Mathematics - 2012-07-19 (page 5 of 6)

Jonas Teuwen

15:00

@HenningMakholm It does, but doesn't that hide where the real problems are?

Zhen Lin

@JonasTeuwen That sounds like two different ways of saying the same thing...

Jonas Teuwen

Kernels of semigroups mix up problems related to measures and operators in one thingie.

Henning Makholm

Problem is there is not much of ZF you can take away without disallowing maipulations that intuitively feel completely unproblematic.

Jonas Teuwen

@ZhenLin Oh, right, let me rephrase.

For example: Say we have a non-doubling measure on $\mathbf R$. One thing we could do is restrict the class of balls to make the measure doubling on this set.

Another way would be to actually change the space $\mathbf R$ that seems to cause these problems.

Henning Makholm

Nobody would ever get much work done if we couldn't trust our intuition about what one is allowed to define, but had to check whether each little step could be formalized in some less-than-ZF.

Jonas Teuwen

15:02

It might end up giving the same thing, but the view can give some other information.

@HenningMakholm Yes, I know, but ZFC is not holy.

Removing something makes it worse, I'm sure. Adding something... I don't know much about that. Other axiom system? Why not? Probably very hard.

ZFC is there "to make things work" and then you get less ideal things like our friend the Banach-Tarski paradox. At least: to me.

I might be misunderstanding stuff, fine. I hope I figure out in the end.

The real problem is probably more that it is way too hard to actually find an axiom system that "fixes" these problems. You fix one problem, get another.

Zhen Lin

It seems inevitable, somehow. I blame $\infty$.

Henning Makholm

Adding something wouldn't work. If a particular set can be proved to exist (and be non-measurable) in ZFC, adding more stuff to ZFC cannot possibly invalidate that proof.

Ragib Zaman

I think its very intuitive that there exists non-measurable sets though.

Zhen Lin

I don't!

Or, to be more precise, I find non-measurable sets non-intuitive.

Ragib Zaman

Well, thinking of Lebesgue measure, it amazes me such a "rough" idea can ascribe measures to so many sets.

Jonas Teuwen

15:06

@ZhenLin So... actually Doron is really brilliant?

Ragib Zaman

Basically filling a set with rectangles

Jonas Teuwen

@RagibZaman Approximation.

Zhen Lin

Indeed. Approximation is all we do in analysis! :p

Ragib Zaman

I know, but it seems intuitive that there should be exotic sets that this simple approach shouldn't work for.

If your approximation is too rough or not suited for the particular object at hand, then it doesn't work.

Jonas Teuwen

@HenningMakholm Oh, could you briefly elaborate on that? We can't add something that excludes the existence of that set? Would that give contradictions somehow?

@ZhenLin Some guy was once claiming too hard and too long that all analysis is is: limits. I felt trolled afterwards that I even responded.

But indeed, approximation is one of the big main tools in analysis.

And the fun thing is... in the "applications" you hardly ever care about the functions that are the extensions of your approximative theorem.

Ragib Zaman

15:09

if we were to tell a lay man the essence to many ideas in analysis, I think I would tell them about limits.

Henning Makholm

@JonasTeuwen If we add something that states that there is no set with such-and-such properties, and we already have a proof that concludes that such a set exists, then there's a contradiction immediately. New axioms can never make old proofs lose their validity.

Zhen Lin

@RagibZaman But what "moral" reason is there for such terrible things to exist?

Ragib Zaman

No good moral reason, just my intuition, which could be quite bad.

Zhen Lin

The real line is an idealised model of a certain geometric notion. Idealised things are supposed to be nice!

Ragib Zaman

Indeed: "In 1970, Robert M. Solovay showed that the existence of sets that are not Lebesgue measurable is not provable within the framework of Zermelo–Fraenkel set theory in the absence of the axiom of choice ."

So theres no simple reason one could give, it will need choice

Jonas Teuwen

15:11

@HenningMakholm Hmm, not sure if I understand. That proof is in a different axiomatic system. I propose to change an axiom (or add, but apparently that does not work) such that that proof does not work.

