Mathematics - 2017-02-18 (page 1 of 6)

Ted Shifrin

00:00

Orientation taken into account.

Danu

@MikeMiller Moduli spaces of what, actually?

Ted Shifrin

We'll check in a bit that what we do is well-defined.

Mike Miller

holomorphic curves

Ted Shifrin

@Danu: Let's wait until you're done with the other convo.

Danu

Okay @TedShifrin

I'm listening

Ted Shifrin

00:00

So $e_1,e_2\in\Bbb R^{k+2}$, I guess.

LegionMammal978

@TedShifrin So would $r_n=(-1)^n(\cos{\frac{\pi n}p}+i\sin{\frac{\pi n}p})$ work better?

Ted Shifrin

Consider $z=e_1+ie_2\in\Bbb C^{k+2}$.

PVAL-inactive

I believe all the stuff in the Big McD-Sala book. So really I believe the case where the curve is rational and the class is J-indecomp.

That's also probably the minimum I need to use...

Ted Shifrin

Huh, @Legion? Just do $n\cdot (2\pi/p)$.

PVAL-inactive

So I guess I have to believe it

Ted Shifrin

00:02

@Danu: What is $z\cdot z$?

(Orthogonal inner product, not hermitian)

Danu

@TedShifrin You mean just $e_1\cdot e_1 +e_2\cdot e_2=2$?

Ted Shifrin

Try again :)

We're complexifying the inner product. So the scalars are still there!

Danu

I don't know what you mean---what is the definition?

Ted Shifrin

If $z=x+iy$, $z\cdot z = \|x\|^2 + 2ix\cdot y - \|y\|^2$.

Everything is $\Bbb C$-bilinear!

Danu

Okay, sure

Ted Shifrin

00:06

So you get $z\cdot z = 0$.

This will give our hyperquadric in $\Bbb CP^{k+1}$.

Danu

I realize

@TedShifrin Ah

Ted Shifrin

Now we just have to check that if we change our orthonormal basis, that we don't mess up the point $[z]\in\Bbb CP^{k+1}$.

leaves it to the reader

PVAL-inactive

@MikeMiller I mean John Baldwin also commented on that article.

LegionMammal978

@TedShifrin Still don't really get this. How does this relate to, say finding the conjugates of $1+\sqrt[3]2$? They seem to be of form $x=1+\sqrt[3]2r_n$ where the $r_n$ are the third roots of unity

Mike Miller

Sure. His comments were good too.

PVAL-inactive

00:10

I'm also likely spending a night in Jonathan Williams house in a couple of weeks

Mike Miller

I don't know him

PVAL-inactive

He's at Binghamton

He was a student of Gompf's.

Mike Miller

Gotcha

Ted Shifrin

@Legion: OK ... And note that if you multiply those two other roots together, it should look a lot like your factorization of sum of cubes formula.

@PVAL: He was at UGA for 4 years or so. Postdoc and then instructor one year.

PVAL-inactive

@Ted ya under Gay.

Ted Shifrin

00:13

I hadn't thought about it, but that makes sense, of course.

Balarka Sen

Ok, gonna try and get a wink

Ted Shifrin

Night, @Balarka.

LegionMammal978

@TedShifrin Ineresting! I get $2^{\frac03}-2^{\frac13}+2^{\frac23}$!

Ted Shifrin

Does that make sense?

LegionMammal978

I see now.

Lemme think for a sec

Danu

00:15

@TedShifrin So just that orthonormal bases of 2-planes are related by multiplication with a complex number?

Ted Shifrin

Right, @Danu :)

Unit complex number.

Danu

Yeah, of course

thanks, that's helpful

LegionMammal978

This would appear to correspond to the formula $1^3+(2^{\frac13})^3=(1+2^{\frac13})(1-2^{\frac13}+2^{\frac23})$

Danu

Any more descriptions flaggy manifolds as projective algebraic varieties? :P

Ted Shifrin

The next thing (which probably doesn't matter to you) is that an oriented surface is minimal iff the Gauss map to this Grassmannian is anti-holomorphic with the obvious complex structures.

Pretty cool, eh, @Legion?

LegionMammal978

00:17

Yup

Ted Shifrin

Um, these are the only real Grassmannians, @Danu (not counting $\tilde G(k,k+2)$, which is the same).

I'm not sure if I remember how to prove that, though.

