Mathematics - 2017-02-24 (page 3 of 4)

Zach Hauk

9:00 PM

hmm @Kasmir?

Kasmir Khaan

perperdicular o AB on A yes

Alessandro Codenotti

$Bbb C^n$and even better $\Bbb P^n(\Bbb C)$ are nice when you talk about algebraic curves, that's all we used them for in last year linear algebra course

Kasmir Khaan

but why we doing this

Zach Hauk

becauswe

this is how you construct an inverse

anyways let me draw it for you...

that line is perpendicular to AB and intersects A

what is it's equation?

in y = mx+b form

actually, that's unnecessary

where does this lines intersect the circle?

Kasmir Khaan

(4,0)

Zach Hauk

9:05 PM

and?

Kasmir Khaan

it is very unclear from question what they mean by Determine the point which is symmetric to 3+i with respect to the circle

Zach Hauk

they meant invert it around the circle

that's what it means

Kasmir Khaan

(0,2)

Zach Hauk

no, it's (2,2), right?

Kasmir Khaan

(2,2) yes

Zach Hauk

9:07 PM

ok and so

now let's do the inversion part

see those two triangles, ABC and ABD?

we want to create a perpendicular to BC

through C

that equation is y = 2, right?

Kasmir Khaan

yes

Zach Hauk

and, also a perpendicular to BD through D

what's the equation of that? x = ?

Kasmir Khaan

x=2

Zach Hauk

not exactly, since remember, we're 2 coordinates to the right

because the circle's center is z = 2

so add 2 to get x = 4

so the inverted point will be the intersection of these two lines, y = 2 and x= 4

Kasmir Khaan

yes z=4

Zach Hauk

9:09 PM

what is the intersection of them?

Kasmir Khaan

(2,2)

Zach Hauk

no...

x = 4 and y =2

those two lines even tell you the coordinates

what's their intersection?

what is the only point with an x-coordinate of 4 and a y-coordinate of 2?

Kasmir Khaan

(4,2) but i dont think that this is how we do it because we doing complex analysis not geometry

hi @Semiclassical

I need your help to clarify a question i got

Semiclassical

complex numbers = pairs of real numbers with a certain multiplication

Zach Hauk

@Kasmir

Kasmir Khaan

9:12 PM

Determine the point which is symmetric to 3+i with respect to the circle abs(z-2) = 2

Zach Hauk

then convert that number to complex

Kasmir Khaan

what they mean by symmetry here?

Zach Hauk

4 + 2i....

@KasmirKhaan i told you, reflection is just inversion. Inversion around a line is just reflection

Semiclassical

Presumably it means symmetric under inversion through that circle.

Kasmir Khaan

hmm okay thanks guys

i dont see any symmetry in the traditional meaning

of 3+i and 4+2i

Zach Hauk

9:14 PM

because symmetry is usually with lines

Kasmir Khaan

this is why am comfused

Zach Hauk

but it has a deeper geometry meaning when we consider mobius transformations

PVAL-inactive

@Hi Ted

@Ted Hi

Kasmir Khaan

so the answer is 4+2i ?

Zach Hauk

correct

Kasmir Khaan

9:15 PM

@TedShifrin Hi Ted :)

Ted Shifrin

Hi @pVAL

Zach Hauk

i can do it with mobius transformations

Ted Shifrin

hi @Kasmir, @Zach

Semiclassical

Suppose you map z->1/z. This maps circles of radius r centered at the origin to a circle of radius 1/r.

Zach Hauk

if that was what you were looking for

Semiclassical

9:15 PM

In particular, the unit circle is symmetric under that inversion.

Kasmir Khaan

Yes we have to practice that =p

Ted Shifrin

@Semiclassic: But that is not what inversion is.

PVAL-inactive

@TedShifrin I am giving my first real research talk in Binghamton next week

Kasmir Khaan

since Ted just came ill post the question again so we all learn :D

Semiclassical

Am I forgetting something dumb, like -1/z?

Ted Shifrin

9:16 PM

Awesome news, @PVAL. Are you YouTube-ing it for me? :)

Kasmir Khaan

Determine the point which is symmetric to 3+i with respect to the circle abs (z-2) = 2

Semiclassical

No, couldn't be -1/z.

Zach Hauk

well

first subtract 2

$1/(z-2)$

PVAL-inactive

@TedShifrin No

Zach Hauk

but then you gotta have a circle of radius 2

Ted Shifrin

9:16 PM

It's not holomorphic, @Semiclassic. Inversion in the circle of radius $R$ is given by $z\rightsquigarrow R^2/\bar z$.

Semiclassical

Hmm.

