Mathematics - 2018-03-03 (page 2 of 2)

0celo7

19:05

@JoeShmo it will be the product of pullbacks since pullback of functions is composition

Balarka Sen

@EricSilva @MikeMiller Alternatively, nail it down with compactly supported Poincare duality

It's isomorphic to $H_1(\Bbb R^2 \setminus 0)$

Which is $\Bbb Z$

I'm a cheater, I know

Joe Shmo

@0celo7 um, (f . g)^* = g^* . f^*; now (f x g)^* = f^* x g^*.. how?

where . is function composition (borrowing from haskell notation)

Vrouvrou

Hello

Balarka Sen

Actually, I just had a godly vision involving the Christ and compact support. I think if $L$ is a noncompact connected $n$-manifold, $H^k_{c}(L)$ is isomorphic to exactly $H^k(L^*, \text{End}(L))$ where $\text{End}(L)$ is the space of ends of $L$ and $L^* = L \cup \text{End}(L)$ is the end-compactification

Vrouvrou

Someone know willem's book : Minimax theorem ?

Balarka Sen

19:11

~~No, uh~~ Ok, no, I think my vision is correct. Lord has saveth my soul.

In this case, for example, $\Bbb R^2 \setminus \{0\}$ has two ends, and the end-compactification is $S^2$ (the end space being the north + south pole). H^1(S^2, (north and south pole)) = H^1(S^2 v S^1) = Z

0celo7

@JoeShmo What?

$\phi^*(fg)=(f\circ\phi)(g\circ\phi)$

Balarka Sen

In fact, in fact, in fact. If the end is nice enough, then $H^k(L^*, \text{End}(L)) \cong H^k(L^+)$ should hold, where $L^+ = L^*/\text{End}(L)$ is the one-point compactification. Eg, if the end has a cone neighborhood.

0celo7

what are you babbling about

Balarka Sen

only rick and morty people will understand

0celo7

@BalarkaSen isn't rick and morty quantum mechanics, which you know nothing about?

Balarka Sen

19:24

lol are you taking the piss for not being called a rick and morty enthusiast

u just got baited son

use homeomorphism to escape

0celo7

I have another MO doubt

Balarka Sen

did you figure out the previous one

0celo7

My advisor and I discussed how one could do it

I have to write up the details

Balarka Sen

mmm i see

0celo7

It's just formalizing that if a nonlinear op is elliptic at one point, it should be at nearby points because Iso is open in Hom

It's not clear that everything involved is continuous, however

So gotta do some work there

@BalarkaSen So right now I don't know if Kazdan-Warner trichotomy is nontrivial in dimensions not 2 or 4

Balarka Sen

19:28

gotcha

0celo7

the KW classes might be empty. The best I can find is a 4-manifold that does not admit metrics of positive scalar curvature, nor a scalar flat metric

$T^4\# K_3$

Balarka Sen

wow

0celo7

or do you write it $K3$

Balarka Sen

K3 I think

0celo7

even in math mode?

Balarka Sen

19:30

mhm

0celo7

$K3$ or $\mathrm K3$

Balarka Sen

the former ideally

0celo7

the wiki page on K3 is pretty bad

I can't tell if it's a class of manifolds in arbitrary dimensions, or what

Balarka Sen

I don't know what it looks like

just know the name

0celo7

where is that german dude

Kodaira showed they are all diffeomorphic

0celo7

19:46

0

Q: Examples of manifolds that do not admit scalar flat metrics

The Kazdan-Warner trichotomy states that for $n\ge 3$, a compact $n$-manifold falls into one of three categories: (A) Every (smooth) function is a scalar curvature. (B) The manifold is strongly scalar flat. (C) The manifold only admits negative scalar curvature metrics. Of course class (A) i...

mago

what's a zero dimensional subscheme of $\mathbb{P^n(k)}$?

Balarka Sen

@mago Are you asking for an example or a definition? A finite number of points constitutes an example, say

mago

initially, a definition would be helpful

Balarka Sen

@MatheinBoulomenos would be able to explain what a scheme is better than me.

mago

my thesiis advisor gave me some lecture from Geramita about Waring's problem (and in the introductory part is all about zero schemes), and I honestly do understand anything ahahaha

thanks anyway @BalarkaSen

MatheinBoulomenos

19:54

@mago Read this post by lush: chat.stackexchange.com/transcript/message/43058565#43058565 an affine scheme $\operatorname{Spec} A$ is a geometric object that contains the information about a commutative ring $A$. A general scheme is a geometric object that looks locally like an affine scheme (similar to how a manifold looks locally like an open subset of $\Bbb R^n$).

