Mathematics - 2018-12-06 (page 2 of 3)

Semiclassical

14:00

@AkivaWeinberger I find myself wondering how they made that. (From the color scheme, I'd say it's mathematica)

creating the half-bagels isn't so hard, I think

Akiva Weinberger

73

A: Cutting bagels into linked halves

Here's one way to slice the donut. To draw one half of the sliced donut I'm using a parameterisation of a torus similar to the one on wikipedia, but with v replaced with u + v and v running from 0 to Pi instead of 2 Pi. This means that the cut is actually a double twist loop. pl = ParametricPlot...

Semiclassical

nice

Akiva Weinberger

Here's a random thing taken out of context

"It turns out that engineers, scientists, and financial folks need to use calculus, but they don't need to understand calculus." Wow, what a clear statement of what seems implicitly to be the dominant position in the contemporary pedagogy of freshman calculus...and which I find to be alarming and pernicious bordering on repugnant. I wish we college math teachers could have an explicit discussion to see whether we really believe this and fully understand its implications. — Pete L. Clark Jan 13 '14 at 13:53

Semiclassical

i mean

Akiva Weinberger

(Some context: They're discussing the epsilon–delta definition of continuity)

Semiclassical

14:10

there's different ways to interpret "understanding calculus"

and I think equating "understands calculus" with "understands epsilon-delta definitions" is not a good idea

elevating that aspect of calculus above all others seems silly to me

Abdullah UYU

Hi all, hi @Semiclassical

Semiclassical

hi

Abdullah UYU

You were a physicist, right?

Semiclassical

ya

Akiva Weinberger

In a sense, calculus is just an easy-to-calculate approximation of discrete phenomena

I mean, it's literally the limit as $h$ goes to zero

Will Njundong

14:16

struggling with simple proofs in my logic course

Akiva Weinberger

Continuous behavior is indistinguishable from discrete behavior as long as the discrete units are too small to be measured

Semiclassical

eh. that's all calculus as a formalism may be

Will Njundong

could someone pleasde help me with showing --> if a | b or a | c, then a | bc

Semiclassical

but in terms of problem-solving, there's a lot more to life than that

Akiva Weinberger

I guess this ignores all the optimization stuff

Semiclassical

14:18

@WillNjundong That's logically equivalent to its contrapositive: If a doesn't divide bc, then a doesn't divide b and a doesn't divide c.

Akiva Weinberger

@WillNjundong That's logically equivalent to "(if a | b then a | bc) or (if a | c then a | bc)"

Semiclassical

(Or in less formal terms: If you can't divide one number by a second, then you also can't divide any factors of the first by the second.")

I'm fine to slag on how finance uses math as a tool without actually understanding what it's doing

Will Njundong

so i can set it up as A = a|b , B = a|c
A --> B

Semiclassical

but stuff like the Black-Scholes equation is certainly calculus

Akiva Weinberger

I don't actually know how finance uses calculus

@Semiclassical What's that?

Abdullah UYU

14:20

@WillNjundong I don't think so

Semiclassical

Black–Scholes model

The Black–Scholes or Black–Scholes–Merton model is a mathematical model for the dynamics of a financial market containing derivative investment instruments. From the partial differential equation in the model, known as the Black–Scholes equation, one can deduce the Black–Scholes formula, which gives a theoretical estimate of the price of European-style options and shows that the option has a unique price regardless of the risk of the security and its expected return (instead replacing the security's expected return with the risk-neutral rate). The formula led to a boom in options trading a...

Will Njundong

whoops made a mistake hold on

Akiva Weinberger

@WillNjundong A = a|b, B = a|c, C = a|bc, you have if (A or B) then C

Semiclassical

(I don't actually know what the Black-Scholes model is, to be sure)

Will Njundong

okay there ^

thank you :)

Semiclassical

14:21

with the Black-Scholes equation being: en.wikipedia.org/wiki/Black%E2%80%93Scholes_equation

Balarka Sen

@BuddhiniAngelika Cayley graphs of semidirect products of groups will look like a nontrivial bundle with basespace being Cayley graph of the normal subgroup and fiber space being Cayley graph of the subgroup that acts on the normal subgroup

Abdullah UYU

@Semiclassical Why I have the impression that you're a British person?

