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Balarka Sen
Balarka Sen
1:05 AM
@JohnRennie
 
 
2 hours later…
Cows
Cows
2:45 AM
Getting motivated!! Epic Programming night coming up ! Going to insert some next level math into my code. Gotta make moolaaa, and change the world at the same time. Going deep in this :P
 
John Rennie
John Rennie
3:44 AM
@BalarkaSen I've read and watched a lot of Beefheart documentaries over the years and they all agree he was a rather weird bloke. He seems to have been a control freak and bullied his musicians to get exactly the sound he wanted. Though members of the band generally expressed respect for him even given his approach so he seems to have been inspiring at the same time.
I wouldn't say Trout Mask Replica is a great album. I quite like it because I love the way he plays with language in the lyrics. But it's far from my favourite and I don't play it that often.
 
heather
heather
is there anyone who's a python programmer in here?
i'm having an odd sort of issue - i set up a file of mine as a package that's accessible anywhere on my system using python setup.py install
(and creating a setup.py file and so forth)
and it works in the windows powershell but not in the IDLE shell or in any of my python files.
what are some reasons this could happen?
 
John Rennie
John Rennie
4:18 AM
@heather isn't there a path variable that tells Python where to look for modules?
I wonder if that isn't being set properly.
 
heather
heather
hmm, i'll take a look at that.
yeah, it's sys.path().
weird, it doesn't look screwed up...
but then i'm not completely sure what is correct =)
 
John Rennie
John Rennie
Missing __init__.py file?
 
heather
heather
nope, got that
i did see something weird just now in the first index of the sys.path(), but deleting that before trying to import my file isn't fixing anything.
 
John Rennie
John Rennie
4:35 AM
If python mymodule.py works from the Windows command line then presumably PYTHONPATH is correctly set up. In that case I'd suspect something weird about the structure of the files in your module.
From the little I remember of it there are all sorts of arcane things about modules that can trip them up.
 
heather
heather
well that's fabulous =)
 
John Rennie
John Rennie
( use echo %PYTHONPATH% at the command line to check it)
 
heather
heather
@JohnRennie that just produced %PYTHONPATH%
 
John Rennie
John Rennie
Ah yes, I've just tried it here and got the same - oops :-)
Let me try and remember how Windows does the environment variables ...
 
heather
heather
hmm, google seems to be telling me to use echo %var%...odd.
 
John Rennie
John Rennie
4:40 AM
I thought there was an environment variable called PYTHONPATH, in which case echo %PYTHONPATH% would show you its value. But it doesn't seem to exist.
Maybe that's unix only.
 
vzn
vzn
@heather recently was dinking around with python, used this site, it was amazing at times, even handled exotic imports correctly/ elegantly/ seamlessly, try it out repl.it
 
heather
heather
@vzn yes, i do use that occasionally, it can be quite nice (esp on a chromebook, but right now i'm using a windows laptop so don't need to)
 
vzn
vzn
@heather my suggestion is to get the code running in repl.it & then if it doesnt work on your local system then its likely a (local) configuration issue
 
John Rennie
John Rennie
If you just type set at the command line it will show you all the environment variables. Have a look through those for anything python related.
 
heather
heather
huh, it's asking me to provide a name...
oh, wait, i'm using the powershell, that's why
 
John Rennie
John Rennie
4:43 AM
Ah, yes. Try the regular command prompt.
 
heather
heather
I see "path" and some "program" stuff, but nothing python.
 
John Rennie
John Rennie
If you just type python does it start the interpreter?
 
heather
heather
yup
 
John Rennie
John Rennie
Then python.exe must be in one of the directories in the PATH environment variable
 
heather
heather
hmm, the pathtext var has a .py ending
let me see if i can find the right one
C:\Program Files\Git\cmd;
if they're in the same order.
that wouldn't make sense, there's a path C:\Program Files (x86)\Python36-32\; which makes a wee bit more sense =)
well, i think i'm going to have to deal with this tomorrow...it's getting rather late here. thank you very much for your help @JohnRennie
 
John Rennie
John Rennie
4:50 AM
@heather sorry I can't help. The python module setup has always been a bit mysterious to me.
 
heather
heather
@JohnRennie oh, no problem! you ruled something out, so that was useful. i.e., i didn't break the whole system, just part of the system =)
 
John Rennie
John Rennie
I didn't know about the imp module
 
heather
heather
that does look very useful...in fact it gives me an idea...
 
