The h Bar

user685252

19:01

The math room owner was banded for a casual mention of a user's "period."

Mark Mitchison

Anyway, I'm just pointing out that @user685252 jumped to some rather odd conclusions there.

0celo7

....

you have to be kidding me

the math room is ridiculous

ACuriousMind

Well, we're not the math room, and the more I hear about it, the less I want to go there :P

yuggib

we will complement here with all the math you need, don't worry

Jiminion

Portions of a hockey game.

user685252

19:02

The English language and usage room is the opposite end of the spectrum

TanMath

Hey @MarkMitchison !!!! long time no chat! nice to see you again!

Mark Mitchison

@TanMath Hi

How's it going

TanMath

@MarkMitchison how is your research?

Mark Mitchison

Coming along OK thanks. Had a paper accepted in NJP recently, now writing a follow-up experimental proposal.

ACuriousMind

@yuggib <3

TanMath

19:04

@MarkMitchison good, busy with college stuff... trying to get back to my quantum mechanics project and work on the simulations...

Mark Mitchison

Sounds good

yuggib

today I used two theorems of Bourbaki's topologie générale in a proof; I can give you all the maths you want :-D

TanMath

@MarkMitchison will you be here more often?

Mark Mitchison

@TanMath Depends, I'll dip in and out.

0celo7

@user685252 really

we're pretty cool in here

Mark Mitchison

19:06

@TanMath It's really just a procrastination tool.

user685252

@0celo7 yup

ACuriousMind

^^someone speaks the truth :P

TanMath

@MarkMitchison because, I would really like to talk to you soon... but right now, I gota study.. I will be back in 1.5 hours or so... will you be there by then?

Mark Mitchison

Yeah potentially, just ping me. I'll be working all night

....until I go to sleep, obviously

TanMath

@MarkMitchison ok then.. talk to you later...

bye!

Mark Mitchison

19:07

alright

yuggib

so by definition the night ends when you go to sleep

0celo7

@user685252 link?

Mark Mitchison

@yuggib Yes. Unless it's morning already

user685252

Go there :-)

Mark Mitchison

If you stay up for three days then the terminology "night" is pushing it a bit

user685252

19:09

English Language & Usage: Multi-Layer…

Not for the faint of heart or those easily triggered by Englis...

5

3

yuggib

:D

DanielSank

@yuggib Link to what?

ACuriousMind

@yuggib That is the correct definition. Just like "morning" is just after I wake up.

user685252

Biologically

yuggib

@DanielSank to your question...ACM provided promptly

Mark Mitchison

19:11

@DanielSank Did you ever find anything out about 3-body interactions in superconducting circuits?

yuggib

@ACuriousMind seems natural

Mark Mitchison

@DanielSank No worries if not. Actually I am pretty sure that such things can be engineered one way or another, particularly if you can couple, say, a superconducting qubit to 2 microwave resonators (which I feel certain has probably already been done).

user685252

Look at the graph of the activities for the room @0celo7

0celo7

@user685252 huh?

yuggib

@MarkMitchison sounds fancy

Mark Mitchison

19:14

@yuggib What does?

user685252

@0celo7 see the bars above?

0celo7

@user685252 yes

user685252

The blue spike is the current time

yuggib

that experimental setup, but just to me since I am on math

Mark Mitchison

ah ok :)

0celo7

19:15

wow, that chat is interesting

Mikhail

Hey guys, I'm a DSP person. How does this notation for temporal averaging in the contex of light look? I=<<U|U>>_t ? Is there some other way to write averaging in bra-ket notation?

ACuriousMind

@yuggib "I am on math" makes it sound like a drug

...which it perhaps is :D

yuggib

definitely is...it makes me flawed

Mark Mitchison

@yuggib Well I don't think it's so fancy actually, compared to what people can do. Experimentalists put in the hours :)

Although perhaps @DanielSank can correct me on that.

