Mathematics - 2012-09-25 (page 1 of 4)

Gustavo Bandeira

00:03

How to center a image?

In a post?

Found.

Thelonius

00:22

hi @anon

anon

hello

Thelonius

i wonder if you might help me with a question about this answer math.stackexchange.com/questions/10603/…

i was told you are good at this stuff

in particular i'm trying to understand exactly why the dual of the adele class group is isomorphic to Q

pete clark argues that that exact sequence is a "self dual sequence" (not sure what that means exactly, unless the point is that all the guys in there are self dual), and seems to suggest that it follows

anon

adele class group is that $\Bbb A_{\Bbb Q}/\Bbb Q$?

Thelonius

yes

and by dual we mean the set of continuous characters

anon

you might find this helpful by the way:
http://www.math.uconn.edu/~kconrad/blurbs/gradnumthy/characterQ.pdf
it glosses over topological concerns though

Thelonius

00:25

sweet

that might be what i'm looking for

@anon the argument in the stackexchange answer i linked to seems more elegant. i just don't get how you can detect the dual from those exact sequences

anon

In AbGrp we should be able detect the middle term of a s.e.s. by just direct summing the outer two things. That seems to work, except for the fact that the adeles are only a restricted part of the sum of the hat{Z} and S^1. I presume there's some obvious modulo considreation I'm not seeing.

Thelonius

@anon i get the philosophy behind that. but how do you detect the duals in this way?

if you want to show that A/Q dual =~ Q

anon

00:43

I'm presuming you find the dual group by dualizing the ses and then solving for the middle term (Q^dual) in terms of the other two terms. note $(\Bbb A_{\Bbb Q}/\Bbb Q)^\vee\cong\Bbb Q\iff \Bbb A_{\Bbb Q}/\Bbb Q\cong\Bbb Q^\vee$.

by pontryagin

Thelonius

that has a ring of truth to it. is this a fact about short exact sequences? something about the dual being a contravariant functor of some kind?

anon

yes, the dual is a contravariant functor

I CBA to do a diagram for chat, but if $f:A\to B$ then $f^\vee:B^\vee\to A^\vee$ is given by $f^\vee:(B\to T)\mapsto (A\xrightarrow{f}B\to T)$, I guess where $T=S^1$ in this context

let me check if it's raining outside

good, not raining. hence $\vee:f\mapsto f^\vee$ is a contravariant functor

3

I'll be back in a bit

Thelonius

ok @anon, thanks.

Parth Kohli

07:00

6 hours later...

Paul Slevin

08:46

Hello

Jayesh Badwaik

@PaulSlevin Hi.

user19161

11:05

@JayeshBadwaik Sent you an email.

user19161

@Charlie Replied to your email.

Old John

@WillHunting Hi

user19161

@OldJohn Hey hey. Old John has awoken after his beauty sleep!

Old John

@WillHunting ... and he certainly needs as much of that as possible :)

user19161

@OldJohn Since he is so pretty!

Old John

11:08

@WillHunting ah - if he were, then he would not need the sleep :)

Alpha

@OldJohn How you are going

Old John

@Alpha Fine thanks - just keeping an eye out for possible flooding round here :(

@Alpha - and welcome! - don't think I have seen you round here

Alpha

@OldJohn thanks

@OldJohn How old are you in stackexchange

Old John

11:24

@Alpha my profile probably gives you the info you want :)

(not sure if you were asking my age, or how long I have been on the site)

John Junior

Hi @PaulSlevin long time no see?

its_me

11:50

Hello. Anyone know a simple way to find what chapters make a particular subject in mathematics? For example, what chapters constitute Arithmetics?

Old John

@its_me I'm afraid there is probably no sensible answer to that one - different authors have very different ideas about what constitutes a subject

I have several books with "arithmetic" in the title, and they vary from elementary school textbooks to postgraduate number theory

BenjaLim

@OldJohn I have a question

It is on algebra

Old John

@BenjaLim I can't guarantee that I will be much help! (But I will try, of course)

BenjaLim

@OldJohn What kind of maps are there from say $\Bbb{Z} \oplus \Bbb{Z}/d \to \Bbb{Z}/n$?

for $d | n$.

Is there a general way to answer a question like that?

or rather what do such maps look like?

Old John

@BenjaLim I would guess we need to look at possible images of generators like $(1,0)$ and $(0,1)$

BenjaLim

11:59

yes.

@OldJohn hmmm

Old John

could the image of $(1,0)$ be anything we want in the co-domain?

