Mathematics - 2017-09-01 (page 3 of 5)

Secret

09:00

possibly more general than that. In phase space, each point is a pair of position and momentum, but time evolution is not explicit

what I want is perhaps something akin to path integration of sorts, except that all possible trajectories are contained in such space for all starting conditions

Or put it in another way, parametrise all possible (classical) trajectories and starting conditions of the double pendulum with the lyapunov exponent

Salt

i was gonna do the same thing, but with my cat

bet your pendulum's exponent can't beat my cat's

Secret

en.wikipedia.org/wiki/Path_space hmm, its literally called path space

@Salt you are going to do some path integration stuff?

(I will assume cat is related to quantum)

Salt

not really, i'm generally stuck on the discrete side of math

however, my cat's attractors are easy to describe

Secret

I see

One reason I choose the double pendulum is because it is one of the uncommon examples (modulo the weather, which is too complex to even start) where the chaotic phenomenon does not involve a strange attractor, hence some notion of fractal like behaviour

I have previously played with the $\lambda$ parameter in the logistic map in mathematica. However the chaotic regime is space filling thus it is not very informative on what really happened

Leaky Nun

$SO(3)/SO(2) \cong S^2$ :O

I do not see how this is true at all

Secret

09:11

1

Q: Logistic map chaos theory experiment, need advice on interpretation of results

Recently while studying this lecture notes to learn about numerical techniques, I performed the following experiment in Mathematica as the lecture note encourage the reader to explore the logistic map (which caused me to first learn and stumble upon chaos theory in the process, hence I am still r...

That's one of my past investigation purely out of interest

But again, I will need to do more chaos theory before I am ready to check these and the double pendulum again

Tobias Kildetoft

@LeakyNun So you want to identify that quotient with rotations in two dimensions

Leaky Nun

@TobiasKildetoft how should I begin?

Tobias Kildetoft

@LeakyNun No idea, that was just my initial idea when I saw it

Leaky Nun

right

SteamyRoot

Prove that $SO(n)/SO(n-1) \cong S^{n-1}$ and then let $n = 3$ :^)

Leaky Nun

09:15

@SteamyRoot nice

SteamyRoot

The standard argument is pretty much that $SO(n)$ acts transitively on $S^{n-1}$ with stabiliser $SO(n-1)$, if I remember correctly

Leaky Nun

@SteamyRoot in English?

Secret

Previous joke sounding question:

4 hours ago, by Secret

[Random]
Demonstration of a group using a group of people in a lecture. But the first question is, what makes a good inverse element in a group of people?

Alessandro Codenotti

If I want to think about this as a quotient of $\Bbb P^3(\Bbb R)$ what should I be quotienting by? An $S^1$ I'd guess, but embedded how?

Secret

(though rubik's cube seemed a better demonstration)

SteamyRoot

09:19

@LeakyNun $SO(n)$ acts naturally on $\mathbb{R}^n$, and the action preserves the length of vectors.

So you can restrict the action to $S^{n-1}$

Secret

~~:39760204 uh, wrong ping?~~

SteamyRoot

Woops, yeah

Leaky Nun

@SteamyRoot and then?

Alessandro Codenotti

As a great circle on the surface of the ball before quotienting, if I'm not mistaken, to answer my own question

SteamyRoot

Then you show that the action is transitive. Meaning for any $x,y \in S^{n-1}$, there exists some $g \in SO(n)$ such that $gx = y$

Which pretty much means the group only has a single orbit.

Leaky Nun

09:22

oooh, group action!

SteamyRoot

Do any other actions exist? :P

Secret

there exists semigroup and ring actions

Leaky Nun

@SteamyRoot nah, I just didn't recognize it as a group action just now

SteamyRoot

And then you just have to calculate the stabiliser of some point and apply the orbit-stabiliser theorem

Tobias Kildetoft

@SteamyRoot There are actions of so many things. It just happens that the term "transitive" may need to have its meaning changed

SteamyRoot

09:28

Well, my view may be heavily skewed, but I've almost never seen anyone use the term "action" for anything other than a group action. Or, at the very least, when I hear someone say "action" without any further information, I assume they implicitly mean a group action.

Why should the term "transitive" have its meaning changed? o.O

Secret

Semigroup action

In algebra and theoretical computer science, an action or act of a semigroup on a set is a rule which associates to each element of the semigroup a transformation of the set in such a way that the product of two elements of the semigroup (using the semigroup operation) is associated with the composite of the two corresponding transformations. The terminology conveys the idea that the elements of the semigroup are acting as transformations of the set. From an algebraic perspective, a semigroup action is a generalization of the notion of a group action in group theory. From the computer science point...

