Mathematics - 2018-05-30 (page 3 of 3)

Nicholas Roberts

19:00

And of course $A_2$ will also just contain the identity permutation

Daminark

How's it going @Alessandro?

Alessandro Codenotti

I have to study differential geometry :/ @Dami

Daminark

Rip

GFauxPas

hey guys

did my message about induction go through? this machine is awful, I'm on a public machine

it'

it's running Linux and the browser is called Tor and i dont understand the buttons

therefore its bad

Daminark

19:17

Maybe try the alternative browser, Ext?

GFauxPas

well whatever I can handle it for an hour, did my induction question go through

Dami were you around when I was asking for help creating a disjoint collection of sets that union to a collection of not-necessarily-disjoint sets

was it you who helped me

Daminark

My "hey everyone!" message was the first one I posted today

GFauxPas

it was 2 days ago I think

anyway

let $\{E_k\}_{\mathbb N}$ be a countable collection of sets not necessarily disjoint

then I make a sequence

$\displaystyle F_k = \bigcap_{j \mathop = 0}^{k \mathop - 1} \left({E_k \setminus E_j}\right)$

and want to argue that the disjoint union of all $F_k$ equals the union of all $E_k$

what type of induction do I need, if any

well-founded?

it doesnt need to be induction in name I just want to be able to pass the union to a limit of sets

Maximus

Is there anyone who can help me with clarifying big omega notation?

GFauxPas

there's a way you can find out if anyone can help you here

Tobias Kildetoft

19:23

@GFauxPas But if all the $E_i$ are equal, then the unions will not be the same

Daminark

I'd be more inclined to phrase it in terms of well ordering of $\mathbb{N}$ than induction actually

GFauxPas

countably infinite I mean

Maximus

If I have x \leq \bigg(2^{-\frac{n}{2}+11} \bigg) can I conclude x \leq 2^{-\Omega(n)}?

GFauxPas

if there's only one set it's a singleton

yeah that's what Im thinking Daminark, it's equivalent ofc

Daminark

So the way I'd argue this would be to say alright, let $x\in \bigcup_{k} E_k$. Then there exists some $n$ such that $x\in E_n$, so choose the smallest such $n$. Then $x\in F_n$

GFauxPas

19:26

I'm unhappy with your notation Max but I think I know what you mean, you're saying you want to replace the index with a function of that order?

Maximus

I can re-write the notation, what part of it makes you unhappy

can I replace 2^{-\frac{n}{2}+11} with omega notation? That is replace 2^{-\frac{n}{2}+11} with 2^{-\Omega(n)}

GFauxPas

$\Omega(n)$ is a set of functions, not a function of $n$, but I think I know what you're saying

well to start with you can factor out $2^{11}$ and ignore it, as a positive scalar multiple doesn't change asymptotic growth in this context

Maximus

isnt it asymptotic decay?

GFauxPas

12 of one, 1/dozen of the other, same thing

Maximus

I see, so I can conclude that by bounding x like that with omega notation, correct?

GFauxPas

19:30

I'm saying you can make the question simpler, I didnt answer anything

I'm just saying you can ignore the 11 part

Maximus

I understand

Thank you very much!

GFauxPas

sure, so, now you want to know

if you can replace $-n/2$ with a function in $-\Omega(n)$

Maximus

yeah

GFauxPas

and the answer is, $2^x$ and $\log_2 x$ are strictly monotonic so you can just look at the powers, youre welcome

Maximus

I see

GFauxPas

19:33

if it's for an assignment you'd want to write out the inequalities involved to show your work but my point is you can take $\log_2$ of both sides of an inequality of positive numbers

Maximus

It's not an assignment, it's for my research

I'm looking into omega notation

and you are using the lim sup definition, correct?

GFauxPas

well, if my argument convinced you, you're good. if you're still unsure, write down some inequalities :)

Maximus

awesome, thank you so much

GFauxPas

I dont remember all the exact definitions of $\Omega$ but I'm thinking of asymptotic comparison of sequences

back to Dami

I like what you're saying Dami

choose the smallest $n$ by well ordering then $x \in F_n$ because

well it's obvious but I have to say why it's obvious

because the set difference removed a set not containing $x$

because otherwise that would contract $n$ being smallest

right?