LegionMammal978

So, now equipped with this $2^{\frac23}-2^{\frac13}+1$, how can I find the other 2 solutions? I guess I could look at it as $x^2-x+1$ with $x=2^{\frac13}$...

Ted Shifrin

What do you mean? No, you want roots of a different quadratic.

But they are just $1+\root3\of 2$ for the two complex choices of the cube root you started with hours ago :)

Danu

@TedShifrin I'm still looking for an explanation why $G_2/U(2)$ is also a complex quadric... Kotschick gave me a lead on that though, I'll check that out soon.

Mike Miller

It's a shitshow out there. Our wifi keeps going down.

Ted Shifrin

00:21

It's gale winds here, @MikeM.

But we've had no water for all day ... goddamn renos.

@Danu: Is it a hyperquadric in quaternionic projective space?

Or in an actual complex projective space?

Danu

@TedShifrin Complex!

LegionMammal978

Okay, so now, setting $r_n=(-1)^{\frac n3}$, the triplet would correspond to the values of $1+r_n\sqrt[3]2$. Now, how would I go about finding radical expressions for $r_1$ and $r_2$? It can be easily be seen that $r_2={\bar{r_1}}$ and vice versa, but I can't figure out much beyond that

Ted Shifrin

Hmm ... Oh, maybe it's formally analogous to what we just did, actually, but you have to complexify $\Bbb C^?$ and act by unitary instead of orthogonal.

The quadratic you wrote down up there ^^^ is almost the right one, @Legion. $x^3-1 = (x-1)(x^2+x+1)$.

Zach Hauk

I've returned for my dinner... a.k.a. jeopardy break

Ted Shifrin

for? from?

I used to love watching Jeopardy. Haven't done it in decades, though.

LegionMammal978

00:27

@TedShifrin So the solutions of $x^2+x+1$ would be $r_1$ and $r_2$, correct?

Ted Shifrin

Yuppers, @Legion.

Zach Hauk

I meant fromn

LegionMammal978

So $r_{1,2}=\frac{1\pm i\sqrt3}2$, lemme now go solve the original problem...

MickLH

Really?

I just saw flag

wutimiss?

@DanielSank effort.

Danu

@MikeMiller Thanks for pointing out the comments---fun read.

MickLH

00:32

@DanielSank I could barely click "reply" these two times

Rand al'Thor

@Jack @Zach There's an "ignore" button you can use in chat so as not to see messages from someone who annoys you.

LegionMammal978

So $\frac1{1+\sqrt[3]2}=\frac{1-\sqrt[3]2+\sqrt[3]4}3$, correct? And now I know all about roots of unity :P

Jacksoja

Thank you rand al thor

Zach Hauk

@Randal'Thor To be honest, he just came around and said I'm annoying. I don't find any reason to ignore him thus far

MickLH

lollll

@Jacksoja Are you sure you're not just projecting?

Danu

00:33

→ 23 messages moved to Trash

Let's move on.

skill patrol

I agree.^

Ted Shifrin

@Legion: To motivate you a bit more, the general way to do things like this is with the Euclidean algorithm for polynomials. Like if I tell you that $\alpha$ is a root of $\alpha^3=2$ and ask you for $1/p(\alpha)$, where $p$ is a polynomial with rational coefficients, you can figure it out :)

@Danu: You toss me so effortlessly in the trash. I'm not sure I appreciate it!

Danu

@TedShifrin Would you prefer it if one half of an argument remains in the chat room? I think that's not a good solution.

Ted Shifrin

I recognize Jacksoja's name, but I can't remember what kind of troublemaker he was in the past.

Well, @Danu, make it look like you expended serious effort, then :D

OK, I'm leaving for now. I just gave Legion a new task. :P I have to make a phone call.

@Danu: You can explain the next level of hyperquadric to me later this weekend when you figure it out.

Danu

@MikeMiller I find it baffling that Lubos somehow finds the motivation to even posts comments like those on the article...

@TedShifrin I wanted to explain the ACS on the bundle of complex structrues today

But oh well...

skill patrol

00:36

Cya prof.

Danu

Time to move on @Jacksoja.

Jacksoja

Sorry again folks

skill patrol

Chill out.