John Rennie
John Rennie
Also look at the site.py file in your Python install directory
 
heather
heather
...site.py?
 
John Rennie
John Rennie
4:57 AM
Apparently when python runs it executes site.py to build the list of directories in sys.path
It'll be in the Lib subdirectory of your Python directory
 
heather
heather
okay
i can't seem to append to sys.path() in my file, which you should be able to do. thus i can't test if adding the home directory to the path fixes anything (i noticed every other path seems to be on the C: side of things whereas my module is in my documents folder, so i dunno.
 
John Rennie
John Rennie
When you've slept on it, I've just been playing with testing some modules of my own inside the python interpreter. Maybe later ...
 
heather
heather
okay, cool.
thank you very much =)
 
Cows
Cows
5:37 AM
Was just reading a simple cs paper. I think there are super humans roaming this planet.
wow!
 
Nick Alexeev
Nick Alexeev
6:12 AM
3
Q: Why is the magnetic field around a bar magnet represented by iron filings?

Michael S.Why would we use iron filings to display a magnetic field interaction and then teach this as the unaffected field of a magnet? Iron affects the field itself. This is analogous to an observer affecting a quantum event through observation. Magnetic mappers that contain iron anywhere near the prox...

magnetics magnetic-flux magnet
 
John Rennie
John Rennie
@NickAlexeev the answers there seem pretty good ...
 
Semiclassical
Semiclassical
the quantitative question it raises for me is how many filings you'd need (per square inch, say) to get an appreciable magnetic field in response
and I suspect it'd amount to the filings not actually being filings
i.e. that they'd have to be densely packed that that the filings would basically have to be pressed together into a unit...in which case you're talking about iron as a ferromagnet
 
John Rennie
John Rennie
6:29 AM
@Semiclassical I have to say that doesn't seem terribly exciting to me. If you know the susceptibility and shape of the iron filings you could calculate the combined field of the magnet + filings. It would be a dull calculation though.
 
Semiclassical
Semiclassical
yeah
good as a test question, maybe
 
Cows
Cows
6:59 AM
Ah let me do some simple calculations with vectors first
can't believe I can't even get this trivial stuff done lol
let me just read first . . . then compute later
 
David Z
David Z
7:24 AM
@heather It's a list, not a function. You should definitely be able to do things like sys.path.append(r'C:\this\is\a\path)` in your Python code.
 
John Rennie
John Rennie
sys.path.append(r'C:\\this\\is\\a\\path') :-)
The years, nay decades, of pain MS have caused by using backslash for directories!
 
 
1 hour later…
ACuriousMind
ACuriousMind
8:55 AM
@JohnRennie Made worse by people writing "clever" file I/O functions that implicitly convert the slashes under the hood, so that the escaped version doesn't work with them...
 
John Rennie
John Rennie
And just don't talk to me about the \n vs \r\n debacle :-)
 
ACuriousMind
ACuriousMind
Ah, yes. Carriage return/line feed. Stuck in the age of the typewriter
 
John Rennie
John Rennie
But then Apple uses \r, which is different again, so maybe I shouldn't be too harsh on MS
 
 
1 hour later…
Emilio Pisanty
Emilio Pisanty
10:19 AM
@DavidZ any idea how this happened?
-1
Q: How to learn Cohen Tannoudji's QM well

Ian_CuiI am studying Cohen's QM, and was puzzled for so many development materials behind the book. How should I go on?

quantum-mechanics
 
David Z
David Z
How what happened?
 
Emilio Pisanty
Emilio Pisanty
I'm on mobile - maybe my client didn't ask whether the question was still open?
@DavidZ it seems to have taken in my answer some four minutes after closure
 
David Z
David Z
Oh, if you're already writing an answer when the question was closed, you still get to submit it up to four hours (?) after closure
 
ACuriousMind
ACuriousMind
@DavidZ But only if you don't reload the page in that time
 
Emilio Pisanty
Emilio Pisanty
Ah
Interesting
thx
 
Balarka Sen
Balarka Sen
10:24 AM
@JohnRennie Yeah he was a loony alright.
 
John Rennie
John Rennie
@BalarkaSen but then, all the best people are :-)
 
Anonymous
@BalarkaSen Holla. I have a naive topology question. You around for a bit?
 
Balarka Sen
Balarka Sen
Yup
 
Anonymous
 
Anonymous
The above curve is a subset of $\Bbb R^2$, say with the standard topology. Say the curve is named $M$ and has a topology inherited from the standard topology. Now it is said that this curve is a topological space but not a topological manifold.
 
