Mathematics - 2017-03-24 (page 5 of 5)

I usually have this stage where I'm too tired to productively do math, but not enough to go to sleep, so I'm kind of roflderping for a little while until I'm done. Other times I just math myself to sleep

SoumyoB

10:17 PM

@Daminark so we're both marked by the same curse?

Sha Vuklia

"math-ing oneself to sleep" XD

Ted Shifrin

Demonark is cursed by being so evil, though ...

Meow Mix

@Ted Different sort of length?

Daminark

Pretty much. Though I think it's really more a function of mismatched sleep and class schedules. I've never had a quarter where I didn't have a class before 10:30 (usually I've got something at 9:30), and yet my natural sleep time is basically 3:30 to 11:30 $\pm \epsilon$

@Sha As Akiva says, anything can be a verb if you verb it

SoumyoB

@ShaVuklia was it insomnia or sleep state misperception? Just curious.

Lanet Rino

10:19 PM

Guys! Ive solved Gabriels horn!

Meow Mix

Solved it?

Sha Vuklia

Hahaha true @Daminark :P

Ted Shifrin

What did you mean, Zach? We have an inner product space, so every vector has a (finite) length.

SoumyoB

is that a millenial conjecture?

Sha Vuklia

whoa, sleep state misperception? :P

Daminark

10:20 PM

@Ted Evil? angelic ring (not sure if commutative) appears above head

3

Sha Vuklia

what do you even mean by thatXD

Ted Shifrin

Do you mean you understand why the paradox isn't a paradox, @LanetRino?

Lanet Rino

I was just being obnoxious. Carry on

Ted Shifrin

Halos needn't be commutative.

5

Meow Mix

@TedShifrin But what about vectors with finitely many zero components?

Ted Shifrin

10:20 PM

Finitely many NONzero?

Namely, vectors that live in some $\Bbb R^n\subset \Bbb R^\omega$?

Meow Mix

Finitely many zero... because that would make them have infinite length

Sha Vuklia

@SoumyoB oh I see it's a real thing (because it's wikipedia-able)

Meow Mix

No, the opposite. Some "infinite vector"

Ted Shifrin

You're confusing yourself, Zach.

SoumyoB

it's type of insomnia where you go into a semi sleeping state after you close your eyes but it's not sleep exactly

it's pretty much useless kind of sleep, you might as well be awake all night

Ted Shifrin

10:22 PM

Our universe is the set of infinite sequences $a=\{a_n\}$ with $\|a\|^2 = \sum a_n^2$ finite by definition.

Meow Mix

Universe??

Ted Shifrin

Yes, universe.

That's the daddy vector space in which $V$ is a subspace.

Is it expanding ?

Oh, alright.

glares at Astyx

10:23 PM

OH, sequences that are eventually zero, so by that you mean finitely many non-zero, right?

Ted Shifrin

Right.

Sha Vuklia

@SoumyoB ah, no, I don't think that was it in my case. I just couldn't sleep a full 8 hours a couple of months, because I was too stressed I guess. So I'd be waking up after only 5 hours of sleep, and wouldn't be able to sleep again

Ted Shifrin

Of course, every element of our universe approaches $0$ in the limit, right, Zach?

Sha Vuklia

but now that I'm learning how to relax better and not to do much the hour before I go to bed, I'm pretty fine

Meow Mix

Well yeah

Otherwise it would be divergent :P

Ted Shifrin

10:25 PM

The series would be ...

Meow Mix

Yeah.

That's what I meant, sorry :P

Ted Shifrin

Remember — try to be careful with the math language.

Meow Mix

like $(1,1/2,1/4,1/8,\dots)$

Ted Shifrin

or $(1,1/2,1/3,1/4,1/5,\dots)$

Meow Mix

Oh, that works because it's being squared, right?

Ted Shifrin

10:27 PM

nods

Astyx

Is there a specific word in english terminology for series of which the general term does not go to zero ?

Akiva Weinberger

Hi. scrolls up You're working in an infinite-dimensional vector space?

Alessandro Codenotti

"Let $X$ be a topological space and $q\ge0$, we'll denote with $S_q(X)$ the free abelian group generated by the singular $q$-simplices, called the group of $q$-chains" (I'm translating from Italian, so I hope the terminology is correct), I don't quite understand the definition of this group

Meow Mix

@AkivaWeinberger Yep.

Ted Shifrin

What do you mean, @Astyx?

