"actually a homomorphism and not just an isomorphism" ...

oh, fair enough.

but yeah

Look what you've done @0celo7

we can't move on until you clear this up

Alright I'm off, cya @all

Does someone know if it's proven whether there is a prime between $n^2$ and $(n+1)^2$ for every $n > 0$ or not?

Could someone of you explain to me why at math.stackexchange.com/questions/1481950/centre-of-the-circle/… the unit normal vector $n(s)$ points towards the circle of the center?

@anon: of course those are the same for manifolds

@NaCl google "prime between perfect squares" --> Wikipedia article on Legendre's conjecture

Thanks @anon, once again :)

@MikeMiller continuous bijections? with boundary that's not true, so does that mean for boundaryless ones?

yeah

Probably true with boundary if you demand that, say, the number of boundary components is the same, but I haven't checked

injective map is homeomorphisn onto image for hausdorff spaces

that's not true

ok, compactness.

$X$ be compact, $Y$ Hausdorff

fine, but I wasn't using compactness above, hence the manifold restriction

eh, no @anon, just not true with boundary. noncompact manifolds with boundary are unethical anyway

Maybe cover up your domain manifold with closed balls of finite radius? Then restriction of the map to each of those is a homeomorphism onto image.

Eh.

good idea but you need to work harder to make it work - there's no reason naively that the map should be locally surjective

yeah, I figured, hence the "eh".

Could you take a look at math.stackexchange.com/questions/1481950/centre-of-the-circle/… and explain to me why the unit normal vector $n(s)$ points towards the circle of the center? @Huy

something I have wondered but never really thought about: if there is a smooth bijection $f: M \to N$, are $M$ and $N$ diffeomorphic?

Only if the inverse is smooth too @MikeMiller

@AlecTeal not the point here.

and how do you prove "only if"?

You find a bijection, name it $f$, pet it, give it love, prove it is smooth, and you see if $f^{-1}$ is smooth, if you're lucky, using similar techniques to that which you tried with $f$ @BalarkaSen

So what's the question?

that's not a proof of "only if".

you have just prove that if there is a smooth inverse, $f$ is a diffeo. that's trivial, follows by definition.

if you haven't read the question, don't try to answer it.

Oh I meant that in the sense of "if what you just said @BalarkaSen you may call it a diffeomorphism only if in addition you have that $f^{-1}$ is smooth"

Your book will have an index, or just click this: maths.kisogo.com/index.php?title=Diffeomorphism

ok, sure, but Mike never asked if $f$ is a diffeomorphism :P

certainly there are smooth bijections whose inverses are not smooth. but the question was whether or not the existence a smooth bijection guaranteed the existence of a diffeomorphism

cool, so i was interpreting the question correctly.

@BalarkaSen I was answering your question

Which question?

Oh also I was talking to Mike

@anon there is another example of the exact same question I asked before except that $f_n(x)=x^n sin (nx)$ on $S=(-1,1)$. As n gets larger and larger the graph becomes $0$ between $-1$ and $1$. So isn't the function going to be uniformly convergent on the set $S$?

Mike asked if $f : M \to N$ is smooth and bijective then whether $M, N$ are diffeomorphic. you said only if $f$ is a diffeomorphism, which you haven't told me the proof of

Hope you didn't remove anything offensive @BalarkaSen

I'd hate to add you to The List

note that (assuming no boundary, which IMO is reasonable) previous discussion implies that $f$ is a homeomorphism, so the question is really about whether there are strange exotic structures that permit smooth bijections $f: M \to M'$.

@AlecTeal i said nothing offensive

i am simply saying that you're misinterpreting the question

@Paradox101 consider g_n(x)=x^n first. is that uniformly convergent on (-1,1)? the sin(nx) complicates things because sin(n) is unruly.

@AlecTeal the point is this : $M$ and $N$ are diffeomorphic does not mean the diffeomorphism has to be $f$. how do you know that there isnot any very weird diffeomorphism $g : M \to N$ even if $f$ is a smooth bijection but does not have a smooth inverse?

@BalarkaSen I think you're reading too much into it, all he said was:

"are $M$ and $N$ diffeomorphic?" is not the same as "is $f$ a diffeomorphism?".

I read up and I cannot see what you guys are posing. I know you know $f$ is just a letter so I wont patronise you and go that route.

Mike confirmed my interpretation, see above.

Yeah, look for continuity (don't worry about partials) it is easy to exhibit such maps (I constantly did this by accident when I first encountered the topic)

um, exhibit which maps?

