Mathematics - 2016-02-29 (page 2 of 4)

Axoren

16:00

Thank you, @BalarkaSen

Balarka Sen

This thread has a few links.

Mike Miller

@Balarka: Yes, there's a pretty nice book on space filling curves that I haven't read.

Has a nice theorem in there (something like "conpact locally path connected metric spaces admit space-filling curves"

Balarka Sen

@MikeMiller Nice. Perhaps you could refer that to @Axoren?

I haven't really read or understood any of this Cannon-Thurston stuff but I try to trick myself into believing that the whole deal comes from limit sets of certain group of isometries of $\Bbb H^3$ being space filling curves on the sphere at infinity.

Mike Miller

Seems you have. Also seems he's aware.

Balarka Sen

I have no idea if that's true.

Mike Miller

16:06

"There's one on springer link, but..."

Axoren

@MikeMiller Lol, so it's that one? At least I have support for it.

Balarka Sen

Right (my no idea comment was a continuation of my previous message about my religious belief)

Axoren

@BalarkaSen I care about the simpler notion. A space filling curve filling $\mathbb R ^2$.

Mike Miller

There is probably only one.

Axoren

After that, it's just mapping tricks back and forth.

Mike Miller

16:08

@Axoreb: Define space filling curve.

Axoren

@MikeMiller Any map $f : \mathbb R \to \mathbb R^d$ is a space-filling curve. To give a more general definition that encompasses other fields and spaces is still beyond me.

Err, hold on.

Left something out.

Balarka Sen

Uh.

Axoren

The map is continuous and onto the whole space.

Mike Miller

OK. I prefer space filling curves to have domain $[0,1]$ and usually cover $[0,1]^d$. or something similar.

Balarka Sen

It's surjective. That's the point.

Axoren

16:11

Yeah.

Mike Miller

You should prove that, given a space-filling curve $[0,1] \to [0,1]^d$, you can set it up so $f(0) = (0,\dots,0)$ and $f(1) = (1, \dots, 1)$. You should then figure out how to patch these together to get a space-filling curve $\Bbb R \to \Bbb R^d$.

Balarka Sen

Anyway, space filling curves on $[0, 1]$, $\Bbb R$ and $S^2$ are not too different.

Axoren

The interval is usually where you start, but then from that you can use composition to fill the whole real plane.

Mike Miller

To find one from $[0,1] \to [0,1]^d$, start with any of the famous ones for $d=2$, and see if you can figure out how, from that, to get one in arbitrary dimension.

J. M.

Any specific curves in mind? (Hilbert, Morton, Peano, etc.)?

Balarka Sen

16:14

If you have a proper space filling curve $\Bbb R \to \Bbb R^d$, then you can extend that to a space filling curve $S^1 \to S^d$, right?

Axoren

@J.M. Hilbert and Peano.

Mike Miller

Sure.

Axoren

I'm more fixing to solve the collision issues in a practical application. A space-filling curve taken from the Hilbert and Peano iterative constructions eventually collides on all rationals $a / 2^n$ or $a / 3^n$ respectively.

Balarka Sen

What do you mean by collision issues?

It can't be injective. No space filling curve can.

Axoren

@BalarkaSen Right, but there are some little tricks that can be applied in practical applications that preserve the other properties and give injective-like-ness.

For example, should a point in $\mathbb R$ map to such a colliding point, shove it somehow and keep note of which original point you shoved.

De Moivre

16:17

I'm asked to prove that $\left\{1, e^{ax}, e^{bx}\right\}$ is linearly independent over $\mathbb{R}$ if $a \ne b$. Google said the easiest way to do this is to something called the Wronskian. Is this how you do it? The concerned matrix is:

$ \begin{aligned} \begin{bmatrix}1 & e^{ax} & e^{bx} \\ 0 & a e^{ax} & be^{bx} \\ 0 & a^2 e^{ax} & b^2e^{bx}\end{bmatrix} =\begin{bmatrix}1 & e^{ax} & e^{bx} \\ 0 & a e^{ax} & be^{bx} \\ 0 & 0 & b^2e^{bx}-abe^{bx}\end{bmatrix}\end{aligned}$

Which is in the upper triangular form, therefore the $\mathcal{W}(1, e^{ax}, e^{bx}) =ae^{ax}(b^2e^{bx}-ab e^{bx})$.

