Mathematics - 2013-02-22 (page 3 of 5)

Peter Tamaroff

18:00

But how do I form $G$?

Using $FG^{(3)}$

I couldn't understand the "relations on".

Tobias Kildetoft

ok, so we want to find 3 generators of $H$ satisfying those relations

Peter Tamaroff

I have this defintion though:

Tobias Kildetoft

$G$ is a quotient of the free group on 3 generators

a good way to think about $G$ is as the set of words in the letters $x_1,x_2,x_3$ and their inverses, and such that you can cancel whenever you have an element and its inverse next to each other

Peter Tamaroff

If $S$ is a subset of $FG^{(r)}$ we say that $G$ is defined by the relations $S$ if $G\simeq FG^{(r)}/N_S$ where $N_S$ si the normal group generated by $S$.

Tobias Kildetoft

also, you can replace anything by something that is equal to it according to those relations

Peter Tamaroff

18:02

Should I find the normal subgroup generated by $\{x_1,x_2,x_3\}$ and quote?

To get what $G$ is?

Tobias Kildetoft

no

you should find generators of $H$ that satisfy the relations given for $G$

Peter Tamaroff

I don't understand what they mean by "defined by the relations in $FG^{(3)}$"

Tobias Kildetoft

then check that mapping those elements to $x_1$, $x_2$ and $x_3$ is an isomorphism

Peter Tamaroff

He defined something saying "defined by the relations in $S\subset FG^{(3)}$ but not about $FG$ itself.

Tobias Kildetoft

@PeterTamaroff they mean the same thing

(he means the given relations)

Peter Tamaroff

18:04

Well, then to find $G$ I can quote like that. I don't mean to show it is iso.

But I need to understand a little how this works, since I have little examples.

Tobias Kildetoft

I thought you did want to show that $G$ and $H$ are isomorphic?

Peter Tamaroff

Yes, sure. But I want to understand how building groups off $FG$ and relations works too.

Tobias Kildetoft

well, they look like I explained

they are words in the generators, and where you are allowed to exchange strings when they are equal according to the relations

Peter Tamaroff

He gives an example about the dihedral group being generated by the relations $x^n,xyxy, y^2$ in $FG^{(2)}$ for example, which I don't understand still.

Tobias Kildetoft

@PeterTamaroff what part of it is giving you problems?

Peter Tamaroff

18:10

@TobiasKildetoft Well, I guessed the relation should be $x^n=1=$,$xyxy=1$ and $y^2=1$, yes?

Because that is what the rotation $x$ and reflection $y$ should satisfy.

Tobias Kildetoft

yes, that is what it means when people just write some words as the relations (rather than writing them as equalities)

Peter Tamaroff

So we get an homomorphism of $FG^{(2)}$ with $D_n$ by the natural homomorphism sending $x\to R$ and $y\to r$ the rotation and reflection resp

That is the basic idea of free groups.

Tobias Kildetoft

right

which is why you want to find generators of $H$ satisfying the relations that the generators of $G$ satisfy

Peter Tamaroff

Now, there is a line where he just writes

$$\tag 3 x^n,y^2,xyxy$$

and he says "Let $K$ be the normal subgroup generated by the elements $(3)$"

He means we should look at the normal subgroup generated by $x,y$ such that $(3)$ is fullfilled right? And it really should read $x^n=1,y^2=1$ and $xyxy=1$

@tobias Yes?

Tobias Kildetoft

no, he means the normal subgroup generated by those elements

Peter Tamaroff

18:16

But aren't they $1$?

Tobias Kildetoft

no

Peter Tamaroff

Oh...

I see.

Tobias Kildetoft

they become 1 in the quotient

Peter Tamaroff

Right.

Tobias Kildetoft

(which is what we want)

Peter Tamaroff

18:17

But why does he now write "by the relations..." and writes a set of equalities.

??

When he talked about relations as just elements we want to be 1?

Tobias Kildetoft

@PeterTamaroff this is because any equality in a group can be rewritten as something = 1

Peter Tamaroff

Now we have $x_3x_1x_2=x_1x_2$, $x_1x_3=x_3x_1$ and $x_2x_3=x_3x_2$

Tobias Kildetoft

sometimes, we like to write it as equalities in one way, sometimes the other

Peter Tamaroff

So we want to look at the group generated by $^{x_1}x_3$, $^{x_2}x_3$ and $^{x_2x_1}x_3$ (the third one is awry)

Tobias Kildetoft

what do you mean by that notation?

usually, it is not practical to try to actually identity the normal subgroup generated by the elements

Peter Tamaroff

18:21

$^g x=gxg^{-1}$

Tobias Kildetoft

since the quotient is what we are interested in, and that is much easier to describe

ahh, that is usually written as $x^g$ (that I have seen)

Peter Tamaroff

@TobiasKildetoft Ah, OK.

