The h Bar

user228700

13:01

@Secret: I don't mean to minimise your pain, but:

user228700

On Middle School Misery

►

Balarka Sen

school is always rather tedious

@0celo7 Hope to talk to you soon.

heather

@Secret i often don't talk about it because i've tried to move on (school switch, not due to what happened, separate issue) but I got bullied a not insignificant amount, especially in sixth grade. but i've never found that talking about it has helped. just sort of letting go of it has done more than anything else.

Secret

Letting go is hard for me due to the way my memories work. They are like incredibly stubborn. They will not give up until whatever their goal is fulfilled, which frankly, I am not very sure what kind of "closure" they are seeking for given how the bullies are no longer relevant after grade 10 and every physical damage is resolved, and given how twisted the dark side had became. It's as if they are chasing for something that is... no longer existed

I sometimes thought I have let go due to how in my daily life and stuff I seemed to be fine without actually actting (I then to be myself in face of most people), but as many of those dream logs have reminded me, they are like those peripheral sensory information I refer to earlier, seemed to be held in the mind for reasons not so certain

Jim

Math question: I know that if you have $n>1$ point charges with the same sign (and let's also specify to equal magnitude of charge) arranged in any non-coincident positions, any gaussian sphere that contains all of the point charges must also contain a point where the magnitude of the electric field vanishes. I know this must be true. However, I had a difficult time proving this to someone mathematically the other day. I'm sure there's a math theorem covering this, anyone know which?

Secret

13:14

Jim: Is that some kind of newton shell theorem?

Jim

It'd be nice to just say a name instead of go through the rigor

@Secret you tell me

Secret

For any arrangement of charges on the surface of a sphere (assuming it is not a conductor nor dielectric), it must corresponds to some effective charge within the sphere when you superimpose all the electrostatic fields they produce. So I guess an argument might be somehow the gaussian sphere never enclose that effective charge?

ACuriousMind

@Jim if the charges are all equal, shouldn't that place be at the center of mess?

Secret

but I don't think that is possible because then gauss theorem in maxwell equation will be violated as the effective charge is always the total enclosed charge within the gaussian sphere?

Jim

@ACuriousMind no. Consider if you had a few closely packed charges and then one far off elsewhere. The center of mass is closer to the clump whereas the point of zero E field is closer to the loner

ACuriousMind

13:35

@Jim Right. How's this: Since the field falls of at infinity, we can view it as a function on $S^3$. It's not defined at the locations of the point charges, so it's actually on an $n$-punctured 3-sphere, which is homotopic to the $n-1$-bouquet of $S^2$s, which has Euler characteristic $2(n-1)$ and therefore for $n \geq 2$ we have to have a zero of the vector field by the Poincaré-Hopf theorem.

Balarka Sen

@ACuriousMind This does not sound right. I don't think P-H works for noncompact fellows. I mean just take a nowhere zero vector field on S^3, then puncture it n times - that's a nowhere zero vector field on the n-punctured S^3.

Irrelevant, but in fact any noncompact manifold should admit nowhere zero vector fields: that's just the self-intersection number of the zero section in the tangent bundle and because everything is noncompact it should just be 0. Intuitively one should be able to "push the zeroes off towards infinity"

I think an ideal P-H for noncompact manifold ought to track information about the behavior of the vector field near infinity (end of the manifold). I dunno how to do it.

Jim

13:51

you all are making this more complex than it should be. I just want know the name of whatever theorem would more succinctly prove that for any arbitrary arrangement of a finite number of similar charges, there exists a set of finite coordinates where the electric field vanishes. Use 2D if it makes it easier.

