Mathematics - 2012-05-20

robjohn

12:00 AM

@BenjaminLim actually what I hear is more like "g'die" :-)

Benjamin Lim

@robjohn yes it's the kinda broad accent that steve irwin has

Zhen Lin

Yeah, Australian diphthongs are a bit odd...

Benjamin Lim

@ZhenLin youtube.com/watch?v=tatl1PBYim4

t.b.

@robjohn Oh, now I get it :) No, I found some raving reviews of them but everyone complained that they're hard to find, so I guess there's not much of promotion for them around here...

Benjamin Lim

@ZhenLin the content of the video may be a bit rude, but it gives you an idea of the stereotypical bogan/aboriginal accent

t.b.

12:03 AM

Why doesn't this guy just open a book on basic operator theory and read like the first five pages?

Mariano Suárez-Alvarez

@tb, that applies to a good 50% of questions :)

Zhen Lin

Agreed.

Benjamin Lim

What is this guy asking? math.stackexchange.com/questions/147194/…

Michael Greinecker

Well, operator theory isn't that popular...

t.b.

Deservedly so, I might add...

Benjamin Lim

12:06 AM

There are a few people I know that do operator theory and one of them is like the biggest bogan I've met, speaks like steve irwin exactly

robjohn

@BenjaminLim I think he is asking the OP for some context. I don't think he is addressing you :-) Pete Clark is no slouch.

Benjamin Lim

@robjohn huh what do you mean?

robjohn

@BenjaminLim About Pete Clark? I have seen is posts on sci.math, and he knows his stuff.

Benjamin Lim

@robjohn No I am asking about the OP and what kind of question he is asking, not about pete :D

Zhen Lin

A couple of the professors in the department are Australian. Though I haven't had the chance to hear the most famous one speak.

Benjamin Lim

12:10 AM

@robjohn whoops somehow the link to the question became a link to his comment!

robjohn

@BenjaminLim Ah! you linked the comment by Pete, so I thought you were asking about that :-D

t.b.

@ZhenLin You've got Crocodile Dundee in your department?

Benjamin Lim

nono somehow the link to the question became the link to the comment

Zhen Lin

John Coates, Ian Grojnowski, and Dan Calegari...

Benjamin Lim

@ZhenLin john coates does not even sound that australian...

Zhen Lin

12:11 AM

He's the only one I haven't heard! :p

t.b.

John Coates was here together with Hartshorne and Cartier about ten years ago due to some anniversary, that was fun :)

robjohn

@BenjaminLim You probably were looking at the question because of the comment in your inbox and then used that URL to refer to it here. I've had to be careful about that myself.

Bill Dubuque

@robjohn Perhaps you're confusing Pete with someone else because afaik Pete participated in sci.math only rarely, perhaps a handful of times at most.

Benjamin Lim

hmmmm yeah somehow I just copied the URL in the address bar and it linked to the comment.

robjohn

@BenjaminLim I have no idea what the OP is asking :-)

Benjamin Lim

12:14 AM

@robjohn He's asked another one math.stackexchange.com/questions/147199/…

robjohn

@BenjaminLim and that would be the link that you followed to look at the comment.

Benjamin Lim

yes

Jonas Teuwen

Holy monkey cows.

Good night guys.

Zhen Lin

Goodnight!

t.b.

good night

Benjamin Lim

12:20 AM

ok guys, good bye I have to go wash up, make breakfast crap it's already 10.20

robjohn

I'm off to walk my dog. I'll be back in a few hours (dinner will be in there somewhere).

jmad

Do you know if that would be better on mathoverflow? (Or maybe it is just not that interesting.)

Hello by the way.

mixedmath

hello

Rajesh D

1:32 AM

Hi @all

leo

hey

What is supposed to be a rotation on $\mathbb{R}^d$?

SO(d)?

Hi @anon

hey

@anon : how do you keep yourself motivated?

anon

1:43 AM

motivated for what, exactly? I tend to learn about things that interest me, so the motivation is naturally present.

Rajesh D

in solving and reading math

ok

leo

@RajeshD Why you are not motivated?

