Mathematics - 2012-03-25 (page 2 of 3)

Rajesh D

14:00

Hi folks

Kannappan Sampath

Hi!

N3buchadnezzar

cya

Rajesh D

howz the long weekend @Kanna

Kannappan Sampath

@RajeshD Has not been productive, but, yes it was fun!

Rajesh D

@N3bu : just a quick question : are you from India ?

@Kanna : fun is what is the product of weekend

ymar

14:03

Hello!

Kannappan Sampath

@RajeshD Remember what TheChaz had to tell you?

@ymar Hi!

ymar

Could somebody please help me with a field theory question?

Rajesh D

@Kanna : I do not

yeah I've seen but i could not understand what he is intending to convey really

Hmmm....

Kannappan Sampath

@ymar My knowledge of FT is zero, a fact you know well! : )

ymar

@KannappanSampath I did not know that!

Rajesh D

14:07

But I do not pester them.....aslong as i do not pester i think its ok....i just leave it to their prerogative @Kanna....

Kannappan Sampath

@ymar Well, you know it now. :-P

@RajeshD But, I would say don't even ask!

ymar

@KannappanSampath :-( Eh... I guess I'll have to fight this myself.

Kannappan Sampath

I guess @anon, can help you! @ymar

anon

probably not, but just ask the question anyway.

ymar

Thanks! I don't think it's difficult. I was reading this

And the second paragraph doesn't make sense to me.

Kannappan Sampath

14:12

Yes, it is written little clumsily, while I think I get what it says.

ymar

I don't get it at all. I think what he's saying is false.

So he takes $F_p$ and asks what are its automorphisms, right?

Kannappan Sampath

Yes.

It's dinner time. I really have to go. I will be back in half an hour.

I am sorry!

ymar

See you!

And he says that every such automorphism raises the generator $r$ to a power that's a multiple of $p$.

That's true in the sense that there is only one automorphism of this field and it is given by raising every element to the $p$-th power.

Skullpatrol

Obnoxious, attention seeking behavior that may use incorrect or upsetting statements.

ymar

But if we take $F_5$, then $x^9$ is an automorphism and $9$ is not a multiple of $5$...

Dylan Moreland

14:18

Why is that an automorphism?

ymar

Because $2$ is a generator of the multiplicative group and $2^9=512\equiv 2 \mod 5$

Dylan Moreland

The multiplicative group is cyclic of order $4$, so I agree that it induces an automorphism of that group.

ymar

It's an identity map in $F_5$

No?

So it is a field automorphism...

Dylan Moreland

$\mathbf F_5$ can't have any non-identity automorphisms, since $1$ generates the field.

ymar

Yes! My goal is to prove that $F_{p^n}$ has an automorphism group which is cyclic of order $n.$ What you said is a special case.

Dylan Moreland

14:24

But you're right, it's interesting that this power is the identity on this field.

ymar

@DylanMoreland Have you seen the link?

Dylan Moreland

I've not seen any link.

In my time here today.

ymar

I'm talking about all of this because I cannot understand what this means. (The second paragraph.)

Dylan Moreland

That site is sort of all over the place.

ymar

Do you mean there are mistakes there?

Dylan Moreland

14:30

I just think that the writing isn't very careful.

There are some gems there every once and a while that come up in a Google search. But I don't think it should be a primary reference for anyone.

ymar

Thanks for letting me know! So where should I look for help with this theorem? (Automorphism group of $F_{p^n}$ is cyclic of order $n$)

I see that it's true for $n=1$.

And I see that every automorphism must be of the form $x^k$

Dylan Moreland

There's always a Keith Conrad handout, or Milne's notes on Field and Galois Theory.

It depends on what you know.

ymar

I just know that $F_{p^n}$ is the set of zeroes of $x^{p^n}-x$ in $\overline{F_p}$

Dylan Moreland

That's a good definition.

ymar

From this I know that the Frobenius automoprhism $x^p$ has order which divides $n$

Dylan Moreland

14:41

Oh, really? You're in really good shape then.