Maybe I should have gone to math.se

Secret

Consider two point charges on a sphere of radius $\rho$ located on the xy plane as follows ($\theta$ is the angle of the position vector of the 2nd charge from the y axis):
$$E_1=k_e\frac{q}{(\vec{r}-\rho\vec{x})^2},E_2=k_e\frac{q}{(\vec{r}-\rho \cos \theta\vec{x}-\rho \sin \theta\vec{y})^2}$$
The total electric field at any point $\vec{r}$ is given as:
$$E_1+E_2=k_eq\left(\frac{1}{(\vec{r}-\rho\vec{x})^2}+\frac{1}{(\vec{r}-\rho \cos \theta\vec{x}-\rho \sin \theta\vec{y})^2}\right)$$
Expanding:

I am however currently too integration symmetry occupied (one of simpleart's projects) to do the n charge case

Balarka Sen

@Jim I wasn't really answering your question, but I don't know the answer. I think It's more likely that a physics person would be able to answer this better than a math person.

Secret

(sorry the stuff above hasa typo, fixing it now...)

Jim

@Secret please note, I did not say the charges were on a sphere. I said that it is true for any sphere that contains the charges. They can be anywhere in the sphere, but all of them have to be in the sphere for it to also contain the zero point. Also, that seems incorrect. For two charges of equal magnitude and sign, there will always be a point of zero field exactly in the middle of them, which is inside the sphere if they are on the surface

good old secants

Secret

Seems that's a sign I have been leaving behind electrostatics for wayy tooo long, sigh...

Secret

14:28

...
How would one proof that the following series
$$\sum_{i=1}^{\infty}\frac{1}{\vec{r}_i^2}$$
subjected to $x^2+y^2+z^2=\rho^2$
must contains a zero...?

Slereah

Does $\delta S = 0$ make sense, mathematically

I'm not sure it does

The well defined operation is $$\frac{\delta S}{\delta \phi} = 0$$

@BalarkaSen He's probably not coming back

Balarka Sen

Perhaps.

Well, depends on what you mean by $\delta S$. $dS = 0$ makes sense in terms of differential forms.

Mike Miller

You can do anything you want towards infinity. Pick your nonzero vector field, make it unit length, then multiply by your favorite smooth function to make it arbitrary length.

There's a version with boundary whenever your vector field is outward pointing.

Sorry, that's only true in even dimensions.

Jim

@Secret one wouldn't. You need a $\hat r_i$ inside the sum. Assuming, of course, that $\vec r_i$ is the displacement from the hypothetical zero point, and not simply the coordinates of the i-th point

Mike Miller

@Jim B and I are from math.SE. Your question says "charges", so I object to calling it a math question. Make it about manifolds and vector fields and i can help.

Secret

14:43

Jim: I am currently thinking about picking an origin, and define all my fields by the ith charge with position vector $\vec{r}_i$. Then the field is zero only when they sum to the zero vector, which is always possible since the zero vector is always linearly dependent. Then it is easy to see that the charges will be distributed in a way that there exists a basis formed by the unit vectors such that it spans either $\mathbb{R}^3$ or its subspaces.

Since the picking of an origin is arbitrary and does not affect the physics, the above arugment should still hold and thus there is always a zero…

Jim

@Secret yes, I know. I want to know the name of the theorem that applies to this

Secret

typo: "charge will be distributed in a way" should be "given any charge distribution made of point charges"

Slereah

14:56

@BalarkaSen The usual variational calculus

Secret

Mike: I think Jim's question is basically asking given any sums of vector fields of this form defined in the 3-ball $$\frac{\hat{r}_i}{||\vec{r}_i||^2}$$ with $||\vec{r}_i||^2 \leq R^2$ for all $i$, the name of the theorem that guarantees the sum always give zero somewhere in the 3-ball of radius $R$

Balarka Sen

@Slereah Oh, then I don't know what it's about.

