Mathematics - 2020-11-29 (page 1 of 2)

1:29 AM

@amWhy The fur collar and the mask were almost too much, so I almost left the fur collar off of the suit while wearing the mask; but that didn't look right either. I kept the collar.

Ted Shifrin

I'll just stay simple and un-Christmas-y. Call me Scrooge.

amWhy

@TedShifrin Welcome Scrooge!

Ted Shifrin

Hey @amWhy.

I could put a menorah if I were observant of anything.

robjohn

@TedShifrin your avatar does look like a dreidel already; not much modification needed...

Mike Miller

@TedShifrin In the end I think my goal of giving tastes of both algebraic and geometric topology was not the right choice. Jack of all trades problem. I think I would probably try to use surfaces (manifolds) as a source of ways to understand the previous constructions until the end of class, when we try to introduce a tool (like homology or pi_1 or such) to distinguish surfaces. But time constraints prevent one from introducing both effectively.

amWhy

1:33 AM

@TedShifrin I'm assuming we'll soon be encountering the Holiday Bash, with hats awarded for certain feats.

Mike Miller

I have outgrown the StackExchange hat event. Instead I write a list of things that I think I deserve a hat for, and send it to my friends and family.

I think one could probably write a great topology course which uses surfaces as its primary examples and motivations. You will occasionally run into walls (like proving that interior points are not boundary points) which you will axiomatize until the last week of the course, when you will prove them with H_1 or whatever.

amWhy

@MikeMiller I like that alternative! Yeah, I've slowly lost interest in wearing such hats. I'd rather have a custom made hat that 10% or more of math.se users are not also wearing!

Ted Shifrin

@robjohn LOL

Yeah, @MikeM, it's always tempting to try to do too much. I never tried to define manifolds in point-set, always saving those for differential topology instead. But doing surfaces and some of the fundamental group stuff could be quite coherent. I think homology is just too much.

Mike Miller

Yes, I completely dropped homology from my plans.

My final "challenge problem" sheet investigates isotopy classes of embedded discs as a way to define w_1: pi_1(S) -> Z/2 and orientations. I expect some 3 people will do it.

I think there's basically no way to make this accessible to everyone within a 1-sem course. Just not enough time.

Ted Shifrin

Wow, that's cool. Challenge problems for the best students (which I'm sure you had) are always fun, if they're into it.

Mike Miller

1:41 AM

Yeah, I have been posting challenge sheets since about the midway mark, called "Curios"

I might do the same for my computational linalg course next semester. It would be nice to give ways to engage with the deeper material for my math major studens

Ted Shifrin

The other good thing is that if your best students work on those (and you are doing the grading), you can write good letters for them if/when they go on to grad school.

Mike Miller

I tend to advise mystudents to ask the tenured faculty for letters

I should maybe polish my materials a bit and post them on my webpage

robjohn

@amWhy The latest post I've seen about a Winter Bash was the Winter Bash 2019.

Ted Shifrin

Yes, that's good advice (and saves you work), but if you teach an advanced course and they engage more with you than with other profs, then maybe not. I think I remember writing for a few people from my MIT postdoc days, and certainly during my early years (pre-tenure) at UGA.

Mike Miller

I think I am unusually accessible, so it is true that I have a good grasp on different skill levels

Ted Shifrin

1:45 AM

Yeah, I believe that.

So your semester ended before Thanksgiving? I guess all schools tried to do that this year because of Covid.

Mike Miller

No, we have two more weeks.

I'm writing up the final set of notes (on surfaces) now.

Ted Shifrin

Ah.

Mike Miller

So I get to get a sense of whether or not the plan works. Conclusion: it doesn't work quite as well as I'd like. :)

Ted Shifrin

It's rare that an experiment on a course works great the first time. It usually took me a few times to fiddle and get it right.

And of course it depends on the students. I guess you have such a large class that you have a decent cross-section anyhow.

Mike Miller

Yes, you're exactly right. I learned a lot from seeing your HWs where you gave different levels of assignment. The idea to try to write "two courses" tailored to different levels of student was fantastic and I internalized it right away.

I'll try to do that in a more formally codified way next time instead of giving non-credit bonus material.