@RagibZaman No, that is a bit a too narrow interpretation.

Ragib Zaman

In my mind, its simply "wow there must be something exotic out there that this so simple idea fails for" and learning about the Vitali set is simply "ahh well there it is!"

Zhen Lin

@RagibZaman You must be a pessimist, or else a combinatorialist! :p

Ragib Zaman

@JonasTeuwen It's clearly not all of analysis, but its probably the single most fundamental idea, no?

Jonas Teuwen

He only proved, that purely within ZF you cannot prove that there exists non-measurable (Lebesgue) sets. That does not mean that "parts" of AC also exclude this.

@RagibZaman Yep, and... if you apply it. You apply it to a nice function as well.

Ragib Zaman

@ZhenLin I've become a pessimist, there have always been counterexamples to so many things I wish were true I have lost all hope...

Henning Makholm

15:13

@JonasTeuwen It's the adding that doesn't work. A (well-formed) proof is valid if and only if the set of axioms being assumed during the proof is a subset of all the axioms of the theory. Adding more axioms makes it easier, not harder to satisfy that condition.

Zhen Lin

@RagibZaman Come escape to abstract mathematics, where we set up our own axioms! :p

Jonas Teuwen

@HenningMakholm Oh, right. Yes, now I understand.

@HenningMakholm So... that makes my "quest" actually way harder.

Ragib Zaman

@ZhenLin Yes I've heard too many jokes about how Algebraists solve the problem by choosing the right definitions!

Henning Makholm

@JonasTeuwen Indeed.

Jonas Teuwen

But maybe I am an idiot that I even think about that, if I can believe these crazy analysts in our group. They seem to think it is useless, but I wonder what goal their esoteric stuff serves. My goal is purely intellectual self-sufficiency.

I should study this set theory in more detail, maybe I can make some toy theory that does not have non-measurable sets but can still integrate many nice functions.

Zhen Lin

15:15

There is nothing wrong in dreaming about alternative mathematics!

Jonas Teuwen

Then the hardest part is probably how that relates to ZFC.

Henning Makholm

For something completely different, I think we should stop providing answers to questions where the number 2012 appears as a seemingly arbitrary constant. At least not until 2013.

5

Ragib Zaman

If you don't have non-measurable sets, you must give up translation invariance or countable additivity. Pick your poison.

Zhen Lin

Or choice!

Ragib Zaman

oh, no, god no!

Henning Makholm

15:18

What is the choice strength required for the existence of non-measurable sets, anyway?

Zhen Lin

Solovay's model has ZF + DC + every subset of $\mathbb{R}$ is Lebesgue measurable + property of Baire

@HenningMakholm Choice for a family of continuum-many countable sets is enough.

Jonas Teuwen

@RagibZaman Well... I might be a retard, but I am not that retarded.

Ragib Zaman

lol

Henning Makholm

@ZhenLin That will be hard to live without, I fear.

Jonas Teuwen

@ZhenLin Aw :-). So no analysis if we drop that.

Zhen Lin

15:20

I hear DC is enough for elementary analysis...

Jonas Teuwen

Well, actually without full AC you already have only some toy analysis.

Yes, DC is enough for toy analysis.

Ragib Zaman

Anything "serious" needs AC critically though

Functional Analysis in particular

Zhen Lin

Sure, but maybe that's because functional analysis is too general.

Jonas Teuwen

@ZhenLin In what sense too general?

So, what would you purpose to cut out?

Which of the following are you willing to give up: 1) Baire 2) Krein-Milman 3) Hahn-Banach 4) Banach-Alaoglu?

Zhen Lin

Well, I would be happy with functional analysis on (smooth, second-countable, Hausdorff) manifolds...

Ragib Zaman

15:22

It's not a choice, all 4 of those are equivalent.

Jonas Teuwen

@RagibZaman They are not.

@ZhenLin Yes, but unfortunately our models of the "real world" are quite ugly. They require these general tools.

At least, the common way of studying (S)PDEs is in a functional analytic framework. There you use basically all of them (in such a way that you cannot avoid them).

So you could of course argue: your models suck. Then you might say: maybe nature just sucks!