Zach Hauk

so you get $4/(\bar{\;z-2\;})$

PVAL-inactive

I think technically the seminar is Geometry and Topology, but I'm mainly assuming a topology background and reviewing the required geometry

Semiclassical

Do you mean, inversion isn't a holomorphic operation?

PVAL-inactive

@Ted do you know what surgery on a knot is?

Ted Shifrin

9:17 PM

@PVAL: If you have presentable notes, I'd love to see afterwards.

Zach Hauk

ok, @Kasmir?

Ted Shifrin

I know approximately, @PVAL. I was around Kirby for years :P

PVAL-inactive

@Ted I'll send you a draft of the paper its about.

Ted Shifrin

Nope, @Semiclassic.

Zach Hauk

so we have $4 / (1 - i)$

is that right, @Ted?

Kasmir Khaan

9:18 PM

z-2 ( bar ?

Ted Shifrin

It's the equivalent of $z\rightsquigarrow \bar z$ if you turn the real axis into the circle, @Semiclassic.

Semiclassical

Then I fail to see how on earth z-> R^2/\bar(z) would be holomorphic.

Zach Hauk

yeah all of that is a complex conjugate, mathjax messed up

Kasmir Khaan

Zack am still trying to read that

Ted Shifrin

@Zach: You need \overline

PVAL-inactive

9:19 PM

I'm wondering if that concept is something that I can assume for people to know. There's so much basics in contact/symplectic geometry i need to state my results (and make them interesting).

Zach Hauk

@Ted oh, :P

i mean the equation

$4 / (\overline{z-2})$

Ted Shifrin

The topologists certainly know, @PVAL. We geometers, not so much, perhaps.

Zach Hauk

@KasmirKhaan so we want the plane to shift 2 units right

Semiclassical

If by inversion one means (for instance) z->1/bar(z) then I don't see how that could be holomorphic.

Zach Hauk

so we subtract 2, so that everything has to be 2 larger to be the same

we have $(z-2)$

we then invert around a circle of radius 2

and our mobius transformation because $4/(\overline{z-2})$

make sense?

Ted Shifrin

9:21 PM

@Kasmir: FYI, it's simple geometrically. You draw a ray from the center of the circle (radius $R$) passing through the point. If your point is distance $r$ along that ray, then you go to the point distance $R^2/r$ along the ray.

PVAL-inactive

@TedShifrin Unfortunately probably much of neither group knows much about contact geometry.

Semiclassical

Which means it has a physics example, hilariously enough.

Ted Shifrin

That is, the product of the distances is $R^2$. So in your case, the point $3+i$ is distance $\sqrt2$ from the center of the circle and the circle has radius $2$, so I go a distance $4\sqrt2 = 2\sqrt2$ along the ray.

Zach Hauk

@Ted by the way, im going to send an application to mathcamp anyways, because apparently even if i don't go, i won't have to take the quiz next year, and i already did this year's quiz and don't want the work to go to waste

Ted Shifrin

@PVAL: You probably need to give some summary background, yes.

That's good, @Zach. And I still suggest the other options, too.

Zach Hauk

9:23 PM

my parents said i could only go if they were free

Semiclassical

Put a charge $q$ at a radius $r$ from the inside of a grounded sphere of radius $R$. Then the electric field is obtained by placing an image charge $q'$ at a distance $R^2/r$ from the center. (I forget what the charge strength should be.)

Zach Hauk

and it seems like those don't offer financial aid to cover it all

Ted Shifrin

@Kasmir: So that lands me at the point $2(2+i)$.

PVAL-inactive

@Ted I probably won't make readable notes specific to this talk, but I'll send you a draft of the work its based on once its written a little nicer

Zach Hauk

which is 4+2i :P

Ted Shifrin

9:24 PM

Thanks, @PVAL. I don't need it to be wonderful. I just would like to get a rough idea :)

PVAL-inactive

I'm mainly working on prepping the talk so the draft is kind of on the backseat

though they're expositons of the same idea.

Ted Shifrin

@Zach: I was trying to make it a bit more geometric. But meh :P

@Semiclassic: Yes, that's the same inversion.

Zach Hauk

@TedShifrin i did a full on geometric explanation

Semiclassical

I imagine there's a simple geometric construction leading to the R^2/r.

Zach Hauk

and he told me he wanted to do it complex analysis-y

Ted Shifrin

9:25 PM

@Semiclassic: It's just scaling $r\rightsquigarrow 1/r$ from the unit circle up to a circle of radius $R$.

PVAL-inactive

@Ted Here's an abstract, Geography and classification of symplectic fillings of Legendrian surgeries

Understanding the smooth topology of 4-manifolds is a notoriously hard problem. Due to a theorem of A.A. Markov, we cannot hope for similar classifications like those in the lower dimensional case. Due to exoticness, we cannot hope to understand 4-manifolds completely just by studying their algebraic topology as we can in the higher dimensional case. However, if we assume a 4-manifold admits a symplectic form and the boundary is a certain contact manifold (with a natural compatibility cond…

Ted Shifrin

I wasn't criticizing, @Zach.

