Mathematics - 2012-04-08 (page 1 of 3)

David Wheeler

12:00 AM

parsing.....

t.b.

what specifically are you trying to parse?

what is classical-analysis?

David Wheeler

i'm reading the link that arturo linked to in his answer that you linked to (which was also written by arturo)

i'm having a bit of difficulty in that he produces a subset A' of H, and presumably the cardinality of A = f(G) and A' are the same, but he doesn't explicitly say so.

i was never aware of how involved the statement epi= surjective actually was in the category of groups....

groups.google.com/group/sci.math/msg/6d4023d93a2b4300 this is what i am reading now

it looks like a proof by contradiction, even though it's not phrased that way, which is throwing me off

perhaps i am missing something, but it looks as if the whole argument rests on a regular action of a group on itself.

well, actually a coset space, but same thing, we get an induced action on a (left or right) coset space from the (left or right) regular action

t.b.

12:28 AM

yes I think that is one of the main points.

David Wheeler

i mean, he appears to be assuming we can pick some element of h NOT in A = f(G), otherwise H/A is trivial.

and so we could take A' = (A - {a}) U {h}, for example, for some element a of A, which can't be a coset of A in H.

t.b.

But A' can be anything. It should just be of the same cardinality as A

(there need not be any relation of A' with H, if I read it correctly. A' is just any set that can be put into bijection with A)

David Wheeler

i get, that, too...he just wants to pick some subset of H that doesn't "line up" with the slicing of H by A, but has the same size as a coset.

what is troubling me, is i don't see how an element of H defines a unique element of Sym(S) = K

robjohn

@JasperLoy It was my wireless router, I believe. :-)

Srivatsan

hi tb, David and robjohn.

robjohn

12:41 AM

I have to go for a walk. I will be back later. Sorry I have been so absent, but I have been fixing up an answer.

@Srivatsan: I am sorry, but I have to leave for a couple of hours.

t.b.

Hey Srivatsan!

Srivatsan

@robjohn It's fine. Take care.

David Wheeler

@Srivatsan salutations

Srivatsan

@tb I need to complete my recent answer by typesetting a small commutative diagram. It's perhaps the simplest possible diagram: a square with two paths from top left to bottom right. How can I make this in MSE?

t.b.

@DavidWheeler I don't think so. H/A is a collection of sets, all of the same size as A. Now add another set A' of the same size as A. An element h acts on H/A and as the identity on A'. The permutation s is a fixed bijection of A and A' which leaves everything else fixed.

David Wheeler

12:45 AM

he writes: "if A' is sent to itself....." well, sure, in that case we do have an element of Sym(S), but how can we ensure that is the case for any h in f(G)?

t.b.

@Srivatsan see here. Zev's answer seems better suited than the accepted one.

David Wheeler

is he just DEFINING an action of H on S?

t.b.

I think so, yes.

David Wheeler

so (A').h = A' by definition, then

t.b.

exactly.

Srivatsan

12:48 AM

@tb I'll try it, thanks. Have you happened to have drawn diagrams in any of your posts?

David Wheeler

so it's almost like he's sneaking in a direct product action, without saying so

WhatsInAName

anyone know chinese remainder theorem

David Wheeler

defining an action (by a group G) on a set C x D, where we already have an action on C by G by letting G act on D trivially.

Srivatsan

@WhatsInAName I suggest that you just post your question. Most basic questions in this chat will get some response from one or another.

WhatsInAName

i am trying to find n choose k mod p*b where p and b are prime

i can find n choose k mod p and n choose k mod b, but need to combine the results

t.b.

12:52 AM

@Srivatsan here. But I prefer typesetting them on my computer and uploading screen shots, as I did here for example.

@DavidWheeler yes.

anon

@WhatsInAName: equation here is the formula you wnat

or rather second to last

Srivatsan

@tb Ah, ok. Thanks.

t.b.

the first poster is one of my favorite users here...

David Wheeler

so then $k(h) = sg(h)s^{-1}$, and for all a in A, $k(a) = g(a)$

WhatsInAName

I don't understand

t.b.

12:56 AM

yes.

David Wheeler

but if i understand what's going on, to prove f is surjective, we can use ANY other group K

WhatsInAName

equation makes no sense to me

i just want to combine two results

anon

@Whats: $[a^{-1}]_b$ stands for the multiplicative inverse of $a$ modulo $b$.

