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3:16 AM
@Mostafa Thanks!
@ACuriousMind Do you know of any work connecting those two ways of framing quantum mechanics in symplectic geometry: the Gromov-Witten invariants approach from string theory and the deformation of Hopf algebras approach from integrability?
@ACuriousMind What I mean by that is that both are somehow an embedding of quantum mechanics in the symplectic framework and they seem to be dual to each other, one being a covariant approach and the other, contravariant. I am curious if there are existing results on such a relation.
 
what on Earth is the sobolev space $H^{1,2}_c$ on a compact manifold
 
3:31 AM
@0celouvsky How do I prove that something is a manifold?
 
Can we please execute authors who do not have a notation index
@BernardoMeurer in general it's pretty hard
most of what you're gonna have will be graphs and preimages
 
Show that the set $M = \{(x,y,z)\in\Bbb R^3 \colon x^2+y^2+z^2 = 1; x=y; z>0\}$ is a manifold and determine it's dimension
@0celouvsky That
 
lol
what
What even is that
It's the intersection of a sphere and a plane
with some other thing
idk man
I'm tired
 
Lol
yeah
Idk this one is fucked up
@G.Bergeron Solve that
 
3:48 AM
@BernardoMeurer well what do you know about manifolds
can you even tell me what shape that is
I'm trying to do other things right now
it's not going to well
 
@0celouvsky I'm not sure I've learned manifolds
It's nowhere in my notes
but it's on the mock test
this is pretty old though
 
I'm trying to track down a sobolev ineq rn, just tell me what set that is and I'll have a better idea
I think it's an open semicircle
so just an interval
there's your proof :P
 
idk what machinery you have
I'm sure there's some perfect way to see it...but your best bet is to describe the set and its embedding
 
I don't even know what an embedding is
This won't be on the exam
so I guess it's fine
 
4:00 AM
hopefully not
that's a hard question
hard because actually writing down the details is hard
it's easy to "see"
@BernardoMeurer proof ^
 
You need a notebook with thicker paper dawg
 
that's not my notebook
it's scratch paper
 
You have ruled scratch paper?
 
Yes
 
4:21 AM
aha
I think I understand analysis
 
4:51 AM
@0celouvsky What?
 
instantons
collective coordinates
ads/cft
orientations et al
life
Such goood
 
@Cows Which school do you go to?
 
I dropped out of school a few years ago (financial stuff)
I have no degree at all
I used to be a double major in physics and computational science
I applied to a few schools
Hopefully I get in somewhere
 
Oh, that's a shame, I was just asking out of curiosity :)
Oh, nice, good luck!
 
thanks
 
4:57 AM
@BernardoMeurer this app ...
 
@JohnRennie Yes, I am still awake
Perfect time to code
 
If you use ffmpeg to do the conversion then you don't need to multithread the app because ffmpeg multithreads conversions.
 
@JohnRennie I thought about ffmpeg. I was thinking of multithreading by getting a thread pool to pick from the file list
 
hey
 
The point is that you don't need to use a thread pool and it's simpler not to.
 
4:59 AM
@0celouvsky I concur.
 
Just call into the ffmpeg libary and a single conversion will use all 8 CPUs.
 
@JohnRennie And I guess it's nice because ffmpeg is a jack-of-all trades
 
Well Kodi uses ffmpeg, which it statically links.
And that works very well.
But I suspect figuring out how to compile and link ffmpeg into your app will be quite hard.
 
There's probably a CMake thingy for it
I'm in love with CMake
 
When you guys get real about what you want to code, you can hire me. I'm a professional
 
5:03 AM
It's an Open Source project
 
oh
 
Anyhow, I'd be inclined to first write the replication bit using regular file copy, and get that all debugged and working. Then all you need to do is replace copying of FLAC files by the convert and copy.
 
@JohnRennie I'm working on the replication right now
I'm suing Boost::FileSystem though
It makes my life easier
 
If you want I have C++ source for the replication bit. It's Windows code, but there isn't much Windows specific stuff in it.
 
Doesn't Boost work on Windows?
I'd love to see the code :)
 
5:05 AM
I've never needed to use Boost on Windows. I just hit the WIN32 API directly.
 
