Mathematics - 2016-02-13

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3:05 AM

@becko Did you get ChatJax working?

2 hours later…

4:40 AM

is there a formal, public service where one could launch a query to find mathematicians interested in our current research other than Google?

hi

can someone help me with planar geometry

user image

5:20 AM

A knockout tournament begins with $2^n$ players and has $n$ rounds. There are no play-offs for the positions $1,2, \cdots, 2^n-1$. What's the set of all possible outcomes? My answer: $\Omega = \left\{1,2^2,\cdots, 2^n\right\}.$

However, my book says this is only if we're interested in the ultimate winner only. Why is this the case?

Morning @Huy.

morning @BalarkaSen

5:47 AM

meh my toasts got over-baked

6:16 AM

I forgot to mention the restriction that there are no play-offs for positions $2,3, \ldots, 2^{n}-1$.

1 hour later…

7:20 AM

Hello@Balarka

Hi.

Hi fiends, I'm reading Folland to brush up on some stuff.

Near the beginning they prove the axiom of choice using Zorn's Lemma

But they assume that the class of well orderings on an arbitrary set X exists and is a set.

How do you prove this in terms of ZF + Zorn's Lemma?

@Albas What are you studying?

Real analysis(sequences and series) and Combinatorics@Balarka

Good. Anything interesting you have learnt so far?

7:24 AM

No not much yet but I will I guess approximately in 2 weeks. I will enter graphs and trees@Balarka

Are those in that combinatorics book?

Yes they are

Cool.

It also has a bit of Sieve theory

I remember trying to learn sieve theory, but the estimations lost me.

Vulgar amount of analysis.

:P

7:26 AM

I have heard it is a very powerful technique in solving problems

Me too. But I don't know anything about it other than that it's inclusion-exclusion on the large.

Yes right. I have only used inclusion exclusion in problems

What are you doing @Balarka? as in studying

Reading.

Reading what?

J. Murayama, "Galois Knot Theory".

7:29 AM

the name sound amazingly cool.

Indeed. But I was merely joking.

That's a mathgen-generated book. Contains a fair amount of mathematical nonsense.

I am studying multivariable analysis right now.

I guess I will do bits of multivariable analysis from rudin in about 2 month or so

Rudin is extremely terse. Not the kind of style I like. But it has good exercises.

Yes it has some very very difficult exercises

I am using rudin as a reference (for questions) and using Tao's book on Analysis

It does. There was the following problem on chapter 2: is there a perfect subset of $\Bbb R$ consisting only of irrationals?

I like this one very much.

7:35 AM

@enthdegree: Do you agree that "orderings on X" form a set?

Cantor set comes to my mind straight away but I dont think so it works@Balarka

Well, the usual cantor set contains a buckload of rationals.

OK, I need to get back to studying.

Bye. Enjoy

7:55 AM

@enthdegree: Ok, I'm going to go to bed. But the point is that the well-orderings on a set is a subset of P(X x X); that is to say, elements are certain kinds of relations (which are themselves certain kinds of subset of X x X).

So you invoke that X x X is a set and restricted comprehension.

@MikeMiller Oh, it's clear now, thanks! I was thinking of relations as elements from the product of \prod_{a in {1,2}) X, which was basically one of the products we were trying to prove was a nonempty set.

Say hi to my friends Will O. and Anton in the program.

Hm, I think I know who the first it is but I'm aware of a couple Antons. But sure, will do, mystery friend.

8:15 AM

How much geometric intuition does group theory require?

None?

One could try developing geometric intuition for group theory. But a good geometric intuition is not a prerequisite for group theory - it's algebra.

Good to know that.

I saw something that I thought was kinda geometric in my assignment.

@user276387 None. when you have to answer questions about symmetry groups during the exam, just rip off an appropriately shaped piece of your exam sheet and flip it around

lol.

@enthdegree lol. That's exactly what I saw that I thought "eek, geometry".

8:20 AM

@user276387 Many groups appear from geometry, e.g., symmetry groups of polygons and polyhedras. That might be on one of those.

@BalarkaSen Yeah, I haven't done any reading on it yet, but it's called dihedral group.