Jim

15:11

@Secret the inequality you give is not really necessary. $\vec r_i$ is the displacement vector between the zero point and a charge, which could have a magnitude larger than R, if the zero point were at one side of the "3-ball" and the charge at the other

Kenshin

sup peeps

who's ready for TRUMP

Secret

I think that is the same as if I set the origin at the centre of the sphere and the zero point will then be some linear combination of the unit vector of the above vector fields that give zero

Kelthar

@DanielSank I'm not saying "requirements" in the strict sense, I'm saying subjects that may be useful, as I have a bunch to choose from, for my master's. Just want to make sure I choose the most useful

@BernardoMeurer just saw that you're from Lisbon, me too! xD Don't tell me you're in the same uni, Instituto Superior Técnico (IST) as well? xD that would make the xilinx issue much more relatable :P

Kelthar

15:33

@BernardoMeurer lol, it was displayed right below your name... feeling stupid now... anyway cool :)

Jim

I'm more looking for the theorem that would prove that for a function $F(\vec{\mathcal R})=\sum\frac{\vec r_i-\vec{\mathcal R}}{||\vec r_i-\vec{\mathcal R}||^3}$, where each $\vec r_i$ is unique, there must exist some value of $\vec{\mathcal R}$ such that $F=0$

Translating the origin to the zero point makes it more difficult because now all $r_i$ are unknown

Bernardo Meurer

@Kelthar Ha

Yeah

IST

That's my school too

Are you at LEIC or MEEC?

@JohnRennie I see you have answered my question, but I cannot find you're answer on the chat log :/

Also, did you see Qualcomms new SoC, the Snapdragon 835, is now using 10nm transistors?

John Rennie

@BernardoMeurer I started a really simple answer, but thought you must already know more than that.

Bernardo Meurer

@JohnRennie I know nothing, Ryan was too busy to explain at all :P

John Rennie

Do ou reall not know why $\nabla\phi$ for a scalar $\phi$ is a vector?

Bernardo Meurer

15:48

Nope

John Rennie

Well $\nabla$ is $(d/dx, d/dy. d/dz)$

So $\nabla\phi = (d\phi/dx, d\phi/dy, d\phi/dz)$ i.e. an object with three components

Bernardo Meurer

Makes sense so far

John Rennie

And it transforms like a vector under a change of coordinates

So if it looks like a vector and transforms like a vector it's ...

Bernardo Meurer

A cow!

John Rennie

:-)

Bernardo Meurer

15:53

But what do you mean by "Transforms like a vector" ?

MartianCactus

he means that its a cow!

its too obvious man......

John Rennie

@BernardoMeurer I'd have to look up the details, but if you do a coordinate transformation a vector transforms in a specific way. $\nabla$ transforms in the same way because it's basically $d\phi/dx\mathbf e_x + d\phi/dy\mathbf e_y + d\phi/dz\mathbf e_z$.

Where the $\mathbf e$s are te unit vectors.

Bernardo Meurer

Hmmm

Sick

John Rennie

Ryan will be able to say the same thing in three hours of incomprehensible maths, but basically it's blindly obvious it's a vector!

Bernardo Meurer

To be fair, he was super helpful when I was studying for analysis :P

You just can't get him started on geometry

qualcomm.com/products/snapdragon/processors/835

So nice

John Rennie

16:02

I'm exaggerating. But I did watch him start to explain what a manifold is using point set topology ...

Bernardo Meurer

lol

Yeah

John Rennie

He didn't finish because the person he was talking to ran away ;-)

It was a load of balls basically :-)

Bernardo Meurer

@0celo7 Please come back

John Rennie

15 days ...

Bernardo Meurer

So close, but so far

John Rennie

16:06

@0celo7 see, I remember your explanation of manifolds even if I didn't understand it!

Bernardo Meurer

@JohnRennie I'm going to stop drinking

and use my money on audio equipment

John Rennie

Cool :-)

I think it's really important to have a setup that you love listening to.

It is a fantastic destressor to be able to put on an album you really love, slump into your chair and just lose youerself.

Bernardo Meurer

I have very good headphones

John Rennie

Let the rest of world go frak itself, at least for 45 minutes :-)

Bernardo Meurer

and I'll be getting some nice speakers as a late xmas gift this month

xmas/birthday

John Rennie

16:09

Or 22.5 minutes if you're silly enough to play it on vinyl ;-)

Bernardo Meurer

Of course I am silly enough :P

John Rennie

What speakers?