David Wheeler

so i could use the A' i proposed, instead of the A' Arturo came up with, all i need is to get "some other group K" to which i can apply the epic property of f to.

Srivatsan

@WhatsInAName Do you know what a "multiplicative inverse of something modulo something else" is, and how to compute it?

WhatsInAName

12:59 AM

So I do a1*n2*[inverse n2 mod n1] + a2*n1*[inverse of n1 mod n2]?

anon

yes

WhatsInAName

in my case what is a1, a2, n1, n2?

n choose k mod p and n choose k mod b

Srivatsan

n1 = p and n2 = b.

And your n choose k mod p is n1. n2 is n choose k mod b.

David Wheeler

and $a_1 = a_2$ = n choose k.

WhatsInAName

i thought n1 was p?

t.b.

1:01 AM

@David: But to do that, you'd need to assume a few things (and make the proof indirect). Doing it the way I said, this always works and you don't have to assume anything and you show that A = H.

WhatsInAName

ah ok

David Wheeler

@tb...assume "what things"?

WhatsInAName

(n choose k)*b*[inverse b mod p] + (n choose k)*p*[inverse of p mod b]

t.b.

@DavidWheeler In your approach you need to assume that you have a subset of H that isn't an A-coset.

WhatsInAName

is a1 n choose k or can it be n choose k mod p / b etc

David Wheeler

1:02 AM

oh, you mean that my way explicitly makes it a proof by contradiction, whereas Arturo is allowing for the fact that A might be H from the outset.

t.b.

exactly.

Srivatsan

@WhatsInAName More generally, your are given two moduli, p and b, and the residues of a hidden number modulo these two moduli. E.g., you know n choose k mod p and n choose k mod b, but not n choose k itself.

WhatsInAName

right

Srivatsan

@WhatsInAName Well, the formula is slightly wrong: you do not know n choose k. How are you going to apply the formula?

David Wheeler

well, there is one assumption in Arturo's argument, namely that H is non-trivial.

Srivatsan

1:04 AM

@WhatsInAName a1 is n choose k mod p.

WhatsInAName

i have absolutely no idea :(

Srivatsan

@WhatsInAName The formula is: n choose k mod (pb) = (n choose k mod p) * b * (inverse of b modulo p) + (n choose k mod b) * p * (inverse of p modulo b) .

t.b.

@DavidWheeler why?

(not that it would matter much :))

WhatsInAName

i just tried that

and for some reason it's not giving me the right value

David Wheeler

because otherwise there is NO such A', because the only subgroup of a trivial subgroup is trivial. the only non-empty subset of H is H itself.

Anna Lear

1:07 AM

@zevchonoles We (well, devs, but whatever :)) can disassociate posts from accounts. Just let us know which one needs to be anonymized.

Srivatsan

@WhatsInAName Give us your values.

t.b.

@DavidWheeler But A' can be any set

WhatsInAName

uint64_t p = 1000003;
uint64_t b = 1000033;
uint64_t x = modBinomial(543, 12, p);
uint64_t y = modBinomial(543, 12, b);
cout << x*b*modInverse(b,p) + y*p*modInverse(p,b);

t.b.

@AnnaLear I've never seen Zev entering this room. Are you sure this comment is supposed to be here?

David Wheeler

oh, you mean A' could be the null set, in which case S = H/A

Srivatsan

1:08 AM

@WhatsInAName What answer do you get? Maybe there are overflow issues?

Anna Lear

@tb Super-ping will notify him of the message, anyway. But thanks for the heads up. I'll try to find a more reliable way to get hold of him. :)

WhatsInAName

i'm using uint64's though

I get 174371283783502760

David Wheeler

so any group acts on the null set trivially, just because there's nothing to act on, so it doesn't really mater HOW we define the action: we could say it turns all members blue.

Srivatsan

@WhatsInAName Well, you need to be able to store the intermediate results without overflow, not just the final answer.

WhatsInAName

pastebin.com/bp37LKUi

everything uses uint64's

t.b.

1:10 AM

@DavidWheeler No, if H = A = {e} then A' is just some one-point set.

(as I said, A' is just some set in bijection with A and A is always non-empty)

David Wheeler

oh! you mean A' could be ANY set, not even having anything to do with H?

