Ah, ok, gotcha. That's pretty neat.

(pinging @Akiva so he can read this)

Thanks for helping out, @MikeM.

Also, which immersion of $S^2$ did you talk about above? I gave two embeddings of $S^2$, one orientation preserving and the other reversing, and gave something which intuitively looked like an isotopy.

That can't be right, @Balarka. An isotopy can't change its spots.

@TedShifrin so the affine span of points in $\mathbb{R}^n$ is the smallest affine subspace of $\Bbb R^n$ which contains all the points?

Yes, @meow.

@TedShifrin Um, what do you mean by "spots"?

and, if the points are not affinely independent, they can be contained by an affine span of lower dimension?

@meow: Just as the span of a bunch of vectors is the smallest linear subspace containing those vectors.

@BalarkaSen I don't see why you think the boundary of a tubular neighborhood of an immersed $\Bbb{RP}^2$ is embedded.

orientation-preserving or orientation-reversing @Balarka

One thing I still need to learn is to not use chat as a crutch and only present a question after it hasn't been resolved with a good amount of time and testing...

But it's so tempting to just ask and not use my brain!

@TedShifrin Ah, yeah, I didn't mean that.

@Brody: Well noted.

DogAteMy must be unlosted by now.

I meant it "should, intuitively, mumble" isotope f to -f. Sorry about that.

Homotopy sounds better than isotopy given my remark, @Balarka, but maybe I'm missing something.

@TedShifrin The term he's looking for is "regular homotopy", aka a homotopy through immersions.

@MikeMiller Just read the stuff about subdividing things. You only do that finitely many times in the eversion, right?

I used that term earlier when talking about Smale.

(Also, I can't read French, re: your link)

@TedShifrin With that said, I now encourage you to not pay me any attention until it's obvious the truck is really stuck in the muck! lol

@Brody: Did I send you exams to do? I vaguely recall that I did.

@TedShifrin $\pi$ is fine

@TedShifrin Think you told me to e-mail you about that. Haven't done that yet

Oh, ok, @Brody.

@AkivaWeinberger Well, it's a little bit tougher than that. What you really do is pick a triangulation of $S^2 \times [0,1]$, on each simplex of which the map is linear.

Uh-huh

But the triangulation of $S^2 \times \{t\}$ is going to be varying as $t$ changes.

Is it possible to use that to answer my original question, though?

I dunno. I didn't really think about your original question.

Where we have a polyhedron made of triangles whose vertices move such that no dihedral angle goes to zero

I'll ask on the main site

It probably does with more care.

You would need to show that you can pick the triangulation to be the standard triangulation of $S^2 \times [0,1]$, where $S^2$ is given some specified polyhedral structure.

Then your dihedral angles not going to zero is the same thing as saying that the map is locally injective.

For multivar math, do you administer exams chapter-by-chapter @Ted?

Hello chat.

@MikeMiller But two faces that aren't adjacent are allowed to pass through each other. Is that still "locally injective"?

Probably, actually

Yeps.

Not quite, @Brody. 6 exams total (plus 2 finals). I also have Spivak exams if you want 'em (ten in total, plus 2 finals).

heya @Fargle

How's it going @Ted?

Note to self, NEVER USE BUSSES AT NIGHT

I mean, the simplest example of an immersion which is not an embedding does that: circle to figure 8.

Trying to chat and work on letters of recommendation simultaneously.

You will miss your stop

You need landmarks, or you need to ask the driver to yell at you, DogAteMy.

@TedShifrin What use are landmarks if I can't see them

It's night

But some things are lit up :P Some.

Use google maps on your phone.

@Akiva Helps if your landmarks themselves are lights.

(Finally, an intelligent suggestion from someone.)

@TedShifrin Oh alright, will keep noted. I'm done chatting too for now...so bye, and lotta thanks

Night, @Brody.

@MikeMiller Good point on the tubular neighborhood thing. I have never actually had a pictorially transparent immersion of RP^2 in R^3. In any case that fails even for the standard embedding of the Klein bottle in R^3 I guess. If you try to thicken up the "outside", it'll invariably hit the thickening of the inside.