no problem Max

Daminark

Yup

GFauxPas

19:42

it remains to show that $F_n$ are disjoint

also obvious :P

why is it obvious tho

well $x \in F_m \implies x \in E_m$

and for $n > m$

Daminark

Assume $x\in F_i \cap F_j$

Okay yeah

You got it I think, continue

GFauxPas

I'm not sure how to say it rigorously

If $n>m$ then $x \in F_n$ and $x \not \in F_m$

but that's assuming the $F$s are disjoint that's circular

Daminark

So assume $x\in F_n$ and $n > m$

Then $x\notin E_m$

So $x\notin F_m$

GFauxPas

ah, I like it Dami. Well ordered sets are cool

If only we could impose a well order on every arbitrary set, wouldn't that be nice

OH WAIT\

meme of head full of stars exploding with light

Did you know Zermelo created the axiom of choice to prove the well-ordering theorem because he intuitively knew the WOT was true but needed an axiom to prove it

not the other way around

he made the AOC to be stronger than WOT, he didn't know it was also weaker

Daminark

Yeah, I heard

GFauxPas

19:55

thanks for the help Dami

are you a student?

Daminark

Yup

You?

konoa

20:16

hi chat
i know it may sound a bit presumptous, because i'm pretty sure i don't have enough breakground to understand why, but I was wondering why $\sigma_s(\nu_2(\mathbb{P}^n)$ does not have the expected dimension for any $n>2$,$s<n$
(recalling $\sigma_s$ to be the $s$-secant variety and $\nu_d$ the $d$- Veronese embedding)

Daminark

20:28

Yo @Mathein, what's up?

Sir Cumference

20:48

Analysis question: Suppose for $f: \mathbb{R} \rightarrow \mathbb{R}$, the fundamental theorem of calculus can be used to evaluate $\int_a^b f(x)dx$ for all $a,b \in \mathbb{R}$. Does that necessarily imply that $f$ is continuous?

I.e. can the fundamental theorem ever hold for a discontinuous function?

Alessandro Codenotti

Hi @Ted

Ted Shifrin

@konoa: In some sense it's because those varieties have a ton of symmetry. There's a beautiful paper of Griffiths & Harris that does all sorts of projective differential geometry relating to this.

Hi demonic @Alessandro.

Alessandro Codenotti

Can I bug you with a couple of diffgeo questions?

Ted Shifrin

@Sir: We already said it holds for some discontinuous functions the other day.

sure, @Alessandro.

Sir Cumference

Wait am I losing it

So what are the necessary conditions for FTC to hold?

Alessandro Codenotti

20:50

All my manifolds are smooth. $X$ is a manifold, $TX$ and $T^*X$ are diffeomorphic, right?

Ted Shifrin

We didn't answer that, and I'm not sure I know a general answer, but it definitely can hold for discontinuous functions.

Sure, @Alessandro.

Sir Cumference

Huh, interesting. I like to explore the important differences between antiderivatives and integrals, which most people shrug off as being "essentially the same"

Ted Shifrin

You need to work lots of examples from a good analysis book like Spivak's Calculus or something else.

Alessandro Codenotti

Ok, I was pretty sure of that, thanks! So we keep them distinct because they're conceptually different, just like $V$ and $V*$ for (finite dimensional) vector spaces

Ted Shifrin

Well, they're dual vector bundles, @Alessandro. So, if you're not in the smooth category, but in an analytic category, they're not generally isomorphic as bundles.

Daminark

20:53

Hey @Ted, @Sir

Sir Cumference

@Daminark Howdy :)

Eric Silva

@TedShifrin whoa what

Daminark

@Alessandro I guess you weren't joking about DG :P

Alessandro Codenotti

Hmm I don't know what the dual of a bundle means

Ted Shifrin

Guess they needn't be even in the smooth category, come to think of it. Even if they're diffeomorphic manifolds. Duh.

Hi Demonark.

Alessandro Codenotti

20:53

@Daminark nah we mostly did curves and surfaces in $\Bbb R^3$ and looked at some general stuff in the last few lectures

Ted Shifrin

Sure you do, @Alessandro. The fibers are the dual spaces of the other fibers :P

Sir Cumference

Just another question, if anyone is willing to answer: math.stackexchange.com/questions/2755711/…

Alessandro Codenotti

@TedShifrin ah, makes sense

Ted Shifrin

@Sir: First, without loss of generality, you can assume the intervals are the same. Now if the images are disjoint, you can certainly find an $\vec F$ where the integrals will be different. You should be able to generalize to see if the images are different on a closed subinterval.

@Alessandro: So what were your questions?

Sir Cumference

@TedShifrin Wait, so there are no disjoint paths for which the integrals are always equal (besides sign)?

