Mathematics - 2017-01-28 (page 4 of 5)

Hiroto Takahashi

17:02

I meant, if we have a sequence ${u_{b}} \in H$ where $H$ is a Hilbert space which converges weakly (i.e. $u_{n} \to u$) and $|u_{n}| \to |u|$, then $u_{n} \to u$ converges strongly? The answer here is false for a Banach space, but what about a Hilbert space?

$u_{n} \in H$*

Secret

I wonder, what does a generic substitution look like in integration. There are so many cases where a substitution works but there seemed to be no easy way to tell why

Meanwhile, for any given integral, one can find its corresponding differential equation that it obeys by letting $y=\int f dx$ and then differentiate and rearrange the integrand

Secret

17:16

These differential equations are then separable by construction

DHMO

In the complex field, are i and -i symmetrical?

it boggles my mind to know that they are symmetrical...

SteamyRoot

What do you mean with "symmetrical"?

Astyx

What do you mean by symmetrical ?

DHMO

it means, for any equality, if we exchange i and -i, we get the same thing

Semiclassical

um, $z=i$ is certainly not the same as $z=-i$.

DHMO

17:21

@Semiclassical well, what i meant is that both statements can be true

Astyx

Well $(-i)^2 = i^2 = -1$ and this is in some way the definition of $i$

Semiclassical

@DHMO Not at the same time, they can't.

Astyx

So yes, you could flip the complex plane around the real axis and do everything you could without any problems

Semiclassical

Then it what sense is that any different from $z=1$ versus $z=2$?

DHMO

@Semiclassical you have a point

Astyx

17:23

But you have to be consistent, that's all

DHMO

maybe I should say that, for any equality without variables, you can swap i and -i and get another equality

and the truthfulness of those equalities will be the same

SteamyRoot

Well, $i$ and $-i$ is just a choice.

Semiclassical

I think what you have in mind is that, as @astyx indicates, one can do complex conjugation of all entries in an equation.

Astyx

That's only because conjugation is compatible with all usual operations

Semiclassical

Right.

DHMO

17:24

@Astyx i'm looking for unusual operations

Balarka Sen

What he probably wants to say that is that complex conjugation is an R-automorphism of C

Indeed, the only interesting one

DHMO

@BalarkaSen what does that mean?

SteamyRoot

The only nontrivial one

:P

Balarka Sen

That's what I said

;)

SteamyRoot

Sneaky!

Balarka Sen

17:25

@DHMO Google is your friend!

Semiclassical

The Schwarz reflection principle comes to mind as well.

Balarka Sen

It's a field automorphism fixing a subfield

DHMO

@BalarkaSen what are the non-interesting ones?

> It is, loosely speaking, the symmetry group of the object.

Astyx

Geometrically speaking, In some way that means you can measure angles in the trigonometric or non trigonometric way without changing anything of your results

Semiclassical

In that it gives a simple condition under which you expect $F(\overline{z})=\overline{F(z)}.$

Balarka Sen

17:26

identity automorphism

DHMO

but is there some unusual... oh!

arg(i) = pi/2

Semiclassical

Yep.

DHMO

why does symmetry break for this one?

Semiclassical

Alternatively, Im(i)=1.

Astyx

In some sense that's the same as asking why multiplying any equality by $-1$ gives you another equality

Because flipping the real line around doesn't change a thing

Semiclassical

17:27

The reason it fails for the arg and im functions is because they're not analytic functions of z.

Balarka Sen

Take the function on C which gives 1 for i and 0 for everything else.

DHMO

@Semiclassical hmm...

Semiclassical

Hence there's no reason that $F(\overline{z})$ should give the same result as $\overline{F(z)}$.

Astyx

And also because these functions are not uniquely defined by the reals @Semi @DHMO

Semiclassical

Point.

Im(z) and -Im(z) give the same result (zero) on the reals.

DHMO

17:29

modulus is symmetrical right

Semiclassical

Right.

