Mathematics - 2012-08-13 (page 1 of 6)

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12:00 AM

ok yeah, was hoping to avoid the quadratic solution

but the efficiency of checking for the condition you want depends on "what" your set of points is.

@DavidWheeler im sorry? my set of points is just a list of (x,y) coordinate pairs

ok, so more or less an arbitrary set of points

@JackSchmidt hey

uh, well im sure there are optimizations that could be made on my set because it does have real world meaning

let's ignore that for now

12:03 AM

@JackSchmidt Do you have an idea on how to repair this?

@JGord Yes, plane sweep algorithms are much faster and explain in many places. I'm feeling lucky gave a decent one: cs.mcgill.ca/~cs251/ClosestPair/ClosestPairPS.html

@BenjaLim Spaces around the Lemma: make the markup work. reading the content now

Ok thank you, will look into it

@BenjaLim I'm a little confused about the question. Since the matrices are permutation matrices, all the entries are at most 1, there are at most k such diagonal entries, but k is only reached on the identity. Hence your arithmetic is just fine, and proves it. Let me consider how to phrase it for $k=3$ and $S_5$.

@BenjaLim YAY! Your answer is right.

oh, I see there is a trick: it is not the identity of $S_5$, it is the identity of $\rho(S_5)$. Let me check if that hurts anything

12:16 AM

i am curious...what does the "k" signify?

@DavidWheeler Just the "case" $k=3$.

@BenjaLim Sorry, I don't see how to fix it: you end up getting $[1,\chi] \leq k - (1-1/[G:K])$ which is not useful.

Like when we say. "Assume it is true for $k=1,\dots,n$. We show it is true for $k=n+1$"

@BenjaLim Also, I think on the second line of the big displayed equation, you use $n!-1 = (n-1)!$.

@PeterTamaroff what i'm getting from the post is k = |X|, in other words, we have a homomorphism from $S_n$ into Sym(X)

12:21 AM

@DavidWheeler I do not know abstract algebra.

Just guessing.

$|X|=k$ would be cardinality?

i can think of one such obvious action, but it's hard for me to imagine them "all"

namely if $X =\{x_1,\dots,x_k\}$ define for $\sigma \in S_n, \sigma\cdot x_i = x_{\sigma(i)}$

hmm...no, that wouldn't work, nevermind

@DavidWheeler it's a neat problem, I think. How do you keep the permutations occupied? S_5 wants to move 5 things, but if you only give it 3...

There is a silly answer if $k=1$: $\sigma \cdot x_i = x_i$.

well, if $k \geq n$ the action i defined above is one, but it's not transitive

@DavidWheeler yup. If $k=n^2-n$ you can label the variables $x_{i,j}$ with $i \neq j$ and let $\sigma \cdot x_{i,j} = x_{\sigma(i),\sigma(j)}$. This is transitve. You can do something similar if $k=\binom{n}{2}$ too. But of course those numbers are not between 2 and n, exclusive.

12:37 AM

so, if we have a 3-element set, we need to figure out how to deal with: (1 4), (1 5), (2 4), (2 5), and (3 5).

and however we "fix (repair)" those, we need to be sure it doesn't ruin the rest of the transpositions

i'm thinking we might be able to leverage the fact (for some n), that Aut(Sn) = Sn

how do we know the embedded S3 inside S5 acts the usual way on the 3-element set?

@DavidWheeler exactly.

unless you meant what anon is worried about :-)

@anon...we don't a priori, it would be nice if it did, but it's not a sine qua non

yeah

If we let $k=n^2-n$ and $n=3$, then we have $X=\{ x_{1,2}, x_{1,3}, x_{2,1}, x_{2,3}, x_{3,1}, x_{3,2} \}$ and $\sigma=(1,2)$ is no longer just a 2-cycle: it takes $x_{1,2}$ to $x_{2,1}$, sure, but it also switches $x_{1,3}$ and $x_{2,3}$. In fact it is a product of three 2-cycles: ( 12, 21 )( 13, 23 )( 31, 32 ).

12:46 AM

well at least it preserves parity...that's a start

well if the questioner is correct...we can't do it

@DavidWheeler David.

Given two spaces $X$ and $Y$ with $X\cap Y=\varnothing$, what is the standard procedure to define a topology on $X\cup Y$?