Axoren

In practical application, I'm never working with points outside of $\mathbb Q$, actually. Do to the limited nature of computers not being able to represent $\mathbb R - \mathbb Q$ anyway but symbolically.

The most important properties I care about is distance locality and the inverse mapping onto $\mathbb R$, which needs the "shove trick" to be possible.

Mike Miller

You have probably lost both of us.

J. M.

Everything's rational in floating point… :D

Axoren

@J.M. gets it, at the very least.

Mike Miller

@Balarka: Here.

Balarka Sen

16:23

Answered (1) in a comment.

(2) has been asked a million times in different places.

J. M.

What I'm hazy about is mapping those curves to arbitrary rectangular domains tho.

Axoren

@J.M. You mean from $[0,1] \to [a,b] \times [c,d]$?

J. M.

I could imagine a(n approximate) SFC possibly entangling itself there.

Mike Miller

@J.M.: Err, scale?

Semiclassical

@J.M. just to make sure i'm remembering right---had i encountered you over at Mathematica SE in comments regarding elliptic integrals?

Axoren

16:26

@J.M. Yeah, simply apply an affine map to the coordinates in the unit square

Semiclassical

(namely, the fact that mathematica isn't any good at simplifying them)

J. M.

@Semi, prolly; elliptic integral simplification is one of my peeves.

Semiclassical

same

Axoren

It definitely doesn't sound fun.

Semiclassical

idg why mathematica is so terrible at them

@j.m. any suggestions what to use instead of mathematica for that?

Axoren

16:28

The bigger issues are with $f : [0, \infty) \to \mathbb R ^2$, anyways. It's obvious that a tiling exists, but ensuring that such a tiling is continuous requires a blend of Hilbert and Peano tiles.

J. M.

@Axoren, you'd think that, but with all that squishing, you're bound to get two different rationals having the same first few (binary or decimal) digits. I don't imagine your system directly manipulating numerators and denominators.

Axoren

@J.M. We already have that issue, we simply have a different basis for collisions.

For larger rectangles, we would simply tile the space.

J. M.

@Semi, it would seem to me that their internal algorithms for those haven't been touched in quite a while, even tho there are newer methods.

Semiclassical

that's annouying

Axoren

Points in the same neighborhood on the line fall into the same tile in the tiled space.

Except for points on the boundaries, of course.

Anyways, I've got to run. Thanks, all.

J. M.

16:33

If you're taking your SFCs piecemeal, the irregularity can work. Long as you keep track of what's kinked where.

@Semi, I really have no software I can recommend in that respect; to this day, I still do elliptic integrals semi-manually.

Semiclassical

i figured as much

any references you tend to use for that? i've typically relied on the wolfram functions site and Gradshteyn+Rhyzik

J. M.

@Semi, Byrd and Friedman's book is usually my first stop. To be fair, it can be modernized in some places, but their tables work.

Semiclassical

mmkay

Mary Star

17:56

Hello!!

I want to show that $A_4$ has no subgroups of order $6$.

The possible cycles of $A_4$ are:
* 3-cycles (order 3)
* 2-transpositions (order 2)
* the identity permutation (order 1)

Since there is no element of order $6$, there is no cyclic subgroup of order $6$, right?

How could we check if there is a non-cyclic subgroup of order $6$ ?

L33ter

hi @BalarkaSen

MATH000

May I ask some quick questions, about sets?

Mary Star

@DanielFischer do you maybe have an idea about my question above?

MATH000

Can we say ∅ ε ∅ ?

Akiva Weinberger

18:11

@MATH000 No.

The empty set has no elements. That's why it's the "empty set."

MATH000

then ∅ subset of ∅?

Akiva Weinberger

Yes.