This guy uses that notation.

Tobias Kildetoft

and no, those are not quite the generators of the subgroup we want

Peter Tamaroff

@TobiasKildetoft Yes, I know. I wrote some wrongly.

Sorry.

I wrote them all wrongly.

They are $x_1x_3x_1^{-1}x_3^{-1}$

Tobias Kildetoft

anyway, the reason they have been written like this is that they tell you how to permute the generators

Peter Tamaroff

18:23

$x_2x_3x_2^{-1}x_3^{-1}$

Tobias Kildetoft

which is nice, because it tells us that we can write everything in a certain way

specifically, we can write any element in the quotient as $x_1^{n_1}x_2^{n_2}x_3^{n_3}$

Peter Tamaroff

But $x_2$ and $x_1$ dont commute.

Tobias Kildetoft

which is good, because it gives us an idea of what sort of generators of $H$ we should be looking for

Peter Tamaroff

$x_2x_1=x_3x_1x_2$

Tobias Kildetoft

no, but you can switch their order by adding an $x_3$

maybe we should have $x_3$ as the first one in those words instead

ahh, no, we can put the $x_3$ at the end as it commutes with both the others

Theorem

18:27

anyone here knows the resource where i can find a detailed treatment of dirichlet integral

Peter Tamaroff

OK, say I give you $x_2^3 x_1^2x_3x_1^{-1}x_2^3$ @tobias

@Theorem $$\int_0^\infty \frac{\sin x}{x}dx=\frac {\pi}{2}$$?

Tobias Kildetoft

@PeterTamaroff rewrite the relation as $x_2x_1 = x_1x_2x_3$

this then shows that $x_2^kx_1 = x_1x_2^kx_3^k$

Theorem

@PeterTamaroff : no , i am talking about $\int_\Omega Du dx$ $\Omega \subset \mathbb R^n$

Peter Tamaroff

@TobiasKildetoft That is not the relation. It is $x_3x_1x_2=x_2x_1$

@Theorem No clue then.

Tobias Kildetoft

@PeterTamaroff but $x_3$ commutes with everything

Peter Tamaroff

18:32

@TobiasKildetoft Oh, right.

@TobiasKildetoft OK, I see that.

For $k>0$.

Tobias Kildetoft

in general, it can be very tricky to work with groups given in this fashion since one can easily overlook some relation that was not obvious

and of course, given some presentation, there is no general way to even tell if the resulting group is trivial

(it is an undecidable problem)

Peter Tamaroff

@TobiasKildetoft I think MJD said that yesterday =P

OK so I gotFUUUUUUU the coffee!

brbr

I'm here.

I got my invented element is $x_1x_2^2x_1^{-1}x_2^3x_3^3 $

I need to get those last elements in order.

Tobias Kildetoft

right

you can just move the $x_3$ to the far right end with no problem

Peter Tamaroff

@TobiasKildetoft Oh, OK.

And now I have to swith that $x_1^{-1}$

Tobias Kildetoft

to see what happens with the $x_1^{-1}$ try to play a bit with the relation involving $x_1$ and $x_2$ to get something with $x_1^{-1}$

Peter Tamaroff

18:42

OK.

On it...

I know!

$ x_1^{-1}x_2 x_1=x_2x_3$

I can use that.

Till I get to the $x_3$ in the vert end.

And use commutativity.

This feels like playing chess, man.

"What move can I make?"

Tobias Kildetoft

indeed

the reason I mentioned that we can rearrange like this is that we then get a "canonical" way of writing elements

also, we can check that this way is actually unique (ie, that the exponents are uniquely determined)

Peter Tamaroff

Right.

Tobias Kildetoft

(this is essentially because no elements have finite order)

Peter Tamaroff

So we can induce the isomorphism via the exponents.

And then we can see the multiplication is the same

And the relations reflec on the exponents.

Yes?

Johan Larsson

What font does Math.Se render its latex in?

Peter Tamaroff

18:48

I think I am starting to see it clear now.

Tobias Kildetoft

I think so (I didn't check if this gives the correct relations)

good

Peter Tamaroff

@TobiasKildetoft I mean, the unusual multiplication of $x_1+x_2+x_2y_3$ in the first coords

Is just the non commutativity we have.