Not that I'm in touch with the current models of speakers ...

Bernardo Meurer

It's a bit of a dilemma right now

I don't have an amplifier

so I need to get powered speakers

which I do not like

but if I buy an amplifier, I get not that good passive speakers

John Rennie

Hmm, I think that really limits your choice.

Bernardo Meurer

which I also don't like

I'm currently thinking of the Klipsch R-15PM

John Rennie

16:11

I wouldn't do it that way. Buy any old cheap amp off ebay just to get you going then save up for a decent amp.

Bernardo Meurer

Problem is I'm only knowledgeable on headphone amps

I know nothing about speakers

nada

zero

so I don't even know what to get

John Rennie

The Q Acoustics speakers get consistently good reviews ...

Bernardo Meurer

Nationalist :P

John Rennie

Do you have to buy new? If you buy from ebay you'll get much better value for money.

DanielSank

@Bernardo this is what I was trying to warm you about

Bernardo Meurer

16:15

@DanielSank Hm?

@JohnRennie Nope, I can buy used

DanielSank

This "transforms like a vector" stuff.

John Rennie

What's the eBay Portugal site?

Bernardo Meurer

Idk if there is one, I use UK one

John Rennie

Or I guess most Spanish sellers would ship to Portugal for a bit extra ...

Bernardo Meurer

or .com

Almost everyone ships here

John Rennie

16:16

From the US you'll have to pay import tax and that can really sting.

It's 20% or something like that.

Bernardo Meurer

Nothing from the US, yeah

Too much trouble

DanielSank

Trying to understand tensors in terms of how their representations behave when you change coordinates is like trying to understand linear algebra by studying how matrices behave when you change basis.

It's ass-backwards.

Bernardo Meurer

@JohnRennie My budget is about 300€ w/o shipping

DanielSank

IMHO

Bernardo Meurer

If we don't account the fact I eat :P

@DanielSank That sounds like a reasonable argument, how would you do it?

John Rennie

16:19

ebay.es/itm/Q-Acoustics-2010i-Main-Stereo-Speakers-/…

Those are allegedly really good entry level speakers ...

59 Euros ...

@DanielSank why is grad of a scalar a vector other than, well, it just is?

Bernardo asked me and I didn't really know because, well, it just is!

Bernardo Meurer

@JohnRennie What about an amplifier?

and a phono preamplifier...

John Rennie

ebay.es/itm/Marantz-PM66SE-KI-Signature-edition-/…

That's a cracking good amp! Google for some reviews

And that's the KI Signature version.

Bernardo Meurer

HOW WIDE IS IT

I need to see if it fits :P

Doesn't have a phono stage it seems

John Rennie

Standard 19" I think

It has a phono stage

Zoom in on the picture

Bernardo Meurer

YES

John Rennie

16:29

But i'm not saying buy that amp. I'm just saying ebay.es will have some good amps and speakers at affordable prices.

I have a Marantz amp (a PM6010) and I really rate it. It's great for rock but also good for jazz or classical.

I don't have experience of other Marantz amps like PM66 but you could Google for reviews.

Annoyingly I have two spare PM6010 amps in a cupboard upstairs, but I suspect the cost of shipping them would be prohibitive because they're big heavy buggers.

Bernardo Meurer

(One second, I have to get popcorn for watching the US inauguration, brb)

John Rennie

US inauguration? Really? You have nothing better to do?

Bernardo Meurer

17:01

@JohnRennie I'm a sadist

and a masochist

John Rennie

Hi Alex :-)

alex

Hey. sorry i'm new to this.

i didn't know hbar was the main chat room

John Rennie

No problem. Have a look at:

-4

Q: How do I access chat?

How do access the chat room to ask a question?

alex

can you tell at which point of my argument i was wrong?

oh thanks for the link

John Rennie

What you calculate in your question is the coordinate time needed to reach the horizon.

And that is indeed infinite.

