The h Bar

@0celo7 I thought you might not like this because - once you have unpacked all the definitions - it allows you to show that two things are isomorphic when all their hom-sets are isomorphic, which is a very non-constructive but sometimes useful way to show isomorphy.

0celo7

what is a hom-set

Why is it non constructive?

ACuriousMind

Well, probably it's not "non-constructive" in a formal sense, but it tells you two things are isomorphic but doesn't really tell you what the isomorphism is (I guess you could unpack all the definitions and the proof to get it, though)

@0celo7 Just the space of morphisms between two objects

0celo7

@ACuriousMind ...I think that's pretty cool.

Why would I not like that?

ACuriousMind

12:22 AM

Dunno, it was a feeling

Apparently it is wrong.

0celo7

You are fallible.

ACuriousMind

Duh.

0celo7

@ACuriousMind I thought you were the pope for roughly 7 months.

@ACuriousMind Still messing with spinors?

ACuriousMind

@0celo7 As we grow older, none of our idols is safe from disillusionment :P

0celo7

@ACuriousMind My idolatry has moved on ;)

To an actual geometer

ACuriousMind

12:26 AM

@0celo7 I'll continue tomorrow, I'll go to bed now. Still haven't found a satisfactory way to get Majoranas

0celo7

Gnight.

NeuroFuzzy

@0celo7 But that doesn't form an error with any of the theorems (in chs. 3, 6,7,8, ignoring homology stuff) as far as I tell, so long as $r$ is high enough. 2 or 3 or 4, right?

0celo7

@NeuroFuzzy No, you literally cannot apply it. If you define your tangent vectors as linear derivations of $C^r$ functions, you cannot apply the derivation rule to functions which are not $C^r$.

The other stuff applies, sure.

But you must define the tangent space more carefully.

Indeed, the only consistent way to do it is like the physicists do.

Most things in geometry are possible on $C^r$ manifolds with $r\ge 3$.

However, there are smoothness assumptions lurking everywhere in "standard" texts.

NeuroFuzzy

@0celo7 I just don't follow. Tangent vectors are linear derivations of $C^r$ functions. If you have a smooth vector field $v$, and you have $f\in C^r$, then $vf$ will be $C^{r-1}$. Where's the problem?

$vf$ being $C^{r-1}$ doesn't pose a problem to the fact that every $v\in T_p M$ will have $v : (f\in C^r) \to \mathbb{R}$ (as a linear derivation)

maybe you could go further and show that these definitions with $C^2$, $C^3$, etc. will all give rise to the same tangent space, and then all future problems have disappeared too.

0celo7

@NeuroFuzzy 2 issues: (1) the derivations of $C^r$ functions form an infinite-dimensional vector space. (2) the partial derivatives do not form a basis so your tangent space does not have the "curve" interpretation.

To show that smoothness is absolutely necessary is what I was talking about earlier with the $g_i$.

A priori, you do not know that derivations have anything to do with derivatives.

It's only true for smooth functions that derivations and derivatives coincide.

NeuroFuzzy

12:38 AM

@0celo7 Okay, sorry, I'm using your word "derivation" without regards to the algebraic meaning

0celo7

My word? I'm looking at your book here.

NeuroFuzzy

I'm really talking about $vf=\frac{d}{dt} \gamma(f(t))|_{t=0}$, which is an explicit reference to first order derivatives

I don't think there's any rigorous inconsistency here...

0celo7

So you're using the equivalence class of curves definition?

NeuroFuzzy

The one used in Renteln, yeah

0celo7

...no he uses linear derivations.

Page 75

NeuroFuzzy

12:42 AM

I am opening up my laptop with the book again

and I'm going to be annoyed if you're wrong :p

0celo7

> To extend these ideas to a general manifold M, we define a tangent vector Xp
at a point p ∈ M to be a linear derivation at p.

Maybe the curves definition (to be called the "kinematical" definition henceforth) is shit, so that's why no one uses it for $C^k$ manifolds :P

Maybe it's consistent.

