Hey chat!

@AkivaWeinberger That it is part of the topology of the space. You can generally make any set open by altering the topology associated with the space.

@Velvet He knows. He was trying to teach socratically.

@DanielFischer Hi, does there exist a holomorphic function on a neighborhood of $0$ such that $\vert f^{(n)}\vert \ge (n!)^2$ for all $n\in J$ where $J$ is an infinite subset of $\Bbb{N}?$ I know the answer is no if we replace $J$ by $\Bbb{N}.$

and Ted applauds DogAteMy for his socratic approaches.

@JeSuis What do you mean with $\lvert f^{(n)}\rvert$? $\lvert f^{(n)}(0)\rvert$, or $\sup \{ \lvert f^{(n)}(z)\rvert : z \in V\}$ where $V$ is - the domain of $f$, or some other set? Something else?

@DanielF !! :)

Hi @Ted.

My personal take is that he meant derivatives at $0$ ... :)

If so one should get a contradicition via Cauchy's integral formula + hol. functions are bounded near $0$

Hmm, maybe not

I expect so too, @Ted, but I'd rather make sure.

@Vrouvrou: You told me that you knew that $N$ was closed. If so, you're done. And I do not know how to prove this.

Greetings @JeSuis and @Balarka

Hi @Ted

Hi @Ted

@TedShifrin by the definition of $N$ i think that it is the inverse image of \{0\} then it is closed

Done with thermodynamics for a while? You'll be amused to know that I volunteered myself to tutor someone (I'm not sure whom) in physics. Yikes.

Lol in high school I used to think physics was my thing

Yeah, it's Cauchy integral formula.

Or rather the M-L inequality

And while my IB background was not the best I still pushed for honors physics

No, @Vrouvrou, you have to think about closed in what. Obviously $0$ satisfies that equation.

And realized that physics very much was not my thing

Heya @Daminark. Did you get that matrices of constant rank thing figured out? That's a very cool and important question.

@Balarka: The term M-L is not universal.

@Daminark: I took a "harder" freshman physics sequence at MIT and absolutely loved it. I still regard Kleppner-Kolenkow as the best mechanics book for someone who knows vector calculus.

@TedShifrin i don't know where it is closed, i just know that $N=\{u\in X\setminus\{0\}, I'(u)u=0\}$ and $I'$ is continuous

@BalarkaSen The Cauchy estimates. (Which are of course consequences of the integral formula.)

So our class couldn't assume people had knowledge of vector calc, since if you had BC Calc in high school, you did Spivak here

So it sorta mixed in the vector calc with the physics

One section used K&K, my section used Morin

@DanielFischer Yeah, that's probably it. I meant the one with $Mn!/r^n$; M-L was something else but related

I don't know, then, @Vrouvrou. You need a proof that it is closed in $X$, not in $X-\{0\}$.

@Daminark: Too bad. K&K is fabulous and has awesome exercises.

@BalarkaSen Yeah, the standard estimate (or ML inequality) is used to get the Cauchy estimates.

@Ted And I hadn't gotten around to it yet, I was in a rush and had pinkeye so at some point I just made this flimsy argument about how if you modified an element in one block, you could get away with only modifying the corresponding element in another, so by witchcraft you had full rank

Right.

Hi chat

@Ted I will say, while that class on the whole went iffy for me since I was in over my head, Morin is a really high quality book

@TedShifrin Was that to me? Not doing much thermodynamics right now, nope.

Good luck with physics :P

But yeah, second quarter of the class was Purcell and third was Morin's draft on waves, along with a couple weeks of thermo from Schroeder

And I found that my ability to visualize insofar as it was necessary for physics was basically nonexistent

Thus I kind of pulled back (under which diffeomorphism?) from the subject

That handwave isn't right, @Daminark. If you look carefully at the formula, you can see that you only need the derivative of an easy piece of it to get surjectivity.

We did Purcell second semester, too, back in 1971. I loved that year of physics, @Daminark.