I guess you only use the Win32API deal with the directories and stuff
 
Yes
 
Since C++ can't do that natively
 
5:10 AM
@JohnRennie Nice. Taking a look at it. Apparently Boost::filesystem::last_time_write I was hoping to use to not re-sync files is only for POSIX-like systems
Oh Microsoft why couldn't you at least try to be POSIX
 
 
The single biggest advantage of Windows for the developer is that there is one, and only one, WIN32 API that is the same on all computers everywhere that run Windows.
 
@Slereah hey
 
It isn't fragmented in the way that unix is, so there has never been a need for unifying third party libraries.
 
POSIX is unified :P
WIN32API is not backwards compatible though, right?
Hmm
Not sure that portrayed my question
 
5:13 AM
The original Windows NT 3.1 had a POSIX v1 subsystem. However POSIX 1 was pretty useless so nothing ever came of it.
 
The Win32API on W7 supports WXP stuff, but not vice-versa
 
@BernardoMeurer I can take code I wrote for Windows NT 3.1 and it will still compile unchanged on Windows 10.
 
The really old UNIX tools (Heirloom) are kind of a mess
I had to compile a project based on Lex and Yacc this week and it wasn't nice
 
Obviously new APIs get added with each new version of Windows, and if you use the new APIs your code won't compile on older versions.
 
happily it was easy to convert to Bison and Flex, because POSIX bless
 
5:20 AM
@Bernardo In fact I can take binaries I compiled on NT 3.1, in 1994, and they still run on Windows 10. How many unix apps can manage the same trick?
 
@JohnRennie Hmm, that is cool, I will give you that
I'm pretty sure things compiled for the old UNIX still run on modern BSD
 
It's because Windows has always been controlled by a single company. That has it's downsides because some of Microsoft's design decisions suck. But it does have advantages. The same applies to OSX of course.
 
OSX is POSIX though
Fully POSIX in fact
which is kind of amazing
 
@BernardoMeurer The thing with equations and all is really because algebraic constraints (equations) leads to an algebraic solution set which is essentially an analytic function on some open set of the surrounding space (R^4 in your case). By the inverse function theorem of analysis in R^n, we know it is locally invertible smoothly and so a diffeomorphism from (an open subset of) the manifold (the solution set) to open sets in R^m, where m is n-(# constraints).
@BernardoMeurer This can be extended to an open cover of the solution set, and thus forms a smooth atlas, proving the object is a manifold (locally R^n objects).
 
@G.Bergeron I am 8 years old
 
5:25 AM
@BernardoMeurer (mental age :-)
 
@JohnRennie That's an overestimate :P
 
@BernardoMeurer after trying for 56 years I have now raised my mental age to 13 :-)
And ... I like having a mental age of 13 :-)
 
@BernardoMeurer In your specific case, you can just say: ''Oh z > 0 so a unique solution to the equation for z on the sphere as $z = z(x,y)$ is given... Which means I have a global chart for the manifold resulting from those two constraints.''.
@BernardoMeurer Adding x=y is just saying that you can find y as a function of x ''y(x)'' everywhere on the final solution set. Hence, again, a global chart is given by, say, the open interval on the x axis from -1 to 1. The maps are smooth and their inverse are smooth, and the global chart is thus a smooth atlas == My object (solution set) is a smooth manifold.
 
@G.Bergeron What does it mean for something to be smooth? What's an atlas?
What's a global chart
 
@BernardoMeurer Smooth: all derivatives are continuous functions on their domains
 
5:30 AM
Ah, nice
Isn't that just being a function in $C^2$?
 
@BernardoMeurer Atlas: a collection of charts that covers the whole manifold
@BernardoMeurer No, $C^\infty$
To any order...
 
Ah, I see
So smooth is a very restrictive requirement
 
@BernardoMeurer Chart: An open subset U of the manifold M paired with an open subset V of $R^n$ together with a smooth map between the two such that its inverse is also smooth.
 
@BernardoMeurer Global: THis just means that I can cover the whole manifold with only one chart, in general this is not the case
 
5:34 AM
So a chart is kind of like a bijective function?
 
@BernardoMeurer Not so much, it is so that we can carry over the differential structures of functions on $R^n$ transparently to the manifold itself
@BernardoMeurer As set yes
 
Cool
Hey, nice things, thanks for sharing man :)
 
Hence why we say a manifold is something that ''looks'' like $R^n$ locally.
or locally euclidean
Analytic functions is the real restrictive criterion: those that equal their Taylor series essentially
 
Or even more, holomorphic functions.
 
5:35 AM
So manifolds are locally euclidean?
 
yes
 
Can they locally be a Hilbert space too?
 