BenjaLim

hmmm

If I wanted the map to be surjective

Old John

ah - ok

BenjaLim

And if I wanted the kernel to be $\Bbb{Z}$

Old John

That makes it interesting!

so the image of $(1,0)$ would have to be 0, wouldn't it?

BenjaLim

12:02

how come

Old John

if the kernel is to be $\mathbb{Z}$, don't we require the image of each $(a,0)$ to be 0?

BenjaLim

why should the second component be zero?

Old John

I am just looking at some elements of the domain at this stage

BenjaLim

ah ok

Old John

if we can decide how the map behaves on $(a,0)$ and on $(0,b)$, then we know how it behaves on the whole domain

BenjaLim

12:05

yes.

Old John

Oh! - I assume we are talking about some kind of homomorphism, and not just any old map of sets :)

BenjaLim

yes that is the thing too

Old John

So - does it have to be that the image of any element of the form $(a,0)$ has to be zero?

BenjaLim

hmmm

Old John

in order to get the kernel you want

BenjaLim

12:07

yes.

I believe so.

well it could be that

say all such $(2n,0)$ map to zero

because the kernel would still be isomorphic to $\Bbb{Z}$.

ah

Old John

But then what would (1,0) map to?

Ah!! - you might be right

BenjaLim

see

Old John

(1,0) could map to an element of order 2 :(((

are we talking about rings or groups here?

BenjaLim

groups

Old John

OK

BenjaLim

12:11

I think I want

to say if $n = dk$

that

all multiples of $k$

$(kp,0)$ all go to zero.

Old John

sounds plausible, yes

so- can we prove it ...

BenjaLim

assume $n$ is not prime

if it is then there is only one homomorphism :d

Old John

aha - anon might help (please!)

BenjaLim

@OldJohn

What if

I said that I send $(0,0)$ to $0$

$(1,0)$ to 1

$(n,0)$ to $n = 0$

Old John

yep

BenjaLim

12:15

and now

$(0,1)$ I send to 1 say.

Old John

maybe sending (1,0) to anything will automatically give a kernel isomorphic to $\mathbb{Z}$

BenjaLim

yeah.

Old John

OK - so we now just need to check conditions for the map to be surjective?

BenjaLim

Well it is right?

Old John

if you have the map you just gave, yes, definitely

I was just wondering if there might be other maps than the one you gave

e.g. map the whole of $\mathbb{Z}$ to 0, and still get a surjective map from the other part of the domain

its_me

12:19

@OldJohn Oh, hmm

Old John

@its_me did you see my other comment a couple of lines further down?

its_me

@OldJohn this?

> I have several books with "arithmetic" in the title, and they vary from elementary school textbooks to postgraduate number theory

Yes, I did. It's true. :)

Old John

@its_me yep - that makes an answer to your question a bit problematic, I fear

its_me

Apparently... I will have to tackle this the other way around. Thanks anyway.

Old John

@its_me OK - no problem

BenjaLim

12:23

/

@OldJohn Don't worry about it

I think I will figure out a way

its_me

@OldJohn Though I think this is a good place for a start: mathworld.wolfram.com/Arithmetic.html

Old John

@BenjaLim I think you are not far off, now - probably some sort of divisiblity condition needed for the case in which all (a,0) map to 0

BenjaLim

@OldJohn Thanks anyway.

Old John

@BenjaLim I am sure you will :)

@its_me a good start, yes

BenjaLim

@OldJohn I think I got it

Old John

12:30

@BenjaLim Great!

BenjaLim

What about this:

@OldJohn

My map $f : Z \oplus Z/d \to Z/n$

takes $(a,b)$ to $ad + b$

@OldJohn It's a group homomorphism

Old John

I'm a bit worried, ....

BenjaLim

how come.

It looks to me that the kernel of $f$ is $\Bbb{Z}$.

and the map is also surjective

Old John

Ah - might be OK - I was forgetting that $d|n$

BenjaLim

yes.

Old John

12:33

if we didn't have that, it would fail

BenjaLim

well we needed it :D

That's how I constructed $f$ really

I thought hmmm

Old John

but I think you are right, in that case

BenjaLim

I can send say $(0,1)$ to say $1$

and then all the guys on the right component will give me all integers from 0 to $d$

and then I could send $(1,0)$ to $d$

and then multiples all the way up to $k$.

remember $dk = n$.

Old John

in fact, you can probably send (1,0) to anything you like, can't you?

BenjaLim

I could

But

I forgot to tell you earlier

I am trying to construct maps to get the following ses:

Old John

12:35

yes ...