@enthdegree You can't cut out an icosahedron from your exam sheet, because $\Bbb R$ is not homeomorphic to $S^2$.

One isn't compact, the other is.

@user276387 Dihedral group is the set of symmetries of a regular polygon of $n$ vertices. You have two sets of symmetries, rotations and reflections. Rotation by $360/n$ degrees is denoted as $r$ and $r$ has order $n$ for obvious reasons. Reflection along an axis is denoted by $s$ which has order $2$. These two generate a group of order $2n$, which is denoted as $D_n$, namely, the dihedral group.

Multiplication is just given by composing symmetries. E.g., $rsr$ is given by rotating once, reflecting, then rotating again.

Once you write down the relators of $D_n$, you can just think about it algebraically and forget about the geometry if you're not comfortable with it.

Cheers for that @BalarkaSen.

@BalarkaSen better get really good at origami then

lol

8:29 AM

Right. Minor typo: I meant $\Bbb R^2$ instead of $\Bbb R$ in the above statement. Making typos on mathematical jokes makes a pretty bad impression, but oh well.

2 hours later…

10:22 AM

Apparently I killed the chat.

10:39 AM

Not the first time.

10:50 AM

Don't worry I kill chats more frequently then you by mistake

1 hour later…

11:53 AM

@Secret I believe you now :D

Hi, by the way.

@Huy Learnt anything new and interesting?

@BalarkaSen: not really, taking a break for a few days. I've felt a bit sick the past days so I think I need to relax a bit

Sorry to hear. What happened? General illness, or something more specific?

cold or flu, not sure

but nothing serious

12:36 PM

@Huy Sorry, was AFK. Good to hear it's not serious.

12:55 PM

no problem, I was AFK too @BalarkaSen. :P

iwriteonbananas

1:48 PM

I just learned another cool way to compute homology of the n-torus

@iwriteonbananas Ah? Which is?

iwriteonbananas

Do you know what Poincare series are?

Yep.

Power series ring with coefficients being betti numbers, right?

iwriteonbananas

Exactly, so poincare series of $X$ is $p_t(X)=\sum_ib_it^i$ with $b_i$ being betti numbers

and they satisfy $p(X\times Y)=p(X)p(Y)$

hence $p(T^n)=p(S^1)^n=(1+t)^n=\sum_{i=1}^n {n \choose i} t^i$

stars and bars, it is nice way of thinking

iwriteonbananas

1:51 PM

using the binomial theorem

How is that computing the homology of the n-torus.

OK, I guess you haven't finished yet.

where bars are representing bags for stars to be, so they are treated like stars because they can be empty or not

iwriteonbananas

well, now we know what the betti numbers are

so we at least know the homology with $\Bbb Q$ coefficients

Oh, so you had finished. Meh. Yes, we only know homology in field coefficients.

This is really the Kunneth formula proof though, isn't it?

iwriteonbananas

Is it? I dunno

forgot the proof

1:56 PM

It's nothing special you just use the Kunneth formula to derive that the cohomology ring of T^n is the exterior algebra and you compute dimension in each grade.

$p(X \times Y) = p(X)p(Y)$ is really a manifestation of Kunneth.

iwriteonbananas

Yeah, true

I'm struggling with exercise 3.3.13

What was that?

iwriteonbananas

Showing there's no retraction $M_g\to M_h'$ if $h>g/2$

$M_h'\subset M_g$ the subsurface of genus h with one boundary circle

I don't think I did this one. Seems interesting.

iwriteonbananas

Yeah, I'm not really sure how to proceed

2:04 PM

My bet would be on retract short exact sequence on cohomology.

That's what I would do first.

Yikes, bananas, you're derailing me from calculus. :P

iwriteonbananas

@BalarkaSen Right I tried that but it doesn't help much

@BalarkaSen great :)

Can we construct a degree 1 map from a smaller genus to a bigger one if we assume there is such a retraction?

@iwriteonbananas Hmm, it doesn't?

