Mathematics

@santimirandarp keep in mind that people knew about parabolas long before they knew about algebra, so the geometric definitions (in one form or another) are the more historical ones

For instance, a parabola is a particular kind of conic section

skullpatrol

\o @Semiclassical not keepin' an eye on the game?

Semiclassical

What game?

(So, uh, no)

skullpatrol

@santimirandarp they were used to calculate the taxes on the amount of area land in ancient Egypt

Leaky Nun

hk01.com/%E6%94%BF%E6%83%85/339557/…

I know you guys can't read Chinese; just click on the big rectangle with big white words

skullpatrol

Game 7 for the Staney Cup: Boston vs St. Louis

Leaky Nun

12:45 AM

that's the live covering of the situation in HK now

Semiclassical

Oh. Yeah no @skullpatrol

skullpatrol

@LeakyNun what am I watching?

Leaky Nun

the HK people are outside the legislative building to stop the second reading of the extradition bill

skullpatrol

students with sticks?

Leaky Nun

that would allow suspected criminals to be extradited to China

Ryan Unger

12:50 AM

@LeakyNun that's a fantastic way to get a virus

Leaky Nun

heh

Érico Melo Silva

@LeakyNun i heard about this, hope the folks from HK are able to fight it

Leaky Nun

yeah

it's 08:51 HKT now, 2 hours until the second reading

skullpatrol

sounds like a good way to get executed, if you get extradited

Ryan Unger

at least you'll provide much needed organs

Leaky Nun

12:53 AM

that's why people are fighting with their last breath

I've never seen this before; they're very disciplined

skullpatrol

yeah, better than the hockey game ;)

Ryan Unger

skull did you expect your team to win

by one point

Balarka Sen

@LeakyNun what's the point of this bill?

Ryan Unger

what about KD man

Balarka Sen

i don't get it. what's the gain in sending suspected criminals to mainland china for trial

Ryan Unger

12:56 AM

I was watching in a bar yesterday, the number of raps fans in america is amazing

skullpatrol

he's getting an MRI today

Ryan Unger

they hate the warriors so much that they forget their DUTY

Leaky Nun

@BalarkaSen that’s why so many people are protesting

it’s the hk govt that wants to push the bill

Balarka Sen

right, i'm just trying to understand what the govt gain here is

Leaky Nun

a government not elected by its people doesn’t care about its people

Balarka Sen

12:57 AM

of course it's a blasphemous policy contrast to individual rights

Leaky Nun

the govt would gain... eh... recognition from china i guess

we still don’t have universal suffrage

skullpatrol

Da raiders are gonna be featured on the HBO series Hard Knocks in August @RyanUnger

Balarka Sen

@LeakyNun oh i didn't know that

Ryan Unger

woo wooooooooooooooo

skullpatrol

:D

Érico Melo Silva

12:59 AM

@BalarkaSen this would let them extradite people on trumped up political charges to take down political opponents

skullpatrol

We took our hard knocks man!

Balarka Sen

That makes sense

Érico Melo Silva

there's clearly a lot to gain if youre selling out your people for more power

Ryan Unger

seems like something that can be turned around

Leaky Nun

there’s a figure of 1/7 of the people that went to protest on sunday

Ryan Unger

1:00 AM

just send everyone to china

skullpatrol

Where they still use the death penalty?

Balarka Sen

i wonder if there are particular individuals that are involved in the political scene that the government would want to take out

Leaky Nun

I don’t know of that then

maybe the people who are still commemorating June 4th

or maybe Christians

who knows

Balarka Sen

@LeakyNun That'd make a lot of sense

Leaky Nun

what if this turns into a second June 4th

skullpatrol

1:04 AM

What happened June 4?

Balarka Sen

god forbid

Tiananmen square, @skullpatrol

Leaky Nun

peaceful protests near Tiananmen square were met with bullets from the Chinese govt

skullpatrol

riiiight :(

Leaky Nun

on June 4, 1989

Érico Melo Silva

there may be some kind of political pressure from china being applied to HK higher officials to weaken HK independence, which this move nakedly accomplishes

Balarka Sen

1:06 AM

I know anticommunists who were involved in the Tiananmen square who are still on the run from the Chinese government eg

Leaky Nun

thats right

skullpatrol

the govt is still chasing 30 years later?

Leaky Nun

yep

I think there are still people detained at home indefinitely

parents of the victims

skullpatrol

>8(

sad world

Akiva Weinberger

What's the deal with the phrase "meteoric rise"

Now I'm no scientist (or so I've been told) but I'm pretty sure meteors do the exact opposite of that

Érico Melo Silva

1:14 AM

meteoric meaning "fast"

like a meteor

Akiva Weinberger

At some point I oughta figure out how to use "Now I'm no doctor (or so the courts determined)" in a sentence

Leaky Nun

I love how they’re passing materials around and shouting their names

they’re very disciplined

Balarka Sen

@ÉricoMeloSilva Xi has been advocating the "one-china policy" since the very inception of his government, of course. It's fun how the national unification game played by China and Russia (with Ukraine say) has been identical

N. Maneesh

b For any closed curve $\gamma$ in unit disc $\mathbb D=\{ z\in \mathbb c||z|<1\}$, $\int_{\gamma}\frac{f(z)}{(z-a)^2}dz=0,\forall z\in \mathbb C$ with $|a|\ge 1/2$. where $f:\mathbb D \to \mathbb C$ ian analitic function. Is the statement true? If we consider the closed, $a=1/2$, $\gamma=\{z\in \mathbb C: |z-1/2|<1/8\}$, we get $2\pi i f'(1/2).$ Right? But in the answer key it is given that statement is true. please help me.