DHMO

Re(z) is also symmetrical

but it isn't analytic

Astyx

Obviously you can define symmetrical functions that are not analytic, eg $1-\chi_{\Bbb R}$ (where $\chi_A$ is the caracteristic function of $A$)

Semiclassical

Yeah.

So the prescription I gave above isn't enough.

Astyx

I doubt there is a converse implication

Semiclassical

17:31

Yeah.

DHMO

any more?

Semiclassical

Depending on what you mean, $f(z)=1+i z$ may work. Then for instance $f(i)=0$ but $f(-i)=2$.

Balarka Sen

@Semiclassical analytical function need not satisfy F(\bar z) = \bar F(z) either? Like, f(z) = iz. f(\bar 1) = i and \bar f(1) = -i.

DHMO

hmm... so we need to find another metastructure

Semiclassical

Yeah. The reflection principle further demands that $f(z)$ be real on the real axis.

Astyx

17:34

I think $\Bbb R$ has to be stable by $F$ probably

Balarka Sen

@Semiclassical That's not what the reflection principle says. It extends f defined on one side of a domain symmetrical about R to the other side.

In any case if the Taylor expansion of f has real coefficients it should work

Astyx

That might be linked that $\Bbb C$ is algebraicly closed for multiplication. In the same way you can chose either $1$ or $-1$ in $\Bbb R$ without changing anything with addition ?

Balarka Sen

At least in a neighborhood of 0. By termwise complex conjugating.

Semiclassical

Eh. Is there an analytic $f(z)$ which is real on the real axis but for which $\overline{f(z))}\neq f(\overline{z})$?

Oh. By demands, I mean "it doesn't apply if f(z) isn't real on the real axis."

Premise, not conclusion.

Astyx

That's an interresting question though

Balarka Sen

17:37

No, by what I said., not by reflection principle. Write down the Taylor expansion.

I guess you could reflection principle it too, along with identity theorem. But this is much simpler.

Semiclassical

Hey, I never said I was being simple :P

Dattier

Hello

Semiclassical

But yes, Taylor series is enough.

Astyx

Do all analytic functions that are real on the real axis have real coefficients ?

Hi @Dattier

DHMO

@Astyx that's interesting...

Semiclassical

17:40

$f(z)-f(\overline{z})$ would be i times a Taylor series with real coefficients in that case.

Astyx

Hullo @Ted

Ted Shifrin

Salut, @Astyx

Semiclassical

In which case I think it could only be real if it's identically zero.

DHMO

@Semiclassical nice

Astyx

I think so too, but I don't know enough about analytic functions to be sure about it

Semiclassical

17:41

It's only nice if i'm not missing an obvious counterexample :P

Balarka Sen

Semi C is correct

Dattier

@Astyx I think yes, you can show that with the conjugate.

Ted Shifrin

Oh, Balarka exists

Balarka Sen

Otherwise it has a non-discrete zero set

Astyx

So the only taylor series that is identically 0 is the series of which the coefficients are 0 ?

Balarka Sen

17:42

Analytic functions have discrete zero set

@TedShifrin I exist, therefore I think.

4

Ted Shifrin

ponders

DHMO

@BalarkaSen why?

Ted Shifrin

@Astyx Yes.

Liad

is it correct that $ Z / Z = Z$ ? where Z is the integer ring

Alessandro Codenotti

Hi @Ted

Semiclassical

17:45

No.

Astyx

@Liad No

Ted Shifrin

heya @Alessandro :)

Alessandro Codenotti

No @Liad

Astyx

Hi @Alessandro

Liad

it is 0?

Alessandro Codenotti

17:45

Hi @Astyx

Balarka Sen

@DHMO Look in any standard complex analysis textbook. It's a version of identity theorem.

Semiclassical

What do you mean by 0?

Astyx

$\{0\}$, yeah in some sense

Semiclassical

What astyx said.

