Mathematics - 2015-10-15 (page 1 of 3)

theDoctor

12:56 AM

So, it seems a lot of you have been geting lost in the honeycombs and byways of mathematics.

Ted Shifrin

@theDoctor: You forgot cobwebs.

anon

@TedShifrin I have a sense that the "accidental isomorphisms" Sp(1)=Spin(3) and SU(2)=Spin(3) are fundamentally different, and should be considered two distinct accidents.

Ted Shifrin

@StanShunpike Mr. Stan, I can't for the life of me figure out what you're talking about here. Oh, you're trying to solve $an^2+bn+c = x_n$ ...

@anon: Well, it's because accidentally $SU(2)=Sp(1)=S^3$. :)

But, actually, I guess it's clear that $SU(2)=Sp(1)$, so maybe no accident.

I guess one is the adjoint representation giving the double cover of $SO(3)$. What's the other?

anon

1:11 AM

To do the Sp(1) one, we let it act on H (quaternions) by conjugation, then restrict it to pure imaginary quaternions. To do the SU(2) one, we act on C^2 in order to act on P^1(C) (= Riemann sphere) which we stereographically project into R^3. While Sp(1)=SU(2), I think the representations H and C^2 are different.

Ted Shifrin

Well, the adjoint representation I wanted to use must be the same as H.

Yeah, I'm pretty sure it is.

Your $SU(2)$ way is the way I've presented it in differential topology courses.

anon

I want to prove SU(2) acting on P^1(C)=S^2 extends to a linear action on R^3 (with minimal effort), and my idea was to see if I could transport the situation over to quaternions where I already know the situation, but that doesn't seem possible

Ted Shifrin

Well, one ought to understand these representations well enough to answer this. One.

anon

now I'm thinking every elt of SO(3) is a rotation around some axis, so upon stereographic projection that axis should go to two points in C and the great circle should go to some generalized circle and we want to find the form of the linear fractional transformation it acts as

Ted Shifrin

Hmm ...

Start with quaternionic way of doing rotation about an axis, instead.

I'm not sure what that great circle has to do with anything.

anon

1:20 AM

well, when you're climbing a wall, you look for things you can grab onto

Ted Shifrin

Wait, what great circle? Where is the angle encoded? Nah, I don't like this.

I think you should go the other way.

I fall off of the things when I'm climbing, so I'd rather not plummet to my death ... yet.

anon

@TedShifrin pick an axis in R^3 and consider the unit sphere. upon stereographic projection (with the plane cutting the sphere in half), that axis corresponds to two points on the sphere corresponds to two points in the plane, and the plane of rotation in R^3 corresponds to a great circle on the sphere corresponds to a generalized circle in the plane

@TedShifrin what's "the other way" refer to?

Ted Shifrin

Trying to go from quaternionic to rotation, which is pretty standard.

anon

I already know Sp(1)->SO(3), it's pretty simple.

Ted Shifrin

I still don't see where the angle of rotation is encoded.

anon

1:23 AM

@TedShifrin if you want to see that angle, then put a tangent vector emanating from one of the fixed points in the plane, and the (mobius map) will rotate it by the angle (I think)

but I need to understand the SU(2)->SO(3) one because I want to generalize it to Sp(2)->SO(5)

Ted Shifrin

I can give you a formula for that map $SU(2)\to SO(3)$ (not off the top of my head, but it's in my diff top notes).

So what do you mean when you say you're trying to decide if $H$ and $C^2$ agree? We have to choose identifications of $Sp(1)$ and $SU(2)$ with the unit sphere and see ...

theDoctor

right...

anon

@TedShifrin I mean if Sp(1) acts on H the same way SU(2) acts on C^2

theDoctor

does anyone have a quick explainations of these new domains su and so, that I hear of?

Ted Shifrin

So you have to pick a point in $S^3$, take the corresponding points in $Sp(1)$ and $SU(2)$ and compute the action. That shouldn't be hard.

anon

1:27 AM

I know SU(2) acts on P^1(C), what I want to see is that if we make P^1(C) the unit sphere in R^3 via stereo projection then that can be extended to a linear action on R^3 (hopefully that can be done with as little formula-manipulation as possible)

Ted Shifrin

I applaud all the geometry you're trying to interject, but wouldn't it be worthwhile just to grunge a particular example or two and see what happens?

anon

@theDoctor they're matrix lie groups. (perhaps don't overuse the english word "domain")

theDoctor

hmm, lie groups, matrix...