NeuroFuzzy

You're totally right about the definition he uses

0celo7

I can read.

NeuroFuzzy

So you're saying the derivations at point $p$ in the sense of Renteln (linear derivation satisfying Linearity + leibniz property) of $C^r$ form an infinite dimensional vector space, and t herefore theorem 3.3 is wrong?

@0celo7 (Re this message)

0celo7

On a $C^r$ manifold? Yes.

It's fine on a $C^\infty$ one.

NeuroFuzzy

12:48 AM

How??

0celo7

Why is his argument wrong or why is it an infinite-dimensional vector space?

NeuroFuzzy

why is it infinite dimensional on a $C^r$ manifold

0celo7

That's...not a nice proof. You basically show that the derivations are isomorphic to a dual space of a vector-space-structure carrying subset of the $C^r$ functions. You then present an uncountable linearly independent set of this vector space. Thus its dual space is infinite-dimensional, and by isomorphism the derivations are.

@NeuroFuzzy It's much easier to point out where Renteln's proof fails.

NeuroFuzzy

@0celo7 where?

0celo7

In his proof of theorem 3.3, he uses these functions $g_i$. Do you know what their regularity is?

NeuroFuzzy

12:54 AM

$C^r$, by assumption, no?

0celo7

No, $f$ is $C^r$.

The actual formula for these functions is $$g_i(x_1,\dotsc,x_n)=\int_0^1\frac{\partial f}{\partial x_i}(tx_1,\dotsc,tx_n)\,\mathrm dt$$

Do you now see what their regularity is?

NeuroFuzzy

Okay, I see that would put them as $C^{r-1}$

0celo7

I think this is the "Lagrange remainder," but I'm not certain.

@NeuroFuzzy Yup, but in the $C^\infty$ case they would still be $C^\infty$.

Now, he applies the derivation $X_p$ to the product $x^ig_i$ and uses the Leibnitz rule.

This is invalid in the $C^r$ case because $g_i$ is not necessarily a $C^r$ function, so technically $X_p$ is not defined on the product $x^ig_i$.

Remember: here the Leibnitz rule has nothing to do with partial derivatives.

NeuroFuzzy

Wow, that's awesome, I never realized that.

0celo7

@NeuroFuzzy I can give a somewhat concrete example.

The function $x^k\sin(1/x)$ is $C^{k-1}$. Suppose we have $x^2$ and $x\sin(1/x)$. The first is smooth and the second is $C^0$. But their product is $C^2$.

That is, $fg\in C^r$ does not imply $f$ or $g$ are $C^r$.

They of course could be, but in general they are not.

NeuroFuzzy

1:06 AM

The uncountability is going to bug me now that I know you're probably right :o

0celo7

Of the derivations?

Let me give you the proof.

NeuroFuzzy

Yeah. I can't see it, but I can see Renteln's proof breaks down.

0celo7

I find weird things like this more interesting than I used to.

NeuroFuzzy

@0celo7 Damn. Thanks a ton for putting up with my skepticism. What book is that?

0celo7

@NeuroFuzzy Manifolds and Differential Geometry by Jeff Lee.

@BalarkaSen That book might be more interesting to you than other Lee.

Balarka Sen

1:19 AM

@bolbteppa It is worth noting that Arnold was exaggerating when he wrote that statement. The student was only joking - clearly Arnold was so against French-styled mathematics that didn't stop to attack everything and all the things in French mathematical education.

That pupil is today a well-accomplished mathematician.

@0celo7 Noted, thanks.

0celo7

@BalarkaSen In short: terser, harder topics, different topics.

The foundations are the same, but Jeff explores connections and Riemannian geometry.