About what people can do, not the hours

user685252

It moves from left to right, so if you wanna get in on the action show up about 3 hours earlier than now @0celo7

0celo7

19:17

@user685252 ah

user685252

Caution be ready for troll lessons ;-)

0celo7

what

user685252

They are masters

With literally years of practice

0celo7

are you implying I'm a troll

because we've been over this

I'm not

Slereah

That's exactly what a troll would say

ACuriousMind

19:20

I'm wondering what exactly the nine stars on "screw you chat star person" indicate, btw.

Do they mean that our star wall is generally crap?

Are they amused about 0celo7's struggle with the astronomer?

Something else?

0celo7

Who starred it?

ACuriousMind

Both?

0celo7

Fess up

I would but can't

user685252

I did it.

Slereah

^the star person

0celo7

19:23

@user685252 then answer @ACuriousMind 's question pls

@ACuriousMind Did you star it?

ACuriousMind

@0celo7 Not all idle wondering must be answered, mind you

Sometimes, the mystery is more exciting than the answer

Slereah

Guys

What's a good feel for "locally lipschitz"

user685252

@0celo7 what question?

Slereah

I am looking for feels

From the definition it seems to be that it can't grow too fast, I guess?

Well vary too fast

ACuriousMind

@Slereah I'd say the feel is the cone picture on the Wikipedia page,

Slereah

19:26

"For a Lipschitz continuous function, there is a double cone (shown in white) whose vertex can be translated along the graph, so that the graph always remains entirely outside the cone."

Many feels indeed

user685252

4 mins ago, by 0celo7

@user685252 then answer @ACuriousMind 's question pls

Slereah

So basically it's a constraint on the derivative except it is also valid when it is not differentiable?

ACuriousMind

@Slereah Yes

user685252

What is he talking about @ACuriousMind?

ACuriousMind

@user685252 He refers to my question why people starred "screw you chat star person".

Slereah

19:28

Thx for the feels

ACuriousMind

I suspect he's more interested in the answer than I am, though

user685252

Because I felt like it. @@0celo7

0celo7

@ACuriousMind what

of course you're more important than a question

ACuriousMind

What I said.

Physics Meta

0

Q: How does copyright here work if I include some of my own posts basically verbatim in a textbook that will be published?

The question basically says it all. I am in the midst of writing a thermodynamics textbook, and there is a question I would like to answer that could be answered by pretty much just copying and pasting the relevant sections of the book. Is this a bad idea? Is it considered copyright infringement ...

discussion

0celo7

19:39

@ACuriousMind ok

@ACuriousMind Why is the mapping $\alpha:R\to R^2,t\mapsto(t^3,t^2)$ not an immersion?

Don't answer

I shall think about this

ACuriousMind

Wasn't going to, but why do you ask me at all if you don't want the answer?

0celo7

why weren't you going to

@ACuriousMind I figured I should think about it some more on my own

ACuriousMind

Because I was going to tell you to try to figure it out yourself first :P

0celo7

well I need to calculate the differential...which is just the thing

the derivative

maybe

John Duffield

@0celo7 What woo do you believe in? This woo.

0celo7

19:51

@JohnDuffield Lol

I think you fail to understand that @Slereah and I don't believe in time travel

We don't

We are interested in it because GR predicts it

(in certain cases)

We want to know why we haven't seen it, if there's some principle that forbids it, etc.

John Duffield

@0celo7 : GR doesn't predict it. If you think that you don't understand GR.

Maiels

Hello! Can someone help me with a quick question?

0celo7

@JohnDuffield Sure.

ACuriousMind

@Maiels Just ask, if someone wants to answer, they will

0celo7

@ACuriousMind So $\mathrm{d}\alpha=(3t^2,2t)$ is clearly not injective

is that all I need to show?

ACuriousMind

19:55

@0celo7 Yes. Why is it not injective?

Maiels

I am currently working on a project about Pendulums. It is about a so called "Mach Pendulum"or "Mach`s Pendulum". I have been searching info about this for about an hour online and I did not even find this name on an article, not even Wiki. Maybe it is called differently?

0celo7

$t^2$ is not injective

ACuriousMind

@0celo7 No, that's not the notion of injectivity that's meant by an immersion.

