Mathematics - 2016-05-03 (page 3 of 3)

Balarka Sen

9:05 PM

@MikeMiller I suppose one could look at spaces which admit open cover by star connected sets.

Can one make it more general than that?

star *convex.

Forever Mozart

how do you achieve a balance in proof writing?

Mike Miller

@BalarkaSen So tell me precisely how to do this for such a space.

Indeed, specialize, and tell me how to do it for $\Bbb R^2 \setminus \{0\}$.

Forever Mozart

too many details and I am afraid the reader will lose track of the 3-4 main ideas

but too few details and they might not consider it a real proof

HELP!

Juan Sebastian Lozano

Inverse pyramid: pretend that your reader will stop half way through the proof, you want them to have a sketch but not be overwhelmed. You should have all the things so that one might be able to see the outline of the ideas, and then circle back and fill in more details, and continue doing this (often it is helpful to say "this lemma will be proven later" or something, so that you can circle back to it) until you have a whole proof.

(and you might say things like "sufficiently nice" if its not actually integral to know at that point in the paper)

Forever Mozart

so i need to summarize the proof first, and then put in the details if they want to read?

Mike Miller

9:20 PM

Surely you've read papers you thought were well-written. Imitate those.

Forever Mozart

usually I dont read the whole paper, I just read the parts im interested in. I will try to think of my favorite...

Juan Sebastian Lozano

Yeah, but, each section should read like a well structured argument, I'm just talking about the relationship between sections.

(well structured with sufficient ambiguity)

Forever Mozart

I sort of know how to organize the sections, I'm talking more about the proof of a specific proposition

Balarka Sen

@MikeMiller This is getting messy quickly. I am trying to piece R^2 up into three pieces of pie (all of them are star-convex with centre the roots of unity) overlapping each other, and joining point $x$ with e.g. the easiest root of unity $(0, 1)$ by throwing ray from $(0, 1)$ depending upon which triant $x$ is in and throwing ray from the centre of that triant joining $x$, and marking the points these two rays hit (it's going to be unique).

And then I have a path between $(0, 1)$ and $x$ by going through the first ray and the second ray.

But the mess makes me doubt if this is what you want.

Mike Miller

@BalarkaSen Since you were apparently upset about a lack of precision. Given a path from $a$ to $x$, you can define $\int_\gamma \omega$. Given, for every $x \in U$, a path $\gamma$ from $a$ to $x$, we get an operator $\int$ that sends $\omega$ to a function $f$. What I want to know is for what $U$ you can choose a collection $\gamma$ so that this operator always spits out a continuous function.

Balarka Sen

9:25 PM

I see, that's much more clear. But I am sorry if I annoyed you with pedantic precisionism, I was just not clear what you wanted.

That's a very interesting question.

Mike Miller

Do you think your above approach will work for this?

Balarka Sen

I am unsure about continuity. Please give me a few minutes.

Juan Sebastian Lozano

Oh, in which case, (in progressing order of precision) you should try to explain the relationships of the statement then move on to explaining the main insights of the proof, then delve into the logical structure of the proof (i.e. the proof proper), separating out important lemmas to either be treated before (if their importance is clear) or after (if their importance is not clear).

In essence, it's designed as sort of a recipe for creating an "image" in your head, where at first its blurry and it gets progressively sharper and better until you can see the whole thing perfectly. However, you want the reader to be able to see the simple outlines of the image with as little work as possible.

(This is based off of my observations of the most helpful things I have read, by the way. I haven't ever actually published anything.)

