The h Bar

ACuriousMind

00:00

@0celo7 You should know by now that I don't care about that

0celo7

@ACuriousMind Yes, what a shame.

Danu

@ACuriousMind My brother does that.

ACuriousMind

@Danu It was not your brother I met :D

But that's nice

0celo7

I just wrote $\in\mathbb{Z}$ on engineering homework

Wonder if I'll get points deducted

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

Aren't you supposed to be in Greece right now?

On holiday or something

Danu

00:03

I am

ACuriousMind

I think they have internet in Greece, too.

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

:)

0celo7

@Danu what arrows

$f:V\to W$

like that?

$\rho_1:G\to\mathrm{GL}(V)$?

@ACuriousMind Are you too pissed at me to answer rep theory questions

ACuriousMind

@0celo7 No, but I don't want to think now, either

0celo7

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ why did you remove that

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

00:14

Because this is better:

in Mathematics, 18 mins ago, by Balarka Sen

And, also, draw pictures instead of thinking symbolically.

0celo7

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ How do I draw a $G$-submodule

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

Try.

ACuriousMind

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ That's a homotopy theorist talking, there are many things you can't draw a proper picture of in other contexts.

Danu

@ACuriousMind I would say a geometer

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

Feynman always tried :P

Danu

00:17

(as opposed to algebraist)

ACuriousMind

@Danu I know BalarkaSen is a fan of homotopy theory/category theory, but you're right, in general, that might be a geometer

0celo7

Ah! If $f:V\to W$ is the desired isomorphism of irreps and $v\in\operatorname{ker}f$, then $f(\rho_V(g)v)=\rho_W(g)f(v)=0$ for all $g$, so $\operatorname{ker}f$ is a submodule, i.e. $\rho_V$ is not irreducible.

And by rank-nullity $f$ is injective!

And surjective.

@ACuriousMind No, that's wrong, one can have a map without a kernel that is not bijective, right?

@ACuriousMind I'm having a tough time proving subjectivity of $f$. Does this work? If $w\in \operatorname{im}f$, then $f(\rho_V(g)v)=\rho_W(g)w$. So the subspace $\rho_W(G)\operatorname{im}f\subseteq\operatorname{im}f.$ But since $W$ is irreducible, we must have equality, implying $f$ is surjective.

No, that doesn't work.

Ah, I've shown that $\rho_W(G)\operatorname{im}f=\operatorname{im}f$, so $\operatorname{im}f$ is an invariant subspace. But $W$ contains no such subspaces, so $\operatorname{im}f=W$.

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ No pictures needed :)

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

00:43

:)

0celo7

Do you want me to try to draw a picture

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

do what works for you

0celo7

damn there's a part ii

What the heck is $\lambda$

Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ

As @ChrisWhite would say "a Greek letter" :D

0celo7

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ Well, it's clearly an eigenvalue of $f$.

So, let $v\in V$ s.t. $f(v)=\lambda v$. Let the space of such $v$s be $U_\lambda$.

Then $f(U_\lambda)=\rho_W(G)f(U_\lambda)$.

Ok, so $f(U_\lambda)$ is an invariant subspace. But such a thing cannot exist, so $f(U_\lambda)=W$.

But $f$ is a bijection, and all of its eigenvalues are the same. Thus $f=\lambda I$.

0celo7

01:26

@DavidZ You answered a question on the theorem of Stokes in GR here. Would you happen to know how to answer this question?

@DavidZ For some reason the "standard" method of obtaining the measure on a hypersurface in Riemannian geometry picks out the outward normal vector. This is apparently not true in GR. I can't figure out why.

@DavidZ Note that this result is "correct" in the sense that three major (one lesser known) textbooks in GR from different eras claim it.

Sigh...I might have found a book that contains a proof, but the libgen PDF cuts off half of it -.-

user507974

01:58

@0celo7 whats your opinion on libgen

0celo7

@user507974 I'll use to it look up one or two proofs in a book.

I have a university library, so it's no different than getting a scan from them.

user507974

@0celo7 your university would do scans?

0celo7

@user507974 yup

user507974

maybe im just not special enough as a lowly undergrad, are you united states?

