@TedShifrin And how do you do that?

Hi guys!

Evening, @TedShifrin

@DemCodeLines I believe the most apparent solutions to your equation there are $\pm i$

@dsillman2000 He's working on $\Bbb Z_{17}$.

Ahhh

I just so happen to be learning about $p$-adic numbers rn

Excuse my ignorance I was coming into this blind

not p-adic

$\Bbb Z/17\Bbb Z$

I'm not sure what that is

@dsillman2000 you don't know what $\Bbb Z/n\Bbb Z$ is?

@dsillman2000 how do you know what p-adics are but not rings of integers mod n?

interesting, 21 minute delay and we both comment at the same time

@arctictern :)

Uhhhhh I was reading a book of brain-teasers that had a question involving $p$-adics so I just looked up what they meant, played around with it, and eventually gave up on that specific brain teaser. I still have a conceptual understanding of $p$-adics.

what does that mean

slightly condescending, but yes

I even know the equation for a sphere!

:P

that's a very strange way to write the circle equation.

why are they called that

@Kaumudi that's (x-g)^2+(y-f)^2 = 2[g^2+f^2]-c^2

so your center and radius seem off

@arctictern agreed.

@arctictern agreed

that's what I just said

it means your center and radius are wrong

Do you all schedule?

shedule time to study?

@Kaumudi complete the square

Find all solutions to the following equation on Z17: x^2 + 1 = 0

That was the question

@Kaumudi looks right

@Kaumudi yes

@Kaumudi probably

so what's the question?

@Kaumudi what you wrote here is different

@Kaumudi yeah ,so it's fine as far as I can tell

which are?

oh

the radius is imaginary!

that's really pretty useless

@Kaumudi nonsense :P

I don't think that makes any real sense

Even in the context of complex analysis

@DemCodeLines: What's the easiest number that's $-1$ mod $17$?

Hi @dsillman

Hi @TedShifrin

Oh boy, I have completely forgotten how to do negatives in modulus. One second.

$16$

Hush, @dsillman! :P

Was I right? xD

1?

@dsillman2000 It doesn't make real sense. That's exactly the point. This makes sense with complex numbers.

Why is it 16?

Because 16 is 1 less than a multiple of 17.

So are 33, 50, etc.

Wait a minute

Couldn't a circle have an imaginary radius in $\mathbb{C}^4$ via quaternions?

@TedShifrin oh ok

A circle can have imaginary radius in $\Bbb C^2$.

Well

Isn't 16 conveniently a perfect square, @DemCodeLines?

Yup

4

And -4, which is what mod 17?

13?

Good.

A radius I always think of as a modulus of a complex number; the distance. Because $|a+bi| = a^2+b^2$, mustn't it be real?

Note $13^2=169$ and that's $17\cdot 10 - 1$.

sorry $\sqrt{a^2+b^2}$

It's not a real circle. :)

Yeah it has an imaginary radius

But a radius is a magnitude

and a magnitude MUST be real

Well, this is a generalization of that notion. You're looking at a complex conic equation.

I'm not able to make the connection, aside from simply saying, "yes, that's true as well"

I'm showing you that $13$ really satisfies $13^2 = -1$ mod $17$.

I think I understand what you mean now

@DemCodeLines: What sort of course are you taking? It seems like you're not much used to mod stuff. So I don't know what you're supposed to know and what you're not.

Because a positive conic with a positive y-offset can only have imaginary solutions, the same plays out for the circle's formula; itself being a conic

Discrete Mathematics. Our textbook is pathetic. At least I find it very difficult to understand.

@dsillman: Working with complex numbers, $z^2+w^2=-1$ is a perfectly fine conic just like $x^2+y^2=1$ is working with real numbers.

Yeah I'm seeing that now

@DemCodeLines: OK, so you don't have much fancy background. You have no way to check that there are no other solutions other than doing all the squares of 1,2, up to 8 (Why can I stop there?) and seeing if they're -1 mod 17.

I do remember the professor talking about being able to stop midway while checking, 8 is midway.

Because $(-a)^2=a^2$.

I just don't know what to do with 0-8 with respect to the equation.

Just compute the squares and see if anything besides 4 works.

poor @Kaumudi

You want to check what the remainder is after you divide $a^2$ by 17.

Like $6^2=36$. What's that mod 17?

2

0celo, did you work out $x^2\sin(1/x)$ and see it's an example of strong diff failing?

@TedShifrin So the whole point about strong differentiability is that if $f$ is strongly diff at $a$, and $df(a)$ is injective, then $f$ is a homeomorphism of some ball onto its image.

Good, @DemCodeLines.

@TedShifrin I did not, but I have 8 hours in a car this weekend and I promise I'll think about it then.

Does that mean that strong diff will allow you to prove the inverse function theorem, 0celo?