At least I can look at the beautiful stars! [cloudy sky] Oh. :(

@Akiva: I'll be in Manhattan from Wed to Sun.

Where in the world are you going, DogAteMy?

(i.e., the standard embedding of K inside R^3 shouldn't have a tubular neighborhood with embedded boundary)

@MikeMiller Oh, cool! @TedShifrin Home

you don't mean embedding of $K$, of course, @Balarka.

Immersion. Ugh.

Can a tubular nbhd of a non-embedded, immersed submanifold ever be embedded?

@Balarka Good.

Good point (even though you removed it). The mapping from the normal bundle isn't an embedding, but the image is an embedded submanifold with boundary.

I hadn't ever thought about that, I guess.

Huh, my route home takes me through that park.

Ya.

@TedShifrin so the binary operation on the projective group is composition of projective transformations, correct?

Yup.

I should really head off to bed now, I have to wake up earlier than usual tomorrow.

Night, @Balarka.

@BalarkaSen, have a good night.

Also if I carry on I'd soon start speaking unadulterated garbage. I can feel it coming.

So, g'night.

I should sleep at some point myself. Still haven't done that.

glares @Fargle

I can feel the Ted-smack coming on...

ducks

tranq-darts @Fargle

@Fargle: Unless you're very short, remember that I'm only 5'6", so ducking is not a smart ploy.

@TedShifrin Damn it, I'm 6'4 1/2"...

wow ... tall guy

jumps

LOL

It's so weird meeting people taller than me. I had an abstract algebra professor at UT--great guy--who was like 6'7".

Now you have another reason to look down on me.

And a physics professor who was about the same height.

@TedShifrin As though I had one to begin with!

Hmm. It's interesting to see how Rudin characterizes connected sets.

Or, well, more precisely, how he defines a separation.

@AkivaWeinberger the statement that $12(x-3)^2 = 12x^2 + 6ax + 2b$ was false

I'm so used to the topological "not the union of two nonempty disjoint open sets" but Rudin uses a little more than that--two nonempty sets $A,B$ are separated iff $A \cap \overline{B} = \overline{A} \cap B = \emptyset$, and then a connected set is one which is not the union of two separated sets.

@Ted: His deleted point was unfortunately not good. If the tubular neighborhood is small enough (small enough that sections are still immersed) the figure 8 cannot give rise to an embedded circle that way.

It makes sense that they're equivalent--it's just odd.

Because the figure 8 has turning number 0.

@MikeM: Oh, duh, I was taking $\epsilon$-nbhd, not mapping the normal bundle.

@Fargle: In your topology course, you should have seen that Rudin's version follows from the usual separation definition.

@TedShifrin I was probably shown that ages ago.

Flipping through Munkres of late has largely been a tour of "oh wow, I guess I knew that at some point".

not sure if this is a stupid question but ive been doing trigonometry using the sin cos tan functions but what exactly do the functions do ? say i had sin(20) and wrote it as f(x) = ? what is it actually doing on my calculator?

Hi @DHMO

That's partly what comes of taking a course before you're quite ready for it, @Fargle ... :)

@Brody: You said you were gone!

@TedShifrin Live and learn. I'll try to get through Rudin as best I can before going through Munkres again.

for now My bad, the "bye" was misleading

uh huh

In fact, the whole comment was misleading...

You should be punished for lying like Trump, @Brody.

@Brody hi

How's it do @DHMO?

well since time is relative @Brody

@WDUK they use a special algorithm, basically a sequence of numbers that converge to sin(20), until the precision required is attained

@Brody fine

why do schools teach us these functions but never going into what they do ? is it just really complicated... one of the few things where we learn to essentially take them for granted

@TedShifrin You should be able to find immersions that are not embeddings but boundaries of tubular neighborhoods are. For instance, push the top and bottom of a sphere into the center.

the algorithm is quite complicated, but there's a simpler version (that calculators don't use) @wduk

@TedShifrin how many hours did you absolutly have to spent on math to understand it at university? and how many hours did ou actually spent? (per week)

but you can use it to understand how they do it @WDUK

@DHMO I realized the dumbness in my statement last (or the other) night in defining $\omega^\omega$. While it is the lim sup of the sequence $\{\omega, \omega^2, \omega^3,\ldots\}$ I obviously did not include every preceding ordinal to establish an identity

do you know a good article that explains it ?