0celo7

Yes.

user507974

02:05

private/public?

i wont get any more specific than that =3

0celo7

@user507974 look at my profile for my school

I'll give you my office number too if you want to visit

user507974

friendly

if i visit Tennessee i may take you up on that offer

0celo7

the library will scan entire books

they have a machine for it

user507974

@0celo7 that immediatelly makes me think two things

a) what machine is that that will scan a book automatically (unless you mean it prints the book automatically)
and
b) they dont have to put up with any copyright crap

0celo7

don't know, don't know

but I've seen it done

color and everything

user507974

02:15

so when you say scan you mean it prints a pdf

0celo7

they send you a pdf

user507974

thats useful

0celo7

@user507974 exactly, so I see nothing wrong with me using libgen

and I also see nothing wrong with having PDFs of books I own

user507974

02:31

@0celo7 agreed

vzn

02:50

@FenderLesPaul cool, thx for sharing; are you interested in that synergy? what do you know about it?

FenderLesPaul

03:16

@vzn yes very interested

it's my primary interest in fact

most of what I know about it comes from my research which is much more on the AdS/CFT side

problems such as how to characterize the computational complexity of entangled states in a CFT and whether or not this is actually dual to the geometry inside a black hole long after it thermalizes due to a perturbation

and how to find an appropriate measure of the information overlap between states that appropriately describes the long time chaotic behavior of a perturbation in a CFT

and what the dual bulk picture of this is

so far those are the only things I've worked on that involve said synergy

Bernard Meurer

04:01

@dmckee Are you around?

dmckee

Sorta.

Why?

Bernard Meurer

I have a simple bra-ket question

|V> - |V> = |0> and not |V> - |V> = 0 right?

dmckee

::darts glances around for fast escape routes::

Bernard Meurer

;-;

dmckee

I'm trying furiously to recall the deep meaning of addition and subtraction in that context.

Bernard Meurer

04:03

I think I'm right

But that shouldn't mean much

dmckee

For instance the state $\lvert 0 \rangle$ of the SHO is not a null state.

That is $ \lvert 0 \rangle \ne 0$.

Bernard Meurer

Thus the first one is correct because subtracting two kets will result in a null ket and not just a zero

since a zero wouldn't be representing a vector

dmckee

Well, what this tells me is that it has been too long since I actually worked quantum mechanics problems.

Bernard Meurer

Well it's the first time in my life tonight

reading Shankar

I'm breaking my quantum cherry

was that overboard?

*popping

0celo7

04:20

@BernardMeurer what the fuck

vzn

04:45

@FenderLesPaul that reminds me, there is some new stuff relating computational complexity to black holes, not sure what to make of it yet, seems a bit outlandish at the moment, have you heard it? eg arxiv.org/abs/1402.5674

user54412

06:10

@0celo7 I think I'm getting something here. That second form that @DavidZ seems to refute is completely standard in my field: $\int_\Sigma \partial_\mu(\sqrt{-g} V^\mu) \, \mathrm{d}^x = \int_{\partial\Sigma} n_\mu V^\mu \sqrt{-g} \, \mathrm{d}^{n-1}x$

user54412

It probably comes down to a case of what ordering is hidden in the implicit wedge product or something. The $\partial\Sigma$'s in the two equations (the above and the $\sqrt{\gamma}$ one are different in this way perhaps?

user54412

I think ACM alluded to this possibility.

user54412

Hmm. Consider a constant-$x^0$ (timelike coordinate) surface. The future-directed unit normal has components $n_\mu = (-\alpha, 0, 0, 0)$, $\alpha = (-g^{00})^{-1/2}$ being the lapse. So then we would have $n_\mu V^\mu = -\alpha V^0$ in $\int_{\partial\Sigma} n_\mu V^\mu \sqrt{\gamma} \, \mathrm{d}^{n-1}x$

user54412

Whereas the $\int_{\partial\Sigma} n_\mu V^\mu \sqrt{-g} \, \mathrm{d}^{n-1}x$ form works with $n_\mu V^\mu = V^0$ on this surface. The factor of $\alpha$ is exactly the difference between $\sqrt{-g}$ and $\sqrt{\gamma}$, but there is still a sign difference -- your sign difference.

user54412

(I'm imagining the "top" constant-$x^0$ surface of a box made from all constant-$x^\mu$ surfaces. If $V^\mu = \rho u^\mu$ this is the continuity equation.)