Why do I have to check against -1?

@TedShifrin That's what I'm thinking my prof's end goal is.

But it's not a diffeo onto its image.

So like a weak inverse function theorem.

Because you're trying to solve $x^2=-1$ mod $17$, @DemCodeLines.

Honestly, I've spent my life doing geometry and I could never have used this, but it's interesting, 0celo.

did that "+ 1" change sides and become "-1"? Is that how?

I moved it to the other side ? ... Like the chicken and the road.

Yup, that's what I meant.

No difference :)

@TedShifrin Well you know the prof who's teaching this and he's more analytic than you were

poor phrasing since you did complex geometry

$x^2 = 0 - 1$

he's an analyst, there

$x^2 = -1$

I'm still curious if there are serious applications, @0celo, but it's interesting.

Right, @DemCodeLines.

@TedShifrin Probably not because he said about 5 times that this stuff never appears in the standard literature

Seems like an amusing fact that couldn't possibly have wide applicability.

But I'm pleased that my favorite counterexample function fails :) So I wonder what an actual example is.

If it did, it'd have to be some sort of very, very careful function space arguments that use the property 0celo7 invoked but don't work with higher regularity, I guess.

@TedShifrin there's another result whose proof is "postponed"

It's some weird form of Hölder continuous, maybe, @MikeM?

How about 9? 9 * 2 - 1

That's what it seems like.

No, is $9^2+1$ divisible by $17$?

Hmm

Remember $9^2 = 8^2$ mod 17.

$df$ is continuous at $a$ iff $f$ is strongly diff at $a$

we proved half of that today

That's bizarre, @0celo. Because that's what $C^1$ is equivalent to.

OR so I thought.

Wait, why? Why does $9^2$ become $8^2 mod 17$?

@TedShifrin no, just at one point

$C^1$ means $df$ is cont. on the whole domain

So $C^1$ at $a$ is not implied by $df$ continuous at $a$?

What does $C^1$ at a point mean?

I've never heard that

Partials are continuous at the point.

Then that's what we mean here, but I don't think many people use that terminology?

Ah, maybe that doesn't prove differentiable at the point.

This is the sort of regularity that loses me.

Holder, Lipschitz, Sobolev, ok...

@DemCodeLines: Because we agreed $9^2 = (-8)^2 = 8^2$.

@0celo: As long as I know partials exist on a little open set, I need only continuity at $a$ to prove differentiability at $a$.

@TedShifrin Sure

So I'm still confused.

I just wouldn't call that $C^1$ at $a$

To me, $C^1$ refers to the whole domain

OK, no point arguing.

Like $C^1(U,\Bbb R^m)$ or something

The question is why this concept is different, then.

It's not, apparently

So maybe it's a convenient way of proving some things, I dunno

I'd love your prof to give us a function that's strong diff'able but not C^1.

His notes end there, he hasn't updated the rest of the week's stuff

@TedShifrin I'll ask him

Keep me posted, 0celo. I'm interested.

@TedShifrin I'm curious where he's getting this material from.

I know he's pulling a lot of the more advanced topics (Stone-Weierstrass ultimate, Arzela-Ascoli ultimate) from Dieudonne.

I don't remember ever seeing this in Dieudonné, but I'll look later.

@TedShifrin He also has this obscure Brazilian text that he says is "way better than anything they bothered to translate."

Do you speak Portuguese?

Nope.

I'm sorry to be repeating myself, but where did we agree that $9^2 = 8^2$?

Does Bourbaki have an analysis text?

Because $9=-8$, @DemCodeLines.

@DemCodeLines 81-64=17=0

@TedShifrin OK, talk written. I can probably tell you the story if you want, though I won't have pictures here.

Would it be easier to scan your notes for me?

@TedShifrin Should the Urysohn lemma be taught in a first course on topology?

Absitively.

First semester

Yup. I did for sure.

@TedShifrin That's what I thought

And my topology prof agrees

but he says it's been an issue in the department for years

some profs refuse to teach it

It and Tietze are the actual non-obvious theorems. :)

I think we're doing that one too

The only part I don't understand is why the dyadic rationals are dense, but he said that's not hard

Same reason the regular rationals are.

I was explaining the proof of that to someone a week ago, and I've already forgotten.

meh

Ah

Same proof

Especially if you just think about dyadic decimals expansions for all reals, @0celo.

In any case, I don't see why one wouldn't do that theorem

they get bogged down and don't cover much?

@TedShifrin I dunno, Heine-Borel kind of blows my mind

get bored

@TedShifrin Uryson took like 20 minutes

Maybe we had some lemmas available

Urysohn was only 26 when he died

ancient compared to Galois

I know 0 algebra, so...

