Mathematics - 2017-09-07 (page 2 of 6)

Danu

12:01

hi

SteamyRoot

ohi

user306715

What is the definition of exterior derivative?

Krijn

@BalarkaSen Hi

Balarka Sen

Hi, @Danu, @Steamy, @Krijn, chat in general

@YCNDM There are textbooks and materials on internet which define the exterior derivative. Why not read those?

Akiva Weinberger

@mago False for $f(z)=1/z$ on $\Omega=\Bbb R-\{0\}$. True whenever $\Omega$ is simply connected ("does not contain any holes").

Balarka Sen

12:14

Or alternatively you could say they have antiderivatives but locally

Akiva Weinberger

(Remember that the complex logarithm is multivalued; we cannot give it a consistent value throughout $\Bbb R-\{0\}$.)

user306715

I have seen two definitions: One of them has an extra term, $\omega ([X,Y])$. I do not know why it appears.

user306715

$d \omega (X,Y)=D_X(\omega (Y))-D_Y(\omega (X))- \omega([X,Y])$

skullpatrol

@BalarkaSen do you know any Sanskrit?

Balarka Sen

@YNCDM You can explicitly take a form $\omega$ and check via computation that it appears (assuming you know the basis-wise definition of $d$). But otherwise I am not sure how to answer "I do not know why it appears". Notice also that $D_X \omega (Y) - D_Y \omega (X)$ is not a linear term; $\omega([X, Y])$ corrects that.

@skulll nope

Danu

12:23

@YCNDM This is not the definition of the general exterior derivative, that's just an identity for the exterior derivative of a one-form

Balarka Sen

@Danu You can extend that to a definition for k-forms though

It involves an ugly sum over the Lie brackets of n-1 of them

user306715

I have seen that as the definition. So what is the original definition?

Danu

@BalarkaSen yeah... but I feel like it's not "morally" the right thing you know

it doesn't tell you what's going on

skullpatrol

:O "morality" from an ex-physics major? :P

joke^

Balarka Sen

True. Actually there are times where it's useful; eg Frobenius theorem is proved using that identity, but I agree; I don't like that formula very much

user306715

12:30

I have also seen the following definition:
$df(x_u,x_v)=\frac{\partial (f(x_v))}{\partial u}-\frac{\partial (f(x_u))}{\partial v}$

user306715

Why are they equivalent to each other?

skullpatrol

What def does your book use?

user306715

I have seen both of them from two different books.

skullpatrol

Ok, how about your prof?

Danu

@BalarkaSen It's definitely one of the most useful formulas... It's not called "Cartan's magic formula" for nothing ;D

(I always saw this as the main application of Cartan's formula)

user306715

12:33

I have no prof in this field.

Danu

Actually, that might not be an accurate statement

The magic formula is overall awesome :P

Balarka Sen

@YCNDM There are multiple definitions, but the one I like is defined basis-wise as $d(f dx_{i_1} \wedge \cdots \wedge dx_{i_k}) = df \wedge dx_{i_1} \wedge \cdots \wedge dx_{i_k}$ and for a 0-form $f$, $df = \sum_{i = 1}^n \partial f/\partial x_i dx_i$.

Krijn

@BalarkaSen How is life?

user306715

So why are the above definitions equivalent to each other?

Balarka Sen

@Danu Hm, I thought Cartan's magic formula was the chain homotopy formula.

@YCNDM It's a computation; check it.

@YCNDM Also this is not a different definition from what you already said. Plug in $\partial/\partial u, \partial/\partial v$ for $X, Y$ in $d\omega(X, Y) = X\omega(Y) - Y\omega(X) - \omega([X, Y])$ and you get that formula

Because $[\partial/\partial u, \partial/\partial v] = 0$

@Krijn Not bad. What about you?

user306715

12:40

Thanks.

Krijn

@BalarkaSen Working on my thesis but can' t get things right

Balarka Sen

How come?

João

Hey people, sorry for interrupting. I am not very active in math.stackexchange.com, and therefore I don't have privileges to comment. It's about this post
https://math.stackexchange.com/questions/1777975/interior-and-closure-of-l1-mathbbn-in-l-infty/1778043#1778043
The comment has a mistake there: where you read -y in B it should be -y in Y

can someone with privileges comment that please?