Forever Mozart

hold on a second let me process that

yes the "image" is very important. I feel like it is a simple image but when I encode it into a proof it might be difficult to see.

or maybe it's just simple to me because I discovered it after months of work

8protons

^ That just popped up on the left banner of the Stack Overflow page I was on. The incorrect grammar made me lol

right* banner

Mike Miller

9:36 PM

maybe, but it's a fair question i bet my students would have appreciated i was more clear about when teaching calculus :)

8protons

It's definitely a great question! Lots of up votes and attention; so much so that it was advertised in my Stack Overflow banner. But "Why do a" was simply humorous :P

Balarka Sen

@MikeMiller No, my approach has no reason to work. E.g., consider $\Bbb R^2$ covered by thickened upper and lower half plane and $(0, 1)$ and $(0, -1)$ be the centre of star convexity. I approach $(0, -1)$ by a sequence moving along $y = x^2 + 1$ towards $(0, -1)$. Fix a ray $\ell$ emanating from $(0, 1)$ to the "left". Then the rays joining $(0, -1)$ with the points in the sequence readily becomes more and more horizontal.

Picking an appropriate form $\omega$ will completely mess continuity up, I think, by exploiting the fact that the length of the path is getting longer and longer.

Mike Miller

:)

Balarka Sen

This is a very interesting question!

Mike Miller

Try to reformulate the question in a way that does not involve forms.

Balarka Sen

9:48 PM

I'll try. I'll have to ponder on this seriously.

On a related note: I think there should be a reformulation as follows: we want to choose a collection of paths $\mathfrak{P}_a$ starting at some given point $a$ so that $\int$ eats a form $\omega$, a path $\gamma$ starting at $a$, and spits out a scalar $\int_\gamma \omega$ in a way so that $\int_{\bullet} \omega : \mathfrak{P}_a \to \Bbb R$ is continuous with $\mathfrak{P}_a$ some reasonable topology given in which paths are close if so are their endpoints.

Juan Sebastian Lozano

Example of what I mean:
Theorem: $\sin^2(\theta) + \cos^2(\theta) = 1$
Proof: [before this, your reader should know what trig functions are, and why they relate to triangles]The relationship between certain transcendental functions ($\sin$ and $\cos$) is quadratic, specifically it resembles a pythagorean identity. [here you could stop reading and move on and use the theorem on faith] More specifically, if we think of these functions as representing the sides of a right triangle with hypotenuse $1$, then this identity is a consequence of the pythagorean theorem. We will show that this is pre…

Balarka Sen

yikes, I forgot \mathfrak P looks that terrible

Mike Miller

Ah, but the function $\int_\bullet \omega: \mathcal P_a X \to \Bbb R$ is continuous (with respect to an appropriate topology on the path space).

Balarka Sen

The topology consists of open sets consisting of paths whose endpoints belong in a ball in $X$, right?

Or well the topology generated by those basic open sets.

Yes, there is more to it than that. I can just pick the whole path space to be my collection.

We want unique path through $a$ and any $x$. One such path for each pair $(a, x)$ in my collection.

Mike Miller

Sure. And what do you want to demand about this collection?

Juan Sebastian Lozano

9:52 PM

You'll notice that this form of proof writing is cumbersome for small proofs that follow from definitions, and that is because it privileges relationships between objects, but thinks of the objects themselves as a lot less important.

Mike Miller

The topology is what I would call the compact-open topology, or the sup norm (if $X$ has a metric), or whatever you like.

Balarka Sen

Did I just answer what I would demand about this collection, or is there anything more to say?

Mike Miller

It still involved forms and integration.

Balarka Sen

Right, I haven't thought about a forms-free way yet. Give me more time, I am excited to think about it more :)

@MikeMiller Ah, you're right.

Mike Miller

I don't think you're so far from it, though

notorious

10:05 PM

Can someone explain this small proof to me? I don't get it :(

Juan Sebastian Lozano

@ForeverMozart I think kind of a central tension in how we write proofs comes from the fact that (ad Nabokov said) "there is no reader, only the re-reader". This means that one can only 'understand' or grasp the meaning of a proof

Mambo

@notorious What are those axioms?

Max

Hi all

i stumbled accross a phrase "finite curve" and I am wondering

what exactly is a formal definition of a "finite curve

ramsay

$\sin x /x$ is not defined at 0, right?