Here the notions "geometric object" and "looks like" are made precise by considering a topological space with algebraic data attached to open subsets

Daminark

Henlo

Joe Shmo

@0celo7 in math.cornell.edu/~sjamaar/manifolds/manifold.pdf in exercise (4.9) (ii), i don't understand what to do with the product of rho and c, also since rho is a closed path, can we assume the winding number of rho is finite? or else presumably something should cancel out

MatheinBoulomenos

You have morphisms of schemes which are compatible with the topology and the algebraic data in a certain way. If you have a morphism of schemes $X \hookrightarrow Y$ that is an embedding onto a closed subset on the topological level, then one calls $X$ a closed subscheme. A closed subscheme is not uniquely determined by specifying a closed subset! For an affine scheme, closed subschemes correspond to ideals, whereas closed subsets correspond to radical ideals

Bogey

Hey, quick question. Let's say there are x number of ways (combinations) to add up 1 penny and 5 penny coins to make 1$. Now, the question is - how many combinations are there to make 2$ out of 1 penny and 5 penny coins? Would you have to re-run the entire calcuation, or would the knowledge of the first question help to answer this? (E.g. if there are XYZ ways to make 1$, then there must be ... different ways to make 2$)

MatheinBoulomenos

@mago It's unlikely that you can grasp this just from a short explanation wirtten in a chat tbh

@Daminark Henlo

mago

20:03

@MatheinBoulomenos totally agree, I'll reutrn asking question after spending some time trying tiio understand this stuff

anyway thanks for the link and the intuitive answer

Lozansky

20:47

Heya @Semiclassical

0celo7

21:45

@BalarkaSen Bah, only garbage answers to my MO question

They gave me more criteria, but how the fuck am I to know which manifolds have which invariants

@JoeShmo do you have a more precise question?

what did you try?

Joe Shmo

tyring as we speak

let me get back to you in a couple hours. i'm on the brink of a breakthrough (famous last words)

i think for (ii) the winding number is still k, and the relevant discussion in the text is EXAMPLE 4.2., or THEOREM 4.3.

(please do let me know if you think im totally wrong :) )

0celo7

I haven't thought about it but $k$ makes sense

Xander Henderson

$k$ doesn't make sense. $s$ makes sense. Well, two of them, as well as two $e$s and an $n$.

Joe Shmo

22:27

@0celo7 i dont know what to do with the multiplication by rho. I want to reinterpret it as composition, but that doesn't help me

anakhronizein

Hola

skullpatrol

Heya

anakhronizein

How are you, @skullpatrol ?

skullpatrol

Fine, thanks @anakhronizein how are you?

anakhronizein

Not too bad.

anakhronizein

22:57

Hi @BalarkaSen

skullpatrol

\o @BalarkaSen

Prototank

If two polynomials in $\mathbb{R}[x,y]$ agree on the $x-$axis and on the $y-$axis, need they be the same?

nbro

If anyone here wants to check the math in the following article, I would appreciate. Let me know if there are any mistakes.

Chan's algorithm

In computational geometry, Chan's algorithm, named after Timothy M. Chan, is an optimal output-sensitive algorithm to compute the convex hull of a set P {\displaystyle P} of n {\displaystyle n} points, in 2- or 3-dimensional space. The algorithm takes O ( n log ⁡ h ) {\displaystyle O(n\log h)} time, where h {\displaystyle h} is the number of vertices of the output (the convex hull...

anakhronizein

@Prototank f(x) = x^2 and f(x) = -x^2 agree on the x and y axes.

Or what do you mean by "agree on the x axis"?

Same intersections?

Prototank

If two polynomials in two variables agree on all pairs $(x,y)$ where one of $x$ or $y$ is zero, are they the same?

Union the axes

Ted Shifrin

23:06

@Prototank: You can easily make up a counterexample.

Prototank

sure, just toss in $xy$ terms into one and not the other....