Assume that it has a smaller order. Say $\varphi^d$ is the identity, i.e. that $x^{p^d} = x$ for all $x$ in the field.

ymar

OK.

Dylan Moreland

You can write down some polynomial now. Polynomials can only have so many zeros.

ymar

Ah! :)

Kannappan Sampath

Back!

Looking at the discussion, everything is way over my head.

What I thought it said was clearly false!

ymar

I understand! $x^{p^d}-x$ can only have $p^d$ zeroes which is less than $p^n$. Right?

@KannappanSampath Welcome back! :)

Kannappan Sampath

14:46

May be we should also have Dylan help us with the confusion over the saturation and the definitions there!

@Dylan Will you help us with a bit of CA?

(of course after this discussion)

ymar

So I think I understand why the Frobenius automorphism has order $n$. So the order of the automorphism group is a multiple of $n$.

anon

@Kanna: How can I see that G-equivariant maps transitively permute orbits of the same type? I think I would have to exhibit an $h$ such that $G_x=G_{hf(x)}$, but I'm not sure how to do that.

Kannappan Sampath

I don't even know how is G-equivariant defined! : (

anon

if G acts on a set X, maps $X\to X$ are G-equivariant if $f(gx)=gf(x)$ for all $x\in X, g\in G$.

Kannappan Sampath

Oh, I see. Let me think then.

Well, can't we do something along lines of this:

We have $\cdot: G \times X \to X$.

A map from $f: X \to X$.

Define another map: $ \circ: G \times X \to X$ by $(g,x) \mapsto g \cdot f(x)$

Now, prove that this defines a group action and that these two actions are equivalent.

I think equivalence of group action follows from the very definition, I guess.

ymar

15:07

Eh, Dylan's gone. No hope for me then. And I have to go. See you later!

Kannappan Sampath

Hmph! Sad! I think he's quite busy these days!

anon

equivalence of actions means one action is the other precomposed with a bijection, right?

Kannappan Sampath

Yes, right!

anon

Okay, so the second part is easy. How do I find the inverse action of $g$? We have $g^{-1}\circ g\circ x = f(f(x))$, which isn't necessarily $x$...

wait, I'm not sure if there's even an identity action, i.e. an $h\in G$ such that $hf(x)=x$ for all $x\in X$.

t.b.

@anon: what do you mean by orbits of the same type?

anon

15:13

I think it means the conjugacy class of the stabilizer of each orbit is the same

Kannappan Sampath

Hmph, Complicated stuff man!

Ignore: What I said blatantly fails!

t.b.

But then I don't understand the statement. Suppose $y$ is a fixed point of your action of $G$. Then the function $f: X \to X$ given by $f(x) = y$ is $G$-equivariant but doesn't seem to transitively permute orbits of the same type.

anon

Sorry, I forgot to mention: equivariant maps are also bijections on the underlying set.

Dylan Moreland

@ymar Well, here's where you would want to know some field theory, or at least some fact that told you that if $L/K$ is a field extension then $\operatorname{Aut}(L/K) \leq [L : K]$.

t.b.

Ah, I would call that an equivariant bijection.

ymar

15:17

I'm back :)

Dylan Moreland

And the key step for that would be showing it for a simple extension: I adjoin a root $\alpha$ of an irreducible polynomial $f(X) \in K[X]$ of degree $n$, so I have an extension of degree $n$. Then the only places I can send $\alpha$ are the other roots of $f$, and there are at most $n$ of these in any extension, not to mention lying in $L$.

anon

@tb: Ah you're right. The original exercise says "group of symmetries of a G-action" which are equivariant maps that are also bijections.

t.b.

As a first simple observation note that $h$ sends orbits to orbits and that $h^{-1}$ is equivariant, too.

Matt N.

Shouldn't the first sentence start with "finite" rather than "cyclic"?

ymar

@DylanMoreland Does $\operatorname{Aut}(L/K)$ mean the group of automorphisms of $L$ fixing $K$?