Astyx

More like $\{\overline 0\}$

Dattier

17:45

Au revoir

DHMO

@BalarkaSen alright

Astyx

Bye

Liad

what confused me is $ Z/Z = \{ z + Z : z\in Z \} = \{Z\} $ because if $g \in G$ then $g + G = G$

Balarka Sen

Zero set of any nonzero analytic function, I meant

Liad

where is my mistake?

Alessandro Codenotti

17:47

$\{Z\}$ is not the same as Z

Astyx

There isn't one

Liad

huh!

right.

it is the zero of the $Z/Z$ qutient ring. thanks

Balarka Sen

@TedShifrin What does the Chern-Gauss-Bonnet theorem say?

Astyx

You're welcome

Ted Shifrin

There's a boundary version, but the compact case says that the integral of the Pfaffian of curvature is an appropriate constant times $\chi$ for an oriented even-dimensional Riemannian manifold.

Balarka Sen

17:50

Hmm

s.harp

@BalarkaSen Are you saying that inanimate objects don't exist?

Balarka Sen

@s.harp I am not an inanimate object, last I checked!

Ted Shifrin

@s.harp is saying inanimate objects don't think.

Balarka Sen

Yeah, but "I" literally means I.

Not everyone and everything in the world

Secret

[Differential equation] Solve for w
$$x^2 w^{(4)}-xw'''+(1-n^2+x^2)w''-2xw'+2w=0$$
To be continued...

Dattier

17:53

I have open a answer, and I want to know, what's the problem with this subject : "Does a compact metric space, been isometric to a part of a finite-dimensional real vector space?"

DHMO

@Secret $w$ is a function of $x$ I suppose

Dattier

a question

Secret

@DHMO It is

Dattier

I like to know

Astyx

You're lacking a verb, are you not @Dattier ?

DHMO

17:54

@Secret that's... interesting...

Dattier

@Astyx Yes I am not english

Balarka Sen

@TedShifrin Is the curvature matrix here the sectional curvature of each coordinate 2-plane in the tangent space?

Astyx

I'm not exactly sure what you mean by that question @Dattier

Secret

That is a linear homogenous ODE of degree 4. I need to check whether I can find an integrating factor for linear ODE of degree n though...

Ted Shifrin

No, @Balarka. That's only the $\omega_i\wedge\omega_j$ component of $\Omega_{ij}$.

Dattier

17:55

there the answer here : math.stackexchange.com/questions/2110783/…

Semiclassical

Boundary values / initial conditions?

s.harp

@BalarkaSen So your existence is conditional on your ability to think?

Secret

no boundary and initial conditions atm, I am interested in the general solution. Still working on it though...

Dattier

@Astyx there the answer here math.stackexchange.com/questions/2110783/…

Balarka Sen

@s.harp That's backwards to what I said!

Dattier

17:56

@Astyx it's Jeopardy

Semiclassical

Mathematica gives an answer for generic $n$ in terms of a multitude of hypergeometric $_1F_2$ functions. But it probably is different for integer $n$.

Astyx

Huh I see your problem

Hiroto Takahashi

just curious, there exist any infinite-dimensional compact space? please, don't judge me

s.harp

It was "I have property: exists $\implies$ I have property: thinks"

Semiclassical

and indeed Mathematica struggles if I take, for instance, $n=0$.

s.harp

17:57

So if at any point you do not have the property: thinks you will not have the property: exists

Astyx

@Dattier Do you by any chance mean : "Can a compact metric space be isometric to a part of a finite-dimensional real vector space?"

Ted Shifrin

@Hiroto: Sure. An infinite product of compact sets is compact (in the product topology).

Dattier

@Astyx thanks

Astyx

Glad to help

Hiroto Takahashi

@TedShifrin Thanks!

Balarka Sen

17:59

@s.harp Ah, I misinterpreted your use of conditional. Sure.

Hiroto Takahashi

that sounds to Tychonoff's theorem

Astyx

How does one prove that the space of bounded continuous real functions is complete ?

Balarka Sen

@TedShifrin Hm, I see.