Rajesh D

i am finding it hard due to a peculiar reason

unlike before

leo

@anon I though so

anon

@leo ba dum tss

leo

1:44 AM

:-)

Rajesh D

I have a math problem in mind, which keeps me demotivated to read about anything else in math

anon

It's good to have an array of mathematical interests, so if you get stuck in one subject you can shift to another :)

Also good if, like me, your attention span has a very small lifespan.

leo

Given some measure $\sigma$ defined on the $d$-dimensional sphere I have to prove invariance by roations in $\mathbb{R}^d$.

anon

Are there no conditions on $\sigma$?

Mariano Suárez-Alvarez

«your attention span has a very small lifespan» is a weird thing to say :D

leo

1:49 AM

I was not sure about what a rotation in $\mathbb{R}^d$ is

anon

Hmmm. If $\sigma$ is an arbitrary measure and $A$ a measurable set, is $\mu(X):=\sigma(A\cap X)$ also a measure? Pretty sure it is.

leo

@anon Given $E\subseteq S^{d-1}$ $\sigma(E)=m(\{x\in\mathbb{R}^d:x/|x|\in S^{d-1}\text{ and } 0\lt |x|\lt 1\})$, where $m$ is the Lebesgue measure

Rajesh D

the worse thing is i have no clue as to how to solve that problem and it also doen't seem to be well defined, I am not exactly sure what it is, I have found no literature on it, but it is just good enough to quickly demotivate me if i read any other part of math, basically i am not acquiring any knowledge in math for the past 1-2 years

leo

@anon yes it is

anon

@leo: Pretty sure you want $x/|x|\in E$

I was going to say, for arbitrary $\sigma$ that seemed impossible. But this $\sigma$ should be symmetrical.

leo

1:57 AM

@anon Indeed must be $E$

anon

Define $F(E)=\{x\in\Bbb R^d:x/|x|\in E,0<|x|<1\}$ (a function mapping sets to sets). Show that if $R$ is a rotation, $F(RE)=RF(E)$. Then $\sigma(R E)=m(F(RE))=m(RF(E))$, and the Lebesgue measure should be rotation-invariant. Here I use the notation $gX=\{gx:x\in X\}$ for $g$ an action on $X$.

That's my personal thought process.

leo

I think the proof will reduce to some computations with determinats

but your approach seems pretty good

prhaps better

Rajesh D

@anon : How is the book by Dummit and Foote?

anon

No idea.

Rajesh D

abstract algebra, which book is good for introductory but with full rigour?

leo

2:09 AM

Herstein seems good

anon

Not sure. My first exposure to aa was with a small, obscure book at my local library that I have since forgot the name of. The only other text I've delved into was Birkhoff-Maclane, which was annoyingly terse.

Neither book did I read much of. Mostly I just pick stuff up here and there.

I still don't have basic things like orbit-stabilizer and sylow theorems memorized. :<

mixedmath

@RajeshD I first learned from Herstein's Topics in algebra. I've since come to adore Dummit and Foote as an intro text. If I were to recommend it now, I'd have a hard time choosing between them for a first glance. But I'd always recommend having a copy of Dummit around if you're going to continue learning

Benjamin Lim

@mixedmath Can I ask you something please?

mixedmath

sure - what's up?

I don't have a lot of time immediately, but I'll be back in a bit if it's too long

Benjamin Lim

@mixedmath Say we have a ring $A$ that is finitely generated as a $k$ algebra

sorry

say we have $A$ that is finitely generated over $B$

Does it follow that $A$ is integral over $B$?

For field extensions one uses the usual dimension counting formula

but that is not available now

mixedmath

2:25 AM

The dimension counting formula?

Benjamin Lim

yeah the one about $[E:F][F : L] = [E : L]$

because

I am looking at a proof of noether normalisation

mixedmath

oh, interesting. Sure

Benjamin Lim

Noether normalization lemma

In mathematics, the Noether normalization lemma is a result of commutative algebra, introduced in . A simple version states that for any field k, and any finitely generated commutative k-algebra A, there exists a nonnegative integer d and algebraically independent elements y1, y2, ..., yd in A such that A is a finitely generated module over, and hence also an integral extension of, the polynomial ring B:=k[y1, y2, ..., yd]. The integer d is uniquely determined by A: it is the Krull dimension of A. When A is an integral domain, d is then the transcendence degree of the field of fractions ...