Dylan Moreland

15:21

@ymar Yep.

Kannappan Sampath

@anon Where is this exercise from? : )

anon

@t.b. What's $h$? What if $G$ is not commutative?

Dylan Moreland

@MattN Does it make a big difference?

t.b.

Sorry $h$ is an equivariant map.

Kannappan Sampath

@MattN But, it seems to make sense to me.

Dylan Moreland

15:22

They're certainly cyclic groups.

anon

@Kanna: Lie Groups: An Approach through Invariants and Representations, second exercise on page 6.

Matt N.

@DylanMoreland No. It doesn't. : )

ymar

@DylanMoreland I think I get it! Thank you very much! :)

anon

@tb Okay, I follow you.

Matt N.

Thanks.

anon

15:22

Ah...

Dylan Moreland

@ymar A pretty comprehensive free source for this would be Milne's notes.

ymar

@DylanMoreland Thanks. I think what you said is enough, right? Because $F_{p^n}/F_p$ is a simple extension. I remember proving that, although I don't remember the proof. But I should be able to recall it.

t.b.

@anon: So let $Gx = \{gx\,:\,g \in G\}$ be an orbit and let $h$ be equivariant. Then $h(gx) = gh(x)$, so $Gh(x) = h(Gx)$. Similar reasoning applied to $h^{-1}$ shows the statement you're after, I think. Note that an equivariant bijection gives a map $\bar{h}: X/G \to X/G$ and this map must be invertible (the inverse is $\overline{h^{-1}}$).

Kannappan Sampath

This looks like what I was talking about, no? @tb

But, yes, not ofcourse, this explicitly. Given I saw the definition for the first time, I think I guessed what was happening, good enough!

ymar

In any case, I'm pretty sure I can do it from here. Thanks again, Dylan and others.

anon

15:29

$G_x h(x)=h(G_x x)=h(x) \implies G_x\subseteq G_{h(x)}$. Same with $h^{-1}$, so $G_{x}\subseteq G_{h^{-1}(x)}$, which is $G_{h(x)}\subseteq G_{x}$. Put these two together and we have $G_x=G_{h(x)}$.

t.b.

yes exactly.

Dylan Moreland

@KannappanSampath Anyway, what's this about commutative algebra?

Kannappan Sampath

@DylanMoreland We have not been able to decide which of the two competing definitions to take for multiplicative set, $S$. One explicitly tells us $0 \notin S$, and the other $0 \in S$.

Both have their own issues!

Zhen Lin

@KannappanSampath: I would allow $0 \in S$, because you could just as well allow any zero divisor to be in $S$.

Once any zero divisor is in $S$, inverting $S$ will annihilate the ring.

Kannappan Sampath

If we assume $0 \in S$, we are forced to conclude that an empty set is the union of all prime ideals.

Zhen Lin

15:34

I said allow, not require!

Kannappan Sampath

Right, so, state the proposition by explicitly saying $0 \notin S$ or something like that?

Zhen Lin

I wouldn't single out $0$ in the context of localisation. $0$ is as bad as any zero divisor.

Kannappan Sampath

Hmph, I do not know what localisation is.

Zhen Lin

Then what are you looking at multiplicative systems for? :-|

Kannappan Sampath

(I regret that I did not read Adkins fully!)

@ZhenLin We proved that lemma that would say any ideal maximal with respect to excluding a multiplicative set is prime. Then, we looked at saturations. We proved that there is a smallest saturation that contains a multiplicative system.

Then, we solved some exercises from Kaplansky regarding these things.

So we have been largely doing it for their own sake rather than for something else.

Dylan Moreland

15:40

You shouldn't require $0 \in S$.

Zhen Lin

Sounds like you're secretly doing localisations to me...

(these exercises make a lot more sense in that context!)

Dylan Moreland

If $0 \in S$, then there are no primes disjoint from $S$, so you can't find maximal elements of the set of primes disjoint from $S$. That's fine.

No contradiction there.