Ted Shifrin

I knew it: theDoctor is mired in a cobweb.

anon

What would the point of computing the action of specific elements of Sp(1) and SU(2) be? I am not sure what you what me to do a particular example of.

theDoctor

1:28 AM

no, I think much of these directions are dead-ends.

anon

you're very trolly/cranky.

theDoctor

calm down anon

Ted Shifrin

You're trying to understand whether the actions are "the same." So see what the action is for a few specific (corresponding) elements.

anon

well, telling me to calm down just made my hormones flare, you see

theDoctor

;^)

Ted Shifrin

1:29 AM

@anon: Don't feed the troll.

theDoctor

they are related to physics though, are they not?

Ted Shifrin

Most of mathematics is related to physics; yes.

Krijn

Most of physics is related to mathematics, I'd say

Ted Shifrin

Both are false, but ... anyhow :P

anon

@Ted is C^2 an irreducible representation of SU(2)?

theDoctor

1:31 AM

most of math is not related to physics actually.

Ted Shifrin

Yes, @anon, I believe so.

Otherwise everybody would have invariant subspaces.

anon

I believe so too. Whereas H is not an irreducible (real) representation of Sp(1) [acting by conjugation]. So, fundamentally different.

Ted Shifrin

But we're talking about two different groups, after all.

theDoctor

btw, I'm not actually a troll -- i really do think math has gone off the deep-end in some places.

(of course, those attached to those areas, think me as such, but oh well)

Krijn

It has, but that's a positive thing

Ted Shifrin

1:32 AM

Well, I have friends come over for drinks and appetizers, so I have to get into the kitchen. Talk to you soon, @anon.

theDoctor

Krijn, I wouldn't disagree, but it's like a old man who's life was built on a particular premise.

and now continues fighting, because his life was dependent on it.

Krijn

Do you mean the mathematicians themselves, or the field on its own?

theDoctor

well, in this area, there seem little difference.

the arena is defended by the old guard.

Krijn

You really speak very vague, which is often not apreciated in mathematics

anon

Nobody knows what it means, but its provocative

►

Krijn

1:38 AM

On another note, is it easy/possible to compute the ideal class group of a number field without computing all of its prime ideals and such?

theDoctor

2:00 AM

you speak very obscure, specialized, language that seems without utility--not often appreciated in the real world.

Krijn

I do.

Its a mathematics chat room

theDoctor

i'm not trying to insult you here.

are you saying mathematics is without utility?

Krijn

I guess some people find applications for it, but that's not why I personally do mathematics.

theDoctor

a bad thing: obscurantism.

what? fart around?

Krijn

What is obscure about mathematical language?

theDoctor

2:02 AM

i mean seriously, if there isn't some hope of utility, what is the possible use for spending the time and making the noise?

Krijn

I like to do it.

theDoctor

it's like trolling within intellectualism.

Krijn

So if I can live my life doing what I like to do, while not caring about the utility of it, I would

theDoctor

that's what you're saying.

but THAT"s what trolling is....boom

Krijn

Trolling is to make a deliberately offensive or provocative online posting with the aim of upsetting someone or eliciting an angry response from them.

theDoctor

2:03 AM

no, that's vandalism.

Krijn

(I did not type that, it was a dictionairy)

theDoctor

trolling is people getting off because they like it, without caring about the consequences.

(YOU)

Hi guys

Hi!

sup

2:04 AM

@anon I bought a black board its on its way to my home :D

anon

cool

Karim Mansour

it will be nice to setup it will be so cool

theDoctor

I suppose you're right about the trolling def.

anon

there is a spectrum of pure to applied, and everyone on the spectrum relies on their neighbors on either side for ideas, inspiration, facts, etc. etc. the pure mathematicians get to play, and everyone else exploits their play for profit, and so the world goes round.

theDoctor

I understand perfectly that distinction.

I'm just not sure that some of this math is really for the ideal of purity, but actually expoits purity.

Antonio Vargas

2:09 AM

That reminds me of something Paul Halmos said: "Applied mathematics will always need pure mathematics just as anteaters will always need ants."

Karim Mansour

haha

their is alot of pure math used in theoretical physics for example

I guess even more applied areas like computer science use pure math

as matter of fact I just attended a talk that uses differential geometry for computer vision

anon

indeed, they do use diff geo

Krijn

Oh wauw, how do they do that?

anon

also fourier analysis

Antonio Vargas

Everybody uses fourier analysis.

Krijn

2:11 AM

Fourier Analysis seems obvious

There's a nice comic (I think from AbstruseGoose) about how pure mathematics is just applied mathematics of the future.

But I can't find it now

Forever Mozart

I work with very large nonmetric spaces. Almost nothing I do will be useful to anyone other than people interested in very abstract properties. If I can make a career out of being practically useless, I win.