@PeterTamaroff umm...i dunno? maybe not possible?

$V \subseteq X \cup Y$ is open iff $V \cap X$ and $V \cap Y$ are open in the topologies on $X$ and $Y$. This is the finest topology in which $X$ and $Y$ have the induced topology. en.wikipedia.org/wiki/Disjoint_union_(topology%29

@JackSchmidt Thanks! The author does that, but as the thing wasn't discussed before, I wanted to know.

@JackSchmidt Topology induced by what?

um...what if the topology on Y and the topology on X are "way different"?

12:56 AM

I guess I should be asking: What is the "induced topology"?

If $A \subseteq X$ and $X$ has a topology, then the induced topology on $A$ (by $X$) is $V \subseteq A$ is open iff there is some $U \subseteq X$ open with $V=U \cap A$.

@JackSchmidt Oh! The subspace topology.

Sorry, I had another name for it.

So the induced topology on $X$ by $X \cup Y$ is all $V \cap X$ for $V \subseteq X \cup Y$ open.

:-)

@JackSchmidt Jack, when I finish this section on identification topologies, could I bother you a second?

@JackSchmidt Seems you have your theory fresh.

Somone help this guy

@JackSchmidt i don't understand...is there any reason to suppose that the sets in the intersection will even work?

12:59 AM

@DavidWheeler Why don't you check it?

@DavidWheeler key thing is that $X \cap Y =\varnothing$ is empty, so that there is nothing to check for compatibility

@anon What is the general formulation?

if they intersect, then I think it is called gluing, and you'd have to assume they are already compatible, i think.

oh! i thought it was NON-empty

induced topology is given on wikipedia

1:00 AM

well, yeah, just "mix and match"

V∩X and V∩Y are gonna be disjoint, so we can use an X-type open set on the former, and a Y-type open set on the latter.

en.wikipedia.org/wiki/Topological_union#Topological_union is the gluing thing. this might be a good example of (a) using functions makes things easy to generalize and (b) formal definitions are already their own generalization (read the two wiki articles, the definitions are the SAME)

@anon Oh, the subspace topology is induced by the inclusion mapping, correct?

yes

And the quotient topology by a quotient map or identification map.

@JackSchmidt that's what i thought he meant

1:03 AM

but it is also a good example of definitions I cannot understand without already knwing what they are trying to define

@anon And in general we induce a topology with a function. OK. Very well then.

and under that scenario, you need some kind of "compatability"

@DavidWheeler yes, definitely

they even assume the intersections are closed

@Peter, other way 'round i think

for disjoint union you want the "dual" of the projection scenario

so you want "injections"

X and Y aren't talking to each other: so you have one messenger carry the X-messages to the disjoint union, and another messenger carry the Y-messages.

@DavidWheeler HAHAHAHHAHAHA

1:08 AM

i think it's a co-product, but i'm never sure of myself with "co-" stuff

yes, it's a co-product. i'm good (whew!)

so the topology is final on the injections

@DavidWheeler You just said it: it is the dual of "product space".

you should notice a pattern...if we have some map f:A-->B and we want to make f continuous

we need to do one of 2 things: make sure A has "enough" open sets, or ensure B has "not too many"

which one we do depends on whether we're free to play around with A, or with B

for some reason, i always find A easier to play around with.

in other words: the subspace topology makes more sense than the quotient topology

@DavidWheeler this is a good way to look at it. I often find it hard to remember if we want the most open sets possible or the least open sets possible.

i'm always getting coarser/finer and weaker/stronger mixed up.

i greatly admire those poeple who can just say: well, just refine the pre-sheaf, and by yoneda you're done! pff! obvious!

and i'm like...wut?

BenjaLim is working out of Hatcher...and i have that book...but i'm not getting it very well. he posted a vanKampen question, and i can't make heads or tails of it.

how do you link to a question so it doesn't display the whole thing?

1:24 AM

@DavidWheeler [like this](you put the link here)

experiment: this question $[here]math.stackexchange.com/questions/180625/…$

ok, obviously that didn't quite work, but i'll try again: [oops]math.stackexchange.com/questions/180625/…

i want to not link the title of the page, but a hot-link to a word....