Kannappan Sampath

YES. Thanks to vacuousness.

@ZhenLin Likely. We did more things there.

Like we explicitly described those smallest saturations.

Chris

Hello, I have a homework on numerical analysis, and I am confused about rounding a number.

I 've found an algorithm about rounding, but I am not sure...
For example, given this number 586, using rounding method of the 3 significant digits, should it be 586 because the 3rd significant digit + 1 is "0" so we leave the significant digit as it is?

Or we should write it as 587, hence 6 > 5 ?

which method should I follow?

anon

I don't understand what the second method is.

Kannappan Sampath

15:47

And, this all arose when we want to define an object to be generated by a set $X$ if it is the smallest saturated set containing $X$. The very existence faced crisis! @DylanMoreland @ZhenLin

Rajesh D

Hey @anon

anon

yo

Rajesh D

how was the weekend

anything special ?

anon

it's still the weekend! but yesterday I helped re-roof and re-shingle a garage. now I'm tired and achy.

Chris

@anon : the 2nd method is what I know in general, we check the digit if it is >5 or <5 and if it's the 1st case, we have this 586 +1, or if it's the 2nd one, we do nothing..in example: 583 -> 580

is that correct? :S

anon

15:52

why +1 and not +4?

also, check which digit? ;)

Rajesh D

@anon : thats a nice way to spend a weekend......doing something constructive

Kannappan Sampath

sigh I am off! I'll figure it out probably later. Thanks @Dylan @Zhen

Chris

Correct, ok to get this straight..we've been given some calculations to do (additions, subtractions)..and we've been told to chop and round the numbers using the 3 significant digits..

So given this one:
$(-0.954 - 24.0) + ( 246.0 +365.0)$ how rounding would be?

anon

Significant figures

The significant figures (also called significant digits) of a number are those digits that carry meaning contributing to its precision. This includes all digits except: * leading and trailing zeros which are merely placeholders to indicate the scale of the number. * spurious digits introduced, for example, by calculations carried out to greater precision than that of the original data, or measurements reported to a greater precision than the equipment supports. Inaccuracy of a measuring device does not affect the number of significant figures in a measurement made using that device, althou...

go to the "rounding" section

Chris

@anon: Thank you :) I am not familiar with the terminology in english and it was hard for me to look for this!

robjohn

16:14

@hhh You might get a better response if you put that into LaTeX

Rajesh D

Hi @rob

how was the weekend

Matt N.

Heh. Iyengar is now into homology and cohomology.

robjohn

@RajeshD It is quite fine. We are expecting some rain later today.

Rajesh D

cool

robjohn

@JonasTeuwen It looks okay.

Rajesh D

16:27

I hope it rains here too...but it will take a month or so for the Monsoon to start..south west monsoon

hhh

$$\partial_{x}(\cos(x)\partial_b)=-\sin(x)\partial_b+\cos(x)\partial_x\partial_b,$$ is this right?

Rajesh D

I just wonder without any specific reason...what is the hottest 'tag' on the site ?

anon

@hhh: Yes, that's correct.

hhh

@anon Thanks, then the bug is fixed :)

hhh

16:49

d<-1e4;R<-9;r<-2;a<-0:(d*pi);b<-0:(d*pi); plot3d(x=(R+r*cos(a))*cos(b),y=(R+r*cos(a))*sin(b),z=r*sin(a))

<-- what is wrong with my torus formula in R?

Jonas Teuwen

@robjohn Good :-).

N3buchadnezzar

BADEKAR

hhh

Must be with the angles...

d<-1e5;R<-9;r<-2;a<-0:(d*pi)/30;b<-0:(d*pi)/40; plot3d(x=(R+r*cos(a))*cos(b),y=(R+r*cos(a))*sin(b),z=r*sin(a))

Now it looks a bit like torus :P

anon

what does d<-1e5;R<-9;r<-2;a<-0:(d*pi)/30;b<-0:(d*pi)/40 mean? the formula inside plot3d looks okay to me.

hhh

d=density

1 000 00 points

a and b are angles

and there is the err because it does not cover all of them easily

0

Q: iGraph or some other Lib to Draw Torus with symbol -markings?