Balarka Sen

i.e., at each point of $\Gamma(N)$, a $\Gamma(H)$ will stick out

Semiclassical

whether or not one takes the widespread usage of B-S as being valid (which is apparently questionable, given the assumptions involved) it's still calculus while being very far distant from any sort of epsilon-delta questions

Akiva Weinberger

I don't know what a European option is

I don't understand options

I should figure this out at some point

Semiclassical

14:23

@AbdullahUYU dunno. though given how random my hours here are, it's not an entirely absurd idea

Balarka Sen

Try drawing it for simple examples, like $D_{2n}$ (semidirect of $\Bbb Z_n$ with $\Bbb Z_2$)

The undirected graphs of $D_{2n}$ and $\Bbb Z_2 \times \Bbb Z_n$ will look the same.

Semiclassical

@AkivaWeinberger finance, as a set of abstract rules and systems, is sorta interesting

Fargle

@AkivaWeinberger I can give a brief explanation---I've been doing research lately on B-S.

Semiclassical

finance as a real-world capitalistic system? pretty damn horrifying

anakhro

@BalarkaSen !!!

Akiva Weinberger

14:25

Most of what I know about the stock market is from browsing /r/wallstreetbets, which is not a good way to learn about the stock market

Balarka Sen

@anakhro !!!

Semiclassical

lol

anakhro

How are you?

Akiva Weinberger

(That's not really true, I know stuff from other places)

Semiclassical

i should probably learn some stochastic calculus

Fargle

14:25

When you purchase an option, you essentially purchase the right to buy/sell a collection of stocks at a pre-agreed price on or by a particular date.

For a call option, you buy the right to buy. For a put option, you buy the right to sell.

Akiva Weinberger

So if it's more than what the actual price ends up being, then you've lost?

and otherwise you've gained

Fargle

You also have the right not to buy or sell.

Akiva Weinberger

so it's only fair if you've managed to predict the eventual cost and if the price of the option agrees with your prediction

Balarka Sen

@anakhro Not bad

How about you

Fargle

@AkivaWeinberger Right, ideally that's what you want. But you are not obligated to transact anyway, so that you don't net an even further loss.

anakhro

14:27

Not horrible. ;)

Akiva Weinberger

Mhm

anakhro

What have you been reading about lately.

Last time we really talked, it was about stratifolds.

Akiva Weinberger

but if I have secret knowledge that says the price is gonna skyrocket and the option is pretty low than I should buy the option

Semiclassical

depends on if its a call or a put option, i guess. if it's a call option, you ideally want the market price to be higher than what you agreed on

Balarka Sen

Singularities, stratified spaces, h-principles, intersection homology. A mix of them

Akiva Weinberger

14:28

How is that different from just buying the stock normally?

Balarka Sen

@anakhro Yeah still sticking to that theme

Fargle

There is further a division between American and European options: all options have an "exercise date". A European option must be exercised on this date and not before. An American option may be exercised at any time up to the date.

Akiva Weinberger

Or shorting it (which is buying negative stock or something)

anakhro

Are you reading Kreck's book?

Balarka Sen

Nah

Kreck's theory is of a different flavor

Fargle

14:29

@AkivaWeinberger The option itself becomes a financial object that can be traded.

Akiva Weinberger

@Fargle Why the names?

Are European options more common in Europe?

Balarka Sen

I should still read it at some point

Fargle

I think they just refer to rules sets used in the stock exchanges in places, yeah.

Akiva Weinberger

Does the NYSE only have American options?

Fargle

I don't know.

anakhro

14:30

I know Mike liked it. He was the one who showed it to me

Akiva Weinberger

@Fargle As does the stock, no?

Semiclassical

@fargle so you end up making wagers on whether the option was accurately priced or not

anakhro

I need to learn at least a little about h-principles.

Fargle

@Semiclassical Essentially.

Akiva Weinberger

OK hold on how about an example

Semiclassical

14:31

(which effectively means you're making a wager on a wager)

Balarka Sen

Yeah Mike is a big fan of Kreck, can confirm

Akiva Weinberger

Let's say Tesla is at 350 and I think that by this time next month it'll be at 420 because I'm a delusional optimist

Balarka Sen

@anakhro Teach me symplectic topology fam

anakhro

I am pretty sure you know as much as me

Balarka Sen

Naw bro

anakhro

14:33

I only have a working knowledge of the basics, because I only ever use those things with contact geometry.