Dattier

@Astyx : an other problem here : math.stackexchange.com/questions/2117467/customized-group and I don't know what is the problem

Balarka Sen

So what's the full thing?

Alessandro Codenotti

18:00

With which topology? @Astyx

Semiclassical

@Secret Actually, it looks like it has at least one nice solution: Namely, $w(x)=1-n^2-x^2$.

Astyx

Oh yes I forgot, infinite norm (is that the terminology ?)

Ted Shifrin

$\Omega = d\omega-\omega\wedge\omega$, where $\omega$ is the matrix of connection $1$-forms, @Balarka.

Daminark

I hear it more often as the sup norm

Ted Shifrin

@Daminark: Cuz you haven't done $L^p$ yet ;P

Semiclassical

18:02

If I'm remembering right, that means you can do reduction of order in order to trade that 4th order ODE for a 3rd order ODE which doesn't have that solution.

Daminark

That is true

Astyx

Then the sup norm, thanks

Ted Shifrin

That's the exercise that Alessandro figured out the other day.

Daminark

It's odd that we're doing functional analysis first, then measure theory

Ted Shifrin

And you never will really learn multivariable calculus well, which makes me very upset.

Astyx

18:03

@Dattier Well it's true it's unclear what you're asking

Balarka Sen

@TedShifrin Makes sense

Astyx

How do you translate from your native language to English ?

Ted Shifrin

I proved Chern-Gauss-Bonnet (probably for submanifolds of $\Bbb R^n$) in several of my grad courses, @Balarka. Notes probably available if you want to learn more.

Dattier

@Asty Jeaopardy 2

Daminark

Yeah, when I'm not completely swamped I need to actually go through the material slowly, especially with integration

Dattier

18:05

@Astyx Jeopardy 2, I have give an answer, and sorry

Daminark

Like, we spent some reasonable amount of time on differentiation

Semiclassical

@Secret Oh, and so is $w(x)=x$. So that gets you down to a 2nd order ODE.

Astyx

Huh, translating sentences word for word is often not the way to go

Balarka Sen

@TedShifrin I'll probably have to get more comfy with moving frames before getting into that stuff. And I'm a lazy bum, so that'd take some time: I'll nag you when I want this stuff!

Semiclassical

On the note of being a lazy bum, I should type up some notes on what I've been doing.

Ted Shifrin

18:06

Well, I taught all the moving frames stuff in the course(s), obviously. But you should still do those exercises on surfaces.

Balarka Sen

Speaking of, I learnt the Frobenius theorem that day (not the proof, reading through Lee to understand it better)

Ted Shifrin

Good, Balarka, now you need to use it to do interesting things.

Dattier

@Astyx yes my english is poor, and I use google

Astyx

You should use sentence translators (even though it's best to know the language you're speaking in)

DHMO

@Dattier tu es francais?

Dattier

18:08

oui

Astyx

Comme tout le monde visiblement

DHMO

@Dattier tu peux parler francais ici... non?

Ted Shifrin

Pas tout à fait, Astyx.

Dattier

Super

Astyx

En première approximation, disons @Ted

Ted Shifrin

18:10

grand erreur :)

DHMO

@Dattier une question... quand on dit "grand erreur", on fait [gran.terreur] ou [gran.derreur]?

Dattier

Le titre que j'aimerais donner est "un groupe sure mesure"

Astyx

Grande erreur @DHMO

DHMO

@Astyx ok

Ted Shifrin

@Astyx C'est ce que j'ai dit au début, et puis je l'ai changé :(

Dattier

18:12

Je ne peux pas parler français ?

Astyx

Et il y a un "e" car c'est "une erreur"

DHMO

@Dattier sure?

Ted Shifrin

C'est une des raisons pour lesquelles le français est emmerdant :) @Astyx :P

Dattier

je suis aussi nul en orthographe

mais aprés pour "sure" c'est une faute de frappe

Astyx

Et le Français aussi ? @Ted

:p

Perturbative

18:14

Hey everyone!