Somehow I think Griffiths is better than Purcell, like it was after reading that book that I had a local understanding of E&M + Vector Calc

Now, it was in the physicist style, and again, visualization $\cap$ Amin $=\emptyset$

Griffiths is meant for a senior-level course, not a freshman-level.

We have to turn you into a geometer so you can visualize, @Damn-ark.

Oh huh, weird, I found it to be somewhat better explained. I used it when I didn't understand what was going on in Purcell

Lol this week's analysis pset is very interesting for sure

I'm excited

It's all on manifolds and the such

Good. I still bitch about your course.

Haha, probably in fairness

In the manifolds part he's really throwing us in the deep end somehwat

*somewhat

I think it's that he really needs to get to functional soon, like the point of honors analysis is to do Rudin-level material fast and get to measure theory and functional

I still suggest some of my lectures might help (same if you'd done more with Stokes's Theorem).

I think it's ludicrous that they think you'll pick up serious multivariable calculus "on your own." I really do not like Chicago's idea of this course.

So we have so far done some examples of manifolds + equivalence of definitions, Schlag mentioned parametrization but said you had to be careful, you needed something like that the subspace topology was equivalent to that generated by the map

There's nothing wrong with taking a graduate course in Lebesgue integration your junior or senior year.

Yes, you do need to be careful about that. The figure 8 is the standard example.

I mean, it's tricky to say, the class changes drastically from year to year

When Schlag taught first quarter, it was mainly multi and linear algebra

But with topology, all they did was prove (and it wasn't well explained) that norms are equivalent on R^n, and do some compactness

The rest was basically linear algebra and multi

I wish someone who loves serious multivariable calculus (+ analysis) did it. ... But I'm obviously biased.

Now, by contrast, Souganidis is very apathetic to the stuff

Now, we did quite a lot of topology-esque stuff

Somehow Chicago has this reputation that you get to do graduate stuff your second year and skip all the "mundane trivial" stuff.

hi @TedShifrin

Hi Karim

I am marking for multi -variable analysis I am really enjoying it

@TedShifrin I really hate the grad students here they are so mean

I was helping this guy in this topology question

Standard stuff in Rudin chapters 2-4, along with Baire Category, Uniform boundedness, absolute continuity, bounded variation, quite some time on uniform convergence

he would have never gotten it, then after that I wanted to check with him something he didn't answer me for 2 days @TedShifrin

@TedShifrin i found in a pdf that $N$ is closed in $X$

I am not gonna discuss stuff with other grad students I will just keep to myself.

Karim: I think that is overreacting.

And multi was just 4 weeks total, with Soug muttering at one point that Schlag would do forms in order to prove "integration by parts" even though they were unnecessary

But do make sure you're careful to think about mathematical correctness with what you say :P

@Daminark: I am not changing my opinion of the course ... or of what I hope you will make sure you sit down and learn. It should not be a point of pride that one can't do challenging multivariable calculus computations.

(or even trivial ones)

Schlag is much more stressful of the multi and geometry, which is probably good. We'll get Marianna third quarter who's much more into the abstraction than either

Oh right so our problem set this week is nice

One is rank theorem for smooth functions between manifolds

anyway @TedShifrin I am enjoying my topology class and commutative algebra

so cool

One is proving that SL(n,$\mathbb{R}$) is a Lie group and computing tangent spaces

My challenge to you (since you don't have enough) is this: Prove that a smooth retract of a manifold (or of $\Bbb R^n$) is a smooth submanifold. Also convince yourself that this is false for continuous stuff.

In our topology class we are gonna go to eventually to homotopical algebra just the surface though.

homological, Karim, not homotopical?

no he said homotopical

I do not know that phrase at all.

Maybe he said some homotopy theory.

