then how will you learn

Hey everyone

Hi @Perturb

@Leaky: Should I be practicing smacking or kicking you out? :)

@TedShifrin :'(

Ted is now a dictator

I learn from the best — I'm almost as bad as the orange narcissist.

lmao

I guess I should remove that.

Anyhow, I agree with Leaky's point, so I'm not smacking him for that.

then what are you smacking me for

General principle.

Congrats on the italics, Ted

I merely asked if I should practice :)

LOL, @Krijn, thanks. It seems I wasn't even consulted to see if I'd be willing.

@TedShifrin In my intro to diff geom class, I saw a really sketchy proof that the Riemannian metric on $\mathbb{R}^n$ is $(dx^1)^2 + \dots (dx^n)^2$. My lecturer was basically like "We start off with Pythagoras theorem in $\mathbb{R}^2$" then at some point later was like "Pythagoras holds in $\mathbb{R}^n$" then I'm pretty sure he was like "taking differentials/infinitesmals we get this"...

Blah. That's nonsense.

We're just writing down the $2$-tensor that gives $\|v\|^2 = \sum (v^i)^2$.

Yeah I figured as much, I just wanted to know how far off from reality that sketchy proof actually was :p

Hi @Ted.

Heya @MikeM.

I am convinced by some stuff on a MathOverflow answer that Gauss-Bonnet does work on non-oriented manifolds. But it's interesting and subtle, and I don't really like the measure stuff. Let me describe how it works.

Yeah, I actually got Dick Palais's 1979 article on it.

You'd like his proof. He does handlebody arguments. :)

I didn't read that. Let me see if my understanding does anything for you.

@TedShifrin do you remember the moment when you hear French and you suddenly realize that you can understand what is being said?

@TedShifrin So $\vec{x}$ is a point on the plane where $\vec{a}$ is a normal vector and the distance to origin is $c/\vert \vert \vec{a} \vert \vert$ and it should also be perpendicular to $\vec{b}$?

The point to me is that on a non-oriented surface S, THERE IS NO FUNCTION K. The curvature required an oriented basis for each tangent space. K defines a function on the oriented double cover which is odd (it negates under inversion).

Nah, you don't need an orientation for $K$. It stays the same if you change orientation.

Really? I am being goofy. But that explains something I was confused about.

My bad.

@Lozansky: Pursuant to our earlier discussion, the cross product equation has more information.

Ok, second go.

LOL

If you consider the 2-form KdA on the oriented double cover, then of course i(dA) = -dA. As you say, K is even. So KdA is odd. This means it doesn't descend to a 2-form downstairs, and doesn't formally integrate to 0.

But it does descend to a 2-form with values in det(TS) downstairs.

Meh now that I see (clearly!!!!) that K is even I care less about this result. It's just saying that volume forms really live in det(TM). Whatever.

@TedShifrin Mhm, $\vec{x}$ is one side (the adjacent is $\vec{a}$) in a parellelogram with area $b$

@MikeM: That's why you want the density/measure $|dA|$.

@Ted But that confuses me: are there really no issues with signs? Taking the integral of my twisted differential form = the integral with respect to |dA|?

Signs scare me.

Your twisted differential form reverts to a usual one if you remove a set of measure 0. :)

Yeah I figured that would work

Measure stuff too irritating

Ok, that's enough facile comments from me today

LOL ... This drove me nuts. I even emailed Palais before I found my error (details on post). Of course, I found the error right after I emailed him.

But he's in Irvine and I need to have dinner with him and his wife sometime — old, old friends.

Always.

@TedShifrin So I guess it is possible to find such an $\vec{x}$ :>

How many $\vec x$? @Lozansky

sup weirdos

is the minkowski inequality made before chosing the norm on L^p or was it after?

because it is just traiangle inequality, why give it such name

heya Eric

@EricSilva am not werido, cannot defind the rest :D

how goes it

@KasmirKhaan it was definitely known before vector space was a formally defined notion even

aha thanks Eric

I donno how that dude derived it ._.'

Last day before I ship out a draft to reference writers

Too much to do

Exciting, @MikeM!

sounds stressful

@TedShifrin I think one

Yup, @Lozansky. The cross product equation gives you a line parallel to $\vec a$. It meets a plane perpendicular to $\vec a$ at, of course, one point.

I'm more with Eric on this one

More with Eric than with ... ?

Stressful > exciting

Ohhh ... I forgot I was the bad one. :)

Maybe I don't know enough, but to me the fundamental group is a really poor way to show two spaces are not homeomorphic even in low dimensions, like for example $\pi_1(\mathbb{Z} \times \mathbb{Z}, (0, 0)) \cong \pi_1(\mathbb{R}^2, (0, 0))$ so to the fundamental groups of these spaces are essentially the same, despite them being very different topologically.

I guess $\pi_1$ wasn't constructed for purposes like this (and we don't even need $\pi_1$ in this case since $\mathbb{Z} \times \mathbb{Z} \subseteq \mathbb{R}^2$ isn't path-connected whereas $\mathbb{R}^2$ is connected) but even so..

$\Bbb Z\times\Bbb Z$ isn't even path-connected. Why are you using algebraic topology?

Right, you even said that.

I wanted to come up with random examples of subspaces of $\mathbb{R}^2$ and then compute their fundamental groups just to get better at computing them

Well, try the figure 8 or the plane minus a few points.

Hmm the figure 8 is the wedge of two circles, $S^1 \vee S^1$, applying van Kampens theorem you get $\pi_1(S^1 \vee S^1) \cong \mathbb{Z} * \mathbb{Z}$

Yup.

I'm not too sure about the plane minus a few points, the plane minus one point is homotopy equivalent to $S^1$ and then we get a fundamental group isomorphic to $\mathbb{Z}$.