And this whole differentiable structure is how you introduce vector fields, forms, tensors, differential equations, attractors, gauge theory, etc...
@BernardoMeurer Well in a certain sense, they trivially are, because they are finite dimensional
 
Oh, they can't have infinite dimensions?
 
A Hilbert space is essentially an infinite dimensional vector space with a topology that makes it so that infinite sequences converges into the space
This is why you can carry over the intuition from R^n to a Hilbert space
Trust me, I tried removing the topology part when younger but you get a nasty, VASTLY larger space.
 
5:39 AM
Yikes
 
@BernardoMeurer Well, yes and no, manifolds are usually defined to be finite-dimensional, you can always try to extend definition but this little thing : $\infty$ likes to sap away at your foundations when you build definitions... a whole other can of worms.
 
Okay, what's the Hausdorffness of a manifold
Because I once filled a form where I put "Non-Hausdorff Manifold" as my gender
and they bought it
 
Hilbert space are why you can write :$\sum_n^\infty a_n \psi_n = \phi$ for instance.
Being Hausdorff is topological
It is essentially separability of points: every points can have a neighbourhood of its own and so ''local'' means something we want it to mean.
 
Hausdorff means your manifold is not dumb.
 
... a neighbourhood of its own that it doesn't share with another point.
 
5:43 AM
@BalarkaSen Are you calling my gender dumb?
 
@BalarkaSen Essentially
 
I'll beat you up kid
 
come at me bro
 
I come from the jungle
 
@BernardoMeurer How can non-hausdorff manifold be gendered?
 
5:44 AM
Brazil, bitch
@G.Bergeron It's my gender
 
Gender actually means something in that context...
@BernardoMeurer He was essentially right, though... :p
 
There can only be 2 Hausdorff 1 dimensional manifolds (upto some equivalence - known as homeomorphism); the real line and the circle. There are ten million 1-dimensional manifolds if you're not Hausdorff
Hence, dumb.
 
How the heck can something not have the concept of neighborhood?
 
@BalarkaSen But I've always seen differentiable manifolds as being defined to be Hausdorff. Are you talking about topological manifolds?
@BernardoMeurer It has
 
@G.Bergeron Yeah, sure. You can also have differentiable structures on non-Hausdorff manifolds.
 
5:47 AM
That's what a topology tells you
@BernardoMeurer It's just that some points are non-local in a certain sense
@BalarkaSen How? A diffeo will be a homeo and it is at least locally Hausdorff, isn't it?
(because R^n is)
 
So you can't have locality in a non-hausdorff manifold?
 
@BernardoMeurer Not absolutly, I guess
 
@G.Bergeron I don't understand your comment. A diffeomorphism is a homeomorphism; what does that have to do with being locally Hausdorff?
I mean, R^n is Hausdorff. The manifold may not be.
 
@BalarkaSen Preserves topology
@BalarkaSen Hence locally Hausdorff, no?
 
@G.Bergeron Sure. For any point there is a small neighborhood in the manifold that is Hausdorff. That doesn't mean the manifold is Hausdorff!
Think of the line with two origins.
 
5:53 AM
@BalarkaSen I know, I was just thinking as to what remains, if you remove Hausdorff from the definition of manifolds and still want a differential structure.
 
The line with two origins has a differentiable structure, for sure.
I don't understand what you're trying to imply.
 
Do non-Hausdorff manifolds admit a bundle structure without too much trouble, tho
 
@BalarkaSen I don't know, exploring, I have not worked with those...
It looks like there is a trap, though
@BalarkaSen It seems like a chart around zero will be troublesome
 
I tried looking up vector bundles for non-Hausdorff manifolds, but
Considering there's only like two books to really tackle the manifolds
It's challenging
Geroch says that you can just define the structure on Hausdorff subsets and make it agree on all Hausdorff intersections
Which sounds reasonable
 
@G.Bergeron There are two charts around the two zeroes; they intersect each other in a R - 0
I have to go now. Talk later
 
5:59 AM
@BalarkaSen Ok
@BalarkaSen Is it possible the ''non-Hausdorff points'' have to be countable?
 
The line with two origins has two non-Hausdorff points
It is very countable
 
@Slereah So you basically ''discard'' the problematic points
@Slereah Of course!
 