@DavidWheeler You're forgetting the parenthesis

And no "$"

ok

i feel better, now

$\text{advanced linkage}$

baby steps, man...i still struggle with latex

1:32 AM

$$\href{math.stackexchange.com}{\LaTeX test}$$

chat parses the link before mathjax parses the latex

interdesting

@DavidWheeler No harm intended but Ben had quite a tough time with that question.

well i was reading the book...but i'm not sure 'slapping on a 2-cell" is standard mathematical terminology. couldbe, tho

@DavidWheeler Slap that, ma' brotha.

@DavidWheeler Though, in Spanish thigns get much more hilarious.

user19161

@peter Which chapter of Mendelson are you at now?

In this question, how is $1+\sqrt{0_R}$ an additive group? By closure under addition it contains $(1+0)+(1+0)=2$, hence $2=1+\alpha$ with $\alpha\in\sqrt{0_R}$, but $\alpha=1\implies1\in\sqrt{0_R}$

or maybe it's suppose to be a multiplicative group and I'm being silly

1:49 AM

yeah, multiplicative. 1+nilpotents is a multiplicative subgroup of the group of units. since it is a commutative ring, it is an abelian group. The exp and log confirm his intention, since they are homomorphisms from + to * and back again.

@JasperLoy Identification topologies, but I'm grabbing stuff from Willard and Munkres too.

my unfamiliarity with commutative algebra is showing

(1+x)*(1+y) = 1+ (x+y) mod (x,y)^2, so "1+" is often shorthand for a homomorphism from + to *.

Gonna cap today.

Oh, no. It is already "tomorrow". Great.

good evening all

2:00 AM

evening, commissioner

2:10 AM

@mixedmath Night.

user19161

@PeterTamaroff It's good not to cap. Capping means rep might be wasted!

@JasperLoy Yes, I know =P

@PeterTamaroff oh, hello good evening as opposed to good bye good evening

cap also goes toward the epic badge though

unless you meant hello night as opposed to goodbye night

;p

user19161

2:11 AM

@anon Or the legendary badge.

@mixedmath Hahah, it meant "Hello, it is night here."

legendary is only for the fantastic four

2

user19161

Olympics is finally over.

2:35 AM

Does anyone know the eymology of the word "torus"?

latin

2:56 AM

"toro" comes from torus as well

however the torus don't looks like a toro, at least for me

@leo Hehehhe no.

N.T.: "Toro" means "bull" in Spanish.

he he

by the way

@leo ?

@PeterTamaroff bull aclaration

@leo I don't get it.

3:06 AM

@PeterTamaroff by the way "toro" means "bull" in spanish

6 mins ago, by Peter Tamaroff

N.T.: "Toro" means "bull" in Spanish.

how is Buenos Aires tonight?

@leo Very nice. It rained like hell all throughout the week, but today, on the Children's Day, the sun came out.

=)

@PeterTamaroff Nice. Here is raining too often. A bit cold right now.

torus is from the Latin for "bulge" or "swelling". the use of the word in mathematics may stem from the fact the the ringed molding at the pedestal of a column was also called a torus.

3:13 AM

@DavidWheeler Thanks.

@leo Oh, it is quite cold here too. There was a hailstorm yesterday.

@JackSchmidt Sorry I had to leave earlier.

@JackSchmidt Would it help to know that the irreducible representation of smallest dimension in $S_n$ apart from the the trivial and sign representation has dimension $n- 1$?

@BenjaLim :-S

:-)

user19161

3:34 AM

@PeterTamaroff etymonline.com/index.php?term=torus&allowed_in_frame=0

user19161

The above is a standard place to look for word etymologies online.

@JasperLoy Thanks.

user19161

@PeterTamaroff There is Children's Day here too!

@JasperLoy "It is....", mein Führer.

user19161

@PeterTamaroff No, I mean there is, not it is. It is on another day.

user19161

3:40 AM

Mwahahahaha!

@JasperLoy "There is a...", mein Führer.

user19161

@PeterTamaroff Actually, the article a is not necessary.

user19161

@PeterTamaroff Also, I think the article the there should be omitted.

@JasperLoy True. =)

user19161

@PeterTamaroff Clearly, everything I say is true, though this world may not understand now...

user19161

3:42 AM

Why are you not sleeping yet?

Pure and true, just like the aryans.