DHMO

@Astyx interesting... in Latin and Spanish, "error" is masculine

@Dattier aprés?

Astyx

Hey @Perturbative

Ted Shifrin

toi, @Astyx, bien sûr :)

Astyx

Bien sûr :)

@Dattier Et donc c'est quoi un groupe sur mesure ?

Perturbative

Secret

18:15

(Ugh, it's 5:15, will continue on the investigation of that ODE later...)

Dattier

J'avais comme titre initial : "un groupe sur mesure" "a cutomized group", qu'est-ce qui ne va pas avec ce titre

Astyx

Ah ok, dans ce sens là

Semiclassical

@Secret Did you see my comments before?

I see two algebraic solutions to that ODE.

DHMO

@Dattier pourquoi "erreur" c'est feminin?

Semiclassical

So you should be able to reduce that to a second order ODE.

Dattier

18:16

math.stackexchange.com/questions/2117467/customized-group c'est un groupe que l'on peut choisir à l'avance et qui permet par exemple de résoudre les équations diophantiennes à puissances

Perturbative

I know how to prove that the given metric induces the usual topology on $\mathbb{R}^n$, take $x \in U$ to be open in $(\mathbb{R}^n, d)$, and show that there exists an open ball with respect to the new metric that lies inside $U$

And do the converse for $(\mathbb{R}^n, d')$

Astyx

@DHMO Because it's feminine in latin I'd say

DHMO

@Astyx it is masculine in Latin

and also in Spanish

decimos "un error"

c'est etrange

Astyx

My bad

Dattier

ne riez pas (svp) et je suis sérieux dans ma question : êtes vous des génies ?

Astyx

18:19

Loin de là

Enfin pour ma part

Perturbative

To do that one needs to relate the two metrics and find an $\epsilon > 0$ such that $B_{(\mathbb{R}^n, d)}(x, \epsilon) \subset B_{(\mathbb{R}^n, d')}(x, \delta) \subset U$ for $\delta > 0$

For the above $\epsilon < \frac{\delta}{\sqrt{n}}$ works

Dattier

@Astyx et @DHMO lors qu'en pensez-vous (de ma question sur les groupes sur mesures) ?

alors

Astyx

Que en soit tu ne poses pas de questions

Dattier

"alors"

Astyx

Tu as un énoncé auquel tu réponds, on ne sait pas ce que tu demandes

Perturbative

18:21

For the converse we need to find an $\epsilon > 0$ such that $B_{(\mathbb{R}^n, d')}(x, \epsilon) \subset B_{(\mathbb{R}^n, d)}(x, \delta) \subset U$ for $\delta > 0$

Ted Shifrin

@Perturbative: Have you said what $d$ and $d'$ are?

Perturbative

@TedShifrin $d$ is the standard euclidean metric

Dattier

J'ai tout précisé, me semble-t-il, dans le lien mis dans la question.

Astyx

Et tu peux modifier tes messages avec la flèche haut de ton clavier au lieu de reposter les mots

Perturbative

But strangely for the converse $\epsilon < \frac{\delta}{n}$ works

Ted Shifrin

18:23

growls @Perturbative ... And $d'$?

Perturbative

I assumed they would algebraically be the inverses of each other.

Dattier

ok, merci

Astyx

Bah non, on sait ce que tu fais, mais on ne sait pas ce que tu veux

Ted Shifrin

This is geometry, not algebra, @Perturbative.

Dattier

je n'ai pas compris

peux tu donner un exemple

Perturbative

18:24

@TedShifrin $d'(x, y) = |x_1 - y_1| + ... + |x_n - y_n|$

Ted Shifrin

Ah, ok.

Perturbative

But relating the two metrics is nothing more than an algebraic manipulation? I assumed they would be algebraic inverses of each other.

Astyx

Grossièrement ton message est structuré de la façon suivante :
Énoncé : montrer que blablabla
Preuve : puisque machin, alors truc, donc blabla

Il n'y a pas de question en soi

Ted Shifrin

Would it help to have $d''(x,y) = \max\limits_{i=1,\dots,n} |x_i-y_i|$?

