The h Bar

0ßelö7

12:00 AM

screw this

Leaky Nun

12:26 AM

4

Level 1, Sherfield Building, Imperial College London

peterh

3

Q: Physics application of $SO(8)$ and Spin(8) triality

Triality is a relationship among three vector spaces. It describes those special features of the Dynkin diagram D4 and the associated Lie group Spin(8), the double cover of 8-dimensional rotation group SO(8). SO(8) is unique among the simple Lie groups in that its Dynkin diagram (below) (D4 unde...

It seems to me a very worthy question, but I have no idea, how to save it.

Anybody has?

Sid

3:36 AM

@LeakyNun that is obviously our headquarters

Prathyush Poduval

@LeakyNun seems like some of the British members have been quite busy :P

wonderich

3:58 AM

I had modified the question just to fit the policy. But it is up to them (those people who are in power). I am not optimistic about it. It just has some strange policy and atmosphere here and there. Physics is just physics. (Physics SE does not represent physics.)

IcE my bReaD

4:51 AM

@peterh let's be clear here, that question is more up @ACuriousMind alley than any of us in the chat, so if ACM decides that it should be put on hold because it is too broad, then he is probably right and there is no point in arguing unless you have a really specific (and reasonable) reason.

Sid

I would say these are some of the problems you face when questions are mod-hammered.

peterh

@IcEmybReaD ACM explained his close reason in a comment. I don't dispute this reason, I am asking how to fix the question.

IcE my bReaD

@peterh sorry, I thought you meant you wanted to save it as in, stop ACM from closing it while keeping it the same.

@Sid of course, but ACM has lots of experience with how this site works and also is very qualified in terms of the physics so his decision based on how that question is in its current form is most likely valid (even if you call it mod-hammering).

Sid

@IcEmybReaD Of course, I don't doubt the experience and qualifications of ACM. I simply say that these stuff will be common whenever there is mod-hammering around

peterh

@IcEmybReaD The close reason was that the question asks for an open-ended list. It is true. But I am personally very curious for an answer.

IcE my bReaD

5:01 AM

@Sid alright fair enough. I see what you're saying. you're looking for a more democratic process for each question. (even though the mods were elected democratically)

0ßelö7

hardly

IcE my bReaD

@peterh alright, that's fine i'm not disagreeing with you! as I said before I thought you meant something else.

also w/e I don't even know why I am commenting on this, it's none of my business, just bored before going to sleep.

@0ßelö7 lol, do you mean because the mods are forced to be elected?

i.e. there has to be mods and there is no minimum vote count OR participation in the election required

Sid

@IcEmybReaD We have a democratic process. Except, it is too slow.

peterh

@IcEmybReaD Not so bad, you are the only one reacted it until now

Sid

Mod-hammering is quicker and usually users don't have a problem with it

peterh

5:07 AM

I think, maybe the list seems open-ended, but so theoretical things aren't very widely used in the practice

This is why this question looks very interesting to me, and also this is why I do thing, maybe the list is open-ended, but probably it wouldn't be a very long list.

The goal would be to find the possible minimal change to the question, which already makes it answerable.

Of course it would be an essential change, thus it would require at least the consent of the OP. Ideally it would be done by the OP.

0ßelö7

5:23 AM

@IcEmybReaD I don't believe in "governments"

least of all the stack exchange governement

peterh

@0ßelö7 In the sense that you don't believe its existence, or that you don't agree it?

John Rennie

@0ßelö7: how did you cook the broccoli?

0ßelö7

@JohnRennie raw as a snack

John Rennie

Aha, yes, I remember the conversation about raw broccoli

It isn't one of my favourite snacks. On the flip side it is virtually calorie free so you can eat as much as you want.

IcE my bReaD

@0ßelö7 ideally how many citations should a very reputable and influential mathematician have by the time they pass the age requirement for a fields medal?

0ßelö7

5:27 AM

how the hell would I know

IcE my bReaD

just random

0ßelö7

@JohnRennie cheap too

$2 for 2 pounds

4 large thingies

IcE my bReaD

because I was comparing the citations of profs at my university.