Krijn

Well after five years I've noticed, mathematics is quite hard

4

Balarka Sen

True

So it's a mathematical trouble you're having?

Krijn

12:45

Mostly getting through an article I can' t seem to grasp

But also some editorial problems

Balarka Sen

ah ok

Krijn

How to (sub)divide chapters, sections

What and what not to scrap

Delete, might be the better word there

Danu

@BalarkaSen Yeah, but you can use it to get that identity, using $da(X,Y)=((i_Xd)a)(Y)=(L_Xa-di_Xa)(Y)$

Balarka Sen

@Danu Ah...

João

Someone here with some knowledge of functional analysis or similar?

Danu

12:48

@BalarkaSen That's how I learned it :)

So Cartan's magic formula expresses $i_X$ as a chain homotopy from $L_X$ to $0$? Never thought about it like that, really.

skullpatrol

I hope you didn't mind me calling you an "ex-physics major?"

Danu

I don't really care what you say about me..

Balarka Sen

@Danu Yeah, that's my usual intuition for it

I like what you said; never realized that formula is a corollary of Cartan

Danu

@BalarkaSen Is it a useful intuition for anything specific?

I never really encountered any use for chain homotopies besides pretty technical stuff so far, that's why I'm asking

Krijn

Why is my pdf suddenly rotated in TexMaker and how do it turn it back

skullpatrol

12:54

Do you know why they made the math mods' office a private room all of a sudden? @Danu

Danu

you switch to TeXStudio, which is the updated version of TeXMaker ;) @Krijn

@skullpatrol No idea and even if I knew I probably wouldn't say :P

Krijn

@Danu :(

João

Never mind, I asked this in the functional analysis chat.

Danu

I'm now pretty weirded out @BalarkaSen. What does the presence of a chain homotopy from $L_X$ to $0$ tell us? That any Lie derivative is null-cohomologous?!

quid

@skullpatrol I was pondering to ask it on meta.

Danu

12:59

^that might be a good idea

skullpatrol

Nah @quid not that important pal.

Balarka Sen

@Danu I am not sure if it's formally useful for anything, but say you integrate $\mathcal{L}_X = i \circ d + d \circ i$ over a small loop $\gamma$, you get $\int_\gamma \mathcal{L}_X \omega = \int_\gamma i_X d\omega$. Think about $\int_\gamma \mathcal{L}_X \omega$ like "$X(\int_\gamma \omega)$"

quid

Well, it's a pretty strange thing to happen.

skullpatrol

Let them have their privacy in peace.

Balarka Sen

So, uh, you're computing a little change in the integral $\int_\gamma \omega$ as you flow along $X$

skullpatrol

13:03

We can make our own room called The Echo Room @quid :-D

quid

or Echo Chamber even.

skullpatrol

Sure, extra rooms are a dime a dozen.

Balarka Sen

@Danu Yeah

Danu

@BalarkaSen I feel like I didn't know that... But maybe I did? Not sure

Kasmir Khaan

hi yall

how do I find a generator of Z/50 *

How do I know it is cyclic in first place?

Danu

13:07

I'm still confused @Balarka. If $\omega=L_X\alpha$ then it's not automatically a coboundary, is it? What about the $i_X d\alpha$ term?

Kasmir Khaan

is it by trial and error? because that can take long time since its order is 20

Danu

Right, @BalarkaSen, $\alpha$ is closed by assumption.

SteamyRoot

@KasmirKhaan $50 = 2 \cdot 5^2$

Balarka Sen

sorry, my internet is being bad

right, $i_X d\alpha$ vanishes

Kasmir Khaan

@SteamyRoot I found the elements but my question was how can we find if it is cyclic or not in an easy way, without rising all elemets to power 20

SteamyRoot

13:16

How do you mean, you "found the elements"?

Kasmir Khaan

The elements in Z/50 *

other notation is U_50

the elements such that gcd(50,k) = 1

@SteamyRoot

Balarka Sen

@Danu I guess "$X(\int_\gamma \omega)$" $= \int_\gamma i_X d\omega$ is like differentiation under the integral sign :P

Kasmir Khaan

{1,3,7,9,11,13,17,19,21,....49}

Balarka Sen

So maybe my supreme conclusion is that Cartan's magic formula is differentiation under the integral sign.

lol

Semiclassical

hi chat

Balarka Sen

13:20

Hi @Semiclassical

skullpatrol

hi pal

Semiclassical

any good math today?