Ted Shifrin

20:59

Nope.

Sir Cumference

Would you happen to have a proof?

Ted Shifrin

Make $\vec F$ that's totally $\vec 0$ on one and has positive work along the other.

You can use bump functions to glue to make a global smooth vector field.

Alessandro Codenotti

Well I wonder how curvature is defined for an $n$-dimensional Riemannian manifold, but I guess that's better answered by a textbook than a discusson in chat!

Ted Shifrin

That's sorta involved, @Alessandro. If you read my moving frames section in Chapter 3 I can tell you quickly how it generalizes.

Alessandro Codenotti

So on to the next question, I have a 2-dimensional smooth manifold. When does it admit a Riemannian metric such that its curvature is constantly 0? (We saw the flat torus in class and it's hurting my brain)

Ted Shifrin

21:01

Compact without boundary?

The flat torus shouldn't hurt your brain ;)

Sir Cumference

@TedShifrin Well, the paths having the same image doesn't necessarily mean the path integrals will be equal, e.g. if one isn't injective

Alessandro Codenotti

@TedShifrin I think I should stick to the syllabus for the moment and come back to the interesting stuff after the exam :D so I'll ask you about that at some point in the future

Ted Shifrin

Gauss-Bonnet gives necessary and sufficient conditions in the orientable case. You need $\chi(M)=0$.

Sir Cumference

I'm wondering what the necessary conditions are for the integrals to be equal, besides the sign

Alessandro Codenotti

@TedShifrin is that needed? I'm pretty sure the flat torus does hurt my mind :/

Ted Shifrin

21:03

@Sir: Even if it's not injective, they can still agree.

Sir Cumference

@TedShifrin Sometimes, but not necessarily, right?

Ted Shifrin

@Alessandro: The interesting result is that $K=0$ iff the surface is locally isometric to the plane.

Right, Sir.

Sir Cumference

Yep, I'm wondering the necessary conditions tho

Eric Silva

@TedShifrin o i c why this is fails if u live in complex land now

Ted Shifrin

You could overlap in a back-and-forth way but still make one net trip from beginning to end.

Alessandro Codenotti

21:04

@TedShifrin the theorema egregium gives one direction, I'm not sure about the other

Ted Shifrin

It's harder, @Alessandro. Shall I refer you to one of my exercises?

You get to use differential forms :P

Alessandro Codenotti

We just defined today what's a 1-form, I guess I should really learn what they are eventually!

Ted Shifrin

Yes, you must.

Alessandro Codenotti

I'll add that to the things I shall learn after the exams then

Ted Shifrin

Basically, it reduces to the fact that on a disk a $1$-form is closed iff it's exact. (No cohomology.)

In general, this is a PDE fact in all dimensions (curvature = 0 iff locally isometric to Euclidean space in the usual metric).

WE have to keep you busy learning interesting stuff so you don't have time to run over us innocent people.

mercio

21:07

oO

Alessandro Codenotti

@TedShifrin hmmm cohomology (and homology) are also on that famous list :P

I guess I'll be busy during the summer

Ted Shifrin

But you should read section 3.3 and see how beautiful differential forms make some of that ugly surface theory you learned ... :)

Are you coming to visit?

Eric Silva

@TedShifrin id call this an ODE fact moreso

Alessandro Codenotti

Ok, I'll read it then!

Ted Shifrin

Well, my proof uses Frobenius, which is PDE for sure. And closed = exact is PDE, not ODE.

Maximus

21:13

When you write up something in Latex and it involves nested proofs, is it best to write \begin{proof} again or is it confusing for the reader? What is the general consensus for when writing a general proof, you'll need to prove something else inside it.

Alessandro Codenotti

@TedShifrin It's not sure yet

Eric Silva

but you can do it w jacobi

Ted Shifrin

I'm sitting here being very stoopid and confused, @EricSilva.

Ah @Alessandro.

Alessandro Codenotti

I think I'm out of diffgeo questions for now, thanks a lot for your help! @Ted

Ted Shifrin

@Maximus: Nested proofs is not good style.

mercio

21:14

^

Ted Shifrin

Happy to continue later this summer, @Alessandro. If something else comes up whilst you're reviewing for the final, let me know.

user193319

Have: If $W$ subspace of $V$, then $\dim V = \dim W + \dim V/W$. Problem: If $\varphi : V \to W$ is a linear transformation, then $\dim V = \dim \varphi (V) + \dim \ker \varphi$. I've tried applying what I have to solve the problem, but I just don't see it. It's clear that the problem I am working on implies what I have, but that isn't helpful...