You don't
 
That's the point
 
The line with two origins still has many problems
Even with just one extra point
 
6:02 AM
I'm not sure a differentiable structure could be defined at all with uncountably many points of ''non-Hausdorffness''
 
Can't define a metric on it, can't define a partition of unity, Stoke's theorem won't work on it
Why not
Just take the line with $\aleph_{556}$ origins
 
@Slereah Stokes is because it is not homotopic to a CW complex, I think
@Slereah Ok, fine, maybe countably in EACH charts
 
It wouldn't be second countable but you can still define smooth maps to $\Bbb R$
There are no non-hausdorff points in a chart
Because $\Bbb R^n$ is Hausdorff
 
NO PARTIONS OF UNITY!!! AAAHHH!!!
:(
 
And Hausdorffness is preserved
 
6:05 AM
@Slereah True
That was my initial point
 
Every non-Hausdorff manifold can be defined as being covered by Hausdorff charts
Which helps if you want to define structures on it
 
The initial question was if a non-Hausdorff manifold can have a gender?
@Slereah Hence what I suspected: locally Hausdorff at least
 
Well, every manifold, Hausdorff or not, is $T_1$
So they are locally Hausdorff
This is due to the fact that $\Bbb R^n$ is $T_1$
So there's always an open set that contains one point but not another
 
@Slereah I work mostly using algebra, and when the topics begins to come out of geometry and into topology I have very much less know-how until the cycle is completed through algebraic topology... but I see your point.
@Slereah This is evident though
but the reflexes are not the same in general
 
I've got a proof somewhere, if you want
 
6:11 AM
@Slereah Proof of?
Come to think of it, I never took a proper topology class -> To be put on my to do list.
@Slereah Isn't is enough to say that a diffeo to R^n will be a homeo to R^n and thus locally Hausdorff?
 
user228700
@JohnR: Morning :-)
 
@Kaumudi.H Morning
 
@JohnRennie So true and why I went with an i3 to, but with a nice gpu for anything parallel...
 
@G.Bergeron Indeed, running parallelised code on GPUs has been something of a gamechanger.
 
6:18 AM
@Slereah thanks
@JohnRennie I felt it was not justifiable to go with more than an i3 (7th gen) with this setup. Also, it came from my own money, so no way I'm buying a 350$ cpu... :)
 
user228700
@JohnR: What's with you and sheep?
 
Some of my code is really serial and the rest, highly parallel...
@Kaumudi.H ?
 
@Kaumudi.H I can't remember how it started.
 
user228700
12 hours ago, by 0celouvsky
can't talk to JR without someone bringing up sheep or his niece
 
@Slereah What the heck?
 
6:22 AM
@Kaumudi.H Somehow the rumour got started that I was in love with a sheep.
 
user228700
x'D Nice.
 
@Kaumudi.H Well, unlimited wool is nice!
@JohnRennie Kind of weird...
 
It's been going on long enough that I don't remember who started it and in what circumstances. But I find it quite amusing so I'm not especially concerned.
 
Why, you're not welsh
 
user228700
It is somewhat amusing x'D
 
6:23 AM
@JohnRennie It would be concerning only if you were concerned :p
 
Unless of course we get divorced, in which case I could get fleeced :-)
 
user228700
Sheesh, you even got married to it, did you?!
 
@JohnRennie I hope you signed a pre-nup!
 
@Slereah Ah good point, my mother's family is Welsh. I wonder if that was the origin of the rumour.
 
@JohnRennie You think people know?
*knew
 
6:25 AM
There have got to be some good sheep puns available, but it's too early in the morning and none come to mind :-)
 
user228700
:-)
 
@G.Bergeron it came up in conversation so it was mentioned in this chat.
 
Anyway, I think I will heed (?) the starred advice of @BalarkaSen and detach myself from the sentient blackhole... It's getting late here!
@JohnRennie ah ok
See you!
 
Some of the Indian students seem fascinated by all things Western - I guess we seem exotic to them :-)
 
It usually goes both ways!
 
user228700
6:27 AM
@JohnRennie ._. Hmm, exactly to what you are referring here, I wonder...
 
@Kaumudi.H For example I've answered a lot of questions about what it was like to be a student at Cambridge.
 
user228700
Ah, yes. We're curious! :-)
 
@Kaumudi.H Trips to Vienna in tailcoat (actually it's mostly Japanese people I've seen) :p
 
@Kaumudi.H I think everyone is curious about what life is like in other countries.
That's normal for humans :-)
 
I know, I'm not a good tourist but I can't help but find Vienna a little cheesy, even though I had some of my best travels there and love the city
@JohnRennie Yes, pretty much my point
Also, they have a great intellectual culture, I think.
@JohnRennie I also think it often is less different than can be imagined
Have any of you noticed the little love story in Austrian Airline's security videos?
 