Maps which are bijective and continuous but whose inverse is not.

Diffeomorphism implies homeomorphism, so if you do what I just said, you've shown it.

@anon for |x|<1 it's 0 and for |x|>1 it's infinty. But shouldn't we also include sin(nx)? How can we accurately say anything if that isn't included too?

@AlecTeal any bijective continuous map between manifolds (without boundary, thanks Mike) is a homeomorphism.

so forgetting about the smooth structure would not help

(without boundary)

@BalarkaSen IF the inverse is continuous too.

NO, @AlecTeal

if your domain and codomain are manifolds, it automatically has a continuous inverse

that is the whole point.

So you think "homeomorphism" means "continuous and bijective" and says nothing about the inverse?

again, for manifolds (without boundary).

@Paradox101 I never said anything about |x|>1. I asked: on (-1,1) is x^n uniformly convergent? That's a warm-up version of your problem. Once you understand the warm-up, then you can try to tackle the original.

@BalarkaSen .... you okay bro?

@AlecTeal: For manifolds without boundary.

@anon yes it is for a large n

Can you prove that claim? (Like what do you count as a manifold, not being funny but there are 2 interpretations, countable-basis/metrisisable and not)

@Paradox101 really? you sure about that? can you force x^n to be arbitrarily close to 0 in arbitrary nbhds of +1? give me an epsilon, and I will show you an x<1 for which x^n is not within epsilon of 0.

@AlecTeal You need to stop pretending like a know-it-all. Seriously.

I'm asking a serious question

Manifolds are second-countable Hausdorff spaces locally homeomorphic to $\Bbb R^n$. Yes, I can prove that claim. :-)

@MikeMiller I would love (an outline of) the proof of "if $f$ is a bijective and continuous homeomorphism between two manifolds (w/ countable basis) then the inverse is also continuous, thus $f$ is a homeomorphism" because I don't think that's obvious and as I mentioned briefly before, codeine

It's not obvious.

@anon yeah i think so. if the n approaches infinity the graph tends to 0 within the interval. If we take epsilon to be 1/2 wouldn't it be uniformly convergent?

Try it for the case where $f: M \to N$ has $\dim M \geq \dim N$ first. This is covered by a classical theorem, usually proved in an algebraic topology course, certainly in Hatcher. When $\dim M < \dim N$ you need to apply the unnamed theorem and some extra cleverness.

oh. invariance of domain.

darn it.

@Paradox101 we're not talking about pointwise convergence, we're talking about uniform convergence. do you understand the difference? True/False: for any epsilon>0 you give me, I can exhibit for you a x<1 for which x^n is not within epsilon of 0. What does that entail? In other words, given an epsilon>0, is there an n for which the whole graph of y=x^n (for 0<x<1 say) is below the line y=epsilon?

To convince yourself it should be true, the case $f: \Bbb R \to \Bbb R$ doesn't need any machinery at all.

@anon in uniform convergence we need to take only epsilon while in pointwise we also need to take an x

@ the original question I asked (if $f: M \to N$ is a smooth bijection between manifolds w/o boundary, are $M$ and $N$ diffeomorphic (not necessarily via $f$)?), there is of course no reason to believe that $f$ is a diffeomorphism; $x \mapsto x^3, $M = N = \Bbb R$ is the standard counterexample. but are they still diffeomorphic?

(as I was pointing out)

by the previous claim about continuous maps, they're necessarily homeomorphic, so the question is whether there are two homeomorphic but not diffeomorphic smooth manifolds $M,M'$ and a smooth bijection $f: M \to M'$. i suspect this is not possible (i.e., that $M$ and $M'$ indeed are diffeomorphic); i'm much more confident about the case when they're compact

@MikeMiller please write me an @comment with the proof

I'd rather not. Balarka seems to get the idea, so maybe he'd like to.

I'll try to formalize my idea after I finish dinner :)

if I have a field $K$, $a\in K^*$ of order 12, i need to prove that $(a+a^{-1})^2=3$

Dumb question: If we equip two copies of $\Bbb R^4$ with two non-diffeomorphic smooth structures, is the identity map from $\Bbb R^4$ with the one smooth structure to $\Bbb R^4$ with the other one smooth?

but $(a+a^{-1})^2=3$ can be written as $a^2+2+a^{-2}=3$, so i need to prove that $a^2+a^{-2}=1$ right?

Does not seem dumb at all to me, because I can't answer it. Is there a classification for smooth structures on R^4? Mike would know, of course.

If I recall correctly, there are uncountably many non-diffeomorphic smooth structures on $\Bbb R^4$.

