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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.
 
how do you achieve a balance in proof writing?
 
@BalarkaSen So tell me precisely how to do this for such a space.
Indeed, specialize, and tell me how to do it for $\Bbb R^2 \setminus \{0\}$.
 
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!
 
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)
 
so i need to summarize the proof first, and then put in the details if they want to read?
 
9:20 PM
Surely you've read papers you thought were well-written. Imitate those.
 
usually I dont read the whole paper, I just read the parts im interested in. I will try to think of my favorite...
 
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)
 
I sort of know how to organize the sections, I'm talking more about the proof of a specific proposition
 
@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.
 
@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.
 
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.
 
Do you think your above approach will work for this?
 
I am unsure about continuity. Please give me a few minutes.
 
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.)
 
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
 
^ That just popped up on the left banner of the Stack Overflow page I was on. The incorrect grammar made me lol
right* banner
 
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 :)
 
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
 
@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.
 
This is a very interesting question!
 
Try to reformulate the question in a way that does not involve forms.
 
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.
 
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
 
yikes, I forgot \mathfrak P looks that terrible
 
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).
 
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.
 
Sure. And what do you want to demand about this collection?
 
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.
 
The topology is what I would call the compact-open topology, or the sup norm (if $X$ has a metric), or whatever you like.
 
Did I just answer what I would demand about this collection, or is there anything more to say?
 
It still involved forms and integration.
 
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.
 
I don't think you're so far from it, though
 
10:05 PM
Can someone explain this small proof to me? I don't get it :(
 
@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
 
@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
 
$\sin x /x$ is not defined at 0, right?
 
Max
is it a curve is finite 1D lebesgue measure?
 
10:08 PM
 
Max
with finite*
 
I wrote it by myself though, so I might have missed some detail
But it is 99.9999999% right if I copied correctly
 
@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?
 
you're a cool dude
@Max Context?
 
Sorry, I had to move away from my keyboard for a bit.
 
10:13 PM
Like I really don't understand the proof. Geometry is so hard...
 
@Juan Interesting analogies indeed.
 
Max
@Mike Miller I am trying to proves some lemmar for covering spaces
 
@JuanSebastianLozano I especially liked this message. It explains my thought process very clearly.
 
@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
 
10:21 PM
@MikeMiller Just to clarify: I have to reformulate your question without saying integration/forms?
 
@BalarkaSen You're not going to get anywhere without doing so.
 
Alright, hmm. I suppose I need to start by looking at the tangent vectors of the paths.
 
Sometimes we can teleport and get somewhere without walking.
 
Please say again the thing you settled on before. You're going further from what you need to do.
 
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.
 
10:27 PM
Sorry, so what should this end up being a continuous map on?
What's the domain?
 
We want $\int_\bullet \omega : \mathcal{P}_a \to \Bbb R$ to be continuous.
 
That's not what you want, no. That's already true. The restriction of a continuous map to a subset is continuous.
 
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
 
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.
 
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.
 
10:34 PM
This is not a helpful reformulation. Abandon it.
 
@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.
 
You were on the right approach. Say the thing you said before, but never saying anything about your bizarrely topologized path space.
 
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.
 
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
 
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.
 
10:57 PM
Go on?
 
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.
 
@JuanSebastianLozano i will read your comments I missed some of them
 
@ForeverMozart No problem, let me know if I was too opaque.
 
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.
 
So, Mike, what's your favorite non-famous open problem?
 
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.
 
Alright, just say it I suppose.
 
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."
 
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.
 
11:22 PM
No I'm saying that the question is literally "When is there a continuous section of the path-space fibration?"
 
Ah, I see.
Right.
 
Do you see any obstructions to that being true?
@AkivaWeinberger Where did you come across the question, anyway?
 
Yes, I suppose I do. $X$ needs to be simply connected for one, right?
 
Why?
 
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 $\pi_1$?
 
11:30 PM
Generalize this obstruction.
 
In fact does $X$ not be contractible also?
 
Da.
 
at least weak htpy equivalent to a point.
OK, that was rather simple. Meh :(
 
We're working with manifolds, so you're fine.
Simple but cute, if one does not spend a half-day on it. :)
 
Which I just did.
 
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.
 
Sounds like a nice result.
 
More or less straightforward. Just wasn't going to ask you to do it.
 
You wouldn't want another half day (or more) wasted.
 
Err, actually because it's a bit less straightforward than the previous, but yes.
 
Nice problem nonetheless. I doubt I could have thought up the section of the path space fibration by myself though - I have forgotten topology.
 
11:37 PM
i'm skeptical, since that's exactly what such a collection of paths is :p
 
(I blame a year of multivariable calculus)
True.
 
shouldn't you blame yourself for a year of multivariable calculus?
 
@JuanSebastianLozano I always like that line from Kierkegaard. It's good one.
 
@MikeMiller well I need to put it in a passive way
one cannot just blame oneself. it has to be done indirectly.
 
I can blame you actively
 
11:41 PM
oh, sure. I was merely talking about self-blaming.
shouldn't you get work done? don't waste the other half of that day that's remaining ;)
I should probably be glad to waste your time, but I am too bitter right now.
 
yeah... I should...
 

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