Anonymous
10:28 AM
I can find an open set about each point of the curve. But it's said that I can't find an invertible continuous map from that open set (the part of the curve within the red circle), to $\Bbb R^2$. But I'm not sure why this is true.
 
Balarka Sen
Balarka Sen
Not to R^2. To R.
A tripod is not homeomorphic to the real line, is the point.
 
Anonymous
@BalarkaSen It's being said I can't find such an invertible continuous map to either $\Bbb R$ or $\Bbb R^2$
 
Anonymous
Lemme link
 
Balarka Sen
Balarka Sen
@Blue I mean, it's outright not a 2-dimensional manifold.
Because every point other than the tripod point has a neighborhood homeomorphic to R.
If it is a manifold, it has to be a 1-dimensional manifold.
 
Anonymous
Ah, makes sense
 
ACuriousMind
ACuriousMind
10:32 AM
@Blue A manifold has constant dimension - and you can easily find charts to $\mathbb{R}$ for parts of that space not including the branch point. So you need to find a chart to $\mathbb{R}$. That there's no chart to $\mathbb{R}$ comes e.g. from thinking about what happens when you remove the branch point vs. what happens when one removes a point of the real line
 
Balarka Sen
Balarka Sen
So how do you prove that the tripod is not homeomorphic to R?
Visually it's "clear" but a proof needs some work. Shh @ACuriousMind
 
ACuriousMind
ACuriousMind
@BalarkaSen Not gonna say more :)
 
Anonymous
@ACuriousMind Right, right, that makes sense. I basically need to show that there's no chart to $\Bbb R$ if the bifurcation point exists. I'll try to write down the proof properly and show it
 
Slereah
Slereah
11:15 AM
Are we talkin' about proving that $\mathbb{R}$ and $\mathbb{R}^2$ aren't homeomorphic
 
ACuriousMind
ACuriousMind
No
 
Slereah
Slereah
o
also hello
 
ACuriousMind
ACuriousMind
Hi :D
 
Anonymous
Holla
 
Slereah
Slereah
@ACuriousMind The empty set has all the dimensions!
 
ACuriousMind
ACuriousMind
11:18 AM
The empty set is an abomination
I prefer my manifolds non-empty
 
Slereah
Slereah
If you want manifolds with branching points they do exist
They are called branching manifolds, I think?
aka train tracks for1D ones
Branched manifold
In mathematics, a branched manifold is a generalization of a differentiable manifold which may have singularities of very restricted type and admits a well-defined tangent space at each point. A branched n-manifold is covered by n-dimensional "coordinate charts", each of which involves one or several "branches" homeomorphically projecting into the same differentiable n-disk in Rn. Branched manifolds first appeared in the dynamical systems theory, in connection with one-dimensional hyperbolic attractors constructed by Smale and were formalized by R. F. Williams in a series of papers on expanding...
It's a fairly annoying definition
 
ACuriousMind
ACuriousMind
Ah, another one of these definitions where not all "<adjective> manifolds" are actually manifolds
I wish people would stop doing that :P
 
Slereah
Slereah
Would you call it a branchifold instead
like conifolds and orbifolds
Manifolds with corners being the cornifold
 
ACuriousMind
ACuriousMind
sounds legit
 
Emilio Pisanty
Emilio Pisanty
@ACuriousMind ah, lighten up
 
Balarka Sen
Balarka Sen
11:21 AM
Branched manifolds appear naturally as leaf spaces of dynamical systems
 
Slereah
Slereah
@BalarkaSen Wrong manifolds!
 
Balarka Sen
Balarka Sen
The terminology is just garbage
 
Slereah
Slereah
Branched manifolds are Hausdorff
 
Emilio Pisanty
Emilio Pisanty
Its impact on COBOL programming is limited
 
Balarka Sen
Balarka Sen
oh
 
Slereah
Slereah
11:22 AM
Although I think they are related
Like
If you take a non-Hausdorff manifold and identify every branch points
You get a branched manifolds
 
Balarka Sen
Balarka Sen
Oh these are manifold with tangential singularities
Train tracks, got it
Fuck these terminologies man
 
Slereah
Slereah
yeah, you identify open sets instead of closed sets
 
ACuriousMind
ACuriousMind
@EmilioPisanty :P
 
Slereah
Slereah
I don't think there's any generalized manifolds with variable dimension, tho
Like just a line tacked on a plane
I'm not sure where it could be useful
 
Balarka Sen
Balarka Sen
stratifolds
i have been thinking about those for a while
 
Slereah
Slereah
11:34 AM
What are they used for?
 