But what the free falling observer records on their clock is the proper time needed to reach (then pass) the hortizon and that is finite.

alex

17:08

no i meant this argument:

one sec

John Rennie

In your linked answer, if you imagine the camera sending pulses of light to you periodically, you will see the camera at closer and closer positions to you as you approach the EH since the constant radius hyperbolas get closer and closer in this region. When you reach the EH, you see the camera in the same position as yourself since the hyperbola at this point is identical to the light trajectory. Is this interpretation correct? does this mean that you, the camera and any other free falling thing are compressed together onto the EH surface for that instant of time? — alex 3 hours ago

This?

alex

yes

John Rennie

You can just post the link to the comment here and it will show the whole comment.

alex

how?

oh sorry ... i guess one question at a time

John Rennie

In your comment where it says two hours ago or the time since you posted the comment, you can click that to get a link to the comment.

Click where I've drawn the circle, the copy and paste the URL here

Anyhow ...

You say you will see the camera at closer and closer positions to you as you approach the EH but that isn't true.

That is true for the Schwarzschild observer watching from far away but not in the rest frame of the freely falling observer.

alex

17:15

oh then what will the observer see?

John Rennie

Which observer? The freely falling one?

alex

yes

John Rennie

For any freely falling observer spacetime is locally flat. So to a first approximation the camera just floats at a constant distance from you.

To a second approximation there will be tidal forces present and assuming you have the camera between you and the black hole then those forces will tend to pull the camera away from you. So you actually see the camera accelerate away from you towards the singularity.

So what you see is completely different to what the Schwarzschild observer sees.

alex

so for the freely falling observer, the EH doesn't exist at all? signals from all parts of space can reach him if sent in the right direction and and were originated at the right time?

John Rennie

Correct.

The freely falling observer sees only an apparent horizon, and that apparent horizon retreats before the observer as they fall inwards so they never reach it. The freely falling observer only reaches the appararent horizon at the moment they reach the singularity.

alex

17:29

oh iv'e never heard anybody mention that. but it seems satisfying. do you have any links where i can read up about this?

the apparent horizon thing, i mean

oh wait wikipedi has an article

John Rennie

Yes :-) I was about to post the link.

Apparent horizon

In general relativity, an apparent horizon is a surface that is the boundary between light rays that are directed outwards and moving outwards, and those directed outward but moving inward. Apparent horizons are not invariant properties of a spacetime. They are observer-dependent, and in particular they are distinct from event horizons. Within an apparent horizon, light is not moving away from the black hole, whereas in an event horizon, light cannot escape from the black hole. It is possible for light to be currently moving away from the black hole (and so outside the apparent horizon), but in...

@alex You remember I said that for a free falling observer spacetime is locally flat, but there are tidal forces present?

alex

yes...

John Rennie

The farther you go from the free falling observer the greater the tidal force, and the apparent horizon is basically the distance at which the tidal forces pull the light away to fast for it to reach the observer.

So it's analogous to an event horizon but it depends on the free falling observers velocity and distance from the singularity.

DanielSank

@JohnRennie it's not. The gradient of a scalar is a covector.

@BernardoMeurer I think it's good to use the proper definition of a tensor. We can try later.

John Rennie

Ah, I was about to ask if you meant a convector :-)

alex

17:40

@JohnRennie ok i think i got the idea. that was really helpful

John Rennie

@DanielSank Hang on, a covector is a one form isn't it?

Ah yes, of course. Oops.

@alex You're welcome. I generally hang around the chat room up to lunchtime UK time if you have any more questions.

alex

@JohnRennie Can i ask another question about GR ? its too informal for the main site but i really want to know the answer

John Rennie

Yes, of course.

DanielSank

@JohnRennie yes

@alex the main site isn't supposed to be "formal", as far as I know.

