Mathematics - 2014-11-21 (page 2 of 3)

11:28 AM

@UserX That's right. Indeed, Map(X) is the permutation group on X. These are "natural groups" meaning that they come up natural from fundamental set theory.

robjohn

@Integrator okay.

It is a (surprising to most beginners) fact that every finite group appears as a subgroup of Map(X), @UserX

So there is no "abstract group" :)

@BalarkaSen Cayley's theorem?

Cayley, dude.

Yes.

Was ready to type him Keyley, so Ceyley is a trivial mistake compared to the former

Integrator

11:30 AM

@robjohn CUL8R

haha

but to the point, there is your motivation for group axioms

@Studentmath That means there are only 5 cosets of G by H.

What's Map(•)? The group of the collection of bijective maps from • to itself?

@Balarka Pedro suggested to look at the homomorphism between $G$ and the permutation group of the set of right-cosets of $H$ (there are 5, so that will be $S_5$)

@UserX yes

@Studentmath i was going to suggest that actually :p

We then see that the $|G:K||K:H|=5$, and thus the only option is $K=H$ and we have that $H$ is normal

K being the kernel

11:34 AM

right

To be honest, group theory is the only subject apart from topology that hasn't come natural to me. When I first learnt calc or analysis, I could guess what an example to a statement would be or what would be taught next. On group theory, I lost that intuition almost immediately

I wonder if there is a way using the fact that $S_5$ has no subgroup of order 15, because for some reason I was heading towards there - and I no longer recall why. I thought there was a hint saying that but I can't see any, so no idea why I was thinking that. Any idea @Balarka?

@UserX you'll get used to it. and topology, even at the point-set level is totally intuitive

what book are you using, @UserX?

@r9m I've done the estimates here, and also found an expression of the Dirichlet series in terms of $\zeta$ (unfortunately, it's an infinite sum, but it's $\zeta$, man).

None. I at first used online sources but then moved to various lecture notes. I haven't found a simple group theory book that doesn't include the rest of abstract algebra yet.

I want a book that's simple and group-theory only

11:37 AM

@UserX Try Dummit-Foote for the formal study. For intuition go through Artin.

Oh you won't find a group-theory only book except graduate level super non elementary ones :P

@BalarkaSen if you were talking about topology I started with Munkres but moved to a less-formal "Topology without tears" in which I like the attitude and teaching style of the Author

I am doing Simmons.

@DanielFischer have you heard about this?

I have Artin's abstract algebra which I intent to read soon

Or not soon, depending on how much work I'll have at school

Oh, I have it @Balarka. If there is such homomorphism, and I know $S_5$ has no subgroup of order 15, I also know $|G|$ divides $5x3$. However, it is not 15 as otherwise there would be subgroup of order 15 in $S_5$ by the homomorphism that injects into $S_5$.

So it must be 5. And that doesn't make any sense, there's a mistake somewhere in the argument

11:49 AM

@BalarkaSen Cayley's theorem includes infinite groups too. How can $(\Bbb R^*,\times)$ be a permutation group of a set? What's that permutation group called and what are it's elements?

Committing to a challenge

@UserX no it doesn't

at least in the usual context

@skullpatrol Well I started listening to children of bodom recently. Which I am sure noone will appreciate

Oh that makes more sense now. What about the generalization of Cayley's theorem on infinite groups? The question "what's that permutation group" still remains...

I can't possibly think of any

It can be done. I'll let you think a bit.

Cool @Committingtoachallenge

11:54 AM

I'll probably fail at finding it. I may know the statement of the theorem but it's not intuitive to me at all.

Committing to a challenge

Shut up @Studentmath

:P

@skullpatrol Do you listen to any 'out there' music?

:(

Oh

Sure, sometimes @Committingtoachallenge

Shizzle

I thought you wrote "let me think for a bit"

11:55 AM

:P

Sorry..

It's OK.

@Studentmath You're supposed to know about graphs aren't you?

Supposed is a good word

Are you familiar with Cayley graphs?

I might need your help then.

I know a bit, yes

I can always try :) Ask

11:58 AM

@Studentmath Consider a finitely generated group $G$ and a generating set $\mathcal{S}$. $\Gamma(G, \mathcal{S})$ be the corresponding Cayley graph.

Okay

@BalarkaSen are there infinitely many answers?