Tobias Kildetoft

@SteamyRoot Because it turns out to be necessary in some contexts to get the "right" behavior. Basically, orbits may need to be "completed" in suitable ways to be well-behaved and this changes the definition of being transitive

This is especially the case in $2$-representation theory, and it then gets a shadow in the study of the actions of certain types of algebras

Secret

For me, I tend to treat action as a map which take an element from an algebraic structure $S$ and some set $X$ and send the result to the set $X$

Leaky Nun

@SteamyRoot no, orbit.

I'm not quite familiar with group action in general

so I only recognized it when you said "orbit"

although I understood perfectly when you said SO acts on R^n

Leaky Nun

09:44

@Mr.Xcoder hi

user84215

Is the projective plane a surface? If yes, why?

Leaky Nun

@MathematicsAminPhysics why do you think it is?

user84215

You think it is not?

Leaky Nun

No, I'm asking you

user84215

I can not show that it is a surface.

user84215

09:47

also can not show that it is not

Leaky Nun

Then did anyone tell you that it is a surface?

> In mathematics, the real projective plane is an example of a compact non-orientable two-dimensional manifold; in other words, a one-sided surface.

Real projective plane

In mathematics, the real projective plane is an example of a compact non-orientable two-dimensional manifold; in other words, a one-sided surface. It cannot be embedded in standard three-dimensional space without intersecting itself. It has basic applications to geometry, since the common construction of the real projective plane is as the space of lines in R3 passing through the origin. The plane is also often described topologically, in terms of a construction based on the Möbius strip: if one could glue the (single) edge of the Möbius strip to itself in the correct direction, one would obtain...

> The projective plane can be immersed (local neighbourhoods of the source space do not have self-intersections) in 3-space. Boy's surface is an example of an immersion.

user84215

Why?

SteamyRoot

You could just start with the definition of a surface and work from there. It's not too hard to find patches that cover the projective plane

Leaky Nun

@SteamyRoot I think he wants to show that it's differentiable

Martin Sleziak

That Boy's surface looks to be quite complicated. People studying differential geometry and related areas must have great visualization skills. (I admire people who are able to see something in pictures like that.)

Leaky Nun

09:52

@MartinSleziak there's a nice story behind it:

> In geometry, Boy's surface is an immersion of the real projective plane in 3-dimensional space found by Werner Boy in 1901 (he discovered it on assignment from David Hilbert to prove that the projective plane could not be immersed in 3-space).

so he proved Hilbert wrong :D

user84215

How can you do that?

Leaky Nun

@MathematicsAminPhysics do what?

user84215

I mean inserting vertical dots in your post.

Leaky Nun

> lorem ipsum dolor sit amet

> lorem ipsum dolor sit amet

Martin Sleziak

It must be quite embarrassing when a Boy proves you wrong :-) I read on Wikipedia that he died relatively young.

user84215

09:55

> jbkwjbsdkjwsdkw

user84215

Thanks.

Leaky Nun

@MartinSleziak can a manifold be not differentiable?

user84215

Why is the projective plane a surface?

Martin Sleziak

@MathematicsAminPhysics If you are curious how something was typed in chat, you can always look at the source of the message. Just go to the transcript, click on the arrow left to the message and click on history. You get something like this - displaying the source: chat.stackexchange.com/messages/39760565/history

user84215

Thanks.

Secret

09:58

what is a surface in the context of diff geom, surely not just anything that can be parametrised with two parameters...?

Leaky Nun

@Secret a differentiable 2-manifold

Secret

though to be honest, I know absolutely nothing in diff geom other than some examples

Leaky Nun

but apparently to me (based on my previous minutes of searching), it seems that every manifold is differentiable

and yes, I've only searched for a few minutes

Kasmir Khaan

Hello yall :D

Secret

Every n-manifold is locally $\Bbb{R}^n$, thus it has to be differentiable

Kasmir Khaan

09:59

I found the flaw in my proof from yesterday =P @LeakyNun

fixing it now :D

Leaky Nun

@KasmirKhaan nice

Kasmir Khaan

Bbtw how to type x'' on latex?

Martin Sleziak

@MathematicsAminPhysics And you can find some basics also in chat help. If I am not mistaken, the subset of MarkDown supported in chat is basically the same as mini-markdown for comments so you might check that, too.