Stan Shunpike

06:46

Hola amigos

@0celo7 thoughts on the Life of Pablo?

0celo7

07:37

@StanShunpike just that I haven't listened to it

@ChrisWhite Hmm, maybe. I'll be bringing Lee to the library tomorrow and will not leave untill I have figured this out. I'll analyze your comment then.

DanielSank

@ChrisWhite How do I make a color plot in matplotlib that doesn't suck?

@BernardMeurer Good for you. That's a good book.

@BernardMeurer actually, you've got it backwards :)

The notation is tricky.

A vector minus itself is the zero element of the vector space, but that's almost never denoted $|0\rangle$.

Usually $|0\rangle$ denotes the ground state of a system.

user54412

07:56

@DanielSank That very much depends on what's making it suck in the first place ;)

DanielSank

08:18

@ChrisWhite I can't discern features.

Even with a log scale some of the stuff at low amplitudes is invisible.

David Z

09:33

@0celo7 no, I wouldn't. Maybe if I had access to my books I could figure it out.

@DanielSank post a picture, maybe we can figure it out

DanielSank

10:02

@DavidZ The problem is just that the default color scheme in matplotlib doesn't spread the dynamic range of the data over an equivalent dynamic range of human perception.

@ChrisWhite knows about this. I was hoping he could recommend color scales that do a better job than the default.

David Z

@DanielSank Use seaborn

or there are several other packages that do roughly the same thing, that just happens to be the one I use

I like seaborn.cubehelix_palette(...) but there are a number of options

user507974

@danielsank oh i havent worked on the mean steps question since i was busy but i realized two mistakes on the initial answer i gave you, i summed the series rep for all integers instead of just odd numbered and I didn't account for the case of the second to last step starting on the desired opposite vertex and jumping back and forth

just thoughts to oneself while cycling around

ACuriousMind

10:19

@BernardMeurer No, $\lvert V\rangle - \lvert V \rangle = 0$ is correct. Substracting a vector from itself gives zero. Not some vector that represents zero of some physical quantity (which would be $\lvert 0 \rangle$), but really the zero vector.

ACuriousMind

Is "numerical" a standard English word for "problem with numbers", or is that another Indian quirk like using "doubt" for "question"?

David Z

What's the context?

Without further information, "numerical" only means "having to do with numbers" as far as I know.

ACuriousMind

@DavidZ Sentences like "Can I use this in numericals?" or "I have this numerical: <problem text> Can you help me?"

David Z

Definitely not standard (American) English.

Well, at least to the best of my knowledge. "numerical" is an adjective, not a noun.

ACuriousMind

Like "doubt" for "question", I keep seeing it from users I think are Indian.

@DavidZ That's what I thought, too

David Z

10:30

@ACuriousMind Probably the same sort of thing. Maybe it is standard in Indian English, but I would edit it.

sharaf zaman

13:04

hey can a question about hybridization be posted on physics SE

0

Q: my question is about hybridization and a bit quantum mechanics?

I have read in many books that hybridisation is the intermixing of atomic orbitals to form molecular orbital of almost same energy. Hybridisation always takes place when orbitals does not have much energy difference like it can't take place between 2s and 5d Now I know hybridisation can t...

here is the question

David Z

I'd say no, but I'm not sure. That being said, definitely don't post it here now that you've already posted it on Chemistry.

sharaf zaman

but i think physicist are more intelligent than chemist

@DavidZ do you know the answer

David Z

Nope

ACuriousMind

@sharafzaman The only way that statement avoids being an insult is by being too absurd to be taken seriously.

sharaf zaman

@ACuriousMind sorry ! i didn't get you

Slereah

13:11

Guys

Does it actually make sense to say "Electrons distribute themselves on energy levels because of the Pauli exclusion principle"

ACuriousMind

@sharafzaman Saying "I think physicists are more intelligent than chemists" is an insult to chemists.