Sid
Sid
I m sed. Comrades lost.
 
Balarka Sen
Balarka Sen
lol
@Slereah Understanding maps. If $f : M \to N$ is an smooth map, the "singularity set" of $f$ (where $df$ has rank less than $\dim N$) is a stratified subset of $M$.
Thom asked if that stratified subset can recover the topological type of maps.
 
Slereah
Slereah
I c
 
Anonymous
Doubt: Does the map from the curve $M$ (excluding the bifurcation point) to $\Bbb R$ need to be onto (to classify as a topological manifold)? That is, does every point in $\Bbb R$ need to have a preimage which is a part of $M$?
 
Anonymous
I'm messing up the definitions, apparently
 
Anonymous
11:52 AM
Nah, they don't mention "onto" as a condition
 
Anonymous
So basically a topological space $(M,\mathcal{O})$ is called a $d$-dimensional topological manifold if $\forall p \in M: \exists \mathcal{u}\in \mathcal {O}: \exists x: \mathcal{u} \mapsto x(\mathcal u) \subseteq \Bbb R^d$. The only conditions are i) $x$ is invertible ii) $x$ is continuous and iii) $x^{-1}$ is continuous
 
Anonymous
 
Anonymous
So, umm, let's first assume we exclude the bifurcation point $B$. Considering the above curve is finite, I could map every point on the upper left branch to an open set in $\Bbb R$ say $(a,b)$ where $a<b$. Then I could map the lower left branch to an open set in $\Bbb R$ say $(c,d)$ where $c>b$ & $d>c$ followed by mapping the right side branch to an open set $\Bbb (e,f)$ where $e>d$ & $f>e$.
 
Anonymous
It seems quite obvious that all the three mappings are continuous since $\forall V\in \mathcal {O}_{\text{standard}}|_{(a,b) \cup (c,d) \cup (e,f)}: \text{preim}(V)\in \mathcal {O}_{\text{standard}}|_M$
 
Anonymous
@ACuriousMind Is this sufficient to say that the tripod without the branch point is a topological manifold?
 
Anonymous
12:07 PM
Also, I guess finiteness of the curve is not necessary?
 
ACuriousMind
ACuriousMind
I'm not sure what you're doing, in particular why there are relations between the $a,b,c,d,e,f$.
For it to be a manifold, there simply needs to be some open set around each point that maps homeomorphically to some open set in $\mathbb{R}$. There doesn't need to be any relation between these maps for different points.
 
Anonymous
@ACuriousMind Say I map the points in the upper left portion of the curve to the open set $(1,2)$ in $\Bbb R$, the bottom left portion of the curve to $(3,4)$ and the right portion to $(5,6)$. I wrote those relations such that there is no overlap between the images of the mappings
 
Anonymous
Non-overlap of the images is necessary, I suppose?
 
ACuriousMind
ACuriousMind
Yeah, but there's no requirement anywhere in the definition of a manifold that you have to map all the points of the manifolds to $\mathbb{R}$ at once without overlap
What you're doing is an embedding, which is a much stricter thing - not every d-dimensional manifold has a $\mathbb{R}^d$ embedding
(in fact, very few do)
 
Anonymous
@ACuriousMind But wouldn't that violate invertibility of the map, if there is overlap between the images of the three portions of that tripod?
 
ACuriousMind
ACuriousMind
12:14 PM
@Blue Again, I think you're looking for the wrong map.
 
Anonymous
I mean, say all the three portions of the tripod got mapped to $(1,2)$
 
ACuriousMind
ACuriousMind
For each point, there needs to be an open set with a homeomorphism to an open set in $\mathbb{R}$.
But these maps are allowed to be different for each point
(But, out of expediency, you would of course pick the same for every point in one such open set)
When you map one of the branches to $\mathbb{R}$, you don't need to consider at all what the other two branches are doing
 
Anonymous
@ACuriousMind But is overlap allowed between those open sets in $\Bbb R$ (which are individually homeomorphic to the open sets of the curve about each point of the curve)?
 
ACuriousMind
ACuriousMind
@Blue Yes.
 