Balarka Sen

Fair enough, I'll have a look.

bolbteppa

1:51 AM

@BalarkaSen Well the kid was joking mathoverflow.net/a/153606/38721 not sure if Arnold knew that but he was making a solid point, a point illustrated via the supposed inability to calculate $\frac{\partial}{\partial r}(x^2 \vec{e}_x + y^2 \vec{e}_y)$ brute force, accompanied by actual comments this is wrong or a bad style of mathematics & that I'm crazy etc... though you find it in good vector calculus books, why? Well a bunch of abstract manifolds concepts 'imply' it's not possible :\

bolbteppa

2:02 AM

It's a known thing that people can get bogged down learning big words and abstract concepts without questioning the necessity for any of it and criticizing the basics, and yeah that's part of the fun at times but if you intend to try to contribute at some stage you have to watch yourself and criticize even definitions, but saying a calculation is flawed, bad mathematics, etc... when you can clearly write down the correct answer is another level

0celo7

@bolbteppa I know how to calculate it, my point is that it means something different than you think.

(The terrible notation does not help.)

bolbteppa

2:23 AM

Ah so you can genuinely calculate it "flat out incorrectly"? What does it mean? What exactly is wrong with the notation in tons of books?

Balarka Sen

@bolbteppa Yes, I am aware of that MO post.

bolbteppa

Cool that he posted on there haha

Balarka Sen

He wasn't really making a point. The kid knew how to calculate 2 + 3.

bolbteppa

Arnold wasn't making a point?

Balarka Sen

No, he was forcing a point out of his distaste towards French-styled abstract mathematics. Forcing a point and making a point aren't the same thing.

bolbteppa

2:26 AM

Ah come on

Ridiculous

0celo7

Wait, is zero a positive number?

According to Bourbaki?

Or according to Arnold?

bolbteppa

You misunderstand the essay clearly

0celo7

I'm reading some Russian thing, it's not working.

Balarka Sen

I think I understand it fine. But whatever, I abandon this conversation.

0celo7

@bolbteppa That partial derivative out front is the Levi-Civita connection of $(\Bbb R^n,\delta)$.

bolbteppa

2:27 AM

@BalarkaSen My god...

2nd time in like 3 days you've done that to people on here dude, not good

haha

Man it's just a derivative you can calculate using limits

You can set up the difference quotient and calculate that limit using the high school definition of a limit

Use the product rule

The basis vectors depend on coordinates, so do the coefficients of the vector field, thus you can calculate those sneaky derivatives using baby calculus, it exists whether you like it or not, your manifolds concepts will bow to this not the other way round buddy ;)

0celo7

Whatever.

bolbteppa

Is $\frac{\partial }{\partial x}(x^2 \hat{i} + y^2 \hat{j}) = 2x \hat{i} = \frac{\partial }{\partial x}[(x^2,0) + (0,y^2)] = \frac{\partial }{\partial x}(x^2,y^2)$ cool or no?

0celo7

2:47 AM

Why would that be cool?

One again, you're implicitly using the Levi-Civita connection.

If you don't see that, go read a geometry book.

bolbteppa

Another general relativity book lying to us all books.google.ie/…

0celo7

Is it so hard for you to accept that these books are wrong?

If you carefully define everything you would see that.

bolbteppa

This is exactly the point Arnold was making, the focus on formalism is literally holding you back from understanding baby calculus ideas and actually calling them wrong

0celo7

I understand what you're trying to do. Doesn't make it correct.

bolbteppa

I mean you just said you had no idea what $\frac{\partial }{\partial x}(x^2 \hat{i} + y^2 \hat{j}) = 2x \hat{i} = \frac{\partial }{\partial x}[(x^2,0) + (0,y^2)] = \frac{\partial }{\partial x}(x^2,y^2)$ means, and are now saying it's wrong since this is a direct example of what the book does

lucas

2:54 AM

@EmilioPisanty Thanks! It wasn’t so important. Sorry for delay, I logged off after that you left the chat room and came back now. Hope a good thesis. I approximately died when I was providing my BS project and I am sure that PhD thesis is more difficult by a factor of infinity! Best wishes for you! And good luck!

0celo7

@bolbteppa I'll bite.

How is that partial derivative defined?

bolbteppa

Do you not know how to define a partial derivative using limits?

0celo7

For a function, sure.

What exactly is that object you're differentiating?