I know I've been silent most of the time here

But I'm going to bed for now

Nighty night!

night, @dsillman. Fare thee well.

bye

@TedShifrin Did you cover all of the metrization theorems?

Nope. Not time in one semester.

Ok, same

He said the proofs didn't contain any essential ideas either

Also, you will never use them.

@MikeMiller Urysohn metrization shows manifolds are metrizable, isn't that useful?

manifolds embed in euclidean space

indeed properly, so they're completely metrizable, which you don't even learn from urysohn metrization

Good point.

Urysohn I at least find interesting, admittedly

Everyone says it's so surprising, but I don't see that

I guess I'm so used to thinking about $\Bbb R^n$ or manifolds

the thing to me is that you actually get a map to R out of a condition about open sets

going from there to actually getting a function to the reals is weird. I mean, I know the construction, but still

@MikeMiller probably a stupid remark but isn't all of topology just conditions on open sets...

lol

fine. a discreteish condition, one that doesn't take a real number's worth of input to state

morning

@MikeMiller Embedability of topological manifolds in R^n is much harder than the smooth version, though, isn't it? I don't know how to do it.

Eh, maybe not.

Well, scrap that, I really don't. The same injective immersion argument might not have an analogue for compact top. manifolds.

Yeah, it's harder. Urysohn is certainly easier than that.

Right.

Wow ... a Balarka sighting during his daytime.

:P

One of my fellow grad students just posted his fifth paper.

Someone I know, Mike?

I think he's going to do fine.

Probably not, @Ted.

No prob.

I think the compact case should be easier. Given a finite cover, one should be able to get a "topological partition of unity" and extend the charts maps on small closed balls to all of the manifold (Tietze says you can do this and "dampen the ends", but I am not really sure if that's sufficient. Then take product, show it's injective, use that compact to Hausdorff injection is embedding.

Balarka, aren't you in school?

School's off :)

@BalarkaSen I am not at all convinced.

Balarka, we've discussed this before. To get an embedding, you need extra functions.

@TedShifrin Sure, I meant all that. I pondered a bit on the topological analogue.

@MikeM: Me neither

@TedShifrin The man is a writing maniac. At least 250 pages all put together so far.

I wish I had his stamina.

Wow. Oh, Trump would love his stamina :P

That might be about the total of all my pages in my career (not counting textbooks, class notes).

Maybe it's enough just to add on the partition of unity functions by themselves, Balarka. I think so.

Because then you can tell when you're in the subsets where the bump functions are precisely 1..

yeah, I guess that is possible.

I wonder how I should utilize the time today now that I woke up early. Certainly not by trying to prove Whitney for paracompact second countable topological spaces of specific covering dimension or something :P

No.

work

vector bundles or one of our flavors of differential geometry

Aha ... I found 0celo's prof's strong differentiabiity in some lecture notes.

Yeah, the point seems to be that you can prove the inverse/implicit function theorem only assuming strong differentiability (i.e., C^1-ness) at a point rather than locally. I wonder if anyone has serious applications of just the pointwise hypotheses.

The author gives my same example, slightly modified to show that the inverse function theorem breaks.

@MikeMiller I wanted to see what $w_2 = 0$ meant yesterday. If $E/X$ is a vector bundle, take it's classifying map $f : X \to BO(n)$. My argument for $w_1 = 0$ was that $w_1(E) = f^*(w_1(E_n))$ is zero and since $w_1(E_n)$ is the nontrivial class in $H^1(BO(n); \Bbb Z/2)$, that implies $f^*: \hom(\pi_1(X); \Bbb Z/2) \to \hom(\pi_1(BO(n)); \Bbb Z/2)$ is zero. Since $\pi_1(BO(n)) \cong \Bbb Z/2$, this means $f_* : \pi_1(X) \to \pi_1(BO(n))$ is zero, hence $f$ lifts to universal cover $BSO(n)$.

If I try to do a similar thing with $w_2 = 0$, the tells me that's zero on $H^2(-; \Bbb Z/2)$; since $H^2(BO(n)) \cong \Bbb Z/2$ I think that would tell me it's zero on $H_2$ (with $\Bbb Z$ coefficients) too. I can't figure out what that'd mean.

I wonder if it tells me something happening skeleton-wise than globally. Hmm.

Yes! $w_1 = 0$ says $E$ is orientable, which happens iff it's orientable over the 1-skeleton hence ultimately trivial over the 1-skeleton. I think together with $w_2 = 0$ it should say $E$ is trivial over the 2-skeleton. Because if I restrict $E$ to $X^2$, that's trivial iff the classifying map $f : X^2 \to BO(n)$ is nullhomotopic. I know that's at least zero in homology (in $H_1$ - in fact $\pi_1$ - by $w_1 = 0$ and in $H_2$ by $w_2 = 0$; the higher $H_k$'s of $X^2$ are zero)... hm.

