The h Bar

@Slereah What's your github name?

Got none

I don't know why if I call it from inside the function it won't work

@Slereah :(

@0celouvsky ...

@BalarkaSen Yes?

3:06 PM

@Slereah How do you not have a GH account?

@BernardoMeurer I don't have one either

why would I

@0celouvsky That's because you want to spite your sister

@Slereah To help me :3

not a very attractive prospect

lol rekt

@BernardoMeurer I don't want to spite her

3:08 PM

@Slereah You'll regret this when I become a gorgeous woman in a couple years

How could you become a woman if you're a man now?

Am I?

We haven't met

Michelle says you're a man

well, a boy

And you trust her?

lol, this paper by Milnor is great: "One first shows that $\Omega_n^\text{spin}$ is isomorphic to the stable homotopy group $\pi_{n+k}M(\mathrm{Spin}(k))$ of a suitable Thom complex; and then determines these homotopy groups by a formidable computation. No details will be given"

3:11 PM

forgot to close your dollars

I am now angry that the authors of that other paper would reference this one if it just states the fact they reference it for without proof or any details

@ACuriousMind Sounds like Thom-Pontryagin

They might as well have just asserted it :P

probably trivial

@BalarkaSen It does - what irks me why they would reference Milnor just vaguely describing the construction instead of something with more...substance. Has no one ever actually published a more precise computation of the spin cobordism groups?

3:14 PM

No idea, actually. Haven't thought about spin cobordisms; what's M(Spin(k))?

stahp with the topology pls

@BalarkaSen I presume it's the "suitable Thom complex"

I walk into physics and there's TEST 2 up on the projector

my heart exploded lol

turns out we're going over test 2, which we took 2 weeks ago

Yashas Samaga

wrong room, sorry

looks relevant: map.mpim-bonn.mpg.de/Spin_bordism

3:17 PM

The assertion that all spin manifolds in 7 dimensions are null-cobordant is kinda important for another paper I'm reading, which in turn is kinda important for another...I'm several levels into these dependencies and at the bottom I find Milnor vaguely waffling about, I'm not amused :P

@BalarkaSen lol, "The spin bordism groups up to dimension 8 are given in [Milnor1963a] without proof. Milnor states that this is the result of a formibable calculation"

Maybe I'm looking at the wrong kind of math but where does this stuff even show up?

lol whoops

@0celouvsky Classifying bordisms (upto structures) is a pretty classic problem

All this algebraic topology is probably not that complicated but I never see algebraic topology more complicated than some homology groups

So it appears quite horrifying to an outsider like me

Oh, it is kind of nontrivial.

It's a hard area of algebraic/differential topology

@ACuriousMind So I guess you probably won't see a lot of places where you will find the calculation done

@0celouvsky Well, in this case it turns out that you can define an interesting invariant of a $G_2$-manifold when it is the boundary of a $\mathrm{Spin}(7)$-manifold, and we kinda want that invariant for all $G_2$-manifolds so we need to know that they all are the boundaries of some spin manifold, which is equivalent to the assertion that the cobordism group is trivial.

3:21 PM

@BalarkaSen I found an alternative proof of that $L^p$ thing. It's actually a really neat thing: if $L^p(\Omega,\mu)\ni f_n\to f$ pointwise, you can't say anything about the $L^p$ limit. But if you know $||f_n||\to ||f||$, then you can say $f\in L^p$ and $f_n\to f$ in $L^p$ norm.

@BalarkaSen Well, at least that page you linked gives a general classification that should give the same result if I understood what it is talking about :D

It's surprising because convergence of norms usually gives you zero information.

I guess I'll file this under "somewhat worrisome black box" and move on.

Ah, yeah, a lot of manifold invariants are defined by fiddling with a manifold it bounds. (eg, all 3-manifolds bound 4-manifolds)

and showing it's independent of the cobordism

@ACuriousMind Yeah.