0ßelö7

eye roll

IcE my bReaD

and I noticed that the citations didn't really correlate with how famous they were.

John Rennie

5:29 AM

I use broccoli a lot in risottos and I really like it. But I've never felt the urge to snack on raw broccoli.

0ßelö7

@JohnRennie I felt the urge. Must be the baby

John Rennie

:-)

IcE my bReaD

@0ßelö7 sorry I don't see why that deserves an eye roll.

3 mins ago, by IcE my bReaD

and I noticed that the citations didn't really correlate with how famous they were.

correction:

it only did in extreme cases, so those with an insane number of citations were really famous as you would expect, and those without many citations were not famous. but for the people closer to the average, there wasn't really a correlation.

user228700

7:31 AM

Hi, everyone :-)

John Rennie

Morning :-)

It's quiet in here right now ...

user228700

So I noticed :-)

John Rennie

It's 08:35 on Sunday morning in the UK. All the sensible physicists are still in bed!

user228700

Sensible, yes! :-)

John Rennie

The silly ones got up at 5 a.m. to check their servers :-)

(all of which are fine)

user228700

7:37 AM

:-) Good.

user228700

Very quiet here too. Haven't installed Python yet.

John Rennie

You could experiment with installing Python in the Linux subsystem if you're feeling bored.

user228700

No, I actually have a bunch of reading to do at the moment. I will get back to it after! :-)

John Rennie

University reading or leisure reading?

user228700

Both.

Sid

7:41 AM

@Kaumudi.H during holidays?

@JohnRennie On Sunday?

John Rennie

@Sid I'm so used to waking at 5 a.m. that I wake at that time even on a Sunday. And since I'm awake checking my servers today means less work tomorrow morning.

Sid

@JohnRennie Wow. The only reason I would be awake at 5 AM is because I am still awake at 5 AM

Anonymous

@Sid Same here. :D

user228700

@Sid Yep.

user228700

@Sid On working days, I wake up at 3 AM :-P

John Rennie

8:02 AM

@Kaumudi.H: is this of any interest?

Anonymous

12

A: Choosing the name of the chat room

I vote that we keep it as the h bar. It has the excellent property that it shares its name with the postgraduate bar back at my uni in London: As you can see, not only is it a very clever name, but it is also thought to be a unique and clever name by a bunch of different institutions (example,...

Leaky Nun

@Blue :o

Anonymous

It seems Emilio was at Imperial too. :P

Sid

Has anyone done that Logician and Hats thing?

Anonymous

@Sid Wut?

Sid

8:14 AM

nvm

Anonymous

"Logicians A, B and C each wear a hat with a positive integer on it such that the number on one hat is the sum of the numbers on the other two. They can see the numbers on the other two hats but not their own. They are given this information and asked in turn if they can identify their number. In the first round A, B and C each in turn say they don't know. In the second round A is first to go and states his number is 50. What numbers are on B and C?"

Sid

@Blue Hmm... interesting

Anonymous

It's easy tho

Leaky Nun

@Blue it's easy once you know the solution

not easy when you don't know

[I'm thinking about it]

Sid

@LeakyNun that. is quite appropriate.

Anonymous

8:21 AM

x+y=z...

Leaky Nun

@Blue please let me think about it

Anonymous

You can close the window and think :P

user228700

@JohnRennie What...

user228700

OK, I've added it to my list.

Leaky Nun

A doesn't know: B and C are not the same
B doesn't know: A is not twice of C
C doesn't know: A is not twice of B

@Blue am I right in the reasoning?

Sid

8:32 AM

Positive means non-zero, right?

Anonymous

@LeakyNun yes

Sid

@LeakyNun Or A is not equal to C. for B doesn't know. Similar for C doesn't know

Leaky Nun

oh right

I'm stupid

A doesn't know: B and C are not equal
B doesn't know: A is not twice of C, A is not equal to C
C doesn't know: A is not twice of B, A is not equal to B

Sid

Or that C is not twice of A. Or B is not twice of A.