Semiclassical

Sure. I mean geometric in the sense of a construction.

PVAL-inactive

Any questions are welcome

Zach Hauk

:P

Semiclassical

9:26 PM

But Wiki already has that

Zach Hauk

oh, good news

Semiclassical

Zach Hauk

i got into all honors classes for 9th grade

PVAL-inactive

Hopefully there aren't any gramatical errors.

Ted Shifrin

@Semiclassic: Not hard. You just need similar triangles.

Zach Hauk

9:27 PM

if anybody cares about an insignificant kid's education

Semiclassical

weird, the wikipedia image isn't loading for me here. oh well: commons.wikimedia.org/wiki/File:Inversion_in_circle.svg#/media/…

skill patrol

Good job @ZachHauk

Zach Hauk

thanku

Semiclassical

Yeah, it's simple. I just wasn't remembering.

Ted Shifrin

Hmm, I never thought about it, @PVAL. When is the boundary of a symplectic manifold w/boundary a contact manifold? I can't contract the symplectic form against the normal vector field?

@Zach: Well, of course you did.

Zach Hauk

9:28 PM

but

i didn't expect it

Ted Shifrin

Duh.

PVAL-inactive

@Ted The case I am talking about is exactly when the normal dilates the symplectic form

Ted Shifrin

Hmm, @PVAL, what precisely does that mean?

PVAL-inactive

Do you know what the symplectization of a contact structure is

You look at Y^3 \time I

Ted Shifrin

I used to :P

PVAL-inactive

9:30 PM

and take $d(\alpha e^t)$

Ted Shifrin

Ah,

PVAL-inactive

Where alpha is the contact form

the conditions is that near the hypersurface, you are locally symmetric to that

Ted Shifrin

Gotcha.

@Kasmir got very quiet. I assume he knows how to do this now.

PVAL-inactive

Due to a prop. of Eliashberg that just meant you have a vector field transverse to it with the contraction of a contact form the symplectic form.

and the flow coordinates in t just become your I

Ted Shifrin

Right.

Explaining why the $e^t$ is in the original rather than, say, $t$.

Alessandro Codenotti

9:33 PM

Hi @Ted

PVAL-inactive

The typical problem in this setting

Ted Shifrin

Heya @Alessandro

PVAL-inactive

is to determine the symplectic manifolds (up to diffeo, symp. deform. or w.e. ) which have boundary a certain contact 3-manifold.

Zach Hauk

@TedShifrin Uhh I had a proof for the first exercise of section 3 but i don't know if it's true

Kasmir Khaan

HI guys

sorry my mom called

And Ted was right !

Ted Shifrin

9:35 PM

OK, @PVAL ... rings bells.

Kasmir Khaan

I know how to do it now :D

thanks Ted semi Zack :D

Zach Hauk

np :P

PVAL-inactive

and now we have topologist-like tools to talk about contact 3-manifolds in terms of surgeries on Legendrian links

due to a lot of peoples work.

So the question becomes given a contact surgery on a Legendrian link, can you determine what the symplectic fillings are?

Ted Shifrin

@PVAL: I think I've heard my former colleagues Matic, Kazez, and their friends, talk about such things.

PVAL-inactive

They certainly would have.

Ted Shifrin

9:37 PM

@PVAL: Just in terms of the link and the surgery type?

PVAL-inactive

Yah

Ted Shifrin

@Zach: That's supposed to be a standard linear algebra exercise everyone does in linear algebra :P

Zach Hauk

i didn't find it hard

PVAL-inactive

so the thing I proved is that if the Legendrian has enough stabilizations (in the front diagram this means there is lots of zig zags), then every filling has the same betti numbers (and hence Euler char.) and signature.

Ted Shifrin

Make sure you did it right :)

Zach Hauk

9:39 PM

by definition those vectors form a symmetric linear map, right?

or am i using a false converse

Ted Shifrin

That makes no sense. Besides, you just have $k$ vectors in $\Bbb R^n$.

What do you mean that vectors form a linear map?

PVAL-inactive

this is for the slope corresponding to a plurisubharmonic function (the slope where the thing is canonically fillable)

Zach Hauk

like

PVAL-inactive

so everything looks a lot like the canonical filling, and I don't know if it always is

Zach Hauk

if we consider the matrix whose column vectors are those set of orthonormal vectors

PVAL-inactive

9:41 PM

@Ted Do you know what the front diagram is?

Ted Shifrin

Oh, @Zach, but you don't necessarily have $n$ of 'em.

Nope, @PVAL.