BenjaLim

$0 \to Z \to Z \oplus Z/d \to Z/n \to 0$.

the first map (on the left) is simply inclusion

the second $f$ say sends $m \mapsto (km,0)$

third say $g$ would send $(a,b) \mapsto da + b$

Old John

then I suspect your map above is just what is needed - modulo the fact that I am not an algebra guy :)

BenjaLim

@OldJohn I am 100% sure it is that :D

Old John

yep - I agree

BenjaLim

My lecturer told me that this is actually homological algebra :D

It's secretly in disguise a compuation of the -

Old John

12:38

yep - it sure is - and right at the edge of my understanding of algebra

BenjaLim

GET READY :

Ext functor

Old John

just a bit unsure if your map $f$ is right ...

BenjaLim

how come?

Old John

ignore my previous comment

BenjaLim

I mean the image of $f$ is all multiples of $k$ in the first component, $0$ in the second

while the kernel of $g$ is precisely that.

Old John

12:39

yes - I was just mentally checking if that made it exact at the centre of your ses

I'm happy again :)

BenjaLim

And $g$ is surjective

Old John

yep

BenjaLim

I can write as a multiple of $d$

plus some integer b with $1 \leq b \leq d$

@OldJohn I knew I would get it :D

Old John

... but if you are going to start talking functors, then I am going to be dumbfounded :)

BenjaLim

@OldJohn I am not so advanced yet, only year 2 at uni :D

Old John

12:40

@BenjaLim me too!

@BenjaLim I never got to the stage of doing functors - my knowledge of things like homology is just what is found in old editions of Fraleigh

BenjaLim

@OldJohn So many sequences

and diagrams

Old John

I remember doing some slightly bewildering exercises on the 5-Lemma with lots of diagram chasing

but that was many years ago

BenjaLim

All this after I procrastinated in playing table tennis :D

Hmm perhaps the follow through in the shots

secretely was producing the maps for me in my brain

@OldJohn I like you very much.

You remind me of my uncle's dad who now lives in sheffield

You seem like gandalf :D

Old John

@BenjaLim I was at Sheffield Uni for a year, doing an education course :)

and by coincidence, the guy who became my research supervisor years later was there at the same time

BenjaLim

@OldJohn I'll tell you something and then I'll delete the comment quickly.

Old John

12:46

OK

OK - no problem

BenjaLim

But I find it crazy

how can someone who deals with lie algebras and algebraic topology

be like that.

Old John

I imagine it might be difficult

Jonas Teuwen

Be like what? 8-).

Old John

@JonasTeuwen Hi Jonas - welcome back

Jonas Teuwen

Hi.

BenjaLim

12:49

@JonasTeuwen Hey

Old John

BenjaLim has just been stretching my understanding of algebra - and I have been pretending to help him :)

BenjaLim

@OldJohn I like algebra very much.

@OldJohn It would be so rad to meet you in person one day :D

Old John

@BenjaLim I like algebra - but I am more of a visual thinker, so I ought to stick things like complex analysis, really :)

BenjaLim

@OldJohn Algebraic topology teaches you to be visual :D

In fact the diagrams may be scary at first

Old John

True!

BenjaLim

12:51

but actually they allow you to prove a lot of things

and with diagrams

a lot of this are in front of ya

like

Old John

I took a book on AT on holiday last week - and it was dreadful

BenjaLim

ya know what I mean yeah?

@OldJohn which one?

Old John

@BenjaLim yep

Jonas Teuwen

@OldJohn Haha, I told you to take something easy!

Old John

a tiny book called "An intuitive approach to AT" I think

BenjaLim

12:52

@OldJohn Actually

I would suggest Bredon

I think a lot of things are laid out in front of your face

For me when I see the algebra

Old John

@JonasTeuwen I broke my rules and also took an easy one : Elementary algebraic number theory - it was a good read :)

BenjaLim

and all the details

I find it extremely satisfying

Old John

@BenjaLim I will look out for that one

BenjaLim

one of the reasons I did maths was

It allowed me to understand and work out everything to the last detail

Old John

at uni (40 years ago) I had a terrible book on AT by Hilton and Wylie

BenjaLim

12:53

I don't know that book

Old John

@BenjaLim you are not missing anything!

BenjaLim

but then that was probably still the beginning

in algebraic topology

Like I bet homology was first introduced not too long before that.

Old John

I am sure more modern AT books are much better than they were in those days

BenjaLim

@OldJohn en.wikipedia.org/wiki/Eilenberg%E2%80%93Steenrod_axioms

Actually

the axioms are like all you need to remember

Once you learn homology

Old John

@BenjaLim Yes, I believe so - one day I must look into AT again - I regret not properly understanding it as an undergrad

BenjaLim

12:55

But once you learn homology

the axioms become very natural

and you remember them of the top of your head.

