Mathematics - 2015-11-07 (page 4 of 5)

Balarka Sen

17:16

hey @Alizter

user147690

@BalarkaSen More difficult than expected :D

Balarka Sen

Indeed.

Agawa001

im having fun solving at-reach combinatoric probs

that reminds me of my first year in college

user147690

@BalarkaSen I'm going to give up for now, but bug me about it tomorrow so I do it

Balarka Sen

Sure.

It is nontrivial, you need semidirect products in it's full power.

(actually, you can do it the bag-of-tricks way Artin does it - which I like very much - but it's roundabout)

Alizter

17:19

Hey @BalarkaSen

Balarka Sen

what's up?

user147690

One question about it, but I might just be too tired.

If I look at $n_3=4$ I get $n_2=1$ to 'fill' the group. But what group has $4$ subgroups of order $3$ and a subgroup of order $4$, normal inside of it. It's $A_4$ right? $V_4$ is normal in $A_4$ and there are $4$ copies of $A_3$ in $A_4$, is that correct?

user147690

I looked up $A_4$ though, and these 4 copies of $A_3$ aren't conjugate, and they should be by Sylow theorem

Balarka Sen

you're looking at order 12?

user147690

Yes

Balarka Sen

17:27

ok, n_3 = 4 gets you n_2 = 1. now you just do semidirect product trickery.

user147690

Don't I obtain $A_4$?

Balarka Sen

yes, I think so.

user147690

Through semi-direct product trickery though?

Balarka Sen

you can actually get an isom this way

@AlexClark huh??

the 2-sylow is normal, while the 3-syow ain't.

user147690

Yep

Balarka Sen

17:28

yes, you need to use semidirect products. but here's another way

Alizter

@BalarkaSen Not much

Preparing a physics investigation

Balarka Sen

recall that sylow's 2nd theorem (I hope?) says that G acts on the 3-sylows by conjugation.

so you get a natural homom G --> S_4 by permutation representation

user147690

Yep

Balarka Sen

ok, you need to show that this is embedding into A_4 somehow.

user147690

hmmm

Balarka Sen

17:30

think about it, I am replying to a mail. It can be done this way, but I need to recall - which I am too lazy to do :P

user143442

@Alizter you're handsome

user147690

Sure haha, I'll talk about it again tomorrow then, since its 3:30

Balarka Sen

k.

user147690

Thanks @BalarkaSen, night

Balarka Sen

g'night

Alizter

17:31

@user I'll take it as a complement, however I am not sure it's the discussion I want to be having now.

user143442

Yes, it's just a complement, not a discussion.

user147690

I have a slight feeling that @user is Jasper and he was throwing us off before, but I must sleep byee

user143442

LOL @Jasper

Alizter

Who does "LOL" remind me of?....

@BalarkaSen Jasper says it too

Balarka Sen

oh heck don't star that Ted'll kill me.

Alizter

17:34

@BalarkaSen You are already a dead man

Balarka Sen

in what sense?

Alizter

The Slappening

user147690

chat.stackexchange.com/…

user147690

I had to come back for that one

user147690

Look at the jasper capital lol ratio

Alizter

17:35

Yeah

Balarka Sen

@Alizter what slappening?

Alizter

@BalarkaSen It was a pun on the film title "The Happening". Because Ted will slap you.

Balarka Sen

Ah, Shyamalan.

user143442

LOL

Balarka Sen

Great films. I didn't watch that though.

user143442

17:36

I'm sorry to disappoint you but I am not @Jasper

Alizter

@BalarkaSen Whatchya upto today?

Did you get better from sickness?

Sarah

hi

user143442

@BalarkaSen nor @TedShifrin

Balarka Sen

Yes, bit better.

Alizter

@BalarkaSen Must be due to a lack of maths

Balarka Sen

17:37

@user I wasn't claiming you were Ted (I mean, of course not). I just said LOL reminds me of Ted.

user143442

ok

Secret

Why there are so few dynamical system background users on this chat?

Sarah

@Jasper

Balarka Sen

@Secret because they're too ergodic to chat on a fixed place for a week.

Sarah

Hello

Alizter

17:39

@BalarkaSen I was about to state something similar myself

Sarah

Hello @Alizter

Secret

@BalarkaSen that's annoying... because I need some help on interpret some chaos theory expeirments results and they are the best people to look for

Alizter

Hi

Balarka Sen

@Alizter Doing calculus, diff geo, etc.