Maximus

@TedShifrin Thank you, then how would you recommend writing?

user193319

I could use a hint.

Alessandro Codenotti

Move the nested proofs outside as lemmas and invoke them in the bigger proof

@TedShifrin sure, thanks

Maximus

21:15

@AlessandroCodenotti What if the nested lemmas are connected with the main proof? Do you re-write all the pre-requisites in the lemma?

Alessandro Codenotti

(Not really related but I never understood "while" vs "whilst" in English)

Ted Shifrin

Sometimes it's preferable to prove the lemmas (or some of them) before, sometimes after.

"Whilst" is old/poetic.

@user193319: $V/\ker\phi \cong \phi(V)$.

Maximus

I see, so I can say "See proof of Lemma x.x"

as a reference

Eric Silva

Frobenius is like an ODE result masquerading as a PDE result

Ted Shifrin

No, it's the result of the lemma, not the proof, that you refer to, @Maximus.

Maximus

21:17

or prove it before and refer to it, correct? @TedShifrin

yes

that's what i meant, sorry

Ted Shifrin

Yes, @Maximus, or even after.

Eric Silva

@AlessandroCodenotti true anglos say whilest

Maximus

yousay x=y (as proved in Lemma x.x)

Ted Shifrin

yeah, @Maximus.

Maximus

@TedShifrin Thank you so much!

@AlessandroCodenotti And thank you as well!

Alessandro Codenotti

21:18

@EricSilva seems legit

Mike Miller

Index is 2 hard imo

Ted Shifrin

@EricSilva: I'm having a total mental collapse. How can two real vector bundles be isomorphic and have different euler characteristics? Did I switch orientations with the isomorphism?

Oh good, @MikeM is here to show me how I'm stooopid.

user193319

@TedShifrin I tried showing that, but I couldn't figure out the isomorphism.

Maximus

It's the first isomorphism theorem

Mike Miller

@TedShifrin Not seeing what you've written above, but Euler class of an oriented vector bundle negates under orientation-reversal, so that must be what happened.

Ted Shifrin

21:20

@user193319: Have you done group theory? It's just like that. But here the proof is just with bases.

@MikeM: I was commenting earlier that for smooth bundles, $E\cong E^*$.

Isa

If I have the solution $u(x,y)=\sum A_n e D_n \sin(x)$ of a PDE where I used separation of variables i.e. $u(x,y)=X(x)Y(y),$ and $X(x)=D \sin(x)$ then I'm able to know the explicit coefficient $D_n$ If I use the sine Fourier expansion, am I right?

Ted Shifrin

(Whereas not true in the holomorphic category.)

Mike Miller

The trick being that when you reverse orientation on M, e(TM) = -e(-TM), but to get the Euler characteristic, you integrate, which uses the ambient orientation of the manifold, so we get chi(M) = chi(-M).

@TedShifrin Ah, yeah, that breaks the orientation.

Maximus

You guys, when someone asks questions here, do you actually read the latex? or do you somehow translate it?

Mike Miller

I think.

user193319

21:22

@TedShifrin Yes, I have. I tried the canonical projection $\pi : V \to V/\ker \varphi$, but this doesn't seem to work.

Ted Shifrin

See "LaTeX in chat" up there >>>>>^^^^^^ @Maximus

Lozansky

@Isa Are you thinking of Fourier's trick?

Ted Shifrin

No, @user193319: Start with a basis for $\ker \phi$ and extend to a basis for $V$.

Lozansky

I.e. identifying Fourier coefficients

Mike Miller

@TedShifrin I think the easiest for me to perform a sanity check is with the recollection that oriented Riemannian vbs of rank 2 == Hermitian vbs of rank 1

Sir Cumference

21:23

@Maximus Use chatjax++

Mike Miller

So O(1)* = O(-1) tells me that this isomorphism swaps orientation

Isa

@Lozansky yes, exactly

Ted Shifrin

@MikeM: I don't see how choosing a Riemannian metric on $E$ to give $E\cong E^*$ messes up orientation. I have taught this before, dammit.

Maximus

@TedShifrin I am unable to find it

Isa

@Lozansky I want them to be in the final answer $u(x,y)=\sum A_n e D_n \sin(x)$ but not sure If it's correct to do so

Mike Miller

21:24

@TedShifrin It is not obvious to me either, but it must happen. Perhaps the explicit case of O(1) over CP^1 will make the computations clear?

user193319

@TedShifrin But my book says that isn't necessary. It says that it is an immediate corollary of the one theorem I already mentioned.

