Mathematics - 2016-10-29 (page 3 of 3)

« first day (2278 days earlier) ← previous day next day → last day (2749 days later) »

7:03 PM

I'm trying to learn something about elliptical curves and have encountered $$\mathds{C}^\ast$$ but I'm not clear on what it means. Is it a ring under complex numbers?

Hello @TedShifrin
Let $u(x) \geq 0$ be harmonic in $\Omega$ and $u|_{\partial{\Omega'}}=0$ where the space $\Omega'$ is a proper subset of $\Omega$. I want to prove that $u(x) \equiv 0$ in $\Omega$.

If we suppose that $u \in C^0(\overline{\Omega'})$ we can deduce that $u(x)=0$ in $\Omega'$.

But how can we deduce that $u(x) \equiv 0$ in $\Omega$ ?

Evinda. You asked me that a week ago and I have no idea.

Hi @Ted

@Edward C* algebra?

Not with elliptic curves. Most likely the dual curve. I need to see context, @Edward.

7:04 PM

@0celo7 Maybe: there's also R*

If it's $\Bbb C^*$ and $\Bbb R^*$, then it's nonzero complex/real numbers.

Probably not then

Hi @Balarka

A ok. @TedShifrin

@TedShifrin Ah, so the $*$ in $\Bbb R^*$ just means essentially "except 0"?

7:06 PM

It could, viewed as a multiplicative group.

Tobias Kildetoft

@Edward in more generality, it means the group of units

That's usually $R^\times$, though, @Tobias.

@TedShifrin That's how I saw it too

Tobias Kildetoft

@TedShifrin I have seen either used about equally often I think

There are dual spaces, dual curves, dual varieties ...

dual elliptic curves ...

Tobias Kildetoft

7:07 PM

But I agree that using a \times is better for that reason

The context here is talking about the reversibility of the discrete log problem in some groups.

That isn't specific enough to help me.

I'm not an expert on cryptography ... at all.

Nor am I, so the question is probably too vague. In short, the DLP is "given an element $h$ in the subgroup generated by $g$, find an integer $m$ satisfying $h=g^m$"

Right, and where is the $C^*$ appearing?

The author notes that this is actually an easy problem to solve in some groups and mentions specifically $\Bbb C^*$ and $\Bbb R^*$.

7:12 PM

Ohhhhh ... Sure, then. Nonzero complex/reals.

Tobias Kildetoft

@Edward then those do refer to the groups of units

hi chat

If it's BlackboardBold C or R, that narrows it down anyhow.

OK, thanks. It was just a footnote, but I didn't understand the notation.

@Balarka: You'll be amused to know I had an email this morning from a mathematician in Saudi Arabia, saying he wanted to translate my diff geo notes into Arabic and have it published at his university.

7:15 PM

Hello again.

hi Mahmoud

Hi TedShifrin :)

@TobiasKildetoft I installed python but it got me nowhere ...

Tobias Kildetoft

@Mahmoud Well, what did you do after installing it?

Tried to install the program but I ... Just don't know.

So my problem is too complex or what?

7:21 PM

@TobiasKildetoft What is ''pip'' ?

Tobias Kildetoft

@Mahmoud I have no idea. I am not very familiar with Python

@TobiasKildetoft Sorry just forget about it I found a video :)

morning sunshine

G'd evening, @MikeM.

the earth says helloooooo

7:33 PM

@TedShifrin Ohh, nice.

LOL, @Balarka, well, I'm not going to do it.

I do believe your multivariable book and diffgeom notes should get more publicity (though I am sure it already has quite some).

well, having a translation I can't vouch for (or update) and a publisher who charges isn't the direction I wanted to go, although probably it would be good for the Arabic-speaking community to have it.

Ah, I didn't notice the publishing part.

7:56 PM

Is $\forall a,b\in\mathbb{N}:b=\lfloor\sqrt{a+b^2}\rfloor\mbox{, if }a<b$ true?

8:29 PM

@NaCl

@Mahmoud

Can you please help me ?

I'm struggling with a python program.

Link?

manim

I'm very new to this (A few hours ago) I installed it with cmd ... but whenever I try to execute a command I get Syntax Error. $:($

What operating system do you use?

8:33 PM

Windows 10 Pro

Oh, so I guess Windows

ye.

Windows is the problem ?

Sorry, I'm not familiar with it

Do you know if you have pip installed or not?

