Mathematics - 2016-03-06

Semiclassical

00:00

@mikemiller as i try to read some stuff on symplectomorphisms, the more i'm faced with the fact that while i may understand exterior algebra, i don't know the ins and outs of differential forms :/

in particular, my appreciation of them as forms is pretty terrible

Mike Miller

00:13

I dunno what you mean by "as forms"

Semiclassical

as things which act on vector fields, basically. and on those same lines, i'm not great with thinking of vector fields as partial derivatives

Mike Miller

What do you think of them as? Infinitesimal flows?

Clarinetist

@user2692669 Not necessarily. At least in interest theory, interpolation is quite common

Semiclassical

00:33

@mikemiller to be honest, i tend to think of them in the most elementary way possible. e.g. the electric vector field in physical space as $\mathbb{R}^3\to\mathbb{R}^3$

AreaMan

@Semiclassical A vector gives you a direction to differentiate in (to take directional derivative). Nothing fancy is going on.

Mike Miller

But that's not correct in most settings, @Semiclassical :/

Semiclassical

and with basis vectors being unit vectors $\hat{x},\hat{y},\hat{z}$

not saying it's fancy. just saying it's not something i've had to properly absorb in practice

not saying it is

AreaMan

Ok. I understand that feeling well. @Semiclassical

Semiclassical

it's 'fine' for multivariable calculus, but not for differential geometry

or, to put it another way: if you asked a physicist what a vector-calculus based intro course looks like, you'd see basis vectors, div/grad/curl, divergence theorem etc.

Mike Miller

00:37

But what does curl or div mean?

Semiclassical

their coordinate representation, really

which, no, is not a good description if you want to do things in a coordinate-independent way

Mike Miller

why should I give a shit about div?

I didn't ask how they're defined, I asked what they mean

AreaMan

Differential topology is locally multivariable calculus, and since most constructions are local (differentiation, degree of singularities) or patched from local things (integration, maps) if you understand the multivariable calculus picture well, then understanding the topological story basically just amounts to understand the calculus and feeling confident that everything glues together. Things gluing amounts to them being geometric, i.e. not dependent on your choice of coordinates.

Semiclassical

mostly they come up in intro physics when people talk about maxwell's equations

ugh, my connection right now is just being crap

Mike Miller

And what do Maxwell's equations tell us?

Semiclassical

00:46

maxwell's equations tell you how sources---namely, electric currents and electric charges (or charge densities)---give rise to electric and magnetic fields which themselves act on currents and electric charges

so, for example one expects that a positive charge acts as a source of electric field. that corresponds in terms of vector calculus to Gauss's law $\nabla\cdot \mathbf{E} = \rho/\epsilon_0$ where $\rho(x,y,z)$ is the local electric charge density

and $\epsilon_0$ is a physical constant

by contrast, the magnetic field $\mathbf{B}$ is divergence-free ($\nabla\cdot \mathbf{B} = 0$) since there's no such thing as magnetic monopoles in classical electromagnetism. in particular, that means that all magnetic fields lines will be closed.

similarly, one has Faraday's law $\nabla\times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t}$ and Ampere's law $\nabla \times \mathbf{B}=\frac{\partial E}{\partial t}+\mathbf{J}$ where $\mathbf{J}$ is the electric current density. (i'm missing a physical constant in front of $J$, but w/e)

which tells you that 1) electric currents give rise to magnetic fields which circulate around the currents, 2) time-varying electric/magnetic fields also produce circulation of magnetic/electric field lines.

(blah blah blah physics) @MikeMiller

Mike Miller

So clearly you have a conceptual understanding of div and curl that aren't just formulas... what are they?

Semiclassical

i guess i'd say it as: the divergence of a vector field is a scalar field which describes how much electric field is coming in/out of the neighborhood of some point. and we can identify that with the local electric charge density.

Mike Miller

So we have a notion that vector fields can go into or out of a point. Do you agree that they encode directions, then? :)

Semiclassical

sure

would be hard to talk about electric flux otherwise

i suppose you're aiming for the fact that, if i have an electric field, i can talk about the electric field lines as being themselves integral curves.

Mike Miller

But that's no different that the idea of them being directional derivatices you claimed not to appreciate - a directional derivative is really nothing more than, well, a direction to derive in.