@ACuriousMind Christ

3:24 PM

@0celouvsky a'right but i dunno analysis

@BalarkaSen The idea is that the only way $f_n\not\to f$ in $L^p$ is if "mass gets pushed out to infinity"

got that

So the convergence of norms is the statement that you don't lose mass, basically

For a counterexample, take $\chi_{[0,1]}$ and scoot it off to infinity

sure, sure. i'm just sayin', i don't particularly care much

so @BalarkaSen do you like my new name?

3:28 PM

it's neat

I should add a german twist. @ACuriousMind ?

0ßelouvsky?

Will PSE let me put a ß in my name?

I think it allows arbitrary unicode?

So a ß in particular will be fine

so I can put the Egyptian hieroglyphic ejaculating penis in my name?

I need an answer

[Caption] I think I need to revise joint probability distributions, cause I need to understand how states go from no correlation to some correlations

@0celouvsky You can, in the sense it is technically possible. However, that'd be somewhat distasteful and we'd probably have to reset it.

3:32 PM

Fair enough

The only states that are uncorrelated in this diagram are the basis states

So that means, given some vanishingly small $\epsilon > 0$ the following state has a nonzero correlation

ugh

$$\frac{\sqrt{\epsilon - 1}}{\sqrt{\epsilon}}\lvert + -\rangle + \frac{1}{\sqrt{\epsilon}}\lvert - +\rangle$$

@ACuriousMind why is the wavelength of fire ~ tungsten filament ~ sun

why can we see fire?

it has to be more than a coincidence

This is so weird, yes, geometrically and algebraically it is easy to understand what is happening because you are essentially rotating the state in the hilbert space $\mathcal{H}_1 \otimes \mathcal{H}_2$ thus nothing is really weird. However, the transition from no correlation to some correlation looks so abrupt

3:40 PM

@0celouvsky It's a miracle

why do people know the refraction index of diamond randomly!?

that is, one only need to rotate the state a little bit off the basis state and suddenly you gain some correlations

why do people know any random values of things randomly

An engineered one in the case of the filament - you make the filament precisely so thick that it reaches the correct temperature at the usual current

the only reason I know the mass of the electron is because of ~~voldemort~~ you know who

other than that I cannot tell you any physical constants!

3:42 PM

voldemort knows the mass of electron?

but these people know specific heats of iron!?

@BalarkaSen yeah, check the 6th book

???

Also, I'm not convinced "wavelength of fire" ~ "wavelength of sun" is really accurate. What's true is they're both visible, but consider that blackbody radiation can't really leave the visible spectrum once it has non-neglegible part above infrared

he uses pair production to make 511keV photons

the magic is not really magic

it's just quantum field theory

So all "hot" radiation will be visible

3:44 PM

The following state, however, has no correlations of any kind because it is a product state
$$\frac{1}{2}(\lvert ++\rangle+\lvert +-\rangle+\lvert -+\rangle+\lvert --\rangle)$$

@ACuriousMind hmm, I didn't know that

what's the precise statement?

@0celouvsky i know the charge of a neutron

@0celouvsky The "peak" can move past the visible spectrum, but at that point the blackbody is radiating so much energy that the tail in the visible spectrum is still, well, distinctly visible

win win

We call that "glowing white-hot", I think

3:46 PM

ahh

interesting

@BalarkaSen I know c to one decimal place

one significant figure I mean

c=1

Also, @0celouvsky, you might be interested to know that the yellow color of the sun is an atmopheric effect

@ACuriousMind I don't know why people say it looks yellow. It looks white to me

Well, but it's yellow-tinted

I'm color blind

In any case, don't look too long into the sun to see if it's yellow or not :D

3:49 PM

what is the atmospheric effect?

@ACuriousMind So is the majority of radiation from the sun in the visible range?

Rayleigh and Mie scattering

I'm assuming we've evolved to see that part

@0celouvsky I believe so, yes.

I just think it's interesting that lots of other things radiate in the same range

And it's a pretty narrow range of the whole spectrum!

@0celouvsky Well, what are those lots of things except for other blackbody radiation?

3:50 PM

I am forgetting, is H_alpha in Balmer series the one where electron jumps from n = 6 to n = 2?

@ACuriousMind spectral lines from atoms are not blackbody, are they?

or n =3 to n = 2?