Anonymous
12:30 PM
Interesting. Now suppose I choose an intersection region between two such overlapping open sets in $\Bbb R$ say (2,3) [which is the intersection region between (1,3) and (2,4)], which will obviously be an open set again. I use an invertible map named $x$ to map all the points in the open set in the blue circle about $A$ to (1,3) and another invertible map named $y$ which maps all the points in the open set in the green circle about $C$ to (2,4).
 
Anonymous
Then, I pick up a point say 2.5 in the open set (2,3), and try to trace back to the point in the original curve whose image is 2.5. Clearly, $x^{-1}(2.5) \neq y^{-1}(2.5)$. That is, there are two such possible preimages of 2.5. Is that allowed?
 
ACuriousMind
ACuriousMind
@Blue Yes.
 
Secret
Secret
1 hour ago, by Slereah
I don't think there's any generalized manifolds with variable dimension, tho
hmm...
I wonder if it is possible to have some kind of generalised manifold such that it is $\Bbb{R}^2$ in one location, and $\Bbb{R}^3$ in another, if that is what "variable dimension" means
More generally, can we have a mathematical space where the dimensionality is "stochastic" in that there is some probability that at some point p, the dimension is n?
I am not sure how messy it is to formulate that
 
Anonymous
1:03 PM
@ACuriousMind Thanks. Got that part. :) Now, I'm trying to see what exactly happens when I include the bifurcation point. The problem as I see is that the point belongs to all the three branches in the original curve. If I consider an open set about $B$ (red circle) and I map all the points of that open set using different maps $x_i$ (where $i$ belongs to an index set) to the region (1,2) on the real line.
 
Anonymous
Now, what exactly is the problem? Why can't the mapping be continuous and invertible now?
 
ACuriousMind
ACuriousMind
@Blue That's for you to find out! As I said, consider what happens when you remove that point vs. what happens to the real line when you remove a point.
 
Anonymous
Okay :P Nice puzzle I'm stuck in
 
Anonymous
Trying to see
 
Shing
Shing
1:21 PM
if I represent position vector as column vector, then I can write : $\text{div} \longleftrightarrow [\text{ }\partial_x\text{ } \partial_y\text{ } \partial_z\text{ }]$ and $\text{grad} \longleftrightarrow \begin{bmatrix} \partial_x \\ \partial_y \\ \partial_z \end{bmatrix}$. where div and grad is transpose of each other. does that mean anything?
 
Secret
Secret
1:49 PM
not really, it just happens that when you have the row vector of derivatives multiplying a vector you get the inner product which is the div.
We knew that div is not contravariant to grad so that their coordinate representation being transpose to each other has no significance
 
Secret
Secret
2:25 PM
0
Q: How to define *dynamical* dimensions?

SomeoneI'm considering a simple toy model. The spacetime is flat with $d$ space dimensions. Using cartesian coordinates, the spacetime metric is Minkowskian : $$\tag{1} ds^2 = dt^2 - dx_1^2 - dx_2^2 - dx_3^2 - \ldots - dx_d^2. $$ A massless scalar field $\Phi$ is propagating in that spacetime accordin...

general-relativity cosmology spacetime space-expansion topology
looks like Slereah discussion on variable dimensions have caused a ripple
The question is not entirely senseless. If spacetime is emergent of some quantum gravity, we will expect dimensionality to not be a conserved quantity
But how to formulate that, I don't have enough quantum gravity background to figure out yet
 
vzn
vzn
3:00 PM
> The operation was earlier called a 'war with water and time' to save the team.
↑ extraordinary/ great/ historic/ rare intl humanitarian "peaceful" rescue mission involving coordination/ cooperation, thai govt, elon musk involved now, 90 divers. but humans just cant avoid using militaristic/ war metaphors for large complex operations. :|
 
ACuriousMind
ACuriousMind
@Secret I have no idea what you're rambling about, but it doesn't sound very nice. Stop it.
4 messages deleted
 
Secret
Secret
done
 
 
1 hour later…
vzn
vzn
4:11 PM
 
 
2 hours later…
Cows
Cows
5:57 PM
I will just ask. I can apply Green's and Stoke's theorem, but I know I don't understand them or at least I have not found an account of what is happening that is satisfying enough. It is not that I am stupid, but I can't explain it.
I don't have that feeling I get when I think I know something. Is there a trivial English derivation of this concept? It turns out that I might need to understand it to add to some logic I am trying to implement to show that something is true. In fact at the moment, I just want to understand (derive
)
without ambiguity green's theorem.
I can't even explain what I want to know, but there is something very strange going on. I know this is trivial stuff.
 