Ted Shifrin

1:56 AM

That diff geo course was actually 3 levels (with all students required to do one medium-level problem each assignment).

But the computational-only level of that course is not present in most advanced courses, I guess.

As I explained to you, though, I was worried about good students who would dumb themselves down to be lazy or ... As I said, that really only happened a few times (one the exceptional case of an A+ student who should have done more medium/challenging ones). It puts the onus on the student to self-challenge, which I suppose is good.

Mike Miller

Yes, I agree. There are multiple problems. Do you focus the course on too advanced topics (suitable for some math majors, but not all), thereby leaving some students behind and giving a warning at the start of the course? Or do you focus on the students who need a lot of patience, thereby leading the advanced students to feel bored? Obviously neither is an acceptable/desirable solution.

This ignores your third level (the grad-student level).

robjohn

@amWhy winterbash2020.stackexchange.com

Ted Shifrin

Well, my third level was intended for the best undergrads too (like ones intending to go to grad school). Obviously I wanted them all to do a blend of levels. ... The problem you state is really a problem with every course, even calculus. My personal taste was to teach to the upper-middle, try to get the bottom to work with me to improve (heavy office hours), and challenge the best with the challenging exercises, not particularly my lectures.

Mike Miller

I think that's the right approach

The problems are actually even more clear at the calculus level

since you have plenty of students who are strong enough to (and want to) become math majors, but those courses are not at all written for aspiring math majors

I'm really looking forward to linear algebra

I think I can write a good course that caters to the weaker student, the strong applied student, and the strong math student

Ted Shifrin

We introduced the computational linear algebra course at UGA as I was retiring, so I never tried to teach it. I never did see a good book for it.

Mike Miller

2:06 AM

I am unsure what I will actually use

I really really like Treil

It is the correct book for a mixed applied-pure course IMO

Also, it makes fun of Axler, which is a plus

Ted Shifrin

I don't know that. LOL, yes, that's fine with me. But this sounds more advanced. Our course at UGA was introductory level. Strang is sort of standard, but there are newer applications than he covers.

Mike Miller

Sorry, I was unclear in my phrasing

What I wanted to say was that it has the right mixtures of applied and pure topics. It's too high-level for the "average" student in a first linalg course.

Ted Shifrin

For the "standard" proof-teaching linear algebra, obviously I'm happy with my own book(s) :P

Mike Miller

right

Ted Shifrin

Ah, so your course is also first. Ours was intended for CS and stat majors, but now most math majors who are "applied" take it instead of the proof-based one (in which a lot of faculty didn't really teach proving much anyhow).

Mike Miller

2:12 AM

The frustrating thing is that while I don't care that a good linalg student understands proofs, I do care that they understand what's going on (especially with change of basis stuff and eigenvector stuff)

And it's hard to imagine a course that gets people to really understand what's going on without proofs

Ted Shifrin

It's a good course for teaching proof-writing, because the proofs are so basic (at least for the first half). Change of basis is not really a proof. It's just an equation that makes perfect sense, although some books do obscure that.

But, yeah, students struggle with change of basis because sometimes the "old" basis is really the new one and the "new" one is really the old (standard) one. But I found it better to always have the old one be standard and just let them rethink the formula.

Mike Miller

I will have to think of this with my prep work, which doesn't begin 'til the break starts.

If you have suggestions for good applications to cover I'd be glad to hear them. I'd like to include the "curio" model I used with topology, but with both pure and applied curios. I think it can be made to work.

Ted Shifrin

I could ask my pal and co-author Malcolm Adams, who's taught the applied course, what applications/sources he decided to include. Or you could email him on my suggestion. :)

Mike Miller

I'll try to remember to do that tomorrow

I took Th+F off, so I am catching up on my work...

My only two days off this semester

Ted Shifrin

Damn. Don't burn yourself out.

Mike Miller

2:21 AM

Well, let me assure you that I'm pretty close to the amount of scotch I can consume while remaining productive, so at least I'm having a good time tonight

Ted Shifrin

Oy. I just ordered more gin and vermouth.

When this is all over, we should do a good meal somewhere :)

Mike Miller

I did that last night with the SO. Thanksgiving prix fixe at a good, well-distanced restaurant. Best squash soup I've had in my life.

We should, though*.