@0celouvsky They ain't, but they are also frequently outside the visible range

Also, obligatory Abstruse Goose

Ok, I guess the latter

I was watching the Joe Rogan Experience with that one guy who used to be married to Katy Perry

Pretty sure he was out of his mind high, but he said some interesting things

Like what if we couldn't smell...it would be impossible to comprehend what smell is

So in out understanding of the universe we're limited by things we can comprehend via our senses

So there might be a lot of stuff we're missing because of that

His solution was take LSD and shrooms

3:54 PM

@0celouvsky Related, What's it like to be a Bat?. Such impossible-to-explain subjective experiences are often called Qualia

So... generalising this observation and using this [PSE] (http://physics.stackexchange.com/questions/164700/how-are-superposition-and-entanglement-related/164716#164716), we could have claimed that the linearly polarised state

$$\frac{1}{\sqrt{2}}(\lvert +\rangle+ \lvert -\rangle)$$

is a correlation between the clockwise polarised state or anticlockwise polarised state

One can demonstrate the mathematical similarities by using the up state and down state projectors, in that one of the basis state will be inner product to 0 and vanishes

As with so many things, philosophers of mind have so far not even agreed on whether they exist in a meaningful sense :P

I was a bat once

It was alright

> Lewy's example led L. HORMANDER [3] to develop a systematic method of constructing linear partial differential equations without solutions.

lol

Sounds rather dickish

What did he do, give them to his students as homework?

3:59 PM

usually

@0celouvsky good idea. i also suggest murder and psychosis

Yosida likes puns

Well I guess, my semi philosophical question is, for any two particle state, it is just a ray in the hilbert space $\mathcal{H}_1 \otimes \mathcal{H}_2$, but why do they behave very differently, e.g. some of these are product states and some of these are entangled states and some of these are something in between, is it because we are effectively only seeing a shadow of them even if we combine results from both subsystems and establish a correlation ?

He defines a fundamental solution of a partial differential operator $P(D)$ as a distribution $E$ s.t. $P(D)E=\delta$

PDE!

@BalarkaSen did you check out r/imgoingtohellforthis?

yeah i did

4:02 PM

[Caption] They all look like rays in here...

Every constant coefficient PDE has a fundamental solution!

Crazy!

dear god

Above is a bad example demonstrating why you should never ever try to draw these spaces. They are huge

@Secret you should draw L^2[0,1]

4:06 PM

Well, I don't even know what $L^2$ look like

also I suspect it will be completely covered with ink once you plot all possible functions in

it is infinite dimensional

aren't we all infinite dimensional, in the end

no

@ACuriousMind is 3-dimensional by AdS/CTF

he's an AI on the boundary of spacetime

PS, don't try to draw the rational line, you will get rekt very quickly by its density

lol

@Secret try to draw a non-Hausdorff manifold

4:17 PM

you can easily do that modulo enlarging the notion of drawing

:P

The split real line is pretty easy to draw

that's Hausdorff if you draw it

you need to draw the topology somehow

that's what the little cup is for

what?

4:23 PM

It's a sketch not an actual picture is what 0celo7's saying. but that shouldn't be a problem

@BalarkaSen the problem is that a subspace of a Hausdorff space (the page) is Hausdorff

go away

so you need to indicate the topology somehow

that's not a problem

You draw a little $\supset$ on the end of $(0, \infty)$

and a little ball on the $0$

to signify the topology

4:25 PM

i like the line with two origins better

The best 1D non-hausdorff manifold is the complete feather

[Caption] Cofinite topology wrapped around a half infinite cylinder:

@Slereah nah, leaf space of reeb

A few open sets illustrated in gray

The complete feather is like

Take $\Bbb R$

Split it at every point $x \in \Bbb R$

Repeat the process

forever

it's a very fluffy manifold

4:29 PM

Is $\Bbb{Q} \times S^1$ a manifold?

What's a bijection between $\Bbb Q$ and $\Bbb R$

There isn't one since $\Bbb{Q}$ is countable while $\Bbb{R}$ is not

there's your answer

@Slereah that does not look like a manifold to me...

