Mathematics - 2015-03-18 (page 1 of 4)

Ted Shifrin

00:00

what you said is true for free abelian groups, @Paul, for sure.

Paul Plummer

@TedShifrin The free group on 2 generators contains any countable free group but it cant have a surjective homomorphism to a free group on more than 2 generators

Mike Miller

No you can't, @Ted. It has those as subgroups, but there's no reason you can surject onto a subgroup.

Ted Shifrin

Oh, right, that's what I was remembering. Even weirder. Of course. Thanks.

<--- old and forgetful

Mike Miller

I find it less weird, personally. I see no reason you can't pick a bunch of "algebraically independent" letters in the free group; indeed I would generically expect that this be true.

@PaulPlummer: Is there an obvious, non-topological reason free groups should only have finitely many subgroups of a given finite index?

Oh, nevermind. Subgroups of index dividing n are the same as the stabilizer subgroups of an action on a finite set, of which there are finitely many (actions are determined by how the generating elements act). Similar to the topological idea.

Paul Plummer

@Mikemiller Well there is for normal subgroups of finite index, and I think one could follow a similar reasoning to get for plain sugroups

and if I thought about it enough I think it would come down to just what you said, studing the actions on finite sets

Committing to a challenge

00:18

Why does noone care to use the sub-rooms?

Mike Miller

Don't see a point.

Semiclassical

as a place to carry out a conversation on a specific topic, they seem useful. the main chat room, by contrast, is more spontaneous and so used more

Paul Plummer

and there doesn't seem to be enough activity in any one chatroom to "force" people to move to a different chat room

Mike Miller

right. the people I like to talk to, as far as I know, come to this room. and they like to talk about the things I like to talk about in tnis room.

and if they stop, I can do something else with my life :)

Alizter

@MikeMiller so he says

Semiclassical

00:25

yeah. i mean, one thing I probably should do is drop my recent bounty on my determinant question on the linear algebra chat

but i prefer this as the place to ask random questions that i don't want to put on the main site

Committing to a challenge

I just think that long discussions on certain topics should go in that topics room, so then multiple discussions on different topics can work concurrently

I know trying not to interrupt a conversation has deterred me from talking in the past

hftf

@Alizter Is it required for question askers to try something?

Mike Miller

shouldn't let it, @Committingtoachallenge

too much of a hassle for me to check other topics rooms. i'd have to add more bookmarks :p

Committing to a challenge

I shall just link to the appropriate room in here when I have a question

Semiclassical

that's the other thing, yeah

though when you're in multiple rooms, you should see the links to the other chats at the same time

Committing to a challenge

00:31

You know if you open up the site rooms page and ctrl click all of the relevant ones and close the tabs all of them stay on the right of the page until you log off?

Mike Miller

meh

Committing to a challenge

I am currently in 6 rooms with only one chat tab

Semiclassical

it's one of those things where, if one was willing to use it, it'd help

but ehh

Mike Miller

i also can't really think of a situation in which i would want to use one of the other rooms, barring the mod room

columbus8myhw

00:51

I never realized that the Well-Ordering Theorem was so easy to prove.

You legitimately choose random elements, one-by-one, until there aren't any left.

(With minor details, of course, since usually the set to be well-ordered is uncountable. Basically, if you've already chosen an infinite amount of elements, and there are still more left, keep choosing elements.)

Committing to a challenge

LMAO

ncbi.nlm.nih.gov/pmc/articles/PMC1311997

Journal: The unsuccessful self-treatment of a case of “writer's block”

ᴇʏᴇs

01:07

lol

A Beautiful Mind

Hi Bart and Alex.

You are my two best friends in this chat.

Committing to a challenge

Read the journal entry above please Jasper, I assure you it is concise

@ABeautifulMind That is good to hear :).

A Beautiful Mind

@Committingtoachallenge It's only one page?

columbus8myhw

I'm trying to get it do something like this:

> Text

but blank

It didn't work.

Committing to a challenge

$\text{}$

Wait what are you trying to do?

$\quad$ This? but without chatjax

ᴇʏᴇs

01:10

A vertical gray line

Committing to a challenge

With or without latex?

A Beautiful Mind

@Committingtoachallenge It is a hoax.

