Hi

hi

How are you @3075?

Why is your name 3075?

I'm fine, and my name is 3075 because I like those numbers in that specific order.

It appears blue to me.

You have synthsomethingsomething?

yeah.

Synesthesia?

Is that the word?

yeah.

Do you like having it?

no.

well idk

because I ocd over blue things.

and I refrain from using certain words or numbers.

makes my diction weird.

So why is your avatar blue, then? :P

because I only like blue xD

includes blue words and blue numbers.

Ah you aren't avoiding blue, you're seeking it out

yeah.

I avoid other colours.

makes life really difficult.

You don't like red?

no.

only blue, and idm neutral colours.

That must be annoying to deal with.

yeah.

@Danu <3

hi Daniel :)

Que pasa?

@0celo7 Embrace it.

@ACuriousMind That's a thing? I didn't know auto-bans were triggered in all chat flagging cases. Huh.

@DanielSank huh?

@BernardMeurer If I have an __init__ file that imports submodules, how do I reload everything in the package?

What do you mean? I get the situation just not the question

@BernardMeurer I mean, how do I reload all of the submodules?

@DanielSank Hey, Sorry, I was out

@DanielSank See if this does the trick

idk if it'll work

@BernardMeurer Decided to just not use import *. I never use it except in packages, but even there it's just such a PITA.

Import * is a bad practice, but hey in case you need it someday there's the function :p

@BernardMeurer It's not terrible in an __init__ file of a package.

It's just annoying if you want to reload.

Some times you want to put a bunch of stuff in one name-space, but it's more than you want in a single file.

I admit this is rare.

Can someone help me with a physics question? I don't know how to do it

@notorious Hi Notorius. Have you tried feeding $N\psi$ into the Schrodinger equation to see what happens?

The N's cancel?

What about the part "How do you determine the specific value for N and why?"

an aside: i finally understand why this chat is called h-bar lol

@notorious have you done the calculation?

Yes I think so. The N's just cancel out and you're left with the original equation

@notorious Correct :-). I don't understand the second part of the question. If you require that the solutions be normalised then $N$ can only be unity otherwise the wavefunction will integrate to $N^2$ not 1.

If you drop the requirement that the wavefunctions be normalised then $N$ can be anything.

ok that makes sense. thank you

...so I have to do a similar problem but I can't figure it out. Am I really that bad at math?

Errr because the Schrodinger equation is linear?

Hm

What is the double cover of a simply connected group?

Is it like two copies of the group

Also if $x$ is in the Clifford group, where does $-x v x^{-1} = v - 2 \frac{v \cdot x}{x \cdot x} x$ come from

@notorious Just like last time, feed the wavefunction $N_1\psi_1 + N_2\psi_2$ into the Schrodinger equation and see what happens.

@Slereah That's always a possible n-fold cover to just take n disjoint copies. But it need not be the only one. Consider e.g. $\mathbb{C}\to\mathbb{C}, z\mapsto z^2$. It exhibits the complex plane as a double cover of itself.

Hm

In particular, "n-fold covering" is a property of a map, not of a space.

Only "universal cover" uniquely defines a space.

Any clue for the Clifford group thing?

The $\cdot$ is the inner product?

I guess I should start from like $-xvx^{-1} = -(x\cdot v) x^{-1} - (x \wedge v) x^{-1}$

Not sure where to go after that though

yes

although

No wait, I think it is

But the inner product of the algebra

What's the $\wedge$, though?

So only $g(,)$ for grade 1 polyvectors

Oh right, wedge is only for n-forms

Not sure what the general formula is for polyvectors

I kinda hate Clifford algebras because there are way too many products and endomorphisms

It's a bit hard to keep track

and subalgebras

(because exams will obviously not accept the geometric solution since we are not supposed to do higher dimensional problems on a oblique projection)

also apparently you can do fermions without the Dirac algebra

And directly with a spacetime algebra

I wonder if it gives much different results

What does it mean to "do" fermions?

well you can write the Dirac equation without the Dirac algebra

as a real theory

...did you just discover Majorana fermions? :P

Not sure!

Are Majorana fermions just spinors with a spacetime algebra?

Being Majorana is just a reality condition on the spinor, so you wouldn't have complex numbers floating around.

Hm, could be then

Majorana fermions are possible in 0,1,2,3,4 mod 8 dimensions, but not generally.

After all the elements of the algebra are the same, pretty much

It's just that it's a field over reals instead of complex numbers

Can't really see anything that just plainly states "Majorana fermions can be done with a spacetime algebra" tho

So I dunno

I don't know what "can be done with a spacetime algebra" means

In the Majorana case, you can find a real subrepresentation of the Clifford algebra

I guess that is correct then

Consequently, this real subrepresentation also forms a representation of the real subalgebra of the Clifford algebra, which is what I think you mean by spacetime algebra

No idea for that $-xvx^{-1}$ business by the way?