Slereah

i mean I can see that is true for some abstract isolated atom in an empty universe

But real atoms are going to be influenced by sources far away from them

Giving rise to very minute splitting of the energy levels

sharaf zaman

@ACuriousMind but i think it is true!

Slereah

If we only consider Pauli, they could all inhabit almost the same energy level

But they do not

I assume because of electron-electron interactions

sharaf zaman

@Slereah to whom you are giving answer

Slereah

13:14

Myself

I am wondering

ACuriousMind

@sharafzaman Intelligence may slightly correlate with your profession, but in general, you can find both very intelligent and very stupid people everywhere. "physicists" and "chemists" are not one faceless entity with one single level of intelligence.

@Slereah :O

You're right, I never thought about that.

"Why are not all hyperfine levels filled if I create hyperfine splitting?" is essentially what you're wondering, right?

That might be a good SE question!

sharaf zaman

@ACuriousMind i think there is no part of chemistry where there is logic to be applied because the inorganic and organic chemistry are full of reactions and no logic the only little of physical chemistry is for intelligents and rest organic and inorganic for those who love to memorize not learn

@ACuriousMind it has been half hour since i asked question no reply so far

ACuriousMind

@sharafzaman So? Half an hour is nothing.

Slereah

@ACuriousMind Basically

Although I guess it's because of electron-electron interactions

ACuriousMind

@sharafzaman Well, you may well think that. Just let me tell you that people do not appreciate being told their field of study doesn't require intelligence, and that you sound incredibly arrogant with what you're saying.

sharaf zaman

13:20

@ACuriousMind thats true ;)

skill patrol

so why say it?

sharaf zaman

@skillpatrol because of my high school where i have to read chemistry tough i like physics, maths and programming]

No answer so far!

skill patrol

it usually takes at least 24 hours to start

sharaf zaman

nah! my experience on physics and maths SE says it take 10-15 minutes

skill patrol

btw do you use "numerical" as a noun?

sharaf zaman

13:34

@skillpatrol No! i don't, but why?

skill patrol

just wondering :)

sharaf zaman

do you use it?

Slereah

0

Q: How important is the Pauli exclusion principle in the distribution of particles on energy levels

It is usually said that the Pauli exclusion principle is the big arbiter of how particles will distribute themselves along energy levels (especially electrons on atomic orbitals), but how accurate is this statement? It's rather easy to see for an abstract atom, floating in the void in an empty u...

quantum-mechanics atomic-physics pauli-exclusion-principle orbitals

there you go

sharaf zaman

@Slereah let me see your research ?

Slereah

wot

sharaf zaman

13:38

i mean let me see what answer you get

Slereah

Well it's on Stack Exchange

You can see it yourself!

sharaf zaman

ya! i know

who is this Ron Maimon, he has given many cool answer but reputation is still 1

1 hour, NO answer

skill patrol

14:14

Tomorrow (Monday) is a leap day, next leap day will be on ...... ? "using calculation only".

sharaf zaman

yess!!

what is difference between meta sites and normal ones

skill patrol

meta is for discussions about "policies"

meta means at a higher level :P

David Z

14:44

Well, meta is for questions about the site. They don't necessarily have to be policy questions.

0celo7

@DavidZ What do you mean by "if you had your books"

Unless you have a magic GR book, the answer is not in any of them :/

Slereah

let's make a petition to switch phi and varphi in latex

0celo7

I think this is one of those things that's in Hawking-Ellis and everyone repeats over the years, but doesn't prove.

Slereah

what thing

0celo7

The thing I've been trying to prove for a week

Where have you been

Slereah

14:52

Working

Well

0celo7

@ACuriousMind What's the canonical reason for why Tarkin can command Vader around in Ep. IV?

Slereah

Pretending to be working

The reason is because Vader wasn't the king of space in episode 4

He was just a random bad guy

0celo7

@Slereah that's a crappy reason

did they retcon his importance in V-VI?

Slereah

I'm not quite sure how important Vader was in episode 4

Do they ever state his rank or whatever

i only remember him being one of the bad guys

Not sure he was the protégé of the Emperor yet

Do they even mention the Emperor in ep IV

0celo7

@Slereah I...don't think so. Been a while since I watched the original trilogy.