@0celouvsky And yet it is

4:32 PM

proof?

It is not a manifold. Any nbhd of any point is disconnected. In the product topology anyway.

I'm assuming it doesn't have the subspace topology from $\Bbb R^2$?

Ah

Apparently it doesn't branch at every point

Only at rational points

books.google.fr/…

Even if it branches at one point it's not a manifold...

@0celouvsky The topology is different.

It's like $\Bbb R \times \{0, 1\}$ with negative real axis squished at both. The neighborhood at the branch point belongs entirely in 1 of the 2 branches, not a tripod

4:34 PM

Only if you assume hausdorfness is part of the manifold definition

Locally modeled on stuff like that

christ, what is that?

I'm a non-hausdorff manifold

I wonder how leaf space of the Reeb foliation looks like. Never gave it a thought.

@BalarkaSen Every time I try to learn about foliations I am turned off by the really quite technical definitions

4:36 PM

I'm guessing it looks like a thing that can't be represented in 2 dimensions

Plaque charts and whatnot

Various submanifold topologies

I want to achieve satori so I can picture what a principal bundle looks like

@0celouvsky Yeah, it looks quite technical because it's really hard to put things in words. Once you get them it's not very hard.

@Slereah just picture a trivial bundle with your favorite 1-d base space and 1-d Lie group

Yeah but

How do u do the structure group

I mean I can picture $(R, S, Z_2)$

But how do you picture $(R, S, S)$

4:45 PM

what is $(R,S,S)$ supposed to be?

Base space $\Bbb R$, fiber $U(1)$ with structure group $U(1)$

or something

Wait, is the structure group of principal bundles supposed to be the same as the fiber itself?

Or am I mistaken

I don't think you can really `see' how the structure group acts

it should be the same picture regardless of the structure group

@BalarkaSen is that right?

Isn't the mobius strip specifically one with $\Bbb Z_2$ as the structure group?

@Slereah Leaf space of a Reeb foliation? I think it's like a line where every point neighbors two points far away somewhere

(the boundary leaves)

@Slereah @0celo I think you can probably see the action of the structure group using monodromy?

IIRC the twisted torus also has a non-trivial group

4:52 PM

@BalarkaSen No, I mean if you just draw the bundle space you wouldn't be able to tell what the structure group is

wouldn't they just all look like the product space if that was true

locally a product space

and that's exactly what they are

Well yeah but the global part comes from the structure group

So

that's a regularity condition, not a pictorial condition...

@Slereah The edges give a $\mathbb{Z}/2\mathbb{Z}$-bundle over $S^1$

But the strip itself is a line bundle

@Slereah yes, in a principal bundle the fiber is isomorphic to the structure group

The crux of the whole construction being that the fiber lacks a distinguished identity element, though, so the isomorphism is not "natural"

4:59 PM

@ACuriousMind what is natural, anyway

@0celouvsky why'd you drop the 7?

@0celouvsky I don't feel like giving the category-theoretic construction that you wouldn't appreciate, anyway :P

because it didn't make sense any more

@ACuriousMind do you mean a natural transformation?

@0celouvsky Not exactly

there's something more natural??

5:03 PM

Without the 7 it's not cool anymore.

I'm not having an old woman tell me what's cool or not.

-_-

::changes username to 777::

you won't do it

no balls

@0celouvsky who are you? XD

what kind of question is that?

5:06 PM

@0celouvsky Ah, sorry, it does rather straightforwardly involve natural transformation. For instance, the natural isomorphism to the double-dual of finite-dimensional vector spaces is just the natural transformation between the identity functor and the double-dual functor evaluated on a certain space.

a guy from the internet

^

@0celouvsky idk who 0celouvsky is but I know 0celo7

@ACuriousMind we've been over this before, so I don't know why you don't think I'll understand it

where is he btw? haven't seen him around lately.

5:07 PM

@OBE then don't talk to me I guess

@0celouvsky I didn't say you wouldn't understand it! I said you would not appreciate it :P

@ACuriousMind category theory is a big part of analysis on Banach manifolds, I guarantee we'll talk about it more one day

5:38 PM

@dmckee Are you around?