Sure, that's another point to be made. If you believe vector fields are just directions at each point (I certainly do) then these other notions - derivative wrt a vector field, integral curves of a vector field - come immediately and naturally.

Semiclassical

01:02

ehhhh. you also get that same notion of direction from thinking of them as maps to $\mathbb{R}^3$

Mike Miller

Not on a manifold.

Semiclassical

sure. but that's not the intro physics context, is what i'm getting at

Mike Miller

The point is precisely that your notion of maps to R^3 is the same as direction (or weighted direction, if you're oicky about the fact that direction usually means unit vector).

Semiclassical

and in general the physics context is on $\mathbb{R}^3$.

back in a bit, have to eat dinner

Mike Miller

Ok, I give up b

Semiclassical

01:30

back

all i'm trying to say is that, within the context of Euclidean 3-space, there's just not the same motivation for thinking of vector fields in the way that one needs for manifolds. and that's the viewpoint i've been exposed to, both as a student and as one who teaches for intro physics courses. @MikeMiller

now, you can certainly argue that that's a limited way of thinking, and i wouldn't necessarily disagree. but that's the convention i'm used to, and that's why the 'proper' way of thinking about vector fields isn't one which comes naturally to me.

nor is it a perspective i'd expect someone who works in physics (and has never taken a course in differential geometry) to possess. (general relativity is an exception to that, but i've also never had a course in that.)

Akiva Weinberger

02:39

@MikeMiller This question (which you commented on) now has an answer.

Mike Miller

I saw and upvoted.

Akiva Weinberger

OK

Doesn't answer if the projection can contain a disk, though.

Semiclassical

those are some darn nice pictures

Mike Miller

meh

Semiclassical

btw, thinking on it more, i think the issue above really comes down to: in physics, we work in euclidean space and so we end up identifying vector functions with vector fields. that's not the right POV for generalizations to manifolds, but we typically don't do those.

PVAL

02:54

I don't think I care about the magnitude of nonzero vector fields.

Semiclassical

then that's your luxury. it'd be hard to talk about particles accelerated by electric fields, for example, if one couldn't talk about how strong those electric fields are.

but at this point i feel like i'm just punching at shadows. shrug

user19405892

03:44

hi

@Semiclassical i have a geometry question

Semiclassical

04:40

@TedShifrin fyi, i went and turned this into a question on the main site

Forever Mozart

05:28

has anyone used one of the packages successfully? multimedia, movie15, media9, animate, etc... I want to put a video in my pdf

@BalarkaSen I generally agree, but in some areas of math there is a great power in being able to think purely formally/symbolically, for example forcing in set theory.

one foot ahead of the other

to the extreme

John Jack

07:55

Hi @DanielFischer

Forever Mozart

09:28

@m0nhawk wow

moonhawk?

je suis un homme @JeSuis

je suis un mathematician

Hippalectryon

mathématici e n :-)

Forever Mozart

oh right

mathematicienne

mathematicien

oui bien sur!

you are like hippopotamus and electricity?

Hippalectryon

Haha not really, more like a horse and a rooster en.wikipedia.org/wiki/Hippalectryon

JeSuis

@ForeverMozart j'en suis ravi

Forever Mozart

oh wow

Agawa001

09:41

hey frenchies

Hippalectryon

@Agawa001 We're taking over the room :D

Forever Mozart

I'm only canadian

Agawa001

@Hippalectryon oui au fait ;D

Forever Mozart

this show is so sad

Six Feet Under

so sad

m0nhawk

@ForeverMozart Hi?

Ilan Aizelman WS

10:50

How can I prove that Prove $u,v$ are linearly independent $C = \frac{1}{2}uu^T + \frac{1}{3}vv^T$, whereas u,v are orthonormal vectors.

@robjohn any ideas? :)

$C$ = (2x2) matrix

0

Q: Prove $u,v$ are linearly independent $C = \frac{1}{2}uu^T + \frac{1}{3}vv^T$, $u,v$ are orthonormal vectors

Prove $u,v$ are linearly independent $C = \frac{1}{2}uu^T + \frac{1}{3}vv^T$, $u,v$ are orthonormal vectors. I'm not quite sure how to approach that. I believe that because uu^T will give me a vector wih size one, and same about $vv^T$, so when multipled with different scalars, we will surely g...

robjohn

11:16

@IlanAizelmanWS It is definitely unclear what you are asking.

what is to be shown?

what does $C$ have to do with anything else?

user46225

are the simplicial homotopy groups of a Kan complex isomorphic to the homotopy groups of its geometric realization?

oh, yes, they are. that's reassuring.