Ragib Zaman

Or maybe that Choice isn't such a bad thing after all?

Zhen Lin

But do we really need to do functional analysis on all topological spaces?

Jonas Teuwen

@ZhenLin No, but do people actually use that or only some crazy French people?

Ragib Zaman

On just normed spaces AC is equivalent to Uniform boundedness, open mapping theorem and Hanh-Banach

Zhen Lin

15:25

shrug

My point is, maybe by restricting the class of spaces we study, we won't need so much AC...

Ragib Zaman

How much more restricted than a plain normed space should we go?

Jonas Teuwen

@RagibZaman No, it is not I believe.

Zhen Lin

Normed space of dimension at most continuum...?

Jonas Teuwen

@ZhenLin Yes, but I argue that that is not true... You even need it for Sobolev spaces on $\mathbf R^d$.

Henning Makholm

@RagibZaman Plain normed space A equipped with a choice function on P(P(P(P(A))))?

Jonas Teuwen

15:27

Hahn-Banach is never equivalent to AC. It is implied by the ultrafilter lemma which is strictly weaker than AC. Additionally, you cannot prove the ultrafilter lemma from Hahn-Banach.

Ragib Zaman

Sorry, I think I mean to replace AC with Baire

then the 4 I listed are equivalent

My bad guys !

This is a good link: en.wikipedia.org/wiki/Axiom_of_choice#Equivalents

Jonas Teuwen

I doubt the one under functional analysis... but I could be mistaken.

@RagibZaman Thanks, nice link. But... what are you trying to say? Some things are a bit wrong and some are like Captain Obvious.

The thing under functional analysis talks about the closed unit ball in a normed space. This space is well, uh, Hausdorff, and that cannot imply AC I think.

Ragib Zaman

Sorry it was just the link I went to to see if my claim about AC being equivalent to the "Big 3" was correct and I realized I mis-remembered

Jonas Teuwen

More something like ultrafilter lemma.

Zhen Lin

Actually, how does one construct a normed space of very high (i.e. greater than continuum) dimension?

Jonas Teuwen

15:34

(which is equivalent with Tychonoff for Hausdorff spaces, I have recently learnt. Tychonoff is equivalent to AC. One direction is Captain Obvious reformulation and the other one is by Kelley in 195x, but the proof contains some errors which can be fixed.)

@ZhenLin That is... a question for Asaf. I believe he wanted to extend the scalar field when I once asked him?

Zhen Lin

Eh, why would you want to do normed spaces over a field other than $\mathbb{R}$ or $\mathbb{C}$...

Henning Makholm

@ZhenLin Which notion of dimension?

Zhen Lin

Algebraic. But analytic would also be interesting.

Henning Makholm

Algebraic is easy enough. Take a large set $X$ and consider the space of functions $X\to\mathbb R$ with finite support, under the supremum norm.

Zhen Lin

Ah, yes, of course.

Jonas Teuwen

15:39

@ZhenLin Eh, why would you want to do normed spaces of dimension greater than continuum?

Zhen Lin

Well, exactly!

Jonas Teuwen

"Because I can!". Exactly, then you can just as well just replace the field.

Zhen Lin

But $\mathbb{R}$ is the unique complete ordered field!

Jonas Teuwen

But we almost always need $\mathbf C$ for nice things to work :-).

Zhen Lin

That's just an algebraic closure...

Jonas Teuwen

15:42

Well, I am not an algebraist. But it is quite cool to write in a harmonic analysis paper: "Now just take the algebraic closure...".

Peter Tamaroff

I'm curious: why do we call the circle a 1 sphere, the sphere a 2 sphere, and not 2-sphere and 3-sphere (according to their "dimension")

Zhen Lin

Because the dimension of a sphere is $2$!

Jonas Teuwen

@PeterTamaroff Volume/surface :-).

Say you are on our friend the ball his surface.

And you want to run, and gravity is like very strong.

How many independent directions do you have you can run to?

Hmm, make sure gravity is not to strong, that you stick to the surface.