@iwriteonbananas Maybe. I don't know! :)

iwriteonbananas

@BalarkaSen Nope, not as far as I can tell

Btw. random fun fact:

compactly supported cohomology is isomorphic to singular cohomology of one-point-compactification

Interesting.

@iwriteonbananas If there is such a retract, then $M_g$ retracts to the wedge of $2h$ circles, right?

Hm. I am pretty sure this can be broken. Hmmm.

So you get an injective map $F_{2h} \to \pi_1(M_g)$, $2h > g$.

I think this is really a group theory problem.

2:35 PM

hi

In $F_{2h}$, product of $h$ commutators of the generators can't be written as product of less than $h$ commutators by 3.3.12. But as soon as you're inside $\pi_1(M_g)$, there any product of $g < 2h$ commutators blows to dust, doesn't it? So you should be able to get a contradiction of some sort, @iwriteonbananas.

hi @BalarkaSen

@iwriteonbananas I am not going to think anymore about this because I have calculus to finish, but I think the way I am thinking about it is on the right track. Oh, and sorry for so many pings, I noticed you left.

Hi @L33ter.

iwriteonbananas

@BalarkaSen Hmm, while I was away I was pursuing a different approach

still trying to construct that degree 1 map

Proof of 3.3.12 uses 11, which is your degree 1 map.

iwriteonbananas

2:42 PM

I think I might have a degree 1 map $M_{g-h}\to M_h$ under the assumption of $h>g/2$ i.e. $g-h<h$

OK?

iwriteonbananas

But i'm not sure. let me ponder for a couple more minutes, lest I embarress myself

Sure.

There is a map $M_{g-h} - D^2 \to M_h - D^2$ by restricting the retract, but I am not sure if you can cap off the disks neither why it is (at least locally) degree $1$. Not sure if this is what you have in mind.

Back to calculus.

iwriteonbananas

2:58 PM

@BalarkaSen Yeah, that's the map

on $M_{g-h}-D^2$ it's the retraction and on $D^2$ it's the identity

it's continuous

and has degree 1

Why is it identity on $D^2$?

iwriteonbananas

that's the definition of the map

It seems to me that the boundary circle of the removed $D^2$ can go anywhere.

iwriteonbananas

no

$r$ is the identity on $M_h'$

that includes the boundary of $D^2$

Isn't $D^2$ the closed disk? So $M'_h - D^2$ doesn't contain the boundary circle.

I have forgotten what's the convention for $D^2$.

If it's the open disk, you're right. If it's the closed disk, I am not so sure.

iwriteonbananas

3:02 PM

I think you're confusing things.

$M_h'$ is genus $h$ surface with open disc removed, so it has a boundary

We assume we have a retraction $r:M_g\to M_h'$

OK, fair enough, I am bad with disks.

iwriteonbananas

So $r$ is the identity on $M_h'$, in particular on the boundary of the removed disk

Then you are of course right.

Yeah, that is obvious.

@iwriteonbananas Whoa, hold on. $M_g - M_h'$ doesn't contain the boundary of the removed disk.

iwriteonbananas

Correct.

So how do you plan to extend $M_g - M'_h \to M'_h - D^2$ to $M_{g - h} \to M_h$?

iwriteonbananas

3:07 PM

By setting it to be the identity on $D^2$

So the map is:

Proof or I don't believe your extension is continuous :)

iwriteonbananas

We want a map $M_{g-h}\to M_h$. Restricted to $M_{g-h}-D^2$ we set it to be $r$ and restricted to $D^2$ we set it the identity

it's continuous because $r$ is the identity on the boundary of $D^2$

No, you're not restricting to $M_{g - h} - D^2$. You're restricting to $M_g - M_h'$.

The former has the boundary circle, the latter doesn't.

iwriteonbananas

@BalarkaSen Neither of them has the boundary circle!

They're the same thing

What? You just said $D^2$ is the open disk. $M_{g - h} - D^2$ then does contain the boundary circle.

iwriteonbananas

3:10 PM

$M_h'$ has the boundary circle and we delete that

Yes, $M_g - M_h'$ doesn't contain the boundary circle. I agree with that. But $M_{g-h} - D^2$ does, and they're not the same thing.