Vrouvrou

@robjohn i have a question please, to prove that $(R,|.|)$ is complete we take any cauchy sequence then it is bounded we apply Bolzano-Weirstrass then we can deduce a convergent sub-sequence

but then why the sequence is convergent ?

user153330

11:38

@user19405892 also the uh..... weirstrass function is everywhere continuous, yet nowhere differentiable

Ilan Aizelman WS

@robjohn I need to show that $u,v$ are linearly independent.

$C$ is a matrix, $2x2$. and it equals to the equation above. and we also know that $u,v$ are orthonormal vectors.

and $u,v \in R^2$

baxx

12:30

Hi I'm struggling with this trig problem : http://i.imgur.com/LPx2ae5.png
Using the value o f k = 4 here's my working - http://vpaste.net/ND0JT

robjohn

14:08

@Vrouvrou I have no idea what $(R,1,1)$ is.

@IlanAizelmanWS what is the first thing you need to assume when showing linear independence?

user19405892

14:49

hi

@robjohn i need help with a geometry question

robjohn

15:03

@user19405892 that's nice.

user19405892

here is the solution

i don't get the prolongation arugment

robjohn

From the statement, this seems to satisfy the constraints

Maybe I have it backwards... The lengths get smaller as the angles get smaller. Never mind.

user19405892

there is no condition on the last two angles they could be concave or convex

$A_0A_nA_{n-1}$ and $A_1A_0A_n$

yes they can't intersect, but you could make that last angle concave

Mike Miller

15:19

@robjohn: It rained pretty hard down here last night.

user19405892

do two things have to be orientable to be similar?

like we can have congruence if two things are non oriented

in other words one of objects, say a quadrilateral is a reflection of the other

Semiclassical

15:47

morning

Mike Miller

15:57

morning

Semiclassical

any nice math stuff today?

Mike Miller

nah, yesterday was a good day though

Semiclassical

ah, cool

sorry if i got a bit ranty re: vector stuff yesterday

Mike Miller

mm, no need to apologize at all

Semiclassical

mmkay

JC574

16:08

Hey guys

Mike Miller

hi

JC574

@MikeMiller how is your french?

Mike Miller

Somewhere around "reading competence".

JC574

ah same

brb

robjohn

16:42

@MikeMiller we were supposed to get an inch last night. It certainly sounded as if we did.

It rained hard from midnight to 5 AM

Vrouvrou

@robjohn i mean $(R,|.|)$

robjohn

16:59

@Vrouvrou It depends on how you are defining $\mathbb{R}$. It is usually defined as the completion of $\mathbb{Q}$.

Vrouvrou

we define R as all real numbers Q and R\Q

Jakob Werner

hello

JC574

sorted out my problem @MikeMiller :)

Jakob Werner

i'm new here

mh not much action here

Puzzled417

17:20

hi

is it possible to have a concave pyramid?

Mike Miller

@JC574: Phew, that was easy.

dash

17:54

If there exists an inverse to a function f, can I say that this function is bijective?

Mike Miller

18:07

If by inverse you mean two-sided inverse (a $g$ such that $f(g(x))=g(f(x))=x$), and such that the domains and codomains match up ($f: X \to Y$; $g: Y \to X$), then yes. That's actually the same as being a bijection.

dash

And if it's not two-sided?

Mike Miller

Then no, and you should be able to construct counterexamples. Try working with finite sets (let one of them be one point, and the other two, say.)

Frank Science

19:13

Hello, all.

Suppose $f\colon\mathbb{RP}^2\to\mathbb{RP}^2$ is a homeomorphism. How can we show that $f$ is isotopic to identity?

@ForeverMozart Je suis <del>un</del> mathematicien.

Mike Miller

@Frank: Start by having it fix a disc. This reduces the problem to "homeomorphisms of Möbius bands that fix the boundary are isotopic to the identity". From here you want to show that you can isotope it to fix a vertical band.

Then, the complement of the fizes region is a disc, whose boundary is fixed, so now just isotope it to be the identity on that disc.