Anonymous
 
Anonymous
So, umm, I've been trying to think about this
 
Anonymous
In the second case, when I include the Bifurcation point
 
Anonymous
I can take branches (1) & (2) of the open set as one component and map it to the real line, such that it includes the bifurcation point
 
Anonymous
Basically I drew the image of the map $f$ i.e. $f((1) \cup (2) \cup B)$
 
Anonymous
6:01 PM
And $f((3))$
 
Anonymous
On the real line
 
Anonymous
But $f$ still looks invertible to me
 
ACuriousMind
ACuriousMind
Well, that's not a continuous map :P
 
Anonymous
@ACuriousMind The definition of continuous which I know is: the preimage of every open set should be a open set
 
Anonymous
Am I missing something obvious?
 
Anonymous
6:05 PM
The preimage of $f((1)\cup (2)\cup B)$ looks open to me
 
Anonymous
And so does the pre-image of $f((3))$
 
Cows
Cows
I mean so "intuitively" . . . . . . big rotating circle = many small rotating circles . . . yes, but . .
lol I can't show
 
Secret
Secret
Perhaps this will help:
I am guessing you are trying to understand how the circulations cancel out into a big circulation
 
Cows
Cows
@Secret yes, I actually have that link open
 
Secret
Secret
Green's theorem
In mathematics, Green's theorem gives the relationship between a line integral around a simple closed curve C and a double integral over the plane region D bounded by C. It is named after George Green, though its first proof is due to Bernhard Riemann and is the two-dimensional special case of the more general Kelvin–Stokes theorem. == Theorem == Let C be a positively oriented, piecewise smooth, simple closed curve in a plane, and let D be the region bounded by C. If L and M are functions of (x, y) defined on an open region containing D and have continuous partial derivatives there, then ...
and there are proofs here also by showing how the curve cancel out
 
Cows
Cows
6:30 PM
hmm, if I actually just follow the math, it shows it works.
now let me just see if i can apply this to some silly proof, . . let me first look at some silly notation for set related stuff
 
Cows
Cows
7:14 PM
back home.
Let me think through this stuff a little deeper( I don't have much choice hehe )
 
 
1 hour later…
Balarka Sen
Balarka Sen
8:16 PM
@Blue u called
 
Anonymous
@BalarkaSen Yeah. I'm feeling too stupid now :/
 
Anonymous
Could you see my question above ^
 
Anonymous
 
Anonymous
I don't understand why this type of a mapping is not continuous
 
Anonymous
For continuity we just need that pre-images of open sets are open sets
 
Anonymous
8:18 PM
Isn't it?
 
Balarka Sen
Balarka Sen
Your map is just pasting the branch (A) and (B) togather?
 
Anonymous
@BalarkaSen The second one (including the bifurcation point)? It's pasting (1),(2) and the bifurcation point B together to form an open set in $\Bbb R$
 
Anonymous
$f((1)\cup (2) \cup B)$
 
Anonymous
(1) and (2) do not include B
 
Balarka Sen
Balarka Sen
I don't see your picture very well. Let's just say $Y$, the alphabet, is the open tripod, which it is. You're defining $f : Y \to \Bbb R$. Could you explain how a bit better?
 
Anonymous
8:21 PM
Lemme draw a nice big picture. Just 2 minutes:
 
Semiclassical
Semiclassical
hi chat
 
Anonymous
 
Sir Cumference
Sir Cumference
So is 0celo going to stay banned for a year?
 
Anonymous
@BalarkaSen ^
 
Anonymous
On the real line, between the two open circles is the open set
 
Anonymous
8:26 PM
on the left
 
Balarka Sen
Balarka Sen
@Blue Is this it
 
Anonymous
@BalarkaSen Right!
 
Balarka Sen
Balarka Sen
Ok, that is indeed not continuous.
Here's why
Consider the green bit, the image of $(3)$ by $f$.
That's an open subset of $\Bbb R$, yes? It's an open interval
 
Anonymous
@BalarkaSen Yep
 
Anonymous
It doesn't include the point B
 
Balarka Sen
Balarka Sen
8:29 PM
Let's call that $U$. $f^{-1}(U)$ is the green bit on your tripod picture, which misses the bifurcation point.
 
