Mathematics - 2016-12-09 (page 2 of 7)

@G.Bergeron just a doubt I got, you can't breath in heigths, like let's says, out of a plane because there isn't enough oxygen right? You breath but you don't get enough oxygen ? Or does it has to do with pressure and the lungs not being able to absorbe it?

meow-mix

@G.Bergeron well

elements of the dual projective plane

are lines in the projective plane

and thus, sets of points in the projective plane

G. Bergeron

the span will be orthogonal to the line that pases through P in $R^3$

meow-mix

1:13 AM

@G.Bergeron and... all of the elements of $\lambda_P$ pass through $P$.... by definition

Null

@KajHansen doesn't matter, what excercises you do? cardio or muscle?

meow-mix

@Null i wish i could stick at a healthy weight

G. Bergeron

ok, you gotta be careful when you introduce things like dual or identification of objects in several space. I was thinking of just more and more possibility of what you meant

meow-mix

i've remained ~80 lbs for the past few years

G. Bergeron

8o pounds!

Kaj Hansen

1:17 AM

Some of both @Null. Mostly "muscle" though. Sets of push-ups, pull-ups, that sort of thing.

meow-mix

@G.Bergeron so think of elements of $\Bbb P^2$ as planes in $\Bbb R^3$

Null

@meow-mix i am 85 kilos (~170 pounds), so feel happy

G. Bergeron

I know what a pencil is

and what you meant by that now

meow-mix

yeah im just trying to clarify what i mean because this stuff gets confusing sometimes :(

G. Bergeron

was just advising

Null

1:18 AM

@G.Bergeron are you astyx?

meow-mix

@Null you think that because of his profile pic, right? :P

G. Bergeron

@meow-mix But see, you placed those elements in $\mathbb{P^2}$ and then in the dual...

Null

no, because of the pencil thing

G. Bergeron

@Null what does that mean

Null

"pen orgy tonight, pencils welcome too!"

or the other way around

really, i just found it funny and consider byuing such a pencil case myself :D

meow-mix

1:20 AM

@G.Bergeron so if you understand me

is my statement correct?

Null

. amazon.com/Happy-Jackson-Pencil-Case-Orgy/dp/B00GYKB3FG

G. Bergeron

@meow-mix Ok as planes I guess you could say they are orthogonal, but not as vector spaces

meow-mix

@G.Bergeron of course

G. Bergeron

Take the sphere in $R^3$

Null

@KajHansen do you smoke or have any other bad habits?

G. Bergeron

1:22 AM

then for P, and his antipode, $\lambda_P$ would be the set of the large circles crossing P and the antipode

meow-mix

yes

G. Bergeron

This is the image under the usual coordinates of The pointed plane that has for normal vector the vector that goes from P to his antipode

So then if you consider the lines in $P^2$

Which are the big circles on the 2-sphere, they can be canonically identified with planes of $R^3$

As the planes that form the big circles by intersection

Then naturally those planes would be at "90 degrees" to the span of the above mentioned pointed plane

fluffy_muffin

Would anyone mind looking over a proof I just completed? I'm not sure my third case follows about there being some element in the positive integers that must exist and I felt that stacking "by the inductive hypthesis" was rather weak. postimg.org/image/okqruh057

stating*

G. Bergeron

but as vector spaces they have a non-zero intersection so they are not orthogonal

Does that answers your question?

meow-mix

yes, thanks

@fluffy_muffin did you really use ms word? :(

G. Bergeron

1:26 AM

@Null why mention that! now I want nicotine...

fluffy_muffin

@meow-mix Onenote. Better than word, but not as time consuming as typing LaTeX.

Null

@G.Bergeron liberte toujours i guess

Ted Shifrin

@meow: Are you still being confuzled?

meow-mix

@TedShifrin no, but i have been confuzling

Ted Shifrin

It better be right this time :P

meow-mix

1:27 AM

what better be right?

Ted Shifrin

Whatever you've settled upon.

Akiva Weinberger

Hi

Thinking about your curvature problem

Ted Shifrin

Cool, DogAteMy :)

Akiva Weinberger

I have an intuitive idea but I'm trying to make it rigorous

G. Bergeron

@Null C'est-à-dire?

@fluffy_muffin There are problems

Akiva Weinberger

1:28 AM

Or, rather, I have an intuitive idea that isn't a proof and probably can't become rigorous, and a separate idea that might be able to become rigorous @TedShifrin

fluffy_muffin

I'm assuming in the third case?

Or in general?