Leaky Nun

@Sid how can you know that?

Prathyush Poduval

8:37 AM

I say we behead those logicians!! Off with their heads!

Sid

@LeakyNun Actually, yes. It's not necessary

Leaky Nun

I'm not getting anything with 50

if it were 33 or 100 then I would know

Prathyush Poduval

If one guy sees the other 2, why can’t he just sum it up and find his number?

Sid

@PrathyushPoduval BEcause only one hat contains the sum of the numbers on the other two hats

Leaky Nun

@PrathyushPoduval because it might as well be their difference

Prathyush Poduval

8:44 AM

@Sid ah okay. The question should have mentioned one and only one

Leaky Nun

@PrathyushPoduval how can there be two lmao

Anonymous

@PrathyushPoduval Even then your logic is incorrect. There can't be any triangle whose every side is sum of other two sides.

Prathyush Poduval

@Blue 000

Leaky Nun

@PrathyushPoduval positive

Prathyush Poduval

I shall think on this while eating

Lunch

@LeakyNun I would have concluded question is wrong :P

Anonymous

8:55 AM

Anyone needs a hint?

Prathyush Poduval

I guess onenumbers is 1

Sid

A=50, B=20, C=30.

(i confess I had to back-solve that)

Physics Meta

1

Q: What does the little rhombus behind the name mean?

Initially, I thought a little rhombus behind the names is a crown of topper in reputation. Then, I thought it signifies the site administrator Thirdly, I thought it is written by the user himself :-) which one is right?

discussion support

Sid

9:12 AM

Yeah, that seems to work.

However, I don't have a rigorous proof for why that is correct. Maybe, I should try to find a proof for that

Anonymous

@PrathyushPoduval Duh

Anonymous

@Sid Think what C would think in the first round. Could $3A=2B$ ?

Sid

Like I said, I need to find a proof for this and a general algorithm hopefully.

I am fully sure my answer is correct.

Anonymous

$C$ knows $A\neq B$, $2A\neq B$ and $2B\neq A$.

Anonymous

Any linear sum of those inequalities also hold true.

Anonymous

9:21 AM

@Sid It's not possible to come up with 20,30 unless you know the algorithm or you saw the answer somewhere...

Anonymous

Or you were randomly putting in numbers

Sid

@Blue Nope. Randomly putting in numbers and one worked

:P

Anonymous

I advise writing down everything in form of equations.

Sid

Let k belongs to N.

A says No: 2k,k,k ruled out.
B says no: k,2k,k and 2k,3k,k ruled out.
C says no: k,k,2k , 2k,k,3k and k,2k,3k and 2k,3k,5k ruled out.

Anonymous

9:41 AM

@Sid Say why is B=1 and C=49 not permissible?

Sid

I don't think A can find out whether his number is the sum or difference of B's or C's for that case.

after the first round, that is.

Anonymous

Duh. I already gave you the hint.

Anonymous

31 mins ago, by Blue

$C$ knows $A\neq B$, $2A\neq B$ and $2B\neq A$.

Prathyush Poduval

@Sid kkk is a possibility

Sid

@PrathyushPoduval what?

Prathyush Poduval

9:51 AM

KKK, google it

Sid

Ku klux Klan isn't something I would bring up in an international chat forum. :P

Anonymous

$50/3$ is not an integer.

Sid

Yeah. I don't see your point

Obviously, for A=50, it can be expressed only as k, 2k, 5k,10k,25k,50k.

Prathyush Poduval

I meant it as a bad joke😅

Anonymous

My vector analysis professor gives stupid homework: $\vec{r}=e^{-t} \hat{i} + \ln(t^2+1) \hat{j} - \tan(t) \hat{k}$. I am supposed to find $[\frac{d\vec{r}}{dt}\space\frac{d^2\vec{r}}{dt^2}\space \frac{d^3\vec{r}}{dt^3}]$. Meh.