Nope

Well the real algebra is the Clifford algebra :p

$Cl(3,1)$ is the Clifford algebra, also called the spacetime algebra

$\Bbb Cl(3,1)$ is the complex clifford algebra, also called the Dirac algebra

books.google.fr/…

someday I should just do a big list of morphisms

I can never remember which is which

@ACuriousMind why

@JohnRennie I'm sure you meant that $N$ is a phase, not unity.

@notorious See ^

@Slereah wtf are you doing

Regarding what

that clifford shit

@Slereah

I am doing some Clifford shit

(because spinors)

why

@DanielSank those nickel ball videos are dope

Well

@Slereah are there noncompact 2-surfaces with everywhere positive curvature

I was wondering what kind of interacting field might violate the quantum inequalities $\rightarrow$ I investigated what kind of interacting field I could extract the stress energy tensor from $\rightarrow$ I tried finding integrable quantum models since it can't be done perturbatively

$\rightarrow$ I had some trouble to find it in dimension $> 2$, even though I know all quantum fields obey the QI in 2D $\rightarrow$ what the hell I'll still do it in 2D, for now, to see how to do a quantum integrable model seriously $\rightarrow$ I investigated the Thirring model since it's the most famous $\rightarrow$ wondering how to get spinors in 2D $\rightarrow$ can't do it via the universal cover of the Lorentz group since it's simply connected

$\rightarrow$ have to do it with Clifford algebras

Also not a clue

I am currently trying to catch a Groudon with good IVs

are there any solve QFT in dimension > 2 I wonder

I know a few are well defined, but are any solved to a decent degree

I know a few that are purported to be somewhat, but they tend to be a bit vague on what "solved" means

@ACuriousMind is "closed form solution" even well-defined?

@0celo7 Believe it or not, it is (I think by convention).

It means the thing you're looking for is on only one side of the equals sign

@DanielSank I think that's "explicit solution"

I think some people also additionally mean that the other side of the equals sign has only elementary functions.

But that's not what most people mean, I think.

@0celo7 oh ok ::shrugs::

@DanielSank yeah

@DanielSank If you got something like $$\int_0^t\mathrm dx\int_0^x\frac{\mathrm d u}{\sqrt{\tanh \cos u}}$$ would you consider it "closed form"?

probably not because it has integrals

+100000 internet points if you can calculate that integral

@0celo7 Yeah that's why I was pointing out the thing about elementary functions.

People seem happy if their integral can be written as some named function.

@0celo7 oops yes.

Which idiot decided to give Groudon rest

this makes catching this thing 10 times worse

@0celo7 It means whatever the person using it wants it to mean, in my experience

@ACuriousMind Awwww, it has some reasonable, albeit roughly defined, meaning.

@DanielSank Well, okay: It means that you can "solve" an equation $f(x)=g(x)$ as $x=h(x)$. And then everyone disagrees what exactly $h(x)$ must be such that this can be called "closed form"

@ACuriousMind Yes.

For instance, if $h$ contains the Lambert W function, is that "closed form"?

@ACuriousMind Which is almost exactly what I said a few comments up.

@ACuriousMind Probably, since that function has a name.

@ramsay That you are taking a limit $x\to 0$ for an equation valid between $1$ and $2$?!

It's the lower bound that's the problem. You can't say 1<x and then send x to 0.

@DanielSank But, well...the Lambert W function is defined to be a function that provides the solution to an equation that has no closed form solution.

Which is why I think the whole idea is a bit ill-defined

@ACuriousMind Ah.

see this equation $$\frac{1}{1+|x|}\leq \frac{e^x-1}{x}\leq 1+|x|(e-2)$$ this is true for $x\in [-1,1]-{0}$ but limit for this at $x=0$ is true why @acu

Rhyming "true" with "acu". Pure poetry.

@ACuriousMind x=h(x)?

$h$ is some expression.

the identity?

@ramsay Because in that case, 0 is a limit point of the interval on which the inequality holds.

should it not be like $x=h(t)$?

@0celo7 What? $h$ is just some function.

@ACuriousMind Hmm, why isn't it the identity

Are there more functions that return $x$ besides the identity?

What?

If $h(x)=x$, then $h=\mathrm{id}$ by definition

I don't see what you're trying to do

Oh

Indeed.

$h$ should just not depend on $x$.

It should depend on $f$ and $g$

@ACuriousMind seems logical! but does this mean sandwich theorem is not true for all inequalities?