Srivatsan

@WhatsInAName Maybe that is not the issue. What is pb? You should mod out the answer by pb at the end; did you do that?

David Wheeler

so A' might be {a*Fred}, where a is an element of A?

t.b.

@DavidWheeler yes... That's what I was trying to say right at the beginning:

WhatsInAName

ah that's what it was!

thank you so much

t.b.

1:13 AM

@DavidWheeler yep.

Srivatsan

@WhatsInAName Because the formula will not recover the hidden number (n choose k) exactly; it will recover it only modulo pb.

David Wheeler

ok, so we're just "padding" A/H by adding some set with |A| elements, and letting h always send A' to A'.

t.b.

as identity.

David Wheeler

g(h)(a') = a' for all a' in A'.

actually, that's not right, g(h)(A') = A', who knows if g(h) interacts with elements or not...

t.b.

No, you really take the coset action and declare g(h)(a') = a'

Then we fix a bijection s between A and A' and fix everything in A/H \cup A' that isn't either in A or A'

Srivatsan

1:18 AM

Ok, see you all. Got to go.

t.b.

that should have been H/A \cup A'

David Wheeler

lol, i get it...i just was struggling as to how in the hell he produced A'. and you're saying it doesn't matter, pick any set with |A| elements except a coset.

t.b.

exactly. I've been trying to get this point through right from the beginning, but I was sorta distracted...

bad teaching, I guess.

David Wheeler

now i'm looking into the schreier refinement theorem...

@tb it's ok, i'm just dense sometimes

it appears that this implies jordan-holder, which in turn implies the structure theorem for f.g. abelian groups

honestly though i got bored reading the proof after "division algorithm" blah,blah, blah "second isomorphism theorem" blah blah blah

t.b.

1:34 AM

where are you getting that from now?

David Wheeler

math.binghamton.edu/mazur/teach/5236.pdf

t.b.

oh, that's the kind of stuff I should read now, because I should actually be sleeping :)

David Wheeler

what i mean is i looked at it, and i was like" oh, right" freaking integers, they're EVERYWHERE....

t.b.

that's basically the way I took it...

David Wheeler

in a vague sort of general way i look at life like this: we have stuff. and things happen. some things that get done to stuff can be un-done, and this is useful for when things aren't going so good. unfortunately, sometimes you have to remember how you got where you were, so you can go back. in a perfect world (read: abelian), it doesn't matter...all roads lead to rome. the end.

t.b.

1:50 AM

I wonder what exactly triggered this, but for some reason it made me laugh. I hope that's okay...

David Wheeler

@tb of course it's OK. i aim to please :)

t.b.

very good then, thanks :)

anon

So two (sub)normal series are equivalent if they have the same factor groups, counted with multiplicity? I think the pdf is missing an apostrophe or something in its definition.

t.b.

I keep hearing and reading the words operator systems and completely positive maps, but I never actually found out why I should care in the first place.

this "survey" is an example that might indicate why I am lost

David Wheeler

it looks like the things quantum physicists care about

t.b.

2:05 AM

It does look like it, yes.

oh no! Victor discovered Wildberger...

David Wheeler

2:20 AM

what?

t.b.

that

David Wheeler

that is a cool video...i've seen it before

t.b.

Well, Wildberger has his own style... Not bad. Don't like the way he pronounces taaaaangent :) But I shudder at the thought of 5 * (how many lectures?) questions on this stuff by V.

David Wheeler

3:08 AM

quick question: how many elements of order 10 in Z12 x Z4 x Z15...i count 12?

anon

The grocery store was almost out of peeps. An outrage, I say.

David Wheeler

3:43 AM

peeps r 3bil

robjohn

@DavidWheeler 3bil?

David Wheeler

eeeeeeeeevillllllll

candies shaped like baby whatever are inherently so

anon

l3375p34k

David Wheeler

i|<n0r1t3?

megatokyo.com/strip/9 classic

anon

I can still read that fluently. Not sure how to feel about that.

David Wheeler

3:55 AM

the reference to "airplane" is too funny

the original metacafe.com/watch/5605341/airplane_oh_stewardess_i_speak_jive

Mariano Suárez-Alvarez

4:12 AM

what's a cool argument that if S is a subspace of R^n of dimension k there exist $1\leq i_1<\dots<i_{n-k}\leq n$ such that $S\cap\langle e_{i_1},\dots,e_{i_{n-k}}\rangle=0$ ?

t.b.