I have it installed

I used it to install other modules

The problem is that all the .py files in the owner's folder gave me syntax errors, they're can't be mistakes, is it because of my Python version ? Namely 3.5.2

Could be, but I don't know. Did you see this? https://gist.github.com/Croxed/f7f451608143f253b50803785f7ce3f0

Might help ya

8:41 PM

I already installed it the problem is I don't know how to use it.

I don't know either

Meh. I just want to animate some videos, :/ Why all of this complexity ?

I'm reading a book that says that $f:\mathbb C^n\to\mathbb C$ defined by $(z_1,z_2,\dots,z_n)\mapsto z_1$ is holomorphic. I agree with this. Then it says that if $M$ is a compact submanifold of $\mathbb C^n$, then $f|_M$ is holomorphic. Why in the world is this true?

$M$ can be a singleton.

9:25 PM

True, @Josué, so the right way to say this is if $M$ is not a singleton you show that in fact it must be :)

$M$ could be closed, too.

Hi @Ted :)

@Josue A composition of holomorphic maps is holomorphic, and inclusion of a complex submanifold is holomorphic. But suppose your compact submanifold is closed. What holomorphic functions are there on a compact manifold?

@MikeMiller "closed" in the sense of has no boundary, not in the sense of point-set topology---perhaps worth saying

I know.

9:36 PM

I didn't say that for your information :P

So I practiced my lecture thingy today---I'll have to shorten it a bit.

I guess I'm going to cut the proof that $T\Bbb RP^n\cong\operatorname{Hom}(\gamma^1_n,\gamma^\perp_n)$ ($\gamma^1_n$ being the tautological line bundle on $\Bbb RP^n$) short.

I see. Then if $i:M\hookrightarrow\mathbb C^n$, $f\circ i=f|_M$ is holomorphic. But $i$ is constant, so $f|_M$ is constant.

If $i$ is constant then $M$ is already a point so it's not surprising that $f|_M$ is constant :P

Hi @Danu, @MikeM

@Danu That is false as stated. You want some tensor powers flying around on the right, and it's only true stably.

@MikeMiller Derp, I messed it up.

The second entry of hom is the orthogonal complement

9:41 PM

Are we getting tensor and tensor flying about?

Okay, corrected it now.

@TedShifrin Jesus.

In any case, the proof is not that interesting and I can just write it down very briefly without losing much insight, I guess.

This question is part of a proof showing that $\mathbb C^n$ contains no positive-dimensional, compact submanifolds. Using this, I can show that each connected component of such a submanifold is a singleton. But I do not know how this implies that the manifold is not positive dimensional.

@Josué Huh? If each connected component is a singleton how could it possibly be positive-dimensional?

9:43 PM

If so, it has a positive-dimensional connected component, @Josué.

@Josué Perhaps recall explicitly what your definition for dimension is?

Presumably something involving the tangent space...

I disagree, @Danu. One way or the other, the Euler sequence is both crucial and very geometric.

hi all

But you can't belabor details in everything ...

@TedShifrin Eh... Do you have a nice proof?

Right now, what I'm doing is just giving a bijection in each tangent space

9:45 PM

who can give us help in calculus?

I told you ages ago when we were discussing cones.

and then hurr durr continuous

I gave you a proof in the holomorphic category (or suggested you work out the details).

@TedShifrin You were talking about... moving frames?

The way I was taught, if the dimension of a complex manifold $M$ is $n$, then that means that $M$ is locally-homeomorphic to $\mathbb C^n$.

9:45 PM

Not for that.

@TedShifrin Also I'm working with $\Bbb RP^n$ of course

I did give an interpretation of the $\text{Hom}(\gamma,\gamma^\perp)$ in terms of moving frames, yes.

In any case, I need to stay more elementary than Euler sequence

Doesn't matter. The Euler sequence is field-independent :)

Didn't we discuss projecting tangent vectors at $x$ and $\lambda x$ down to the sphere?

Yes, that was you.

Yeah, sure we did

9:46 PM

Well, that was the map with the twist in it.

i have a problem in calculating the normalization cst. Anyone could give me help

The explanation you should give in seminar is this: If you want to see how you're moving in the space of lines, you choose a local section of $\gamma$ and differentiate it. The part that moves you radially along the line descends to the $0$ tangent vector on the space of lines. So you care only about the orthogonal part.

But I like my $\pi_*$ argument cuz it's totally geometric and general. :)

@Student404Mus Are we supposed to know what you're talking about? :)

@Josué And a point is locally homeomorphic to $\Bbb C^n$ for what $n$?