I need to draw this torus here (where you connect the horizontal and vertical sides to get the 3D object. The iGraph package has at least random geometric graphs something, more here, but apparently not for drawing this kind of detailed torus below. R: a bit torus-looking but without labels an...

In ubuntu, one needs the RGL lib to draw things...

...but there is surface3d --command which is probably better here...

I am just trying to get random numbers between 0 and 2pi there with d, a and b...

ymar

17:27

I've just reached 2000 rep! :)

Matt N.

@ymar Yay! Congrats : )

ymar

Thanks!

Matt N.

@tb But for a finite cyclic group $G$ it's not $G \leftrightarrow S^1$, it's $G \leftrightarrow \mu_n$ where $\mu_n$ is the group of $n$-th roots of unity, right?

(Assuming $|G| = n$.)

N3buchadnezzar

@ymar How is it going ?

=)

ymar

@N3buchadnezzar Everything's fine, thanks. :)

N3buchadnezzar

17:40

=)

ymar

Still integrating?

N3buchadnezzar

I have to do philosophy!

It blows!

ymar

Philosophy is the most boring thing people have invented.

N3buchadnezzar

@ymar Can one invent philosophy?

ymar

@N3buchadnezzar Sure, I can but I don't want to!

N3buchadnezzar

17:45

I want to differentiate philosophy out of my course list... Because differentiation is the opposite of integration.

...

Kannappan Sampath

Since I live in India, I forsee a danger of "iyen" bug biting me! Why the hell is this guy crazy trying to but together some words in logical fasion and succeeding almost every time luckily?

hhh

$$\partial_{x}(\cos(\theta)x)=\cos(\theta),$$ is this right? (I am practising polar coordinates...) @anon

Matt N.

@ymar : )

Kannappan Sampath

BSD, Hodge and all of that.

N3buchadnezzar

???

Kannappan Sampath

17:48

@N3buchadnezzar Do you know that user named Iyengar?

anon

@hhh: is $\cos\theta$ independent of $x$?

N3buchadnezzar

@KannappanSampath no?

hhh

@anon I don't know, I just switched from Cartesian $x,y,z$ to $r,\theta, \phi$ -- is it or not?

Kannappan Sampath

@N3buchadnezzar You'd be good if you just saw his profile and no more!

anon

@hhh: If they aren't independent, then $\cos\theta$ is not a constant, no?

hhh

17:51

$$\partial_{x}(\cos(\theta)x)=\cos(\theta)-\left(\sin(\theta)(\partial_x\theta)\right)x,$$ or this?

@anon If dependent, then $\cos(\theta)$ is not constant. That is correct. So we have to have the other term -- but I am unsure here, well I will have the other term nevertheless or? (it is just a zero term in independent case)

anon

well, that is correct, but not fully simplified, and probably not the right tact. just curious, was the original expression $\partial_x r$?

hhh

@anon No, I can show original expression later when I have done some calculations more -- this was just a subproblem...

anon

If you don't see that $\theta$ varies with $x$ then I'm not sure you understand polar coordinates. If you rotate something about the origin then of course it's x-coordinate will vary along the angle.

hhh

@anon Yes and that is why need also the second term, yes that is right -- good to get it clear :)

N3buchadnezzar

@KannappanSampath I read some of his answers and questions, he does not seem that bad no ?

anon

17:56

anyway, $x=r\cos\theta$, so $\partial_x(\cos\theta)=\partial_x(x/r)$, which you can do with the quotient rule and $r=\sqrt{x^2+y^2}$

you could also do that prior to the product rule, i.e. for $\partial_x(x\cos\theta)$.

N3buchadnezzar

Although he seems strange, I liked his question about Galois theory !