Alizter

@Secret Just ping 'em

@BalarkaSen Do you have any good examples of continuous non-homotopic functions?

Balarka Sen

17:41

non-homotopic function does not make any sense

Sarah

Hello @BalarkaSen

Balarka Sen

hi

Alizter

@BalarkaSen Like between two spaces

But they look like they should be homotopic

but they are not

Balarka Sen

Oh, you mean non-homotopic function_s_

user143442

@Alizter can I have your email?

Balarka Sen

17:42

sure, lots

Alizter

@user erm depends what for?

Balarka Sen

$S^1 \to S^1$, the zero map and the identity map

user143442

just to know you better

Alizter

@BalarkaSen But that is an obvious one, do you know any non-obvious ones?

Balarka Sen

the identity map and the map $z \mapsto z^2$

Alizter

17:43

@user You can chat to me through here

@BalarkaSen Complex plane?

Balarka Sen

@Alizter And it's not obvious.

Proving it requires nontrivial amount of algebraic topology

Alizter

Thats a good one

Balarka Sen

@Alizter S^1 is considered as the unit circle in C

Alizter

oh ok

Balarka Sen

@Alizter ???? look at Hatcher chapter 1

it requires covering space theory

user143442

17:44

@Alizter but this is a maths chat, I'd like to talk with you about other things

Alizter

@BalarkaSen I have a severe lack of Hatcher here

It only came up in discussion

I haven't studied it in detail yet

Balarka Sen

can you clarify what you meant by "Thats a good one"?

Alizter

@user What would those things be?

@BalarkaSen I liked it?

Secret

@Alizter How to search for stack exchange users with a known field of study?

Balarka Sen

@Alizter You need to know point set topology to study algebraic topology

@Alizter Liked what?

Alizter

17:45

@Secret You can look at the users with highest reputation on a certiain tag

user143442

@Alizter You and me

Balarka Sen

If you're being sarcastic, it really is a nontrivial fact that $\pi_1(S^1)$ is nontrivial

Secret

@Alizter thanks, will try

Alizter

@BalarkaSen There is no sarcasm here, I guess I am just easily impressed

Balarka Sen

Were you referring to the example id and $z \mapsto z^2$?

Alizter

17:46

@user I am not sure there is much to talk about there

Balarka Sen

I'm confused otherwise.

Alizter

Yes

Balarka Sen

OK.

user143442

@Alizter I'm sure there's a lot to talk about

Alizter

Don't worry about it though, I haven't studied these things yet so I have a tenancy to talk nonsense.

Balarka Sen

17:47

If you want to study some algebraic topology, study point set topology from Munkres first.

Alizter

@user Sure, then invite me to a new chat room if you really want to.

Balarka Sen

A piece of advise : don't feed the troll

Alizter

@BalarkaSen Depends on whose the troll here.

Balarka Sen

user is a well-known troll.

user143442

@BalarkaSen you're difaming me

Sarah

17:48

@user I'm sorry but you sound super creepy

user143442

@Sarah I didn't ask for your opinion, did I?

Sarah

@user No you didn't, I didn't ask for your permission either.

deostroll

how to convert a 2d figure to a 3d projection?

Alizter

@deostroll There are a lot of 3d objects with the same shadow

user143442

@Sarah Well, you should give your opinion only when you're asked to give it

user143442

17:50

If not, keep it to yourself

Sarah

@user Sorry I dropped my bucket of free speech

deostroll

I have to draw it on a 2d graphics surface

Alizter

@deostroll Do you mean a 3d object to a 2d one?

user143442

Your rights end where mine begin

Alizter

Because the other way round is a bit weird

Sarah

17:51

@user You have the right to remain silent

deostroll

I mean, say an equilateral triangle projected on a YZ plane...

user143442

You have the right to give your opinion, but that doesn't mean I care about it

deostroll

But I have to do all that using a 2d graphics program...

Alizter

@deostroll Have you tried Wikipedia?

Sarah

@user Well you do or else you would not have replied to my opinion.

Alizter

17:53

They have some good resources on the mathematics behind computer graphics

user143442

I'm just telling you to keep your opinion to yourself when you're not asked to give it

user143442

cause if none asked for it, that means that nobody cares

Sarah

Does anybody else find @user's opinion intolerable?