Sorta.

I'm allegedly working.

5

@dmckee Quick C99 question

If I have a 2-D array foo[MAX][MAX]

I'm a native speaker of the 89 standard, but OK.

and then some function that takes in a 1-D array bar(foo[MAX])

I can pass the "2nd dimension" of the array to the function easily with bar(foo[x]) where x is some int

Is there a similarly easy way to pass a "1st dimension" of the array?

A "2D array" in c is more properly conceived of as an array of arrays.

5:41 PM

::nods::

What you do when you call bar(foo[0]) is formally passing the first element of bar.

Which is usually written as a row.

As I understand the question, you are asking about passing a column?

If so, then no. There is no easy way.

Yep, that's what I was asking

Sigh

Here I go allocating more memory

If you have to have that feature than you may want to implement the "2D array" in some other data structure.

I want to change this thing to some nice struct eventually

but I'm a lazy bastard right now

A 2D linked structure or a big hash-map where the index is formed by, say, concatenating the row and column values.

What's best depends on what other featuers you need the thing to have.

O(1) random access lookup or O(N) walking and so on.

5:45 PM

@dmckee Have you played the game 2048?

@BernardoMeurer If you can afford the memory then storing the 2D array in both row-major and column-major is an option.

@BernardoMeurer Yeah.

That's what I'm doing, in C with SDL

OK, so there is a trick code-golfers use.

the 2-D array in question hold the 5x5 board, and each cell has the index of 2^n of the square in there

where 0 means no square

Rather than passing a row or a column, do this....

* Store the cells in a 1D array.

* You'll have to do the row-column math yourself, but that is easy.

* Pass an element and a skip-distance.

5:47 PM

    for (int row = 0; row < MAX_BOARD_POS; ++row){
        int buf[MAX_BOARD_POS] = {0};
        for(int i = 0; i < MAX_BOARD_POS; ++i) buf[i] = _board[i][row];
        collapse(buf, false);
        for(int i = 0; i < MAX_BOARD_POS; ++i) _board[i][row] = buf[i];
    }

* Make the skip distance 1 to walk a row.

I did that

* Make it row-length (i.e. 5) to walk a column.

Not exactly a good solution lol

JR told me to use a 1-D array too, but it seemed unnecessary

There is, of course, some fussing stuff around the limits.

And you have to decide if you want to walk it circularly (use modular arithmetic) or not.

@BernardoMeurer Well, it's mess, but you should encapsulate all that in a set of access functions to that you just present a opaque object to the rest of the program.

5:51 PM

@dmckee Oh, that is inside a function

If you have a GH account I can give you access to the repo

Much as I'd like to play with this notion, I'm afraid I have too much on my plate right now.

@dmckee LAAAAAAME

:)

Thanks for the help, I'll use the crappy for loops for now, probably move onto something better once basic functionality works

^ That's the way to "just get it working".

Is that a good thing or a bad thing?

DanielSank

Hi, everybody.

5:58 PM

@DanielSank <3

DanielSank

<3

@0celouvsky </3

@BernardoMeurer It's a good thing for turning in assignments, and may be a good or bad thing for maintenance depending on how messy the code is. It's a bad thing from the point of view of elegance and is often less than optimal from a performance point of view.

@dmckee You had me at the first sentence

@DanielSank :( what

DanielSank

@0celouvsky I like teasing you.

6:11 PM

@dmckee What naming convention people use in C?

for like functions and variables and so on

All of them. Depends who you ask.

I'm thinking camelCase for functions and under_scores for variables

@DanielSank hmm

I loathe underscores.

@dmckee :(

I like it

They're nice to read

6:13 PM

I also find that I rarely need multi-word variable names if (a) I've partitioned the code into enough functions and (b) I think carefully before choosing them.

So I use all-lower for variables and camelCase for functions.

@dmckee I have board_size and square_size for example

I also use a prefixedCamelCase when I'm writing an encapsulated objects to make up for the lack of namespaces.

and some #defines like MAX_BOARD_POS

I have to say that namespace encapsulation is something that c++ does much better than c.

::nods::