Ted Shifrin

I am, sadly, avoiding restaurants until the vaccine. Good thing I'm a good cook :P'

Mike Miller

Us too, mostly. First time out in months.

But indeed safety first

Ted Shifrin

I am so sad that so many of my favorite good restaurants will not come out the other end of this.

Not to mention the hundreds and hundreds of thousands of people ... but we know why that is.

Mike Miller

2:29 AM

indeed

I'm about two scotches too deep for serious conversation though, I'm at the intellectual level required to write calculus notes and no higher

Ted Shifrin

LOL ... I'm about to reheat some of my chicken pot pie with buttermilk biscuit topping.

Mike Miller

that sounds great

I cooked chicken scarpariella the other day

I was happy with that

Ted Shifrin

Immodestly, it was.

Oooh, I don't know that one.

Mike Miller

Chicken, sausage, onions, peppadews, potatoes.

Saw it in the window of an italian deli and it looked good, so decided to try it myself

Ted Shifrin

Nice. Herbs, presumably.

The peppadews don't sound too Italian.

Mike Miller

2:33 AM

Sure, of course. Standard seasonings and some sprigs of rosemary in the skillet.

Ted Shifrin

So nice acid from the peppadews and maybe more.

Mike Miller

I think the peppadew was just a suggestion for vinegared pepper.

I'm sure pepparoncini would have done the job with a less pleasant color

Ted Shifrin

Yeah, I'm just thinking splashes more of vinegar or lemon.

Sounds good!

Mike Miller

Pardon the typos --- again, there is a non-trivial obstruction class to achieving perfect spelling here.

Ted Shifrin

LOL, one scotch is the obstruction class, and we've exceeded that.

I actually had not noticed typos, so maybe my one gin martini is my obstruction class.

Mike Miller

2:39 AM

pepparoncini was the one I caught right now, and many others caught before I hit the "send" button...

Ted Shifrin

Oh, yeah.

Mike Miller

OK, now that I have finished that glass, it is probably time for me to stop doing math for the night. Ciao!

Ted Shifrin

LOL, night! Ciao.

skillpatrol

cya

Ted Shifrin

2:56 AM

Oh, hey skill.

I mean skull. Damn you.

3

skillpatrol

:D

Mike Miller

@TedShifrin you've had some drinks too then.

robjohn

Holiday garland for anyone who wants it

skillpatrol

A new commutative diagram editor for anyone who wants it.

5

Ted Shifrin

No, @MikeM, he entered as skill, damn him.

2

Mike Miller

3:17 AM

Starred for belligerence

Ted Shifrin

Grrr

robjohn

@TedShifrin displaying your bearicose nature?

copper.hat

a little off topic, but who reads the college application essays?

(i mean other than parents like myself :-))

skillpatrol

3:33 AM

Administrators?

copper.hat

Really?

That's a lot of reading!

skillpatrol

yup

Ted Shifrin

No, admissions staff, occasionally some faculty.

@robjohn Indeed.

skillpatrol

3:51 AM

a Grrr essive by nature...

...in a competitive sense :-)

copper.hat

thx!

my daughter is going to college in the uk, so i avoided the process with her. unfortunately for my poor 17 yo. son...

skillpatrol

4:08 AM

np

love_sodam

5:36 AM

How can I show that a geodesic on a surface of revolution could not be asymptotic to a latitudinal curve unless the latitudinal curve is itself a geodesic?

Leaky Nun

5:49 AM

rogā gallīs, omnēs revolūtiōnum sciunt

Ted Shifrin

@love_sodam See Exercise 27 on p. 78 of my differential geometry text.

copper.hat

6:05 AM

that's gaulling...

Ted Shifrin

Oh?

copper.hat

sorry, it was a reply to @LeakyNun

Ted Shifrin

Oh.

Leaky Nun

^^

love_sodam

6:57 AM

@TedShifrin I saw exercise 27. Taylor expansion of exercise 23 shows something?

Terran Swett

7:12 AM

Just wondering, is there a term for an open set which is the interior of its own closure? Or for a closed set which is the closure of its own interior?

Mary Star

9:37 AM

Hello!! I have written the formula of Newton's method to appoximate $a^{1/n}$.
Now I want to show the inequality $x_{k+1}-a\leq \left (1-\frac{1}{n}\right )(x_k-a)$.

