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ACuriousMind

00:03

:25044167 I'm not telling you what to do, I'm not qualified for that

0celo7

I realized that.

I don't know what to do though.

I'm supposed to have "grown out of" ADHD

And there's no way to actually test it, really

I need to talk to my doctor when I go back home

In any case, @ACuriousMind, I've been looking back at the "all Galilean spaces are isomorphic" thing

and that took so long, the proof is spread out over hours of chat :D

ACuriousMind

Hehe...yes, I recall that that didn't go quickly

0celo7

00:20

my mission this weekend is to piece together scraps of paper and chat to write down the proofs from Arnold so I don't forget them

Dr. Dre - Still D.R.E. ft. Snoop Dogg

►

@ACuriousMind Classic.

@dmckee I have a question about terminology

This problem involves a man throwing a box up at an angle

I'm supposed to calculate the "vertical reaction force" of his feet against the ground

So I can get a recoil force using impulse

is the vertical component of that what they want?

(or the negative of that) or do I also add his weight

ACuriousMind

00:35

Does the box contain fruit for the weird people in those basic math word problems who are always buying and sharing fruits?

0celo7

...huh?

I know how to do the problem, just don't know what they want

So I think I know how to do it :D

ACuriousMind

Nothing, I'm just wondering what is in the box the man is throwing

0celo7

probably useless math books

abstract algebra, topology, etc.

@ACuriousMind you're free to interpret the problem too, but there might be a language barrier

"Assume he completes the throw in 0.25 s. What is the vertical reaction of both his skates (i.e., the sum of the two) onto the ice during the throw?"

it's normal force

so that's dumb

more work :'(

ACuriousMind

I would interpret that such that you should indeed calculate the vertical component of the recoil force.

0celo7

does that include the weight?

assuming you defined it as I did, it doesn't

ACuriousMind

00:41

No, I think it's really just the reaction force to the force the man exerts on the box

0celo7

but that's what the profs want

recoil + weight

i.e. the force his feet are exerting on the ground during the throw

ACuriousMind

Well, I'd call that recoil/reaction + weight, not merely reaction. But, well, I never had mechanics in English, so I don't know about proper terms.

0celo7

I don't either

dmckee

@0celo7 I'm uncertain. If I was writing a question of that kind I would either want the whole normal force acting on the thrower or would ask explicitly about the change of the normal force acting on the thrower.

Unfortunately I don't know that the language there is universally understood to have a unique meaning.

0celo7

@dmckee someone asked one of the profs on Blackboard and he clarified as normal force

@ACuriousMind are all linear transformations invertible?

mapping from n dimensions to n

ACuriousMind

00:55

What have you tried in finding the answer to that question? :P

Because there is a completely equivalent question I'm quite sure you know the answer to

0celo7

uh

I don't really know how to approach it

I told you, my linear algebra class is a joke

ACuriousMind

Do you know that there is a bijection between linear maps $\mathbb{R}^n\to\mathbb{R}^m$ and $n\times m$ matrices?

I.e. every matrix defines a linear map and every kinear map has a matrix?

0celo7

yes, yes, of course

that we did cover

ACuriousMind

Are all $n\times n$ matrices invertible?

0celo7

::smashes head against keyboard::

uh, just forget I asked that

ACuriousMind

00:58

Told you you knew the answer :P

0celo7

ok, so in the "all inner product spaces are isomorphic" proof

we set $\phi(e_i)$ equal to the vector in R^n with 1 in the ith slot, right?

how does one show that $\phi$ has an inverse

ACuriousMind

You, uh, just define it in the same manner?

0celo7

what guarantees that the matrix of $\phi$ is invertible

ACuriousMind

@0celo7 Why are you worrying about that? Just define the inverse $\psi$ by $\psi(x_i) = e_i$, where I denoted by $x_i$ the vector with a one in the ith slot.

Now it's evident that $\psi\circ\phi$ and $\phi\circ\psi$ are both the identites, and thus $\phi$ is an isomorphism.

0celo7

I just am!

ACuriousMind

01:09

@0celo7 Well then, strictly speaking, you don't have the notion of "invertible matrix" since the map is not from a vector space to itself.

And the "matrix" of $\phi$ is rather silly: In the natural way to write it down, it looks like the identity matrix (but it isn't, because it's between different spaces).

0celo7

alright, I get it

all inner product spaces are isomorphic, proof recorded in the holy notebook

what should I write down next?

ah, the unitary matrix thing

@ACuriousMind So I tried checking what matrix commutes with the complex structure

and I found it's just GL(n,R)

I think

I commuted a matrix with four blocks ABCD with the complex stucture and got A=B=D=-C

so that's one block dof, so unless the blocks are complex I got GL(n,R)

Oct 21 at 16:00, by ACuriousMind

As for why those are those are those who preserve the complex structure, just write the complex structure as $\left(\begin{matrix}0 & -E_n \\ E_n & 0\end{matrix}\right)$ and just examine which block matrices $\left(\begin{matrix} A & B \\ C & D\end{matrix}\right)$ with $A,B,C,D\in\mathrm{GL}(n,\mathbb{R})$ commute with it.