Jacksoja

Roger that

skill patrol

:-)

MickLH

00:37

aw did I miss an implicit answer to my question

Zach Hauk

hmm

euclidean geometry is not one of my strengths

LegionMammal978

00:51

@Ted But yeah, thanks for the free math lesson :D

@TedShifrin Hmm... I'll be thinking about this, Euclidean algorithm is probably implemented in a surprisingly subtle way ;)

Ted Shifrin

01:45

@Legion: Not so subtle. But if $p(\alpha)\ne 0$ (which we certainly want to assume), this means that the polynomial $f(x)$ of which $\alpha$ is a root (which we should assume cannot be factored) cannot divide $p(x)$, and so their gcd is $1$. :P

Semiclassical

hi @ted

Brody

Hey @Ted @Semiclassical

Zach Hauk

welcome back, @Ted

Daminark

Hey everyone!

anon

been awhile since both of us here at the same time @robjohn

at least me as anon

@Daminark hello

Daminark

02:05

How's it going?

arctic tern

meh

Daminark

Darn, what's wrong?

Semiclassical

@ZachHauk Finally got a good picture:

And, yes, AD+BC=AB. (I used the tangency of the circles centered at A,B in order to construct it, actually)

Brody

Exercise: Suppose $x,y\in\Bbb R\setminus\Bbb Q \;\; | \;\; (x<y)$. Construct (a specific) $z\in\Bbb R\setminus \Bbb Q \;\; | \;\; (x<z<y)$.

Zach Hauk

@Semiclassical hmm

how should i got about proving that those circles are tangent?

Semiclassical

02:08

shrug

I'm just happy to have the picture.

Zach Hauk

:P

Semiclassical

One thing to note, though, is that the angle bisectors of BCD and CDA both meet at O.

Ted Shifrin

Good picture, @Semiclassic.

Zach Hauk

hm that's odd

how did you arrive at that conclusion?

Semiclassical

It's because of the circle-tangent-to-other-sides condition.

Ted Shifrin

02:10

If you can prove that, Zach's theorem is done.

I think.

Remember Ted's favorite result about angle bisectors ...

Daminark

For a second I thought you said "Zorn's lemma" instead of "Zach's theorem"

Ted Shifrin

glares @Daminark

Zach Hauk

angle bisector theorem?

Semiclassical

Well, the point is that the triangle BCD should have the same incenter as CDA.

And one constructs the incenter as the intersection of the angle bisectors.

Zach Hauk

yeah, i found a kite

Ted Shifrin

02:12

@Zach: I love this theorem. Wonderful vector proof, or a classical Euclidean proof.

Zach Hauk

and realized that's why it occurred

Ted Shifrin

The angle bisector divides the opposite side in a ratio given by the adjacent sides.

Zach Hauk

which theorem?

Ted Shifrin

I done just telled you.

Semiclassical

A quick discussion on how I constructed it, for clarity.

Daminark

02:13

Oh so we're going to do Hahn-Banach on Monday

And it uses axiom of choice, which Schlag hates

Ted Shifrin

Monday is a national holiday?

Zach Hauk

@Semiclassical i noticed that if you let $X$ and $Y$ be the points of tangency on $CD$ and $BC$ respectively, then $OXCY$ is a kite

Ted Shifrin

LOL ... I don't understand this obsession with AC.

Semiclassical

Oh, neat.

Daminark

Really? We're having school on Monday lmao

Ted Shifrin

02:14

Well, UGA never observed it, either.

Zach Hauk

(and thus, that $OC$ bisectt $XCY$

Ted Shifrin

When I was a kid (in CA), Lincoln and Washington both had days off.

Zach Hauk

cuz $CX = CY$ by power of a point and

Ted Shifrin

I think Zach is gonna figure this out. I'm no longer paying attention.

Zach Hauk

a ngles $OXC$ and $OYC$ are both 90 cuz tangent lines are perpendicular

Daminark

02:15

But yeah, I actually told Schlag about Norman Wildberger

Ted Shifrin

Oh, ugh.

On second thought, with your judgment, don't tell him about me. :D

Semiclassical

I started with the segment AB, then picked a point E on there as where the two circles would be tangent. I then drew those circles, and arbitrarily chose a point on the second one to be C.

Daminark

raises hands innocently

I mean I told him in jest

Semiclassical

With ABC in hand, I constructed the circumcircle and chose D as the intersection of that with the other tangent circle.