Akiva Weinberger

So what would an option look like in that scenario

Semiclassical

In that case, you'd value any call option which would promise you the right to buy at a price less than 420.

Akiva Weinberger

What sort of option would I want to buy and what sort of option would I not want to buy

anakhro

If $\ker\alpha = \xi$, your hyperplane field for your contact manifold, then $d\alpha\vert_{\xi}$ is a symplectic form.

Akiva Weinberger

@Semiclassical Mhm

But if I buy a stock today and hold on to it for a month

anakhro

14:34

Then symplectic stuff pops up.

When you are merely working with contact manifolds.

Akiva Weinberger

then I'm essentially buying a stock for 350

so why would that be different than buying an option that lets me buy Tesla stock for 350 in a month

Balarka Sen

@anakhro $\alpha$ is a contact 1-form for $M^{2n+1}$ if $\alpha \wedge (d\alpha)^n \neq 0$, right?

Akiva Weinberger

I'd also get the benefit of getting dividends for that month

Fargle

@AkivaWeinberger The difference is that if you outright buy the stock for 350, you may not be correct in that the price will go to 420 next month.

anakhro

Yes.

Only locally defined usually, though.

Akiva Weinberger

14:36

So, if the stock actually tanks to 200 or whatever and I had bought the stock today, I'm stuck with the stock?

Fargle

Yeah. Good luck selling it for the same price you bought it for, at least.

anakhro

You can then show that $d\alpha\vert_\xi$ is non-degenerate.

Balarka Sen

So yeah, $(d\alpha)^n$ is nonvanishing on $\ker \alpha$

Akiva Weinberger

Whereas if there was an option and I bought it what would happen

Balarka Sen

Meaning $d\alpha|_{\xi}$ is symplectic on $\xi$

Fargle

14:37

If you buy the option, and then the price tanks and you don't exercise, the only hit you've taken is the price of the option itself.

Akiva Weinberger

So like I'd buy a 400 dollar option for next month because that's less than the 420 I think it'll be, yeah?

Wait hold on

anakhro

Yuppers.

Akiva Weinberger

Is the price of the option the same as the price it lets me buy the stock at?

Semiclassical

I think that's where it gets tricky.

Akiva Weinberger

If the option lets me buy next month's stock at 400, does the option cost me 400?

anakhro

14:38

And if I am not blinded to something trivial, I think this is equivalent to xi being a contact hyperplane field.

Semiclassical

there's no obligation for the person selling you the option to set the price there

anakhro

So it goes the other way.

Semiclassical

they can sell it at whatever price they think they'll make money at

Akiva Weinberger

So it's cheaper if no one else thinks that the stock will be above 400

Fargle

@AkivaWeinberger The price of the option is the estimated value of the risk that the option seller is taking by possibly being forced to obligate himself, I believe.

Akiva Weinberger

14:39

and it's more expensive if everyone else is a delusional optimist like me?

Semiclassical

It's a wager on a wager.

Akiva Weinberger

That's very Hofstadtery

Fargle

Maybe more accurately, the price of the option is what Black-Scholes and other models are trying to figure out.

Akiva Weinberger

(Incidentally, hofstadterophobia is the fear of hofstadterophobia)

Semiclassical

This can all get pretty byzantine

Balarka Sen

14:41

@anakhro What's the best way to understand that $\alpha \wedge (d\alpha)^n \neq 0$ is equivalent to "complete nonintegrability" of $\xi$?

Fargle

Okay so: there are two components to the valuation of the option. One is the "intrinsic value"---this is just the difference between the current value of the stock and the value you're buying the stock at (the "strike price"). If, say, you're doing your call option, and you agree to buy at 400 later and it's at 350 now, the value of the option is -50 for you or 50 for the seller.

Balarka Sen

I know integrability means $\Omega^\bullet(\xi)$ is $d$-closed

Fargle

Then there's the "time value"---this refers to the risk incurred by the seller. Basically it's "the rest of the premium the buyer has to pay that isn't factored into intrinsic value".

That's the part that things like B-S are designed to approximate.