Ted Shifrin

Holomorphic category the bundles aren't isomorphic!!!

Maximus

@TedShifrin never mind I found it

Lozansky

@Isa You need boundary/initial conditions to determine the coeffiecients

Maximus

@SirCumference Thank you

Mike Miller

21:24

@TedShifrin I'm just abusing notation to denote the underlying oriented real vector bundle

Ted Shifrin

Hmm.

Mike Miller

And in holomorphic category I believe O(1)* = O(-1) is true

Which is the "oriented statement" I think you seek

The transition function for O(1) is $z \mapsto z$ on an annulus, right? We could see how that changes using the metric trick.

Ted Shifrin

I don't know, @user193319. I have no idea what your book knows and what it doesn't.

user193319

In my first post, I said I have the following: If $W$ subspace of $V$, then $\dim V = \dim W + \dim V/W$. Problem: If $\varphi : V \to W$ is a linear transformation, then $\dim V = \dim \varphi (V) + \dim \ker \varphi$...And my book says it follows immediately from this.

Ted Shifrin

Of course, @MikeM, I know how connection and curvature "transform" when I go to the dual, so I could cheat that way, too. But ...

Isa

21:27

@Lozansky yeah I forgot about it.. duh

Ted Shifrin

@user193319: But it does follow immediately if you've already proved what I said. I can't do math this way.

Mike Miller

So I'm thinking, @Ted, that the identification $E \cong E^*$ restricts to the trivial bundle as the transpose, right?

I'm too slow for this right now.

Ted Shifrin

Me too.

Transition functions for the dual are the inverse transpose.

Lozansky

@Isa Also, $u$ looks to be only a function of $x$? And I would rewrite $A_nD_n = C_n$ for simplicity

Mike Miller

So we want $z^{-1}$ transpose, which should just be $z^{-1}$.

That's O(-1), right?

Ted Shifrin

21:29

LOL ... one way or the other.

Mike Miller

I would have said O(n) is the guy with transition function $z \mapsto z^n$.

Ted Shifrin

Your point is that $z^{-1} = \bar z$ reverses orientation on the circle.

It depends on whether you're transforming frames or local sections in a trivialization.

Leaky Nun

hi @ted

Ted Shifrin

hi Leaky.

@MikeM: I still don't see how if all the transition functions for $E$ are in $GL^+(k)$, then the same fails to hold for $E^*$. Grr.

Mike Miller

@TedShifrin I think that is true. That guy is still orientatable, just with a different orientation.

Ted Shifrin

21:35

But what does that even mean? Oh help and bother.

Mike Miller

I think you swap the orientation on some basic chart, or something like that, I guess

and then the transition functions say this "new orientation" is the "new orientation" on other charts as well

I'm making this up as I go.

Ted Shifrin

This is deplorable that I'm stuck on this.

Mike Miller

That's what I should say about my research for the past two weeks as I face a deadline.

Ted Shifrin

Well, ignore me and work on your own stuff. We'll sort this out.

Mike Miller

@TedShifrin Catch up with me elsewhere sometime.

Ted Shifrin

21:44

Okey dokey. The key point is that the underlying real bundles of the complex duals are not isomorphic even if the real bundles are isomorphic. ... Talk soon.

Maximus

22:22

@TedShifrin are you familiar with Omega notation?

can someone explain to me intuitively what omega notation means?

Big O notation

Big O notation is a mathematical notation that describes the limiting behaviour of a function when the argument tends towards a particular value or infinity. It is a member of a family of notations invented by Paul Bachmann, Edmund Landau, and others, collectively called Bachmann–Landau notation or asymptotic notation. In computer science, big O notation is used to classify algorithms according to how their running time or space requirements grow as the input size grows. In analytic number theory, big O notation is often used to express a bound on the difference between an arithmetical function...

Their definition uses lim sup of some 2 functions

Mary Star

22:45

Let $f:\mathbb{R}^n\rightarrow \mathbb{R}$ be twice continuously differentiable and homogeneous of degree $2$.

I want to show that $f$ has its possible local extrema at its roots.

I have shown that (using Taylor expansion around the point $0$) that $f(0)=Df(0)=0$.
This means only that the function can have a local extrema at the root $0$, or not?

Mary Star

23:19

Or do we have to take the expansion at an arbitary point a?

Mary Star

23:40

Does someone have an idea?

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