Old John

@BenjaLim that sounds good :)

BenjaLim

I guess a lot of maths is like that.

Old John

But, I must go and do some other stuff - I will be back later

BenjaLim

Ok

Old John

@BenjaLim yep

BenjaLim

12:57

see ya

John Junior

later

Old John

Have fun, folks :)

user19161

@BenjaLim Now you sound like an expert.

BenjaLim

as they say @OldJohn

@OldJohn come hang man

that's what ma mates say

Old John

later!!

BenjaLim

12:58

@WillHunting I saw so many sequences the past few days

user19161

@BenjaLim Spectral sequences?

BenjaLim

I need a rest from all the diagram chasing

user19161

Oh, exact sequences.

BenjaLim

rather I mean facts like to every ses of chain complexes you get a les in homology

user19161

How is the Munkres AT book?

BenjaLim

12:59

I am using bredon heavily now

user19161

Ha! I told you dude!

BenjaLim

I find it easy to understand

user19161

Bredon is the best book on AT. It even gives you GT and DT!

BenjaLim

yeah

nite

user19161

But I will be studying from the Lee books instead first.

user19161

13:02

@BenjaLim Night! See you in your dreams!

Jonas Teuwen

I love spectral measures.

Jayesh Badwaik

@JonasTeuwen Is spectral measures related to this ?

The spectral treatment is one hell of a converging solution method for DEs.

Jonas Teuwen

13:21

Yes, it is.

The expansion is just integration wrt the spectral measure.

Jayesh Badwaik

@JonasTeuwen Okay. We used a finite element implementation of it from a paper to do our stuff for Bachelor's. I always used to wonder, how come people think up all such things. Now,I am starting to see behind the curtains.

Parth Kohli

14:07

Quod erat demonstrandum.

Is @JohnJunior the same as @OldJohn?

Jonas Teuwen

14:20

No, more like a son he never had.

MJD

Is it most correct to spell Weierstrass as "Weierstrass" or as "Weierstraß"?

Helmut

Wirestross

MJD

@Hamlet Thanks!

Helmut

people who use beta for the ss annoy me

must we write chinese authors names in the original ideogram?

MJD

That is not beta; it is LATIN SMALL LETTER SHARP S.

HTH

Helmut

14:32

it doesn't help, because i have no use for small latin letters not in the English alphabet when writing people's names

MJD

How sad for you to live in a world where not everyone uses your preferred language!

Old John

It annoys me when people insist on Gauß - just because I don't know how to do it on my keyboard

Helmut

when I'm in Old Germany I do as the Old Germans do. i haven't been there in awhile, though

MJD

It annoys me when people insist on "L'Hôpital", which is an anachronism. "Weierstraß" might be a similar anachronism, but I am not sure.

Actually it annoys me whenever anyone insists on any particular orthography. But I would like to be correct in my own writing without insisting on anything.

Helmut

i insist on Григо́рий Я́ковлевич Перельма́н for Perelman

user19161

14:38

@helmut Hey are you the author of the LaTeX book?

Helmut

i will walk out of lectures that do not correctly attribute his name, and instead replace it with a vulgar Americanism. :snob:

i'm the author of the local-global principle. i don't do latex

MJD

That's funny, Perelman's middle name is the same as Khinchin's.

Nick Kidman

hi

MJD

I suppose Я́ковлевич is fairly common.

Nick Kidman

In a computer, addition is appearently implemented via some bit tricks, i.e. if you have 001 and 001 is sees that it makes 010 and so on. How does this relate to primitive recursive functions though? I though you need these for computation.

MJD

14:41

If I changed my own name to Mark Я́ковлевич Dominus, I could write my initials as MЯD, which would be pretty cool.

Old John

But you should only do that if your father were called Я́ков - or something like that :)

user19161

@MJD In any case, you can be recognised by the potato.

MJD

Come to think of it, I could spell my middle name to "Яасон" and use Я legitimately.

Helmut

Is the space of all maps from a group to the circle compact? Is there some natural topology in which to make sense of that?

MJD

Ahem. "Ясон"

user19161

14:43

Since there is a potato in this chat, shouldn't there be a tomato as well?

MJD

@NickKidman That is a really strange question.

Old John

Hmm - John in Russian comes out as Джон - not that great :(

Nick Kidman

@MJD: ...

MJD

@OldJohn Or as Иоанн or Иван.

Old John

@MJD yep

My surname in Arabic is a bit interesting: ووردزورث

@robjohn Exactly - actually the famous guy's name :)

robjohn

14:56

@OldJohn I would be surprised if mine translated into Arabic.