31.98, 32.06, 36.99, 19.10, 29.02, 27.32, 30.10, 32.50, 22.27, 28.38, 30.58, 35.74, 31.24, 33.71, 31.91, 28.19, 31.52, 29.46, 27.30, 21.72, 32.74

New personal best with one hand =)

Kannappan Sampath

@N3buchadnezzar ??? What are these array of numbers?

N3buchadnezzar

rubiks cube, one handed

Kannappan Sampath

I see.

@Matt How far is Additive Combinatorics done?

Hope to see you here as often as a normal week would bring you here! : )

t.b.

@MattN I don't see a duality here. You can send a cyclic group $G$ of order $n$ to $S^1$ by choosing a primitive $n$-th root of unity $\mu_n$ and sending a generator of $G$ to $\mu_n$. But neither the choice of generator nor the choice of $\mu_n$ are canonical.

Kannappan Sampath

18:12

Meta discussions shed more light.

t.b.

?

Kannappan Sampath

Hmm, MO-meta sheds more light. Is that better?

Well, ignore. I am grumpy today, I guess! I am irritated by what I see as happening....

t.b.

Okay, I'm ignoring :) There isn't much going on on meta.MO, these days...

Matt N.

18:31

@KannappanSampath Sorry, was afk. I don't know yet. I've got quite a busy term this term. But I'll be doing CA again soon.

Kannappan Sampath

@MattN Hope you're OK. Please join us sometime. We have been doing some multiplicative sets stuff and found things interesting! : )

Matt N.

I will don't worry : ) Just need to get some stuff done.

Kannappan Sampath

:-)

Matt N.

:,( But for $G = \mathbb Z / n \mathbb Z$ I can define an isomorphism from $G \to \widehat{\mu_n}$ as follows: $k \mapsto (\omega \mapsto \omega^k)$, no?
Also, $\mu_n \to \widehat{G}$ via $\omega \mapsto (1 \mapsto \omega)$.

Yes, the latter is what you mention there.

I don't understand "canonical".

hhh

@anon Perhaps this q shows the idea I cannot understand here, I cannot understand the premise about the $z$ there with $\hat e_R$. I am uncertain whether it is a geometric interpretation or some other way?

Matt N.

18:40

Ello : )

t.b.

Hi

@MattN yes, you can. The point is that there are choices involved in this isomorphism. There is no distinguished element $1$ in $G$ and there is no distinguished primitive $n$-th root of unity. You have to pick them. That's what is meant by "not canonical".

Matt N.

So what I wrote works?

And gives me the "duality"?

t.b.

No, the duality is given by $\hat{G} \times G \to S^1$, $(\gamma,g) \mapsto \gamma(g)$.

In this duality there's no choice involved. By definition an element of $\hat{G}$ is a homomorphism $G \to S^1$, so you can evaluate that homomorphism on $g \in G$. *This* is the pairing (analogous to the pairing you have between a vector space
and its dual space).

Matt N.

Thanks. Poor me. You just turned my productive day that I thought I had into a puff of smoke. :,(

@tb Thank you.

Asaf Karagila

One... last... diagram chase...

Matt N.

18:47

I thought duality meant that I can map a homo in one thing to a thing in the other thing and vice versa via an isomorphism.

Asaf Karagila

Must... not give up... set theory lies beyond!

t.b.

But you can show that there is an isomorphism between a finite abelian group $G$ and its dual group, but there is no distinguished such isomorphism. The point is that you can reduce to the case of a cyclic group and the map you wrote down can be interpreted as giving such an isomorphism. (by fixing a generator of $G$ and declaring that it is sent to $\omega^k$ - this determines a unique homomorphism $G \to S^1$ and every homomorphism $G \to S^1$ is of this form).

@MattN No, that's not the case. The dual group of $S^1$ is $\mathbb{Z}$ and the dual group of $\mathbb{Z}$ is $S^1$. You cannot relate those groups directly via an isomorhpism, but they are related by the duality.

Matt N.

@tb Yes sorry, that's what I meant. I know that, since I proved it the other day.

And that's why I thought I could just do the same for a finite cyclic group.

t.b.