Since we subtract at the left side $a$ instead of $a^{1/n}$ we cannot consider it as the error, right?

I achieved to show the inequality but for $a\geq 1$:

$$x_{k+1}=\left (1-\frac{1}{n}\right )x_k+\frac{a}{nx_k^{n-1}} \leq \left (1-\frac{1}{n}\right )x_k+\frac{a}{n\left (a^{1/n}\right )^{n-1}} = \left (1-\frac{1}{n}\right )x_k+\frac{a}{na^{1-\frac{1}{n}}} = \left (1-\frac{1}{n}\right )x_k+\frac{a^{\frac{1}{n}}}{n} $$
Then subtracting at both sides "$a$" we get $$x_{k+1}-a\leq \left (1-\frac{1}{n}\right )x_k+\frac{a}{n}-a=\left (1-\frac{1}{n}\right )x_k-\left (1-\frac{1}{n}\right )a=\left (1-\frac{1}{n}\right )(x_k-a)$$

We have that $x_k\geq a^{1/n}$.

Could you give me a hint for the case $a<1$ ? Or is there an other way to show the inequality without taking cases for $a$ ?

Mary Star

10:12 AM

I have also an other question... I have posted it in the main, here is the link:

0

Q: Existence of LU decomposition

I saw the below sentence in some notes: Let $A\in \mathbb{R}^{n\times n}$ be a not necessarily symmetric, strictly positive definite matrix, $x^TAx>0$, $x\neq 0$ und $Q\in \mathbb{R}^{n\times n}$ an orthogonal matrix, then $B=Q^TAQ$ has a LU decomposition. I want to understand this implication,...

Does someone of you have an idea?

skullpatrol

12:34 PM

\o @nbro wazzup bro?

nbro

fine, what about u?

skullpatrol

fine, thanks

is that the world cup on your avatar?

Maradona?

The hand of God

"The hand of God" was a phrase used by the Argentine footballer Diego Maradona to describe a goal that he scored during the Argentina v England quarter finals match of the 1986 FIFA World Cup. The goal took place on 22 June 1986, at the Azteca Stadium in Mexico City. Under association football rules, Maradona should have received a yellow card for using his hand and the goal disallowed. However, as the referees did not have a clear view of the play and video assistant referee technology did not yet exist, the goal stood and Argentina led 1–0. The game ended with a 2–1 win for the Argentines, thanks...

Thorgott

1:01 PM

@TerranSwett they're called regular open/regular closed

Mike Miller

Or "open domain" / "closed domain"

nbro

@skullpatrol yes, exactly, it's MARADONA raising the world cup in 1986

skullpatrol

one hand of god helped him raise that cup, pal

nbro

yes, he inspired me so much when I was a kid/teenager

skullpatrol

1:18 PM

coolio

Arief Anbiya

1:49 PM

Hello, just want to share a mathematical prank : youtu.be/nvDVs2_qo-Q

Mats Granvik

2:45 PM

Is the Riemann hypothesis equivalent to showing:
$$\lim_{n\to \infty } \, \left(\frac{1}{1-\frac{\sum _{k=1}^n \frac{(-1)^{k-1} \binom{n-1}{k-1}}{\zeta \left(\frac{k}{n}+s-\frac{1}{n}\right)}}{\sum _{k=1}^n \frac{(-1)^{k-1} \binom{n-1}{k-1}}{\zeta \left(\frac{k}{n}+s\right)}}}+s\right)=\lim_{n\to \infty } \, \left(\frac{1}{1-\frac{\sum _{k=1}^n \frac{(-1)^{k-1} \left( \begin{array}{c} n-1 \\ k-1 \end{array} \right)}{\zeta \left(\frac{k}{n}+s\right)}}{\sum _{k=1}^n \frac{(-1)^{k-1} \left( \begin{array}{c} n-1 \\ k-1 \end{array} \right)}{\zeta \left(\frac{k}{n}+s-\frac{1}{n}\right)}}}-s\r…

Reduce[rho == s + 1/(1 - A/B) && s + 1/(1 - A/B) == Conjugate[-s + 1/(1 - B/A)], Re[rho], Complexes]

@epic_math

Sha Vuklia

Is Varadarajan the guy who invented $k$-spaces?