Balarka Sen

Hi @anakhronizein! I saw your overleaf page. I'll check it out soon

Ted Shifrin

Right, or more generally, terms with $x^jy^k$, $j,k>0$.

Hi Balarka

Balarka Sen

Hey Ted

Prototank

Full disclosure, I'm looking for an algebraic map between algebraic subsets $X\to Y$ that is an epimorphism but not a surjection. My first guess was to include the coordinate axes into the plane, but this doesn't work

Ted Shifrin

23:10

I don't think categorically enough to distinguish between epic and surjective.

Balarka Sen

I should finish Taylor expanding the Riemannian metric upto second order in geodesic coordinates someday

Daminark

> that feeling when Ted isn't a category theorist

Balarka Sen

> that feeling when geometry hasn't been approached properly

Ted Shifrin

Demonark: If you're so smart, explain the difference to me in the category of algebraic subsets (of affine space or projective space).

Daminark

Wait I never said I was smart

anakhronizein

23:12

Why is geometry one of the worst subjects to read sometimes. :(

skullpatrol

because you don't "read" it, you "absorb" it :P

Balarka Sen

@TedShifrin I do think epimorphisms in the category of affine k-algebras are more complicated than cokernel zero maps

Ted Shifrin

"geometry" is awfully vague and vast.

Balarka Sen

Maybe I should say monomorphism because I dualized

And kernel zero thereof

PVAL-inactive

whats a map of affine subsets?

is that like a map of affine varieties?

Ted Shifrin

23:15

I would have thought a backwards map on coordinate rings

Balarka Sen

regular map

Same as what Ted said

Ted Shifrin

But I don't understand epic in any category but sets, I suppose.

Balarka Sen

Me neither.

Joe Shmo

@TedShifrin is there a formula for a pullback of a product of two functions? i.e., (f x g)* = ...?

Ted Shifrin

Hi @JoeShmo. Give me more context.

skullpatrol

23:16

How's the neck Professor?

Ted Shifrin

Domain? Range?

Joe Shmo

question (4.9) (ii) in http://www.math.cornell.edu/~sjamaar/manifolds/manifold.pdf

:)

Ted Shifrin

Still messed up, skull, but the physical therapy is helping.

Joe Shmo

hey are you ok?

Ted Shifrin

I wouldn't call that a product mapping. Very misleading.

Just degenerating disks and vertebrae.

Joe Shmo

23:18

f: [0, 1] -> R^2, g: [0, 1] -> (0, infinity)

oh man. sounds painful. how are you doing?

Ted Shifrin

When you say product mapping I think mapping from product of manifolds to product of manifolds.

Joe Shmo

no no no. im not that smart

Ted Shifrin

My suggestion for questions like that is to start by drawing pictures.

PVAL-inactive

I think any open immersion will do.

Joe Shmo

i have an intuitive understanding of what's going on. rho scales the path t in an orientation-preserving fashion.

rho != 0, which is good.

Balarka Sen

23:19

Bullshit

xy = 0

project

ok

Yes, that seems right

Joe Shmo

but how do i rigorously show that the winding number of ~c = k, still?

Ted Shifrin

@JoeShmo: So if you deform a curve without crossing the origin, what happens to the winding number?

Joe Shmo

nothing

the winding number remains the same

Ted Shifrin

I would suggest doing (iii) before (ii). You're defining winding number by pulling back $d\theta$ and integrating?

Balarka Sen

By xy = 0 I meant xy = 1. Sigh.

Joe Shmo

23:21

he's referring to some formula (as in, "...by using the formula") and i strongly suspect he's referring to theorem 4.2

PVAL-inactive

my intuition is that unique continuation should imply that the inclusion of any affine open will behave categorically like a surjection.

I mean like xy\ne 0

included into A^2 for instance

Prototank

@BalarkaSen What do you mean here?

Joe Shmo

winding number of a path c about the origin = w(c, 0) = integral along c of (alpha_0) where alpha_0 = (-ydx + xdy)/(x^2+y^2)

Balarka Sen

@PVAL-inactive That's not an affine varieties, that's a quasi-affine thing. I am thinking of the hyperbola xy = 1 projected to the x-axis

The image is x-axis minus the origin

Adeek

hi @TedShifrin

Ted Shifrin

23:22

I didn't look, @JoeShmo. OK, so what happens to that form if you scale vectors by a factor $\rho$?

