Mathematics - 2013-03-01 (page 2 of 4)

awllower

13:00

Hm, I am ot so familiar with gragh thery, and know not too much about well-founded relations, either.

Haha

I think my bed is waving to me now, and it says: come to me---
Hm, I suppose that the only regular path is to go to bed now...
In any case, I am to sleep now, byebye!
:-D

user58512

@awllower, bye bye!

sorry I went awya to get a cup of tea

I really hope you will feel better tommorow

awllower

That is fine.
Have a nice cup of tea!

user58512

see you :)

awllower

Thanks for that. :-)

user58512

13:18

@TobiasKildetoft, hello I got to use that nicely theorem about G/N being cyclic ! thank you

to show that |G|=p^2 is abelian

Tobias Kildetoft

ahh, yes

that too follows just by noting that the center cannot be maximal. But also from the quotiernt not being non-trivial cycic

user58512

yeah :D

then we get the fact about |G|=p^3, must have center p^2 or be abelian

using that result

wait I think I got that wrong

Tobias Kildetoft

yeah, it has center of order p or is abelian

user58512

13:34

ah rthe only difficult case is when Z(G)=C_p and G/Z(G)=(C_p)^2 but then we just repeat the same type of proof, every element of g is of the form x^i y^j c and these types of things commute

Tobias Kildetoft

@user58512 ?? not all groups of order $p^3$ are abelian

user58512

I messed up :S

Tobias Kildetoft

there are 2 distinct non-abelian group of order $p^3$ for any prime $p$

how to describe them depends on whether $p$ is $2$ or not

user58512

if $|G|=p^3$ and $|Z(G)|=p$ then $G/Z(G) = C_{p^2}$ or $(C_p)^2$. In the first case $G$ will be abelian - but how do we discharge the second case?

Tobias Kildetoft

@user58512 we can't

user58512

13:38

oh!

Tobias Kildetoft

as I said, there are in fact two different non-abelian groups like that

user58512

I think would the group be C_p x C_p x C_p in that case?

Tobias Kildetoft

that is one possibility, but there are also non-abelian ones

user58512

oh no I mixed up p and p^2

if $|Z(G)|=p^2$ then we have an abelian group

Tobias Kildetoft

what are you trying to show?

right

user58512

13:42

so I can actually say that if |G|=p^3 then G is abelian or Z(G)=(C_p)^2 !

hi @amWhy, sorry I disappeared when talking to you last my internet stopped working for some reason

Tobias Kildetoft

@user58512 no, if the order is $p^3$ and the group is not abelian the the center is $C_p$

amWhy

@user58512 Oh! no problem...people come and go here...no worries!

2

user58512

but if the center is C_p then we have G/Z(G) = C_p so G is abelian by the lemma?

lemma G/N=Q cyclic implies G abelian

Tobias Kildetoft

@user58512 no, we had order $p^3$ now, not $p^2$

the quotient by the center will be as you wrote

(ie, $(C_p)^2$)

user58512

im going to write it out all formally so I get everything correct

this is weird

if $|G|=p^2$ then $G$ is abelian because - either G=Z(G) or |Z(G)|=p and then G/Z(G) = C_p so abelian by the lemma

Tobias Kildetoft

13:51

right

user58512

but ... if it's abelian then it's center is the whole group!

so it's not C_p

Tobias Kildetoft

yeah, that way of stating the result is a bit strange

but not wrong for that

user58512

is there a way to state this proof that less jarring? :D

Tobias Kildetoft

note that I formulate it as "the quotient by the center can never be non-trivial cyclic"

rather than as "if the quotient by the center is cyclic, then the group is abelian"

user58512

ah, that is a lot better!

thank you :D

Tobias Kildetoft

13:53

also, note that it can also be stated as (and this is a bit stronger) "if the quotient by a subgroup of the center is cyclic, then the entire group is abelian"

user58512

hmm

I better work out that bit you mentioned earlier, about the case |G|=p^3 and G/Z(G) = C_{p^2} or (C_p)^2

D_8 and the quaternion group :D

Tobias Kildetoft

right, when $p = 2$

user58512

@TobiasKildetoft, I have a question

@TobiasKildetoft, some people use $|D_{2n}|=2n$ and others use $D_n$ is 3d rotations $n$-gon.

why don't we call one D and the other Dih?

so there's no confusing

Tobias Kildetoft

@user58512 because then you would have to convince one camp to change their notation, which is not going to happen

user58512

I would change my notation

Tobias Kildetoft

14:01

I usually write "the dihedral group of order ..." if I want to make sure there is no confusion

user58512

that's a good way

Tobias Kildetoft

(and yet, somehow that still caused confusion at one point)

user58512

is there a geometric realization of the klien four and quaternion groups?

lol

$$\langle i,j,k \mid i^2 = j^2 = k^2 = ijk \rangle $$

hm that's confusing

this is i^2 = 1 not -1

oh nevermind, I was reading the presentation wrong

wow groupprops.subwiki.org/wiki/Category:Particular_groups

Tobias Kildetoft

there is a different presentation of the quaternion group which makes it more clear how it fits into a larger class of groups

Charlie

@skull Hello

skullpatrol

14:11

@Charlie Hi.