Meow Mix

10:27 PM

I don't know why, but I also started a cookie clicker game.

Ted Shifrin

DogAteMy: He's supposed to be understanding $V$ versus $V^{\perp\perp}$ a bit better.

Akiva Weinberger

I remember hearing about a story where Hilbert attended a lecture which involved Hilbert spaces, and at the end, he had to ask the lecturer what a Hilbert space was

'Cause they named it after him without him knowing

Ted Shifrin

You can take all integer linear combinations of $q$-simplices, @Alessandro. What do you not understand?

Akiva Weinberger

@TedShifrin Ah. (I'll point out that I don't think I understand it completely either)

Ted Shifrin

DogAteMy: My adviser was S-S. Chern, after whom Chern classes are named. He only started calling them that after about 25-30 years.

Astyx

10:29 PM

In french we have "$\sum u_n$ est grossièrement divergente lorsque $u_n$ ne converge pas vers 0"

Alessandro Codenotti

So the operation is pointwise addition?

Akiva Weinberger

What's the subspace?

Ted Shifrin

Oh, I see, @Astyx. Merci bien. We have no such term in English.

Meow Mix

@AkivaWeinberger Vectors with finitely many nonzero components

Ted Shifrin

No, @Alessandro.

Astyx

10:30 PM

Right, thanks

Akiva Weinberger

Ah.

Ted Shifrin

Think of the 4 faces of a tetrahedron. Their union is a $2$-chain.

Akiva Weinberger

What would the orthogonal space even be there?

Ted Shifrin

Did you see what the giant vector space was, DogAteMy?

Akiva Weinberger

It's not the zero space, is it?

Astyx

10:30 PM

It's polynomials, really

Akiva Weinberger

@TedShifrin Things with finite magnitude, yeah?

Meow Mix

Mhm.

Ted Shifrin

$\ell^2$

Not Euclidean magnitude!

Well, OK, limiting Euclidean magnitude.

Akiva Weinberger

9 mins ago, by Ted Shifrin

Our universe is the set of infinite sequences $a=\{a_n\}$ with $\|a\|^2 = \sum a_n^2$ finite by definition.

@TedShifrin Ah, OK

Ted Shifrin

OK, OK. :P

Akiva Weinberger

10:32 PM

(Fault in our Stars?)

Ted Shifrin

So the underlying inner product, as Zach said earlier, is $\langle a,b\rangle = \sum a_nb_n$.

Akiva Weinberger

Makes sense

Alessandro Codenotti

Hm, wait, a singular $q$-simplex is a continuous function $\sigma:\Delta_q\to X$ where $\Delta_q$ is the standard $q$-simplex, right? I still don't understand what the group operation is supposed to be, what is meant with a linear combination of simplices

Akiva Weinberger

So things orthogonal to $(1,0,0,\dots)$ would need to have no first coordinate?

And similarly for all the others

Meow Mix

Yeah

Akiva Weinberger

10:33 PM

so the orthogonal space is the zero space??

Ted Shifrin

You literally take the union of the images, with integer multiplicities, @Alessandro.

Like a sum of paths ...

You don't add pointwise.

Akiva Weinberger

@AlessandroCodenotti Formal linear combinations (you make a free abelian group out of them)

Ted Shifrin

You're in a general topological space, so there is no addition :P

DogAteMy, he's trying to understand that.

Akiva Weinberger

In any case, the sum of two simplices isn't another simplex

Ted Shifrin

You actually take unions and then allow integer multiplicities.

Meow Mix

10:34 PM

OH

Yeah @Akiva you spoiled it! /s

Ted Shifrin

DogAteMy likes to spoil it.

Akiva Weinberger

@TedShifrin I saw the f.a.g. (sorry!) defined more generally

Ted Shifrin

He does that frequently.

Unfortunate abbreviation!

Akiva Weinberger

@MeowMix Oh, sorry, I didn't realize you didn't solve it yet!

Meow Mix

@AkivaWeinberger I'm only joking with you :]

Ted Shifrin

10:35 PM

DogAteMy often jumps in and spoils it — he doesn't mean to show off, but ... well, he does. :P

Alessandro Codenotti

Aha, it's formal combinations! So the group of 2-simplices in, say $\Bbb R^3$ is made of triangles and unions of triangles?

Akiva Weinberger

Sorry!

Ted Shifrin

Like the examples I gave you, which you ignored, @Alessandro.

Images of triangles.

Akiva Weinberger

But different chains can have the same image…?