ᴇʏᴇs

With

Committing to a challenge

The unsuccessful self-treatment of a case of "writer's block"

The unsuccessful self-treatment of a case of "writer's block" is a humorous academic article by psychologist Dennis Upper about writer's block. It contains no content outside title and journal formatting elements, including a humorous footnote. Published in 1974 in a peer reviewed journal, Journal of Applied Behavior Analysis, it is recognized as the shortest academic article ever and a classic example of humour in science, or at the very least among behavioral psychologists. It has been cited at least ten times. The article received a humorous positive review which was published alongside the...

columbus8myhw

> test

> test

Committing to a challenge

01:12

Not a hoax, but it is funny

columbus8myhw

That's how you do it.

ᴇʏᴇs

@columbus8myhw How

columbus8myhw

Type this:

> test

@ᴇʏᴇs

Committing to a challenge

> meow

$\begin{array}{c|c} &\text{Meow}\end{array}$

columbus8myhw

> > meow

rebel <

So... Likud or Labor?

(If you could vote, that is.)

Committing to a challenge

01:14

Liberals

columbus8myhw

That's not a party.

Committing to a challenge

I don't know what likud is

We have liberal and labor in straya'

columbus8myhw

It's basically Joint Arab List, Labour, or Likud. Everyone else is too tiny to matter...

@Committingtoachallenge Dude, you don't know what today is?

(Was)

Committing to a challenge

@columbus8myhw ???

A Beautiful Mind

@columbus8myhw Today is my appointment with the doc.

columbus8myhw

01:15

Election day in Israel

It's a really big one.

Netanyahu might finally lose power, but they're not sure yet.

*Bibi

A Beautiful Mind

Just use batteries.

Committing to a challenge

Australia doesn't care much about this stuff sadly

columbus8myhw

OK. Basically, it's between two people named Bibi and Buji.

(They have real names, but that's what everyone calls them.)

Bibi is, well, you know, Bibi.

Buji is more center-left.

A Beautiful Mind

And Baba?

columbus8myhw

(And by "Bibi" I mean Benjamin Netanyahu.)

One of his campaign ads had a hilarious pun on his name.

> You wanted a babysitter? You got a Bibi-sitter!

A Beautiful Mind

01:22

@Committingtoachallenge @ᴇʏᴇs I am afraid I may never ever get well. Then I am screwed for life.

Committing to a challenge

@ABeautifulMind If you don't get well soon I'll have to visit you

And maybe you can learn to cook so you can cook me a meal lol

A Beautiful Mind

@Committingtoachallenge What upsets me most is that if I had reacted differently to the abuse, I would not have gone mad. I blame myself for that.

Committing to a challenge

@ABeautifulMind It's not your fault

@ABeautifulMind You can't choose how you react, we aren't machines

Why do I suck at real analysis so much :'(

A Beautiful Mind

@Committingtoachallenge Drawing pictures help.

Committing to a challenge

I am going to try 'Stephen Abbott's Understanding Analysis'

A Beautiful Mind

01:31

@Committingtoachallenge What problem do you have actually? Maybe I can help.

Committing to a challenge

Based on this

@ABeautifulMind It's not a problem set that I am having trouble with, it is everything :\

I have trouble building actual proofs

A Beautiful Mind

@Committingtoachallenge That's what I am asking actually.

Committing to a challenge

I can't seem to solve any problems is the heart of the problem

A Beautiful Mind

@Committingtoachallenge Drawing pictures help to motivate your proof. I see a sequence converging to a point as dots getting closer and closer to some point.

Committing to a challenge

Like a question $\{a_n\}, \{b_n\}$ are cauchy, prove that $\{a_n + b_n\}$ is cauchy

Seems obviously true, but I don't know how to write it rigorously

A Beautiful Mind

01:34

@Committingtoachallenge Well, the a's get closer and closer and the b's get closer and closer right?

Committing to a challenge

And helping me with that won't help me with my lack of proof abilities unfortunately

A Beautiful Mind

So that means somehow the sums also get closer and closer.

Committing to a challenge

I can understand exactly what it is saying, but I can't write my epsilons and M,N's

A Beautiful Mind

And you start by writing down the distance between the sums.

And try to express them in terms of the distances.

Bingo.

See this natural thought process?

That is math in progress.

Committing to a challenge

These are my thoughts meta.math.stackexchange.com/a/5101/142198

Mike Miller

01:36

@Committingtoachallenge What you want to be small is $|(a_m+b_m)-(a_n+b_n)|$. Does that inspire any thoughts?