I don't understand the sentence "such that A is a finitely generated module, and hence also an integral extension of, the polynomial ring $B$".

mixedmath

anyhow, when you say that A is f.g. over B, do you mean it's a f.g. - B-module?

if so, then yes - that means it's integral

Benjamin Lim

yes

The thing is .....

say for example $A = B[k_1\ldots k_n]$ where the $k_i's$ are the generators

don't I need for each of the generators to be integral over $B$?

mixedmath

2:28 AM

whoa, is A over B or B over A?

Benjamin Lim

Sorry

A over $B$

@mixedmath yeah It's clear to me that if each of the $k_i's $ is integral over $B$ then i'm safe

but nowhere in the definition of finitely generated do we need that the generators be integral over $B$.

mixedmath

so in particular, we can say something a bit stronger

Benjamin Lim

that is

@mixedmath How do I know that $A$ is integral over $B$?

It is not necessarily true that a submodule of a finitely generated module is finitely generated no?

mixedmath

we can say that an element a in A is integral over B iff the subring B[a] in A is finitely generated as a B-module

Benjamin Lim

yes

mixedmath

2:32 AM

I had to make sure I had the statement right - sorry

Benjamin Lim

but now

we don't know that $B[a]$ is finitely generated over $B$

mixedmath

well, this is equivalent to B[a] being contained in a subring of A that is finitely generated over B

Benjamin Lim

yes

oh but that subring is just our $A$?

mixedmath

which here, is in fact A itself

Benjamin Lim

OMG facepalm.....

OMG OMG OMG OMG

@mixedmath it's so clear now

mixedmath

2:35 AM

great!

Benjamin Lim

BIG FACEPALM

@mixedmath I'm used to dealing with field extensions it feels kinda weird that with ring extensions not everything transfers over

mixedmath

that's funny - I don't have very many field tidbits near the surface of my memory. I really only remember the ring ideas

Benjamin Lim

yeah for me it's the complete opposite

I think I can continue now, thanks for your help.

mixedmath

great - good luck

Benjamin Lim

bye!

Antonio Vargas

3:13 AM

I'm amazed there's a limsup tag.

robjohn

3:24 AM

I'm as close to 25K as I can get without a question downvote or downvoting someone

@AntonioVargas that seems a bit ridiculous

Antonio Vargas

@robjohn I'd be happy to! Would you like to choose the target?

robjohn

@AntonioVargas No, but thanks :-)

Antonio Vargas

;)

robjohn

I can live with 25,001

Antonio Vargas

dupe

Brian M. Scott

3:43 AM

@AntonioVargas But with a successful answer, so no harm done.

anon

5:36 AM

Is there an elegant generalization of $$|\mathrm{End}(G)|=\sum_{H\le G}|\mathrm{Aut}(G/H)|$$ to more than just abelian $G$?

Matt N.

That's a nice answer.

Rajesh D

@robjohn : I've found a webpage of you on the internet!

robjohn

7:12 AM

@RajeshD Really?

Jonas Teuwen

7:26 AM

It is bloody early!

@robjohn I can downvote you if you want 25000 8-))).

robjohn

@JonasTeuwen No, really, but thanks.

Jonas Teuwen

@robjohn There are probably many robjohn's crawling on the surface of this earth... It sounds very generic to me. Do you know the math genealogy project?

If you do: Are you aware that you're not in there (I think)?

robjohn

@JonasTeuwen I think I've heard of it.

Jonas Teuwen

@robjohn Here it is!. You can add your name with your thesis title, then people can find who your "parents" are. Looks fun to me 8-).

So, if I understood you correctly, Terence and Charlie are your brothers...?

@robjohn The site tells you that teddy had a burger as advisor 8-)!

Matt N.

7:59 AM

Anyone have a minute to talk about commutative algebra?

Never mind : )

Jonas Teuwen

@MattN I do, but I know monkeys about it! 8-))).

Gigili

8:19 AM

@MattN Do you have the pdf version of the book?

David Wallace

8:31 AM

@JonasTeuwen Yeah, some of the people I know are monkeys too.