@Studentmath Define $M$ to be the minimal genus surface on which $\Gamma$ can be embedded without any intersection of edges.

@UserX answers to what?

I.e. be planar?

robjohn

@Studentmath $\boxed{\bbox[5px]{\text{will think for food}}}$

12:02 PM

right

@Studentmath the genus of $M$ can be computed by computing the euler characteristic $\chi(\Gamma)$

Right

@Robjohn haha

@Studentmath I am trying to inspect the object $M$.

That's probably a bit out of my knowledge now, but go on

clearly $G$ acts on $M$ by acting on the vertices of $\Gamma$ freely and properly discontinuously. an interesting question would be what is $\pi_1(M)$.

What does $\pi_1$ stands for?

12:07 PM

the fundamental group.

i should ask Mike.

It's much more group-theory than graph-theory I am afraid, but I think I did read something about it, let me see if I can find it.

@BalarkaSen to my question.

@UserX sure

@Balarka for $M\ge 2$ it has equivalent in topology iirc

equivalent?

12:10 PM

Then I guess we can define our group operation as a function f: R to R , find a map that will satisfy the group axioms and we have a permutation group right?

sure

Equivalent known problem

Okay thanks for everything. I have to go.

@skullpatrol Yes, I've already used it. Great news it was.

Very cool.

12:13 PM

Though @Balarka, isn't $\pi_1(M)$ dependant soley on $M$?

@MikeMiller $\Gamma = \Gamma(G, \mathcal{S})$ be the Cayley graph of $G$ with generating set $\mathcal{S}$. $M$ be the smallest genus surface on which $\Gamma$ can be embedded planarly. Genus of $M$ is equivalent to $1 - \chi(\Gamma)/2$. What can we infer about $M$?

@Studentmath it's homeomorphism type, yes.

the point being?

Well, using Cayley graph the fundemantal group of surfaces with genus at least 2 was studied around 1900

oh?

Indeed

reference?

12:16 PM

I think the name of the researcher is even in Wikipedia

Max Dehn

@DanielFischer AWESOME answer !!! :D ..

@Balarka I may be wrong, but a bit of googling leads me to think it has to do with the small cancellation theory. You will probably understand it all much better

Oh it's just a tiny bit of knowledge about these surfaces it seems...

I'll go back to my problems..

@Mike @Studentmath Something interesting must happen when $G$ is hyperbolic. I guess $M$ must also be hyperbolic in that case.

@Balarka all finite graphs are hyperbolic iirc

12:31 PM

@Studentmath no

that is so false

$\Bbb Z^2$ for example is not an hyperbolic group

Hrm

I recall the argument went along the lines of "relative to any finite set of generators, the cayley graph of a finite group is hyperbolic as it is bounded"

hyperbolicity in groups is a nontrivial concept found by gromov.

@Studentmath well i have a counterexample

so any proof you might have seen is false or is a proof of something else

Probably

Oh, $Z^2$ is not finite

I'm talking about finite groups

I wrote finite graphs

no idea why

then it's trivial

any finite graph is quasi-isometric to a point

so it's trivially hyperbolic

Yeah. I probably won't contribute anything non-trival to this conversation though :P

But Mike will probably be of better use here, I really barely touched cayley graphs as I didn't know enough about groups to make any good use of them

12:38 PM

ah i see

If you need anything with Random Graphs, I will probably be much more useful

what's a random graph?

1:12 PM

@Balarka there are two main models, G(n,p) and G(n,m) (there are others). G(n,p) is the undirected graph on $n$ vertices, so that $P(ij\in E(G))=p$ for every two vertices independent of the others.

$G(n,m)$ is the graph on $n$ vertices chosen uniformly from the graphs with $m$ edges (and $n$ vertices)

For example, by studying the properties of G(n,1/2), you are basically studying the properties that apply to almost all graphs

1:43 PM

I wonder, I know isomorphism doesn't change the order of an element. Can I conclude from that, that if there is isomorphism $G\to H$, then $o(G)|o(H)$?

