean, if you make the surface of revolution

Ted Shifrin

Sphere bits?

It's closer to a torus.

Akiva Weinberger

How?

It's two concentric spheres, but you punch a circle out of each sphere

and connect them with tubes

Ted Shifrin

Oh, that's what you meant.

Akiva Weinberger

17:36

The y-axis is the line of symmetry

Ted Shifrin

OK

Akiva Weinberger

So yeah, if you rotate the inner sphere, the cylinder goes through the same transformation we did above

(hyperboloid, double cone, hyperboloid, back to cylinder)

(if you do it 360 degrees)

Ted Shifrin

OK.

Akiva Weinberger

And then we can nullhomotope our two-bulbs-connected-by-a-tube thing to zero

So that's the animation I drew above. A sphere starts as a point, turns into two bulbs with a tube, the inner bulb twists, and then it goes back to a point.

And this is happening in $\Bbb R^3\setminus\{p\}$.

This sequence of maps $S^2\to\Bbb R^3\setminus\{p\}$ starts and ends as a constant map, so you can think of it as a sequence of slices of a map from $S^3\to\Bbb R^4\setminus\rm line$.

(And since $\Bbb R^3\setminus\{p\}\cong\Bbb R^4\setminus{\rm line}\cong S^2$, you can think of this as a map from $S^3\to S^2$.)

Ted Shifrin

You don't have the constant map in this picture.

Akiva Weinberger

17:41

And I claim, but will not prove, that that map is homotopic to the Hopf map.

@TedShifrin At the very start and end of the animation

Oh, well, I didn't draw it

I drew it as a small ball

Ted Shifrin

I see that when the rulings all cross at a point, the resulting shape is contractible, but I don't see a constant map.

Akiva Weinberger

Well, it starts as a point (aka a constant map), then grows to a small sphere, then grows to the two-bulbs-and-a-tube shape

Ted Shifrin

You don't need to keep repasting the picture. :)

Akiva Weinberger

Sorry

Ted Shifrin

So what's the point of the transformation with the crossings in the middle? That seems contractible to me. But what's the relevance?

Akiva Weinberger

17:44

You know how a 4D object can be visualized as a movie of 3D slices?

Ted Shifrin

@EricSilva: You missed my whole stooopidity about non-orientable Gauss-Bonnet.

Sure, DogAteMy.

Akiva Weinberger

This is a movie of 3D slices. The instant with the double-cone is the middle slice.

Joe Shmo

whos dogateme?

Akiva Weinberger

Me

Leaky Nun

8

Q: Uses of ordinals

Are there any interesting theorems outside of set theory that use ordinals in their proofs? The only example I know of is Goodstein's theorem, and I haven't been able to find anything else. In other (more vague) words, what is the use of ordinals? (Other than Goodstein.) Theorems that use the w...

Ted Shifrin

17:45

But why do we need that middle slice?

I'm totally missing the point.

Pun perhaps intended.

Leaky Nun

@AkivaWeinberger I would never have guesssed that you posted it

Perturbative

Hey everyone

Ted Shifrin

hi Perturb

Eric Silva

@TedShifrin what entailed the stoopidity

Perturbative

Hi @TedShifrin

Akiva Weinberger

17:46

If the animation was just a sphere appearing, growing a bit, shrinking a bit, and disappearing, that would also correspond to a map from $S^3\to S^2$,

but it would be a boring map 'cause it'd correspond to the zero element of $\pi_3(S^2)$

It'd be nullhomotopic

Somehow, it has to get tangled with $p$

Ted Shifrin

Eric: I always thought orientability was necessary. I thought I'd done the computation for the Möbius strip in $\Bbb R^3$ that contradicted G-B. But I finally realized yesterday that I was off by a sign on the geodesic curvature on part of it. See this.

Akiva Weinberger

($p$ being the point I deleted from space, shown in blue)

Ted Shifrin

DogAteMy: I don't offhand see what this has to do with the Hopf map.

Joe Shmo

@Ted, I'm plotting sech, tanh, but i have no intuition

Eric Silva

lol i’ll bookmark this for later

i still have travel brain

Ted Shifrin

17:48

@JoeShmo: Forget about $\tanh$. What is the plot of $\sech$?

Oh, maybe my solution uses something you haven't done yet.

Joe Shmo

its a smooth bump that peaks at y=K

Ted Shifrin

This is why out-of-context questions on MSE suck.

Right, @JoeShmo: It's a bounded function.

Akiva Weinberger

@TedShifrin The Hopf map is a map $S^3\to S^2$ that can't be homotoped to zero. This is a map $S^3\to S^2$ that can't be homotoped to zero.

Ted Shifrin

@JoeShmo: I guess you weren't asking for a global holomorphic function. The question was to show no local function? Or no global function?

Akiva Weinberger

You can show that they're the same map (up to homotopy) but it's more pictures, and I don't know how to draw them

Joe Shmo

17:49

Show that there does not exist holomorphic functions of z = a + ib whose modulus is equal to K/ cosh a,
where K != 0 is a constant

Ted Shifrin

Maybe if you think of the linking circles description of the Hopf map you can show the connection, DogAteMy.