@Charlie How are you?

Tobias Kildetoft

@user58512 After you asked something about the definition of transitive at some point, I got to thinking about this and I have just asked a question math.stackexchange.com/questions/317878/…

user58512

hmm nice question!

Charlie

@skullpatrol I'm good, and you?

user58512

from my notes I think that A_5 is transitive on the sets {1,2}, {1,3}, {1,4}, {1,5}, {2,3}, {2,4}, {2,5}, {3,4}, {3,5}, {4,5}

yes I have proved that

skullpatrol

@Charlie Fine, thanks. I was just reading an interesting answer here. The number of up-votes is amazing.

Tobias Kildetoft

14:16

@user58512 $A_5$ is 3-transitive

hence also 2-transitive

user58512

haha.. well that proves it much easier

I did it explicitly

Tobias Kildetoft

@user58512 in general, $A_n$ is $(n-2)$-transitive but not $(n-1)$-transitive

Charlie

@skullpatrol interesting !

Charlie

14:33

@skull look this

It's a character from a famous brazilian comic book

Dominic Michaelis

huhu

skullpatrol

@Charlie nice

Charlie

Haha

skullpatrol

Hoho

Charlie

@skullpatrol he's just a skull

Hehe

skullpatrol

14:36

Hihi

Charlie

Sometimes hyhy

skullpatrol

Indeed a semivowel

Charlie

@skullpatrol ;)

skullpatrol

@Charlie ;-)

Charlie

@skullpatrol ;D

skullpatrol

14:38

@Charlie ;-D

Charlie

@skullpatrol ;DDD

skullpatrol

@Charlie ;-DDD

Charlie

@skullpatrol ;*

skullpatrol

@Charlie ;-*

Charlie

:-D

skullpatrol

14:45

:D

Charlie

(-:

skullpatrol

(:

Charlie

(-;

skullpatrol

(;

user58512

14:56

actually here's a new question: which symmetry groups arise as twisty-puzzles?

Tobias Kildetoft

what are twisty-puzzles?

user58512

let me find a pic to show

Tobias Kildetoft

ahh, so generalized versions of Rubik's Cube?

user58512

yeah

I can't even solve it :D

Tobias Kildetoft

the cube? Neither can I

user58512

15:02

what if you use group theory on it? :D

Tobias Kildetoft

too much work (as someone else has already done it)

user58512

it seems like even the symmetry group of the 2x2x2 is absolutely huge

3674160 permutations

Tobias Kildetoft

15:17

@user58512 but considering it is naturally a subgroup of $S_{24}$, it is not that big

user58512

15:28

I'm still confused about the presentation of quaternion group

groupprops gives $$\langle i,j,k \mid i^2 = j^2 = k^2 = ijk \rangle $$

but rschwieb wants = (-1) there math.stackexchange.com/questions/317880/…

I don't actually see how to show $i^4 = 1$ here

I can get (ijk)^2 = iiijk = ijkkk

it's not even clear this group is finite

anon

@user58512 if you look closely, GP links to this groupprops.subwiki.org/wiki/…

user58512

how does it know z^2 is the identity?

oh it explains

@anon, very nice!

Charlie

15:46

@dominic how are you ?

Dominic Michaelis

i am fine

i finished my first tikz graphic latex.informatik.uni-halle.de/latex-online/temp/…

Charlie

@DominicMichaelis good good

Charlie

16:13

Hello @arkamis

Arkamis

morning @charlie

Charlie

Everything good, @arka ?

Arkamis

sore

played full contact last night. Took a couple of hits to the ribs. 80% sure they're broken now

Charlie

Oh my god! Did go to a hospital?

Dominic Michaelis

does anyone konw from where i can get handbook of categorial algebra from corceux or mac lane categories for the working mathematician ?

as a free pdf

Tobias Kildetoft

16:30

Does anyone know a good reference for getting used to extracting information from first quadrant spectral sequences?

Arkamis

@Charlie No, there's nothing that I can do about it.

An X-ray may or may not show a break, and the only way to tell is wait a few weeks and see if it feels better

Eric Gregor

if $A$ is a square matrix, and we have $b=Ax$, where $b,x$ are column vectors (and $b$ has norm 1), why is it true that $|x_j|^2=|b^T a_i|^2$?

a_i is the i-th column vector of $A$

Peter Tamaroff

@JayeshBadwaik Good! You're back.