The hard part here is the vertical band.

Semiclassical

19:50

hi @ted

Ted Shifrin

hi @Semiclassic

You get an answer yet?

Semiclassical

to the asymptotic one?

Tobias Kildetoft

@TedShifrin Hi

Ted Shifrin

Uh huh

hi @Tobias

Semiclassical

nope

i'd be curious if you see any obvious counterexample to the question as i formulated it :)

Ted Shifrin

19:52

I upvoted the question :)

Semiclassical

yaaaay

Ted Shifrin

This is not the sort of mathematics I think about, but I hope you'll get some experts responding.

Semiclassical

internet validation woohoo

yeah, same

Ted Shifrin

When you say the coefficients are comparable, you mean $\lim a_n/b_n = 1$?

Semiclassical

I wouldn't be surprised if the question, as posed, has a reasonably simple counter-example. but i also wouldn't be surprised if strengthening the conditions a bit makes it work (i.e. validate the 'proof' for the hypergeometric series example)

yeah

presumably one could generalize that to something like $a_n/b_n$ always being 'constant enough' but eh

not my ballgame

Ted Shifrin

19:56

Well, be careful. Seems that could give you something like $e^{2x^2}$ versus $e^{x^2}$, whose ratio is far from polynomial.

Semiclassical

good point.

i had more in mind something like $a_n/b_n$ always being 'almost equal' to 1

the funny thing is i'm grading the extra credit assignment which inspired that question right now

Frank Science

@MikeMiller Thanks. I'll spend time thinking. It doesn't seem easy to me.

Semiclassical

which means i end up crediting people for an argument which on its face just isn't sensible

the joys of being a TA :)

Frank Science

Another question: suppose $C$ is a connected topological curve in $\mathbb{RP}^2$. Then the homeomorphic type of $(\mathbb{RP}^2,C)$ is determined by the image of $C$ in $H_1(\mathbb RP^2,\mathbb F_2)$.

These statements are described without proof on Viro's book.

Semiclassical

i am being pretty generous on that part for that very reason, though. not much point being strict about something which isn't really rigorous.

Mike Miller

20:01

@Frank: That one should be easier to prove. Think of intersection numbers with the obvious curve in $\Bbb{RP}^2$.

Ted Shifrin

Well, it's not the place of the physics course to be rigorous. But it would be nice to know if the result is even correct. :)

Semiclassical

quite!

hah, here's something funny

Frank Science

@MikeMiller The problem is converse. When the images are the same, I need to show that they homeomorphic.

Semiclassical

most people stated the pseudo-argument just in terms of "oh, higher powers look the same for both series"

Mike Miller

I know. I mean to say use the intersection number and reduce the actual intwrsection number to 0 or 1 depending on parity.

And do that by hand. You will probably prefer to think of smooth curves here.

Semiclassical

20:03

which is kind've nonsense to me. but on this guys paper, they did the same truncation argument i came up with myself. (which isn't rigorous either, but w/e). so that's pretty neat :)

Ted Shifrin

Well, keep me posted, @Semiclassic. I'm heading out ...

Semiclassical

aye aye

Frank Science

$C$ splits $\mathbb{RP}^2$ into one or two parts. If it's two parts, one is inside and the other is outside. Maybe we can apply Riemann's mapping theorem to identify the inner parts.

JeSuis

@TedShifrin bonsoir

Frank Science

Bonne soirée.

GGG

20:18

Let $T$ be the linear operator on $\mathbb{R}^3$ defined by

$$T(x_1, x_2, x_3)= (3x_1, x_1-x_2, 2x_1+x_2+x_3)$$

Is $T$ invertible? If so, find a rule for $T^{-1}$ like the one which defines $T$.

Prove that $(T^2-I)(T-3I) = 0.$

I've found that $T$ invertible, and that the rule is $$T^{-1}(x_1, x_2, x_3) = (\frac{1}{3}x_1, -x_2, -\frac{2}{3}x_1+x_2).$$

But to show that $(T^2-I)(T-3I) = 0$, am I supposed to use linearity/earlier part, or just the matrix?

Frank Science

These methods are equivalent.

GGG

Cool!

Does anyone know sources for linear transformation problems at level of the one above?

Frank Science

I don't know what that level means, but Linear Algebra Problem Book by Halmos is valuable.

GGG