Ragib Zaman

The confusing point is that laymen think of "sphere" to refer to the solid ball, while mathematicians mean just the surface.

Jonas Teuwen

You are like a monorail which can go everywhere!

I am starred... too often.

I tend to star retarded and funny things. I hope it is the second in that case 8-).

Peter Tamaroff

15:48

@RagibZaman Oh, right, here it says $x_1^2+\cdots+x_{n+1}^2\color{red}{=}1$. Not $\leq$!

Ragib Zaman

@PeterTamaroff What are you reading atm?

Peter Tamaroff

@RagibZaman Introduction to Topology by Bert Mendelson, a little of Rudin's Principles and Naive Set Theory by Halmos.

But mostly the first one.

Ragib Zaman

That's great!

Mendelson is a great introduction to topology.

Peter Tamaroff

@RagibZaman You used it?

Ragib Zaman

Well, I've skimmed over many portions.

It's not very often I "use" a book where I read it cover to cover, I usually work out of my lecturers course notes and refer to books when I have trouble.

It's much quicker learning that way, books often have lots of extra material that's more interesting (and makes more sense) after a second reading when the key ideas have solidified in ones head and its been seen how the ideas apply in other fields.

Zhen Lin

15:55

Yes, definitely.

Ragib Zaman

For example, in my first course in real analysis we learned about absolute continuity (defined it, proved some basic theorem about it) but there were no questions about it, examples and it didnt appear in the test, and I totally forgot about it.

But then later when I was learning about the Lebesgue theory of integration it comes up and only then you see some importance in the concept.

(This is more than a year later)

Peter Tamaroff

@RagibZaman That's true. Well, before properly studying from a book I read it a couple of times. (Obiously the first few sections which make sense, not "Homotopy and the Fundamental Group" section) That way I skim through it a little and see if I'm ready to start studying from it.

Zhen Lin

I've never grasped absolute continuity...

Ragib Zaman

To be honest, me neither!

But now I know why it is important!

lol

Jonas Teuwen

@ZhenLin Uh... did you even try?

Zhen Lin

16:01

Sure, it made no sense!

Jonas Teuwen

Now I am curious. Why?

Zhen Lin

Who knows. I don't even remember the definition now.

Ragib Zaman

Can you describe a graphic image that gives the essence of the idea geometrically?

Jonas Teuwen

@ZhenLin Basically it is Radon-Nikodym, you seem to like abstraction.

Zhen Lin

Actually, maybe I'm not thinking of absolute continuity. That seems like something straightforward.

Jonas Teuwen

16:04

@ZhenLin Yes, that was why I was so surprised.

@ZhenLin A measure $\mu$ is absolutely continuous wrt the Lebesgue measure if and only if $F(x) = \mu(-\infty, x]$ is locally ac.

Not so profound...

Zhen Lin

Oh, it's uniform integrability that I got confused about!

Ragib Zaman

That is absolutely continuity of measures!

I was talking about absolute continuity of a function

Jonas Teuwen

@RagibZaman That is the same basically.

That is what I write there.

Zhen Lin

Absolute continuity of measures is a very easy definition.

Jonas Teuwen

A function is locally absolutely continuous if and only if its (distributional perhaps) derivative is absolutely continuous with respect to the Lebesgue measure.

Zhen Lin

16:07

Ah, yes, I remember now. I learned it in Stochastic Financial Models...

Ragib Zaman

That is quite a nice characterization.

Far better than thinking about pairwise disjoint unions of intervals with total length less than delta blah blah blah.

Zhen Lin

I feel I should learn stochastic calculus properly someday.

Jonas Teuwen

@ZhenLin You need uniform integrability to get equivalence between convergence of expectations and probability right? I think it is just some technical condition...

Zhen Lin

something like that

it was very technical

Jonas Teuwen

That's analysis. Full of noninsightful technical conditions. Nice results.

I would like to compare it with the concept of a "field" in physics (electric, whatever). You don't know what it is, but you surely know what you can do with it.

So I guess, that the technical condition is just something we need to do certain things which actually are quite ... reasonable to want.