Balarka Sen

No, because I ignore trolls.

Alizter

@BalarkaSen What happened to that chat room you created a while ago?

user143442

17:55

don't share the cake

Balarka Sen

@Alizter Which chatroom?

Sarah

@user Why not it's delicious

Balarka Sen

There's an algebraic topology chatroom I own, but I did not create it.

Alizter

@BalarkaSen I can't remember what it was called but we had a lot of discussion in it at the start

Balarka Sen

Maybe it just froze.

I forgot about it too.

Alizter

17:56

You told me a lot about Gal in there

Balarka Sen

galois groups are fun.

Alizter

yes they are

I am trying to develop my knowledge of group theory and fields before I play with them again though

Balarka Sen

you need rings more than groups

historically, rings, fields and galois groups came before groups

but knowing a bit of group theory won't go useless, certainly :)

Agawa001

i thought that user is sent just to me specially but i was wrong

Alizter

@BalarkaSen It's just that I would like to eventually use it to show solvable Galois groups

Balarka Sen

18:06

Solvability comes from Galois theory, not group theory.

Alizter

@BalarkaSen But is there not a notion of a solvable group?

Balarka Sen

Chain of Kummer extensions <---> Solvability of Galois groups.

Alizter

Yes

Balarka Sen

@Alizter Motivated from Galois theory. Galois first discovered that Galois group of splitting field of certain quintics are not solvable.

So solvability, historically, was a notion for Galois groups, not general groups

Alizter

@BalarkaSen I know thats what solvability comes from

But I thought it was a property of groups that could be studied as a property of groups without having to know about Galois theory

Balarka Sen

18:09

Yes, you can. But if your ultimately want to study Galois theory, why study it abstractly with groups first?

I never had any feel for solvable groups until I learnt Galois theory.

Alizter

I guess I don't want to learn how to use a hammer on a nail but rather how to use a hammer and then learn that it can be used on nails.

Balarka Sen

Hammer was discovered to be used on a nail, though. That was the motivation of Galois on the first place.

But if you want to, ok.

Sarah

@Alizter user is trying to flirt with you

Alizter

@BalarkaSen I know but perhaps it can be used for other stuff

Or I am inefficient with my hammering education

Balarka Sen

Yes it can be used on other stuff.

Alizter

18:12

Maybe I am trying to eat too much cake at once

Balarka Sen

(Don't ask me what, I have never ever studied abstract group theory)

user143442

@Sarah none asked you

Sarah

@user Sorry just expressing my opinion again. Oopsy daisy

Alizter

@BalarkaSen OK

Balarka Sen

But I encourage you to learn group theory if you want to :)

It's a cool branch of math which I wish I knew more about.

Slereah

18:14

Hey the maths

I am having trouble comprehending the method used here :

Isn't the derivative of $y'$ wrt $y$ supposed to be 0

Sarah

user is getting frisky

Alizter

@BalarkaSen have you ever looked at dual numbers?

Balarka Sen

$k[x]/(x^2)$?

Alizter

Yes

Balarka Sen

I kind of have no idea why people call those dual numbers.

Alizter

18:17

Because Wikipedia says so

Balarka Sen

They're just ring of infinitisimal vectors of $k$ at $0$.

Alizter

haha

@BalarkaSen I found them very interesting

Is there another way of constructing them other than a quotient ring?

Balarka Sen

They are interesting, but for high-brow Grothendieck algebraic geometry reasons I don't understand.

Why do you find them interesting?

Alizter

@BalarkaSen Non-standard analysis

Slereah

nvm found the problem

Alizter

18:20

I read a paper on it a while back

They used dual numbers to describe some things

Balarka Sen

@Alizter Global section of the structure sheaf of the variety defined by $X^2 = 0$? lol.

Alizter

@BalarkaSen I have lost you there

Balarka Sen

@Alizter ah, nice. I don't know nonstandard analysis =)

@Alizter was a joke. I am pretending to be a French algebraic geometer.

what I said is essentially the same as the quotient ring defn

Alizter

What is a sheaf?

Balarka Sen

do you know what a topological space is?

Alizter

18:22

I remember it being a manifold with more structure or something

Balarka Sen

no.