Null

@G.Bergeron its just a slogan from Gauloises cigarettes

Ted Shifrin

DogAteMy: That sounds promising.

meow-mix

@TedShifrin i settled that on page 344, (in $\Bbb R^3$) the span $\mathrm{Span}(\xi,\eta)$ is orthogonal to all the planes in $\lambda_P$

and i believe that those vectors uniquely determine the conic

G. Bergeron

Hey ted, is the answer saying for any $V$ around $L$, there is a $U$ around $\infty$ such that $f(U) \in V$ given by any $U$ around $\infty$ minus the Integer points. REmoving points is substracting closed set so still an open then the limit is 0

Ted Shifrin

1:30 AM

That can't be right, @meow. I didn't even look. A plane can only be orthogonal to a line in $\Bbb R^3$.

meow-mix

@TedShifrin well

Akiva Weinberger

So the maximum curvature gives a minimum circumference of this loop, and I need to show that it gives a minimum diameter as well

G. Bergeron

@Null Français?

meow-mix

i just asked if planes can be "perpendicular" and people said yes, so :/

Ted Shifrin

You shouldn't trust too many people, @meow.

Akiva Weinberger

1:31 AM

and that, just like the minimum circumference, the minimum diameter is achieved in all directions in the circle

G. Bergeron

Not as vector space! You have to mention that

Ted Shifrin

Planes can be perpendicular, but people use that word even when they intersect.

G. Bergeron

otherwise people will assume that like I did

Ted Shifrin

DogAteMy: Of course such intuitions require justifications.

Null

@G.Bergeron i can understand only few things of it. i'm german

Ted Shifrin

1:32 AM

But I gave you this question partly because I like it and partly because I wanted to see what you'd figure out.

Null

@TedShifrin why do people call planes perpendicular if they only intersect?

Akiva Weinberger

I haven't thought about the rigorous definition of curvature in a while

G. Bergeron

@Ted He means perpendicular as in the dot of the normal vectors is zero

Akiva Weinberger

and then I only used it to calculate a few things

G. Bergeron

@AkivaWeinberger commutator of the covariant derivative

Akiva Weinberger

1:33 AM

Magnitude of the derivative of the vector with respect to arclength, right? @TedShifrin

Ted Shifrin

@meow: The pencil of lines through $P$ corresponds to a circle's worth of planes containing a fixed line in $\Bbb R^3$ [like a revolving door rotating about the axis]. These planes fill up everything.

Akiva Weinberger

@G.Bergeron I don't know these words

meow-mix

@TedShifrin yes, they fill up everything

Ted Shifrin

DogAteMy: Yes, rate of change of the angle the tangent vector makes with a fixed direction.

fluffy_muffin

@G.Bergeron Were you referring to the conclusion at the end that a^m = 0. Just realized that was wrong. Should have been a = 0.

meow-mix

1:34 AM

but i believe that, using @G.Bergeron notion of "perpendicularity" between planes, the span must be "perpendicular" to all the planes in $\lambda_P$

Ted Shifrin

BTW, DogAteMy, feel free to download my free diff geo text if you're interested :)

G. Bergeron

@TedShifrin I just illustrated this all

meow-mix

simply because all the elements of the span are normals to the planes in $\lambda_P$

Ted Shifrin

So something's still fishy, @meow.

Akiva Weinberger

@TedShifrin Whoah, I completely forgot you had a free text. Either that, or I never knew in the first place?

G. Bergeron

1:35 AM

@meow-mix that is not true

Ted Shifrin

All the elements of that span are orthogonal to the vector in $\Bbb R^3$ corresponding to the point $P\in\Bbb P^2$.

Akiva Weinberger

This? alpha.math.uga.edu/~shifrin/ShifrinDiffGeo.pdf

G. Bergeron

@fluffy_muffin nope, how do you know elements have inverses?

Akiva Weinberger

And is curvature the magnitude of the derivative of the tangent vector with respect to arclength? @TedShifrin

Ted Shifrin

Yup, DogAteMy.

Yes, that's right, provided you mean unit tangent.

G. Bergeron

1:36 AM

@AkivaWeinberger Differential geometry

meow-mix

@Ted sorry for being a hassle, i just want to have the right intuition of these ideas

Akiva Weinberger

@TedShifrin Right, yes

Ted Shifrin

I'm not complaining, @meow. This is new stuff for you.