And, without the boundary circle, I don't know how you're planning to extend.

iwriteonbananas

$D^2$ is the closed disk, I have no idea where I said it's the open one

OK, let's get this disk out of the way.

iwriteonbananas

LOL

You have $r : M_g \to M_h'$. You restrict to $r : M_g - M_h' \to M_h'$.

3:13 PM

hi

iwriteonbananas

Exactly

can someone help me with understanding a planar geometry question

Then you define $f : M_{g - h} \to M_h$ to be $r$ on the complement of the closed disks on the edge, and identity on the closed disk, right?

user image

Is this question in 2D or 3D?

That is, let $D^2 :=$ closed disk. Then define $f : M_{g - h} \to M_h$ to be $f(x) = r(x)$ for $x \in M_{g - h} - D^2$ and $f(x) = x$ for $x \in D^2$?

iwriteonbananas

3:15 PM

YES!

exactly

Now what is your proof that $f$ is continuous?

It seems to me that $r$ can be wild on one side of the boundary circle and identity on the other side, making $f$ very discontinuous.

iwriteonbananas

$r$ is the identity on $\partial D^2$

I am not sure how continuity follows from that.

Please explain. And do have patience with me! :)

iwriteonbananas

Sure

Oh, I see what you mean.

iwriteonbananas

3:18 PM

So we got a map $(M_{g-h}-D^2)\sqcup D^2\to M_h$. On the first summand it's $r$ and on the second one it's the identity

$r$ is good on $\partial D^2$, so it can't be wild on one side yet identity on $\partial D^2$ because $r$ is continuous.

iwriteonbananas

Now $r$ is the identity on $\partial D^2$

You win.

iwriteonbananas

ok, great

:D

Good map.

OK, why is it degree 1?

iwriteonbananas

3:19 PM

I'm flattered

Consider any point inside the disk

Ah, right.

iwriteonbananas

Actually I'm not 100% sure about proving that it has degree 1

Well, any point inside the disk has exactly one preimage, and restricting to a small nbhd we see that $f$ is orientation preserving there.

So by local degree formula, degree is $+1$.

iwriteonbananas

right, that's what i figured

i guess that's fine

You're good.

I do hope my skepticism above was constructive skepticism. :)

iwriteonbananas

3:23 PM

For sure

it helps with the learning

I still think 3.3.12 should tell you there is no injective map $F_{2h} \to \pi_1(M_g)$ for $2h > g$.

iwriteonbananas

Yeah, the hint in hatcher says to use that problem

i haven't done 3.3.12 yet however

It said previous, so it can be anything before 13 :D You never know if it's that problem he's referring to.

iwriteonbananas

Haha, you're right

Could be something from chapter 0

@iwriteonbananas It shouldn't take you long. It's not hard.

@iwriteonbananas Well, we're definitely using something from chapter 0...

iwriteonbananas

3:28 PM

I think i'll do problem 14 first

But apparently you need to use 12 :D

iwriteonbananas

Great, that shall provide perfect motivation to do 12 afterwards

can anyone help me with a planar geometry question?

morning chat

hi

iwriteonbananas

3:33 PM

Hey semi

user image

Seems like you need to use the retraction from the Hawaiian earring onto the wedge of $2k$ circles at each stage.

is this in 2D or 3D?

why would they talking about a line dividing the plane into two half-planes if they were in 3D?

hmm yes but why would they be talking about a plane in $2$D besides the euclidean plane?

3:35 PM

well, even if the plane lies itself in 3D, the points lie in that 2D subspace

oh ok so they are talking about a plane in 3D space but then talking 2 dimensionally about the points on the plane?

well, they're talking about the two-half planes determined by $g$ on $E$

yes, but is this plane $E$ they are talking about in 3D? Otherwise they must be just talking about the euclidean plane 2D

why would it make a difference?

there are infinitely many planes in 3D right?

and 1 plane in 2D?