0celo7

@ACuriousMind ok, then I'm not sure what notion of injectivity they mean

ACuriousMind

The differential itself as a map $T_p \mathbb{R}\to T_{\alpha(p)} \mathbb{R}^2$ has to be injective for every $p\in\mathbb{R}$.

0celo7

19:57

well...didn't I just show that?

ACuriousMind

Saying "$t^2$ is not injective" says that the map $\mathbb{R} \to \mathrm{Maps}(\mathbb{R},\mathbb{R}^2), t\mapsto \mathrm{d}\alpha_t$ isn't injective.

But you want to say that $\mathrm{d}\alpha_t$ isn't injective for some $t$, not that the assignment $t\mapsto \mathrm{d}\alpha_t$ isn't injective. Do you see the difference?

This is a really common confusion, btw

John Duffield

@0celo7 : I kid ye not.

0celo7

So I need to calculate $\mathrm{d}\alpha_t(v),v\in T_t\mathbb{R}$?

ACuriousMind

@0celo7 Well, $\mathrm{d}\alpha_t$ is a matrix. When is a matrix not injective?

(You could do what you say, but there's an easier, more general way)

0celo7

@ACuriousMind Is it maximal rank?

ACuriousMind

20:01

@0celo7 Exactly, if it's not injective, it doesn't have maximal rank.

0celo7

Oh

At $t=0$ the things are linearly dependent

ACuriousMind

(And if it doesn't have maximal rank, it's not injective)

@0celo7 Exactly.

0celo7

Gotcha

Slereah

you know

0celo7

@ACuriousMind So an interval is not homeomorphic to two intervals in an X shape, right?

Slereah

20:09

The first time ever I saw the words "star shaped neighbourhood"

It was a friend who was writing some Warhammer flavor text

bc he was a math guy

0celo7

I'm trying to show this without waving hands

ACuriousMind

@Slereah We should call balls "death star shaped neighbourhoods".

Slereah

But @ACuriousMind

That's no star

0celo7

What is an open set on the X shape

ACuriousMind

@0celo7 Well, you need to define that in order for the sentence "an interval is not homeomorphic to two intervals in an X shape" to have any meaning

But whenever it's something you can embed in $\mathbb{R}^n$, if nothing else is said, it's just the subspace topology

0celo7

20:11

@ACuriousMind It's in the context of "embeddings have no self intersections"

@ACuriousMind So the intersection of the X and a ball in $R^2$?

ACuriousMind

Yep

open ball, of course

0celo7

yes

ACuriousMind

It's always annoying when you talk for minutes only to find out one person's balls were closed and the other one's open.

And yes I am fully aware that that sentence sounds weird.

0celo7

had to read it twice :P

Ok, I can't construct the homeomorphism but I'm not sure how to prove it can't be done.

ACuriousMind

There's a trick. :)

0celo7

20:16

I'm sure there is...

ACuriousMind

Genuine one, I'm not sure how many people come up with it themselves

Homeomorphisms preserve connectedness. Removing one point from the target space and its preimage from the source space thus need to make both spaces fall apart into equally many connected components

0celo7

ah

1 turns into 4 for the X

ACuriousMind

Exactly

0celo7

You know what my next question is, right?

ACuriousMind

sigh...you want a proof for the assertion with the components?

0celo7

20:18

You know me so well

I mean, I knew that, I think. But I'm making notes in my book of proofs now

planetmath.org/homeomorphismspreserveconnectedcomponents

ACuriousMind

It's not hard, fortunately: Take the preimage of every connected component in the target space. Every one of these has to be a connected component of the source, and since the morphism is a bijection, it has to be a different one for each.

So you need to have a connected component in the source for every of the target, and conversely every component in the source maps to one of the target, so you get a bijection between the connected components, meaning there are equally many of them.

0celo7

Every one of these has to be a connected component of the source

bolbteppa

" The point about homology is that it measures defects/obstructions. Suppose I have a map of vector spaces, $f:V\rightarrow W$. What is the obstruction that prevents that being an isomorphism? There is the kernel of $f$, stopping it being injective and the cokernel of $f$ stopping it being surjective. Homology measures defects. Snake lemma tells you how those defects are related, it tells you that if you were looking at n-dim holes, you next need to take into account the n-1 dim holes."

https://www.physicsforums.com/threads/exact-sequences.164213/
Why aren't math books writtten like this :\

0celo7

thinking about that

ACuriousMind

@0celo7 Well, the link you gave is the formal proof.