Balarka Sen

So the map is k[x] included in k[x, y]/(xy - 1). Maybe some algebraist can tell us if that's an EPIC MORPHISM :boom:

Adeek

hi everyone

Ted Shifrin

hi Karim

Have you paged @Mathein? @Balarka

PVAL-inactive

the complement of a hypersurface should be affine..

Balarka Sen

No, that's not what an affine variety means. You're describing a quasi-affine variety

PVAL-inactive

23:23

xy\ne 1

should work then

it isn't affine as a subset of A^2

Balarka Sen

It can be realized as image of a map from an affine variety if you mean that

Those are precisely what quasi-affine varieties are, actually

@TedShifrin Good point. @MatheinBoulomenos, come to the rescue

Prototank

Okay this makes sense

Balarka Sen

Is k[x] --> k[x, y]/(xy - 1) a monomorphism?

Joe Shmo

@TedShifrin i dont know. there's funny business going on with the pullbacks

Adeek

@TedShifrin can I ask quick question in complex analysis ?

Balarka Sen

23:25

Forgot to dualize.

Ted Shifrin

No, no funny business, @JoeShmo.

Prototank

Okay, thanks for the example

Ted Shifrin

Write down the pullback by $c(t)$ explicitly, and then write down the pullback by $\rho(t)c(t)$ explicitly. First think through what happens when $\rho$ is constant.

PVAL-inactive

xy\ne 1 has coordinate ring K[x,y,z]/((xy-1)z=1)

Ted Shifrin

What's the question, Karim?

Joe Shmo

23:27

@TedShifrin i tried that. let me try again

PVAL-inactive

or in other words its the localization of xy-1 in K[x,y]

this is clearly affine

Ted Shifrin

@JoeShmo: Honestly, (iii) is telling you what's going on geometrically. Projecting the curve to the unit circle gives the same $1$-form when you pull back.

Joe Shmo

yes, right, i get that

Adeek

@TedShifrin I was wondering if there is some kinda of generalization of Great Picard Theorem to complex manifolds ?

Joe Shmo

but the question is asking to use "the formula" (although it doesn't specify what "the formula" is)

Ted Shifrin

23:28

Hyperbolic complex manifolds is all about that, Karim. There are various books on this.

Adeek

oh very cool.

Ted Shifrin

The formula has to be $d\theta$, which is the $1$-form you wrote down, @JoeShmo.

PVAL-inactive

the immersion is the obvious map K[x,y] to its localization.

Balarka Sen

True I see what you mean now

Prototank

So the locus of $xy-1$, sent to the x-axis is an algebraic map

Balarka Sen

23:30

Yes, @Prototank

Prototank

but any two functions which agree on everything, except zero, must be the same

Balarka Sen

Quite. Sorta a density argument but not quite.

Prototank

polynomials*

Adeek

anyway I am going to work

cya @TedShifrin

Joe Shmo

@TedShifrin ok. let me write down the explicit formula of the pullback by c and rho

Prototank

23:31

Thanks @BalarkaSen and @PVAL-inactive

Adeek

@TedShifrin Btw The other day I was teaching the class about differential forms in my vector calculus

PVAL-inactive

I guess its easier to do this a dimension down

Ted Shifrin

You shouldn't do that, Karim. I already warned you about that.

Adeek

I had few minutes on my time I thought it would be nice to give students some idea about them

Joe Shmo

@TedShifrin meaning, i know what the winding numbers are, and i can prove what they are in each case, but i can't compute them explicitly, which is alarming to me anyway.

PVAL-inactive

23:32

A^1 minus the origin has coordinate ring C[x,y]/xy=1

Balarka Sen

Localization of C[x] at (x) again isn't it

Ted Shifrin

@JoeShmo: It's almost impossible to compute explicitly.

Joe Shmo

but they're asking me to compute

Ted Shifrin

Unless you wind around the circle a certain number of times. But maps are homotopic to such things.

PVAL-inactive

and the map C[x] to C[x,y]/xy=1 is an open immersion.

Ted Shifrin

23:33

No, they're asking you to find the pullback and see its integral is no different. They're not asking you to compute the integral.