Kasmir Khaan

(x^-1 )-1

Leaky Nun

@KasmirKhaan just type literally x''

or (x^{-1})^{-1}

Kasmir Khaan

10:00

does x'' work?

Leaky Nun

@MartinSleziak mini-markdown does not support quotes

Kasmir Khaan

second one is better :D

SteamyRoot

You may want to look into what "differentiable" means with regards to manifolds first...

Leaky Nun

@SteamyRoot I think it's something you impose on a manifold if I'm not getting it wrong

Martin Sleziak

@LeakyNun It depends on the definition of manifold or on the convention, doesn't it? Some people study something called topological manifolds.

SteamyRoot

10:01

^

user84215

Are you answering to my question?

Leaky Nun

so, if I'm correct, @MathematicsAminPhysics, RP2 is a surface because it is a 2-manifold and every manifold can be equipped with a differential structure.

> Differentiable manifolds, for example, are topological manifolds equipped with a differential structure.

Martin Sleziak

@LeakyNun Ok, so I was not exactly right. But still I think most of the things supported in mini-markdown can also be used in chat. (So they might be occasionally useful.)

SteamyRoot

There are manifolds that do not allow any differential structure, though.

Mr. Xcoder

@LeakyNun Hello!

user84215

10:03

2-manifolds are surfaces?

Leaky Nun

@SteamyRoot :o for example?

@MathematicsAminPhysics for an immersion of RP2 in R^3, see Boy's surface

I think this is enough proof that RP2 can be equipped with a differentiable structure, hence a surface

SteamyRoot

No idea. The examples aren't exactly "standard" objects

Mr. Xcoder

What are "manifolds"?

Leaky Nun

@Mr.Xcoder n-dimensional "surfaces"

for 1-dimension it's a curve or a line or a circle

for 2-dimensions it can be the surface of a sphere or the surface of the torus

we call a line a 1-manifold

Mr. Xcoder

Um, interesting. i didn't know of that denomination

Leaky Nun

10:05

@Mr.Xcoder it's studied in topology

it's where people get coffin mugs and doughnuts confused

because their surfaces are the same in a topological viewpoint

(both have a hole)

Mr. Xcoder

Brb solving something

user84215

My question is in the context of elementary differential geometry. It does not have an elementary proof?

SteamyRoot

Just work with the definition...

If you want me to do half the work for you: use the patches $(x,y) \to (1,x,y)$, $(x,y) \to (x,1,y)$ and $(x,y) \to (x,y,1)$.

user84215

I think the projective plane is the affine plane plus the points at infinity. Right?

SteamyRoot

Yes, and the points at infinity form a projective line $\mathbb{R}P^1$, actually.

Leaky Nun

10:13

@SteamyRoot the RP2 is R^3\{0} quotient what?

user84215

then what should I do?

SteamyRoot

@MathematicsAminPhysics For the third time: work with the definition of a differentiable surface.

Leaky Nun

@MathematicsAminPhysics do you know the definition of a surface?

user84215

I think I know.

SteamyRoot

@LeakyNun You quotient by the relation $x \sim y \iff \exists \lambda \in \mathbb{R}: x = \lambda y$

Leaky Nun

10:15

@SteamyRoot oh, just like RP1

SteamyRoot

Yup, it's the same for any dimension.

Leaky Nun

and I represent (x,y,z) by (x/z,y/z,1)?

SteamyRoot

What do you mean with "represent"?

Leaky Nun

I mean, the equivalence class

phi([(x,y,z)]) = (x/z,y/z) where z is non-zero, and (x/y,infty) where z is zero but y is non-zero and (infty,infty) where y and z are zero?

SteamyRoot

Well, if $z$ is nonzero, then those elements are in the same equivalence class, yes.

Not sure if it's a good idea to use $\infty$

Leaky Nun

10:17

I thought RP2 has infinity points

SteamyRoot

Yes, but that doesn't mean using the symbol for infinity is a good idea

It's not that it "naturally" has infinity points.

Leaky Nun

then how is it usually represented?

and what is the analogue of mobius transform?

SteamyRoot

For some coordinates $x,y,z$, all points are of the form $[(x,y,z)]$

You can make a choice for what the points at infinity are, e.g. $z = 0$

Leaky Nun

oh i found a good representation

SteamyRoot

Then you can identiy all points of the form $[(x,y,1)]$ with the affine plane

Leaky Nun

10:20

S^2/~ where x~y iff x=-y or x=y

user84215

I can not understand your proof.

SteamyRoot

and $[(x,y,0)]$ become the points at infinity.