Semiclassical

Which requires you make certain assumptions about how the stocks themselves are liable to vary

anakhro

Well yeah it's just linear algebra with Frobenius's integrability theorem

Fargle

14:43

Right.

Semiclassical

e.g. volatility and the like

anakhro

Like the case for n=1 can be seen easily from:

Akiva Weinberger

So you'd need to know things about the company and how well they're likely to do

Fargle

So for example, if you knew something about how the stock price is changing and so on, then you might be aware of that this stock is going up, and might price the option you sell in such a way that covers you in case you end up selling the stock at a loss to yourself.

Akiva Weinberger

and like Tesla just made a new line of cars (I think) so they're doing pretty well now I think

Fargle

14:45

"You" being the option seller now, rather than the option buyer.

anakhro

$$d\alpha(v,w) = v(\alpha(w)) - w(\alpha(v)) - \alpha([v,w])$$

Akiva Weinberger

Mhm

anakhro

For integrability

Akiva Weinberger

and so if it all goes sideways, I've just transferred my losses to whoever bought the option

but if it all goes well then I don't make as much money as I would have otherwise

?

Balarka Sen

@anakhro That's the one which shows algebra of vector fields being bracket-closed is the same as algebra of differential forms being d-closed, I believe

Akiva Weinberger

14:47

So it's like I'm smoothing the stock price

Semiclassical

that doesn't sound right, since whoever bought the option still has the right to not exercise their right to purchase

Fargle

Fun fact: even though you can exercise early in an American call option, it is never optimal to do so unless the stock pays dividends.

Akiva Weinberger

@Semiclassical Oh that's true

@Semiclassical They still have the loss from the price of the option

anakhro

@BalarkaSen involutive iff integrable

Semiclassical

yeah

Fargle

14:49

@AkivaWeinberger They do, but something to consider is that the money that they would actually pay for the stock can be invested in a place and accruing interest.

That seems to be a big assumption in a lot of these models: that money not being spent is usually having interest earned on it meantime.

anakhro

I guess I should say completely integrable, but I think that's understood. :P

Akiva Weinberger

I've been told not to invest in individual stocks anyway since averages smooth out randomness

so groups of stocks are safer

(unless you just want to support the company)

Will Njundong

I think im up against a trick question here...

Fargle

It depends, as in all things, but that's probably a decent rule of thumb.

Will Njundong

"WHat does a 60 second stopwatch read 82 seconds after it reads 27 seconds?"

am i supposed to assume it winds back to zero?

Akiva Weinberger

14:51

Stopwatch vs timer

Timers count up, stopwatches count down, right?

Semiclassical

^

Will Njundong

oh i was thinking of it as a timer haha

Semiclassical

I mean, the question is still the same: What happens when you hit 0 (for a timer) or 60 (for a stopwatch)

Akiva Weinberger

So the stopwatch would say "zero" or "time completed" or "beep beep beep" or whatever

Will Njundong

They have 2 such similar questions. Still unsure if i should respond to both with 0 lol

sorry 60

Semiclassical

14:53

I suspect what they mean is that the number wraps around to 0 after 60

Akiva Weinberger

Wait wait

I might have had it backwards

Semiclassical

so it'd go ...58, 59, 60, 1, 2, 3,...

Will Njundong

well it says atfer it "reads" 27 seconds

Akiva Weinberger

and stopwatches count up and timers count down

Will Njundong

@Semiclassical thats what i though initialy but not confident about it

Akiva Weinberger

14:54

That's how Apple's clock app does it anyway

Semiclassical

yeah, I'm not either

i think a real-world stopwatch would be as likely to stop at 60

a better question imo would be: Suppose your stopwatch displays times as hours:minutes:seconds

Will Njundong

Its a standalone question. Just as is

Balarka Sen

@anakhro Right, so $\alpha \wedge d\alpha = 0$ is equivalent to integrability of $\ker \alpha$. Because that means $d\alpha =0$ on $\ker \alpha$, so for all $X, Y \in \ker \alpha$, $0=d\alpha(X, Y) = X(\alpha(Y)) - Y(\alpha(X)) - \alpha([X, Y]) = -\alpha([X, Y])$.

So $[X, Y] \in \ker \alpha$ as well.