Joe Shmo

Determine the
winding numbers of the following paths c˜: [0, 1] → R
2
\ {0} by using the formula, and then
explain the answer by appealing to geometric intuition

yes yes. that's what i mean

Ted Shifrin

Given that one integral comes out $k$, you can tell the other one does also, without doing any explicit integration.

Joe Shmo

yes

Adeek

brb to work.

Balarka Sen

@Daminark this is what happens when real men do real math for too long. they start calling maps between commutative rings open immersions

Joe Shmo

23:34

my idea of a computation

Balarka Sen

(I hope you don't mind the memes @PVAL)

PVAL-inactive

open immersion is a term i got from Hartshorne

Balarka Sen

Oh really

Ted Shifrin

@Balarka: I see no evidence of "real men" in this chatroom.

Balarka Sen

@TedShifrin Except the 16-18 year old real men who infest it, you mean?

Daminark

23:36

@BalarkaSen such is life

Ted Shifrin

No comment @Balarka

Balarka Sen

LOL @Ted

PVAL-inactive

since the ring map is an injection and injections for rings are automatically monomorphisms it easily follows that this map is an epimorphism.

Kasmir Khaan

AM A MAN ! AM AN ANCHORMAN , I ANCHOR THE NEWS

what movie ?

is that quote from ?

Joe Shmo

ANCHORMAN

BOOM

Kasmir Khaan

23:36

haha

Ted Shifrin

@PVAL: That's what started this whole thing. I don't know the difference between monic/injective and epic/surjective in the category of rings as opposed to the category of sets.

damn, now Kasmir is sounding as bothersome as Demonark and Balarka.

Balarka Sen

@PVAL Wait, monomorphism or epimorphism?

PVAL-inactive

the map on rings is a monomorphisms

Kasmir Khaan

Ted! one time i had to make a quote from movie

dont put me in that category :D

PVAL-inactive

so the reversed category map will be an epimorphism

Balarka Sen

23:38

OK great

@Ted So the map on affine varieties need not be surjective

to be an epimorphism

There's your difference

Ted Shifrin

So what is the definition of epic?

Prototank

But the map you guys came up with is awfully darn close to a surjection

PVAL-inactive

gf=gh \implies f=h

Balarka Sen

aka right-cancellative

@Prototank Yeah I'd want to know if all epimorphisms arise this way

Prototank

if the category is concrete surjection implies epic, obviously

Balarka Sen

23:40

As open immersions

Prototank

but the converse doesn't hold always

Balarka Sen

Only @Mathein can't tell that

Ted Shifrin

PVAL is confuzling me.

PVAL-inactive

whoops

i did it backwards

Ted Shifrin

I see. Thanks.

I guess I'm not sure I ever grokked this when I took 5 quarters of Hartshorne. I must have.

The Testosterone Fanatic

23:41

YAS YAS YAS Dragon Ball Super's next mega episode is about to air in 20 minutes!!

and hi chat

PVAL-inactive

I think this should morally be unique continutation of holomorphic functions on an open set.

The Testosterone Fanatic

holomorphic functions care about morals?

Balarka Sen

holomorphic functions don't have hollow morality

2

ba dum tss

Lozansky

Let $\epsilon>0$ and $x>0$. Then what is the even extension of $\delta_{\epsilon}(x)$ on the whole of $\mathbb{R}$, in the limit as $\epsilon \to 0$?

Ted Shifrin

hollow morphality, Balarka

@Lozansky: Was that question supposed to make sense to us?

Lozansky

23:45

@TedShifrin :(

Does the term "even extension" not make sense?

It is a direct translation from my language

Daminark

@TedShifrin for reference, my above comment was intended is a joke, but if the jokes are actually that annoying I can stop

Ted Shifrin

Demonark: When almost everything coming out of your "mouth" is de facto a joke, it's gotten too annoying. I'm happy to discuss actual mathematics.

@Lozansky: What is $\delta_\epsilon(x)$?

Daminark

Alright, I'll keep that in mind

Lozansky

@TedShifrin Dirac-delta distribution centered at $\epsilon$

Ted Shifrin

But this is, of course, not an actual function, unless you're doing an approximate one.

But the usual ways of thinking about it give something on all of $\Bbb R$ that is even with respect to $x=\epsilon$.

Akiva Weinberger

23:48

Hello

Lozansky

No, still a distribution

Balarka Sen

@TedShifrin As the hip kids alternatively say nowadays instead of "Indeed": "Holomorphic, my dude".