Érico Melo Silva

@BalarkaSen i thought the one-china policy specifically refers to the relationship between the PRC and taiwan

hong kong is autonomous but still "part of the PRC"

Balarka Sen

1:17 AM

Ah

Érico Melo Silva

now of course there are lots of reasons for the PRC to want to coarsen the autonomy of hong kong and make it bend the knee

Leaky Nun

also they’re setting up barricades with metal fences like wtf

skullpatrol

I saw that^

no sign of riot police...yet

Balarka Sen

@ÉricoMeloSilva A pun you might enjoy

Érico Melo Silva

lol i hate this prick

Balarka Sen

1:25 AM

Buckley?

Érico Melo Silva

yeah

Balarka Sen

yeah he's a dick

the very model of american conservative media :3

listen to my man Kerouac

he's drunk like nobody's business

"You might have to... in due time... because of the Atom. ite. BOMB! HAH"

Érico Melo Silva

lol it was good

anakhro

hi all

@BalarkaSen would you be interested in doing a 5 minute calculation for me to tell me whether I am an idiot or not?

Balarka Sen

Ya ok tell me

I might turn out to be the idiot though

anakhro

1:31 AM

Okay on the sphere $S^2\subset \mathbb R^3$, take the vector field $X = (xz-y)\partial_x + (yz + x)\partial_y - (x^2+y^2)\partial_z$ and the area form $\Omega = x\,dy\wedge dz + y\,dz\wedge dx + z\,dx\wedge dy$. Calculate $d(\iota_X\Omega)$.

Balarka Sen

Oh that looks too annoying

I pass

anakhro

The answer allegedly is: $y\,dy\wedge dz - x\,dz\wedge dx + 2\,dx\wedge dy$.

However, I do not get to the answer ever.

Heh, that's okay.

My calculation isn't really that long. I just do $\iota_X\Omega$ and then take $d$ afterwards.

imgur.com/2MxuJ9a

I am pretty sure there is a typo in my solution here.

Since I have not since gotten the same answer when I do it.

But none of my other answers are the same either.

I think it is maybe a problem with $d$

What does $d_{S^2}$ look like on $S^2$?

That is, can I just do the usual local coordinate definition, but with $(x,y,z)$ because I am in R^3?

Balarka Sen

1:51 AM

No, that is the problem. $d\Omega = 3 dx\wedge dy \wedge dz$ on $\Bbb R^3$ for example, but $d\Omega = 0$ on $S^2$ (or rather, the pullback of $\Omega$ to $S^2$ is closed)

anakhro

Alright so I have to find out how to represent $d_{S_2}$ in $\mathbb R^3$ coordinates?

Balarka Sen

You have to do this in spherical coordinates or something. That's why I said it's too annoying.

anakhro

I don't know why the author would not have done it in spherical coordinates to begin with if that was the only way to do it.

Balarka Sen

You have to parametrize to take the exterior derivative in the first place.

anakhro

I think he has some other way in mind.

Because admittedly the entire problem would have been easier in spherical coordinates.

Balarka Sen

1:55 AM

OK but I am not going to think about it lmao

There might be some insightful way by seeing exactly what flow $X$ generates, and then noting that your thing is just $\mathcal{L}_X \omega$ by Cartan's magic

skullpatrol

@BalarkaSen do you know of any good cricket World Cup feeds?

Balarka Sen

nope dont follow it myself

skullpatrol

oh, i guess i'll go with the bbc then

thnx

anakhro

lol

user10478

Hello, it has been my understanding that if at least one integrating factor depending only on $x$ exists for an ODE, $M(x,\ y)\ dx + N(x,\ y)\ dy = 0$, then the formula $u(x) = e^{-\int \frac{2D=curl\begin{bmatrix}M(x,\ y) \\ N(x,\ y)\end{bmatrix}}{Q}\ dx}$ will find some of them. However, I appear to have come across a counterexample. Is the above a correct principle?

Leaky Nun

2:47 AM

Secret

4:01 AM

Hmm...

f(h,h')=0, f(h,h')<0?

What about soulmates, is f continuous, what is the induced topology of F[H]

user131753

Is it true that any group $G$ is isomorphic to a subdirect product of simple groups?

user131753

It can be shown that any ring can be represented as a subdirect product of subdirectly irreducible rings. (In fact this was the original motivation for asking the question.)

user131753

If $G$ is an additive abelian group then $G$ can easily be considered as a ring by defining multiplication "$\cdot$" on $G$ as the the map $\cdot:G\times G\to G$ defined by $a\cdot b=0_G$ for all $a,b\in G$.

user131753

Denote this ring on $G$ by $R(G)$.

user131753

4:18 AM

Observe that if $S$ is a subgroup of $G$ then $S$ is an ideal of $R(G)$ and conversely for any ideal $I$ or $(R(G),+,\cdot)$, $(I,+)$ is a subgroup of $(G,+)$. In other words the ideals of $R(G)$ are precisely the subgroups of $G$.

user131753

From this observation it follows that $G$ is isomorphic to a subdirect product of commutative simple groups.

user131753

But what about the case when $G$ is not abelian?