Nick Kidman

@MЯD: ...

robjohn

@OldJohn Hmmm... it says جونسون

Old John

@robjohn Yes - that is pretty close :)

Arabic does not have an "o" vowel, so uses "u" instead (it is sometimes pronounced "o" in some dialects)

robjohn

@OldJohn the last "o" is pronounced as in sun anyway

Old John

some names are harder as Arabic doesn't have things like a letter for "p" :(

@robjohn ah! - true

Jayesh Badwaik

15:13

@NickKidman You can define additions as recursive. Not necessarily adding incremental recursive, but bitwise recursive.

Nick Kidman

@JayeshBadwaik: I don't necessarily understand what you mean or how it answers the question. like...with register machines, or turing machines, you have a general framework to do any computation. If for addition (and maybe other arimentrical operations) you use the implementation via bits like that, you might be more efficient, but then what guaranies you that you don't lose any computational power?

What is the set of all things implemented into the computer, so that all in all, you are sure to be turing complete?

I mean why do you do implement specific taskes like addition in such a hardcoded sense?

Jayesh Badwaik

@NickKidman Efficiency? Else, our computers will be so so slow.

If you have seen a turing algorithm for simple addition, you must know how long it is.

Nick Kidman

Shouldn't the universal machine be able to do anything, for example addition. If it's just about efficiency, what is the list of things which are hard coded like that?

Jayesh Badwaik

@NickKidman Okay. So, basically, things which are hard coded is mainly influenced by market demand.

Are you familiar with the RISC/CISC debate?

Nick Kidman

addition seems to be trendy

nope, elaborate on the debate

Jayesh Badwaik

15:23

I would rather you read about both the systems on wikipedia.

Reduced instruction set computing

Reduced instruction set computing, or RISC (), is a CPU design strategy based on the insight that simplified (as opposed to complex) instructions can provide higher performance if this simplicity enables much faster execution of each instruction. A computer based on this strategy is a reduced instruction set computer also called RISC. Various suggestions have been made regarding a precise definition of RISC, but the general concept is that of a system that uses a small, highly-optimized set of instructions, rather than a more specialized set of instructions often found in other types of ...

Complex instruction set computing

A complex instruction set computer (CISC, ) is a computer where single instructions can execute several low-level operations (such as a load from memory, an arithmetic operation, and a memory store) and/or are capable of multi-step operations or addressing modes within single instructions. The term was retroactively coined in contrast to reduced instruction set computer (RISC). Examples of CISC instruction set architectures are System/360 through z/Architecture, PDP-11, VAX, Motorola 68k, and x86. Historical design context Incitements and benefits Before the RISC philosophy became promin...

Apart from that, you may also want to read about SSE instructions.

The theory of computation is good, but the implementations are very very convoluted.

For example, even now, engineers hand-design a lot of transistors in the intel processor chip to get things right. Even after so much automatization.

Anyway, coming back to the topic,

the basic thing is a CPU can process one string of 64-bit at a time. If you have instructions such that you can obtain any $2^64$ numbers, then your machine is effectively turing complete.

So, basically, the only thing you need is Load/Store and Increment/Decrement.

However, doing so would be foolish, since it is so so inefficient.

So, all the normally used tasks are hard coded.

So, for multiplication/division, you have floating point units, there are special units for cryptography and special units (SSE) for vector addition. There is also a concept of redundant register sets, to minimized the time of a context switch in the processor.

Parth Kohli

This thing shows that I have a +64 change. What does that mean?

@Jayesh: How did you embed that?

Jayesh Badwaik

@ParthKohli Wikipedia and Youtube are automatically embedded.

Parth Kohli

Oh, gotcha.

Jayesh Badwaik

@ParthKohli This week rep change is given there.

Parth Kohli

@Jayesh: My week reputation is different than change.

Jayesh Badwaik

15:34

@ParthKohli Yup, your reputation is last week rep + change.

Parth Kohli

So week rep is last week's rep?

Jayesh Badwaik

No wait, I am not sure.

Change is the rank change.

And week rep is the rep gain in the week.

math101

Hey guys

@JayeshBadwaik are you familiar with finding error bounds for polynomial interpolation

Parth Kohli

@math101: Hey!

math101

Hey Parth

Jayesh Badwaik

15:46

@math101 I was, I may not remember exactly now. Also, I am somewhat busy right now. Sorry?

math101

awww too bad

robjohn

Time to take Lilly to the vet for a checkup. bbl

math101

Hope everything goes well with Lilly :)

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$Mathematics$ Mathematics