@JasperLoy Please, never say "This is a lie" or the world would collapse over itself.

user19161

@PeterTamaroff Never say never: Justin Bieber.

if i said that what i said would only be true if i never lied about telling the truth falsely, would you believe the alternative?

@DavidWheeler An embedding from $X$ to $Y$ is a continuous map from $X$ to a subset $A$ of $Y$, correct?

3:44 AM

well it's raining outside but I don't believe it, does that answer your question?

@anon...."mu"

@DavidWheeler [brain:hurts #shut down immediately]

@PeterTamaroff usually embeddings are injective

user19161

@PeterTamaroff See youtube.com/watch?v=_Z5-P9v3F8w

user19161

My favourite is still youtube.com/watch?v=kffacxfA7G4

3:49 AM

@DavidWheeler Yes, an embedding is an homeomorphism that is not onto.

there is a difference between "injective" and "not onto"

user19161

@PeterTamaroff He was talking about injectivity, not surjectivity.

@JasperLoy I don't know what your point is.

@DavidWheeler I know that.

justin bieber is just so..."precious". do you think he uses hair spray?

user19161

@PeterTamaroff The point is, an embedding is usually injective and not neecessarily surjective, but in the definition you gave, it mentions nothing about injectivity but insists on surjectivity.

3:52 AM

@JasperLoy An homeomorphism is one one, silly.

Can user t.b. be pinged here? I remember that for a while, he couldn't, and then there was some discussion about it, but I don't remember the outcome.

originially you wrote "homomorphism" which is quite different than "homeomorphism"

@PeterTamaroff if so, then an embedding is not necessarily a homeo, right?

user19161

@PeterTamaroff Oh OK, I was talking about the term "embedding" itself in the entirety of mathematics.

i prefer the term "topological isomorphism", but history is not on my side.

user19161

3:53 AM

@DavidWheeler That sounds unnecessarily verbose.

@anon I wrote: "...an embedding is an homeomorphism that is not onto." So yes, it isn't an homeomorphism.

user19161

@PeterTamaroff a, not an

An apple is a fruit that is not green. So yes, an apple isn't a fruit. Wat.

@anon What?

user19161

Geezis, I am getting confused!

3:55 AM

You just wrote "an X is a Y that is not Z." so yes, X isn't a Y with X=embedding, Y=homeomorphism, Z=onto

Willard says: "If $f$ is everythign but onto we call it an embedding..."

an embedding (in topology) is a map f:X→Y such that f is a homeomorphism of X with f(X).

@DavidWheeler Precisely.

but then f:X->Y is only a homeo if f(X)=Y, ie if it is onto, no?

such a map is necessarily injective, and may or may not be surjective (although the cases one is usually interested in are the ones where f is NOT surjective)

3:58 AM

@anon Yes.

user19161

Well, different authors use different definitions. Just read on more to figure out, that's all.

user19161

Sometimes the wording is not crystal clear and we just have to figure out from later pieces of the writing.

user19161

We can't expect everyone to write like Bourbaki, can we?

if such a map exists, X is said to be embeddable in Y. this is an intrinsic topological property of X (but not necessarily of Y).

user19161

I am going to bed, good night bros!

4:04 AM

for example, a simple curve in the plane, parameterized by t in [0,1] is an embedding of the unit interval in the plane.

often such a curve is just called $\gamma$

@DavidWheeler Right.

Why $\gamma$?

For $\gamma$urve?

note that such a mapping need not be open, or closed, but if an injective continuous map is either open or closed, it is an embedding

@DavidWheeler Yes.

i don't know why $\gamma$ is used...i just know it's fairly common notation, in line integrals, for example.

so one might see the notation: $$\int_\gamma f = \int_{\gamma(0)}^{\gamma(1)} f(\gamma(t)) \cdot |\gamma'(t)| dt$$ (crossing my fingers the latex comes out right)

@DavidWheeler Note that $\gamma$ is "c", the initial of curve. It is very customary to do that in mathematics, isn't it?

@DavidWheeler In fact, I do that most of the time. If I change my notation I usually get confused,

4:15 AM

well, greek is funny...gamma is actually sort of halfway between a "k" sound and a "g" sound, and both roman letters derive from it

the greek alphabet doesn't parallel the modern one, very well

@DavidWheeler Well, in Spanish $c$ sounds actually like $\gamma$. Go to the google translator, and write "cama" in spanish. Then make it say it.