Anonymous

10:02 AM

I should skip such problems

JiK

11:36 AM

physics.stackexchange.com/questions/360211/… The question wrongly states that $x^2+y^2=l^2$ implies $x=\sqrt{l^2-y^2}$ and asks whether you can use only $y$ as the generalized coordinate for a pendulum. All except one answer seems to ignore (a) the actual question, and perhaps even more importantly, (b) the error that lead to the actual question. So now there's 4 upvoted answers that don't address the question.

What should really be done in these cases? I decided to go the ugly brute force way and downvote and comment on those answers, but that feels suboptimal and impolite...

Prathyush Poduval

@Blue good for you, it means you don't need to learn for theexam and can do what you want :P

JiK

@Blue I think usually the point of such questions is that it's then easy to see if the student has understood the point of elementwise derivative wrt time, but wow, I don't think even finding the third derivative of $\ln(t^2+1)$ is good homework, it's just testing your brute force skills.

Anonymous

@PrathyushPoduval Ya....that's what I'll do

Sir Cumference

1:19 PM

Morning

Anonymous

Morning. How's life going? :D

heather

morning

my sleep schedule is very messed up, but I think i've got it back on track now.

Anonymous

@heather I bet it is not as bad as me. I wake up in the afternoon. :'Dd

Anonymous

Something like 1 pm

Anonymous

And go to sleep at 6 am

heather

1:23 PM

oh, yep, not nearly as bad.

Prathyush Poduval

1:39 PM

I'm more of a daytime sleeper

during school days, I take lot of naps in class, and work in the night

Anonymous

@PrathyushPoduval Same here

Anonymous

I don't like daytime

Prathyush Poduval

I like daytime, but i don't like class

on holidays, I sleep properly

Anonymous

No one likes class

Prathyush Poduval

that depends on the class

hopefully next year, things are gonna be much better

Anonymous

1:41 PM

Yeah, but interesting classes are rare

Prathyush Poduval

there are 0 interesting classes in school

I'm attending only the 75% required

Anonymous

@PrathyushPoduval Uni classes are mostly boring everywhere. You might luckily get 2 or 3 good professors though. The problem with the top uni professors is that they are far more interested in research and don't care about teaching well, much.

Prathyush Poduval

Yeah, hence the "hopefully" :P

though what i've heard is many of the 1st and 2nd year prof's at iisc is quite good

Anonymous

You can get into real research work quickly, at IISc. Some awesome physics research goes on there all the time.

Anonymous

Try to develop good relations with the top researchers

Prathyush Poduval

1:45 PM

yeah, there are also cumpolsary projects in the summer to continue getting the fellowship

@Blue Yeah I will

Anonymous

@PrathyushPoduval What fellowship?

Prathyush Poduval

@Blue KVPY

Anonymous

Oh, I see

Prathyush Poduval

I might get a taste of their teaching this dec, during the kvpy camp in iisc

Anonymous

Are you gonna appear for KVPY this year, too?

Prathyush Poduval

1:46 PM

yeah, but not gonna prepare much

chemistry is the only trouble, which i'll do in the last few days

Anonymous

@PrathyushPoduval Yeah, true. Good luck with that.

Prathyush Poduval

@BalarkaSen

My Left Ting Just Went SKRRRA

►

Just ketchup no sooce

epic style

Balarka Sen

2:12 PM

@PrathyushPoduval the ting memes are stealing the show nowadays

0ßelö7

@BalarkaSen lookie i.gyazo.com/afc610b415aa43027cf018bbc0923b68.png

Balarka Sen

why are you rickrolling me with algebra

0ßelö7

@BalarkaSen I think I want to be an algebraist

Balarka Sen

Write $M_n(k) \times k^n$ you hipster

0ßelö7

@BalarkaSen oh that's what I meant

oops

@BalarkaSen does the proof look good otherwise?

Balarka Sen

2:26 PM

i think so

0ßelö7

sites.math.washington.edu/~pollack/research/…

ayyyyyyyyyy

lecture notes from a course Schoen taught on scalar curvature

Balarka Sen

oh oh

I remember the Kazdan-Warner theorem

0ßelö7

@BalarkaSen I want to do a presentation on it

lots of analysis

but I'm stuck with this impossible one

@BalarkaSen if you see positive mass, shout out

I'm looking but could miss it

Balarka Sen

Ok

Prathyush Poduval

@BalarkaSen yeah, have been seeing many of it

His sooce has swag

Balarka Sen

2:41 PM

@PrathyushPoduval I think it's a bit overrated personally

when you already have guys like MC Mental and 7K Tha God...