For $n\neq 4$ there is only one smooth structure on $\Bbb R^n$ however (up to diffeomorphism).

@iwriteonbananas: I don't know! This is one of the things I was guessing about. I would guess the answer is no.

What I mean is whether here is something like "such an such smooth structures are the only ones on R4"

@BalarkaSen can you help?

I know (as a fact) that there are uncountably many of those.

@Balarka: With uncountably many of them that's probably a hard statement to write down.

@MikeMiller Fair enough

@MikeMiller Hmm, true.

There is, I think, a universal one into which all the others smoothly embed, so that would I suppose give an answer to your question. But "open submanifolds of a manifold I find hard to deal with that are homeomorphic to $\Bbb R^4$" is perhaps not what you want.

@anon graphically at least it appears that we can find an n for which x^n is below y=epsilon

@MikeMiller Oh, I am curious about that universal thing you have mentioned. And no, haha, the latter is not what I was looking for.

I have neither a reference nor a proof sketch.

Probably any reasonably general book on 4-manifolds has a reference in it somewhere. Try Scorpan, say.

@BalarkaSen (By the way, I'm pretty sure we're going to prove the Acyclic Model Theorem tomorrow in my alg top class, if you've heard of it :P)

er, no @iwriteonbananas. I have never heard of it.

@BalarkaSen It actually looks somewhat similar to the fundamental lemma of hom. algebra: en.wikipedia.org/wiki/Acyclic_model#Statement_of_the_theorem

Let me write the name of the theorem down somewhere, I'll have a look later.

Thanks.

Sure. It provides a way to prove the Eilenberg Zilber theorem with purely abstract nonsense, so far as i can tell.

ok.

Is this correct?

Nope. Try plugging in $-1$.

In general, $|x+y|$ is usually not $|x|+|y|$.

@BalarkaSen No.

I had a specific thing in mind.

There was some fun times with a discontinuity involved.

then "homeomorphism and not just an isomorphism" does not make sense to me.

Well that's too bad.

@MikeMiller ok I got it. thanks :)

can you elaborate on the specific thing?

I dunno, I use the word "isomorphism" for isomorphism of groups/modules/rings/fields or algebraic structures.

Show that Sp(2) is homeomorphic to the interior of a 3-dimensional solid torus.

ah, you're on Lie groups :)

no idea what those are, so I'll shut up :D

There's an angle involved.

And it's not immediately clear if you can choose it continuously.

So it's easy to write down the isomorphism.

But to show homeomorphism is a pain.

ok, I see.

Thanks.

np

@0celo: Why don't you just show diffeomorphism? Is the map you're using not smooth? The IFT means it's usually easier to demonstrate a map is a diffeomorphism (when it is).

Oh, I see, you're not even sure it's continuous. Fair enough.

tbh I'm still not even sure how to show it's continuous

but I'm not hung up on it

so whatever

They are definitely isomorphic and people smarter than me say they're homeomorphic. So that's good enough.

I'm not quite sure what isomorphic means here.

Isomorphic as groups, presumably.

@AntonioVargas sorry, I did

I could put a Lie group structure on the interior of a solid torus, but certainly i can't think of a canonical one.

You first show $\mathrm{Sp}(2)\cong \mathrm{SL}(2,\mathbb{R})$

Ok, got you.

Then you do something called the "Iwasara decomposition"

It's basically modified QR for a 2x2 special matrix

And you get a rotation, and two other things

And in the end the solid torus falls out.

Is this the correct expansion? The $x+1$ term in the denominator of the last term seems strange.

@0celo: It will probably be easier for you to show the map from the solid torus to $SL_2$ is a diffeomorphism, as opposed to the other way around.

@MikeMiller Perhaps.

@Paradox101 We know that $\sum_{n=0}^{\infty} x^n=\frac{1}{1-x}$ for $|x|<1$

@Paradox101 Replace x with -x.

@Paradox101: What you wrote down earlier, by the way, is usually known as $-\sinh(x)$, the (negative of) the hyperbolic sine function.

@evinda for the expansion in the image i uploaded this is correct right : $\sum_{n=0}^{\infty}(-1)^n x^n=\frac{1}{1+x}$ ?

@Paradox101 Yes.

@evinda but then according to the image above where did the denominator in the last term in the expansion appear from?

@MikeMiller ok

@Paradox101 It should be a typo. You are looking for a closed form of $\frac{1}{1+x}$. Then at the closed form $\frac{1}{1+x}$ shouldn't appear.