@ramsay are all those functions continutious at $x=0$?

or however you spell that word

Okay, "closed form" is when $f(x)=g(x)$ is equivalent to $x = h[f,g]$ for $h$ a "nice" functional of $f$ and $g$. What's "nice" is up to the person using the expression "closed form".

$C^0$

@0celo7 no they are not "continuous" at $x=0$

@ramsay What?

@ramsay The squeeze theorem only works for continuous functions.

The sandwich theorem is true for all functions.

@ACuriousMind How can it be true for all functions

Limits do not exist for all functions

@0celo7 $\frac{e^x-1}{x}$ is not continuous but squeeze theorem works

@0celo7 It works for all functions for which its premises (the inequality and the existence of the limits) are true.

There is no need to assume continuity.

@ACuriousMind hmm, I forgot what continuity means

carry on

@ramsay: I don't know what your question is. The squeeze theorem has clear hypotheses (the inequality holds on an interval, and the limits of the upper/lower function towards a limit point of that interval exist). Whenever these hypotheses are true, you can apply it. If they are not true, you cannot apply it. What confuses you about that?

@ramsay what is your question

my question came all way from MSE

@ACuriousMind wait! what is limit point wikipedia gives hard definition

For an interval $(a,b)$, the limit points are simply $a$ and $b$ (and every point in the interval).

@ramsay $c$ is a limit point of a set $A$ if there is a sequence in $A$ converging to $c$, which never takes the value $c$

by the definition of ACM.....

0 is a limit point of that set

take the sequence $a_n=\frac{1}{n}$

@ramsay Note that my "definition" doesn't apply to $[-1,1]-0$ because that's not a single interval

the message i posted in it $x\in [-1,1]-\{0\}$

whatever

but 0 is excluded hence it is not a limit point

wrong

but I don't care

good luck

@ramsay How did you arrive at that conclusion?

you mean that inequality

No

then?

oh! sorry

I mean how did you conclude that "0 is excluded hence it is not a limit point"

No one said anything about limit points that would allow you to conclude that it is not "because it is excluded".

For an interval

$[-1,1]-0$ is not an interval

I like how even ACM ignores my correct comment

Hm, Sandwich theorem

As delicious as the Hamburber problem

0celo7 gave the correct definition for a general set.

@0celo7 What am I supposed to do with it? :P

@ACuriousMind tell this guy that I'm right and he's not

make me feel good about myself

lots of things, really

ok is this interval an $[-1,0) \cup (0,1]$

40 fucking pokeballs later...

@0celo7 I was getting to that :P

I bet it's not even the right nature now

WHY ARE THERE UNSKIPPABLE CUTSCENES IN POKEMON GAMES

@ACuriousMind Are the local vector fields a sheaf

More importantly why are you playing a pokemon game

Even though you are a grown man

probably the best RPG series

can i have a bounty start on MSE?

lol

Nature good, IVs decent.

@Slereah should I keep it

No, euthanize it

:(

@ACuriousMind Do you know of any noncompact manifold with everywhere positive curvature

De Sitter space

Riemannian

Hmm

Maximally symmetric $\Bbb R \times S^3$, then, maybe

I guess something like...yeah

How about in 2 dimensions

Dunno

I'll keep this Groudon

Not like I battle Ubers for a living, anyway

I got a bunch of lecture notes binded.

To read over the summer.

@DanielSank I'm a bit hardcore with this namespace stuff, I'm an advocate for it not being used 99% of the time

like using namespace std in C++; nope!

same with Python, import the package and refer to its methods as package.foo.bar()

It helps the code maintainability a lot

@Danu You! Riemann surface person!

@0celo7 Eh...

I'm not even able to attend any of the lectures, though I do go to the exercise sessions...

@BernardMeurer You may not realize that lots of projects you use do import * in their __init__ files.

@DanielSank I know that, which doesn't mean I have to do the same :)

It's just how I like to write my code, I'm not preaching

So... what about it @0celo7?

@BernardMeurer Scala is the only language I know of where the package system makes sense.

The package structure there is not constrained to be the same as the file system tree.

@DanielSank Doesn't Scala use Java-esque system?

@BernardMeurer Not at all.

In Java you have one class per file and the package tree is the same as the file system tree.

In scala you can decide what package your modules go in.

Hmm that sounds cool

I've been promising to pick up Scala for a while now

ended up picking C++ first

now I want to check Haskell, but maybe I'll skip to Scala

so we can be Scalamates

@Danu Can we classify the noncompact 2-surfaces?

I'm looking for a 2-surface with everywhere positive curvature.

@0celo7 as I've said a couple of times already, I dont know shit about diff. geometry, but would a paraboloid work?

@AccidentalFourierTransform Probably!

Can't believe I didn't think of that :/