Alessandro Codenotti

@ShaVuklia I think they were introduced by Hurewicz?

Mike Miller

3:04 PM

@AlessandroCodenotti This would make sense as he invented the compact open topology

Balarka Sen

Hi, all

Alessandro Codenotti

I'm not sure I'm seeing the connection @Mike

Oh exponential objects

Hi @Balarka

Balarka Sen

I've never heard of Varadarajan in the context of $k$-spaces. He did probability and Lie groups.

Mike Miller

The whole point of k-spaces is that you get an exponential law of sorts

The compact-open topology has an exponential law for locally compact spaces and the idea of a k-space is there to generalize this

He preferred to think of homotopy groups as fundamental groups of loop spaces, which is why he wanted the adjunction

Balarka Sen

Fact. Given any triangulation of the sphere, you can come up with a circle packing on the sphere whose nerve (vertices are centers, geodesic edges between vertices of tangential circles, faces made up of triples of tangential circles) is isomorphic to it.

Mike Miller

3:14 PM

No assumptions?

Oh of the 2-sphere

Balarka Sen

Right. You can delete the face at infinity, and make it a thing about finite planar triangulations if you want.

Kind of clear in retrospect, isn't it? Like, draw small circles around each vertex of the triangulation, slowly bloat them and see where tangency is achieved.

Take any planar domain $\Omega \subset \Bbb C$. Pick a small radius $\varepsilon > 0$ and consider the standard hexagonal circle packing on $\Bbb C$ by circles of radius $\varepsilon$. Draw all the circles which appear inside of $\Omega$ in this packing. This gives a finite planar triangulation; by Fact you can construct a circle packing of the unit disk whose nerve is this triangulation.

This gives a map $\Omega \to \Bbb D$. Thurston: Let $\varepsilon \to 0$, and you converge to the Riemann map.

Mats Granvik

Clear[f, A, B, x, n, k, s, rho];
s = 1/3 + 14*I;
Reduce[rho == s + 1/(1 - A/B) &&
s + 1/(1 - A/B) == Conjugate[-s + 1/(1 - B/A)], Re[rho], Complexes]

$$\left(\left(\Re(B)<0\land \Re(\text{rho})=\frac{1}{2}\right)\lor \left(\Re(B)=0\land \left(\left(\Im(B)<0\land \Re(\text{rho})=\frac{1}{2}\right)\lor \left(\Im(B)>0\land \Re(\text{rho})=\frac{1}{2}\right)\right)\right)\lor \left(\Re(B)>0\land \Re(\text{rho})=\frac{1}{2}\right)\right)\land A=\frac{3 B \text{rho}+(-4-42 i) B}{3 \text{rho}+(-1-42 i)}$$

$$f(\text{x})\text{=}\zeta (x)$$

$$A(\text{n},\text{s})\text{=}\sum _{k=1}^n \frac{(-1)^{k-1} \binom{n-1}{k-1}}{f\left(\frac{k}{n}-\frac{1}{n}+s\right)}$$

$$B(\text{n},\text{s})\text{=}\sum _{k=1}^n \frac{(-1)^{k-1} \binom{n-1}{k-1}}{f\left(\frac{k}{n}+s\right)}$$

Mike Miller

3:30 PM

@BalarkaSen Ridiculous

This can't say anything about behavior on the boundary I guess, since the Riemann map usually only extends as a map from D^2

Mats Granvik

So if the first equality above^^ is true, and the starting point is s=1/3+14*I then there is no scenario where the nearest Riemann zeta zero rho (closest to s) is not on the critical line.

Balarka Sen

Yeah I guess that's correct. So eg the circle packing on $\Omega$ never reaches $\partial \Omega$, it just gets very close to it, whereas in $\Bbb D$ the corresponding packing actually have horocircles; it's an honest packing of the closed disk.