What is the retract of a manifold?

yeah maybe like higher homotopy or something

btw check this question was very cool to figure out

$f\colon X\to Y$ is a retraction if $Y\subset X$ and $f|_Y =\text{identity}$.

it is about how $I \times D^n$ retracts onto

$\{0\} \times D^n \cup I \times S^{n - 1}$

very cool question

@Daminark: An easy point-set topology exercise is to prove that a retract must be a closed subspace (assuming Hausdorff, if you've even done what that is). Give me a subset $Y$ of the plane and a retraction to it ... with $Y$ not a submanifold. (Hint: What's the easiest example of something 1-dimensional that's not a submanifold?)

Karim: I think I last thought about that in grad school.

OK, lunch time for this Bonzo. Bubye.

brb school time too

@TedShifrin When we prove that $N$ is closed on X how to continue ?

@MikeMiller Oh my bad. Couldn't tell from his messages.

Also, hi

@Vrouvrou "$N$ separates $X$ into two components" is the same as "The complement of $N$, $X\setminus N$, has two connected components"

ok

Since $0$ isn't in $N$, it's in its complement. I've fairly certain that, if $X$ is connected, and $N\setminus X$ is open, then the connected components of $N\setminus X$ are also open

which means the component containing $0$ is open

@BalarkaSen can we discuss a problem really quick

which means there's a small open ball around $0$ contained in that component.

@Adeek OK

because I found a mistake in something I did

Do all mathematicians use Graph Paper to write math?

Probably not? @Lelo

@BalarkaSen so I showed surjectivity

using the universal property of quotient spaces

Not a bad idea, though, I guess

now I thought I can just use the fact that q has a set theortic right inverse which

Yea I wondering the same thing

we can use

but that doesn't work

Perhaps I'll make the switch

so I think I have to use the univerisal property again to show injectivity

@BalarkaSen Suppose $[f_1 \circ q] = [f_2 \circ q]$

Then there exists a homotopy $H : I \times (X,A) \rightarrow (Y,y)$.

@Vrouvrou Actually, I meant that if $X$ is locally connected, and $X\setminus N$ is open, then the connected components of $X\setminus N$ are also open. But I have a suspicion that the $X$ in your problem is $\Bbb R^n$, which is definitely locally connected.

if Q is a quotient map from $(X,A)\ to\ (X/A), \{*\}$ do we get a quotient map $s I \times (X,A)\ to\ I \times (X/A,\{*\}$ ?

@BalarkaSen ?

No.

:S

Assuming by $I \times (X, A)$ you mean $(X \times I, A \times I)$.

right

how can I show injectivity then I think I have to factor the homotopy H somehow

using universal property

If $f_1 \circ q$ is homotopic to $f_2 \circ q$ as maps $(X, A) \to (Y, y)$, there is a sequence of maps $\varphi_t: (X, A) \to (Y, y)$ defining that homotopy. These factor up into maps of the form $\psi_t : (X/A, A/A) \to (Y, y)$ - a potential candidate for a homotopy between $f_1$ and $f_2$. Now all you have to do is to show that's continuous in $t$ variable.

oh this is continous by the universal property again

Which is?

the maps $\psi_t$ is continous iff the maps $\phi_t$ is continous

But that doesn't say if it's a continuous sequence of maps in the $t$ variable, which is the nontrivial bit.

oh :S

Showing continuity in $t$ variable is a little more work. Your initial idea was good, but note that $X \times I/A \times I$ is not homeom to $(X/A) \times I$ contrary to what you thought. However, $X \times I/A \times I$ is a quotient of $(X/A) \times I$.

does these maps $\psi_t$ satisfy the property that $\psi_0 = f_1$ and $\psi_1 = f_2$ ?

I see

Of course they do! Eg $\varphi_0$ is $f_1 \circ q$, so if it factors into $\psi_0 \circ q$, by universal property $\psi_0=f_1$.

Because factorization is unique.