Akiva Weinberger

01:49

is what I'd say

Semiclassical

If your stopwatch displays 27 seconds, what will your stopwatch display for seconds after waiting 82 seconds?

Akiva Weinberger

14:57

Hm, it might be paused :P

anakhro

Yep!

Will Njundong

The second similar one is "What does a 60-second stop watch read 54 seconds before it reads 19 seconds".

Now one scenario i have in mind is: it was just sitting there at 0 for 54 seconds before it was started then looked/stopped at 19 second?

anakhro

And that convinced me enough that the general case works and I never did work out the linear algebra for that.

Will Njundong

Ambiguous stuff here and very annoying

Semiclassical

Yeah, that's even worse (for just the reason you said)

anakhro

14:57

Still on my list of things to write down.

Balarka Sen

I guess the problematic this is to figure out, if $\omega$ and $\eta$ are $k$ and $\ell$-forms respectively, what $\omega \wedge \eta$ evaluates to on a $(k+\ell)$-tensor field

Secret

I need a stopwatch that count time nonlinearly, as recently I have strange memory problems where spacetime is completely scrambled

Semiclassical

A non-ambiguous version: You leave your stopwatch running for at least 10 minutes. When you come back, it displays 19 seconds. How many seconds did it display 54 seconds prior to that?

For a logic course, these problems are remarkably lacking in clear assumptions.

Will Njundong

yea i wish

Balarka Sen

I think $(\omega \wedge \eta)(v_1, \cdots, v_k, w_1, \cdots, w_\ell)$ should be a signed sum of $\omega(-, \cdots, -)\eta(-, \cdots, -)$ where the dashes are cyclic permutations of $v_1, \cdots, v_k, w_1, \cdots, w_\ell$

Will Njundong

15:00

Its just those 2 before the chapter relevant questions begin. I guess theyre supposed to be warm up or brain teasers lol

Akiva Weinberger

I guess you should probably say 19-54 mod 60

Balarka Sen

Because $\omega \wedge \eta$ is the antisymmetrization of $\omega \otimes \eta$

Akiva Weinberger

whatever that happens to be

Semiclassical

@akiva more broadly, there's a question of rational pricing

Akiva Weinberger

25?

anakhro

15:01

What's your favourite way of defining differential forms, @BalarkaSen?

Akiva Weinberger

@Semiclassical When it's an integer multiple of 1/n for some n?

(Sorry)

Balarka Sen

@anakhro Basis-wise, with the $dx_I$'s

Semiclassical

who knows, that might be more rational than what they actually do

('Rational pricing' is only so good as your assumptions about future behavior. And, well, human beings are pretty bad at predicting the future.)

Akiva Weinberger

So it's the assumption that things cost what they should

and the "law of one price" that they mentioned says that things cost the same everywhere

Semiclassical

the prlblem which the section at the bottom emphasizes is systemic risk due to financial interconnectedness

Akiva Weinberger

15:04

and that's not true if there are, like, tariffs, but the hope is that it holds in the absence of things like that

anakhro

How do you define the exterior product @BalarkaSen

Balarka Sen

$dx_I \wedge dx_J = dx_{(I, J)}$ :3

Semiclassical

well, I think that "the law of one price" is only ever intended to hold for a given market

if two markets don't interact, then there's no direct mechanism for their prices to be related

Akiva Weinberger

Things cost differently in different neighborhoods if the demographics have different amounts of money

Semiclassical

they may be correlated, but there's no feedback

The tricky thing these days, of course, is that everything is so globalized

Akiva Weinberger

15:05

'cause the assumption is the richer people won't go to the poor neighborhoods for cheaper goods because inconvenient

Will Njundong

@AkivaWeinberger thats actually... a very nice approach

Akiva Weinberger

So I'm guessing stuff will be cheaper in Bet Safafa than Talpiot but I haven't checked

Semiclassical

I feel like the dream of financial math (or of people who use it, anyways) is to find a way to value objects independently of human beings

which seems like a mass delusion tbh

Akiva Weinberger

I've heard that stuff is cheaper in the Muslim Quarter of the Old City but I don't spend a lot of time in the Old City anyway so I dunno

(Talpiot was a bad comparison anyway 'cause it's not a residential area but whatever)

Balarka Sen

@anakhro The relevant $h$-principle in the contact world is that any open manifold admits a contact structure, right?

anakhro

15:12

I have no idea anything about h-principles

Balarka Sen

I think this theorem is true

anakhro

Hmmm.