MATH000

Thanks. Is there a difference between { ∅} and ∅?

Akiva Weinberger

Yes. The former has one element (namely, Ø), and the latter has no elements.

Think of it it terms of bags:

MATH000

so we can say ∅ ε { ∅}, correct?

Akiva Weinberger

18:13

Ø is an empty bag, and {Ø} is a bag containing an empty bag.

Yes.

MATH000

Thanks a lot for the help!

Mary Star

@AkivaWeinberger do you maybe have an idea about my question above about the subgroups of $A_4$ ?

Akiva Weinberger

@MaryStar I haven't thought about it

Mary Star

@AkivaWeinberger Ah ok...

MATH000

18:30

Can we say: {1,2,{1,2}}-{1,2}={1,2}?

Daniel Fischer

@MaryStar If $A_4$ had a subgroup of order six, that subgroup would be generated by an element of order three and an element of order two. Since the naming of the symbols on which $A_4$ operates is arbitrary, you can assume that the element of order three is the cycle $\sigma = (1\, 2\, 3)$. Now look at the elements of order two, and look what subgroup is generated by $\sigma$ and the chosen element. You will see that it's the whole group $A_4$.

Balarka Sen

@L33ter Hello.

L33ter

So, I have idea of what is happening for the triangulation that hatcher expalins

explains

Akiva Weinberger

@MATH000 I think that would be {{1,2}}

@MATH000 Since it's the set of everything that's in {1,2,{1,2}} but not {1,2}

In other words, it's the set containing things that are 1, 2, or {1,2} but aren't 1 or 2

Balarka Sen

Self-reference. Ugh.

Akiva Weinberger

18:39

So it's the just set containing {1,2}

Meaning it's {{1,2}}

Mary Star

If $A_4$ had a subgroup of order six, would that subgroup be generated by an element of order three and an element of order two, because $2$ and $3$ are the divisors of $6$ ?

The element of order two will contain at least one common symbol with $\sigma$.
Suppose that the element of order two is $\tau =(3\, 4)$.
We have that $\sigma \tau=(1\, 2\, 3\, 4)$, $\tau \sigma=(1\, 2\, 4\, 3)$, right?
How can we see that the subgroup that is generated is the whole group $A_4$ ? @DanielFischer

Paradox 101

Order of an element is the same as the order of a group?

Balarka Sen

@Paradox101 Not necessarily. Every element of Z/2 x Z/2 has order 2 whereas order of the group is 4.

Paradox 101

Ok so then can we say that order of an element divides order of a group?

Balarka Sen

Yes.

This is Lagrange's theorem.

L33ter

18:42

let us look at the triangulation idea of hatcher

suppose we have $\Delta^1 \times \mathbb{I}$

if we look at the simplex at the bottom [v0,v1].

suppose we lift v1 continously to height 1

MATH000

@AkivaWeinberger thanks!

L33ter

it will trace triangle [v0,v1,w1] right?

@BalarkaSen

Balarka Sen

$\Delta^1 \times I$ is a square, and you're cutting the square along the diagonal. That's it.

L33ter

yes

but I actually I was thinking of doing it

systematically

for any n

Balarka Sen

(I am going to remind you that you still haven't explained the idea of the proof)

Mary Star

18:45

Do we have to check also the elements $\sigma^2 \tau$, $\tau \sigma^2$, $\sigma^2$, $\sigma^3=\tau^2=1$ ? @DanielFischer

L33ter

we subdivide $\Delta^n \times \mathbb{I}$ into (n + 1) simples

s.harp

I want to draw $$\frac{\overline{\Xi}}{\Xi'}$$ on a blackboard

L33ter

so now we will have a boundary operator on prism $\Delta^n\times \mathbb{I}$

which will be the top - bottom + sides(modulo signs)

Balarka Sen

Huh? Subdivision of the prism hasn't got much to do with the boundary operation...