Krijn

Ah found it: abstrusegoose.com/504

There is no branch of mathematics, however abstract, which may not some day be applied to phenomena of the real world. (Nikolai Ivanovich Lobachevsky)

theDoctor

i suppose that's inarguable.

Forever Mozart

right

theDoctor

has anyone really thought about the metaphysical or phenomenal aspects of $i$, speaking of physics.

btw, what is a "nonmetric" space?

does that mean spaces which do not have unit and consistent measurement properties?

Forever Mozart

2:19 AM

at the very least I'd say nothing I do will be used in applied math or physics in my lifetime

Krijn

What do you do, then?

theDoctor

computes the omega number...

Forever Mozart

I work with big spaces

cardinalities much greater than $\omega$

or the cardinality of $\mathbb R$

Krijn

What do you do with the space?

What's your field, your interest?

Forever Mozart

and there is no metric topology which coincides with the topology of these spaces. So there is no way to assign distances between pairs of points in a meaningful way

Krijn

2:27 AM

Are they Hausdorff?

Forever Mozart

yes

theDoctor

what's bigger than the cardinality of R?

Krijn

$2^\mathbb{R}$

Forever Mozart

$2^\omega$

oh bigger

theDoctor

that doesn't make sense.

Forever Mozart

2:28 AM

yeah $2^{2^\omega}$ would be bigger

theDoctor

2^R is nonsense, it's still R

Forever Mozart

power set of a set has larger cardinality

Krijn

Give us an argument why it would still be $\mathbb{R}$?

theDoctor

you made the claim, you give it. otherwise I'm thinking of the power set q.

Krijn

Cantor's theorem

In elementary set theory, Cantor's theorem states that, for any set A, the set of all subsets of A (the power set of A) has a strictly greater cardinality than A itself. For finite sets, Cantor's theorem can be seen to be true by a much simpler proof than that given below. Counting the empty subset, subsets of A with just one member, etc. for a set with n members there are 2n subsets and the cardinality of the set of subsets n < 2n is clearly larger. But the theorem is true of infinite sets as well. In particular, the power set of a countably infinite set is uncountably infinite. The theorem is...

Does that suffice?

Forever Mozart

2:30 AM

Suppose there are as many subsets of $X$ as there are elements of $X$

theDoctor

but there's no intrinsic notion of sets in R, it's uncountable.

being innumerable, you can't make a set out of it.

Krijn

Why not?

theDoctor

it's uncountalbe.

uncountable.

a set relies on discrete, discontinuous members.

Krijn

So why can we not take a subset of an uncountable set?

theDoctor

dont' you know these things?

Forever Mozart

2:32 AM

what we mean is that there is no one-to-one correspondence between real numbers and the set of subsets of real numbers

there are different uncountable sizes

theDoctor

you can take a subset of an uncountable set, but you can't make a powerset out of it.

Forever Mozart

uncountable just means infinite but no one-to-one correspondence with $\omega$

theDoctor

no, it means no one-to-one correspondence with domain of natural numbers.

why bring omega into it?

Forever Mozart

yes that is an axiom: the power set axiom. If $X$ is a set then the collection of subsets of $X$ is a set.

theDoctor

for example {sqrt(2), pi} is a subset of the reals.

Forever Mozart

2:34 AM

$\omega$ is the set of natural numbers

theDoctor

but you can't specify R as a set.

Forever Mozart

0,1,2,...

theDoctor

not the whole domain R.

so you're not talking about Chaitin's omega as I was a few lines ago?

Krijn

Okay, then I take $X$, the powerset of $\mathbb{N}$ as my set

And take the powerset of $X$

theDoctor

hm

Forever Mozart

2:36 AM

$\mathbb R$ is definable using set theory axioms

you define natural numbers, then integers, then rationals, then reals

theDoctor

no, how do you get from the rationals to the reals?

Krijn

You use the notion of Caucy-sequences

theDoctor

i means it's a nice convenience, but it's not pure.

Forever Mozart

yes, or Dedekind left sets

suppose the rationals have been defined

let $\mathbb R$ be the set of all subsets which are closed downward

theDoctor

chaitin aruges that almost all reals are unnamable and unfindable.

all subsets of Q?

Forever Mozart

2:39 AM

yes

Mike Miller

@anon: I'm not sure if you should consider either of those accidental isomorphiisms the same but you should consider SU(2) = S^3 to be both true and obviously true, in the same way that SO(2) = S^1 is obviously true

theDoctor

and what is "closed dowwards? if you can indulge me for a bit.?