3:40 PM

yes, and? if the points all lie in $E$, then the question of what the rest of the space is like is entirely irrelevant to whether $g$ properly divides $E$.

any more than taking a piece of cardboard and moving/turning it in space changes its shape

then why couldn't they have just said "points in the plane"?

shrug

iwriteonbananas

@BalarkaSen Yeah that seems about right. However I can't make it work

maybe they are trying to be more general

it applies equally well to a 2D plane in 3D space as to the 2D plane itself---or 10D space for that matter

3:44 PM

Were they trying to be more general? Because if so they aren't since any plane E in 3D they talk about we could just consider to be the euclidean plane in 2D

it may be an obvious generalization, but it is a generalization

@iwriteonbananas no you're not going to distract me anymore lol. I am trying to make the constraint second derivative test work.

good luck on your trip to hawaii.

:P

iwriteonbananas

@BalarkaSen haha! i wish

@Semiclassical If you are curious about the solution to the question here is the official one

nah, it's fine

3:48 PM

user image

my guess would have been induction.

have you ever been to hawaii? do people there really were such earrings?

back later

iwriteonbananas

@BalarkaSen I haven't been there, but I would think they only wear earrings which are homeomorphic to a finite wedge of circles

I would like to see someone wearing an actual Hawaiian earring

@BalarkaSen Problem 14 is bugging me. I really don't know how to proceed.

@iwriteonbananas Me too. Re: the exercise. Maybe you can try attaching a cell along $[f_1, f_3][f_2, f_4]\cdots$. If this loop is zero on $\pi_1$, then that weird beast is going to be homotopy equivalent to sphere wedge earring, as homotopic attaching maps imply homotopy equivalent spaces. I dunno, just try to hammer it somehow.

iwriteonbananas

3:59 PM

I think it would be easier to see that the loop must be nonzero in $\pi_1$

But it's somewhat remarkable that it's nonzero in $H_1$ since finite commutators would of course be trivial

My original levelwise idea probably wouldn't work. Lots of loops on hawaiian earring which is locally trivial (i.e., trivial at each level), but globally nontrivial.

@iwriteonbananas OK, you have a point.

iwriteonbananas

How the heck do we use exercise 12?

Uh.

OK, so note that $[f_1, f_2][f_3, f_4] \cdots [f_{2n-1}, f_{2n}]$ cannot be written as fewer than $n$ commutators, because $i : \vee_{2n} S^1 \to H$ induces an injection on $\pi_1$, and that product is not $0$ on $\vee_{2n} S^1$.

This happens for each $n$.

iwriteonbananas

4:15 PM

injection, not isomorphism

Typo.

iwriteonbananas

@BalarkaSen right

Horrible case: I try to find a mistake for one hour or so and I can't find it (through a proof of 15 pages). There is a tiny thing somewhere, well hidden from me.

That's ridiculous.

It ruined my day.

iwriteonbananas

@BalarkaSen I really don't know where to go from there lol

Do you seriously think I do either?

iwriteonbananas

4:25 PM

Nah, i guess you're doing calculus. sorry for distracting you

is there someone like me or m i the unique existence in the universe who thinks better in bathroom and dreams about mathematical solutions ?

Nope, I am thinking about this, thanks to you.

iwriteonbananas

hahah

And I still don't know how to do it, even after 30 minutes of active thinking :|

iwriteonbananas

my apologies

I've more or less just been staring at a blank piece of paper the past 30 min

4:30 PM

@iwriteonbananas This seems like a hard problem. I'd have to think more, but I really need to get a few more chapters revision on calculus done before the day ends. Very sorry :(

Let me know if you come up with something.

iwriteonbananas

@BalarkaSen No problem, I'll let you know if I make any progress.

Cool. I will pace up a notch because I want to catch up with you before you reach chapter 4 :)

I'm leaving the chat because I want to concentrate on calculus. See you later.

iwriteonbananas

@BalarkaSen Great to hear. Laters

FOUND IT! - A WRONG SIGN

5:00 PM

hmm i think i grasped now on this combinatoric form, after it slipped out my hand

5:29 PM

LOL, I just realized it's 19:30. The last 2 hours passed with a very high speed.

Amazing!

6:28 PM

@Semiclassical It would be fun to see one day a game on PS4, say, with integration stuff. It would be an amazing game.