Slereah

14:59

I mean IIRC Grand Moff Tarkin was the guy ordering Vader around

Alderaan's Destruction - A New Hope [1080p HD]

►

Holding Vader's leash

But nobody remembers Tarkin because he doesn't have a bucket on his head

0celo7

I feel like Leah was not distraught enough for the destruction of her home.

Slereah

I am kind of sad that the prequels never included Tarkin

Tarkin seems like an interesting guy

0celo7

The prequels needed to explain midichlorians.

They just replace one mystery with another.

Not a very good explanation if you ask me.

Slereah

Eh

Midichlorians didn't fit the tone of Star Wars much

0celo7

@Slereah Know any good references on complex GR?

Slereah

15:04

the force is just Magic

Penrose, I guess?

I don't know if he's good but he was the big guy on it

0celo7

@Slereah I'm saying if they want to explain it, the midichlorians were a shite way to do it.

Slereah

Maybe they planned to

0celo7

@Slereah His two volumes?

Slereah

But then noticed that it did not play out well in SW1

0celo7

How good is Penrose and Rindler anyway?

Slereah

15:05

Dunno

They don't even mention midichlorians outside of SW1, I think

0celo7

I think in Ep. 3.

Slereah

Oh yeah, maybe in the Palpy speech

0celo7

When Palpatine is talking to Anakin about Palagius.

^can't spell that shit

Slereah

Good old Palpy

0celo7

@Slereah When you use the divergence theorem on a timelike hypersurface, why must you take the inward pointing normal?

Slereah

15:08

iunno

0celo7

I guarantee it's because the induced metric $\iota^*g$ is Lorentzian. But I have no clue why.

Wow that might be the nicest picture explaining charts.

skill patrol

this from a guy who hates diagrams :P

0celo7

@Sᴋᴜʟʟᴘᴇᴛʀᴏʟ I always draw a picture when proving geometry things.

skill patrol

@0celo7 have you seen this one?

FenderLesPaul

@vzn that paper is what motivated my project

on using computational complexity to describe interior black hole geometry long after the thermalization time of the black hole

Bernard Meurer

15:23

@ACuriousMind But doesn't any operation in a vector space yield another element of that same space? And $0$ wouldn't be in a vector space, but the null vector $|0>$ obeying $|V> + |0> = |V> would$

FenderLesPaul

@vzn also if you're interested look up the arxiv papers of Patrick Hayden and Daniel Harlow

particularly Daniel Harlow

he mainly works on fundamental relationships between quantum info and black hole geometry

he has some papers on black holes as quantum error correcting codes

He was a student of Susskind's in fact, but he's a post-doc at Harvard now

@vzn arxiv.org/abs/1503.06237

Slereah

@BernardMeurer : use \vert and \rangle

Bernard Meurer

@Slereah Yeah I just went back to check on how ACM was doing it

FenderLesPaul

@vzn there are also some papers on using tensor networks to construct emergent space-time geometry

Bernard Meurer

there should be just a \ket

Slereah

15:28

$\ket{butt}$

nope

ACuriousMind

@BernardMeurer You usually don't write the null vector as $\lvert 0 \rangle$.

Bernard Meurer

@ACuriousMind Shankar does :(

ACuriousMind

$\lvert 0 \rangle$ should be a non-zero vector labeled by $0$ (e.g. because it is the eigenvector of an operator with eigenvalue $0$)

Bernard Meurer

Yeah now you lost me on the eigen stuff

ACuriousMind

It's pretty common to write the vacuum or ground state $\lvert 0 \rangle$ (e.g. of the harmonic oscillator).

The null or zero vector should really be just $0$. It isn't a state, and kets are supposed to correspond to states.

Bernard Meurer

15:35

That's why I got confused

FenderLesPaul

That's terrible notation

What book is that?