No, @Perturbative. It really is geometry, not algebra.

Dattier

j'ai motivé mes questions par ce texte :

I prepare a difficult competition (for me), the strategy I decided to use, is to prepare a lot of exercice, to be able to propose them, to the juries.

For now, my level for problem solving is not up to the challenge, and I would like to improve this level by the solutions you will want to bring to these exercices. Indeed to know a solution even intuitive, allows me to better understand the solutions proposed, and even in some cases to have solutions I would never have thought otherwise.

The diverse origin of the mathematicians of this site, allows me to hope that this phenomenon (to have …

Ted Shifrin

18:26

Hi @AndrewT

Andrew Thompson

Hi @TedShifrin. How are you?

Ted Shifrin

Pretty well, thanks, @AndrewT, and you?

Astyx

Je n'ai vu ce texte nulle part dans ton post

Dattier

je l'ai mis en lien

Ted Shifrin

@Astyx: nulle? :P

DHMO

18:27

@Dattier alors que doit-on faire?

Perturbative

@TedShifrin I wouldn't think so, it's not that I'm stuck on it or anything, it just seemed weird to me that the algebraic inverse of $\epsilon$ didn't pop up for the converse. But as you said it's geometry not algebra

Andrew Thompson

@TedShifrin Quite well. Been stuck on the same problem for a few days now which is always a tad depressing.

Astyx

@Ted nulle.

Perturbative

@DHMO @Astyx Lol, are you guys still discussing linguistics?

Simply Beautiful Art

@AndrewThompson Aw :-(

Ted Shifrin

18:28

@Andrew: I spent most of my graduate life and career stuck on things for months on end.

Dattier

Il me semble que tout est expliqué dans mon petit texte

Ted Shifrin

@Astyx: We're even now :D

Simply Beautiful Art

@AndrewThompson In the end, it feels so much better though, trust me :-)

DHMO

I just can't figure out why "erreur" is feminine.

Astyx

Ah oui, les gens ont tendances à ne pas cliquer sur les liens. Autant demander directement des conseils sur la façon de prouver les choses

Ted Shifrin

18:29

@DHMO: As I complained earlier, there are some rules, but often it's random.

Astyx

@Ted Since when did this become a competition ? :p

DHMO

@TedShifrin no, often it corresponds to the Latin gender

Ted Shifrin

@Astyx: Since your regained your health :)

@DHMO: Since Latin has neuter, that won't go far.

DHMO

@TedShifrin but "error" is masculine the last time I checked

Dattier

@Astyx ok admettons le, et là quel est le problème : math.stackexchange.com/questions/2113635/…

Astyx

18:30

So I should have told you every single mistake you ever made in French since then ? If only I'd known .. :p

Ted Shifrin

@Astyx: Gender issues concern me less than syntax/grammar, but in general I would like to know if I mess up.

Astyx

Tu t'es contenté de mettre un énoncé sans expliquer ce que tu as tenté ni ou tu en es dans le problème

Dattier

Mais c'est des énoncés que j'ai fabriqué

Je le dis bien dans le petit texte

sauf si google m'a trompé

Astyx

@Ted very well then :)

En soi je ne sais pas trop pourquoi les questions sont fermées, personnellement je n'ai jamais eu de problème avec ça (cela dit j'ai posté seulement deux questions)

Ted Shifrin

@Astyx: On ne discutera plus de mathématiques.

Dattier

18:34

??

Je ne comprends pas ta réponse (si elle s'adresse à moi ) @Astyx

Hiroto Takahashi

Guys, there are any subspaces for $\ell^{2}$ such that $H_{s} \subset H_{t}$ such that $s < t$ for $0 \leq s, t \leq 1$

?

Astyx

@Dattier Je suis en train de dire que je ne peux pas vraiment t'aider parce que je ne sait pas ce qui motive la fermetures des questions (sinon que certains OP se foutent vraiment du monde parfois)

@Ted C'est un peu triste en soi, non ?