While reading this essay I'm seeing why his book is the way it is.

bolbteppa

If $\vec{A}(x,y) = A^1(x,y) \vec{e}_1(x,y) + A^2(x,y) \vec{e}_2(x,y)$ then $\vec{A}(x + dx,y) - \vec{A}(x,y) = A^1(x+dx,y)\vec{e}_1(x+dx,y) + \dots$ so that...? (exercise: form difference quotient then directly copy the proof of the baby caculus product rule, voila)

Just do what you do when doing single-variable calculus, treat it like a function

0celo7

is that supposed to be a vector field on a manifold

or a vector valued function on the plane

bolbteppa

3:09 AM

I wrote a vector field in the plane mapping vectors to vectors

0celo7

that doesn't answer my question

bolbteppa

Have you taken multi-variable calculus or vector calculus yet?

It is both

0celo7

@bolbteppa No

bolbteppa

Ok treat it as a single-variable calculus problem, that's what you do in those classes anyway

0celo7

I don't know that either...

bolbteppa

3:14 AM

Do you know how to prove the product rule in calculus?

0celo7

...no

bolbteppa

I am not sure if you're mocking me or serious, it's fine if serious

0celo7

why should I know it if I can look it up

bolbteppa

Genuine question: Why prove Sard's theorem and not the product rule?

0celo7

I proved the product rule in real analysis

@yuggib Is it true that "French mathematicians" dislike diff geo?

bolbteppa

3:27 AM

I've given you a perfect example of using the product rule, something you could repeat the proof of with verbatim, an example of an exercise you can do to refresh your memory on the proof of that theorem too or a chance to go out into the wilderness and try to recreate the proof to prove you actually understood it.

0celo7

The proof is not particularly hard IIRC

you have to add and subtract in the numerator of the difference quotient in a clever way

bolbteppa

You just add $0$ so that you can factor nicely

0celo7

Correct.

bolbteppa

Use the terms in the numerator to remember them too

0celo7

3:54 AM

@bolbteppa Hmm?

The way I would prove it is work backwards, actually

turning $f'g+fg'$ into $(fg)'$ is simple

Balarka Sen

@bolbteppa I duck out of conversations when I have nothing positive to contribute or share or see no gain in the discussion.

0celo7

5:27 AM

@JohnRennie good morning

John Rennie

Morning. Gosh, you're up late ...

0celo7

5:39 AM

@JohnRennie Neighbors are turning up

Watching a new show

It's about the Colony

just watched a squad of lobsters get destroyed :)

It's called Turn

I'm hoping it's a series about the rev. war that doesn't glorify the rebels

@JohnRennie you wouldn't happen to speak technical Norwegian, would you

John Rennie

@0celo7 sadly not

0celo7

I have to order this special rhenium heat shield for a device at work from this Norwegian company

I thought I might get a better deal if they think I'm Norwegian

exchange student or something

Fox news people sound...overly American

John Rennie

6:05 AM

I wonder why rhenium is used. As far as I know it doesn't have any unusual thermal properties.

0celo7

doesn't melt

at least...I think it's rhenium

John Rennie

wow, melting point 3182C

that's high :-)

0celo7

ok, since the chemist is suprised it's probably rhenium

John Rennie

What are you doing that uses such high temperatures?

The highest temperatures I had to use were around 1200C to make silicon disulphide, and that was a bit scary.

0celo7

I'm not using high temperatures like that

our lab specializes in ceramics

1700C is common

I opened a 1000C furnace the other day while it was on

very cool

also quite warm

;)

John Rennie

6:12 AM

At 1200C the heat blasts out at you the moment you open the door. I'd guess at 1700C you simply don't open the door while it's on!

That pesky $T^4$ term :-)

0celo7

at 1000C it was vaporizing dust in the air

we don't open them at 1700C

John Rennie

We used to use the furnaces for making glasses. They needed rapid quenching, so we used to quench them in molten lead!

0celo7

...what

why not water

John Rennie

Water didn't cool them fast enough because it simply vaporised and you got an insulating gas layer.