Anonymous
Yes
 
Balarka Sen
Balarka Sen
That's not an open subset of the tripod under the subspace topology from $\Bbb R^2$ :)
 
Anonymous
@BalarkaSen Umm, why not? This is probably the exact point I'm missing
 
Balarka Sen
Balarka Sen
Wait, no, that wasn't correct, sorry. That is open.
Here's what is not open.
Consider the blue+red bit, including the image of the bifurcation point.
Call that dude $U$. That's an open interval in $\Bbb R$, so is open.
$f^{-1}(U)$ is the bluebranch+bifucrcationpoint+redbranch in your tripod. THAT is not open
 
Anonymous
@BalarkaSen Okay, why so?
 
Anonymous
8:33 PM
I can find a soft ball for every point on that bluebranch+bifucrcationpoint+redbranch
 
Balarka Sen
Balarka Sen
Remember that subspace topology means intersection of your tripod with open subsets of $\Bbb R^2$ are the open subsets of the tripod.
So if $f^{-1}(U) \subset \mathsf{Y}$ is open, it has to be of the form $\mathsf{Y} \cap B$ where $B \subset \Bbb R^2$ is an open subset of $\Bbb R^2$.
Agree?
 
Anonymous
Got to revise the definition of subspace topology. Gimme a minute
 
Anonymous
I'm confusing things
 
Balarka Sen
Balarka Sen
You can tell me the definition you know. Your soft ball thing would also work.
 
Anonymous
Say $S$ is a subset of topological space $M$
 
Anonymous
8:39 PM
The topological space has a topology $\mathcal {O}_M$
 
Anonymous
Now $\mathcal{O}|_S$ is the topology inherited from $M$, restricted to $S$
 
Anonymous
Which is defined as:
 
Anonymous
$\mathcal{O}|_S := \{U\cap S | U \in \mathcal{O}_M \}$
 
Balarka Sen
Balarka Sen
In short open sets of $S$ are of the form $U \cap S$ where $U$ is an open set of $M$.
Ugh, small u for open set
Your notation sucks :P
 
Anonymous
Okay now? :P
 
Balarka Sen
Balarka Sen
8:43 PM
Ye
This red subset of the tripod $\mathsf{Y} \subset \Bbb R^2$ cannot be written as $\mathsf{Y} \cap U$ where $U \subset \Bbb R^2$ is an open set.
Do you see this?
 
Anonymous
@BalarkaSen Agreed now!
 
Balarka Sen
Balarka Sen
The point is if it was of the form $\mathsf{Y} \cap U$, then it would be a subset of $U$ (because $\mathsf{Y} \cap U\subset U$). But any open set of the plane containing the bifurcation point of the tripod contains a little bit of all three branches. The red set doesn't contain the entirety of one branch. Contradiction.
 
Anonymous
@BalarkaSen I need to get one thing cleared before answering that. Does it matter whether the branches of the tripod are finite or infinite?
 
Balarka Sen
Balarka Sen
Naw
 
Anonymous
@BalarkaSen Oh. So if I take all possible softballs in $\Bbb R^2$ there's no such softball such that it contains only portions of the red branches but not of the third branch (full black). That's the point, right? BTW all open sets in $\Bbb R^2$ with standard topology have to be softballs, right?
 
Balarka Sen
Balarka Sen
8:49 PM
Yup! (As in, that's indeed the point)
Well... what's a softball? :D
Is it what normal people call disks of finite radius?
 
Anonymous
$$B_r(p) = \{(q_1,q_2,...,q_d)|\sum_{i=1}^d (q_i-p)^2 < r^2\}$$
 
Anonymous
Where $r\in \Bbb R^{+}$
 
Anonymous
And $(q_1,q_2,...,q_d) \in \Bbb R^d$
 
Balarka Sen
Balarka Sen
Balls of finite radius. No, not all open sets of $\Bbb R^2$ are balls of finite radius. Consider the open annulus $\{(x, y)\in \Bbb R^2 : 1 < x^2 + y^2 < 2\}$.
But for any open set $U$ of $\Bbb R^2$, for any $x \in U$, there is a ball of finite radius $B$ containing $x$ such that $B \subset U$
In particular, any open set of the plane can be written as a (possibly uncountable) union of balls of finite radius.
The slogan is "balls of finite radius form a basis of the topology of the Euclidean plane (or space in general)"
 
Anonymous
I learnt that the standard topology is defined in terms of soft balls. That is, $U \in \mathcal{O}_{\text{standard}}$ if $\forall p\in U \exists r\in R^+: B_r(p) \subseteq U$
 