N3buchadnezzar

@r9m Heya

$$ 3<\int_0^\pi \frac{x}{\sin^{\sin \cos x}}\,\mathrm{d}x < 4$$

Or something like

$$\frac{5\pi}{4}<\int_0^\pi \frac{x}{\sin^{\sin \cos x}}\,\mathrm{d}x < \frac{4\pi}{3}$$

That's a stupid question, I meant homomorphism

2:08 PM

@DanielFischer I suspect the guy you just commented to is not asking in good faith. Look at his question about whether $\nabla$ is a vector and look at the comments under the answers.

@MikeMiller I suspect that too, but for the moment, I gave him the benefit of the doubt. Haven't yet seen the other, going to look at it.

@MikeMiller Have you seen my question above? What are your thoughts on it?

@DanielFischer In a vacuum, I would do the same, yeah.

But I generally try to stay out of vacuums as they're cramped and dirty.

I don't really understand your question @Balarka. But I also don't know anything about the stuff you're thinking about. I keep telling you this!

That question was generally silly, I take it back.

@MikeMiller If you're interested, I can elaborate.

2:16 PM

Is the genus of a graph not defined by the least genus of a surface it can be embedded in?

I am interested in the surface, yes.

What's the fundamental group of that surface?

The free group on $2n$ generaters, mod the relation $[a_1,b_1]\cdots [a_n,b_n]$.

eh. ok it's just homeomorphic to a surface of g genus. bah.

That comment doesn't seem to make sense. What's homeomorphic to a surface?

@MikeMiller The fundamental group is definitely not interesting enough to ask at this point. Here's an example of what I have in mind : PSL(2, 5) is just a dodecahedra which embeds in a sphere. the dodecahedra thing makes me suspsect that Cay(PSL(2, 7)) embeds in the klein quartic

an evidence is that by the 84(g-1) theorem, the klein quartic has 84(3 - 1) = 168 size automorphism group, and order of PSL(2, 7) is precisely 168

@MikeMiller well the smallest genus surface on which some Cay(G) is embedable is homeomorphic to genus 1 - \chi(Cay(G))/2 genus surface so the fundamental group is not at all interesting w.r.t G

it's just the same as asking the fundamental group of an n-torus.

2:28 PM

n-torus means the product of $n$ copies of $S^1$. You mean "the aueface of genus $n$". And I know it's not; I even gave you the fundamental group up above :P

hm.

Do u guys know a proof of the whole -1, 0, 1 automorphism thing of Klein?

Uniformization theorem!

@Mike @Balarka mind seeing if I am on the right path with some proof in Abstract Algebra?

ok

Uniformization theorem

In mathematics, the uniformization theorem says that every simply connected Riemann surface is conformally equivalent to one of the three domains: the open unit disk, the complex plane, or the Riemann sphere. In particular it admits a Riemannian metric of constant curvature. This classifies Riemannian surfaces as elliptic (positively curved – rather, admitting a constant positively curved metric), parabolic (flat), and hyperbolic (negatively curved) according to their universal cover. The uniformization theorem is a generalization of the Riemann mapping theorem from proper simply connected open...

How about the Schwarzian derivative, nice derivation of that from nothing?

Here is intuition for the proofs!!!! YAY! mathoverflow.net/a/10548/38721

2:37 PM

@Mike Just an unrelated question : is there a way to homotope two maps by geodesics on some space? i mean, there you have the straightline homotopy which is like homotoping two maps through the geodesics on $\Bbb E^n$, so there must be some kind of generalization or something.

Balark it sounds like you want to homotopically deform the metrics on a space so that the geodesics in each metric deform into each other homotopically

oh?

You want to take a map connecting two points, call it the geodesic for some metric, then instead of deforming the maps deform the metrics so that their geodesics are the curves you would have deformed the original curves into

I am just guessing, it sounds right to me, I have never done it and have no idea how to do it in practice

It sounds cool though :D

sure does, but i have to formalize what i mean first before even using it

seems like what you're doing is close enough

Wow

So lets say you want to homotopically deform curve A into curve B. Instead of deforming A into a bunch of random curve in between, you could use this idea select a specific set of maps in between, namely those maps which are geodesics of a specific family of metrics

In other words I guess you could use this idea to choose specific homotopically equivalent maps

2:46 PM

@bobby I don't think he wants to change the geodesics. I think he wants to push points along the geodesics.

right

like slide A along the geodesics to homotope to B

What are A and B?

So, $|G|$ is odd, $|G:H|=5$. I thus know there is a homomorphism from $G$ into $S_5$. Thus $O(G)|O(S_5)$, and thus $O(G)|15$.