This might be a very interesting project.

A knockout tournament begins with $2^n$ players and has $n$ rounds. There are no play-offs for the positions $2, \cdots, 2^n-1$. What's the set of all possible outcomes? My answer: $\Omega = \left\{1,2^2,\ldots, 2^n\right\}.$

However, my book says this is only if we're interested in the ultimate winner only. Why is this the case?

6:45 PM

read the second sentence

2

Q: Another way of doing integration

What's your option for calculating this integral? No full solution is necessary, it's optional as usual. Calculate $$\int_0^1 \frac{2 \zeta (3)\log ^3(1-x) \text{Li}_2(1-x) }{x}-\frac{2 \zeta (3) \log ^2(1-x) \text{Li}_3(1-x)}{x}+\frac{ \log (x) \log ^5(1-x)\text{Li}_2(1-x)}{x}+\frac{\log ^4(1-...

calculus real-analysis integration definite-integrals polylogarithm

^^^

wow iv just aquiered the yearling badge

The meaning of the title is integration at a different level. Just to know it is still an easy level compared with the stuff I'm working on now.

i though i had lapped a flat year long ago

hey guys, i am having a problem

let us say there is a game of chance

and there is 45% chance of winning one game

how can i calculate the chance of winning n times in a row?

I tried doing (9/20)^n

but that did not work. Any suggestions?

7:02 PM

why did that not work

well, let us say n is 7, that would be 9/1280000000

0,00000000703125 would be the answer

what is that in percentage? Multiplying that by 100 did not give me a percentage

The same formula I tried doing here worked when applied to a 1/2 chance of winning by multiplying (1/2)^n by 100 I got the correct percentage

????????????????????????????????????????????

why does it not give you a percentage

0,00000000703125 * 100 = 0,0000000703125% ?

so winning 7 times in a row at a 45% chance is 0,0000000703125%?

That answer does not seem logical to me, that is why I came here to get my question answered, to see if I did it correctly

what is your logic

I am sorry, I don't quite understand you. I just wanted to know if I did it correctly or not

7:14 PM

(9/20)^n is correct

ah ok, so the answer I also submitted was correct?

Thank-you

Not that good at mathematics, sorry if I offended anyone by my inferior knowledge in mathematics :P

(9/20)^7 is not 9/1280000000.

4782969/1280000000 ?

@Huy I was wondering what the purpose of the second sentence is - what's the point of the restriction that there are no play-offs for positions $2, \ldots 2^n-1$?

Yes, that is equal to (9/20)^7. What you wrote before wasn't.

7:17 PM

@user276387: the purpose is precisely that only the winner matters

Yeah Balarka, there was something wrong with my answer seeing as it was not logical

Thank you :) I now understand it

also the problem is written very poorly

So the fact that we're only interested in the winner was already in the question. Thanks @Huy.

@Huy you should see our exam questions.

you should see mine

7:23 PM

no thanks

seeing one is enough.

@Huy $Q(x) = x^tAx$ be a quadratic form. I want to extremize this subject to $g(x) = \|x\|^2 = 1$. Lagrange says $DQ = \lambda Dg$. Differentiating says this means $2Ax = 2 \lambda x$. I.e., $Ax = \lambda x$. So constrained extremums of $Q$ are precisely the eigenvectors of $A$.

as $\|x\|^2 = 1$ defines a sphere, which is compact, max value theorem implies any symmetric matrix $A$ has at least 2 eigenvalues.

I think this is a pretty cool application of Lagrange in linear algebra.

7:40 PM

@BalarkaSen: yes, I've seen that before. maybe even in chat here, years ago.

Ah. It's in Ted's book, so it's likely that Ted might have said it.

probably. :D

8:12 PM

Hi. Is every irreducible polynomial also irreducible element over some ring?

Why there are two different names for the same thing?

@dash: "irreducible" seems like one name to me. As you say, an irreducible polynomial (with say rational coefficients) is an irreducible element if $\Bbb Q[x]$.

I'm looking on the wikipedia definitions of irreducible element and polynomial and for example 5x+10=0 in Z is irreducible element, but not irreducible polynomial. Am I correct?