Jonas Teuwen

@DavidWallace In a good way?

Matt N.

10:18 AM

@Gigili No, I have a hard copy from the library : ) I also might have a djvu version somewhere...

anon

Which book?

Limitless

10:39 AM

@MattN, why is .djvu so popular for math texts? Is this format the "original format"?

Ananda

@ Limitless , i think thats true.

Limitless

It's such a shame, @Ananda. I love pdfs.

Gigili

11:02 AM

@MattN How hard exactly?

@anon Introduction To Commutative Algebra by Michael Atiyah and MacDonald.

Matt N.

11:23 AM

@anon What Gigili said.

@Limitless Dunno,I don't like soft copies anyway.

@Gigili Hard enough to cause head trauma to a burglar. But I haven't experimentally verified it.

Not very heavy though so not that much use.

@Gigili Why actually? Would you like to study it?

Ananda

@Limitless @ Limitless , me too . You can try converting DJVU format to PDF .

Matt N.

"Teddy effect" : )

SauravTomar

Hi guys, has anyone of you enrolled in any open online course like udacity coursera or MITx in mathematics?

Jonas Teuwen

12:25 PM

@MattN I don't know how good Teddy is at research, but he certainly is a great (and patient) expositioner!

Matt N.

12:42 PM

@JonasTeuwen I know that. From first-hand experience. : )

Gigili

@MattN Uhum, I want to read a chapter to see if I like the topic.

@MattN Beri nice.

Matt N.

@Gigili Sure. Tell me how I can share it with you. (Assuming you don't want to share your email address in here.)

@Gigili It's not a good choice though if you want to end up liking the topic.

No one I've talked to about the book so far thinks it's a good text book.

Gigili

@MattN Stare at the screen, I'll give you my email address and will delete in within s few seconds. (Yes, everybody wants to have my email, yes)

Matt N.

Also, it's riddled with typos.

k, I'm staring.

Gigili

Oh?

Matt N.

12:45 PM

Got it.

Gigili

Uhum, thank you.

So what else do you recommend? If the book is not a good one?

Jonas Teuwen

@MattN 8-).

Zhen Lin

I like Eisenbud's Commutative algebra.

Jonas Teuwen

@ZhenLin What are the applications of commutative algebra (outside algebra, I mean)?

Zhen Lin

Algebraic geometry, algebraic topology...

Jonas Teuwen

12:49 PM

It looks interesting, but I need some motivation for myself why I'd like to know this as there are many other interesting things too :-).

Zhen Lin

I don't find it interesting in its own right.

Matt N.

@Gigili I don't know. I have to stick to this for now because I'm taking an exam for a course that more or less followed this book.

Anyway, now you have this for starters : )

Gigili

@MattN Thanks a lot.

Matt N.

@Gigili Np : )

@JonasTeuwen And many more know, too. : )

Jonas Teuwen

@MattN Teddy is one of the bros...

Sb Sangpi

1:33 PM

what is the different between Flag & vote down? nearly the same?

Jonas Teuwen

They map... into the same space?

SauravTomar

@SbSangpi flag is when you want to report something, and down vote is when you are disagreeing with something mildly

flag is useful for comments as you cannot downvote them

Sb Sangpi

@Sauravtoma thx for that, appreciate it!

SauravTomar

anytime

my MSE rep is now of 4 digits

Shane

1:51 PM

Would anyone who is familar with Cholesky reduction for matrices be willing to help me? I have an exam coming up, had a terrible lecture and his notes are contradicting resources I am finding online

Jonas Teuwen

@ZhenLin Okay, then why do you find it interesting? :-).

Zhen Lin

I don't. The parts I like are basically algebraic geometry, and the parts I don't like are basically the dirty work that goes into the foundations.

Jonas Teuwen

That does not really make me want to study it... 8-).

Zhen Lin

2:06 PM

shrug

I'd rather promote category theory. :p

Jonas Teuwen

Hmm... Will that make me understand tensor products of Banach spaces better?

Dylan Moreland

2:45 PM

@Jonas Grothendieck was a functional analyst at one point, right? But the Wiki page makes it seem like there's a lot of choices involved, which does not appear to be in the spirit of category theory.