@BalarkaSen I should probably know what Cartan's magic formula is, sigh

SteamyRoot

What you want to use is that, if $n = p_1^{e_1} p_2^{e_2} \cdots p_k^{e_k}$, then $(\mathbb{Z}_n)^* = (\mathbb{Z}_{p_1^{e_1}})^* \times (\mathbb{Z}_{p_2^{e_2}})^* \times \cdots \times (\mathbb{Z}_{p_k^{e_k}})^*$

Semiclassical

something something interior product / exterior derivative / Lie derivative

skullpatrol

How goes your TA situation?

Balarka Sen

13:22

I think Ted likes to say it's like an infinitisimal version of Stokes' theorem but I'm having trouble remembering why

Kasmir Khaan

@SteamyRoot Euler phi function right? but my question was how do we know it can be generated by any element ? ie cyclic

Semiclassical

Good.

Balarka Sen

@Semiclassical yeah it says $\mathcal{L}_X \omega = i_X d\omega + d(i_X\omega)$

Semiclassical

It got resolved.

Kasmir Khaan

Sup semi

Semiclassical

13:23

hi @KasmirKhaan

@BalarkaSen Neat.

skullpatrol

Excellent.

Semiclassical

My main problem with Cartan is that, of the three operations there, I only know how to use $d$ fluently

SteamyRoot

@KasmirKhaan If you calculate what group $\mathbb{Z}_{50}$ is isomorphic to, you'll know whether it's cyclic...

Semiclassical

$\iota_X$ I know at a manipulation level but not intuition

and $\mathcal{L}$? nope

Kasmir Khaan

@SteamyRoot hmm so without isomorphism it cnt be done in an easy way ?_

Balarka Sen

13:24

I never gave interior product much thought. Lie derivative is cool.

SteamyRoot

What do you mean?

Using the formula I just gave you, it's really easy...

Kasmir Khaan

one way is to rise all elements to power 20

Semiclassical

interior product satisfies $i_X \omega(X_1,\ldots,X_n)=\omega(X,X_1,\ldots,X_n)$?

Balarka Sen

Right

SteamyRoot

Well, have fun with that :^)

anon

13:25

@SteamyRoot $\Bbb Z_{50}^\times$

Semiclassical

right.

but while I can write that down, I don't have an intuition for what it says

Danu

@BalarkaSen Hahaha

Kasmir Khaan

@anon Thanks for clarifying things =p

Semiclassical

with $d$ I can say "oh, it's basically a coordinate-independent way of expressing div/grad/curl"

Dattier

Hi, why I can't read @SteamyRoot

Semiclassical

13:27

but with $i_X$ I only have the formal meaning of it.

Dattier

I don't see his commenter

Balarka Sen

@Semiclassical yeah I don't really understand what interior product does. but $\mathcal{L}_X \omega$ is literally $d/dt \phi_t^* \omega|_{t = 0}$

modulo a sign perhaps

Semiclassical

eh, and that doesn't mean much to me since I haven't ever absorbed the 'tangent vectors = equivalence classes of curves' notion

or whatever it's supposed to be

Dattier

Sorry, but why I can't read @SteamyRoot

SteamyRoot

I don't know why you can't read...?

Balarka Sen

13:30

@Semiclassical why do you want to think about tangent vectors like that?

I literally think about tangents as tangents

skullpatrol

@Dattier check your ignore user list

Semiclassical

isn't that what $\phi_t$ is supposed to be communicating here?

but I'm probably misremembering

Balarka Sen

oh it's the flow corresponding to $X$

Dattier

@skullpatrol Thanks

Semiclassical

ah.

skullpatrol

13:31

np

Balarka Sen

any vector field on a compact manifold gives rise to a 1-parameter group of diffeomorphisms "tangent to $X$"

user8469759

hi guys

SteamyRoot

Heh, I was ignored? :P

Danu

@BalarkaSen Seems to be the only way to efficiently compute differentials n stuff to me

user8469759

Not sure you're familiar with it

but say you have a red black tree

Dattier

13:32

@SteamyRoot maybe a mistake of me

user8469759

each leaf is red

skullpatrol

Accidents happen :P

user8469759

and I wonder if the sibling of each leaf is also a leaf

Semiclassical

The main place in physics where I think Lie derivatives show up is in the symplectic manifold formulation of Hamiltonian mechanics.