Things can overlap and stuff

Ted Shifrin

Parametrized images.

Akiva Weinberger

10:36 PM

OK. I think.

Ted Shifrin

I'm emphasizing that they are not necessarily geometric triangles in the space (which may not even be an affine space).

Alessandro Codenotti

@TedShifrin right

Meow Mix

anyways, so that means $V^{\perp\perp} = \text{our universe!}$

Ted Shifrin

Right, Zach.

Akiva Weinberger

I had it defined as the set of functions* from the integers to the singular simplices

Ted Shifrin

10:37 PM

Question for you: Do you think $V$ is closed?

Akiva Weinberger

(which in fact can be used to define the free abelian group on any set)

*such that only finitely many terms are nonzero

Alessandro Codenotti

@TedShifrin so even a straight segment can be a 2-simplex in $\Bbb R^3$ if its the continuous image of a triangle (projection on a side or something like that)

Ted Shifrin

Right, @Alessandro.

There are $10$-chains in a $2$-sphere. They're just not very interesting.

Akiva Weinberger

@AlessandroCodenotti Did you learn about boundary maps yet?

Sha Vuklia

How can I show that the directional derivate is a linear combination of the partial derivates? Say we consider $\mathbb R^2$, then we have:
$$
(D_{(u,v)}f)(x,y)=u(D_1f)(x,y)+v(D_2f)(x,y)
$$
I know that the following holds: $(D_{\alpha u}f)(a)=\alpha(D_uf)(a)$. I feel like I should be using that. But how can I split the directional derivative in a sum?

Ted Shifrin

10:38 PM

You need to know $f$ is differentiable for that to be true, @Sha, so you need to use the definition of differentiable.

Akiva Weinberger

@AlessandroCodenotti I don't actually know where you are. Do you already know other types of homology, or is this the first one you're learning about

Ted Shifrin

I think he's just starting, DogAteMy.

Alessandro Codenotti

@AkivaWeinberger They where introduced with chain complexes, I still have to find out what singular simplices have to do with this

Akiva Weinberger

Arright

Alessandro Codenotti

@TedShifrin I am

Ted Shifrin

10:39 PM

So you are starting with singular chains, not simplicial chains?

(For the latter you require that the mappings be homeomorphisms to their images.)

Sha Vuklia

@Ted So if $f$ is differentiable, then all partial derivates exist. Should I write out the definition on either side of the equation?

Ted Shifrin

@Sha: It's much stronger than that.

Akiva Weinberger

(Simplicial chains are where you triangulate your space and call the triangles "simplices", IIRC)

Sha Vuklia

@Ted yea ok, but I haven't had that yet

Akiva Weinberger

(well, $n$-dimensional triangles)

Ted Shifrin

10:41 PM

BTW, at some point, my lectures on YouTube might help you with what you're doing.

Sha Vuklia

I'm only introduced to directional and partial derivates for the moment

Alessandro Codenotti

@TedShifrin I think so, there's only singular homology in my professor's notes

Sha Vuklia

oh really @Ted

which one?

Ted Shifrin

That formula doesn't hold if just the partial derivatives exist, @Sha.

Alessandro Codenotti

@TedShifrin is that message for me or Sha?

Ted Shifrin

10:41 PM

That formula for the directional derivative will fail in general, @Sha.

for @Sha, @Alessandro. Sorry.

No lectures on homology. :)

Sha Vuklia

@Ted So it's better to leave the proof for now, and first read on about differentiability?

Ted Shifrin

Yup. Not better. You must :)

Sha Vuklia

haha alright

Ted Shifrin

My lectures are linked in my profile, btw.

Alessandro Codenotti

Ah, I would have liked to see them

Ted Shifrin

10:43 PM

You knew about that, @Alessandro, I'm pretty sure.

Alessandro Codenotti

I know you have multivariables calculus videos, not sure about the rest

Ted Shifrin

@Meow: Did you ever tell me if $V$ is closed in the big space?

That's what we're talking about, @Alessandro.

Sha Vuklia

oh cool! you're treating the things I have to know for my test. I will watch those tomorrow!

I'm off to bed now, see ya!