@DavidWheeler Here

Bloody monkey...

@DavidWheeler Did you listen?

i think some of spanish pronunciation is "older" than english

the pronunciation of the letter "v" (in words) for example, is closer to when "u", "v" and "w" were all the same letter.

now, the french, on the other hand, have made a concerted effort to get rid of as many consonants as possible (except when they spell, where they throw in a few extra just for fun)

@JonasTeuwen hey

I am trying to show the indicator function

4:26 AM

...what to indicate? is it picking it up?

isn't it "constant" on [a,b]?

on an interval, the indicator function is just a "step-function"...conversely, step functions can be defined as (formal) sums of pairwise disjoint indicator functions (given a suitable partition of whatever)

there is a nice discussion of integration here

4:42 AM

Lol wut?

@JonasTeuwen I have just constructed a sequence of Riemann integrable functions

such that their mean squared norm goes to zero

Okay. Isn't that easy?

but for no $\theta \in [0,2\pi]$ does $f_k(\theta)$ converge to anything.

I wouldn't know; I didn't sleep.

That's an exercise from S.-S. right?

@JonasTeuwen It suffices to work on $[0,1]$

yes

So you choose $I_k \subseteq [0,1]$ like this:

4:46 AM

Yes, I know how to do it. But tell me, I will check it.

$[0,1/2], [1/2,1], [0,1/4],[1/4,1/2], [1/2,3/4],[3/4,1]\ldots $

Say it in words please.

Divide $[0,1]$ in half, then quarters, then eights, etc

So for each $k$ you split up $[0,1]$ into $k$ even length subintervals?

Fine.

Yes.

Let $f_k = \chi_{I_k}(x)$

Then it is clear that $$\int_0^1 |\chi_{I_k}|^2 dx \rightarrow 0$$

4:48 AM

Just the length of the interval.

Yes, and no convergence because it runs away.

@JonasTeuwen as $k \to \infty$

Wait, the $I_k$ is the running interval right?

You pick the next one.

yes.

I see, yes. That's the idea.

Yes but the point is the lengths are going to zero.

4:49 AM

Yes.

That's the typical idea for such examples. Did you figure it out by yourself?

@JonasTeuwen Well there was a hint in stein and shakarchi. But I constructed the intervals myself :D

Well "typical" - apparently not many people know it.

@BenjaLim Very good then!

@JonasTeuwen Well first of all you had to pick closed intervals

Why?

Because for open intervals this is not possible I think by compactness of $[0,1]$

Your $I_k$ would be an open cover for $[0,1]$

Oh not wait I am getting mixed up.

forget it.

4:51 AM

It does not need to be. Does not really matter either.

@JonasTeuwen Right. But we still need to understand why for no $x \in [0,1]$ does the sequence of real numbers $f_k(x)$ converge.

@BenjaLim That is easy, right?

It can only converge to $0$ or $1$.

And both suck for larger $n$.

Yes I think the problem is that the sequence is alternating

@JonasTeuwen My proof is by a picture :D

@JonasTeuwen Wait let me upload the picture :D

@JonasTeuwen Are you there

I am taking the picture now.

@BenjaLim I am.

the picture is coming.

I will show this to pierre on wednesday.

coming.

@JonasTeuwen

4:57 AM

@JackSchmidt Jack, are you there?

user image

@t.b. is in the house. @tb

@PeterTamaroff DOn't see him

@JonasTeuwen the picture is loading

"Please show a little bit of your own efforts and read meta.math.stackexchange.com/q/1803/5363 – t.b. 1 min ago"

@BenjaLim Waiting... 8-).

4:58 AM

@JonasTeuwen see the pic

@BenjaLim Bizarre picture. Works I guess.

The idea is that the verticle line cuts through 0 and 1 always

@tb Here? No sleep either. Oh man.

no matter how large $k$ is

@JonasTeuwen You can always find $n>k$ such that $f_k =1$ at the point $\theta$

or $f_k = 0$

Yep.

Kannappan Sampath

4:59 AM

@JackSchmidt Thanks for pointing me to the relevant exercises. I will get back in a few days if I face some problems solving them. Thank you and regards.

This was a nice problem.

@KannappanSampath Hey

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