Jonas Teuwen

@DylanMoreland That was his initial topic. Then he went all crazy. You know what happened eventually right?

Dylan Moreland

@Jonas He disappeared into the south of France?

Jonas Teuwen

@DylanMoreland To farm apples, so have I heard.

Moshe

3:38 PM

Anyone here?

anon

vaguely

Moshe

@anon - Cool, I'm studying for a precalculus final. I need a little help.

I've done this example and I'd like some verification if I've done it correctly.

The problem:

anon

alright. do you have the mathjax working?

and/or know latex?

Moshe

@anon Yup, I do, but I don't really know it. I can try.

Find an equation for a line going through the point (4,-1) and parallel to the line $2x-3y=9$.

My solution:

anon

oh yeah I remember you

Moshe

3:48 PM

Heh :-)

$y=2/3x+8/3$

Gigili

Your solution is changing!

Moshe

@Gigili No, I changed because of MathJax. It's finished now.

Gigili

$y+1=2/3(x-4)$

Or wait ...

Anon weisst mehr!

anon

@Moshe This can't be correct. Your proposed solution has the same exact slope as the original line.

Moshe

@anon But it's parallel, it should have the same slope.

I'm not looking for a perpendicular line.

anon

3:52 PM

Oh sorry, I was confusing it with perpendicular from the other day

I'm still not fully awake :)

Then yes, that works.

Gigili

@anon How so?

He should just use the $a$ from the first line, or I'm not fully awake too?

anon

no wait, plugging in (4,-1) doesn't work

Hint the equation parallel to ax+by=c that goes through the point (u,v), where u and v are given, will be ax+by=au+bv (note the RHS is a constant).

Gigili

@Moshe: Make $y$ alone, then find $a$ when it's in the form of $y=ax+b$

Moshe

Can I use the point slope formula here?

Gigili

Then plug in the point in to $y-y_0=a(x-x_0)$

anon

3:55 PM

@Moshe Sure. (That's equivalent to my hint.)

Moshe

@anon Ah, thanks.

anon

(The reason I gave my version is that the point-slope formula (on y=mx+b) won't work when it's a line like x=b, you would have to switch to x=ny+c form.)

Bill Dubuque

5:02 PM

@anon You might be interested to know that every answer in the thread you mentioned yesterday has now been downvoted twice (mine 3 times). This frivolous voting behavior is disturbing. It has been mentioned in passing in this meta question.

anon

@BillDubuque I did notice that just awhile ago. Note that the first downvote on my answer was from JonasTeuwen, not some random user, after I complained that it superficially looked like I might have been "framed."

I think I've managed to find a group-theoretic generalization of the cyclotomic polynomials, but so far they are only defined for abelian groups, and in particular the identity $x^n-1=\prod_{d|n}\Phi_d(x)$ only has an analogue for abelian Aut-abelian groups.

Actually, all of its subgroups need to be aut-abelian

For all the group theory I know that might be equivalent to being cyclic, which means this only works for cyclo polys anyway. Bah.

Bill Dubuque

5:30 PM

@anon You might find this of interest: Ayoub, On the group ring of a finite abelian group, BAuMs, 1969,245-61.

anon

thanks

Interestingly, you seem to have anticipated that I was defining the polynomials through the group algebra...

whoah, I just realized that paper is by a couple. (either that or brother/sister.)

Ilya

6:07 PM

@robjohn: a new starred reminder of Chat Rulz is needed! :)

leo

||x||

nice

mixedmath

8:25 PM

good afternoon, oh mighty slumbering chat

Michael Greinecker

good what-seems-to-be-afternoonto-yo, mixedmath

mixedmath

I suppose we could start the convention that we greet everyone based on some preestablished timezone

or always say good afternoon-to-me to you

Michael Greinecker

I think the convention to greet a person according to ones own time is a good one, it informs others about where someone is coming from. good night here

Limitless

8:54 PM

@All I'm curious if there's any chance what-so-ever of getting the chat's mathjax to work on my iPad. Thoughts?

Limitless

9:06 PM

Oh my
I got it to work. :)

Matt N.

9:28 PM

@JonasTeuwen ayt?

Jonas Teuwen

@MattN How rare. I just arrived.

Matt N.