which would make sense---Hamiltonian flow and all that

user8469759

(question about graph theory/computer science)

Semiclassical

13:34

hah, I just checked my copy of Arnold and he doesn't call it "Cartan's magic formula"

he calls it the 'homotopy formula'

SteamyRoot

Cartan's magic formula is something I saw in symplectic geometry

Semiclassical

He relegates it to a problem, though he may use it again later.

he also seems to avoid using the phrase "Lie derivative"

he acknowledges it, but he first labels it as "$L_X$ is the differentiation operator in the direction of the field $X$."

Kasmir Khaan

Guys can you please clarify something for me ?

I am trying to find the geneators of U_50

the order of U_50 = 20

<a^k> = < a^gcd (20,k) >

{1,3,7,9,11,13,17,19 }

Those must be the power of what?

Balarka Sen

@Semiclassical right, that's what it is

Semiclassical

He also terms it as the "fisherman's derivative"

"the flow carries all possible differential-geometric objects past the fisherman, and the fisherman sits there and differentiates them."

skullpatrol

13:47

nice visual

Semiclassical

The exercise he gives for that has the following hint. "We denote by $H$ the "homotopy operator" associating to a $k$-chain $\gamma:\sigma\to M$ the $(k+1)-$chain $H\gamma:(I\times \sigma)\to M$ according to the formula $(H\gamma)(t,x)=g^t \gamma(x)$ (where $I=[0,1])$." ($\{g^t\}$ is the phase flow of the field $X$.)

"Then $g^1\gamma-\gamma = \partial(H\gamma)+H(\partial \gamma)$."

So that maybe means something to you, even if it doesn't mean anything to me. :P @BalarkaSen

Balarka Sen

right, he's explaining chain homotopy

Semiclassical

ah.

neat

skullpatrol

Would you @Semiclassical consider
A. P. French's physics textbooks as outdated for today's undergrads?

Semiclassical

never read it

so could not say either way

Kasmir Khaan

13:54

<3^k> = < 3^gcd(20,k) >

Why does gcd (20,k ) have to be k < 20

Balarka Sen

@Semiclassical I have explained the chain homotopy formula here (from the topological point of view) if you ever want to read it.

Kasmir Khaan

?

Semiclassical

Neat.

skullpatrol

What textbook did you use in your first year at uni?

Semiclassical

good question. i'm not sure i remember now, tbh

that would've been back in 2006

we currently use Mazur in the version of intro physics which physics majors take

skullpatrol

13:59

I see.

Semiclassical

@BalarkaSen not going to try to understand that right now, but i can see a bit of what's going on

in that the LHS of what Arnold asks you to prove tells you about how the chain $\gamma$ changes in one second of time evolution, and the chain homotopy stuff is telling you how that prism subdivision works

Balarka Sen

mhm

Semiclassical

you can tell I understand this very precisely :P

Balarka Sen

different contexts; same formula

Semiclassical

right.

I still find it weird that Arnold doesn't really seem to use Cartan's magic formula at all in his text

on an entirely unrelated note: f*** seasonal allergies ugh

i took my allergy med two hours ago, and I can still feel the effect of said allergies

Kasmir Khaan

14:13

@Semiclassical can you help me with my abstract algebra exercice?

Semiclassical

the U(50) one?

Kasmir Khaan

Its not about that number but the concept

I found out that 3 generates U50

I know this formula

<a^k > = < a^gcd (n,k)>

Semiclassical

tbh, to the extent that I know anything about this is because of google-fu

Kasmir Khaan

where the order of a = n

Semiclassical

Right.

Kasmir Khaan

14:15

Ill show u where i was stuck

Semiclassical

I'm not hearing a question yet.

ok

Kasmir Khaan

<3^k > = < 3^gcd (20,k) >

so why do we take all the k's that are relativly prime to 20 and less than 20

the less than 20 part I did not get

anon

<3^(k+20)>=<3^k>

Kasmir Khaan

What modulus are we working on ? 20 or 50 ?