Right?

@JohanLarsson I guess it is the default tex font.

Tobias Kildetoft

@PeterTamaroff yes, but I am not sure if $x_1$ should correspond to $(1,0,0)$

Peter Tamaroff

@TobiasKildetoft Oh. But the iso is just mapping generators to generators.

Tobias Kildetoft

@PeterTamaroff well, suitably chosen generators

Peter Tamaroff

18:51

We have to see what the gens are. Yes.

Tobias Kildetoft

so we want to find generators of $H$ that satisfy our 3 relations

Orange Harvester

@JohanLarsson STIX if installed on your computer, else default TeX, as given here.

Tobias Kildetoft

probably $x_2$ and $x_3$ can be chosen to be $(0,1,0)$ and $(0,0,1)$

Peter Tamaroff

@TobiasKildetoft RIght.

@TobiasKildetoft Yes.

@TobiasKildetoft It is pretty amazing how we can present some groups by using $FG$ and few relations.

It is the basic idea under presenting subspaces and linear equations right?

With vector spaces and subspaces.

Rings a bell, at least.

Tobias Kildetoft

@PeterTamaroff we can get all groups this way actually

Peter Tamaroff

18:56

@TobiasKildetoft WOAH!

Tobias Kildetoft

it is in fact really trivial to do. Take the free group generated by the elements of the group you want (need not be finite), and take the normal subgroup generated by all the relations satisfied by that group (might be an enormous amount)

this is of course completely impractical

Peter Tamaroff

@TobiasKildetoft Wait.

But I think all groups aren't "finitely presented".

Right?

Tobias Kildetoft

right, we might need a non-finitely generated free group to start with

but this is ok

for a countably generated one, take the commutator subgroup of the free group on two generators

(this is a strange thing about free groups. Any subgroup is also free, but might be generated by a lot more elements)

Nimza

Hi! If $f \colon \mathbb{R}^n \to f(\mathbb{R}^n)$ is such that $f''(x)>0$ is it true that $f$ is diffeomorphism?

Tobias Kildetoft

in fact, any free group that is not cyclic contains the free group on any number of generators (at most countably many) as a subgroup

Peter Tamaroff

19:09

@Nimza Looks good. If $f''>0$, $f$ is striclty increasing, whence $f$ is one one. Note now that $(f\circ f^{-1})'=1$ that is $f' \circ f^{-1}\cdot (f^{-1})'=1$

user19161

@PeterTamaroff So which classes are you doing this semester?

Peter Tamaroff

This means that $(f^{-1})'=1/(f'\circ f^{-1})$

You must now check that $f'\circ f^{-1}$ never vanishes, @nimza

Can you deduce it from $f''>0$?

@JacobBlack Analisis 1 and Algebra 1

:8228794 Oh, didn't look that, hehehe

I'm still in my $\Bbb R^1$ paradise.

Nimza

@PeterTamaroff yep, the problem is with injectivity when $n$ is general

@PeterTamaroff why are you sitting here?) in line)

Peter Tamaroff

@Nimza Ah?

Nimza

@PeterTamaroff why are you still on $\mathbb{R}$?)

Peter Tamaroff

19:13

@Nimza Oh, I haven't taken any $n$-dimensional analisis yet.

Nimza

@PeterTamaroff aa

Peter Tamaroff

@Nimza I actually start this cuatrimester.

The program is:
Topology in $\Bbb R^n$
Functions $\Bbb R^n\to \Bbb R^k$
Differential calculus in many variables
Extrema of multivariate functions
Double and triple integrals

I actually know all of the first part. (topology)

Nimza

:) general topology of course (you know)?

user58512

hello

Nimza

@user58512 hi

user58512

19:17

hi @Nimza, do you know about elliptic addition theorem

Nimza

@user58512 I had it in my course the last semester, but I don't remember details

Peter Tamaroff

@Nimza Right. But it is not much. It is: completitude, supremum and equivalent defs, distance, open and closed disks, interior points, interior, closure, closed sets, bounded sets, limits, limits in coords

Dunno if we'll see something about compactness for example

I hope so.

@JacobBlack Dude. I think I know most of what is given in Algebra I.

user19161

@PeterTamaroff Well done Pedro, I have high expectations of you. Don't forget me when you win the Fields medal.

Peter Tamaroff

@JacobBlack Haha, you'll jinx me.

Orange Harvester

@PeterTamaroff I wish I could say the same about Algebra Chapter 0.