What's $O$ here?

A and B are paths from [0, 1] to your space M (manifold?) with geodesics defined with fixed endpoints

2:48 PM

Furthermore I know $S_5$ has no subgroup of order 15, and thus $O(G)$ can't be 15. Thus $O(G)$ is 5 or 3. It must be 5 as otherwise $|G:H|=5$ would lead to contradiction.

O is the order

I.e. | |

$S_5$ definitely doesn't have order 15... oh, never mind, I see

And thus $H$ is normal in $G$, as the kernel of the homomorphism is a subset of $H$, and $H$ is one-element

I.e. $e$

I don't understand, you said you wanted to homotope two maps by geodesics in some space, if you picture the top half and bottom half of the circle in the R^2 plane, connecting the points (1,2) to (3,2), and you want to deform the top map into the bottom map by saying each map in between is a geodesic then you have to say which metric it's geodesic to! This means you have to invent a metric for which it's a geodesic for, no?

Actually not necesserily $e$

But the point still holds

@Balarka Maybe it would help if you could draw a picture. The point of the straight-line homotopy is that it shows that $\Bbb R^n$ is contractible. Manifolds in general won't be...

@bobby I don't think he's interested in changing the structure of geodesics. He's interested in exploiting the existing structure...

2:53 PM

Yes, but if you are in the plane and the top half of the circle and the bottom half of your circle are the two maps you are homtopically deforming between, you cannot use the word geodesic because the only geodesic is the straight line splitting the circle in half. If you want "to homotope two maps by geodesics on some space" then you will have to call every map in between a geodesic... The only way to do this is to invent ad-hoc metrics and use embedding and all this stuff

I could be wrong or misunderstanding :)

N3buchadnezzar

Hmm

How can I find all the integer solutions to $f(n,m) = (n-m)^2 - (n + m)$?

3:24 PM

@Balarka @Mike does it make sense, what I said?

@Studentmath Why should the homomorphism be injective?

(I promise there are lots more groups than $\Bbb Z_5$ that have a subgroup of index 5)

I know that if $G$ is a group, $H$ a subgroup of $G$ and $S$ the set of right-cosests of $H$, than there is homomorphism from $G$ into the set of permutations of $S$, and the Kernel of that permutation is a normal subgroup of $G$, and it is also the maximal one that is subset of $H$.

So I have $G$ as a group, and $H$ as a subgroup. $|G:H|=5$, so I know there is homomorphism into $S_5$, by that theorem - right?

Yes. Existence is fine.

Your order considerations seemed to assume that the map was injective... but it doesn't need to be. You need to be a teeny bit more careful.

Ah, right.

$o(G)\le o(S_5)$ is the right statement.

Huh? No, its image will be a subgroup.

But consider $\Bbb Z_5 \times \Bbb Z_{175648}$. This has a normal subgroup of index 5... but certainly doesn't have order less than 120.

3:33 PM

Ahahhh

Okay

Do you see the factor you're missing?

I think I should look at $G/Kernel$

I know it's isomorphic to $S_5$.

Noo

Why not?

You know it injects into $S_5$.

Let $G = \Bbb Z_5$. Is any quotient of that isomorphic to $S_5$?

3:36 PM

Not even a bit

I see now

I will have to prove it injects into $S_5$, though.

No... it's automatic

First isomorphism theorem

States that if there is epimorphism from $G$ to $S$, then there is isomorphism between $G$ and $G/K$

Oh.

I am silly :) I get it now

If I phrase it as homomorphism of $G$, then $G/Kernel=Image$

So I know $o(G/K)|o(S_5)$, Also $o(G/K)$ is odd and thus $o(G/K)|15$

Yep...

It can't be 15, as I've shown $S_5$ has no subgroup of order 15..

And I'm done :) Few more arguments and I show $Kernel=H$

Thanks a lot @Mike, it really clarified all the ideas and definitions

@N3buchadnezzar Hello :)

3:44 PM

No prob

@r9m Ullo

@Hippalectryon yes sir .. Hello to you too sir !

@r9m Just saying hello -__-

Were you expecting... HULLO ヽ( ◔◇◔)人(◔◇◔ )ﾉヽ( ◔◇◔)人(◔◇◔ )ﾉヽ( ◔◇◔)人(◔◇◔ )ﾉ ?