4:22 AM

Don't know if that's cool enough: pass to the quotient which has dimension $n-k$. The images $f_1,\ldots,f_n$ of $e_1,\ldots,e_n$ span $\mathbb{R}^n/S$. Take a linearly independent subset $f_{i_1},\ldots,f_{i_{n-k}}$ and look at their preimages. The intersection $S \cap \langle e_{i_1}, \ldots, e_{i_{n-k}} \rangle$ is zero because $f_{i_1},\ldots,f_{i_{n-k}}$ are linearly independent.

Mariano Suárez-Alvarez

cool enough :)

thanks

anon

5:27 AM

@tb What a gentleman you are.

t.b.

I wrote $\int_{n}^x |f(t)|\,dt \leq \int_{n}^{n+1}|f(t)|\,dt$, the right hand side of which is independent of $x$. Now integrate this over $[n,n+1]$ to get the desired inequality.

where $f$ should have been $f'$ of course... :)

The question comes from here btw.

user19161

5:47 AM

@DavidWheeler Oh I would not dare to eat them.

Benjamin Lim

hello

t.b.

Am I missing something or is this a non-issue about a far too generic concept?

Benjamin Lim

$(1)$ here is the trivial group?

t.b.

6:08 AM

yes.

got involved in null-o-logy again. bleeeh

Matt N.

Snoooow!!!

Must be my birthday today. : )

Mariano Suárez-Alvarez

6:23 AM

can everyone see the history of removed comments or does that come with modhood?

anon

modhood

t.b.

Modhood privilege.

Mariano Suárez-Alvarez

aha

anon

it certainly wouldn't be everyone.

t.b.

also on the main site. You can see all deleted answers with 10k, but comments are gone.

Mariano Suárez-Alvarez

6:25 AM

yup

Matt N.

What the...?

It's... unicorns!

anon

oh boy, it's the musical topology dude

Matt N.

: D

I think it's unicorns.

anon

"Don't you get it? Music is semi-algebraic sets."

t.b.

I really fail to understand the point of this. I would insist that a property is closed under isomorphisms anyway. But I wouldn't call a property of modules over a ring a module property provided the zero module satisfies it. I wouldn't call a property a pointed space property provided the one-point space has it.

Benjamin Lim

7:43 AM

@tb Hey are you around?

I am trying to understand why if $k \otimes_A (M \otimes_A N) = 0$, where $k = A/a$

$M,N,k$ are $A$ - modules

Then $M_k \otimes_k N_k = 0$ where $M_k = k \otimes_A M$, $N_k = k \otimes_A N$

I can see that $a \subset \operatorname{Ann}(M_k)$ so that is how $M_k$ and $N_k$ become $k$ - modules

The confusion that is coming is that because the "base ring" has changed from $A$ to $k=A/a$

Any kind of universal property of tensor product or whatever that I want to use is rendered useless

Sorry I forgot to say that $A$ is a local ring with maximal ideal $a$

If I write out the definition of $M_k \otimes_k N_k$

It is $(M \otimes_A k) \otimes_k (k \otimes_A N)$

Can I say that this is isomorphic to $M \otimes_A \bigg( k \otimes_k (k \otimes_A N) \bigg)$ ?

Kannappan Sampath

8:40 AM

Morning folks.

Benjamin Lim

@KannappanSampath Want to try to answer the above?

Kannappan Sampath

@BenjaminLim Hey, I have not learnt or come to tensor products as yet.

Benjamin Lim

@KannappanSampath It is really confusing!!

Kannappan Sampath

Yes, I have been told it is, by you and others but I am keeping my head cool for now, ;)

Benjamin Lim

@KannappanSampath Brace yourself

Kannappan Sampath

8:46 AM

:)

Jonas Teuwen

9:20 AM

Good morning.

t.b.

10:17 AM

There. Labeled equations with links :)

Daniil

10:36 AM

Hi guys

aw, this is quite good: automata.rwth-aachen.de/teaching/videos.en it's a shame it's not open to public

Benjamin Lim

10:47 AM

@tb No worries I got it now it comes from extension of scalars :D :D :D

@KannappanSampath Wait till you have to deal with tensor products over different base rings, a lot of fun :D

user19161

11:04 AM

@BenjaminLim Wow, three consecutive smiles. You must be very happy this Easter!