@TedShifrin So, my audience is not so strong in terms of background so they won't find something like that satisfying, but I can maybe say it anyways while I write out the proof I have now. Thanks :) It's definitely a nice way to think about it for me!

I think that it's good to get people thinking about choosing a function that lifts the curve in $\Bbb P^n$ and differentiating it.

That came up recently with @Balarka and he had not developed that intuition (until I poked him in the eye with it).

9:52 PM

It's interesting.

Then I can say something like: since every connected component of such a submanifold is a singleton, each chart of the manifold is homeomorphic to $\mathbb C^0$. Contradiction.

@Danu: If I may presume to pontificate one sentence more — In seminar talks, it helps to help people develop intuition rather than giving dreary formal arguments.

Some of the latter are, of course, necessary :)

Most things are pretty nice in what I'm talking about

the most technical thing is perhaps this thing.

Or maybe the application of SW classes to division algebras

it's just a problem that got me a lot of time

I've already said I do not think about characteristic classes the way Milnor develops them.

But, @Student404Mus, we don't know what you're talking about!

9:55 PM

But the application of SW classes to division algebras is awseom :)

So you are an algebraist after all, @Danu :D

sould i explain?

@TedShifrin Lol.

If you want help, you'll have to explain, yes, @Student404Mus.

If I get into Bonn, I'll spend a qualifying year taking mostly algebra :P

9:56 PM

Not mostly, but some, yes ...

It's really nice that they offer this PhD-after-qualifying-year type deal

chicago in the US does that, too

Neat

every student starts with first-year sequence, period

But Chicago notoriously demolishes people anyway, @MikeM :)

9:57 PM

It's good for me because I need some time to catch up

@TedShifrin Shh.

One of the best undergraduates I ever taught almost didn't survive ...

@ShayBenMoshe We can just chat here :)

I actually only came here to see if you had done any computation using KO

It's 1 am here so I should probably go to sleep though

What does "PhD after qualifying year" mean?

Is that like some sort of conditional acceptance?

10:01 PM

the problem consisting calculation of the normalization constant K from a function described as:
psi=K(x+y+2z)exp(-alpha*r)

K is the normalization cst alpha is real

i done the following transformation (catesian to spherical coords.)

psi=Krexp(-2alpha*r)(sin(theta)*cos(fi)+sin(theta)*sin(fi)+2cos(theta))

then to find K

integral of |psi|^2=1 the last expression gives

|K|^2* integral r^3*exp(-alpha*r)dr * integral d(fi) * integral (sin^2(theta)*cos(fi)+sin^2(theta)*sin(fi)+2cos(theta))

@ShayBenMoshe I don't really have time to do any computations, sorry. I'm also not that great at that sort of thing.

@Student404Mus: If you're going to participate much on MathStackExchange, you need to learn some basic MathJax ... This kind of stuff is almost unreadable.

@PVAL-inactive Kinda. You just do courses for one year, but already get some money (800 euros per month, IIRC). Then you're expected to start the PhD, but if you don't do well I guess they won't take you after all.

ok

So this is a probability distribution on all of $\Bbb R^3$ and you're trying to choose $K$ to make the integral over all of $\Bbb R^3$ equal to 1?

10:02 PM

When I was in SK's alg. geom. class, he would tell non-flattering stories about JPM

yes

lol

exactly @Ted Shifrin

Is there really $|\psi|^2$ when $\psi$ has no complex numbers in it?

it's the probability that i spent a lot of time to ket k

10:03 PM

50 bucks says this is a QM problem

Wouldn't I expect an $e^{i???}$ for QM, @Danu?

your question is my question

@TedShifrin No, because you usually just do separation of variables.

the problem is i found it in theprob. set

The time part is not interesting in static problems, so you just study the time-independent equation.

10:05 PM

I see.

This is typically what students are doing, of course.

So I'm looking at $\iiint_{\Bbb R^3} (x+y+2z)^2 e^{-2a\sqrt{x^2+y^2+z^2}}\,dV$?

Reminiscent of the hydrogen atom, but not quite...

of course it's time independent

actually we are learning angular momentum

@Student404Mus: By symmetry, the $xy$, $xz$, and $yz$ terms will all integrate to $0$. So I look only at $x^2$, $y^2$, and $z^2$ with the exponential, agreed?

10:07 PM

And then you separate da integral into factors

or not

jk

Actually, we'll have $(x^2+y^2+z^2)e^{-2ar}$ and then an extra bunch of $z^2e^{-2ar}$. So I have to do $\int r^4 e^{-2ar}\,dr$ ... Did you get to that, @Student404Mus?

just a moment

ok @Ted Shifrin

then ?