But maybe I am just inverse eating. Perhaps there is a nice intuitive interpretation, but then for me, it would need some rephrasing to something... nicer.

@ZhenLin I know quite a bit about stochastic calculus. Are you looking for the application oriented side or the pure math side (I am more experienced with that).

Zhen Lin

16:14

shrug

I would like to the main definitions and results, perhaps.

Jonas Teuwen

Seeing my monologue, I need to start working.

For normal SDEs?

Zhen Lin

I don't know what that is, even!

Jonas Teuwen

There is a nice book by Steele, stochastic calculus with financial applications or something like that. The title is deceptive. It is quite rigorous actually.

Ah! Then certainly this book. If you know probability theory and real analysis you can get all the main ideas and theorems in like two days.

Zhen Lin

I'll have a look.

Jonas Teuwen

If it is too easy, I surely have much harder things you can read. But those... cloud the ideas which are actually quite easy.

Zhen Lin

16:19

"... stochastic integrals are only defined with respect to Brownian motion ..."

is that a bad thing?

Jonas Teuwen

@ZhenLin Nope. You can transform Brownian motion into something more profound.

Plus it is not so bad to restrict to Brownian motion to start with, you do the same with Lebesgue integrals.

Zhen Lin

hm

I suppose so. I'm not sure I've even seen a rigorous definition of Brownian motion, hah.

(other than Lévy's)

Jonas Teuwen

@ZhenLin The book has it!

Not even a definition??

The book has a proof (constructive) that it exists.

@ZhenLin For your question about "is that a bad thing?" you could see this.

Zhen Lin

Ugh, I vaguely remember something like that. I had to appeal to some kind of Cameron–Martin result or something to solve a problem...

Jonas Teuwen

This was not a mathematics course, I presume? 8-).

Zhen Lin

16:27

Stochastic Financial Models!

Jonas Teuwen

I mean, even we, an applied math department do this course much more rigorously.

Yes, we have that course as well.

Zhen Lin

It was an undergraduate course though. There's not enough time to do all the required stochastic analysis properly!

In fact, officially, students are not even required to know measure theory...

Jonas Teuwen

Ah, here it is after stochastic analysis. Which already requires measure theory.

Zhen Lin

I still remember that in first year, there was an elementary probability problem that I solved using optional stopping theorem...

I wonder what the intended solution was.

Jonas Teuwen

Haha.

Zhen Lin

16:35

"A gambler plays the following game. He starts with $r$ pounds, and is trying to end up with a pounds. At each go he chooses an integer $s$ between $1$ and the minimum of $r$ and $a - r$ and then tosses a fair coin. If the coin comes up heads, then he wins $s$ pounds, and if it comes up tails then he loses $s$ pounds.

"The game finishes if he runs out of money (in which case he loses) or reaches $a$ pounds (in which case he wins). Prove that whatever strategy the gambler adopts (that is, however he chooses each stake based on what has happened up to that point), the probability that the game finishes is $1$ and the probability that the gambler wins is $r/a$."

When I mentioned it to a friend, she pointed out that it was a martingale and the result immediately followed from the OST...

but this was a first-year question!

Chuck Fernández

what is a first year question?

Zhen Lin

A question for first-year undergraduate students.

Chuck Fernández

and what is a martingale?

Zhen Lin

A stochastic process $X_t$ satisfying various conditions, but most importantly, such that $\mathbb{E}[X_t] = \mathbb{E}[X_0]$

Jonas Teuwen

@ZhenLin Yes, I would use a martingale as well.

Chuck Fernández

16:39

do first year undergraduate students know stichastic processes in cambridge?

Ragib Zaman

Most don't.

Zhen Lin

Hell no.

Jonas Teuwen

@ZhenLin Uh... $\mathbf E[X_{n + 1} \mid X_n, \dots, X_1] = X_n$?

Zhen Lin

Yes, indeed. But to solve this problem really all one needs to know is that the expectation at the stopping time is the same as the starting value...

Jonas Teuwen

Oh, that is what you mean. Yes, so maybe the question is: prove OST (in this special case).

Zhen Lin

16:42

Wasn't there some terrible technical condition involving boundedness?