This is what comfusing me

anon

3^(k+20) = 3^k mod 50

because 3^20 = 1 mod 50 no?

Kasmir Khaan

14:17

oh right order of 3 is 20

so 3^20 = e

anon

so the residue (3^k mod 50) only actually depends on the residue (k mod 20)

Kasmir Khaan

by residue you mean modulus?

anon

no, residue

3^k mod 50 is a residue, 50 is the modulus

Kasmir Khaan

Ah we call that remainder =p

Okay so if I understood it correctly

Semiclassical

residue, remainder

Kasmir Khaan

14:19

given any U_i

Semiclassical

not much difference even at the level of the popular meanings of the words

Kasmir Khaan

first we try to find a generator

then from that generator call it a

<a^k > = <a^(gcd(n,k) > where n is the order of a

user8469759

Let $f_1,...,f_n$ be $n$ continuous functions in a interval $[a,b]$

Kasmir Khaan

the set gcd( n,k ) are the exponents we rise a to

to get all the generators

Is that right?

user8469759

in $[a,b]$ I define $b(x) = \min(f_1(x),f_2(x),\ldots,f_n(x))$

Kasmir Khaan

14:21

Thanks alot guys ! :)

anon

@KasmirKhaan wait, no

the other generators will be a^k where k is coprime to n

when k is coprime to n, <a^k>=<a^gcd(n,k)>=<a^1>=<a>, so a^k will also be a generator

user8469759

If $A_j = b^{-1}(B_j)$ where $B_j = \left\{ x : b(x) = f_j(x) \right\}$

what can I say about $A_j$?

Kasmir Khaan

@anon yes that was i meant , in case of U50 , order of it is 20, and 3 is a generator

so we rise 3 to phi (20)

to get all geneators

user8469759

assuming $B_j \neq \emptyset$, can I say $\mu(A_j) \neq 0$?

Kasmir Khaan

@anon because in this case gcd(20,3) = 1

anon

14:23

@user8469759 if $B_j$ is a set of $x$-values then why are you taking $b^{-1}$ of it? wouldn't you take $b^{-1}$ of a set of values in the range of $b$, not in the domain of $b$?

user8469759

sorry

I meant to write the counter image

anon

?

user8469759

Let me start over

anon

yeah, I'm asking you why you're taking the preimage of a set of domain values. don't we normally take preimages of sets of range values?

user8469759

I wrote it wrong...

I meant $A_j = \left\{ x : b(x) = f_j(x) \right\}$

so counter image of $b$ when $b = f_j$

can I say the lebesgue measure of $A_j$ is not 0?

you can assume $A_j$ is not empty

so for all $j$ there's an $x \in [a,b]$ where $b(x) = f_j(x)$

anon

14:29

Suppose $\alpha\in A_j$. Let $i$ be a different index. Then $f_i(x)-f_j(x)>0$ when $x=\alpha$. Since $f_i,f_j$ are continuous, this must hold in some $\epsilon_i$ interval around $\alpha$. Then $A_j$ must have size at least $\min\limits_{i\ne j}\epsilon_i$.

user8469759

can we work in another way instead

$f_1,\ldots f_j$

are continuous

(assume from $[a,b] \to Y \subset \mathbb{R}$)

since they're continuous they're measurable

can I use some measure theory argument for the same conclusion?

I guess the argument would be kind of the same

Let $A_j = \left\{ x : b(x) = f_j(x) \right\}$

by definition we have

$A_j = \left\{x : f_j(x) - f_i(x) > 0, i \neq j \right\} = \left\{x : g_{ji}(x) > 0, i \neq j \right\}$

hence $A_j = \bigcap_{i\neq j} g_{ji}([0,\infty))$

because each $f_i$ is continuous so are $g_{ji}$

we have $[0,\infty)$ measurable, so the counter image for each $i\neq j $ is measurable

and finite intersection is also measurable

because $A_j$ is open in $\mathbb{R}$ its measure must be $\neq 0$

is this argument correct?

anyway it should actually be $A_j = \bigcap_{i\neq j } g_{ji}((0,\infty))$

I mean opensets in $\mathbb{R}$ have as basis openintervals $(a,b)$ with $a < b$ so the measure should be not 0

does this work?

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$Mathematics$ Mathematics