0celo7

20:26

I now need to show that since $f$ is continuous, $f(X_i)$ is connected...I will think about this.

DanielSank

@MarkMitchison I don't remember off-hand whether that exact experiment has been done, but it wouldn't be hard at all.

Andrew Houck (sp?) probably already did it.

Mark Mitchison

You mean an SC qubit coupled to 2 waveguides?

DanielSank

@MarkMitchison Actually, if you're interested I can put you in contact with just the right person (and as a bonus, he's actively looking for theory input).

@MarkMitchison Yes.

Mark Mitchison

@DanielSank That would be fantastic, since I'm currently looking for experimental "output", if you catch my drift.

As in, I want someone to do some experiments with the theory that we're working on

I don't know how interested in quantum heat engines your colleague will be though...

DanielSank

Probably very.

Mark Mitchison

20:36

OK. I think at microwave frequencies it might be possible to do some cool stuff; so far we've only looked at optical frequencies.

It would be great to speak to your friend in any case!

DanielSank

Done.

Check your inbox.

Pedram is pretty good at linking theoretical proposals to the capabilities of our system.

Mark Mitchison

Great, thanks a lot!

TanMath

Hello @MarkMitchison @DanielSank .. Nice to see both of you together...

0celo7

@ACuriousMind The inverse image of an open set under a continuous map is open...what about the inverse image of a closed set?

ACuriousMind

@0celo7 Think about the complement.

TanMath

20:51

@MarkMitchison do you know about FRET?

0celo7

@ACuriousMind Ok, I have an idea

TanMath

@MarkMitchison my not there?

0celo7

@ACuriousMind My thoughts on that: Let $f:X\to Y$ be continuous and $Y=Y_1\cup Y_2$. Assume that $Y_1\cap Y_2\ne \emptyset$. By definition of continuity, $f^{-1}(Y_1\cap Y_2)=f^{-1}(Y_1)\cap f^{-1}(Y_2)$ is open. Now take the complement and use a de Morgan law: $[f^{-1}(Y_1)\cap f^{-1}(Y_2)]^c=f^{-1}(Y_1)^c\cup f^{-1}(Y_2)$. Suppose now that $Y$ is not connected, i.e. $Y_1$ and $Y_2$ are clopen. Then $[f^{-1}(Y_1)\cup f^{-1}(Y_2)]^c$ is closed.

By hypothesis $X$ is connected, so $f^{-1}(Y_1\cap Y_2)$ must either be the empty set or $X$. There are no maps into the empty set and it cannot be $X$ because that would require $Y_1=Y_2$. Thus $Y_1\cap Y_2$ cannnot be empty and $Y$ must be connected.

damn, that edit is wrong

wait I don't even need a de Morgan law

do I?

$f^{-1}(Y_1\cap Y_2)$ is automatically clopen...

or not...now I'm not sure :/

ACuriousMind

Wait, what do you want to show?

0celo7

21:06

Continuous map on a connected space maps into a connected space

Screw the editing time limit

ACuriousMind

So your conclusion must be that $f(X)$ is connected, not that $Y$ is connected.

Because that's false if $f$ is only assumed to be continuous.

It has to be surjective for that to follow

0celo7

Well I'm taking $Y$ here as $f^{-1}(Y)=X$

I use that property in the second to last line, where I say $Y_1=Y_2$

ACuriousMind

Okay, so you want to show: If $X$ is connected and $f: X\to Y$ is continuous and surjective, then $Y$ is connected.

0celo7

Yes

I would fix that proof but editing time limit

I want to show that $f^{-1}(Y_1\cap Y_2)$ is clopen if $Y$ is disconnected

so it is either $\emptyset$ or $X$

surjectivity means it cannot be $X$

hmm, not entirely sure why it can't be $\emptyset$

stupid question, is $\emptyset\subset X$?