What is an h-principle, exactly?

Akiva Weinberger

Economics (slash mathematical finance) is all fun and good until you realize it actually affects people's lives and then it's less fun

Semiclassical

yep

It's all fun and games until someone loses their house

(I mean it's not really fun and games up to then either)

Secret

The best fun is something that never affect people

anakhro

15:19

@BalarkaSen I think that might be just for "almost" contact structures.

Secret

War is fun in computer games, but horrifying in real life

anakhro

I think in general it is hard to say whether a contact structure exists.

Or that might just be the case for tight contact structures.

Semiclassical

@Secret A thermonuclear weapon is like the sun brought to earth, which is neat up until you remember why it's a good thing the sun is 93 million miles away from earth

Secret

indeed

anakhro

@BalarkaSen it does have something similar to symplectic geometry on the 1-jet space

That is,

$$\omega\quad\colon\quad T^*M\quad\colon\colon\quad\alpha\quad\colon\quad T^*M\times\mathbb R$$

Heh, that looks horrible.

Akiva Weinberger

15:27

Same thing with politics

I was in the middle of the book The Dictator's Handbook which is about the political theory of leaders and why they do horrible stuff so often

Balarka Sen

@anakhro So suppose I have a 3-manifold $M$. Consider the bundle $\Lambda T^*M \oplus \Lambda^2 T^*M$ on it. A point in the total space of this bundle is a 3-tuple $(p, \omega_p, \eta_p)$ consisting of a point in the base, a 1-form at the tangent space of that point, and a 2-form at the tangent space of that point.

Akiva Weinberger

and it's interesting, it makes the case that there's no line between autocracies and democracies, only a vague sliding scale, and the leaders of those countries all operate under the same rules

but a lot of the examples it brings up are horrible incidents in horrible places

I should finish reading it

(C G P Grey did a video inspired by that book a little while ago)

The Rules for Rulers

►

This thing

Buddhini Angelika

Thank you @BalarkaSen so then what I thought is correct,

Balarka Sen

Consider the subset $\mathscr{R} \subset \Lambda T^*M \oplus \Lambda^2 T^*M$ cut out by the equation $\omega \wedge \eta \neq 0$. This is an open subset of the total space of the bundle consisting of "formal contact structures", i.e., a section $s : M \to \Lambda T^*M \oplus \Lambda^2 T^*M$ such that $s(M) \subset \mathscr{R}$ determines a 1-form $\omega$ on $M$ such that there is a 2-form $\eta$ on $M$ with $\omega \wedge \eta \neq 0$.

Akiva Weinberger

Economics and politics are kinda the same thing anyway, they both study power (the ability to make other people do what you want them to do)

I say this despite knowing very little about both fields

so I'm probably very wrong

Balarka Sen

15:34

Call the space of formal contact structures $\Gamma(\mathscr{R})$. Note that a contact structure on $M$ is a formal contact structure which is integrable, i.e., $\eta = d\omega$. Let's call $\text{Hol}(\mathscr{R})$ the subspace of integrable formal contact structures.

My guess is that the inclusion $\text{Hol}(\mathscr{R}) \to \Gamma(\mathscr{R})$ is a homotopy equivalence if $M$ is an open 3-manifold.

'Cuz this is the kind of setup which occurs in $h$-principles. $\mathscr{R}$ is like a differential relation on 1-forms on $M$, and we're saying if that differential relation has a formal solution, it also has an actual solution.

@anakhro What's an almost contact structure?

@anakhro Oh tell me about that

anakhro

Okay, I will have to double check it later, but I think just having a non-degenerate 2-form on a hyperplane field.

Balarka Sen

Oh, that's precisely a formal contact structure

Because $\omega \wedge \eta \neq 0$ implies $\eta$ is nonvanishing on $\ker \omega$

anakhro

But then this calls into question whether I had the reverse of the other thing before wrong

Balarka Sen

So my claim is that on an open 3-manifold every almost contact structure can be C^0-small perturbed to get an actual contact structure

Every odd dimensional manifold has an almost contact structure, right?

anakhro

I am not sure.