Leaky Nun

@MathematicsAminPhysics did anyone give you any proof yet?

SteamyRoot

And, since you can mod out those points by multiples again, you'll notice that it consists of the points $[(x,1,0)]$ and $[(1,0,0)]$, i.e. the projective line.

user84215

As I said, the projective plane is the affine plane plus the points at infinity.

Leaky Nun

10:21

@MathematicsAminPhysics what were you referring to when you said "your proof"?

user84215

Then you use (x,y) -> (x,y,z) patches?

user84215

Then how can you cover the points at infinity?

Leaky Nun

Mr. Xcoder

@LeakyNun Too similar to adnan's photo lol

And I wouldn't have picked 9 anyways... looks around

user84215

You are disappointed in me?

Leaky Nun

10:33

2

user84215

That is, the image of patches can be any objects?

BAYMAX

those are some nice memes

is there some page where i can get them

?

especially the first one!

Leaky Nun

@BAYMAX I got both from 9gag

BAYMAX

nice

user84215

10:50

It seems that no one interested in my questions.

user84215

I think questions are not answered here; questioners are answered.

Secret

[Random]

Some random semigroup out of the blue:

$a^2=b^3$

cb=c

->

Leaky Nun

@Secret I'm interested

Secret

I have no idea. Right now the thought process is too random. sure it has one absorber, c, and we have squares and cubes like a^2 and b^3, but I don't have other ideas on what else to add to it yet

I think I should do some chemistry in the meantime...

Balarka Sen

@MathematicsAminPhysics No, it's just that you make absolute 0 effort to understand people's answers here.

I don't see any other person complaining.

user84215

11:03

As I said, I know little mathematics.

user84215

Please assume that you are answering to a high school student.

Balarka Sen

A high school student doesn't know what RP^2, or a surface, is. In any case, what is your definition of a surface?

I think we have asked you that for multiple times.

To answer the question "why is RP^2 a surface?" we need to know how you define a surface.

Leaky Nun

@MathematicsAminPhysics whining is never the way to get more answers

It just makes you look like an attention-seeker

user84215

You want to cover the projective plane with (x,y) -> (1,x,y) , (x,1,y) , (x,y,1) patches?

Balarka Sen

Correct.

user84215

11:14

And these patches cover all the projective plane because of the definition of the projective plane as the quotient of the M^3 space?

Balarka Sen

I don't know what M^3 is; it's a quotient of the (deleted) Euclidean space R^3 - 0, yes.

Every point on RP^2 is of the form [x : y : z]

Leaky Nun

well, to be pedantic, R^3 \ {0}.

Balarka Sen

Thanks.

Here [x : y : z] represents the homogeneous coordinates; it's the equivalence class of (x, y, z) in R^3 under the identification (x, y, z) ~ ($\lambda$ x, $\lambda$ y, $\lambda $ z)

So the open sets given by $x \neq 0$, $y \neq 0$ and $z \neq 0$ cover RP^2

user84215

So we should prove that the map between any two patches is differentiable?

Balarka Sen

If say $x \neq 0$, then such a point is just of the form [x : y : z] = [1 : y/x : z/x] in homogeneous coordinates

@MathematicsA Yes, you should.

user84215

11:18

We should

Leaky Nun

I wonder what would happen if we keep 0 as its own equivalence class @BalarkaSen

Balarka Sen

@MathematicsA There's no we. You should take it as an exercise to prove it. It's a one line proof.

@LeakyNun You don't really get a manifold

user84215

I am lazy.

2

Leaky Nun

@MathematicsAminPhysics then don't learn.

Balarka Sen

@MathematicsA So why are you complaining about people not answering your questions?

Leaky Nun

11:21

@BalarkaSen let's just use S^2/~ where x~y iff x=-y or x=y :P

Balarka Sen

We are not going to do your work.

4

If you're lazy, stop asking questions and posting those shitty workshop and MSE university ideas.

Don't pretend you are an expert academician if you don't want to work.

Leaky Nun

@MathematicsAminPhysics the first time we saw you here, you were challenging us

user84215

The projective plane is equal to the the S^2 sphere with being identified its antipodal points?

Leaky Nun

@MathematicsAminPhysics yes

user84215

Why?

Abdullah UYU

11:26

Hi, do you think this sentence makes sense? "Draw a line passing through $C$ such that $\measuredangle{BCD}=30^{\circ}$ where $D$ is the intersection point of $AB$ and that line."