Prathyush Poduval

I like that quote, gives me the feels when I hear it

0ßelö7

@BalarkaSen someone needs to make this into a book

Slereah

4:23 PM

Articles on my site I tend to say "It's hard to define exactly" for weird GR things

Damn spacetime constructs and their vagueness

vzn

4:38 PM

in theory salon, yesterday, by vzn

Quantum Weirdness Once Again Shows We're Not Living in a Computer Simulation / sciencealert

Anonymous

@BalarkaSen How does Laplacian being $0$ at a point imply that the value of the function at that point is approximately average of the neighboring points? They mention that in a Khan Academy video. I'm not sure how they concluded that.

Anonymous

They are considering analogy with second order derivatives...

Anonymous

Harmonic Functions

►

Balarka Sen

@Blue Non-obvious

Anonymous

Is there some big derivation for that stuff?

Balarka Sen

4:41 PM

This is called the mean value theorem for harmonic functions

0ßelö7

It's called the mean value theorem for harmonic functions

Balarka Sen

@0ßelö7 #rekt

Anonymous

Harmonic function

In mathematics, mathematical physics and the theory of stochastic processes, a harmonic function is a twice continuously differentiable function f : U → R (where U is an open subset of Rn) which satisfies Laplace's equation, i.e. ∂ 2 f ∂ x 1 2 ...

Balarka Sen

I only know the proof for dimension 2

Which follows from Cauchy mean value theorem for complex functions

Anonymous

Woah...I will probably need hours to understand the proof

0ßelö7

4:42 PM

@Blue See Gilbarg and Trudinger "Elliptic Partial Differential Equations of Second Order" Ch. 2

Balarka Sen

@Blue The general theorem is a nontrivial PDE result

Anonymous

@0ßelö7 Okay, bookmarking it. :P

Anonymous

Thanks

0ßelö7

@BalarkaSen It's nontrivial but the proof is clever. Not hard.

Balarka Sen

I should learn it at some point

0ßelö7

4:44 PM

But again, Blue asked you and I had to give the answer. Am I supposed to not be upset?

Balarka Sen

lolol

Anonymous

@0ßelö7 Sorry :'D From next time I'll ask you every analysis doubt I face

bolbteppa

@Blue this is nice too math.stackexchange.com/a/50285/82615

0ßelö7

@Blue You need to first learn the Gauss-Green theorem.

It's one of the most important theorems in analysis.

Balarka Sen

@Blue I can tell you the proof for 2 dimensional harmonic functions if you want.

0ßelö7

4:47 PM

@BalarkaSen does that proof use complex analysis?

Balarka Sen

Yes

It's a direct consequence of Cauchy MVT

0ßelö7

That's too far into Conway, I never got there

@Blue If you want an actual PDE book, Evans is better. He has the proof too.

The section on harmonic functions is elementary and he has a review of everything in the appendices.

Balarka Sen

@0ßelö7 Conway? Stein-Shakarchi is the book for an analyst like you

0ßelö7

Conway is a functional analyst

I live and die by his FA book, which references his CA book

@BalarkaSen American Conway, not British Conway.

Balarka Sen

Ah-hah

0ßelö7

4:54 PM

@BalarkaSen Li and Schoen got a mean value theorem on complete manifolds

it has a ridiculous looking constant

$$\sup_{x\in B_p(\rho)}f^2\le C_3(1+\exp(C_4\rho \sqrt{R}))V_p(4\rho)^{-1}\int_{B_p(4\rho)}f^2$$

Balarka Sen

Sid

@BalarkaSen You need to work on those eyebrows. :P

0ßelö7

If the Ricci curvature is nonegative it looks a lot better

Balarka Sen

@Sid It's made by a friend. The eyebrows are fine

It's thonking

give it a break

0ßelö7

@BalarkaSen My advisor worked with Schoen and Yau back in the day. Should I see if he can get them send me the proof .-.