I am trying to figure out the definition.

Balarka Sen

15:46

If $M$ is 3-dimensional, it's just a pair $(\omega, \eta)$ of a 1 and a 2-form such that $\omega \wedge \eta \neq 0$, equivalent to what you said.

I think it should follow from making a general section of $\Lambda T^*M \oplus \Lambda^2 T^*M$ transverse to the subset $\{\omega \wedge \eta = 0\}$.

anakhro

Geiges uses the definition that an almost contact structure is a hyperplane field xi with a complex bundle structure J and a choice oriented line sub-bundle complementary to xi.

Oh he has the full definition earlier.

And he assumes coorientability.

So the line subbundle trivializes defines the coorientation.

Balarka Sen

@anakhro This should be equivalent because the space of fiberwise symplectic structures is homotopy equivalent to the space of fiberwise complex structures

An almost symplectic structure naturally gives rise to an almost complex structure, by using the symplectic matrix as your $J$, and vice versa

In appropriate coordinates the symplectic matrix looks like $[-I, O; O; I]$ by Darboux

anakhro

I think the best you can get in cases is overtwisted contact structures.

Balarka Sen

I had written this up earlier here

@anakhro What's that again? There's an immersed disk with boundary transverse to the contact field such that the characteristic foliation looks like spirals emanating from the origin?

I think you told me before once

anakhro

Yes.

They are the ugly contact manifolds.

Well, the characteristic foliations can look quite complex and nice, but are hellish to deal with.

Balarka Sen

16:06

@anakhro What's special about overtwisted contact structures?

They have some sort of stability, I presume? If you perturb an overtwisted contact structure it should still be overtwisted?

anakhro

They are easier to deal with.

But for me in particular, they have singularities.

Abdullah UYU

Given such a collision, how can we determine the type of collision?

In the case of a perfectly inelastic collision, the particles stick together. That's not the case, here.

But other than that, how can I say that it is elastic or inelastic, right? It can be either of those. I think it has to be said for us to determine it.

Semiclassical

as ever in a collision, the relevant quantities are momentum and kinetic energy

the latter may or may not be conserved, but the former certainly is.

Abdullah UYU

I realized that I should simply check whether if the kinetic energy is conserved or not.

Semiclassical

yep

Abdullah UYU

16:22

But, there is a moot point in that procedure, I think.

But I can't properly phrase it.

Ryan Cameron

What happened in the last line? My math box says that rewriting is based on the rule:

$$(a+b)^2=a^2+b^2+2ab$$

The equations: (these are from my math book)

$$(x^2+6x)+(y^2-2y)=-7$$
$$(x^2+6x+9)+(y^2-2y+1)-9-1=-7$$
$$(x+3)^2 +(y-1)^2=3$$

Semiclassical

not really. you're given enough info to deduce whether it's elastic or not

Ryan Cameron

The left side does not satisfy either $(a+b)^2$ or $a^2+b^2+2ab$, what am I missing?

Semiclassical

@RyanCameron the entire left side doesn't, but $(x^2+6x+9)$ and $(y^2-2y+1)$ certainly do

Ryan Cameron

I don't get it what is a and b equal to ? - if you were to apply the rule mentioned above

Abdullah UYU

16:27

I am aware of that. That's not what I'm uneasy about.

Semiclassical

Look at it and figure it out. Your book is telling you that $(x+3)^2=x^2+6x+9$.

Ryan Cameron

uh ...

a=x, and b=3

Semiclassical

And there you are.

Ryan Cameron

Thank you Semiclassical, now I can happily continue my math journey :)

Semiclassical

@AbdullahUYU Then I'm really not seeing what you're worried about.

Abdullah UYU

16:30

OK. I concluded that it's not elastic because kinetic energy is not conserved.

Semiclassical

Correct.

An interesting exercise is the following: If it were an elastic collision, resulting in particle 1 moving off at an angle of 90 degrees, what velocity would particle 1 end up with?

Abdullah UYU

Effectively, in the first case, it's $v_1^2$ and after it's $\frac{7v_1^2}{8}$

Yeah, that's what I'll do immediately

Semiclassical

(My intuition is that you'd end up with $v_{1x}'>v_1/2$)

none

@BalarkaSen I'm not sure if you saw my discussion with Mike yesterday, but to make things brief I was wondering if there are any nice representations of $Homeo^{+}(S^{1})$, the group of orientation preserving homeomorphisms of $S^1$.