Jayesh Badwaik

@PeterTamaroff YES!

@Arkamis what did you play full contact?

Dominic Michaelis

@eric it isn't let A=0 and b=0 than every x fullfills the equation

ah sry b has norm 1

i still think you need something like A ist invertible or something like that

Eric Gregor

16:38

$A$ is orthogonal! sorry should have said that

or at least unitary

Jayesh Badwaik

ha, that solves everything I suppose.

Eric Gregor

that is an important detail...

Dominic Michaelis

yeah it is !

Eric Gregor

wait, so why does that make it solve everything?

i'm just trying to follow a proof of something that uses this

Dominic Michaelis

cause orthognal matrices are isometrics

Eric Gregor

16:39

can you translate?

Dominic Michaelis

linear functions induced by orthogonal matrices are isometrics

Eric Gregor

what does isometric mean?

Dominic Michaelis

if A is orthogonal it say $\| A x \|= \| x\|$

Eric Gregor

but this is saying something a little stronger. that $|x_i|^2=|b^T a_i|^2$

Dominic Michaelis

you have A x = b multiply with A^T implies x=A^T b

Eric Gregor

16:43

ah, and $|A^t b|=|b^T A|$?

i guess that doesn't seem obvious to me

Dominic Michaelis

be carefull with absolute value and the norm, A^T b is a vector

Eric Gregor

the proof uses absolute value

claiming that $|x_i|^2=|b^Ta_i|^2$

Dominic Michaelis

yeah notice that the first component of A^T b is the same as the scalar product between b and the first row of A

so in fact you can say x_i = b^T a_i

which is stronger than |x_i|^2 = |b^T a_i|^2

Eric Gregor

i see, thank you!

Arkamis

@JayeshBadwaik Hockey

Jayesh Badwaik

16:55

@Arkamis Ice Hockey? Oh...

Etiquette Guidelines | LATEX support for chat

6

star this and you would not regret it tomorrow!

Charlie

17:13

@jayesh

Jayesh Badwaik

@Charlie yes?

Charlie

@JayeshBadwaik start it, for the next week

;)

Jayesh Badwaik

:P

Charlie

:P

Jayesh Badwaik

@Charlie my gtalk is refusing to connect

Charlie

17:21

Ooh! I realized something is wrong @jayesh

Jayesh Badwaik

actually I should say, my kopete is refusing to connect

;-)

Charlie

Jayesh Bedwaik

Jayesh Badwaik

what?

Charlie

@JayeshBadwaik hahaha

@jayesh almost bed time

Jayesh Badwaik

@Charlie yes

Eric Gregor

18:32

what is a "leading minor" of a matrix?

Carpediem

18:50

Hello

Can someone help me with this ?math.stackexchange.com/questions/318099/oscillating-function

@EricGregor Can you help me with this by any chance math.stackexchange.com/questions/318099/oscillating-function

Gugg

19:16

@mjqxxxx Calling mjqxxxx

Anyone, if I leave a message for someone like above, will they always see it? Or is there another way to send a message?

Carpediem

yes

@gugg

@PeterTamaroff Hello

Gugg

@mjqxxxx Hi. Sincere apologies for deleting my question on Bell, taking your fine answer with it. The problem was that the question was MUCH more false from even its outset, and was extremely misleading.

Carpediem

@I changed my username

@PeterTamaroff

Peter Tamaroff

@Carpediem Good.

Carpediem

@PeterTamaroff how are you ?

Peter Tamaroff

19:26

@Carpediem Tired. Didn't sleep well last night, and had a long day today.

Carpediem

@PeterTamaroff Can I ask you a question about differential equations?

I read something in my lesson I was curious about..

I am considering and ODE of the form x'+kx=k cos wt

I was curious why the input and response curves intersect only at the maximum and minimum points of the response curve

Is there a way to show this without actually solving the equation ?

Peter Tamaroff

@Carpediem Let me read.

I take it $k,w$ are free.

Carpediem

@PeterTamaroff yes

Peter Tamaroff

@Carpediem The "input" would be the solution $x(t)$?

Carpediem

Yes

Charlie

19:32

Hola @peter

Peter Tamaroff

And the response curve is $k\cos\omega t$?

Charlie

Hi @Carpediem

Carpediem

@PeterTamaroff yes

@charlie hello

Peter Tamaroff

@Carpediem OK.

@Carpediem Any initial condition?

@Charlie Hey

Carpediem

@PeterTamaroff no..

Charlie

19:38

Everything good with you guys?

Peter Tamaroff

@Carpediem But you don't have a unique solution without an initial condition

Carpediem

@PeterTamaroff I think x(t=0)=1

@PeterTamaroff I am not sure because nothing is written apart from what I told you

Peter Tamaroff

@Carpediem OK, and how do you know that the input and response intersect at such places?