OK, good. What about minus 2 points?

What's that homotopy equiv to?

For some reason I'm thinking that $\mathbb{R}^2$ minus $2$-points might be homotopy equivalent to the wedge product of $2$-circles (figure 8 space), and thus would have fundamental group $\mathbb{Z} * \mathbb{Z}$.

OK, figure that out with a good picture. What about minus 3 points?

It would be homotopy equivalent to the wedge product of $3$-circles and have fundamental group $\mathbb{Z} * \mathbb{Z} * \mathbb{Z}$ (and the same would hold for all $n$)

Ah, OK. Good.

You'll need subspaces of $\Bbb R^3$ to get non-pathological things that are more interesting.

hey @TedShifrin

@TedShifrin Should I be worried that I can't write down explicit homotopy equivalences for these spaces at the moment? Because like I can draw a picture and be like "This seems like this is homotopy equivalent to that" but it feels like I'm not giving a complete answer

You don't need precise formulas as long as it's clear from your drawing what the mapping is and why it's a deformation retract.

Hi Piggy

What do you mean by "You'll need subspaces of $\Bbb R^3$ to get non-pathological things that are more interesting."?

Are there "reasonable" subspaces of $\Bbb R^2$ with fundamental groups other than free products of $\Bbb Z$'s?

He's not talking about $2$-D CW complexes here.

i see

Oh oh ... MikeM is back.

@TedShifrin Ah okay that gives me a bit of relief. Also, quick question, why do most authors introduce retracts, deformation retracts, strong deformation retracts as sort of seperate concepts, when they could just say those are special cases of homotopy equivalences?

But they're not all homotopy equivalences.

They're important concepts in their own right.

Well, for example, one uses the deformation retraction in an explicit form to prove the Poincaré lemma that a closed $k$-form on a star-shaped region is exact.

I mean, you don't need the formula to know the result, but you might want to construct the $\eta$ so that $d\eta = \omega$.

hmm

And certainly there are places in differential topology where we want to see how to get the homotopy equivalence by following the flow of a vector field ... or something.

I'm sure there are better examples. @MikeM can probably oblige when he has time.

thanks

Okay I get what you're saying @TedShifrin. Thanks!

Sure thing.

I'm off to bed now, night everyone!

Night!

Hello!

hi Demonark

Is your brain back to its usual addled state? :)

Heh, yeah I already forgot what partial derivatives are :P

I'm not sure you ever knew :D

rehi Eric

wonders when Eric will get back to geometry :D

i’m taking geometry in fall!

mumbles Bryant :D

What are you taking in fall? I've forgotten.

Oh you're taking the class on complex manifolds?

Oh right

complex geo w webster

I hope he does a decent job.

ya me too

My mathematical brother had better not let you down.

its a 3 quarter sequence so i hope its cool

Has he announced anything about it?

he posted the course description

More details than the usual catalog nonsense?

something about inhomogeneous CR equations, hermitian and kahler geo and complex monge ampere or smth

no more details than that

oh, so definitely PDE slant .... so I won't have to help (cuz I can't).

lol

That's definitely more than what the catalog says because it seems that class isn't in the catalog :P

i’m p excited about it

I hope you learn about curvature and Chern classes too ...

Though the catalog doesn't seem to have that many of the second year+ topics classes

i’m not doing anymore math classes than 1 per quarter this year

Oh that's the only one you're doing?

Smart move.

You want to be eager when you get to grad school.

yeah i’m donezo with taking loads of math for now

i’m taking lit classes and history classes to fill it out

Cool ... And don't forget you're cooking me dinners.

oh man yeah i wanna cook

Deal. :)

I went to a "bring food from your heritage or childhood" potluck last night. I made stuffed cabbage (a bit French-ified).

Sounds fun

Traditional Russian/Polish dish.

oh that sounds dope

oh yeah, I talked about this the other day, didn't I, Demonark? Cuz we were talking about raisins and Moroccan ...

Ah yeah true

And lol I just checked Canvas, the only class that's up for me is rep theory

Who's teaching it?

Someone named Maxime Bergeron, a postdoc

Ah ...

rep was a good class

I sat in briefly on his algebra class (non-honors) second year and he was quite good there so I'm definitely excited. Also the content is stuff I've been meaning to learn for a while

There isn't much info yet, just a link to Piazza, the book ("The Symmetric Group" by Sagan), and a link to the website where psets will be uploaded

Hi guys

I blew away my AoPS kids talking about the rigorous definition of limit and doing examples. They're totally skeered. I wonder if they'll all drop.

Rip, hopefully not

welcome to epsiland

Yup. Even my smart kid who's got good intuition was confused about working with the logic and the inequalities. But I sent them my notes (that eventually ended up in Spivak) with more examples written clearly. I hope they look.

I don't want it to be a crappy course. 4 of the 5 are already taking a high school calculus course (AP mostly).

It is good to see it early.

Well, I figure only one or two of them might end up doing math, but still ... it'll help when we get to convergence of sequences/series, etc.

As they say

I'm old enough that my math texts were written in stone tablets. You didn't realize?

i think in my hs class two of us are tryna go to grad school for math

Yeah quantifiers are tricky at the start. I like thinking about them as a game

I came from a phenomenal hs class. I was the only math PhD, but there were various doctors and musicians/playwrights.

Yeah, Demonark. I was trying to emphasize just the basic manipulate-the-algebra-and-stipulate-if-necessary game. This is far from a complete Spivak approach.

quantifiers are still tricky tbh

As can be the ambiguity in "any" in mathematics — some get that it's "every," others think it's "some."

spooked

who's spooked?

me

why?

eh you know this and that

sorry if I was the cause

naw