Abdullah UYU

16:49

$v_1'=\pm\sqrt{\frac{v_1}{3}}\hat{j}$

How's j hat written, by the way?

Rithaniel

Should be \hat{j}

Test: $\hat{j}$

Hmmm, nope, that makes it offcenter

Abdullah UYU

$||v_1'||=\pm\sqrt{\frac{v_1}{3}}\hat{\jmath}$

I gave up :)

Oh, what a shame!

Ok, you're right.

Semiclassical

@AbdullahUYU that looks right

Abdullah UYU

Ah, no. It's not comparable.

Semiclassical

well. $\|v_1'\|=v_1/\sqrt{3}$ anyways

the magnitude is just a positive number, and $v_1$ is positive

and $1/\sqrt{3}=0.577\ldots >1/2$, so that lines up with intuition

Abdullah UYU

17:02

I'm ok with taking the positive one.

squaring them gives: $\frac{v_1}{3}$, $\frac{v_1^2}{4}$

Doesn't it depend on $v_1$?

Semiclassical

What?

I'm comparing $\|v_1'\|=v_1/2$ with $\|v_1\|=v_1/\sqrt{3}$

Abdullah UYU

Little correction made

Semiclassical

for that, it suffices to note that $1/2<1/\sqrt{3}$

Why are you squaring $v_1$ in the second one but not the first?

Abdullah UYU

It's not $\frac{v_1}{\sqrt{3}}$ but $\sqrt{\frac{v_1}{3}}$

Semiclassical

No, it isn't.

That wouldn't be a velocity, and $\|v_1'\|$ certainly must have units of velocity

Abdullah UYU

17:06

Ah yes you're right again.

Ok, that looks fair.

Thanks @Semiclassical

Semiclassical

np

Will Njundong

17:23

relatively prime

whats that?

Akiva Weinberger

Two numbers are relatively prime if the largest factor they have in common is 1

4 is relatively prime to 9, but 4 is not relatively prime to 6

$n$ is always relatively prime to $n+1$

Another word for "relatively prime" is "coprime"

You'd write it as $\gcd(a,b)=1$

@WillNjundong

Will Njundong

So as my exercise says, list all positive integers less than 30 that are relatively prime to 20, is my work correct?:

1, 3, 7, ,11, 13, 17, 19

ÍgjøgnumMeg

You're missing some

Will Njundong

it cant be more than 20 can it?

ÍgjøgnumMeg

Yeah, it says positive integers less than $30$ that are prime to $20$

Eran

17:32

I'm trying to show a counter example of Converse lagrange's theorem in group theory.
i.e i need to give an example of a group G such that |G| = n and k|n and there's no group of size k , I thought about taking A_5 and k=30, I've seen before that A_5 is simple so if there a subgroup of order 30 , it means that it's index is 2 which implies that it is normal , contradiction. is my counterexample correct?

Akiva Weinberger

You're missing 9, as well as 21, 23, 27, 29 (since it says "all positive integers less than 30" and not "all positive integers less than 20")

@WillNjundong

Will Njundong

oookay, got it thank you. man this stuff is so simple but confusing

Balarka Sen

18:06

@none Oh, I don't know much about this. @PaulPlummer might

I thought it's more interesting look for representations of finite groups to Homeo(S^1) than representations of Homeo(S^1) though

There are interesting subgroups of Homeo(S^1) which has interesting representations, I suppose. PSL_2(R) = Isom(H^2) sits as a subgroup of Homeo(S^1), as every hyperbolic isometry acts on the boundary of the Poincare disk by homeomorphisms

Lecter

18:38

0

Q: Problem about constructing extended ternary golay code from ternary golay code

I have one problem about some construction I'm about to explain. I'm asked to construct $\mathcal{G}_{12}$ from $\mathcal{G}_{11}$. In my notes I'm using those generator matrices (respectively): $G_{12}=\begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 & 1 \\ 0 & 1 & 0 & 0 & 0 & 0

coding-theory

Someone can help?

Rando Hinn

Hi everyone... how can I get $y^3=x$ into Matchad's plotter?

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