Akiva Weinberger

The nice thing about singular simplices, chains, and boundaries is that if the image of an $n$-dimensional chain $a$ lies in the union of two open sets $U\cup V$, and if each simplex in the boundary of $a$ ($\partial a$) is either in $U$ or $V$, then there exist $n$-chains $u$ and $v$ and an $n+1$-chain $x$ such that the image of $u$ is in $U$, the image of $v$ is in $V$, and $a=u+v+\partial x$

Ted Shifrin

Night, @Sha. :)

Astyx

10:44 PM

Bye @ShaVuklia

Alessandro Codenotti

@TedShifrin good point :P

Akiva Weinberger

Oh, god, that looks much more convoluted than it was in my head

Ted Shifrin

Convolution is a math word, DogAteMy. Don't confuzle us.

Akiva Weinberger

It looks more confuzzling than it was in my head

Meow Mix

@Ted Closed as in, complement in our universe is open?

Daminark

10:45 PM

See you @Sha!

Ted Shifrin

That'll do, Zach, yes.

Daminark

And lol @Ted I made that pun a good deal when we were talking about it

Akiva Weinberger

What's the complement of the universe even mean?

The complement of it in what?

Meow Mix

I said "complement in"

Ted Shifrin

No, no, I asked whether our subspace was closed in the big space.

Akiva Weinberger

10:45 PM

Ohh, sorry, I misunderstood you

Derp

Astyx

What are we doing to that poor subspace ?

Ted Shifrin

Asking if it's closed.

Meow Mix

Wow, thanks Windows 10. For crashing the taskbar

Akiva Weinberger

Right, we have a metric defined by $\|x-y\|$, so we have distances, so we can talk about closedness

Ted Shifrin

Yup, normed vector spaces have a topology :P

Astyx

10:47 PM

Is that for a greater cause ? Or just for the fun of it ?

Ted Shifrin

I'm leading Zach to make a conjecture, with luck.

Meow Mix

What do we call a vector space with norm and inner product?

Normed inner product space?

Akiva Weinberger

Just inner product space, I think

Alessandro Codenotti

Inner product space

Akiva Weinberger

because if you have an inner product you have a norm

Ted Shifrin

10:47 PM

The norm comes from the inner product, so you don't need that repetitive redundancy.

Daminark

Repetitive redundancy... I love it

Akiva Weinberger

(defined by $\sqrt{\langle x,x\rangle}$)

Astyx

Prehilbertian ?

Eric

hi chat

Astyx

Oh, not in english, shame :(

Ted Shifrin

10:48 PM

I never say that.

Akiva Weinberger

@Astyx Isn't that the one before the Jurassic

Ted Shifrin

Hi @Eric.

Alessandro Codenotti

We started talking about Hilbert spaces in the analysis class recently, to get to Fourier series, interesting stuff

Daminark

Hey @Eric!

Astyx

Hi @Eric

Eric

10:48 PM

@Astyx, I have never used this word but I have actually heard someone say prehilbert space out loud

Ted Shifrin

Zach gets side-tracked sooooo easily ...

Akiva Weinberger

Is he still here

Meow Mix

YHes

Ted Shifrin

Well, "preHilbert" is easier than "not necessarily complete."

Meow Mix

Someone invited me to a word game

Astyx

10:49 PM

In french it is standard to call it "espace préhilbertien"

Ted Shifrin

Zach, I'm leaving in a moment, so concentrate.

Meow Mix

Okay, I'll quit it for now.

Akiva Weinberger

@MeowMix Word game $\to$ word dame $\to$ ward dame $\to$ ware dime $\to$ …?

Astyx

That's one way of concentrating

war time

Meow Mix

Ok, so right now we are dealing with the set of infinite sequences whose series are finite but have infinitely many non-zero components

Akiva Weinberger

10:51 PM

Ooh, fun fact, "hymn" is the only four-letter word that can't be made into another word by changing one of its letters.

Ted Shifrin

OK, you mean the complement of $V$, Zach?

Meow Mix

Oh wait,

I could just prove it contains all its limit points.

Ted Shifrin

OK ...

(My definition in Chapter 2 is that it is closed under limits of convergent sequences, yes.)

Meow Mix

Suppose there existed a point in the complement that was a limit point

Akiva Weinberger

I think I have to leave now

Ted Shifrin

10:54 PM

LOL, DogAteMy.

Bye, have fun!

Bye :)

Bye

hi @Ted

Bye @Akiva

10:55 PM

Hi @Danu. ...

Astyx

I'll probably go too

Danu

I'm reading Milnor's Morse Theory

Astyx

Bye chat, good days to you

Ted Shifrin

I'm leaving in a moment, actually. I have a torte to put chocolate on.