@WDUK the simpler version or the actual version?

simpler will do

i probably wont understand the complex one yet

@Brody so you have learnt more about ordinals

@MikeM: At this point I'm not sure :)

@Fargle I'm guessing it's to avoid stuff about "open in the subspace topology"

@TedShifrin: Think about that example! The appropriate side of the tubular neighborhood separates the middle.

@WDUK it is just the Taylor series: sin(x) = x - x^3/3! + x^5/5! - x^7/7! + ...

(The wrong side makes the immersion worse.)

@AkivaWeinberger Yeah, I guessed it was just to avoid discussion of general point-set matters.

@Null: That was a long time ago. I took a number of graduate courses when I was in college, so I put in a lot of time.

@Fargle: I think Rudin does way too much point-set as it is.

@Brody so do you understand why it is countable?

@MikeM: I can't stop to think now. I'm trying to get a recommendation posted to the web.

@TedShifrin Oh no, don't tell me there's a better intro analysis book. I already got all the way to chapter 2!

@DHMO I think $\omega^\omega :=$ $\{0,1,\ldots ; \omega, \omega+1,\ldots $ $;\omega\cdot 2, \omega\cdot 2 +1,\ldots ; \omega ^2, \omega^2 + 1,\ldots\}$ is the one. I mean to include all $\omega ^ p \cdot m + n$

@Brody \text{blub}

Nothing as infamous as Rudin, @Fargle.

Haha, just making sure.

@Brody And things like $\omega^2+\omega+1$...

Ugh, right. So they look like polynomials

@TedShifrin so, 50 hours is not unusual to spend?

Nevermind about earlier then

@Brody polynomial of finite degree

@Brody What are you trying to prove? That it's countable?

50 seems way excessive until graduate school in this country, @Null.

I definitely didn't do that much in undergrad.

But I was a pretty bad undergrad for most of it.

let's just say i'm bad haha. i somehow had the misconception that math=calculating. poor ed-system.

@AkivaWeinberger I'm just trying to define it, lol. @DHMO asked how to show it was countable, but he might know already

Ah.

What's wrong with $\omega\cup\omega^2\cup\omega^3\cup\dotsb$, assuming you have those defined?

I said differently, i.e. $\omega^\omega :=\{\omega, \omega^2, \omega^3,\ldots\}$ which is certainly incorrect I think

It is

@AkivaWeinberger isnt there some other notation? a big cup?

@Brody just add sup and it becomes correct

It's the limit of the sequence, but not the set itself of course

@Null Yeah, $\displaystyle\bigcup_{k=1}^\infty\omega^k$

(Doesn't much matter where $k$ starts)

$k$ can be arbitrary?

should start at 0

no

@Brody the starting point of k can be arbitrary, not k itself

@DHMO That's what I meant. My bad

yep. your bad

:p

and a big cap means the intersection of all those sets?

yes

@Brody prove that A\B is countable where A and B are countable

Removing a countable amount of stuff from a countable amount of stuff... leaves a countable amount of stuff :P

thanks

that's the proof i have been waiting for

it's so neat and elegant and precise and concise

@Brody (at most countable)

Yep, I know. Don't even mention it ;)

i detect sarcasm at levels we shouldn't been able to detect!

@TedShifrin Does countable mean countably infinite or at most countably infinite?

Depends on the text. :)

@Null can you prove?

If finite is included, then is the above still right? ~.~

@Brody What Ted just said--Rudin takes "countable" to mean "countably infinite" and "at most countable" to mean "finite or countable".

@DHMO no, my line of thinking is as brody's

@Fargle Oh, okay. Thanks

come on people

i know it isn't taught in undergrad

altho i find it not striking, as removing R from R leaves {}

but i think it's good to have some knowledge