Chuck Fernández

are you yand zehnlin?

yang*

Zhen Lin

nope

Chuck Fernández

whan?

i mean wang

Zhen Lin

good god, are you going to guess all the surnames?

Chuck Fernández

are you the imperial overlord of the srudent room?

no

I ask because i found this website thestudentroom.co.uk/member.php?u=152723

2

If 13 people have different cards is it possible for all of them to have the same cards they began with after an odd number of trades? where a trade is when one person trades his card for the card of another person?

Ragib Zaman

16:55

No

Chuck Fernández

what about 12 people?

does the number of people matter?

Ragib Zaman

Oh wait sorry I answered too quickly

We are allowed to trade back?

Chuck Fernández

yes

Ragib Zaman

Oh actually it doesnt matter

Zhen Lin

@RagibZaman The answer is no, plain and simple. The identity permutation is even.

Ragib Zaman

16:56

Do you know about cycles and transpositions?

Chuck Fernández

does it have to do with the 7 bridges problem?

Ragib Zaman

Not quite.

Chuck Fernández

No but i have the diestel in my house

Ragib Zaman

Very good.

Chuck Fernández

this is graph theory right?

Ragib Zaman

16:57

Yes

Zhen Lin

I'd call it group theory, really...

Ragib Zaman

The 7 bridges problem?

Chuck Fernández

Seven Bridges of Königsberg

The Seven Bridges of Königsberg is a historically notable problem in mathematics. Its negative resolution by Leonhard Euler in 1735 laid the foundations of graph theory and prefigured the idea of topology. The city of Königsberg in Prussia (now Kaliningrad, Russia) was set on both sides of the Pregel River, and included two large islands which were connected to each other and the mainland by seven bridges. The problem was to find a walk through the city that would cross each bridge once and only once. The islands could not be reached by any route other than the bridges, and every bridge ...

Zhen Lin

The original problem with the cards.

Ragib Zaman

The problem with the cards could be considered group theory

but I'd put it in elementary combinatorics

we just happen to see that when we start group theory

Chuck Fernández

16:59

why isnt it possible?

nevermind i figured it out

is it possible for an even number of cards to be in different owners hands?

in an odd number of steps

anon

What does it mean to say "a linear order on $\Bbb N$"; does that mean a subposet of $(\Bbb N,\le)$ (which would make the comment false)?

Zhen Lin

@anon He probably means that every maximal chain induces a unique linear order and vice-versa.

Chuck Fernández

is anyone looking into my question?

anon

@ChuckFernández See Ragibs question to you. Each one-on-one trade is a transposition permutation of the 13 hands, which means the sign / parity of resulting permutation (all trades combined in order) will be a permutation with $(-1)^{13}=-1$ sign / odd parity. The identity permutation is even parity, so this cannot be the identity permutation (all hands go back to where the originated).

Chuck Fernández

17:17

oh, i see it so clearly know

Mark Dominus

It's funny that parity arguments like that are ubiquitous, but there's nothing analogous for dividing permutations into more than 2 sets.

anon

(I can't tell if you're being sarcastic or not.)

Chuck Fernández

me?

Mark Dominus

Yes, you.

Chuck Fernández

no, i wasnt

Mark Dominus

17:22

For example, suppose instead of swapping one-for-one, we allow only triangle swaps, where A gives a card to B, B gives one to C, and C gives one to A. Is there any analogous theorem? I think there isn't.

anon

@MarkDominus Wouldn't there one for every character of $S_n$ (partition into distinct preimages)? Parity is just a one-dimensional rep.

Mark Dominus

Isn't parity a zero-dimensional character?

Zhen Lin

Aren't characters one-dimensional by definition?

Mark Dominus

I think it has more to do with $A_n$ being the only interesting normal subgroup of $S_n$.

Zhen Lin

Well, actually, $S_3 \triangleleft S_4$...

Chuck Fernández

17:24

isnt there a proof that parity 1 cant be parity -1?

anon

Parity is the character of the 1-dim rep $S_n\to F^1$. Characters are the traces of reps, and of course reps can be >1 dim, but I the simple multiplicative properties don't carry over (so they are no longer group homomorphisms).