PVAL-inactive

If we take $\alpha= dz+ydx$ and look at the projection to the xz plane

Zach Hauk

oh

Ted Shifrin

@Zach: Just write a standard linear independence proof. "Suppose $\sum c_iv_i = 0$. We must show that $c_i=0$ for all $i$."

PVAL-inactive

we get a projection of Legendrian knots where the crossing is determined by the slope

Zach Hauk

9:42 PM

alright

PVAL-inactive

and there aren't any vertical tangencies

so we have these cusps on the left and the right

Zach Hauk

assume that they are linearly dependent

Ted Shifrin

NOOOO @Zach.

Get your logic right.

Alessandro Codenotti

what are you trying to prove @Zach?

Astyx

Hi chat

Alessandro Codenotti

9:43 PM

Hi @Astyx

Astyx

Again to some

Ted Shifrin

@PVAL: Slow down. What am I projecting to the $xz$-plane?

Zach Hauk

i was doing reverse-contradiction

haha

PVAL-inactive

@Ted If you look at the first figures here arxiv.org/pdf/math/9803019.pdf

A knot which is an integral curve of the contact structure

TK \subset of the 2-plane field

Zach Hauk

so we have tha

Ted Shifrin

9:44 PM

@PVAL: Ah, it is a Legendrian knot.

Hi @Astyx.

I see @PVAL.

@Zach: You still need to learn linear algebra better :)

PVAL-inactive

Yah, and so there is (really are two) a natural operation (s) on legendrian knots called stabilization, where you add a zig zag into a horizontal arc in this diagram

Zach Hauk

there exists some $c_i$ for which one of them are non-zero, such that $c_0\vec{v}_0 + c_1\vec{v}_1 + \dots = 0$

PVAL-inactive

It's a theorem of Fuchs and Tabikanov(? I don't know the second name)

that any two Legendrian links which are topologically isotopic

Ted Shifrin

No, the best form is to remove all contradictions, @Zach. Just start with my sentence. And argue directly that all the $c_i$ must be $0$.

PVAL-inactive

become Legendrian isotopic after enough of these moves

Ted Shifrin

9:47 PM

Oh, cool, @PVAL. Yeah, I know none of this.

@Zach: BTW, there is a short appendix on linear algebra stuff.

PVAL-inactive

@Ted Figure 7 of the Gompf paper I linked shows stabilization drawn for you.

Ted Shifrin

OK.

Zach Hauk

Subtract one of the non-zero $c_i$ from the right hand side

say it's, $c_k$

PVAL-inactive

and now my theorem is that after enough stabilizations the symplectic fillings of a contact surgery on something after a sufficient stabilizations (this is the surgery corresponding to a plurisubharmonic function on the 4 manifold which naturally bounds the surgery)

have a unique set of characteristic numbers

Ted Shifrin

@Zach: No. You're still assuming someone is nonzero. I want none of that.

Zach Hauk

9:50 PM

we get $c_0\vec{v_0} + \dots = - c_k\vec{v_k}$

PVAL-inactive

(betti numbers euler characterisitic, and signature)

Ted Shifrin

Try to be as simple as possible.

PVAL-inactive

up to blowing up and blowing down

Ted Shifrin

That's pretty cool, @PVAL.

Zach Hauk

hmm how should i make this simpler

Ted Shifrin

9:51 PM

Do a $1$-line proof to show that $c_1=0$, @Zach, from the original equation.

PVAL-inactive

so it seems they could all be the same symplectically (and diffeomorphic), but I think the techniques to prove that would be orthogonal to the techniques I got the result with

Ted Shifrin

Well, @PVAL, if you get to show off different techniques, all the better for you :)

PVAL-inactive

Well I don't know much about the techniques that seem like they could work.

Ted Shifrin

Ah, time for a coauthor ? :)

PVAL-inactive

I am happy with the result as is

Ted Shifrin

9:53 PM

LOL, sure. I'm thinking future ... :)

PVAL-inactive

@TedShifrin That would be nice

Most of the younger mathematicians in my field have almost all collaborations as their work (besides their thesis)

I feel like people generally though when they hear about my problems tell me that they will make sure to avoid them

Ted Shifrin

There's no question that the collaborative stuff I did was far better than the solo stuff. Not that I'm totally stupid. It depends on personality, of course, but I truly enjoyed the collaboration and it was much more fulfilling.

PVAL-inactive

as to not mess up a possible thesis at this point.

Ted Shifrin

Yes, of course. That's very kind of them, too. But ... later on ...

Astyx

Is there a time when you feel you know maths well enough to consider yourself good at it ?

Ted Shifrin

9:59 PM

@Astyx: There's always more to learn/understand :P

PVAL-inactive

@Astyx How much money did you get because of your skill?

Zach Hauk