:DDD

what are those ? peanuts?balls?buttons?eyes?

( ' ヮ')ノ.・ﾟ*｡・.・ﾟ*｡・.・ﾟ*｡・(>⊆< )

3:51 PM

Naruto ended !! relief !!!!

Lol

@r9m Don't forget to watch all those I have sent you :D

@Hippalectryon irk ! I haven't seen/read the last few episodes of Naruto .. so it isn't the end for me just yet !

@r9m I haven't finished Naruto at all

@JasperLoy /o

Jasper Loy

@Hippalectryon I just had 300g ribeye steak, lol.

@JasperLoy :O Lucky

3:54 PM

@Hippalectryon okay ! :) good

@JasperLoy YUM !!

@r9m There are several anime I haven't finished yet xD

I've started yet another one

I haven't been watching anime/reading mangas much ! ... I need to reduce the load of the ecchi genre first before I start with Naruto again :P LOL

@r9m (•̀～•́ ；)

URMMMM

John Doe

aaaaaaaaaaaaaaaaaaaaaaaahhhhhhhhhhhhhhhhhhhhhhh!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!‌!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

@Hippalectryon wat ? u don't like those ? :P or are you under age ? (in which case I apologise)

3:58 PM

@r9m I'm 16 -___-

I find them too stupid to watch

@Hippalectryon hah ! okay ! :P

@Hippalectryon u boring :P

@r9m Lol :D

@Hippalectryon wait ,, I'll add a Q on anime and manga SE .. which are the best ecchi mangas out there ! :P

@r9m No -__-

@r9m That's a bad question

@r9m It's not the kind of question you would ask on a SE website

@r9m I know some, gimme some examples of those you have watched I'll see what I can find for you

@Hippalectryon then is a good question ? u have any suggestions ?

@Hippalectryon wait I'll browse SE first and see if it was already asked ! :P

4:04 PM

@r9m -__-

@r9m ?

@r9m Strike Witches maybe

@Hippalectryon okay lemme check

@r9m myanimelist.net/anime/3667/Strike_Witches

@Hippalectryon anime.stackexchange.com/search?q=ecchi see this

@r9m What about it ?

@Hippalectryon people have asked questions on the topic and were answered :P its not uncommon

@Hippalectryon duh .. kids stuff

4:09 PM

@r9m They haven't asked an opinion post 'what is the best .....'

@r9m What about Highschool of the dead ?

@Hippalectryon ah right !

@Hippalectryon lemme check

@r9m Or, Senran Kagura myanimelist.net/anime/15119

okay _/_ I surrender ! :P enough !!

@HatMan meta.math.stackexchange.com/questions/17368/…

:D

@r9m xD

Hat Man

@Hippalectryon I don't need one.

Hats are for lesser mortals.

4:14 PM

@HatMan my gravaters wearing one .. and its an immortal Vampire !

@r9m But @HatMan is an immortal hat !

Hat Man

@r9m I mean, I don't need an extra hat.

@HatMan well neither do I .. ;)

@r9m What about an umbrella ? ( ／ ´·︿·`)／｀、、ヽ｀☂

@Hippalectryon it doesn't rain here very often .. :P

4:17 PM

@r9m It rains tables

( ᕗ ˘ω˘)ᕗ︵ ┻━┻

@Hippalectryon well we don't have Hippas roaming about the skies here ! (clear sky) .. so it doesn't rain table either :P

@r9m xD

4:29 PM

@r9m Op of what I am watching atm youtube.com/watch?v=iVBmh1EmC5M

@Hippalectryon okay !! good ! enjoy !!

was @HatMan always your user name ? :) I certainly remember your blog ! but I don't recall you as HatMan !

@r9m "boywholived", I think.

@DanielFischer wow !!! you remember !! ?! :D or you found out from old comments ? :-)

@r9m No, I checked a couple of answers and the associated comment @-pings. But also here we have confirmation.

@DanielFischer the Detective !! Cool ;)

pfft !! this one's crazy !!! I'll share this on fb ! :P lol

Sep 24 at 8:39, by r9m

@Moron hey ! nice blog there !! :-)

4:59 PM

@r9m Good digging. Now you mention it, I remember that username too.