Plus, drop something at 1200C into water and you need to stand a loooooooooong way back :-)

0celo7

hmm

suppose I have a 3L bucket of LN2

and I drop a 1700C rhenium pellet in

what happens?

John Rennie

6:17 AM

I run away screaming

It wouldn't be that bad. You'd get a minor explosion, but I don't think you're talking about major devastation.

Liquid nitrogen has a low thermal conductivity so you'd get immediate formation of a gas pocket and that would insulate the pellet.

If you're actually contemplating doing it I'd be inclined to start at a lower temperature and work upwards.

But you may find older members of your research group have already done it and they'll know far more about it than me.

0celo7

I just want to do it for the lulz

I'm not actually going to do it

John Rennie

Liquid nitrogen is surprisingly boring. We all have fun freezing rubber then smashing it, but the novelty wears off fast. Because liquid nitrogen has a low specific heat and low conductivity it's not that spectacular.

If you put your hand into liquid nitrogen for a moment it wouldn't do any real damage, though it hurts (I speak from experience)

0celo7

I froze a banana once

John Rennie

Takes a while though doesn't it?

0celo7

it was too bruised

@JohnRennie Yeah

I think the skin was frozen

not much more

we use LN2 for cooling our spectrometer

John Rennie

6:25 AM

When I was a kid I thought LN would freeze things instantly, but sadly not.

0celo7

there's always some left over to play with

It will freeze metals p. quick

John Rennie

I guess metals have a high conductivity, but even so I think only small samples would freeze quickly - where quickly means a few seconds.

yuggib

6:49 AM

@0celo7 well, they don't dislike diff geo, but probably like more abstract math more ;-)

0celo7

you can make it as abstract as you want

Yashas Samaga

7:36 AM

$\frac{a}{b}$

e.e it doesn't work in the chat

yuggib

@0celo7 then it becomes category theory ;-P

I am just joking; simply in the bourbaki there is very few diff geo

but I don't think it is disliked in france

Yashas Samaga

9:14 AM

1

Q: Why does static charge build up and how do I prevent it?

I bought my 11-month-old son a tricycle from Smart Trike (official website) and I have noticed that after a few laps indoors on my marble floors static charge would build up in him as I can see his hair standing up, and when I touch him I get zapped. This does not happen when we play the tricycle...

Will connecting a wire between the handle and the floor even work?

My idea was to connect a wire to the metal handle and leave the other end lying on the floor so that any excess charge build up gets transfered back to the ground.

The boy gets his hair charged up so the charge must be moving from the wheels to some metallic object then through his hand or legs.

Danu

9:47 AM

Man, if I'd get a dollar for every time I had to Google "Pauli matrices"...

David Z

At first I read that as "Paul matrices"

...actually, that should be the name for the real version of the Pauli matrices, e.g. as they're used in quantum information theory

Danu

hehehe

ACuriousMind

10:04 AM

@Danu I hope you're doing that for all the relations they fulfill, not to remember what they are ;P

Danu

10:16 AM

@ACuriousMind The structure constants get me every time. 2x2 matrix multiplication = too much to ask.

ACuriousMind

@Danu Just remember the anticommutation relations as them being the 2D Euclidean Clifford algebra :P

I can't remember whether it's -i or i or i/2 or whatever in the commutation relations either

Danu

^

ACuriousMind

Not helped by the fact that some authors do scale them by a factor of 2

Danu

...it's $2i$

(for the unscaled ones)

David Z

10:57 AM

0

Q: How long does it take for an Astronaut around the Moon to receive a spoken message from Houston?

How long does it need till an Astronaut gets a spoken message from Houston and then replies to it? What happens if there is an emergency situation where a solution has to be suggested within a time that is less than the time the message needs to be transmitted?

radio-frequency

Thoughts on whether that's on topic?

Danu

11:26 AM

Borderline...

skill patrol

12:35 PM

\o @yuggib

yuggib

o/

skill patrol

How are you?

DanielSank

@ACuriousMind As they damn well ought to!