Dattier

C'est l'impression que j'avais (pour les OP)

Ted Shifrin

Moi, je ne suis pas triste.

Danu

o/

Krijn

18:38

o/

Ted Shifrin

Greetings, @Danu.

Simply Beautiful Art

o/

Astyx

@Ted C'est gentil ça ..

Ted Shifrin

And @Krijn.

Astyx

o/

Simply Beautiful Art

18:38

@TedShifrin D: You broke the trend!

Krijn

Hey @Ted

Ted Shifrin

Good, @Simply.

Simply Beautiful Art

T^T

Krijn

I got through most of my work for the weekend today already :D

So I can take the evening of

Simply Beautiful Art

@Krijn Congrats!

Astyx

18:39

@Dattier :p

Simply Beautiful Art

I don't know how you manage to work

Dattier

:p = lol ?

Danu

Speaking of work... Do you wanna help me understand something small, @Ted?

Ted Shifrin

I'm not necessarily good for much, @Danu. But you can try.

Astyx

yup @Dattier

Danu

18:40

It shouldn't be too crazy

Dattier

yup=yes ?

Astyx

yup

Danu

So consider an almost Hermitian manifold $(M,J,g)$. Usually one talks about the form $\omega(-,-)=g(J-,-)$ but the paper I'm reading (this one) instead uses the bivector $F$ defined by $g(F,\alpha\wedge\beta)=g(J\alpha,\beta)$ (so the dual of $\omega$).

Astyx

Gotta go now, bye all

Ted Shifrin

Bye, Astyx.

@Danu: Why do they mess with inducing $\tilde g$ on $\Lambda^2$? Seems unnecessarily abstruse.

Dattier

18:43

La réponse yup=non marche aussi (avec tes seuls réponses pour l'instant)

Danu

Now consider the Levi-Civita connection $\nabla$. It's a short calculation that for any $\alpha,\beta\in T^{1,0}$ (the $+i$ eigenspace of the complexified tangent bundle $TM_{\Bbb C}$) and $X\in TM_{\Bbb{C}}$ $g(\nabla_X F,\alpha\wedge\beta)=2ig(\nabla_X \alpha,\beta)$ and $g(\nabla_X F,\alpha\wedge\bar\beta)=0$.

Dattier

@Danu Fluo ?

Danu

@TedShifrin ...I guess we'll get to that soon enough.

Ted Shifrin

I'm not going to think about this, @Danu. I just don't have the desire.

Danu

I hadn't posed my question yet! :P But okay.

Ted Shifrin

18:45

I can't think through all this.

Danu

What's up?

Ted Shifrin

Nothing. Just no patience for thinking.

Probably time to take a break from here.

Danu

Hmkay..

Dattier

@Danu : c'est bien toi (Fluo) ?

Danu

@Dattier Non.

Dattier

18:46

Ah

Bonne soirée (même à Fluo... :p)

Robert Frost

@Wojowu you there?

Liad

hi, in ideals, is it correct that if $R $ is commutative then $(a,b) = (c) $ iff $c = r_1 a + r_2 b $ i need it for something, and i almost sure it is correct, could someone say if im right and explain?

DHMO

it turns out, trying to prove that $\forall n \in \Bbb N:\forall S: |S| = n \implies |\mathcal P(S)| = 2^n$ is a pain in the butt

Why does $1+z+z^2+\cdots$ converge whenever $|z|<1$?

Simply Beautiful Art

@DHMO Because the partial sums converge

DHMO

@SimplyBeautifulArt how do you know?

oh, limit!

Simply Beautiful Art

18:59

$$1+z+z^2+\dots+z^n=\frac{1-z^{n+1}}{1-z}$$

O_O lol

DHMO

thanks

Tobias Kildetoft

@Liad No, that only implies that $c$ is in that ideal

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$Mathematics$ Mathematics