Bernard Meurer

Shankar

FenderLesPaul

Ew

Bernard Meurer

Principles of Quantum Mechanics 2nd Edition

FenderLesPaul

But yeah that's just really bad notation dude

Bernard Meurer

15:39

You people should create the notation to rule them all and make my life easier

Slereah

It is not a state because it cannot be normalized

ACuriousMind

Not only is it terrible notation, it's also terrible grammar :P (should be "exists", not "exist")

FenderLesPaul

@BernardMeurer is this your first time learning QM?

Bernard Meurer

@FenderLesPaul Uhum

FenderLesPaul

Have you thought about using a different book?

One with a lot more problems in it that you can work through

and that isn't as confusing

Bernard Meurer

15:45

@FenderLesPaul I just picked this one up, I'd be totally fine with using a different one

FenderLesPaul

Like Griffiths

Bernard Meurer

Getting it's pdf now

FenderLesPaul

Also any news from schools?

Bernard Meurer

Interview with UPenn was stellar, same for Minnesota

The answer for the latter should come around this week

1 rejection 9 TBD

FenderLesPaul

dope, best of luck :)

0celo7

15:51

@FenderLesPaul Where's the solutions?

Bernard Meurer

@FenderLesPaul Thanks man :)

FenderLesPaul

@0celo7 working on it I still have to make sure I'm not missing subtelties

0celo7

@ACuriousMind why is it terrible notation

@FenderLesPaul ok

ACuriousMind

@0celo7 I already wrote why.

0celo7

@ACuriousMind he has not gotten to the section on eigenvalues yet

@ACuriousMind and I'm telling you it makes sense if you read the book

ACuriousMind

15:53

18 mins ago, by ACuriousMind

It's pretty common to write the vacuum or ground state $\lvert 0 \rangle$ (e.g. of the harmonic oscillator).

0celo7

$|V\rangle=\vec V$ at this point

ACuriousMind

17 mins ago, by ACuriousMind

The null or zero vector should really be just $0$. It isn't a state, and kets are supposed to correspond to states.

0celo7

no eigenvalues

@ACuriousMind I can repeat myself too

Bernard Meurer

Aren't kets vectors?

0celo7

48 secs ago, by 0celo7

@ACuriousMind he has not gotten to the section on eigenvalues yet

43 secs ago, by 0celo7

@ACuriousMind and I'm telling you it makes sense if you read the book

ACuriousMind

15:53

@0celo7 None of my two statement there refers to eigenvalues

0celo7

He has not gotten to states

ok then he has not gotten to the section on states yet

the notation makes perfect sense if you read the book

later on he uses $0$ for the $0$ ket

so calm your tits

2

Slereah

kets are vectors yes

ACuriousMind

@0celo7 that's not what I call "perfect sense".

Slereah

In a Hilbert space

0celo7

@ACuriousMind orly

ACuriousMind

15:55

Introducing the notation $\lvert 0 \rangle$ for the zero vector and then switching to $0$ is silly; why not write $0$ from the beginning.

0celo7

@skillpatrol yes

@ACuriousMind don't know

@BernardMeurer is the first one to be confused

Slereah

$\vert 0 \rangle$ is kind of a bit overloaded with symbolisms

0celo7

that's not a good thing

Bernard Meurer

I'm confused because the axioms are now infused in my bloodstream

Slereah

To be fair though

You basically never use the zero vectors in QM

ACuriousMind

15:56

@Slereah coughcontinuouseigenketsdontlieinahilbertspacecough

Bernard Meurer

@ACuriousMind You're going to make me cry

0celo7

@ACuriousMind really?

Slereah

BUT THEY LIE ON THE EXTENSION TO DISTRIBUTIONS

so there :p

ACuriousMind

@BernardMeurer Ignore that last message

0celo7

make it rigged

ACuriousMind

15:57

@Slereah Which isn't a Hilbert space

Slereah

What is even a Hilbert space

Bernard Meurer

Am I on a Hilbert space?

Slereah

Nothing but a miserable pile of vectors

ACuriousMind

@Slereah Complete innner product space.

@BernardMeurer I have no idea what you're on ;)

Slereah

I know :p

Bernard Meurer

15:59

@ACuriousMind I'm smoking manifolds & hamiltonians

0celo7

@Slereah is "ge tit" something in French

Slereah

no

Bernard Meurer

nopé

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