Balarka Sen
Balarka Sen
8:55 PM
Yup
 
Anonymous
And that "open sets = elements of standard topology"
 
Anonymous
But it's interesting to note that their unions
 
Anonymous
Can also qualify as open sets
 
Balarka Sen
Balarka Sen
Yup
They "generate" the topology
@Blue In particular, if $p \in \mathsf{Y}$ is the bifurcation point, and $U$ is an open subset of the plane containing $b$, then there is a ball $B_r(p)$ of finite radius such that $B_r(p) \subset U$, and any ball of finite radius around the bifurcation point contains a bit of all three branches.
So neighborhoods of the bifurcation point of the tripod are also tripods
Your V-shaped bipod is, unfortunately, not open
 
Anonymous
@BalarkaSen Right! This makes sense now, finally!!!
 
Balarka Sen
Balarka Sen
8:57 PM
Cool!
 
Anonymous
Phew phew phew phew
 
Anonymous
Spent a whole day on this :P
 
Anonymous
But was worth it
 
Balarka Sen
Balarka Sen
It's good. I remember being confused a lot about subspace topology when I learnt it
The red set immediately feels open because it's homeomorphic to an open interval
But it's not an open subset of the space!
 
Anonymous
I tried to read from Munkres and got bored soon :/ So I'm learning DG+Topology together in the context of GR. Probably I'm missing out some important parts, but I guess I can get back to those skipped parts later on
 
Balarka Sen
Balarka Sen
9:00 PM
@Blue But you see why your function $f$ SHOULD not be continuous, right? It has a "hump discontinuity" at the bifurcation point.
Point-set topology needs some push. Munkres is a very rigorous text, I don't recommend it as a thorough reading
When you have gathered some momentum go through the first few things
 
Emilio Pisanty
Emilio Pisanty
@rob Thanks for the name-drop. Or was that a subtle dig to get me to write an answer?
 
Anonymous
@BalarkaSen I'm not yet able to relate the two different definitions of continuity. The one we learnt in high school (discontinuity of graph) and the topological notion continuity i.e. pre-image of open sets should be open sets
 
Balarka Sen
Balarka Sen
That's a very cool exercise.
You have to factor through the epsilon-delta definition of course.
That's basically a restatement of "preimage of open intervals is open"
 
Anonymous
Ah, I was expecting that. I'll try it out tomorrow then. Too sleepy now :P
 
Balarka Sen
Balarka Sen
go to slep
 
Anonymous
9:05 PM
Thx a lt :P
 
Anonymous
Gdnt
 
Balarka Sen
Balarka Sen
nt
21
Q: Why should I care about the Jones polynomial?

JuliaThe invention of the Jones polynomial led to hundreds of papers and a Fields medal. However, as far as I can tell it had few consequences in topology. After all, after Thurston’s work we already had algorithms to completely classify knots, so by itself a new invariant seems to be of limited va...

gt.geometric-topology knot-theory
Thought the physicists might find this relevant
 
Cows
Cows
9:50 PM
From looking at the above ^
I am somewhat puzzled. From the response with 21 upvotes, it seems people were more or less more impressed by the fact that it arose naturally in physics via some path integral and operator type voodoo.
This made math people excited and began to dedicate more time to think about this
very interesting
mathematics out of physics huh
Would one say that taking the path integral approach is the way to go for most problems and to accidentally discover hidden connections?
as far as i know PI does not mean much to math peeps unless some new developments happened while I was asleep
btw I am just an interested observer on the matters and hope to understand this stuff someday
*btw just just downloaded @Jonny Evans visual guide to math on github via his site. Very informative
 
 
2 hours later…
Emilio Pisanty
Emilio Pisanty
11:52 PM
113
Q: We need to improve the quality of our spam!

UndoRecently, I've noticed a downhill effect in the quality of spam posted on Stack Exchange websites. Take this as an example (found on Space.SE): There are a great many things wrong with this artifact: There is not one capitalized letter in the entire post. (-1 grammar point.) The only pun...

discussion spam fun
wha...?
@ACuriousMind whatcha make of that?
 
Mikhail
Mikhail
n o i c e
So, I got a boring Green's function of free space in 3D. What are some fun basese to decompose it into :-)
Like it probably decomposes into a Fourier series, but I want something that is more compact, like maybe a Zernike series or something
 

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