Which explains why $\Bbb D \to \Omega$ would extend to the boundary, but not the other way

It's like, if you had "flexible" disks whose radii would shrink if you applied force to it (whilst still remaining a circle), and if you tried to squeeze the boundary of the domain in the left to a circle, you'd get a picture on the right, and that'd be approximately biholomorphic

Mike Miller

3:47 PM

@BalarkaSen This all makes perfect sense actually

After all a holomorphic map is a map which sends tiny discs to tiny discs

Balarka Sen

Yup

That's how one proves this works. Show that the map is $(1 + \varepsilon)$-quasiconformal

Mike Miller

@BalarkaSen Is this somehow secretly normal families again

After all you need to justify that your family of quasiconformal maps ought to have a limit

Yep

4:02 PM

hey guys i want to have a bump oscilate in a half circle im using the equation

exp( ((x-x0cos(pi t))^2+(y-y0cos(pi t))^2) / (2) )

problem is that that makes the bump just go round in a circle in time, any ideas?

Balarka Sen

@MikeMiller Yeah

The family seems to be clearly equicontinuous though

Any family of uniformly quasiconformal maps are, right?

Mike Miller

You need some local boundedness condition

I assume uniformly quasiconformal is a bound on the (1+varepsilon) term

But when vareps = 0 you still need a condition for Montel

Balarka Sen

Hm, I'm confused. The maps you get by circle packing are not holomorphic, how do you plan to use Montel?

I am saying by uniform quasiconformality, they are equicontinuous and clearly bounded, because maps to D. By Arzela-Ascoli you get a convergent subsequence

And thus the limit function. Now the limit is (1+vareps)-quasiconformal for all vareps

So 1-quasiconformal, i.e., conformal

I assume that's how it would go at least

Secret

4:18 PM

Why is trying to find the equation form of the integral surfaces of the ultrahyperbolic equation so hard

I am not interested in its initial value problem (so whether it is ill posed don't matter), I only want to check what its general solution look like

Mike Miller

@BalarkaSen I'm saying that equicontinuity is not obvious to me

Since in the holomorphic case you need to make assumptions for that to hold

I'm not using Montel but using the assumptions it requires as a way to cast aspersion

AttractorNotStrangeAtAll

Hello!

I would like to know what you guys/girls think of starting a textbook by its exercises, and then going back to theory to learn how to solve each problem.

Balarka Sen

@MikeMiller Oh, got it. OK, I meant uniformly equicontinuous on compact subsets.

schn

6:08 PM

Suppose $x_1$ can take $n_1$ values and $x_2$ can take $n_2$. Then how many values can $x_1-x_2$ take?

Astyx

it depends

schn

How?

Astyx

Somewhere between $n_1n_2$ and $n_1+n_2$

schn

How do you reason?

Astyx

Look at $x_1\in\{0,4\}$, $x_2\in\{1,2,3\}$

And $x_1\in \{1,2,3\}$, $x_2\in \{1,2,3\}$

schn

6:11 PM

Right. Suppose $n_1$ and $n_2$ are coprime.

Astyx

It still depends

Chaneg the second example to $x_1\in \{1,2\}$, $x_2\in \{1,2,3\}$

schn

Right. Suppose instead of $x_1-x_2$, one has $x_1/n_1-x_2/n_2$.

Astyx

Multiply the sets by $n_1$ and $n_2$ respectively and you get what you started with

schn

How did arrive at $n_1n_2$ and $n_1+n_2$?

Astyx

the first one is if you can get a different result for each pair $(x_1, x_2)$

The second one I'm not too certain about, but that's what my intuition tells me

I'm gone for now

schn

6:24 PM

Suppose $x \in 0, \dots , n_1$ and $y \in 0, \dots , n_2$, with $n_1,n_2$ coprime, then how many values can $x_1/n_1-x_2/n_2$ take?

monoidaltransform

6:49 PM

0

Q: finding terms in ODE

Consider the equation $\frac{\partial{y}}{\partial{t}}+y\frac{\partial{y}}{\partial{x}}=\Gamma \frac{\partial^2y}{\partial{x}^2}$ where $\Gamma$ is a constant term. Given that $y=\frac{y_0}{2}-\frac{y_0}{2}tanh(\frac{x-ct}{L})$ satisfies the equation above, find $L$ and $c$ in terms of the const...