ACuriousMind

01:30

@0celo7 I get $B=-C$ and $A=D$, so two $\mathrm{GL}(n,\mathbb{R})$ matrices.

0celo7

...

how did I get D=-C?

that's not even what I have written in my matrices?

smh

see what I mean D:

@ACuriousMind other problem, why do these have to be GL(n,R)?

isn't GL(n,R) invertible? it has to be by defintion of a group

Slereah

it is

0celo7

but I see no reason why A,B have to be invertible

ACuriousMind

@0celo7 Yes. Non-invertible matrices have no chance of preserving the structure

0celo7

@ACuriousMind ok, why?

@ACuriousMind the hermitian inner product defined under that exercise is not antilinear in one of its arguments, is that an issue?

ACuriousMind

01:53

@0celo7 Uh. Thinking about it, we're actually only interested in the group of matrices preserving the structure. Sorry, of course non-invertible matrices can commute with the structure.

0celo7

@ACuriousMind ok, I figured something like that

ACuriousMind

@0celo7 Why would it be an issue?

0celo7

the unitary group preserves a Hermitian inner product

the one he defines isn't a hermitian inner product

it's not antilinear in an argument

ACuriousMind

Is it linear in that argument?

0celo7

it's linear in both

@ACuriousMind I can't get this to work with that antilinearity in the "standard" inner product!

@ACuriousMind yeah, I'm having trouble with this proof :(

ACuriousMind

02:12

@0celo7 Did you remember that the complex scalar multiplication is defined as $(a+b\mathrm{i})x = ax + bJx$ for $J$ the complex structure?

0celo7

complex structure outta nowhere

uh, I did not

first I'll prove "all unitary groups are isomorphic"

then do that one

notation should be obvious: $$\langle x,y\rangle_V=\langle U_V,x,U_Vy\rangle_V=\langle\phi U_Vx,\phi U_Vy\rangle_W=\langle U_W\phi x,U_W\phi y\rangle_W$$

$\phi:V\to W$ was proven to exist earlier this evening

@ACuriousMind make sense?

so $U_V=\phi^{-1}U_W\phi$

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0celo7

@ACuriousMind sorry, I don't see how that helps

my proof of "all unitary groups are isomorphic" never uses the properties of unitarity, antilinearity or hermiticity

ACuriousMind

@0celo7 Write the Hermitian product as $y^T x + \mathrm{i} y^T J x$. Now using $(a+b\mathrm{i})x = ax + bJx$ and $J^T = -J$, you can show it's antilinear in the first and linear in the second argument.

Also: We are under ATTACK?!

2

0celo7

@ACuriousMind ok

DanielSank

02:23

@ACuriousMind Yeah.

Happens all the time.

0celo7

that's not clear at all from the defintion though

wait, if $Jx=\mathrm{i}x$, then $y^Tx-y^Tx=0$ o.o

ACuriousMind

@DanielSank First time I hear of it

DanielSank

@ACuriousMind I mean, websites are attacked all the time.

ACuriousMind

@0celo7 No, you are not allowed to do that. In $\mathrm{i}y^T J x$, the $y^T,J,x$ are treated as a real matrix and as real vectors. Inside that formula, complex scalar multiplication is not defined.

0celo7

o...k, so not sure what to do now

also my unitary isomorphic proof is flawed somewhere

I think...

I never use any of the properties of the complex inner product in the proof

all I use is the general invariance group defintion

ACuriousMind

02:29

@DanielSank Hm, I guess so. Didn't notice anything odd on the site, though

@0celo7 That's because the same proof works to show that all groups that preserve the same kind of inner product are isomorphic, e.g. it also shows all orthogonal groups are the same.

0celo7

> it also shows all orthogonal groups are the same

yes, that's true

ACuriousMind

(of the same dimension)

0celo7

@ACuriousMind that's implied

so, about this antilinear thing

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Tweeted by StackStatus on October 30, 2015 at 2:23 AM

0celo7

@ACuriousMind ok, I need a bigger hint for this problem

I'm not sure what I can and can't do now

ACuriousMind

02:43

Just compute $\langle (ax+bJx,y) \rangle$ and $\langle x,ay+bJy\rangle$ where $\langle -,-\rangle$ is the Hermitian product.

0celo7

NO

I WAS DOING THAT

DON'T HELP ME

;_;

I'm a confused kid

ACuriousMind

You, uh, told me you needed a bigger hint?