ACuriousMind

That proof looks quite convoluted to me. How about this: Suppose $Y$ is disconnected, i.e. $Y = Y_1\cup Y_2$ with $Y_i$ clopen and disjoint. Then $f^{-1}(Y) = f^{-1}(Y_1) \cup f^{-1}(Y_2)$ would be the union of clopen sets, and $f^{-1}(Y_1)\cap f^{-1}(Y_2) = f^{-1}(Y_1\cap Y_2) = f^{-1}(\emptyset) = \emptyset$, so the union is disjoint and $f^{-1}(Y) = X$ is disconnected. Contradiction.

0celo7

21:14

what

Ah

Ahhhh

ACuriousMind

I wasn't quite sure what you were trying to do with the intersection - the preimage of the empty set is the empty set

0celo7

@ACuriousMind Proof?

ACuriousMind

@0celo7 Suppose $x\in f^{-1}(\emptyset)$. Then $f(x)\in\emptyset$. Contradiction.

0celo7

what

oooooo

ACuriousMind

Shall I elaborate (that is admittedly very terse) or have you got it?

0celo7

21:19

There exists no such $x$.

So the set of all such $x$ is empty.

ACuriousMind

Yep, that's it

0celo7

So why does a continuous map send closed sets to closed sets?

I'm not entirely sure what to do...

Mark Mitchison

@TanMath I'm here

0celo7

Let $f:X\to Y$ and $U\subset Y$ be closed.

ACuriousMind

@0celo7 For $f: X \to Y$ and any $U\subset Y$, $X - f^{-1}(U) = f^{-1}(Y-U)$, where the minus is the "without" operation usually written with a slash (but I always forget in which direction it is slanted)

0celo7

21:25

@ACuriousMind I prefer the minus over the slash tbh.

ACuriousMind

What is it with all the people and $\pi$ today?!

0

Q: How accurately have we measured the effect of $π$ in electromagnetism?

Assuming the universe had a discrete structure at the quantum scale, then it seems $π$ would be an uncomputable value for all practical purposes. That is, if $π$ is approximated by the number of sides in a polygon, then we could approximate pi with as low as 3 sides, to billions of sides or more....

electromagnetism waves

TanMath

@MarkMitchison so do you know about Forster resonance energy transfer?

0celo7

$f^{-1}(Y)=X$?

Mark Mitchison

@TanMath I can't say I know much about it, I'm afraid.

Certainly not the details.

0celo7

If $f$ surjective here?

ACuriousMind

21:26

@0celo7 No.

That always holds.

0celo7

@ACuriousMind Ok, that's where I got stuck when I tried to do it.

Mark Mitchison

@TanMath Although from what I have read and heard, I thought FRET is usually modelled as an incoherent process.

Pies

huh, my question is popular

cool

ACuriousMind

@0celo7 $f^{-1}(Y) = X$ is by definition of a function, no need for surjectivity.

TanMath

@MarkMitchison really?

Mark Mitchison

21:28

But perhaps I am totally wrong about that

0celo7

@ACuriousMind :(

ACuriousMind

Surjectivity would be $Y = f(X)$.

0celo7

@ACuriousMind Right...

Mark Mitchison

@TanMath e.g., wiki states that Forster transfer is usually modelled as incoherent in a biological setting

But this information might easily be out of date

Probably FRET is just an effective model for treating energy transfer by dipole-dipole coupling

ACuriousMind

"$\pi$ is essentially not real"...why are people so obsessed with not-computable numbers? Reality doesn't have to be described by computable things unless you are sure we're living inside a simulation.

0celo7

21:31

Uh, we are.

TanMath

@MarkMitchison but for some reason, FMO complex dynamics isn't modelled with FRET...why?

Mark Mitchison

@TanMath Where have you read or seen that?

TanMath

@MarkMitchison nowhere have I found FMO complex modelled with FRET...

ACuriousMind

@0celo7 Well, you may well believe that, but I don't, and there's nothing either of us can really do to convince the other.