L33ter

yeah

it has to do with the details

which I don't want to get into

Paradox 101

18:49

Oh ok. I have to prove that a group $G$ of finite order that has two subgroups with different orders is cyclic. So then I'm going to assume that the order of $G$ is either a prime, a product of two primes, or a prime raised to the power $3$ (as the subgroups have different orders) and then generalize it from there? @BalarkaSen

L33ter

Since, it will require referencing algebra

Pedro Tamaroff

@L33ter

L33ter

@PedroTamaroff

Pedro Tamaroff

What's your mathematical background?

Like, what courses have you taken?

"And stuff?"

L33ter

General topology,Complex analysis,real analysis, linear algebra, matrix analysis,algebra I,algebra II,algebra II, applied algebra, linear algebra 2, all the calculus.

set theory

number theory

Pedro Tamaroff

18:50

Cool.

Balarka Sen

@L33ter You're starting off from the wrong place. Taking boundary of a prism per se doesn't make sense. We have a singular prism, i.e., a map $\Delta^n \times I \to X$. We're taking boundary of that after realizing that as an $(n+1)$-chain.

Daniel Fischer

@MaryStar Yes, but if we had a target order that isn't squarefree, it would not be so simple. A group of order six contains elements of order two and elements of order three. The subgroup generated by such a pair has an order divisible by $2$ and by $3$, so by $6$, hence must be the whole group.

Your choice of $\tau$ isn't right, since a single transposition is an odd permutation, and thus doesn't belong to $A_4$. The elements of order $2$ in $A_4$ are the permutations of the form $(a\,b)(c\,d)$, where $a,b,c,d$ are all distinct.

L33ter

I m currently taking this algebraic topology. I am having troubles when proof relies on some visual ideas.

Pedro Tamaroff

@L33ter Well, I could tell you how I understand the proof of this theorem, I you let me.

L33ter

yeah

that would be great

Balarka Sen

18:52

@PedroTamaroff I think letting him figure it out would be better, but what do I know.

Pedro Tamaroff

Well, given any space $X$, there are inclusions of $X$ into the top and bottom pieces of $X\times I$.

Namely, $i_1:X\to X\times I$ that sends $x\to (x,1)$, and $i_0$ that sends $x\to (x,0)$.

Note that if $f,g;X\to Y$ are homotopic, then there is a continuous map $H:X\times I\to Y$

Such that $Hi_0=f$, $Hi_1=g$.

Agree?

L33ter

yes

Pedro Tamaroff

OK. Now when we consider the maps $f_*,g_*$ induced on homology, we have that $f_*=(Hi_0)_*=H_*( i_0)_*$ and the same for $g$.

Because homology is a functor, so it respects composition of maps.

L33ter

yes

Pedro Tamaroff

Now, since $H_*$ is obviously homotopic to $H_*$ (I mean, homotopic as chain maps of complexes), then the problem reduces to showing that for any space $X$, $i_0,i_1$ induce homotopic maps from the singular complex of $X$ and that of $X\times I$.

Because as we have shown above, $f,g$ homotopic always factor as $Hi_j,j=0,1$ for a suitable $H$.

Balarka Sen

18:57

@L33ter doesn't know chain homotopy.

Pedro Tamaroff

Hatcher uses that language.

MATH000

U{{5},{1}} can someone please explain what that means? does it mean: ∅U{{5},{1}} ?

Pedro Tamaroff

(Let him decide what he know and doesn't)

Balarka Sen

Yes, but @L33ter hasn't read that bit yet, I think.

L33ter

no I haven't read it yet

Pedro Tamaroff

18:58

Then how are you planning to read the proof? =/

L33ter

I don't understand the statement $H_{*}$ is homotopic to $H_{*}$

Balarka Sen

Hatcher doesn't actually explicitly say the word "chain homotopy" except once in the whole proof.

Pedro Tamaroff

Yeah, that's silly.

Balarka Sen

(I'd disagree, but again, what do I know! I mean, homotopy came before chain homotopy, not the other way around. The idea of chain homotopy emerged from the objective to try to reproduce a relation between chain-level maps of two homotopic maps $X \to Y$.)

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