6:33 AM
@0celouvsky "Notice $M$ needs not be Hausdorff. The concept of Hausdorffness is irrelevant for much of local differential geometry. It becomes relevant in passing from a Riemannian metric to a distance function." -Hicks
I guess you can totally define a metric tensor on a non-Hausdorff manifold
 
@Slereah But then it's not really a metric for anything...
 
Who cares
The Lorentz metric isn't a metric either
 
@Slereah Well since it lives in flat space, its not the same
@Slereah Variational calculus cares
 
It's fairly hard to guess what is affected by Hausdorfness until you work out everything really
 
@Slereah The real answer right here...
 
6:44 AM
"It is also customary to require a semi-Riemannian manifold to be Hausdorff; however, as far as the local differential geometry is concerned, this is irrelevant so the restriction is not enforced at this time."
Man Hicks is probably the only book to even mention this
Ah, apparently the trick is that, in general, the Riemannian metric tensor defines a pseudo-metric on the manifold
But it only becomes a metric if the manifold is Hausdorff.
On the other hand, MSE says you can't pseudometrize a non-Hausdorff manifold
Argh
Because a pseudometric on a $T_0$ space is also a metric
Also the reference Hicks gives is in German
Variationsrechnung im Grossen
 
7:08 AM
@BernardoMeurer Good news: I'm not gonna have classes next week because of the cruise trip. Teachers won't bother with the remaining students
The travellers woke up at 5 AM today
 
Anonymous
7:24 AM
10,000 starred messages :-)
 
I noticed
Do I get a prize or something
 
Anonymous
@JaimeGallego I should get one. I was the starrer of the 10,000 th message :P
 
Anonymous
And maybe you too ;)
 
7:50 AM
 
 
1 hour later…
9:20 AM
@G.Bergeron Nah. You can analogize the line with two origins constructions to line with two cantor sets.
$\Bbb R \times \{0, 1\}$ with $C \times i$ identified in both for a fixed Cantor set $C$.
 
@BalarkaSen can you solve this enigma
 
which
 
Hicks says non-Hausdorff manifolds admit a pseudometric from a metric tensor which generates the topology of the manifold
But manifolds are T0 , and T0 spaces have pseudometrics being metrics
And obviously non Hausdorff manifolds admit no metrics
Riddle me this, @BalarkaSen
 
I don't think T0 is sufficient for being metric. You need Hausdorff.
 
Well the theorem is apparently that the existence of a pseudometric + T0 means the existence of a metric
 
9:37 AM
Sounds believable.
 
But then what is up with Hicks
Cf chapter 6, p. 69
Especially the theorem p.70 and its corrolary
 
Strange; I don't know how to construct a pseudo Riemannian metric on non Hausdorff chaps
 
@G.Bergeron I'm afraid I don't know much about either of these approaches
 
Apparently the bundle business doesn't depend at all on the hausdorffness of the manifold
It just has different global consequences
also the metric is riemannian
It's just that it doesn't define a metric space
 
I see.
 
9:46 AM
Due to branching points having a distance of 0
 
ya, gotcha
Nice picture.
Does the partition of unity argument require Hausdorffness? I don't think so actually
You should be able to just do that to get a Riemannian metric on the manifold.
 
It does apprently
No partition of unity on non Hausdorff manifolds
 
Oh, wait, I guess
 
Also more generally no metrics on them, from what's his name metrization theorem
 
Yeah, I take back what I said on that.
@Slereah What do you mean by metric? That there is no metric on non-Hausdorff chaps is easy to prove.
(otherwise it'd be Hausdorff!)
It's not obvious to me how to get a Riemannian metric on non-Hausdorff manifolds.
 
9:51 AM
Well non-Hausdorff manifolds admit the usual tensor bundle, on which you can define a metric tensor, according to Hicks
From those, you can define a distance function by the usual integral definition
Which is obviously not a metric, but is a pseudometric according to Hicks
 
Strange.
 
But then the whole T0 thing and so I dunno
 
This is too hard for me :P I never thought about the Hausdorffness technicality
 
I don't know, maybe you can't define a global section?
I should read the full proofs
 

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