Forever Mozart

$A$ is closed downward if $x\in A$ whenever there exists $a\in A$ with $x<a$.

theDoctor

isomorphism? there is none (supposedly) in R to Q

Krijn

He's on another topic

anon

2:41 AM

@MikeMiller well, first one writes down what form elts of SU(2) take (depending on two complex numbers with |a|^2+|b|^2=1), and then it's obviously S^3.

Forever Mozart

then you order the set of these "left" sets using inclusion

what you have is a complete linear order extending the rationals

and this is called the set of reals

Mike Miller

@anon: sure, the same up to conjugation if you'd like I guess

there is an unfortunate fact that we don't have SSp(2) = S^7 because SSp does not exist :(

though I guess maybe one should say that's why S^7 is not a lie group

theDoctor

ah @Krijn:

sounds specious.

anon

@MikeMiller if one thinks of the "special" prefix as forcing preservation of (hyper)volume and orientation (in euclidean space), I think Sp(n)s are already SSp(n)s.

user143442

What does $\subset \subset$ mean?

anon

2:46 AM

For (certain?) topological rings R, one can take the maximal ideal of cauchy sequences within the ring of sequences of elts of R, and then quotient. Given a partially ordered set X, one can form Dedekind cuts as partions (L,U) of X s.t. a<b for all a in L and b in U, order under inclusion and thereby form a complete lattice (this is also called a completion).

@user relatively compact in

(i.e. closure of subset is a compact subspace)

theDoctor

here's a question: if you order Q, is it continuous?

anon

define continuous

theDoctor

R is "obviously" continuous, but Q does not seem so.

Forever Mozart

$Q$ is a totally disconnected space

$R$ is connected

(in the usual topologies)

theDoctor

yeah @ForeverMozart that is my sense of it.

anon

2:50 AM

two notions: linear continuum and continuum (topology). neither has Q being a continuum.

theDoctor

continuous = connected.

right, but R is continuous.

we enumerate an "element" from R for our utility or convenience, yet it is never exact.

Krijn

We do not enumerate elements of $\Bbb R$

theDoctor

we use Q, however, from a point of contact where it is exact, methinks

anon

he means enumerate digits of a given element in practical applications

theDoctor

we approximate on R, but get exactitude on Q.

right

like physics, fro example

which is hardly (never) in Q.

Krijn

2:53 AM

What's exactitude? (I'm not native)

theDoctor

good question. exactitude:

exactitude is something we define, by stating it from first principles. Like stating the first unit, 1.

the distance from 0, from there we can get the whole numbers, fracitons, and Q.

Krijn

No really, I mean like a dictionairy definition is fine.

theDoctor

we use the symbol 1 to mean exactitude, for example, but we use the term 1.0, to say approximation in R.

user143442

@anon but they're not topological spaces in this case

user143442

they're tangent spaces

theDoctor

2:55 AM

it means, complete precision, purity.

user143442

vector spaces

anon

@user I have no idea what "this case" is

Krijn

Ah okay @theDoctor

theDoctor

something you don't see from the real world, but get in mathematics (often).

user143442

im telling you

user143442

2:56 AM

this case is tangent spaces

anon

vector spaces over R or C are automatically topological spaces.

user143442

hmm i didnt know that

user143442

how

anon

picking a basis and transporting the topology over from R^n or C^n

theDoctor

strange you say vector spaces over R...

anon

2:56 AM

the resulting topology is independent of choice of basis

theDoctor

in almost all of physics, I think you use vector spaces IN R.

anon

@theDoctor R^n is not in R

theDoctor

ah, sorry, i meant R^n

unless we have a 1d world :)

user143442

ok than you @anon

anon

yes, R^n is over R, C^n is over C, Q^n is over Q, etc.

bad timing Krijn

theDoctor

2:58 AM

i think it's misleading to think of topology in R or C.

anon

I doubt you know what topology means

theDoctor

i don't think you should think of R^n over R at all.

Krijn

?!

anon

nor know what "over R" means

theDoctor

but a 3-d space composed of three orthogonal dimensions with R.

Krijn

2:59 AM

3D?!

Forever Mozart

it is a vector space with scalars in R... "over R"

theDoctor

R^3

sorry

anon

vector spaces do not come equipped with any notion of orthogonality, that's part of being an inner product space. the two concepts are often conflated.

theDoctor

R^n was automatically transcribed into R^3 for me.

user143442

@anon why are you so upset?

anon

2:59 AM

@user ?

Karim Mansour

Hilbert spaces though are example of inner product space that is complete

its nice

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$Mathematics$ Mathematics