Mark Dominus

Right, I chose my word "interesting" carefully. That corresponds to the permutations in which one player sits out all of the trades. not interesting.

Zhen Lin

No, that's not how you embed $S_3$ as a normal subgroup.

Oh wait, I mean $V_4 \triangleleft S_4$.

anon

I hate my connection, I'm always a minute behind.

Chuck Fernández

why cant a parity -1 move be made in an odd number of moves?

Mark Dominus

17:26

Well, that corresponds to the situation in which the four players are in two separate leagues of two players each, and can trade only with the other player in their own league. Also not interesting.

Zhen Lin

@anon Ah, right. I forgot the definition of character...

Jonas Teuwen

@anon Don't worry. We know you're the fastest.

Chuck Fernández

can a permution have both parities?

Mark Dominus

No.

Chuck Fernández

why not?

is there a proof?

Mark Dominus

17:30

Yes, but I can't think of a sufficiently simple one offhand.

Chuck Fernández

which one?

can someone please prove it to me?

Mark Dominus

@ChuckFernández Try this: secure.wikimedia.org/wikipedia/en/wiki/…

Chuck Fernández

theres one part of the proof i dont understand

how does he know k-k`andm-m`is even?

Mark Dominus

k' is an odd number multiplied by k.

So k' has the same parity as k: odd if k is odd, even if k is even.

Since k and k' have the same parity, k'-k is even.

Chuck Fernández

isnt he supposed to prove that k`and k have equal parity?

Mark Dominus

17:42

No, he wants to prove that k and m have the same parity.

Chuck Fernández

oh, right

oh, i get it, thanks

Mark Dominus

Sure.

I am glad you asked. It is not an obvious thing.

Gigili

Ignoring users feels great. It's like putting your hand over their mouth while seeing how hard they try to talk.

Mark Dominus

There's a guy I've been ignoring on IRC for years. Mostly I have forgotten him, but every so often someone addresses him and that reminds me he is still there.

"Oh, yeah, that guy I ignored."

Henning Makholm

Seems like Makoto Kato is just now busy deleting all his self-answers. I don't know whether to rejoice or to cry foul.

Mark Dominus

17:53

That seems a shame.

Why rejoice?

I thought the problem was that he edited them too much, not that they were there at all.

Henning Makholm

@MarkDominus Mostly because he was getting quite annoying, and the deletion spree could be a sign that he's packing up and leaving for ever. Which wouldn't be quite as good as just learning the community norms already, but a workable second-best option.

Kannappan Sampath

@anon Your definition of parity needs a lot of sanity check. For instance, one needs to say why the parity is constant on conjugacy class; why is identity a even permutation among a few others. I'd prefer to define the sign of the permutation to be the determinant of the matrix associated to it under (left) regular representation.

Zhen Lin

A high-powered definition to be sure, but at least it's well-defined!

Mark Dominus

Is there something that is like the natural density of a set of natural numbers, only instead of being linear (and so assigning a density of 0 to the set of squares, the set of cubes, etc.) it is exponential (and so assigns them different densities)?

It's sort of using an arithmetic mean of the characteristic function of the set, but I can't think of a way to get it to use a different mean.

Kannappan Sampath

@ZhenLin commenting on whose defn, if I may ask?

Zhen Lin

18:07

Yours.

anon

While I said parity is the trace of a 1-dim rep of Sn, I didn't mean to define it one way or another.

Kannappan Sampath

@anon Sorry, then, for making noise. :)

Zhen Lin

... actually, that's circular, depending on how you define the determinant...

Kannappan Sampath

@ZhenLin I was about to say this. But, yeah, we have a "sign-free" definition for determinant.

Zhen Lin

Yes, I suppose one can prove that the top exterior power of a finite dimensional vector space is one-dimensional without invoking the fact that there is such a thing as parity.

But, well, it seems a bit unusual to go to the trouble of developing linear algebra like that before defining the parity of a permutation...

Matt N.