Alizter

No ok

Yes, i know what a space is

Balarka Sen

if $X$ is a topological space, then a presheaf is an assignment of each open set $U$ of $X$ to a, say, abelian group $\mathcal{F}(U)$ which is functorial (google that term).

A sheaf is a presheaf with a certain desirable property which makes the assignment "continuous" in a sense.

Alizter

So it describes the structure of how to traverse the space?

deostroll

Nope...I still can't figure it...whats the math behind project a 2d figure on a 3d coordinate system...like say, an equilateral triangle on the YZ plane

Alizter

18:26

@deostroll What exactly do you mean?

Balarka Sen

@Alizter uh, no.

it's a generalization of ring of functions on $X$.

Alizter

@BalarkaSen Then I haven't understood it, don't worry about it then

I will get there when I get there

Balarka Sen

ok.

Agawa001

im getting stch at this question math.stackexchange.com/questions/1517189/…, symmetricity is confusing my mind

deostroll

@Alizter I mean to draw a figure like this...

user143442

18:30

@Alizter :(

Sarah

@deostroll Could you try the figure again

user143442

why did you leave me?

deostroll

https://commons.wikimedia.org/wiki/File%3AKoch_Cantor_cartesian_product.png

Sarah

@user because you kept asking him about his sexuality on a math forum

deostroll

I guess you'll have to copy and paste in browser

user143442

18:30

that wasn't a math forum, that a was a chatroom for the two of us

Alizter

@deostroll oh I see

So you want a 2d object embeded in 3d space?

Then you need a plane of sorts

evinda

Is anyone of you familiar with the simplex algorithm?

I have this question:

Agawa001

lil bit

Alizter

Which is just two vectors

evinda

0

Q: Why are the quantities equal to 0?

I am looking at the general form of the Simplex algorithm with the use of tableaux. $\overline{x_0}$ is a basic non degenarate feasible solution and thus the columns $P_1, \dots, P_m$ are linearly independent. The first step is to create a $(m+1) \times (n+4)$ matrix as follows: At the first ...

deostroll

18:32

I have to draw that stuff using a 2d graphics program...

Sarah

@user Still what were you hoping to get out of it?

Do you think Alizter will just fall in love with you through an internet chatroom?

Alizter

Guys can you discuss this somewhere else?

Balarka Sen

Create a chatroom for that crap, Sarah, user.

user147690

Just block them, they'll keep trolling

user143442

I'm not discussing, it's her the one who's bothering me

Balarka Sen

18:33

Good idea.

@AlexClark Shouldn't you be asleep?

OK, ignored.

user143442

she's a troll

Alizter

Much better

user147690

@BalarkaSen I wish :D

user143442

LOL everyone ignored @Sarah

user143442

I'm so glad

Sarah

18:34

@user You are the troll silly

user147690

@BalarkaSen I had something I promised myself would be done today, then I started doing Sylow revision

Agawa001

@BalarkaSen ignoring is better than renaming this chatroom anything you want

Balarka Sen

@Agawa001 :D It was a joke though.

But 8 people agreed with me!!

Sarah

@user I think they ignored you as well

David Zhang

Hey guys, I have an algebra question I'd like to run by you.

Does every infinite field (of any characteristic) contain a countably infinite subfield?

3

Alizter

18:35

Shoot

user147690

@DavidZhang Yes

Balarka Sen

What about $\overline{\Bbb F_p}$?

Alizter

@DavidZhang yes

user147690

@DavidZhang $\Bbb F_p$ or $\Bbb Q$ are the prime subfields

Balarka Sen

oh, I misread countably by uncountably.

:P

user147690

18:37

I was worried for a second, I didn't have latex on and I saw the bar and thought wtf

Balarka Sen

hehe

David Zhang

Hm, ok. So what's the argument? I get that every field contains either $\Bbb F_p$ (characteristic p) or $\Bbb Q$ (characteristic 0)

But if you contain $\Bbb F_p$ how do you get to a countably infinite subfield?

Balarka Sen

@AlexClark Wait a second. F_p is not countably infinite.

user147690

I just thought that

Alizter

Its countably finite

David Zhang

18:39

Right. I'm specifically interested in countably infinite subfields

Balarka Sen

Not so hasty. Hm.

I think \bar F_p works.

user147690

That's the algebraic closure?

Alizter

yes

Balarka Sen

nod.

user147690

That does work since any algebraically closed field is an infinite set

David Zhang

18:39

So every infinite field containing F_p also contains its algebraic closure?