Akiva Weinberger

I know

Ted Shifrin

@JoeShmo: Again. That's vague. Does the holomorphic function need to be defined on $\Bbb C$?

Poline Sandra

@TedShifrin bonsoir, s'il vous plait y'a t'il une définition pour la distance induite comme pour la topologie induite

Ted Shifrin

If so, it cannot exist.

Joe Shmo

17:50

suppose all of C

Akiva Weinberger

cosh(a) is zero sometimes

$\pi i$, no?

Joe Shmo

that's the problem verbatim ^. i know that its vague

Daminark

Hello everyone

Akiva Weinberger

No, half

Ted Shifrin

@Poline: Bien sûr. On emploie la même formule.

Perturbative

17:52

Hey @Dami

Ted Shifrin

DogAteMy: We're saying $|f(z)| = K/\cosh(\Re(z))$.

Akiva Weinberger

Ohh

Ted Shifrin

@JoeShmo: So if it's a global question, I don't need Cauchy-Riemann or anything.

Akiva Weinberger

So $f(z)=r(z)e^{i\theta(z)}$

Ted Shifrin

If you've proved Liouville's Theorem.

Joe Shmo

17:53

nope

Leaky Nun

@Daminark were you doing gre?

Joe Shmo

Note: Try and solve this problem without using theorems of complex analysis we will see later in the course.
Lectures 1, and 2 are sufficient here.

Ted Shifrin

I presume the question means there's no holomorphic function on any disk around $0$.

Oh, well, that was useful information, BTW.

Daminark

Yeah I did it this morning. Could've gone either way

Joe Shmo

complex complex analysis theorems

Ted Shifrin

17:54

So, Demonark, you're still alive!

No, very simple theorems, JoeShmo. Complex analysis is very elegant.

Poline Sandra

donc s'il vous plait, $|.|_{\mathbb{Q}}$ est la distance discrète ? ou c'est juste la différence entre deux nombres rationnelles @TedShifrin

Akiva Weinberger

Obviously, we can assume $K=1$

Ted Shifrin

La deuxième, bien sûr.

Joe Shmo

@Ted, yeah, so far the statements i'd seen can be enigmatic, but the proofs are straight forward.

Ted Shifrin

So, how can a holomorphic function have a modulus that's constant on all vertical lines?

We need something like $e^z$ for that. Aha.

Poline Sandra

17:55

mais la topologie induite sur Q est la topologie discrète? @TedShifrin

Akiva Weinberger

$f(z)=e^z$ does

Joe Shmo

how do you know that?

Ted Shifrin

Non, @Poline. Vraiment pas.

Akiva Weinberger

Grr

Ted Shifrin

LOL @Grr

Daminark

17:56

I hope I'm safely 700+, I dunno if I'm 800 or not

Akiva Weinberger

So look at $\ln f(z)$ (noting that $f(z)\ne0$)

Ted Shifrin

When I took it it was out of 990. How confuzling.

Akiva Weinberger

3 mins ago, by Akiva Weinberger

So $f(z)=r(z)e^{i\theta(z)}$

Poline Sandra

@TedShifrin l'intersection de chaque intervalle avec Q c'est les points de Q

Ted Shifrin

DogAteMy: They know basically only Cauchy-Riemann at this point.

Daminark

17:57

It still is but they often curve it so the max score is less on any given test

Akiva Weinberger

$\ln f(z)=-\ln(\cosh a)+i\theta(z)$

Ted Shifrin

How ridiculous, Demonark.

Akiva Weinberger

@TedShifrin So no $\ln$? Fine

Poline Sandra

@TedShifrin comme la topologie induite sur N

Perturbative

Do those scores for the GRE equate to like percentages on a test? @Dami

Akiva Weinberger

17:57

Remind me what C–R actually says?

Joe Shmo

statement about the equality of partials

Ted Shifrin

@Poline: Comment pensez-vous que la topologie soit discrète?

Joe Shmo

Cauchy–Riemann equations

In the field of complex analysis in mathematics, the Cauchy–Riemann equations, named after Augustin Cauchy and Bernhard Riemann, consist of a system of two partial differential equations which, together with certain continuity and differentiability criteria, form a necessary and sufficient condition for a complex function to be complex differentiable, that is, holomorphic. This system of equations first appeared in the work of Jean le Rond d'Alembert (d'Alembert 1752). Later, Leonhard Euler connected this system to the analytic functions (Euler 1797). Cauchy (1814) then used these equations to...

Ted Shifrin

DogAteMy: It says the derivative matrix is a scalar times a rotation matrix.

Akiva Weinberger

Clearly, $\frac{\partial u}{\partial y}=0$

Joe Shmo

17:58

yeah

but that doesnt take me far

Ted Shifrin

You're writing $f = u+iv$?

I'm confuzled.

Akiva Weinberger

Yeah

Joe Shmo

or.. i arrived at (1/v) du/db = tanh(a)

yes

Akiva Weinberger

Er, $x+iy=a+ib$

Joe Shmo

let me share work

Ted Shifrin

17:59

So $u^2+v^2$ doesn't depend on $y$.

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