Maybe it's weaker than saying it's trivial on the 2-skeleton.

$x+2$

hmmm mathjax doesnt work

@thunderbolt There is a link on the right on how to make it work in chat

remind me what it means for an action to be effective

the homomorphism G -> Homeo(M) is injective

(1) is equivalent to Homeo(M) being torsionfree, (2) means Diffeo(M) is torsionfree.

@abenthy I am skeptic if every effective action can be made smooth.

@BalarkaSen Well, that is not quite necessary for this statement

For example it's not true that every free action on a smooth manifold is smooth. Eg, Z/2 acting by reflection on a specific exotic S^7 (there exists one such, as the group of exotic 7-spheres is Z/28 which admits an element of order $\neq$ 2)

Nice, I get what you mean. Some people call a manifold for which (1) or (2) holds asymmetric and I was wondering if this is the same. But it seems also to me like one should call (2) "asymmetric in the smooth category" or something then.

@TobiasKildetoft I agree, I was just being skeptic in general.

My example above was probably too hard, there should be a much more easier one. But my example also says this Z/2-action on this exotic S^7 is not homotopic to one which is smooth.

btw. is it common to call the category of smooth manifolds shortly the "smooth category" or did I just take that up somewhere wrong?

That's the right terminology.

You should ask your question somewhere else, I don't know a definite answer to it.

Or wait for the topology people to arrive.

You already helped me, thank you :) But I should probably really ask it somewhere.

If X is a projective variety is it true that $ \text{Proj}(S(X)) \cong X $?

ah never mind -

Hello, to prove that $f(E\setminus A)= F\setminus (f(A))$ where $f: E\rightarrow F$ is the bijectivity of $f$ is important ?

@Vrouvrou No, but injectivity is

can i say that let $y\in f(E\setminus A)$ then there exists $x\in E\setminus A$ such that $f(x)=y$ then as $x\notin A$ , $y=f(x)\notin f(A)$

@Vrouvrou Were you not the one who asked about images of intersections recently?

no

ahh yes

i asked about a counter example

Yes, but you also mentioned being aware that it held for injective functions

yes

And you can translate this to that

but why i can't say this : let $y\in f(E\setminus A)$ then there exists $x\in E\setminus A$ such that $f(x)=y$ then as $x\notin A$ , $y=f(x)\notin f(A)$

@Vrouvrou is the function injective here?

no without injective

why i can't say this

then the last part does not follow, as something else might also map to y

but we have $y\in f(A) \Longleftrightarrow \exists x\in A, f(x)=y$

yes, but the "exists" does not require uniqueness

nor that no element outside $A$ can also map to $y$

If $f(z)$ is analytic in $D: 0<|z| \leq 1$ and $z^{k} \cdot f(z)$ is unbounded in $D$ for every integer $k$, how can I prove $f(z)$ has an essential singularity at $z=0$?

@TobiasKildetoft i don't understand the last sentence

@Vrouvrou I mean that just because something in $A$ maps to $y$ does not mean that nothing outside $A$ can map to $y$.

Is it related to the fact that there is no $k$ such that $\lim_{z \to z_0 } (z-z_0)^kf(z) = 0$?

0k thank you , the injectivity is important here because x is note unique

Actually that's incorrect

There is no integer $k$ such that $(z-z_0)^kf(z)$ has a finite nonzero limit as $z \to z_0$

@Lozansky Do you understand why it has a singularity at $0$ in the first place?

Or more specifically, why is the singularity not removable?

@BalarkaSen If it were removable then $$f(z) = \sum_{j=0}^{\infty} a_j z^j$$ but then $$z^kf(z) = \sum_{j=0}^{\infty}a_j z^{j+k}$$ but we know that's not possible because $z^kf(z)$ is unbounded and therefore there's no Taylor series that converges around $0$?

Actually I'm not sure

@Lozansky Yes. If it were removable then $f$ would extend to a holomorphic, in particular continuous function on $|z| \leq 1$. $z^k f(z)$ would also extend to a holomorphic in particular continuous function on $|z| \leq 1$ by the power series you wrote down. But continuous functions on compact sets are bounded, contradicting hypothesis.

Now, can you tell me why that singularity is not a pole?

@BalarkaSen Then there would be a $k$ such that the power series I wrote down is bounded, but there's not?

Indeed, tautologically so.

What do you conclude?

$f(z)$ must have an essential singularity at $z=0$

Da.

^^

Thanks

Sure.

The point is near a pole a holomorphic function looks like $a_{-k}/z^k + \cdots + a_{-1}/z + a_0 + a_1z + \cdots$ for some $k$.

@Vrouvrou Ohh, and actually you also need surjectivity for the original claim (as can be noted by trying with $A = \emptyset$).