Balarka Sen

But how do you know it does not have any countably infinite subfield?

Alizter

However is it a subfield of every infinite field?

Balarka Sen

I don't have an argument.

I just think it doesn't.

user147690

Well we only care about characteristic $p$, and I think I might be too tired :D

Balarka Sen

Someone needs to summon anon.

David Zhang

18:42

Hm, so perhaps the question is not as easy as I first imagined

Alizter

That would mean $\bar{\Bbb F_p}$ can be extended to any infinite field

Balarka Sen

I think this is a good question, @DavidZhang.

user147690

Maybe @MartinSleziak can answer?

Martin Sleziak

And maybe not.

David Zhang

I think I'll post it to Math.SE, then.

user147690

18:43

7 mins ago, by David Zhang

Does every infinite field (of any characteristic) contain a countably infinite subfield?

Balarka Sen

@DavidZhang You may want to try asking it to @anon first.

I am certain he can answer.

user147690

We know if it has characteristic $0$ it has prime subfield $\Bbb Q$, so we only need to show it for characteristic $p$ @MartinSleziak, do you know?

Martin Sleziak

Cannot we simply add some elements by transfinite induction?

Balarka Sen

well, ok.

user147690

I don't know transfinite induction unfortunately, maybe David does though

Balarka Sen

18:45

any subfield of \bar F_p is an alg extension of F_p, right?

and any alg extension of F_p is of the form F_p^n

which is finite

ehh, no.

just pick an infinite subgroup of \hat Z

then you can FTGT to get an infinite extension of F_p. obviously.

Martin Sleziak

Ok, can we create some countable chain of subfields $G_0\subsetneq G_1\subsetneq G_2 \subsetneq \dots$?

Each of them by adding one algebraic element, so they will be finite, but the cardinalities will be increasing.

And then take their union.

Balarka Sen

that'll give me the algebraic closure itself

well, if G_n = F_p^n

because you're taking direct limit, right? direct limit of F_p^n's is precisely \bar F_p

Martin Sleziak

So then it is union of countably many finite sets... ?

I am probably missing something.

Balarka Sen

oh. heh.

Martin Sleziak

Or maybe not. Wikipedia claims the same thing: The algebraic closure of a field K has the same cardinality as K if K is infinite, and is countably infinite if K is finite.

Balarka Sen

18:50

\bar F_p is countble.

right, @Martin. I was being silly.

so no way positive char works.

because every field of positive char is a subfield of \bar F_p, hence countable.

that settles @DavidZhang's question, I guess.

David Zhang

Wait, so every infinite field of characteristic $p>0$ is already countable?

I don't know much algebra, but somehow that doesn't sound right

Martin Sleziak

Why is every field of characteristic p subset of $\overline{\mathbb F_p}$?

David Zhang

Right, surely there are transcendental extensions of F_p which are not countable?

Balarka Sen

is F_p(X) a subfield of \bar F_p?

of course not. ok.

that's troublesome.

Martin Sleziak

But the answer is still the same.

We have some field of characteristic p.

Then it contains $F_p$.

If it is algebraic, then it contains $\overline{F_p}$, which is countable.

Balarka Sen

18:55

but what if it's transcendental?

Martin Sleziak

If not, take any transcendetal element $x$ and take $F_p(x)$; i.e., the smallest field containing $F_p\cup\{x\}$.

Balarka Sen

@MartinSleziak Then it's *contained in \bar F_p.

Martin Sleziak

@BalarkaSen I don't think that's true, but you probably know more algebra then me, ....

But let's talk about the transcendental case first, ok?

So all I need to do is to show that $F_p(x)$ is countable, and I am done, right?

Balarka Sen

no alg extension of k contains \bar k other than itself.

(by defn)

@MartinSleziak uh, no. say K/F_p is a transcendental extension. then F_p(x) is contained in K. if F_p(x) is countable, that does not say much about K

David Zhang

Well, in any case, I've posted the question to Math.SE here: math.stackexchange.com/questions/1517718/…

Martin Sleziak

18:59

But the question was whether there is a countable subfield.

And F_p(x) is a countable subfield.

David Zhang

I'm afraid I've got to leave at the moment, but if you work out the answer here, please do post it there.

Transcript for

$Mathematics$ Mathematics