It's just hard to keep up with hundreds of lines of conversation :)

G. Bergeron

@meow-mix Only the normal vector of each plane will be orthogonal NOT the vector spaces, because they actually intersect at non-zero points

fluffy_muffin

@G.Bergeron Oh. Right... Well no idea how to prove that then, without that assumption :/

G. Bergeron

1:39 AM

@TedShifrin But you didn't tell me if my answer was good :(

Ted Shifrin

I can't keep track of everything in here.

Akiva Weinberger

Sorry ^

meow-mix

@TedShifrin do you agree that each element of $\lambda_P$ has a normal vector of the form $s\xi + t\eta$?

Ted Shifrin

DogAteMy: Interesting where you got that link. The one on my profile page goes elsewhere.

G. Bergeron

each x is the identity over itself...

Ted Shifrin

1:41 AM

Yes, @meow, where both $\xi$ and $\eta$ are orthogonal to the vector $\mathbf p$.

G. Bergeron

to all the planes in $\lambda_P$
and i believe that those vectors uniquely determine the conic

G. Bergeron
G. Bergeron
Hey ted, is the answer saying for any $V$ around $L$, there is a $U$ around $\infty$ such that $f(U) \in V$ given by any $U$ around $\infty$ minus the Integer points. REmoving points is substracting closed set so still an open then the limit is 0

@meow-mix when seen as planes in $R^3$

Ted Shifrin

@G.Bergeron: We're talking only about integer points. You can't remove them.

Akiva Weinberger

Hm. If $T$'s range is the unit vectors and $\theta$ is the angle of $T$, is $dT/d\theta$ a unit vector?

G. Bergeron

But this changes everything!

meow-mix

@TedShifrin ok then i understand :P

Ted Shifrin

1:42 AM

Given a nbhd $U$ of $L$, there's $N$ so that $f(n)\in U$ for all $n>N$.

OK, @meow. Sorry this is so crazy :)

Akiva Weinberger

$d(\cos\theta\hat\imath+\sin\theta\hat\jmath)/d\theta$

Ah, so yes

meow-mix

@TedShifrin no it's fine, really. i'm glad i have the support, otherwise self-learning would be a nightmare :P

Ted Shifrin

no, DogAteMy, not a unit vector :P

G. Bergeron

well then you remove the even points

Akiva Weinberger

No? @TedShifrin

Ted Shifrin

1:43 AM

Oh, yes, but you want $d\theta/ds$ to differentiate with respect to arclength.

I misread.

Akiva Weinberger

@TedShifrin So instead of $|dT/ds|$ I can use $d\theta/ds$

Theyre equal

Ted Shifrin

@meow: I think self-learning is not so easy at all. But one reason I suggested this stuff is that I was willing to answer questions. (Although I'll be traveling at the end of the month for 2 weeks and less around.)

Right, DogAteMy. I said that somewhere up there ....

Akiva Weinberger

Oh, must have missed it

Ted Shifrin

That's the trouble with not being able to ping DogAteMy :D

Akiva Weinberger

You could actually ping me if you wanted to. You know that, right? :P

G. Bergeron

1:45 AM

@fluffy_muffin The idempotency induces a modular structure in the product.... hence you can define inverses

but you have to prove that

meow-mix

@TedShifrin would you like me to read sections 3, 4, & 5 once i'm done with the exercises and everything?

G. Bergeron

@TedShifrin Are you teaching a class here?

Ted Shifrin

@meow: If you're enjoying learning this stuff, keep going. There's cool stuff comin' up.

meow-mix

@Ted ok :)

Ted Shifrin

But there's a lot of exercises in section 2 !!

Nope, @G.Bergeron ... not that I'm aware of :P

Akiva Weinberger

1:46 AM

And then $d\theta/ds\le\kappa_{max}$ means that $s\ge\int_0^{2\pi}d\theta/\kappa_{max}=2\pi/\kappa_{max}$, which gives us our minimum circumference

which is achieved in the circle

meow-mix

Artin's should arrive sometime in the next week or so

Ted Shifrin

Very cool, DogAteMy.

@meow: You've got plenty to do.

G. Bergeron

who is DogAteMy?

Ted Shifrin

Don't forget school, too :D

Akiva Weinberger

Me

meow-mix

1:47 AM

@TedShifrin just saying

Akiva Weinberger

His nickname for me @G.Bergeron

meow-mix

my current math class is a breeze

Ted Shifrin

I'm not worried about your math classes, @meow.

meow-mix

:P

G. Bergeron

@AkivaWeinberger by the way ain't curvature referring to several concepts

Gaussian curvature

Mean curvature

Akiva Weinberger

1:48 AM

Don't those refer to surfaces?