@evinda $\sum_{n=0}^{n} x^n=\frac{x^{n+1}-1}{x-1}$

good evening everybody

@Ted are you online by any chance?

@evinda yes I was wondering whether there was some mistake in it. Thank you :)

and $x^{\infty}-1$=-1 ok ok you right

$x^\infty -1=-1$

what

@Paradox101 You are welcome! :)

@MikeMiller can you explain how to prove $|sin x| \le 1$? I first thought about expanding sin into it's power series and then somehow untangle 1 from it but it's not working

Do you consider the equality $\cos^2 x+ \sin^2 x=1$ as given? @Paradox101

@evinda nothing is assumed in this question so I suppose I can use whatever I can. But how do I prove that using this identity?

@0celo7 |x|<1

$\sin^2 x= 1- \cos^2 x \leq 1 \Rightarrow \sin^2 x \leq 1\Rightarrow -1 \leq \sin x \leq 1 \Rightarrow |\sin x| \leq 1$ @Paradox101

Oh so if we just use this and prove for both cos and sin that is sufficient? @evinda

Yes. @Paradox101

@Agawa001 ahh

@AlecTeal Here you are.

@evinda if we were to approach $|sin x| \le |x|$ then we need to use the power series expansion right?

Yes. @Paradox101

Thanks @robjohn was there an answer?

@evinda ok thanks

:) Btw... In which semester are you? @Paradox101

@evinda I'm actually in my 5th semester, but we don't start studying math until our second year

are you an undergraduate?

So don't you study mathematics? Yes, but I will finish after this semester and continue by doing my master. @Paradox101

@evinda yes I do. We do introductory bio, chem cs etc in our first year. And choose our majors in sophomore year. Are you planning on continuing with masters in pure math?

The master is called Mathematics and its applications. @Paradox101

@evinda do you have master degree of two years curriculum ?

It is 3 semesters @Agawa001

@evinda any recommendations for a good analysis book for self-study?

3 semesters back-to-back sounds intense!

I don't know. I have studied analysis from the notes of the prof... @Paradox101

Are you sure @evinda by that logic you've been doing Analysis for.... at least 2 years

Why? What do you mean? @AlecTeal

@Alessandro: You paged me?

yes, do you have 10 minutes for a question? @Ted

(good evening by the way)

Well, it's good afternoon here :) I need lunch soon, but what have you got?

Oh ok. I haven't found my professor's notes to be very useful in that they mostly only contain definitions which aren't enough to really grasp the topic @evinda

I'm trying to show that if $V$ is a finite dimensional vector space then $\dim V/ W=\dim V-\dim W$ @Ted

(what's the proper way to write a quotient space in latex?)

@Alessandro You want the slash in the other direction

Ah, just $V/W$.

What is your understanding of $V/W$? What, for example, would be a basis for it?

hi @Tobias :)

@TedShifrin Hi

My idea was to start with a basis of $W$ and complete it to a basis of $V$, I think that if I take the equivalence classes of the vectors that I add to the base of $W$ they should be a base of $V/W$, but I'm not sure about that

You're right. Do it :)

Much less concretely, you can also use rank-nullity.

Better to actually understand what the equivalence classes mean at this stage, so I censor that suggestion.

ok, so when I add the first vector $v$ to the base of $W$ I know that it can't be in the class of $0$ (since the elements of $W$ are), so $[v]$ is linearly indipendent in $V/W$, now I'm not too sure what happens when I add more vectors

we haven't talked about rank-nullity yet @Balarka

@TedShifrin I agree with you, hence I have added "less concretely"

Maybe I should edit it to include "much".

Ah, @Alessandro. OK.

Just take your basis $w_1,\dots,w_k,v_1,\dots,v_\ell$ and show that $v_j+W$ span $V/W$ and are linearly independent (by the standard definition).

I'd rather have an inner product and use perp, @Balarka :)

hm, ok, let me think about it

OK, @Alessandro, I'll check on you after I eat my lunch. :) Don't run too far.

hah.

Bon apetit, @TedShifrin

appétit, you mean?

buon appetito @Ted

Gracie, @Alessandro :)

rolls eyes

but realizes he does not have 7 of them

Lol

That's the only thing Balarka has actually learned from me ... how to roll his eyes.

Alright, this geodesic ball thing is driving me nuts

(it's grazie with a "z")

I'm taking lunch break. ... Oops, thanks, @Alessandro. :)

is (-infinity, 1] considered a closed or open interval?

Use the definition of a closed set and tell me what it should be.

(It's not a matter of convention)