Cool, @Danu.

Danu

Give me one second to ask a question

Alessandro Codenotti

10:55 PM

I'm going as well, thanks for your help @Ted @Akiva

Ted Shifrin

Night, @Alessandro.

Alessandro Codenotti

good night/day everyone

Danu

Does $S^2=\Bbb CP^1$ embed into $\Bbb C^n$ as an (affine) algebraic variety?

Daminark

Lol everyone's leaving... I guess I should get back to Rotman soon anyway, see you around guys!

Ted Shifrin

NOO ...

You know that.

See ya, evil Demonark.

Meow Mix

10:56 PM

Hmm, how should I go about doing this in a reasonable time

Danu

@TedShifrin It doesn't? Hmkay

Meow Mix

I don't wanna piss @Ted off.

:]

Ted Shifrin

Good idea, Zach :P

@Danu: The only compact, connected complex submanifolds are points.

Danu

Hahaha

Ted Shifrin

You've done this.

Danu

10:57 PM

derp

thanks

complex manifolds so crazy

Ted Shifrin

Holomorphic functions so rigid :P

Danu

so what the heck do I care about affine algebraic varieties that embed into $\Bbb C^n$, actually?

Which ones are there even?

Ted Shifrin

Those are called Stein manifolds. That's what @PVAL spends his life thinking about.

Any algebraic curve in $\Bbb C^2$?

Generalizations thereof ... ?

Danu

Ah, okay!

Ted Shifrin

Non-compact things are important, too, you know!

Danu

10:59 PM

I did want to know what Stein manifolds are supposed to be

@TedShifrin I haven't thought about those in a long, long time...

Ted Shifrin

That's what they are. Things that embed as closed submanifolds of $\Bbb C^n$.

@Danu: Soon I'm gonna start thinking more about specifics about my trip, so I may be in touch ...

Meow Mix

@Ted In order for it to be a limit point

Danu

@TedShifrin A'ight!

Meow Mix

(Say the limit point is $\vec{L}$) We would need for any $\epsilon > 0$, there exists $\vec{v}$ such that $|\vec{L}-\vec{v}| < \epsilon$

Ted Shifrin

Sure thing, Zach.

Meow Mix

11:01 PM

But then, if $\vec{v}$ is finite

Ted Shifrin

There's an obvious way to create $\vec v$.

Meow Mix

this can't be possible, because $\vec{L}$ by definition has infinitely many non-zero elements

Wait, that doesn't sound right.

Yeah, it isn't

Uhh, it doesn't look like it contains all it's limit points...

But I'll look a little further

Ted Shifrin

OK. I'm not giving anything away. But I will pester you about this. (Balarka did something similar where he promised to figure something out and I kept pestering until he did.)

Meow Mix

I mean, look at $(1,1/2,1/4,1/8, \dots)$

Ted Shifrin

Yeah?

What do I do with it?

Meow Mix

11:07 PM

For any $\epsilon$, can't we just find some $n$ such that $\sum 1/i^2$ to $n$ is less than $\epsilon$, then just take the first $n$ epsilon of our sequence?

Wouldn't that satisfy the definition of closed?

Ted Shifrin

Better to use a different letter, but you are thinking the right thing.

So what do you now think?

Meow Mix

I think it's open...

Ted Shifrin

Oh, you mean from $n$ to $\infty$ is less than $\epsilon$?

Why open?

Remember that open is different from not closed.

Meow Mix

No, I mean, well for example

OH, not closed I mean

Ted Shifrin

So do you have a conjecture on what the limit points are?

Meow Mix

11:09 PM

Uhh, it seems like pretty much all the points in the complement of $V$ in our universe are limit points, but it sounds weird.

Ted Shifrin

So the word for this is: $V$ is dense.

And the smallest closed set containing it is the whole space.

Meow Mix

OH that's what dense means!

Ted Shifrin

So something for you to ponder over the weekend: Is this related to $V^{\perp\perp}$ being the whole space?

Meow Mix

Alright.

Ted Shifrin

OK ... happy dinner and games!

Meow Mix

11:12 PM

And to you as well. :]

nbro

@TedShifrin It turns out that $||u \times v||$ is equal to the length of the projection of $u$ onto the plane to which $v$ is the normal whenever $||v||=1$

If you want to see more details: math.stackexchange.com/questions/2201735/…

indeed my idea was correct

I simply didn't have a good knowledge to manipulate trigonometric identities