Max

is it a curve is finite 1D lebesgue measure?

notorious

10:08 PM

I think these i.imgur.com/M1pvgnK.png

Max

with finite*

notorious

I wrote it by myself though, so I might have missed some detail

But it is 99.9999999% right if I copied correctly

Juan Sebastian Lozano

@ForeverMozart once we have read it and reflected on it ('re-read' it in our mind). It's similar to Kierkegaard's notion that life must be understood backwards but lived forwards. A proof is like experiencing the mathematics, but only once one has experienced it can you understand it. However, in life, as in math, we experience in moments, not in decades, so a proof must necessarily be disjoint-

it must necessarily be broken up into lemmas and so on, otherwise it would be like experiencing a decade of life without stopping and the all of a sudden stopping and trying to reflect on the whole of the experience.

Does that make any sense?

Mike Miller

you're a cool dude

@Max Context?

Balarka Sen

Sorry, I had to move away from my keyboard for a bit.

notorious

10:13 PM

Like I really don't understand the proof. Geometry is so hard...

Balarka Sen

@Juan Interesting analogies indeed.

Max

@Mike Miller I am trying to proves some lemmar for covering spaces

Balarka Sen

@JuanSebastianLozano I especially liked this message. It explains my thought process very clearly.

Mike Miller

@Max That doesn't help me see what a finite curve is, sorry.

Max

Oh, I am sorry, i mistranslated a part of the textbook

it should be finite line segment, not finite curve

Balarka Sen

10:21 PM

@MikeMiller Just to clarify: I have to reformulate your question without saying integration/forms?

Mike Miller

@BalarkaSen You're not going to get anywhere without doing so.

Balarka Sen

Alright, hmm. I suppose I need to start by looking at the tangent vectors of the paths.

user174558

Sometimes we can teleport and get somewhere without walking.

Mike Miller

Please say again the thing you settled on before. You're going further from what you need to do.

Balarka Sen

I reformulated it as follows: Given a domain $X$, I need to choose a collection $\mathcal{P}_a$ of paths starting at some $a \in X$ consisting of a unique path for each pair of points $(a, x)$ with those endpoints, so that the operator $\int$ which eats (path, form) and spits scalar $\int_\gamma \omega$ is continuous on 1st variable, $\mathcal{P}_a$ given compact-open topology.

Mike Miller

10:27 PM

Sorry, so what should this end up being a continuous map on?

What's the domain?

Balarka Sen

We want $\int_\bullet \omega : \mathcal{P}_a \to \Bbb R$ to be continuous.

Mike Miller

That's not what you want, no. That's already true. The restriction of a continuous map to a subset is continuous.

Balarka Sen

Sorry, I think I am a bit confused. I defined the topology on $\mathcal{P}_a$ by declaring the basic open sets to be collection of paths with endpoints lying inside a ball in $X$. Is that the same as restriction of the compact-open topology on $\mathcal{P}_a X$? Then why isn't everything trivially true?

I am confusing myself :S

Mike Miller

That is obviously not a good topology.

Not Hausdorff.

You think the path that goes from $(0,0)$ to $(n,0)$ then back to $(0,0)$ should be arbitrarily close to the constant path?

There is no reason to be thinking about that particular topology... I don't know why it came up.

Balarka Sen

There is no loop around $a$ if the constant path is already in $\mathcal{P}_a$, as we demanded that there is a unique path with endpoints $(a, x)$ for every pair.

But in general it is not nice, yes.

Mike Miller

10:34 PM

This is not a helpful reformulation. Abandon it.

Balarka Sen

@MikeMiller Hmm. We want $f(x) = \int_a^x \omega$ to be a continuous function of $x$. I guess I mistranslated that into a bad topology.

Horrible typo, sorry.

Mike Miller

You were on the right approach. Say the thing you said before, but never saying anything about your bizarrely topologized path space.

Balarka Sen

I want to say "choose some collection of paths $\mathcal{P}_a$ with a unique path for each pair of endpoints so that $\int_{\bullet} \omega : \mathcal{P}_a \to \Bbb R$ is continuous" but I suppose this is not helpful as my topology on $\mathcal{P}_a$ was not good? am I supposed to think what's the right topology should be?