18:12

@t.b. Well I guess I was thinking of $S^1$ as $[0,1)$ which also must be false since $S^1$ is compact. Now I'm not sure why I thought of it that way. Anyway: guess in this question I am trying to ask "why does Pontryagin duality tell me that the dual of a compact group is discrete". And you tell me "it tells you that" : )

anon

(Also, the fact that parity under the character definition is a class function is easy: every group homomorphism with abelian image is a class function, and 1-dim reps are group homomorphisms. Similarly an identity is always in a kernel.)

Matt N.

Not sure why my link is broken. shrug

anon

Put a [ on the left of the first $

Matt N.

No, I needed to escape "[" with backslash.

anon

wow, it was way out there

Matt N.

18:15

Huh?

Kannappan Sampath

@anon Hmm, seems reasonable.

Matt N.

Anyway, I have to go again. Byee!

Michael Boratko

18:37

@anon Thanks for leaving the comment on my question - if you had a second I would like to pick your brain further...

anon

sure

Michael Boratko

Here's the WolframAlpha output for the standard Legendre equation (with alpha replaced by d)

Note particularly that the solution is given as a linear combination of $P_d$ and $Q_d$, the Legendre functions

Now here's the WolframAlpha output for the second equation in my question - the one we are to transform into the Legendre equation

Note that this time the solution is similar, with $\frac 1 2 (\sqrt{4c+1}-1)$ taking the place of $d$

this seems to point toward the idea that the only restriction required is $4c+1>0$, as was indicated in the question by Apostol

anon

oh, my comment is incorrect then. I think I see what the problem is.

Michael Boratko

Also, although $a=1/4$, $b=0$, and $c=-1/8$ doesn't seem to give an appropriate $\alpha$ in my work for the transformation, it does have as it's solution a perfectly reasonable (and real valued) Legendre function (see the WolframAlpha output as well)

anon

y' has to transform with the change of variables too, I believe

hey look it's tb

Michael Boratko

18:46

ahah - that's right, because this didn't add up

anon

@MattN Is going to be furious.

t.b.

@MattN. Assuming that you have normalized Haar measure on $G$, the orthogonality relations tell you that $$\int_G \gamma(g)\,dg = \begin{cases} 1 & \text{if } \gamma = 0 \text{ is the trivial character}\\ 0 & \text{otherwise.}\end{cases}$$ But that integral is nothing but the Fourier transform of the constant function $f(g) = 1$, so $\hat{f}(0) = 1$ and $\hat{f}(\gamma) = 0$ otherwise.

But this tells you that $\{0\}$ is an open set of $\Gamma$, so $\Gamma$ is discrete.

Matt N.

Teddy!!! : ) mwah

t.b.

Hi, Matt, mwah, back :)

@anon phew

Matt N.

Meh I was planning to not sit in this chat room for the rest of today. : )

t.b.

18:53

I'm not going to stay very long...

Matt N.

I assumed.

t.b.

But I have a few minutes.

Matt N.

Man, I don't know what a Haar measure is. Been meaning to learn about it but nasty homework sheets etc have been keeping me from learning anything I wanted to learn : )

Ok, thanks, let's leave it at that for now. I'll read about Haar measure and stuff.

anon

just know it's a (left or right) translation-invariant measure on certain topological groups unique up to normalization.

Matt N.

It would take too long otherwise.

t.b.

18:55

@anon exactly. Locally compact is what you need.

@MattN. On the circle it's just the usual angular measure (the same as Lebesgue measure on the interval)

Michael Boratko

@anon so $dx/dt=A$ removes the $A^2$ term completely, should I reply to my own question with that as an answer or would you like to?

Matt N.

Ok. But I need some time with this. I feel too pressured into understanding it because you don't have time now.

anon

@MichaelBoratko when was the last time you checked your question/inbox ;)

Matt N.

@t.b. Can I just take this comment and think about it?

Michael Boratko

@anon Very good! Thanks again!

anon

18:59

no, thinking is not allowed, you know that Matt

t.b.

heh :)

@MattN. what do you mean? sure, that's why I wrote it...

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$Mathematics$ Mathematics