This is a curve

G. Bergeron

Yes

well then torsion

Akiva Weinberger

A plane curve @G.Bergeron

G. Bergeron

-_-

Ted Shifrin

applause for DogAteMy

Akiva Weinberger

@meow-mix Let me guess: The times when you learn something you're not already somewhat familiar with are few and far between?

G. Bergeron

1:49 AM

You asked for curvature and I proposed the most overkill definition ever...

meow-mix

@AkivaWeinberger in math class? yeah

G. Bergeron

What is middle school?

meow-mix

in new jersey, 6th to 8th grade

Ted Shifrin

@G.Bergeron: Just so you know ... I am a differential geometer.

meow-mix

ages 11 to 14

G. Bergeron

1:50 AM

What age is that?

ah ok

Akiva Weinberger

@G.Bergeron Here's a nice question: What's the 4D version of a helix? Which has constant curvature, torsion, and whatever's after torsion

Ted Shifrin

LOL, DogAteMy. Sounds familiar.

G. Bergeron

damn! nice you're not looking at cats video on the internet

meow-mix

(maybe i am)

Akiva Weinberger

@TedShifrin I know you know it

And the solution seemed so obvious in retrospect…

G. Bergeron

1:51 AM

@TedShifrin I am doing my PhD in mathematical physics

meow-mix

if you look at "DogAteMy" without capitals, it becomes "dogatemy", which kind of sounds like a government ruled by dogs

G. Bergeron

diff. geometry is more for fun

Ted Shifrin

cool, @G.Bergeron.

Akiva Weinberger

Wouldn't that be a dogocracy

Ted Shifrin

I don't see it yet, @meow. But that's because your catty meow got in the way.

G. Bergeron

1:52 AM

@AkivaWeinberger what would count as an answer?

Ted Shifrin

DogAteMy: dog-go-crazy?

Akiva Weinberger

Uh, parametric equations @G.Bergeron

Sophie

@AkivaWeinberger that seems more fitting than communism for a shoe banging incident

and for "we will bury them" too now that I think about it

Akiva Weinberger

I don't get it

Shoe-banging incident?

Ted Shifrin

Krushchev, DogAteMy: ancient history.

meow-mix

1:54 AM

actually

it sounds like "dichotomy"

Sophie

@AkivaWeinberger in 1956 Khruschev banged a shoe in the UN and told he was going to bury capitalism

meow-mix

"dogatemy" = dog-ah-te-mee

Akiva Weinberger

Oh, OK, need to brush up on my USSR history

since I completely missed all of it

(in the sense of being born after it)

meow-mix

in fact its the same word except for vowel change, and sounds similar because "g" and "c" are sounded and unsounded versions of the same consonant

Akiva Weinberger

Currently learning US history. Jackson. We'll get to the Cold War at some point

*voiced *voiceless @meow-mix

Ted Shifrin

1:55 AM

Usually high school terms run out before anything close to the modern era.

meow-mix

@AkivaWeinberger IPA nerd >.>

how ironic, a nerd calling a nerd a nerd

Akiva Weinberger

does a linguolabial trill

I've always felt like more languages should have a rolled 'b' sound.

Ted Shifrin

How do you roll a 'b'?

Akiva Weinberger

Since rolled 'r's are so common and since it's fun

@TedShifrin I'd have to show you, it's hard to describe

Ted Shifrin

Only Spanish really rolls an 'r'.

Akiva Weinberger

1:57 AM

Bilabial trill

That's one of my favorite consonant sounds, after the (I think it'd be called) retroflex ejective

A nice hollow 't' sound

meow-mix

@Ted i believe i understand how conics work, but i'll leave it for tomorrow; i've done enough today

Ted Shifrin

calls linguistics expert for phone-a-friend

Best to work out example(s) for that, @meow.

Night!

meow-mix

(what i'm thinking is that $\xi$ and $\eta$ form a projective transformation that acts on an element of $\Bbb P^1$ $[s,t]$ to form a point of $\Bbb P^2$, and the line $L_[s,t]$ is normal to that transformation)

anyways, thanks for the help as always. bye :]

Ted Shifrin

Night :)

Akiva Weinberger

Another great consonant sound is the bidental fricative

Ted Shifrin

2:02 AM

(gesundheit)

Akiva Weinberger

Sounds somewhere between an s and an f

Ted Shifrin

DogAteMy: my linguistics friend at UGA would love you as a student :)

Null

what is the maximum sidelength of a cube inside a tetrahedron with sidelength 1?

Akiva Weinberger

I'm sure I would love him as a teacher

Ted Shifrin

He's a smart dude.