Peter Tamaroff

19:20

@OrangeHarvester What is that?

user19161

@OrangeHarvester That book is becoming very popular, I don't know why...

Peter Tamaroff

@OrangeHarvester But it says it is graduate level!

Orange Harvester

@PeterTamaroff Before Algebra I comes Algebra chapter 0. :P

user58512

I am very frustrated I can't find a development of some results I am curious about

Orange Harvester

@PeterTamaroff I have high expectations (and no hard work) from myself.

user58512

19:21

I spent a long time in the library today looking in the books for it

Peter Tamaroff

@OrangeHarvester So you'll have ODing expectations if you worked hard!

Orange Harvester

@PeterTamaroff Yes!

Peter Tamaroff

You'd have to freaking stick a needle full of adrenaline in their hearts if you worked hard.

Damn you!

Orange Harvester

@PeterTamaroff Yeah! IV would be my right hand.

user19161

@user58512 What result?

Nimza

19:23

I don't know how do people find motivation to learn algebra, it's so dry in comparison with topology or some applied fields like optimal control

Orange Harvester

@Nimza Learn some computer science dude!

Nimza

@OrangeHarvester what do you mean?

user19161

@Nimza Well, to find motivation, connect it to geometry, like algebraic geometry or transformation groups.

Nimza

@JacobBlack aha, algebraic geometry is the point but one doesn't tell about it when one speaks about algebra in uni!

user58512

@JacobBlack, addition theorem for trig integrals

user19161

19:24

Many things become more beautiful when we see it geometrically, after all we are all visual creatures.

Orange Harvester

@Nimza Well, computer science is heavily invested in semigroup based algebra. But, you can analyze complex systems in computer science using algebra.

Nimza

@JacobBlack yes, I'm agree! And it's my point of view too

Orange Harvester

@Nimza Here

Nimza

@OrangeHarvester hm, I was linked almost only with computer science first 3 years, and I didn't use algebra to write such things as http servers of javascript interpreter. O thanks for the link

Orange Harvester

@Nimza That was programming, not computer science. ;-)

Nimza

19:26

@OrangeHarvester ah)

user58512

can you help me

Ilya

hi everybody

Nimza

@Ilya hi

Peter Tamaroff

@tobias

Ilya

@Nimza watched any good movies recently?

user58512

19:30

hello

Orange Harvester

@JacobBlack It introduces category theory form the outset, something of interest to those more inclined towards computer science.

Peter Tamaroff

In the group $\Bbb Z^3$ with multiplication as usual but in the 2st coord $x_1+y_1+x_2y_3$ two elements will commute $\iff x_2y_3=y_2x_3$

Nimza

@Ilya watched Kevin Smith's 5 movies from the Askewuniverse

Ilya

@Nimza have no idea what is it. Sci-fi?

user58512

:(

Nimza

19:34

@Ilya no, usual films about life. First - Clerks - is about 2 vendors. All the action takes place in their shop :) It isn't well known film as I know but I liked it so much. As the others

@Ilya Chasing Emy - is about a commicwriter who loved lesbian

Ilya

aaaaa

Jay and Silent Bob

Nimza

@Ilya aha, "Strike Back" is the worst one for me

Ilya

I've seen only that one

Dogma is from the same series?

Nimza

it's a pity :(

aha, Dogma is one of the best

Ilya

I was told that a good one

I'll maybe watch it soon

I watched Moon 2112 today, quite a good one! @Nimza

Nimza

19:40

@Ilya thanks, will add it to my empty watchlist

Ilya

@Nimza I can send you mine

Nimza

@Ilya go gtalk

Ilya

I'm there

Peter Tamaroff

@tobias I think we can choose the generators as $x_1=(0,1,1),x_2=(0,0,1) $ and $x_3=(1,0,0)$; in that order.

FUUUU

Once again.

I wrote an eqn wrongly.

OK $abc=cb$ forces $a=(a_1,0,0)$ in the first place

While $bc=cb$ and $ac=ca$ forces $a_2c_3=c_2a_3$ and $b_2c_3=b_3c_2$

The first coord in $abc=cb$ gives after use of $a_2,a_3=0$ that $c_1=0$.

Dominic Michaelis

19:55

@charlie hi sweetie

Peter Tamaroff

@DominicMichaelis "Oh, we are sweetie now?"

Dominic Michaelis

only charlie and me

Charlie

@DominicMichaelis hi Domy

@PeterTamaroff hola Pedro

@DominicMichaelis how are you?

Transcript for

$Mathematics$ Mathematics