Balarka Sen

5:03 PM

lol

worth a try

0ßelö7

I have to believe someone has written up the proof since 1996

that's when the last big survey was

I might be missing something

@BalarkaSen have I actually told

holy fuck

My phone is crazy

@BalarkaSen have I actually tolf

:(

how is this happening

It just sends!

Balarka Sen

you have tolf

0ßelö7

:(

@BalarkaSen do you actually know what the positive mass theorem is?

There we go

Balarka Sen

no

0ßelö7

5:20 PM

@BalarkaSen it says that if the scalar curvature is nonnegative, then a certain asymptotic integral is nonnegative. The manifold needs to be asymptotically Euclidean and that's a pretty technical condition

Balarka Sen

ok

0ßelö7

It turns out that a certain estimate in the Yamabe problem requires on a quantity being nonnegative

And that quantity is exactly that integral (the "mass") on a related manifold

The second part says that if the mass is zero, the manifold is actually flat R^n

Balarka Sen

thats pretty cool

0ßelö7

@BalarkaSen Witten, Parker, and Taubes have a good proof for any dimension on spin manifolds

its something you should learn

Slereah

The Harambe problem you say

Balarka Sen

5:35 PM

lmao

0ßelö7

6:20 PM

@BalarkaSen what is TQFT about

Balarka Sen

it's a formal way to say "bordism-invariants associated to manifolds"

ie you can distinguish non-cobordant manifolds using them

0ßelö7

@BalarkaSen is that only of interest to topologists?

Balarka Sen

vOv

it's got the name "QFT" in it

0ßelö7

neither of us are physicists, so that doesn't mean much

Balarka Sen

tru

0ßelö7

6:24 PM

@BalarkaSen there's more algebra

I think I should give up.

the next one involves irreducible polynomials, which I have no clue about

@BalarkaSen I think I'll give a counterexample for when the Yamabe theorem fails (in the noncompact case)

hope it's not too hard

Mithrandir24601

Reading through one or two MSE meta posts just made me realise that I'm definitely very glad I did physics/CS and not maths

Balarka Sen

normie

0ßelö7

@Mithrandir24601 why? links?

Mithrandir24601

@0ßelö7 One thing is that I'm rolling my eyes that this is unacceptable as it's supposedly a "circular proof"

Balarka Sen

It indeed is depending on what your definition of $\sin$ and $\cos$ are.

0ßelö7

6:39 PM

...that is very likely circular

Balarka Sen

The point is in deriving $d/dx \sin(x) = \cos(x)$ you need that limit.

Unless, of course, you have some smartass definition of sin and cos

0ßelö7

The correct definition of sine is via a series, and then one can compute the derivative termwise because the formal derivative series converges uniformly.

Balarka Sen

Like solutions of $y'' = y$ with the appropriate initial conditions

0ßelö7

@BalarkaSen I consider that "smart" not "smartass"

Balarka Sen

@0ßelö7 yeah yours is smartass

0ßelö7

6:40 PM

But I'm a smartass so what so I know

Mithrandir24601

@BalarkaSen Yeah, that's fine. At some point in my life, I have probably proved that. If I want to do it now, I already know that $d\sin x/dx = \cos x$

0ßelö7

"proved"

Maybe some bullshit geometric proof

But fact is the only Bourbaki definition is via a series

Balarka Sen

@Mithrandir24601 That's fine. But to use L'Hopital to compute that limit is, regardless of the circularity issue, indeed a big nuke.

If you take $\sin(x)$ to be defined completely trigonometrically, it's a simple pictorial argument.

0ßelö7

>.>

Mithrandir24601

@BalarkaSen And? So what if it's a massive shortcut, using something that's already known? <- this is my viewpoint. Evidently, I do not care about maths in the same way as a mathematician

0ßelö7

6:46 PM

I have no idea what you're even saying

CooperCape

'Cause what's the point in proving it using something that requires you to prove it?