I see

OK, continuity in $t$ variable. Let $F : (X \times I, A \times I) \to (Y, y)$ be defined by $F(x, t) = \varphi_t(x)$ as you were doing before. Then this descends to a map $G : (X \times I/\sim, A \times I/\sim) \to (Y, y)$ where $\sim$ identifies $(x, t)$ and $(y, t)$ for each fixed specific $t$ whenever $x, y \in A$.

right

right @BalarkaSen

@AkivaWeinberger no $X$ is a Banach space

$X\times I/\sim$ is exactly $(X/A) \times I$ and $A\times I/\sim$ is exactly $(A/A) \times I$.

oh cool

So written otherwise $G$ is a map $((X/A) \times I, (A/A) \times I) \to (Y, y)$. Aka, $G:(X/A, A/A) \times I \to (Y, y)$. It is now a simple matter of checking that $G(x, t) = \psi_t(x)$.

That's the homotopy, continuity all corollary of the universal theorem.

very cool

thanks a lot @BalarkaSen

I should really have a deep check for my arguments before I accept them

I thought that because q is a quotient map

then it has set theortic right inverse

I finally found something I've been looking for for years--one of my old textbooks referred to $\forall$ as a "super and" and $\exists$ as a "super or", and I found that really illuminating, but I couldn't remember where it came from.

then I used that to induce a map g which i used to proved injectivity

@BalarkaSen

@TedShifrin My understanding when people use these words they mean algebraic K-theory or some generalization of it.

@Adeek Quotient maps most of the time doesn't have an inverse. You are sending a lot of points to the same point; where will that "squished point" go back to?

@PVAL Something something Quillen

model categories something

oh right right @BalarkaSen

@Adeek Ah, no, wait. I mistook right for left. Then sometimes it does (easiest examples are sections of a vector bundle); but there are counterexamples.

You don't need model categories to talk about K-theory.

Eg, [0, 1] \to [0, 1]/(0 sim 1)

thanks @BalarkaSen for your help I am off to class

@MikeMiller I was mumbling vaguely since I heard of model categories in the context of homotopical algebra.

(PVAL's K-theory comment was in reply to Ted's comment about homotopical algebra above)

I know nothing of this though.

Oh yeah I guess when people say homotopical algebra they probably mean K-theory, Quillen stuff, model categories, operads, whatever. I don't think this has any place in a first or second or third algebraic topology course.

Looking back up there just gave me an "oh duh" moment: a retraction $r : X \rightarrow Y$ is a category-theoretical retraction of the inclusion $i : Y \rightarrow X$. That's why it's called a retraction in category theory.

That kind of things scare me off.

@Fargle Mhm

They're not actually scary but there's no reason to bother with them before you need them.

I'm just bugged I didn't see it before, haha.

Hello there.

Heya @Mahmoud

@MikeMiller If $E/X$ is a bundle with structure group $G$, then that gives a map $\pi_1(X, p) \to G$ modulo choice of a trivializing atlas of $E/X$: for any loop $\gamma$ cover it by charts $(U_\alpha, \varphi_\alpha)$ with $U_0$ containing $p$, say, then look at the transition functions $h_{ij} : U_i \cap U_j \to G$ and let $\gamma$ be sent to $h_{01} h_{12} \cdots h_{k0}$ in $G$. The classifying map $X \to BG$ is the delooping of this, yes?

I guess it's not entirely clear if this thing preserves homotopies.

Hi !

Isn't $\lim_{x\to \infty} \frac{4x^2-5x}{1-3x^2}=\lim_{x\to \infty} -\frac{4x^2}{3x^2}=-4/3$ ?

I saw someone use L'Hospital's rule for it :P

Unless, I'm actually wrong.

multiply by $\frac{1/x^2}{1/x^2}$

So my method is wrong ?

Its not wrong but not very clear

It may not always be obvious to just remove lower order terms

If you multiply the top and bottom by 1/x^2 however it becomes much more clearer what happens as x gets bigger

@Balarka That is going to depend on your choice of sequence of charts.

@Balarka You need to choose a connection on the bundle. That will give you a monoid homomirphism $\Omega B \to G$. When the connection is flat it descends to a map in pi_1

hey @MikeMiller have you ever heard of Topology,geometry, and gauge fields by Gregory ?

it is very cool book.