What do you mean by "the same"?

Matt N.

There is an isomorphism from $\widehat{\mathbb Z}$ to $S^1$ and from $\widehat{S^1} $ to $\mathbb Z$. So I thought by giving an isomorphism from $\widehat{G}$ to $\mu_n$ and from $\widehat{\mu_n}$ to $G$ I get my duality.

t.b.

18:55

Well, that is true but maybe I misread your notation and you were right all along. So you write $\mu_n$ for $\widehat{\mathbb{Z}/n\mathbb{Z}}$?

Matt N.

No, I'm using $\mu_n$ for the group of roots of unity.

I also replaced $S^1$ with $\mu_n$.

Because I can't map $\widehat{\mathbb Z / n \mathbb Z}$ bijectively to $S^1$.

Asaf Karagila

Well, the dream of getting both silver and bronze specialist badges at once is dead. No way I can fill 15 answers to previously unanswered questions in the next few days.

Matt N.

Thanks Teddy, I think I need to think about it some more.

hhh

@anon I tried other problem, I feel I am doing things the hard way -- is there any simpler way for this?

t.b.

@MattN I think you got it quite right, but you're putting it a bit confusingly. Fix a a primitive root of unity $\omega$ in order to identify $\mu_n = \langle \omega \rangle$. Now a homomorphism $\mathbb{Z}/n\mathbb{Z} \to S^1$ is determined by $1 \mapsto \omega^k$ and every such homomorphism is of this form.

hhh

19:02

@anon I am trying to calculate $\nabla\cdot\bar{F}$ in the paper with polar coordinates.

t.b.

Conversely, every homomorphism $\mu_n \to S^1$ is determined by what it does on $\omega$ and it must send $\omega$ to $\omega^k$. If $k \equiv l \pmod n$ then $\omega^k = \omega^l$, so such a homomorphism is determined by an element of $G$. This gives you an identification of $\hat{\mu}_n$ with $G$ (or $\hat{\hat{G}}$, rather).

hhh

I think I could reuse $F=|\bar x|^2 \bar x$ somehow, thinking...now I am doing it in manual spaghetti, ideas to improve?

Matt N.

@tb Yes. Except for I write $1 \mapsto \omega$ for any old $\omega$ in $\mu_n$. (instead of $\omega_k$, does it matter?)

@tb That's also what I have!

You just saved my day. : )

t.b.

pheew

As I said, you've got it quite right. :)

Matt N.

Thanks : ) Now I don't have to cry myself to sleep.

Asaf Karagila

19:05

@tb I think you'll enjoy this one.

Matt N.

@tb But then what was this about $\hat{G} \times G \to S^1, (\gamma,g) \mapsto \gamma(g)$?

t.b.

@MattN that is the actual duality. By picking generators of $\hat{G} = \langle \omega \rangle$ and $G = \langle x \rangle$ (here $G$ is finite cyclic) you sort of "coordinatize" this duality.

Kannappan Sampath

@AsafKaragila I think, I would like that to be granted!

Matt N.

@AsafKaragila Supersized font is also quite annoying. It makes my chat move around. Of course, now that I said this you'll have to reply to this message with a supersized font message.

Asaf Karagila

@MattN Hah, no time for that. Must chase this diagram until the sun goes down in Japan.

Supersized font, however, is a product of LaTeX so the chat has no control over that.

Plus, it's a way for me to express myself when I have finished with the Whitehead assignment. :-P

t.b.

19:09

@MattN what is really going on is that the duality gives you a map $G \to \hat{\hat{G}}$ and Pontryagin's theorem tells you that this is an isomorphism. You make this isomorphism explicit by fixing generators.

Kannappan Sampath

ymar has a nice rep. 2000 rep points! But I am not sure if this qualifies as nice!

Matt N.

Thanks. I need to let this sink in. I don't really see what the actual duality is doing yet.

I mean I see what it's doing: it's evaluating a character.

But I don't yet see what this has to do with the two isomorphisms.

t.b.