John Rennie

@DanielSank I did know that. Honestly!! :-)

Bernardo Meurer

17:47

@JohnRennie Depending on how heavy the amp is it might be cheaper for me to pay you for the shipping than to get one and pay shipping anyway

John Rennie

@BernardoMeurer Let me have a look around and see what I've got. IIRC the spare amps have minor faults e.g. some of the connectors don't work but that may not matter for your purposes. I'll see what postage costs are like to Lisbon for heavy packages.

alex

what type of observers do these different coordinate systems correspond to? like KS, Eddington Finklestein, etc. ? is it some observer with a really weird trajectory with some weird type of acceleration ? or do they not correspond to any observer at all ?

John Rennie

@BernardoMeurer You can have an amp for free if you want. I can't be arsed to list them on eBay as they aren't worth much.

@alex They don't correspond to any observer, that is no observer would every measure space and time in those units. They are purely mathematical devices invented to make some calculations easier.

@alex but even apparently simple coordinates, like the Schwarzschild coordinates, can be more subtle that you think. For example the Schwarzschild $r$ coordinate is not radial distance.

alex

ya . i think the only solid quantity which is "real" and not a construct is the proper time, right?

John Rennie

The proper time is an invariant, which means all observers in all coordinate systems will agree on it's value. And as you say it has significance because it's the time record on a clock carried by an observer.

But there are other invariants in GR.

alex

17:57

I also wanted to ask what is the general view of the physics community on the AMPS paradox? although its still controversial, is the firewall solution or ER=EPR solution considered the more serious mainstream answer?

John Rennie

I suspect most physicists think all that stuff is pure speculation.

It's impossible to say what if any nuggets of truth lie within it.

alex

Any invariant which a non physicist like me can attempt to understand?

is the paradox itself taken seriously though?

John Rennie

@alex actually I guess most of the invariants are somewhat abstract, like the scalar curvature or Kretschmann scalar.

Maybe you should stick to the proper time :-)

alex

or do they dismiss the paradox itself as frivolous and not worth the worry?

John Rennie

I would guess most physicists just assume we don't yet understand quantum gravity so anything involving it is speculative. Most of us just get on with life and don't worry about the information paradox at all.

But that's just a guess. I can't speak for all physicists :-)

alex

18:10

I see. Well thanks for all the help and showing me the chat :-). it's bed time in India so I guess I'll sleep now. goodbye !

John Rennie

Goodnight :-)

obe

hi

Swapnil Das

Hi

Bernardo Meurer

@JohnRennie I love you

DanielSank

18:58

@JohnRennie Nobody said otherwise.

John Rennie

20:03

@BernardoMeurer actually shipping to Portugal looks like it's under 20 Euros, even for an item as heavy as the amp (about 7kg). I don't have a box suitable for shipping the amp, but it's about the same size as a PC and the company I work for probably have old PC boxes I could use.

Leave it with me and I'll see if I can get a box from work. If I can then I'll ship the amp over to you and you can see if you like it. I think some of the inputs don't work because they're old and corroded, but that's easily fixable if you can use a soldering iron.

Bernardo Meurer

20:35

@JohnRennie I can use a soldering Iron :)

And soon enough I'll have access to a lab

Now I only need to get speakers!

JamalS

20:58

Can anyone explain what is meant by $\mathcal{O}(-2)$ in the context of resolving singularities (in this instance, the fixed points in $T^4/\mathbb Z_2$ under the action of $\mathbb Z_2$)?

ACuriousMind

21:39

@JamalS I know that notation only for the twisted structure sheaf

JamalS

@ACuriousMind Turns out it is the cotangent bundle over $\mathbb P^1$, or equivalently the tensor product of the tautological line bundle with itself.

It has self-intsersection number -2, hence the name.

Bernardo Meurer

@JohnRennie Why do you have three of the same integrated amp?

ACuriousMind

@JamalS Ah, then it is the twisted sheaf! The tautological bundle is $\mathcal{O}(-1)$ because it's the inverse/dual of Serre's $\mathcal{O}(1)$, and since $\mathcal{O}(n)\otimes \mathcal{O}(m) = \mathcal{O}(n+m)$ the notation fits.

JamalS

@ACuriousMind Oh, I have not done sheaves at all yet.