0celo7

@ACuriousMind I know, but I had figured it out

now the victory is less good

actually I got antilinear in the second

ok, after a bit of derping I figured it out

ACuriousMind

As usual, then ;)

0celo7

$\langle x,y\rangle =(x,y)+\mathrm{i}[x,y]$. $\langle \eta x,\beta y\rangle=\eta\bar\beta\langle x,y\rangle$

You agree?

ACuriousMind

02:55

yes

0celo7

but...

you said antilinear in the first

that was my derping

ACuriousMind

It, uh, doesn't matter in which it is antilinear

0celo7

;_; I spent so long trying to reproduce your result

@ACuriousMind I'm flipping through problems in Arnold

"find 1-forms on the plane that are not differentials of any function"

I thought the plane had trivial cohomology

ACuriousMind

So?

0celo7

I can't read :)

thought it said closed

0celo7

03:18

@ACuriousMind how exacly is the limit on the top of page 81 chart independent?

ACuriousMind

@0celo7 The "prove this!" there is meant to say you should figure that out yourself :P

0celo7

@ACuriousMind honestly, I've tried

I don't see what property of charts is needed

ACuriousMind

That they're diffeomorphisms.

0celo7

ok, so the charts are diffeomorphisms between the domains

is $\phi$ the image of a curve in $M$ under a chart?

ACuriousMind

Yes.

0celo7

03:27

well, no

it says on the previous page near the bottom $\phi(t)\in M$

ACuriousMind

@0celo7 On the previous page, $M$ is embedded. Here it is not, hence the need for a chart.

0celo7

ah

ok, notation

$\phi$ is the curve

chart is $h$, other chart is $h'$

ugh no clue what to do :/

@ACuriousMind I don't see what diffeomorphism has to do with anything

0celo7

03:51

@ACuriousMind are the charts somehow linear or something?

I'm completely lost

ACuriousMind

04:06

@0celo7 Observe that $\lim_{t\to 0}\frac{(h\circ \phi)(t)-(h\circ \psi)(t)}{t} = (h\circ \phi)'(t)-(h\circ \psi)'(t)$ and the use the chain rule for $(h'\circ \phi)' = (h'\circ h^{-1}\circ h\circ \phi)'$

(where primes except for the one on $h'$ denote usual differentiation)

0celo7

ok, I figured out I need to get the equation after "observe that"

but I don't see how I get it

ACuriousMind

$0 =t-t$

0celo7

o.o

what

ACuriousMind

Add $+t-t$ to the numerator.

0celo7

I know that's what you meant

I still don't see it

ACuriousMind

04:10

You do realize that $(h\circ \phi)'(0)$ is defined as $\lim_{t\to 0}\frac{(h\circ \phi)(t) - t}{t}$?

0celo7

@ACuriousMind no

ACuriousMind

But that is the definition of a derivative!

0celo7

@ACuriousMind huh?

I don't see how that is of the form $[f(x)-f(0)]/x$

ACuriousMind

No, wait :D

Godamnit, why it Arnold define this so weirdly

0celo7

Russian

ACuriousMind

04:12

If he had just taken the equality of the derivatives as the definition instead of that limit, it'd be so much easier.

0celo7

:)

I have 5 books here that do it that way lol

Arnold is the ugly duckling

ACuriousMind

Well, we can always throw the towel and just Taylor expand

0celo7

D:

HERESY

this isn't physics, taylor expansions are forbidden

@ACuriousMind ok, then what?

ACuriousMind

No, it's rigorously allowed.

0celo7

are chart transtitions just linear maps on vectors?

@ACuriousMind is that it? just taylor expand?

ACuriousMind

04:21

We Taylor expand $f := h'\circ h = f(h(\phi(0))) + Df(h(\phi(0))(h(\phi(t))-h(\phi(0)) + r(h(\phi(t))(h(\phi(t))-h(\phi(0))$ where $\lim_{t\to 0} r(h(\phi(t))) = 0$ by Taylor's theorem.

0celo7

lol screw that

ACuriousMind

Hm, few brackets missing. Anyway, doing that carefully yields the chart independence for Arnold's weird definition, since a bunch of terms drop out due to $\phi(0) = \psi(0)$ and the Jacobian $Df$ is linear

0celo7

aight

thanks pal, will look at it in physics lecture tomorrow

@ACuriousMind shouldn't you be asleep?

is this a record?

ACuriousMind

What? No, this is quite far from a record :P

In autumn/winter the sun comes up much later to remind me to go to sleep :D

0celo7

you're a daemon

dude, your sleep schedule is identical to mine and we live 6 hours apart :D

diff eq test on Tuesday is gonna suck

Variation of crap and Laplace transforms and reduction of order

and pursuit curves isogonal trajectories

gulp this might be pretty tough

@ACuriousMind I'm off, I still have a reflection ;P

ACuriousMind