Carpediem

It is a question in the lesson

It is a reading question asked by our professor

I think I have it

The derivative at the maximum and minimum points are zero

so we have kx=k\coswt

ie input equals response

@PeterTamaroff

@PeterTamaroff right ?

Gigili

19:57

Left.

Carpediem

@Gigili why ?

Charlie

@Carpediem you said "right"

Gigili

@Carpediem I don't know the exact reason. (I'm joking, a bit)

Carpediem

@Gigili -.-

:)

Peter Tamaroff

@Carpediem Oh, flawless.

Good work.

Carpediem

19:59

@PeterTamaroff I guess you really are tired today

Ethan

Prwtijghgy jhggdhhdmhmd

anon

22 secs ago, by Ethan

Prwtijghgy jhggdhhdmhmd

Charlie

fgh sdfg gkjhkushfkjhhkjahkghhgç

çççççç

Ethan

Tfhfjktjggyvhjgjhmnggkhgndtigujgvyvrcyevfhugyufvyuyjyj

Charlie

nhysahj uushuju hasuy ????

Ethan

20:06

Pcfynethwrtbrshyreh!

Charlie

hgsdgfkusadhgkufow8eyb :)

Ethan

Fhteethgdhtdjdgjgd..

Charlie

gffvtrcty

user58512

hi

Charlie

Hi

You'll never know how I watched you
from the shadows as a child
you'll never know how it feels to be the one
who's left behind
You'll never know the days, the nights,
the tears, the tears I've cried
but now my time has come
and time, time is not on your side

...now I've got you in my sight
With a Goldeneye, golden, goldeneye
with a goldeneye, goldeneye.

Ethan

20:26

Horrible, just horrible

Charlie

@Ethan How dare you????

user58512

finally got i^4 = 1

Charlie

clap, clap, clap

if i^2=-1

Ethan

Boo

user58512

hi @Ethan

starting with i^2=j^2=k^2=ijk we get i=jk and k=ij by cancelling, together that gives i=jij squaring j^2=i^2=ji^2ij=ji^4j so 1=i^4

that proves the quaternion group is finite too

Charlie

20:31

good, good

user58512

is there a shorter derivation? I didn't use the fact that powers of elements commute i.e. the element z in z=i^2=j^2=k^2=ijk is central

hi @mick

mick

@Ethan what is horrible ?

user58512

:(

Charlie

@mick My song

mick

@user58512 Sorry ! Hi user 58512 !!!

user58512

20:34

:)

Charlie

Hi @mick

mick

No this is horrible song

http://www.youtube.com/watch?v=ypaCKPOSlRA

Charlie

@mick yeaaah...

mick

I like goat editions :)

Tfhfjktjggyvhjgjhmnggkhgndtigujgvyvrcyevfhugyufvyuyjyj ??

plz guys stop typing with one hand , its disgusting and there are kids here :)

user58512

mick what are you studying

mick

20:38

economy

Charlie

@mick And you complained when I called you kid

mick

@Charlie Its only funny when I say it ;)

@Charlie Btw to be brutally honest , I felt a bit belittling.

►

Charlie

@mick OHH!

@mick So sorry then

user58512

I wanna tlak about math

Peter Tamaroff

@Carpediem I am.

mick

20:42

omg I got 1857 !!

Riemann started to think about his zeta function and published the RH 2 years later !! :D

user58512

really? only 2 years?

Charlie

@mick Oh my God!

@user58512 what branch?

user58512

@mick, do you like the ordinals? e.g. $\omega^{\omega^{\omega^{\omega^{\omega^{\cdots}}}}}$

mick

►

user58512

:(

Charlie

20:45

user58512 I'm not invisible, ok?

mick

@user58512 again ??

Why do goat editions use sheep ? philosophy at the highest level :)

Charlie

Aristotle would be amazed

mick

@Charlie :)

Charlie

:D

Not just Carpediem, Inutilia truncati too

Carpediem

@Charlie I'm back

Charlie

20:49

@Carpediem :)

mick

@Charlie whats so great about this guy anyway ?? Im not a fan so far !

Charlie

@mick Aristotle?

He had some thougts which I was a little against

but

that was in his time

anyway, I'm not a philosopher

mick

@Charlie I study the masters , not their students.

Charlie

@mick Every master was once a student, so...

mick

@Charlie Not true

Charlie

20:54

@mick defend your thesis

mick

one word : autodidact @Charlie

Charlie

@mick even then, is a student

Dominic Michaelis

does anyone have mc lanes category theory for working mathematicans ?

Charlie

no :(

You must put yourself in the condition of lerner to be able to teach, and to teach is to learn twice

Transcript for

$Mathematics$ Mathematics