ACuriousMind

I probably should've done more "actual" geometry instead of all that sheaf theory, too :P

Monad

21:52

Where can I find out the speed necessary for a hypothetically indestructible object covered in nothing but body fluid to set ablaze when travelling through Earth's troposphere?

unless someone knows an average for anything similar to my proposed scenario?

bodily fluids*

ACuriousMind

...covered in bodily fluids? What are you thinking about?

Monad

just a fantasy question, to be simpler, at what speed would an indestructible fist have to move to set ablaze?

JamalS

@ACuriousMind I'm sometimes tempted to switch from physics to differential and algebraic geometry :)

Because literally everything I enjoy in physics is strongly related to them.

Monad

I'd like to find out where I can find out about this unless someone could give an accurate guess.

ACuriousMind

@JamalS Oh, I know that feeling; @Danu already succumbed to its siren call ;)

JamalS

21:56

@ACuriousMind Oh yes, I'm aware. I read a tad of his thesis :)

ACuriousMind

@Monad I'm pretty sure there is a way to compute the temperature of an object subject to air friction but I can't tell you how to find it, sorry. You could ask this on the main site, though.

JamalS

@Monad Search for 'atmospheric entry' in NASA's archive of papers and technical reports.

@ACuriousMind @Monad Seems there is an answer here already: physics.stackexchange.com/questions/43768/…

Monad

I was just reading Wikipedia about atmospheric entry but can't pinpoint anything because in cases like meteors, flames highly depend on the meteor itself and the fact that it is a meteor.

David Z

@ACuriousMind With the caveat that you (@Monad) should ask "how to compute the temperature of an object subject to air friction", and make sure you do some research on your own first - don't ask "how fast would an indestructible fist have to move to be set ablaze"

Monad

Yeah Ik thanks

ACuriousMind

21:58

@JamalS Good find!

Monad

Seems about mach 3

Thanks all! :)

Bernardo Meurer

@JohnRennie "You really should know the proper definition of a manifold, I'm very disappointed" - Ryan

Danu

22:14

@JamalS O(-2) is not the cotangent bundle of P^1 by definition though; that's a result. The definition is as ACM said, the tensor product pf the tautological line bundle with itself

See section 2.4 of Huybrechts' book for a proof.

...or is it 2.5?

It's worthwhile to understand the Euler sequence on P^n

JamalS

@Danu Have you read any of Husemöller by the way?

DanielSank

22:35

@BernardoMeurer that sounds wrong.

Danu

@JamalS Nope

Skyler

23:30

howdy folks

Bernardo Meurer

@Skyler Hallo

Skyler

@DanielSank lol I think your icon just disappeared as mine came in

hows your day Bernardo

hi @heather

nice timing

Bernardo Meurer

@Skyler It's been alright, how about you?

Skyler

@BernardoMeurer pretty good, I kinda stalked Dan inadvertently a few hours ago

heather

hello all

=)

Bernardo Meurer

23:34

@Skyler I stalk him weekly

It's fun

Skyler

@BernardoMeurer but like in person

Bernardo Meurer

@Skyler Sweet, I've only done that twice :P

Skyler

He's in the middle of work and I sent him a message with the clothes he's wearing

I had a meeting with the senior director of the company that shares office space with him so it just kinda workked out that way

Bernardo Meurer

Sweet

He's a male fashionista

Skyler

@BernardoMeurer totally

Bernardo Meurer

23:48

@DanielSank When is the Sank collection for Winter 2017 coming out?

We all want the hottest socks and sandals combinations of the season

3

Skyler

I once wore dark blac toe sockks with black flip flops. People wouldnt notice and when they did it really messed with them

Bernardo Meurer

Lol, that's awesome

It's too cold here for that

in the winter at least

Skyler

@BernardoMeurer I remember going out thanksgiving 11pm in the desert with shorts and flip flips (I had a warm jacket to be fair)

but my uncle just looked at me in the most incredulous way, and said "you Santa Barbarian"

Bernardo Meurer

Ha, great

Santa Barbara is really cool

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