Maybe I;ll think for a minute.

Mike Miller

I honestly have no idea why you're trying to topologize it. Use the same topology as on the whole path space.

Like you're Dale Earnhardt in the NASCAR race of your life and you're about to cross the finish line but suddenly you turn right

Balarka Sen

I am not sure if I am to laugh at that or cry at that! But give me a few minutes, I am a slow thinker.

I see, I was being silly.

Mike Miller

10:57 PM

Go on?

Balarka Sen

Hold on, I see why I was silly, but I still need a forms free way of saying this. It's coming to me, I'll have to think a bit harder.

Forever Mozart

@JuanSebastianLozano i will read your comments I missed some of them

Juan Sebastian Lozano

@ForeverMozart No problem, let me know if I was too opaque.

Balarka Sen

11:09 PM

We don't want $\phi := \int_\bullet \omega : \mathcal{P}_a \to \Bbb R$ to be continuous: that's obvious from giving it the compact-open topology (which really is the most natural topology here). We want $f := \int_a^x \omega : X \to \Bbb R$ to be a continuous function, i.e., $f^{-1}(-a, a)$ to be open.

Now we know that $\phi^{-1}(-a, a)$ is open in compact open topology, and I think there should be a correlation between them (look at endpoints of paths in an open set in compact open topology?), but I can't pin it down. I'll probably need more time.

Akiva Weinberger

So, Mike, what's your favorite non-famous open problem?

Mike Miller

You're working too hard. This should have not taken you more than a couple minutes, @BalarkaSen. Should I just say it?

@AkivaWeinberger I'm pretty sure I posted an answer to that where it was written.

Balarka Sen

Alright, just say it I suppose.

Mike Miller

What is a collection of paths $\mathcal P_a$ other than a section $X \to \mathcal P_a$ of the path-space fibration?

Sorry, I should say "what is a collection of paths $\{\gamma_x\}$ other than a section $X \to \mathcal P_aX$ of the path-space fibration? You already know that the map $\int_\gamma \omega: \mathcal P_aX \to \Bbb R$ is continuous; so what you want is for the section to be continuous."

Balarka Sen

Are you asking me of a different way to interpret collection of paths or is that nudge towards looking at it as section of the path space fibration?

I suppose the former in which case I can't think of one immediately, if I know one.

Mike Miller

11:22 PM

No I'm saying that the question is literally "When is there a continuous section of the path-space fibration?"

Balarka Sen

Ah, I see.

Right.

Mike Miller

Do you see any obstructions to that being true?

@AkivaWeinberger Where did you come across the question, anyway?

Balarka Sen

Yes, I suppose I do. $X$ needs to be simply connected for one, right?

Mike Miller

Why?

Balarka Sen

The path space is contractible.

And if there is a section $X \to P_aX \to X$ is identity. When can identity be 0 on $\pi_1$?

Mike Miller

11:30 PM

Generalize this obstruction.

Balarka Sen

In fact does $X$ not be contractible also?

Mike Miller

Da.

Balarka Sen

at least weak htpy equivalent to a point.

OK, that was rather simple. Meh :(

Mike Miller

We're working with manifolds, so you're fine.

Simple but cute, if one does not spend a half-day on it. :)

Balarka Sen

Which I just did.

Mike Miller

11:31 PM

I don't expect you to prove the converse, but the converse is true. If your domain is contractible, you can choose a set of paths so that the output $f(x) = \int_\gamma \omega$ is continuous (smooth, even).

So while this is not possible on a punctured $\Bbb R^3$, it is possible on the Whitehead manifold.

Balarka Sen

Sounds like a nice result.

Mike Miller

More or less straightforward. Just wasn't going to ask you to do it.

Balarka Sen

You wouldn't want another half day (or more) wasted.

Mike Miller

Err, actually because it's a bit less straightforward than the previous, but yes.

Balarka Sen

Nice problem nonetheless. I doubt I could have thought up the section of the path space fibration by myself though - I have forgotten topology.

Mike Miller