Akiva Weinberger

2:10 AM

So if the curve is $\gamma$ then $\gamma=\int Tds$?

Which means, the "diameter" in the $x$-direction (the difference in $x$-coordinate between the points with vertical tangents) is $\int_0^\pi\sin\theta ds$

which I need to compare to $2/\kappa_{max}<2ds/d\theta$

Correction: $\int_{\theta=0}^{\theta=\pi}\sin\theta ds$

So $\int_0^\pi\sin\theta/\kappa~d\theta$

Ahh. OK. I see how to do it. I'll show Ted tomorrow (since I probably have to leave chat now for tonight).

G. Bergeron

2:34 AM

$(t,\cos (t),-\sin(\frac{\pi }{6}-t),-\sin(t+\frac{\pi
}{6}))$

would be one suggestion for several reasons... But then again In $R^4$ you cannot define curvature and torsion and etc. as you did

Something special happens in $R^3$ which gives you the cross product

Akiva Weinberger

I don't think that works

G. Bergeron

Otherwise it is just the external product, some I am nut sure what you meant by beyond the torsion

My other suggestion would be noting

Oh and I forgot to normalize that thing

...

Akiva Weinberger

Not torsion, then. Normal vector I think?

G. Bergeron

you have the tangeant vector

and thenyou could take a derivative of that

Akiva Weinberger

I'm confused now. There is a thing, I'm forgetting what it's called

G. Bergeron

2:39 AM

It's called the Lie derivative

so the normal vector is related to the Lie derivative

Akiva Weinberger

Wait, no, it is still torsion, there's a way to define it in 4D

G. Bergeron

there is a way to define in any dimensions

they are not vectors anymore though

Ted Shifrin

DogAteMy, there's a "next" torsion for the extra dimension. There's no other name for it.

@G.Bergeron: You still recursively define an orthonormal frame, assuming non-degeneracy.

Akiva Weinberger

Whatever. Ignoring all derivatives and stuff: This is the most general curve in 4D such that there's an isometry of space that takes it to itself.

Is that the right word? Compositions of rotations?

G. Bergeron

A symmetry

Akiva Weinberger

2:45 AM

It continues this pattern: Line, circle, helix, ___

It's the curve that does not fit into three-space but does fit into four, such that every point looks like every other point.

G. Bergeron

@TedShifrin Define it how, by taking derivatives of derivatives of ...?

@AkivaWeinberger any non-coplanar 4D curves do not fit in 3D

Akiva Weinberger

Right, but I want it to satisfy the condition that there are symmetries of space taking it to itself that take any point to any other point

Like translations for the line, rotations for the circle, and screw motion for the helix

which just moves every point further along in the curve

G. Bergeron

A corkscrew symmetry does that for the curve I gave

Akiva Weinberger

You sure?

G. Bergeron

Yes, you translate in the first coordinate while rotating along a circle that is at 45 degree to the three remaining coordinates

So rotation + translation = corkscrew

Akiva Weinberger

2:53 AM

@G.Bergeron Wouldn't that mean that it's actually 3D?

Fitting in a 3D hyperplane

and congruent to a helix

G. Bergeron

A projection certainly is but no

like the helix being projected on the circle

What did you have in mind

?

Akiva Weinberger

If you project it onto the last three coordinates do you get a circle or some weird 3D thing

If it's the first, then the whole curve is just 3D. If it's the latter, I don't know what rotation you're using.

In any case, I have in mind $(A\cos a\theta,A\sin a\theta,B\cos b\theta,B\sin b\theta)$, and I'm pretty sure it's the only solution

Well, it and things congruent to it

In particular, it's bounded.

G. Bergeron

That's a torus

Akiva Weinberger

The symmetry of space is rotating the first two and the last two coordinates independently.

@G.Bergeron A geodesic on it, yes

G. Bergeron

yes

Akiva Weinberger

2:58 AM

And the torus is a "Clifford torus," a subset of the three-sphere with no curvature

Well, no Gaussian curvature

G. Bergeron

wait the torus does not embed in the 3-sphere...

Akiva Weinberger

It divides the 3-sphere into identical halves, and thus has no clearly defined inside or outside. Like the equator on the 2-sphere.

G. Bergeron

Oh but it is not exactly a normal torus

Mike Miller

Huh? Sure it is.

Akiva Weinberger

Yes it does. $(\cos\theta_1,\sin\theta_1,\cos\theta_2,\sin\theta_2)/\sqrt2$

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$Mathematics$ Mathematics