@TedShifrin You're being used for a calculator :P

Then you're almost done. So you have to do several integrations by parts for that.

Nah, @Danu, I'm just directing traffic, not micromanaging.

ok @Ted Shifrin

then ?

sorry my connection has problems

yes but the problem is not in the radial part

10:14 PM

Then you're almost done. Except there's $\int_0^\pi\sin\phi\,d\phi$ (or $\theta$, most likely, if you're a physicist) and for the $z^2$ integral, there's $\int_0^\pi \sin^3\phi\,d\phi$ ... and then there's a $2\pi$ in front of everything.

the problem of 0 i got is in angular part

What do you mean by problem ?

right?

It seems like you're not remembering $r^2\sin\phi \,dr\,d\phi\,d\theta$ ...

But I have no idea what 0 you're talking about.

we i integrate with respect to (sin^2(theta)*cos(fi)+sin^2(theta)*sin(fi)+2cos(theta))

=0

10:16 PM

Note that I simplified the analysis above before switching to spherical coordinates. You should go back and understand that.

knowing that: theta: 0<pi and fi: 0<2pi

OK, so my $\phi$ and $\theta$ need to be switched, which is what I figured, but, even so ...

ok

Hmm, @Balarka is awake past his bedtime again.

And I'm doing algebra. What has the world come to?

10:19 PM

Happiness.

I had to spend hours writing garbage for my school project. Ugh.

I hope it wasn't garbage.

And now I have to head to bed, so not much done today.

Your high school sounds intense :P I never worked hard in high school...

Wish I'd discovered my academic interests earlier though

Nah, @Danu, I just complain.

10:21 PM

and boy! does he ever ... complain and procrastinate.

you spent high school and college being a party boy, @Danu?

Oh, wait, you're still one of those :D

@TedShifrin Forreal? :P

@TedShifrin I didn't party much. I did a lot of sports, though. In high school I was just generally bored out of my mind, so I spent a lot of time playing video games.

Right now, I don't think I do much besides working & spending time with my lady friend---which does admittedly take up a lot of time.

I was kiddling.

Anyways, I try to keep in mind that relationships are more important than math.

meh

@Danu, well that's not true in general

10:24 PM

if you really want to do a Ph.D., that balance may have to change a bit.

If I get a PhD spot I'll have to leave her (unless it happens to be Munich).

well ... not worth worrying about it yet ...

I have had a happy balance.

you mean you support yourself well enough with tutoring and poker, @MikeM?

That's more like it

10:27 PM

By relationships, I mean romantic ones by the way. I'm not sure if you guys misunderstood that part.

I understand what you meant.

My net gain from poker has been minimal. And that's not what i meant.

@Danu: I knew, but your statement rings true either way.

I have maintained a number of close friendships for 30-40-50+ years, which often means a lot more than romantic relationships that fizzle after a few months or years. But I do not disparage trying to find happiness, not in the least.

@TedShifrin What about romantic relationships that last that time?

That's fantastic if they do.

10:30 PM

My grandparents were married 64 years and dating 67. That's not so common.

I know some folks, both str8 and gay, who have very long-lasting relationships. Very touching and wonderful. Just isn't super common. (For example, my dad died when he was 53, much to my mom's sorrow.)

Anyhow, we don't need to make this Dear Abby.

Why do you always add the 8?

^

Huh? You mean rather than typing out 5 extra letters?

But you don't consistently stick to that principle

10:32 PM

It's quite common in today's world. I don't get what your problem is.

Just curious :)

"always add the 8"? rather than subtracting something?

I'm not offended. Just curious.

I don't understand the issue.

u certainly dont skip most ltrs

10:34 PM

Ah.

h8 it

It's a very common abbreviation in today's world.

It must be because I secretly hate all straight people.

^

secedes

Or because you love Avril Lavigne.

10:35 PM

I have no clue who that is.

she dated a sk8r boi

she sed cya l8er boi

shrug

"these kids" :D

« first day (2278 days earlier) ← previous day next day → last day (2749 days later) »

Transcript for

Oct '1629

$Mathematics$ Mathematics

Associated with Math.SE; for both general discussion & math qu...

join this room
about this room

00:00

06:00

12:00

18:00

all times are UTC

$Mathematics$

site design / logo © 2024 Stack Exchange Inc; legal

mobile