Daminark

"Yeah there's this guy in New South Wales who doesn't even like infinite sets, it's pretty funny"

Ted Shifrin

02:16

@Daminark: You may be many things. But innocent is not one of them.

Daminark

What crimes have I committed?

Semiclassical

To verify that there's a circle which is tangent to the sides, I constructed the angle bisectors of BCD, CDA and found that they indeed intersect on AB.

Ted Shifrin

Bannon told me to keep quiet, @Daminark.

Semiclassical

Bam.

Hmm.

If Cheney was Darth Vader, what does that make Bannon?

MickLH

"Bannon" is that like ban cannon?

Semiclassical

02:18

I really want to go with Ganon (but then I am a giant video game nerd).

Ted Shifrin

um, no, @Mick. If you were in the US, you'd know.

MickLH

I don't want to think about it :'(

Ted Shifrin

Nor I.

Daminark

wonders why Steve Bannon knows of my existence

Ted Shifrin

@Daminark: Intellect is verboten. Remember 1984?

Daminark

02:19

Right...

Semiclassical

Ignorance is freedom and so forth.

Ted Shifrin

Or Fahrenheit 451, for that matter.

Zach Hauk

Ganon?

Semiclassical

Cheney:Darth Vader::Bannon:Ganon.

Zach Hauk

as in, Ganondorf with the power of the triforce of power?

Semiclassical

02:20

Right.

aka the giant pig villain at the end of most Zelda games.

Daminark

Lol I've only ever played the original Legend of Zelda

Oh yeah I remember Fahrenheit 451 from a long time ago

Ted Shifrin

OK, I'm gone. Time to cook dinner. Be bad, y'all.

MickLH

peace

Semiclassical

later @ted

Zach Hauk

ok so

OD and OC bisect angles $ADC$ and $DCB$ respectively

Semiclassical

02:21

Right.

Zach Hauk

ok well now im feeling stupid cuz im not sure what to do

Semiclassical

I suspect there's a simple statement there: Two segments AB,BC are mutually tangent to a circle whose center lies on their angle bisector.

I guess the point is that the points of tangency are equidistance from B, so one gets a pair of congruent right triangles.

Daminark

Alright @Ted

Semiclassical

As to how it helps you...oh, look, a convenient distraction! exit stage left

Daminark

I shall spread mischief in the land, but only on your request

Zach Hauk

02:26

hmm

Daminark

Motion to have all schools replace computational integral calculus with convergence theorems in Lebesgue integration

Semiclassical

get out

Daminark

raises hands What?

Zach Hauk

i found SOMETHING

umm, can you draw 2 things on your geogebra?

MickLH

@Daminark he said "get out", it means get out

Zach Hauk

02:28

1. segments $AC$ and $BD$

2. angle bisectors of angles $ADC$ and $BCD$

Semiclassical

I adjusted the placements of points so as to remain clear.

Daminark

I mean more like, what did I do that makes you want me to get out?

Semiclassical

Well, considering computational integral calculus is really important to anyone who actually wants to -use- calculus

Yeah, I'm not exactly in favor of that.

Brody

Oh no, I botched this problem on the exam :/

25 mins ago, by Brody

Exercise: Suppose $x,y\in\Bbb R\setminus\Bbb Q \;\; | \;\; (x<y)$. Construct (a specific) $z\in\Bbb R\setminus \Bbb Q \;\; | \;\; (x<z<y)$.

Zach Hauk

oh boy

Brody

02:33

...just realized. At least it was just extra-credit

Semiclassical

Given two distinct irrationals, construct some irrational between them?

Brody

Yep.

Semiclassical

Hmm, that's a neat one.

Zach Hauk

well

i think Ted was talking about the angle bisector theorem?

Semiclassical

Yeah.

Brody

02:36

I said if they have the same sign, take the average. If they have opposite signs, just halve one of them.

But obviously, the mean can be rational.

Semiclassical

1+sqrt(2), 1-sqrt(2)

Zach Hauk

in that case all we find is that, when $X$ is the intersection of $CO$ and $DB$ and $Y$ is the intersection of $DO$ and $CA$, then $DX/DB = BC/CD$ and $YA/YC = DA/DC$

Semiclassical

Yeah.

I wonder if the arithmetic-geometric mean if two irrational numbers can ever be rational.

I'm guessing the answer is yes, alas.