0ßelö7

^ this

Balarka Sen

@Mithrandir Nuking simple exercises does not really reveal anything about the exercise, except perhaps your knowledge of the nuclear bombs, is my viewpoint

0ßelö7

@BalarkaSen what's the biggest nuke?

Computing Euler characteristics of polygons using Atiyah-Singer?

Actually that seems pretty hard

Mithrandir24601

@CooperCape My point is that $d\sin x/dx = \cos x$ is an already known result. the limit of $\sin x/x$ may be part of that result, sure, but that doesn't make it invalid. I mean, do you start every proof with the fundamental axioms?

0ßelö7

6:49 PM

No, but you should make sure that your proofs aren't circular

I mean, what are you even arguing here?

Balarka Sen

@Mithrandir For a mathematical coherent proof, you have to make sure there's a proof of $(\sin(x))' = \cos(x)$ which does NOT use that limit exercise.

If you don't care about giving mathematical coherent proofs, I mean that's fine. But that's how basic logic works.

theDoctor

okay, let's cut through all the bullshit

Balarka Sen

Until and unless you have made sure there's such a proof, you have merely observed $(\sin(x))' = \cos(x)$ implies $\lim_{x \to 0} \sin(x)/x = 1$, not proved $\lim_{x \to 0} \sin(x)/x = 1$.

theDoctor

SPACE: Is it real?

Has anyone touched space or really SEEN it?

0ßelö7

@BalarkaSen just like all these Yamabe papers basing everything off of the unproven $n=3,4,5$ positive mass theorems...

theDoctor

6:53 PM

cut the crap

0ßelö7

I see it right now

CooperCape

@Mithrandir24601 Yeah I guess it's a proof... just kinda requires that proof... and then that requires the proof. Absolutely wiiiiiiiilllld.

theDoctor

if space isn't real, then all of astrophyics and much of cosmology is unfounded.

Balarka Sen

lol

theDoctor

there's ZERO reason to suppose some 3+1 substructure to space

CooperCape

6:55 PM

Actually I do really like those pictures which have that picture within that picture etc. They kinda cool. Until you get to a point where image resolution is like 1 pixel (usually about 2 pictures in)...

Phase

Ok so you know how when working in finite vector spaces with quantum mechanics the algebra isn't guaranteed to be right, since [A,B] = ihI makes no sense due to the traces being cyclic, how is it that the riemann zeta function extended to -1 (-2 or -3 depending on dimensions) describes the Casimir effect?

0ßelö7

what

dmckee

Hey, @BenCrowell! Tag! You're it.

Balarka Sen

@0ßelö7 So any complex structure $J$ on a complex vector space $V$ decomposes it as $V = V^+ \oplus V^-$ where those are the eigenspaces of the eigenvalues $\lambda = i, -i$ of $J$.

Phase

Well the sum of natural numbers doesn't converge to -1/12

Mithrandir24601

6:57 PM

@CooperCape I get that, from a maths perspective, you need to prove it (without L'Hopital) first, but I don't get what's wrong with the idea of "OK, there's a different way to prove [x], which is used to prove [y]. We now know [y] is true, but I've forgotten what the result [x] was. I'll just use [y] because I don't want to go through that proof, or I just want an example of how this works"

Phase

but summation methods can redefine convergence and assign spooky values to divergent sums right?

CooperCape

actual @Phase I got the T-shirt for that so ummm I think you're wrong there.

0ßelö7

@Phase it converges in some bullshit sense

Phase

The T-shirt for what?

0ßelö7

@BalarkaSen yep

Phase

6:58 PM

I'm just confused as to why it works

Balarka Sen

So if you take a Riemann surface $X$, $TX$ decomposes as $T^+X \oplus T^-X$. I think $\partial/\partial z$ and $\partial/\partial \bar{z}$ are sections of those subbundles.

But I am not 100% sure

CooperCape

@Phase A T-shirt that says $$\sum_{n=1}^{\infty} n = -\frac{1}{12}$$

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