ordinary-differential-equations contest-math

geocalc33

7:46 PM

let's say you have a matrix transformation acting on points in $(0,1)^2$, given by $h_s$=$\begin{pmatrix}
e^{-e^{s}} & 0 \\
0 & e^{-e^{-s}}
\end{pmatrix}.$

and another matrix transformation acting on points in $(0,1)^2$, given by $g_s$=$\begin{pmatrix}
1-e^{-e^{s}} & 0 \\
0 & e^{-e^{-s}}
\end{pmatrix}.$

obviously these transformations "overlap." So how exactly would you combine $g_s$ and $h_s$ to recover a net transformation? I think $g_s \circ h_s$?

love_sodam

8:21 PM

I heard that the monkey saddle is an example which shows that the converse of Gauss remarkable theorem does not holds in a sense that rotation along $z$-axis does not change the Gaussian curvature but not isometry.

But rotation along $z$-axis is isometry isn't it? It's a rigid motion

Ted Shifrin

9:00 PM

@love_sodam They mean a $2\pi/3$ rotation? If it maps the surface to itself, then, yes, it's an isometry.

love_sodam

@TedShifrin I mean any rotation.

Ted Shifrin

There's a standard example of a diffeo between two surfaces that preserves curvature but is not an isometry. Any rotation will not map the surface to itself. Please give the exact, precise statement and the source.

Balarka Sen

Longpills, @TedShifrin?

Ted Shifrin

Huh?

Balarka Sen

The diffeo which preserves K but is not an isometry :D I call it "long pills"

Ted Shifrin

9:06 PM

I don't think we're thinking of the same thing. Mine is #12 on p. 65 of my text.

Balarka Sen

Oh, let me see. I was thinking of taking a pill (cylinder with two ends capped off) and elongating the cylindrical flat part.

love_sodam

It's a book exercise. just a second

Balarka Sen

Ahh right I forgot about this example

love_sodam

@TedShifrin Exercise 5.40 is my question

Ted Shifrin

I think he wrote another crap exercise.

Balarka Sen

9:11 PM

Yeah the exercise makes no sense

love_sodam

lol

I don't understand why this book is a main textbook

Ted Shifrin

Rotating the $uv$-plane does not give a rotation of the surface in 3-space.

It looks beautiful, the book. But I'll stick with mine.

C.F.G

Hi dear friends.

love_sodam

I'll skip that exercise

Ted Shifrin

I was shocked when he didn't even recognize that ruled surface question. I recognize most every exercise in my 4 books.

love_sodam

9:16 PM

Getting hard to believe the content or exercise in this textbook

C.F.G

I have a question on isomorphic vector bundles. How continuous maps that maps fibers to fibers isomorphically then it has to be a homeomorphism. In the Milnor-STASHEFF's book the authors proved it locally and concluded that bundles should be homeomorphic. But why?

Ted Shifrin

I think he gets somewhere with the diffeo given by rotating in the $uv$-plane. If we trust that horrible formula. We have to see what the first fundamental form is.

love_sodam

Actually I solved 4.45 and that I remember that formula is right

Ted Shifrin

And did you compute I?

CFG I no longer have the book. Part of the definition of a bundle morphism is that you have a continuous/smooth map. So I do not follow.

love_sodam

what is I?

Ted Shifrin

9:23 PM

First fund form

Mike Miller

@love_sodam deep question...

Ted Shifrin

LOL

Mike Miller

@C.F.G Did they prove this for vector bundles or for fiber bundles? One needs a small amount of care here

C.F.G

@MikeMiller Hi. For vector bundles

Ted Shifrin

Soon we'll be back to our discussion of non-iso bundles with homeo total spaces.

love_sodam

9:27 PM

$\sigma_u = (1,0,3u^2-3v^2),\sigma_v = (0,1,-6uv)$ So, $E = 1+(3u^2-3v^2)^2, F = -6uv(3u^2-3v^2),G =1+(6uv)^2$

Mike Miller

So the point is that you need f^{-1} to be continuous as well. It suffices to check this locally because it suffices to check continuity locally in general.

Your map f is locally of the form (x, v) mapsto (x, A_x(v)) where A_x is a family of matrices, continuously depending on x. Here I have chosen trivializing charts for both E and F.

Ted Shifrin

OK, @love_sodam, so rotation of the uv-plane induces a diffeo of the surface that is not an isometry but preserves curvature. But nothing to do with rotating in space.