Akiva Weinberger

So do Carmo (book) has been trying to teach me about connections and I think I sorta get them now

Balarka Sen

Reference:

@Akiva Huzzah

Akiva Weinberger

Also I get why $\nabla_XY-\nabla_YX=[X,Y]$ is a reasonable assumption

Balarka Sen

23:49

Do you now

The Testosterone Fanatic

@Daminark I suffer the same fate as you, especially because I specialize in stochastic stuff, and most discussions here are never relating to that topic

Ted Shifrin

So distributions only operate on compactly supported smooth functions, @Lozansky. So what do you mean by "extend to all of $\Bbb R$"?

heya DogAteMy

Balarka Sen

@The What fate? Daminark actively engages in a lot of conversations, mathematical or non-mathematical, in this chat.

Lozansky

It would be like $f^+(x) = \cases{\delta(x-\epsilon), & x>0 \\ \delta(\epsilon-x), & x<0}$

But they are the same, $\delta$ is even

Ted Shifrin

But a distribution is NOT a function, @Lozansky. It operates on functions.

You can't write stuff like that.

The Testosterone Fanatic

23:51

@BalarkaSen well, my bad, I extrapolated too much from Ted's comment

Akiva Weinberger

@BalarkaSen Incidentally

Balarka Sen

@Akiva Oh god

Ted Shifrin

DogAteMy has been totally memefied.

Balarka Sen

We're all influenced by Daminark

I welcome it

Akiva Weinberger

I linked to a Google image search

Ted Shifrin

23:52

I don't.

Lozansky

@TedShifrin Sorry I am informal. We could say $\delta^+_{\epsilon} = \delta_{\epsilon}+\delta_{-\epsilon}$ maybe?

Ted Shifrin

@Lozansky: So you mean even NOT with regard to $\epsilon$ but with regard to $0$?

Lozansky

Then we get $\delta^+ = 2\delta$ as $\epsilon \to 0$. Good result?

Well, $\epsilon \to 0$

Ted Shifrin

I still don't know what even means for a distribution.

Lozansky

Is there subtle difference?

Ted Shifrin

23:53

(I mean, I do, but you haven't said it.)

Adeek

@BalarkaSen

Balarka Sen

Adeek

lol

Balarka Sen

Oh jesus christ

That Morrisey image is killing me

Akiva Weinberger

Who is that

Adeek

23:54

@BalarkaSen In my talk which is next month I was gonna mention how things in complex analysis the genetic code of holomorphic function is determined at very small level

hahaha

Balarka Sen

Morrissey, the lead singer in Smiths

Alex

new phone, who dat

Lozansky

@TedShifrin We have only talked about even/odd extensions for functions (not distributions), but we often treat $\delta$ like a function

Akiva Weinberger

Are those lyrics

Alex

@AkivaWeinberger Well, I'm still fond of you.

Ted Shifrin

23:55

@Lozansky: Well, it's not.

Lozansky

I know

Ted Shifrin

So think about what "even" should mean.

Balarka Sen

@Alex LMAO

Actually, they are lyrics. :)

Alex

Damn

Akiva Weinberger

I'll take that as a "yes", despite the image saying the opposite

Alex

23:56

Even the 'meat is murder'?

Balarka Sen

Very much

Akiva Weinberger

https://en.wikipedia.org/wiki/Meat_Is_Murder

Apparently it's an album

Balarka Sen

It's their second studio album iirc

sniped

Alex

Not sure if I should or shouldn't listen to this

Balarka Sen

Smiths have some interesting songs but Morrissey is a dick

A very well known one at that

Lozansky

23:58

@TedShifrin I only know of the definition $U_f[\varphi] = \int_{-\infty}^{\infty} f(x) \varphi(x) dx$ for distribution $U$ that can be represented by locally integrable function $f$ and $\varphi \in \mathcal{D}$

Ted Shifrin

But the delta distribution isn't represented by an $L^1_{\text{loc}}$ function.

Lozansky

True

But still it makes sense

To define $\delta[\varphi] = \int_{-\infty}^{\infty} \delta(x) \varphi(x) dx$

Ted Shifrin

@Lozansky: I'm suggesting that you try to define evenness by applying to test functions $f$ and $g=f\circ -$.

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