Sorry, it should be other way around... it is irreducible polynomial, but not irreducible element

8:29 PM

is a line that lies in a plane parallel to the plane?

Hello @DanielFischer
Could I ask you something?

I want to find the center $Z(S_n)$.

It holds that $Z(S_n)=\{ c \in S_n: cg=gc, \forall g \in S_n \}$, right?

How can we continue?

maybe start with the first few $n$ to get a feel for it? @evinda

$S_1$ and $S_2$ are abelian and so not very interesting. so probably $S_3$ is a good place to start.

(and $S_3$ is a subgroup of all higher symmetric groups. not sure, but that may be helpful)

8:48 PM

@Semiclassical Doesn't it hold that $cg=gc, \forall g \in S_3$ for any $c \in S_3$ ?

so $S_3$ is abelian? that's what that statement would amount to.

@Evinda $S_n$ is the symmetric group on $n$ symbols?

@Evinda $S_{3}$ isn't abelian. It's isomorphic to $D_{6}$ where it's easier to see why the elements don't commute.

9:09 PM

in geometry if we say we have 3 points X,Y,Z in space, does that imply the points are distinct?

usually yes

@Max are there cases where they don't imply distinctness?

i think every time i've seen that, they meant distinct points

also if we have 4 points in 3 space X,Y,Z,T, does there always exist a plane through XZ parallel to YT?

an obvious counterexample is if the lines XZ and YT are perpendicular

9:27 PM

@s.harp: Nice question. No idea why it has a close vote.

Hey can anyone maybe help me with this following question
http://math.stackexchange.com/questions/1653706/finding-the-limit-of-recursive-sequence

9:44 PM

So, did any finish my integral in the spirit of the art?

user image

Don't miss it, it's very nice.

I'm out with math today (keeping some energy for the great Sunday).

@SanicHodgeheg Isn't $S_3$ isomorphic to $D_3$ ?

can you factor constants out of cross products?

@Evinda Sorry, I meant $D_{6}$ in the $D_{2n}$ notation.

@DanielFischer Yes, it is.

do constants factor out of cross products?

9:59 PM

I'm not really sure which notation is more common, $D_{2n}$ or $D_{n}$? I'm used to the former but if the latter is more widely used let me know.

@Evinda Okay. Then look at what commutes with transpositions. Since $\tau^{-1} = \tau$ for a transposition, we have $\tau g = g\tau \iff g = \tau g \tau$. Let $\tau$ run through the transpositions $(1\,2),\, (1\,3),\,\dotsc,\,(1\,n)$.

@SanicHodgeheg You can bet that whichever book/paper you read always uses the other notation.

@SanicHodgeheg Annoyingly, both are used equally often, as far as I have seen.

@DanielFischer @BalarkaSen Wonderful!

10:30 PM

Why does it hold that $\tau^{-1} = \tau$ for a transposition? @DanielFischer

@Evinda Just compute $\tau^2$ for a transposition $(k\,m)$. You will find that it is the identity.

11:10 PM

@DanielFischer And now we have to check for which $g \in S_3$ it holds that $g = \tau g \tau$ ?

11:32 PM

@Evinda Since you want the centre of $S_n$, that must hold for all transpositions ($S_n$ is generated by transpositions, so that is also sufficient). You won't find many such $g$.

11:42 PM

@DanielFischer It holds for g=(1) and g=$(\tau)$, right? How can we find all the g s for which the equality holds?

it works for that particular $\tau$

@Evinda You look at a fixed $g$. Then you see whether $g = \tau g \tau$ holds for all transpositions $\tau$.

How can we check if it holds for all transpositions $\tau$ given that $g$ is fixed? @DanielFischer

@Evinda You take one transposition after the other and check. Until you find one that fails the test, or have exhausted all transpositions.

So do we pick g=( 1 2), (1 3),.... , ( 1 n) and so on? @DanielFischer

11:59 PM

@DanielFischer Doesn't it hold that ( 1 2) (1 n ) (1 2)=( 2 1 n 1) ? Or am I wrong?

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