Mike Miller

Then f^{-1} is locally (x,v) mapsto (x, A_x^{-1} v). If you can show that this is continuous, then f^{-1} is locally continuous, hence continuous everywhere.

But as composites of continuous maps are continuous, it comes down to showing that the inversion map GL(n) -> GL(n), A mapsto A^-1, is continuous

Let me point out here that the authors did not show that "locally, E and F are isomorphic". They already have a purported isomorphism from E to F. They just need to check that its inverse is continuous.

And continuity of a given map may be checked locally.

love_sodam

@TedShifrin That rotation is $(u,v,z)\mapsto (ucos\theta-vsin\theta,usin\theta+vcos\theta,z)$ isn't it?

C.F.G

@MikeMiller Do you have the book? see here

Balarka Sen

9:35 PM

Adam Curtis makes fantastic documentaries

Ted Shifrin

That doesn't map the surface to itself unless $\theta = \pm 2\pi/3$.

love_sodam

Oh, that.. I see

Mike Miller

I'm not going to click the link, I have other things to do if you can't say why I didn't already answer your question

Sorry

f^{-1} is a function, which you are trying to show is continuous. For any function between topological spaces, if that function is locally continuous (for every x in the domain, there is an open set around x so that the function is continuous on that open set), then it is globally continuous.

That is all they're using here

@BalarkaSen Do you happen to know an example of a meromorphic function which has an antiderivative, but for which it isn't obvious?

Something meromorphic at 0 where you can check that the integral around 0 is 0 perhaps

C.F.G

@MikeMiller Well, Lets to apply this proof to mobius strip and cylinder. I think One can find a local continuous map between them that maps fibers to fibers. isn't?

Mike Miller

Go ahead and tell me your continuous bijection from the cylinder to the mobius band

C.F.G

9:44 PM

@MikeMiller locally not globally.

Mike Miller

I don't care. That's not the statement of the theorem.

You are given by hypothesis a continuous map from one bundle to the other which is an isomorphism on each fiber

That is, in particular, a continuous bijection. It has a well-defined inverse. They are showing that inverse is continuous.

Balarka Sen

@MikeMiller Hm, I don't know. $1/\sin^2(z)$, perhaps?

Mike Miller

That seems a good idea

Actually wait the contour integral you'd want to do seems awful

Ted Shifrin

@Balarka What are you guys trying to do?

Balarka Sen

I wouldn't do contour integral, I would note it looks like $1/z^2$ around $0$

Ted Shifrin

9:48 PM

Easy, Mike. 0 residue.

Mike Miller

I am covering basic complex analysis for calculus students. No residues.

Balarka Sen

Actually $\cot(z)$ is the antiderivative haha

Mike Miller

Hmm

Ted Shifrin

But that is good, because cslculus tells you the antiderivative over $\Bbb R$. Hence ...

Mike Miller

I guess this is silly

Ted Shifrin

9:49 PM

Oh, Balarka beat me.

Mike Miller

Thursday we will talk about the FTC from real calculus and how it holds over in complex calculus; the contour integral is still an antiderivative evaluated at endpoints, but now you only get an antiderivative if your contour integrals are path-independent. It would be nice to illustrate the latter point with an example.

Ted Shifrin

Maybe negative.

Balarka Sen

Negative, yeah

Mike Miller

On the other hand, "residue theory" is no more than the observation that your function is a polynomial in 1/z plus a holomorphic function, so that you can just apply Cauchy

So I can pretend we do know residues I guess

Ted Shifrin

Or write down Laurent series and integrate.

love_sodam

9:54 PM

Is there any general terminology about this: The distance circle of radius $r>0$ about a point $p$ in a regular surface $S$ is $C_r(p)=\{q\in S: d(p,q)=r\}$ where $d(p,q)$ denotes the intrinsic distance (or geodesic distance I think)

Ted Shifrin

Geodesic circle

love_sodam

Oh ok.

This strange book always makes strange terminology

Balarka Sen

There should be plenty of meromorphic functions with antiderivative but no closed-form antiderivative as such. $e^z/z^2$, maybe

@MikeMiller

WolframAlpha says I need the $Ei$ function lol, so there you go

Mike Miller

That's a good one

I gave e^{-z^2} but that's a relatively trivial example