Well, it makes some stuff easier, anyhow.

skill patrol

\o @DanielSank

ACuriousMind

@DanielSank Oh, is this one of the topics one can start a holy war over? :)

skill patrol

12:44 PM

Is there seriously a political party who wants to put the Berlin Wall back up?

ACuriousMind

22 hours ago, by ACuriousMind

(They're a satirical party)

skill patrol

Sorry, missed that.

skill patrol

1:00 PM

thanks

John Rennie

1:13 PM

phdcomics.com/comics/archive.php?comicid=1888

Not that I've ever had a letter like that you understand.

skill patrol

1:28 PM

Have you ever seen a molecule like that @JohnRennie

DanielSank

1:40 PM

@ACuriousMind probably

John Rennie

1:51 PM

@skillpatrol No

DanielSank

@skillpatrol yo

Physics Meta

2:04 PM

1

Q: Very Naive Questions

I am troubled by the comments on this question and many others like it. The (obviously very naive) questioner posits a physically impossible situation (often FTL travel as in this case, but there are other examples), asks what it would imply, and gets jumped on for positing something impossible ...

discussion

Emilio Pisanty

2:21 PM

facebook.com/Dar0s/videos/743288175773777

0celo7

3:02 PM

Hello

@JohnRennie great show

The revolutionaries aren't saints

I've been waiting for a show like this for a long time

I'll break my "no TV" rule for it.

0celo7

3:17 PM

Although the Brits are truly insane ;)

vzn

3:37 PM

@JohnRennie (lol) reminds me, has anyone seen this? bought it at caltech bookstore gotta watch sometime soon phdmovie.com ... (reminds me of silicon valley show also... geekz rulez!)

Secret

Efavirenz

Efavirenz (EFV), sold under the brand names Sustiva among others, is a non-nucleoside reverse transcriptase inhibitor (NNRTI). It is used as part of highly active antiretroviral therapy (HAART) for the treatment of a human immunodeficiency virus (HIV) type 1. For HIV infection that has not previously been treated, the United States Department of Health and Human Services Panel on Antiretroviral Guidelines currently recommends the use of efavirenz in combination with tenofovir/emtricitabine (Truvada) as one of the preferred NNRTI-based regimens in adults and adolescents. Efavirenz is also used in...

user116211

@JohnRennie Reality sucks ;)

Secret

---
(If time) I will be posting some preliminary interpretation thoughts shortly. But first I must present the result of that exercise

vzn

@JohnRennie geez man dont give anyone any ideas around here, lots of impressionable young minds and even ~½-hoodlums o_O

Secret

I have tasted liquid nitrogen before (for minimal chance of killing myself, as a 0.5 mm diameter droplet caught from the making of liquid nitrogen ice cream)

It has a taste best described as concentrated water

of course, my colegues said don't talk about that again because it is reckless

As an aside I am still trying to replicate that taste in cusine so people can experience that taste in a safer manner

(wihtout using actual liquid nitrogen)

Concentrated water taste like this: We all knew that any drinking water has a funny earthy taste. Now imagine this taste being made 10x more intense and persist as an after taste for 10 minutes

(That's what my tongue said liquid nitrogen taste like)

ACuriousMind

3:48 PM

@Secret Water does not have "an earthy taste" unless it's run through peat or something like that.

Secret

I don't really know how to describe that watery taste. The closest thing is the taste of mineral water. But I also found a similar taste drinking distlled water (which has very low concentration of dissolved minerals) also

In short, water is not tasteless to me

One of the h barer said that taste is best described to be "earthy", like rainwater

It might be just a subjective thing, I don't know, perhaps water is really tasteless to most people

ACuriousMind

@Secret It's rarely truly "tasteless", but its taste depends on where it comes from, i.e. the minerals and other stuff in it. There's no one "water" taste.

Secret

hmm, in that case a taste from distilled water might be from some dissolved impurities from the preparation process or the bottle that is holding it

0celo7

What does distilled water taste like

We have some in my lab

But I'd probably die if I ate it

*drank

The star wall is kind of messed up right now.

Transcript for