Leaky Nun

@MathematicsAminPhysics you can easily view this from its fundamental polygon

user84215

fundamental polygon?

Leaky Nun

@MathematicsAminPhysics bfy.tw/Dgz6

user84215

That is very sophisticated for me.

Leaky Nun

@AbdullahUYU yes, but I'd say "Draw a line passing through $C$ which intersects $AB$ at $D$ such that $\measuredangle BCD = 30^\circ$"

@MathematicsAminPhysics what is?

this is the fundamental polygon of RP2

user84215

11:31

understanding fundamental polygon

Abdullah UYU

aha, that's more clear @LeakyNun

SteamyRoot

@MathematicsAminPhysics I literally mentioned the patches before.

1 hour ago, by SteamyRoot

If you want me to do half the work for you: use the patches $(x,y) \to (1,x,y)$, $(x,y) \to (x,1,y)$ and $(x,y) \to (x,y,1)$.

Leaky Nun

@SteamyRoot oh, you missed this part:

14 mins ago, by MathematicsAminPhysics

I am lazy.

Context:

17 mins ago, by MathematicsAminPhysics

So we should prove that the map between any two patches is differentiable?

17 mins ago, by Balarka Sen

@MathematicsA Yes, you should.

17 mins ago, by MathematicsAminPhysics

We should

SteamyRoot

Heh

user84215

What is your goal by posting my messages?

Leaky Nun

11:36

@MathematicsAminPhysics whatever you think the goal is

user84215

Maybe people do not think good things about it.

user84215

I still can not understand why the projective plane is equal to the the S^2 sphere with being identified its antipodal points.

Leaky Nun

you just need to normalize each equivalence class in RP2

SteamyRoot

The projective plane is $\mathbb{R}^3 \setminus \{0\}$ with all multiples identified.

You could also first identify vectors equal up to a positive constant, to get $S^2$.

Or, another (more graphical) explanation: $\mathbb{R}P^2$ is the set of all lines through the origin in $\mathbb{R}^3$.

And any line through the origin intersects $S^2$ in a pair of antipodal points

Leaky Nun

@SteamyRoot RP2

SteamyRoot

11:44

Woops. Fixed.

user84215

Thanks. I have gotten it.

user84215

Do you think I will become a great mathematician?

Leaky Nun

not if you're lazy

SteamyRoot

Mathematicians are insanely lazy, but in a different way. Not in the way that we let others do our work for us.

Leaky Nun

@MathematicsAminPhysics maybe you can be a physicist instead.. oh wait they're the same

user84215

11:48

By Amin's theorem

user84215

So I will never become a mathematician?

SteamyRoot

Who knows.

If you want, then work for it.

Also, it's an unwritten rule to not name something after yourself

user84215

What do you mean by saying "to not name something after yourself"?

user84215

I have done it?

Leaky Nun

Amin, do you seriously believe that Mathematics is Physics?

user84215

11:54

I have proved it.

SteamyRoot

You say "Amin's theorem"

Tobias Kildetoft

@MathematicsAminPhysics He was referring to you saying "Amin's theorem"

SteamyRoot

You named it after yourself.

... does this count as both of us being half-sniped ?

Tobias Kildetoft

You need to somehow make others name things after you

user84215

I have proved it in my profile.

Tobias Kildetoft

11:55

@SteamyRoot Not sure

Leaky Nun

(Context for other people: from Amin's profile:)

> Amin's Theorem: Mathematics = Physics .

Proof: Mathematics is more beautiful than or as beautiful as physics. Physics is more beautiful than or as beautiful as mathematics. As we know, the relation "more beautiful than or as beautiful as" is antisymmetric. Hence, Mathematics = Physics . Q.E.D

Kasmir Khaan

@LeakyNun Should I use that theorem on my abstract algebra homework ? :D:D:D

Leaky Nun

totally, if you want to flunk it.

Kasmir Khaan

@LeakyNun Almost done with the second exercice btw , shuld I send it via email?

Was just joking =p

Leaky Nun

have you done the first?

Kasmir Khaan

11:58

well I have fixed it

it is not 100% like you want it

let me just sedn it

Leaky Nun

alright send whatever whenever

Kasmir Khaan

so you see what I mean =p

Ill send you the first one now

not done yet with the second :D

one second and look ur email =p

Leaky Nun

ok

Tobias Kildetoft

@SteamyRoot Now, if only someone would choose to name one of my more interesting results after me, rather than one that is just a small improvement over something known.

SteamyRoot

:P

Transcript for

$Mathematics$ Mathematics