Mark Mitchison

@TanMath But isn't FRET in the end just the mechanism by which excitons hop from one chromophore to another?

0celo7

21:34

@ACuriousMind Back to the X

Mark Mitchison

Many of the abstract theoretical models do not depend on the microscopic details of this coupling

0celo7

Doesn't that (removing a point) just show that 2 intervals --> 4 invervals is not a homeomorphism?

ACuriousMind

@0celo7 Does it mark a spot?

0celo7

@ACuriousMind My ignorance :(

ACuriousMind

@0celo7 That's why we've shown now, correct.

TanMath

21:36

@MarkMitchison true, but I think maybe since it is coherent.. But they do have a mulricjromophoric FRRT used on other photosynthetic systems...

ACuriousMind

But this is the crucial argument to see that interval->X cannot be a homeomorphism, either.

Slereah

This is America @ACuriousMind

Freedom is everything

it can be anything

0celo7

@ACuriousMind On a first examination I understand that

not so sure now

ACuriousMind

@0celo7 Formally, what's left to use is that if $f : X \to Y$ is a homeomorphism, then so is $\bar{f} : X - p \to Y - f(p)$.

Because we've shown that $\bar{f}$ isn't one, and using that we get that $f$ isn't one, either.

@Slereah ...since when is this America? oO

0celo7

@ACuriousMind Sigh, another proof to think about

Physics Meta

21:39

0

Q: What makes a question too broad?

I recently answered -How much forward force is exerted by an idling automatic car- but it was closed as too broad. I dont particularly care either way, and I respect the moderators judgment, but its closure confused me. I understand it is a general question, broad because it does not specify ...

discussion close-reasons

ACuriousMind

@0celo7 Well, those are all the little things where the lecturers and authors always say "It is now obvious that..." ;)

Slereah

Rammstein - Amerika (Official Video)

►

0celo7

@ACuriousMind Yeah...

@ACuriousMind So the proof is obvious?

ACuriousMind

@0celo7 Well...as obvious as the last few ones we did, I'd say, since usually the whole "the interval isn't homeomorphic to X" is called obvious.

0celo7

It is obvious, just hard to prove

@ACuriousMind Can I have a hint?

ACuriousMind

21:46

@0celo7 It's a specific case of "The restriction of a homeomorphism to a subset is a homeomorphism onto its image (in the respective subset topologies)".

0celo7

...which I don't know either

ACuriousMind

That this is a bijection is really obvious, the slightly non-trivial part is that it's still bicontinuous.

0celo7

@ACuriousMind Yes to the first part.

ACuriousMind

C'mon, the second part isn't hard, either - a restriction is not really a different map, it still does essentially the same!

0celo7

I know...

Oh, I really have to go now.

I'll think about this.

Sir Cumference

21:51

Sigh...how do I delete my area 51 account?

ACuriousMind

10

Q: How to delete my Area 51 user account?

I have accounts in Stack Overflow, Programmers, Unix & Linux and Area 51. I don't want to continue my Area 51 account, so I planned to delete my Area 51 account. I have performed the following steps to delete my account, but I couldn't! Case 1 If your account has votes or posts do the follow...

support area51 profile-page user-accounts deleted-accounts

Slereah

You must ask the secret government

Sir Cumference

thanks

Slereah

Oh, that area 51 you mean

Forget what I said then

ACuriousMind

@Slereah lol

Slereah

21:53

From what I've heard the actual area 51 is pretty pedestrian

In the 90's the local workers there sued the government for unsafe work conditions

I assume a UFO fell on them

Toxic chemical from what I recall

"In July 2013, CIA released an official history of the U-2 and OXCART projects that officially acknowledged the existence of Area 51"

A fancy plane

Gee wizz it's like all our Christmases have come at once

0celo7

I think Area 51 is just a distraction, the aliens are really kept under the shack down the road.

ACuriousMind

@Slereah Your Christmases must be pretty sad affairs :P

Slereah

Like the aliens would live under such conditions

They didn't travel all that way just to stay at a motel

The OXCART project is pretty phallic

English Language & Usage: Multi-Layer…

Transcript for