Benjamin Lim

No I'm not. Too many people getting drunk, this pisses me badly

user19161

Why do drunk people piss you badly? Is it because they piss on you?

Benjamin Lim

@JasperLoy No pun intended before.

user19161

@BenjaminLim Wait do you mean pun?

Benjamin Lim

yes

user19161

11:08 AM

I asked because t and n are quite far apart. It was a weird typo.

Benjamin Lim

i'm really tired got really maggot yesterday :D

user19161

@BenjaminLim Congrats! Do you have a girlfriend?

Benjamin Lim

no never had one in my life

@JasperLoy do you know what getting maggot means?

user19161

Same, haha.

user19161

@BenjaminLim I read the transcript. It is some kind of stupid drunk dance or something.

Benjamin Lim

11:12 AM

no it means getting drunk

@JasperLoy where are you now in the world?

user19161

@BenjaminLim Singapore.

Benjamin Lim

are you singaporean?

user19161

Yes.

Daniil

What do you think?

user19161

@Daniil Looks deep.

Daniil

11:15 AM

infinite words & automata is probably the only "research-level" area I've been studying :(

@JasperLoy nice userpic you've got

user19161

@Daniil So you are into logic?

Benjamin Lim

hahahaha

user19161

@Daniil Oh I am a Belieber.

user19161

@BenjaminLim I guess that was for the JB pic.

Daniil

@JasperLoy I am into intersection of logic and CS, really.

belieber, lol

user19161

11:17 AM

Perhaps he represents the childhood I missed...

N3buchadnezzar

How is it going ? =)

user19161

@N3buchadnezzar Hey! Did you get drunk over the weekend?

N3buchadnezzar

@JasperLoy Totally yesterday, I am taking it slow today.

Benjamin Lim

@JasperLoy Cody Simpson 143 :D :D youtube.com/watch?v=MYRNGeIJMiQ&ob=av2e

N3buchadnezzar

My friend cam back from Russia with Energy Drinks with 8% alcohol...

user19161

11:19 AM

@N3buchadnezzar Good. Just remember: if you drink, don't drive. If you drive, don't drink.

Matt N.

Why don't you insist on using wrong English then. I don't care so: troll fail.

N3buchadnezzar

Don't drink and drive, think and derive =)

user19161

I just had some tequila with my friend the past few days. We mixed it with fruit juice.

Nimza

Hello

N3buchadnezzar

We had 1l litre of Vodka, but it was plenty with all the energy drinks and cider.

user19161

11:21 AM

@BenjaminLim Ah, I think I only heard about him once.

Benjamin Lim

143 = I love you

user19161

@N3buchadnezzar That deserves a star but I don't do stars.

N3buchadnezzar

It is impossible for me to get drunk on beer, haha.

Energy drinks with 8% ftw

Benjamin Lim

you russian?

N3buchadnezzar

Norwegian

user19161

11:23 AM

Someone said with reference to vodka in the other room: bucketwise is not wise.

Benjamin Lim

@N3buchadnezzar ah ok

N3buchadnezzar

@JasperLoy A bucket of Vodka is always wise.

user19161

So is that you in the pic @ben?

N3buchadnezzar

Benjamin Lim

@JasperLoy yes at royal national park, sydney

user19161

11:25 AM

@N3buchadnezzar That's for vomiting after drinking.

N3buchadnezzar

@JasperLoy Then you need to empty it for vodka first =)

user19161

@BenjaminLim I had a prof from Australia. He drinks apple juice during lectures. He has a whole carton in his office.

N3buchadnezzar

I love apple juice, and orange juice.

Btw, is there anyone here good at using maple ?

user19161

I remember the three m's are not free: maple mathematica matlab

user19161

Good thing the two l's are free: linux latex

N3buchadnezzar

11:30 AM

At university they are free. We need to learn to use maple and matlab.

user19161

@N3buchadnezzar Of course, because the university has paid for it!

N3buchadnezzar

@JasperLoy ^^

user19161

TV and dinner time!

N3buchadnezzar

If you visit a shady pub in a pirate bay, I think you can find a free version too,

Benjamin Lim

@JasperLoy Your surname does not sound singaporean

Transcript for

$Mathematics$ Mathematics