Remember what is going on for vector spaces: you have a pairing $V^\ast \times V \to \mathbb{R}$ given by $(\varphi,x) \mapsto \varphi(x)$. This gives you a map $V \to V^{\ast\ast}$ by sending $v$ to evaluation at $v$.

Matt N.

Yes.

Oh.

@tb Thanks : )

leo

nice one

t.b.

19:17

@MattN: I guess you can see now that you do exactly the same thing when you identify $\mathbb{Z}/n\mathbb{Z}$ with $\hat{\mu}_n$.

@leo heh, thanks :)

leo

:-)

(-:

Kannappan Sampath

@tb I started doing the section called $\S$Diagrams from Bourbaki. I have a doubt. Can you help me delete a brain glitch, I have had?

t.b.

delete the glitch?

But shoot!

Kannappan Sampath

If I have a two squares sharing an edge as my diagram, I would like to clarify what would it mean for it to be commutative.

So, I'll call the maps like $f,g$; $a,b,c$ and $l,m$

anon

@hhh: Sorry I was eating breakfast. I'll take a look.

Matt N.

19:21

Something like
$$ \begin{array}{c}
\widehat{\mu_n} \times \mu_n \to S^1 \\
(\omega \mapsto \omega^k, \omega_0) \mapsto \omega_0^k
\end{array}$$

t.b.

@KannappanSampath so, $f$ and $g$ are horizontal and composable, as well as $l,m$ and the vertical maps are $a,b,c$?

(and point downwards, say?)

Kannappan Sampath

@tb Yes. Right.

Matt N.

And then something similar for $\mathbb Z / n \mathbb Z$.

t.b.

@MattN this looks good :)

Matt N.

phew

Thanks : )

t.b.

19:24

@KannappanSampath then this is just saying that $af = lb$ and $cg = mb$.

(the two inner squares commute).

hhh

@anon I have forgot $R^2$ in front of everything but small thing...I think there could be easier way to calculate the dot-product in polar coordinates...

anon

what do $e_R, e_\theta, e_\phi$ all stand for?

Kannappan Sampath

@tb This is exactly where I would like clarification, So, I do not have worry about maps like $mbf$, right?

anon

and are you sure that $\text{div}=e_R\partial_R+e_\phi\partial_\phi+e_\theta\partial_\theta$?

hhh

@anon Unit vectors in polar corodinates?

leo

19:26

If I recall correctly, if $A$ is a matrix $n\times k$ and $B$ is a matrix $k\times n$, the matrix $n\times n$ $AB$ is never invertible. Am I right?

Kannappan Sampath

Well, did I make a fool of myself? I think I did! I got that!

t.b.

@KannappanSampath No, and I think I messed up the first relation: $bf = la$. Well, you can deduce that $cgf = mbf = mla$. In a planar diagram you just have to check the "faces".

And you did not make a fool out of yourself :)

@leo what if $k = n$? :)

anon

@hhh Sorry, I'm not familiar with those. What are they in Cartesian coordinates?

leo

@tb yes you right.

Kannappan Sampath

@tb Yeah, right. Thank you.

I learnt about that faces only now. (through you, I mean.)

hhh

19:30

@anon I know the mistake, $\nabla=\bar{e}\partial_r+\frac{1}{r}\bar{e}_\theta \partial_\theta+\frac{1}{r\sin(\theta)}\bar e_\phi\partial_\phi$ (according to my book but probbaly myself to dedude this p.817)

t.b.

Okay, I have to go for a bit. BBL

Matt N.

See you later.

Well, hopefully. : )

anon

@leo If $k<n$, yes.

leo

@anon indeed! :-)

Kannappan Sampath

I have got an integral calculus quiz coming up while I have barely managed to prove the theorems done in class. I have to read up hard and may not be around as often. So, bye for now!

leo

19:33

@KannappanSampath bye!

N3buchadnezzar

19:51

Did someone mention integral quiz?

:D:D:D

Transcript for

$Mathematics$ Mathematics