MickLH

construct it

I want to see this lol

Brody

Arithmetic-geometric mean? Let me look this up

Semiclassical

02:37

Arithmetic–geometric mean

In mathematics, the arithmetic–geometric mean (AGM) of two positive real numbers x and y is defined as follows: First, compute the arithmetic and geometric means of x and y, calling them a1 and g1 respectively (the latter is the principal square root of the product xy): a 1 = 1 ...

MickLH

btw AGM is awesome for computing some useful things

Semiclassical

It really is.

Brody

Neat. Had no idea this was a thing

Zach Hauk

HMM LET ME SEE

MickLH

You get a thing kinda like that with any two means

Zach Hauk

02:38

oops

MickLH

Though the rest seem harder to extract delicious fruit from

Semiclassical

Yeah.

Brody

Like an arithmetic-(arithmetic-geometric mean) mean?

MickLH

I tried to study it before, but I didn't get very far

Zach Hauk

$DB/DX=CD/CB$, $YA/YC=DA/DC$, multiply them to get $DA/CB = (DB * YA) / (DX * YC)$

Brody

02:40

So the solution is not super obvious?

Zach Hauk

that doesn't help much...

Brody

I want to add some proportion of the difference to the lower quantity, but I don't know if it guarantees irrationality in general

Zach Hauk

:/

MickLH

@Brody The only way I've made any progress on the topic is squeeze theorem lol

Semiclassical

I suspect one could use the idea of the arithmetic-geometric mean.

Namely, you get a sequence of pair a_n,b_n

They'll be between your initial a<b. So all one then needs to argue is that there's no way for those pairs to always be rational.

MickLH

02:43

There was one delicious fruit I found though: Arithmetic-Harmonic Mean of $1$ and $x$ is $\sqrt{x}$

Which is super useful for doing arbitrary precision rational computations with interval arithmetic

Brody

@Semiclassical Hmm... alright.

Interesting thought

Semiclassical

As an example, 2-sqrt(2), 2+sqrt(2) -> sqrt(2), 2

In that case, the geometric mean is sqrt(2) which is irrational and you're done.

Zach Hauk

idk what to do :I

Semiclassical

If it weren't and neither was the arithmetic mean, then repeat.

Daminark

If someone is thinking of using calculus, why not crush that desire and merely engage in pointless abstraction as an alternative?

Also sorry I was out lol

Semiclassical

02:46

That's a suggestion, though, not a proof. I dunno what the best approach is.

Hmm, now I find myself wanting to construct an example of two irrational numbers whose geometric and arithmetic means are both rational.

Brody

When is $\sqrt{u}$ rational?

MickLH

when $u$ is a perfect square?

Zach Hauk

no

MickLH

I guess you need something stronger there

Semiclassical

I think I'd need to find $a^2-cb^2$ to be a perfect square for $c\neq 1$.

MickLH

02:49

@ZachHauk Yes. It obviously is. That doesn't mean it describes every rational $u$.

Zach Hauk

oh

i thought you were describing all cases

when $u = a^2/b^2$ for some relatively prime $a,b$?

Semiclassical

$\sqrt{u}=p/q\implies u=p^2/q^2$, yeah.

Okay, $4-2\sqrt{3}$ and $4+2\sqrt{3}$ have geometric mean $\sqrt{16-12}=2$ and arithmetic mean $4$.

MickLH

(just to clarify, I usually put a question mark after things when I mean to imply "This isn't strong but could provide at least one example")

Semiclassical

Of course, then the next geometric mean is 2sqrt(2) which is an irrational between 4-2sqrt(3) and 4+2sqrt(3).

But now I've got myself wondering about this.

Zach Hauk

time to try a different imo geometry problem

Brody

02:53

Backtracking a bit... I was thinking about $x+\frac{y-x}{r}$ for some particular $r$ transcendental. But idk :/

MickLH

@Semiclassical can you construct a pair that alternates rationality exactly $n$ times through the AGM process?

Semiclassical

I was wondering about that as well.

MickLH

At first I thought of $n$ at infinity, but then I thought it's more interesting finite

Balarka Sen

what's $\nu$

Semiclassical

My $\hat{\nu}$, that's what's $\nu$

4

(I don't actually have a new hat.)

Brody

02:58

This shiny $\overline{crow}$

MickLH

hl3 confirmed?

Brody

had nothing less awful to offer, but I'm not ashamed ;P

Transcript for

$Mathematics$ Mathematics