Committing to a challenge

Put the cauchy sequences together, and get them less than \epsilon /2?

A Beautiful Mind

Can you express the distance between the sum as the sum of the distances or something like that?

Mike Miller

Reasonable idea. Use facts you already know to turn that into a proof.

Committing to a challenge

$|a_m - a_n|\lt \frac\epsilon2 + |b_m - b_n|\lt\frac\epsilon2$

A Beautiful Mind

Yes, go on.

Of course, for the a's and for the b's there will be different M, so you take the max of the two M's.

Committing to a challenge

01:40

$\text{above}\lt\frac\epsilon2 + \frac\epsilon2=\epsilon \implies \text{above} \lt \epsilon$

Oh okay, that makes sense

A Beautiful Mind

You need to turn all these into a coherent proof.

I am not sure you know the proof as yet.

Committing to a challenge

That actually doesn't seem too bad, thanks!

A Beautiful Mind

Some people think they get it but they don't.

Can you try writing it down in full properly?

That is when we will see the mistakes, lol.

Committing to a challenge

@ABeautifulMind I will in a min, just need to print some bank statements

Trying to get a place in the city

So we don't need to commute for two+ hours a day

and get up at 530

Actually now that I have done that I have to go to class, so I will do it when that finishes!

Thanks for your help Jasper & @Mike

@Abe Actually skipping class because I will be late and I have anxiety for late entrance

A Beautiful Mind

02:00

@Committingtoachallenge It is not your fault.

Committing to a challenge

@ABeautifulMind xD

Well I could have left earlier and it wouldn't have happened so in some sense it is my fault

I took beta blockers for a week and I felt really really relaxed and had no anxiety, but I felt like a different person so I stopped taking them

I don't know if I should retake them or not sometimes, I can't make or take phone calls unless I prep a few hours before hand

A Beautiful Mind

@Committingtoachallenge Scared to talk to who?

Committing to a challenge

@ABeautifulMind Anyone that isn't direct family or my girlfriend

David Wheeler

02:15

Most proofs involving sums and limits usually wind up invoking the triangle inequality at some point.

Semiclassical

Fun triangle inequality fact: when people talk about the twin paradox in special relativity, it pretty much amounts to the reversal of the triangle inequality in minkowski space

Committing to a challenge

@Semiclassical Fun

" "

David Wheeler

That is because the minkowski metric is only a pseudo-metric

Semiclassical

right. one has to be careful that one is talking about timelike paths (which is physically reasonable, since observers don't move faster than the speed of light)

Committing to a challenge

02:34

I have no idea how to spend my time studying right now...

David Wheeler

neither do I, for I know not what's at stake

Committing to a challenge

I can do a textbook and I will learn it properly, or I can do an assignment and get the grades, or I can read the course work, which seems to have holes(but is most related to grades), then even if I do the textbooks I don't know which is best for me to do

Then I have 5 courses all with different expectations and I have no idea how long each are meant to take me, so I can't dictate my time well

Maybe this is why people focus on one field of math......

If I just do the assignments I will do bad on the exams, and if I do the textbooks I will run out of time for the assignments and do well on the exams

ᴇʏᴇs

What's an example of a set in a topological space that's compact but not bounded and not closed

A Beautiful Mind

Metric space then, since you have bounded.

ᴇʏᴇs

We have not bounded

Committing to a challenge

02:40

Pretty sure a compact set is closed

Also I thought they were bounded necessarily

A Beautiful Mind

A compact metric space is closed.

Committing to a challenge

Maybe this is just for metric spaces

ᴇʏᴇs

No, the Sierpinski topological space has not closed but compact sets

A Beautiful Mind

BUt boundedness implies metric right?

ᴇʏᴇs

I don't know

A Beautiful Mind

02:41

I think so. Where did you get question from?

ᴇʏᴇs

I'm just thinking

A Beautiful Mind

I have said everything above, so question is wrong.

Wait, maybe I did not read carefully...

Hmm, seems I am right...

ᴇʏᴇs

Oh yea, I think every non-empty compact set is bounded in any topological space

A Beautiful Mind

Boundedness implies metric right?

ᴇʏᴇs

I don't know about that

I wouldn't know how to prove it

Paul Plummer

02:44

@ᴇʏᴇs I think abeutiful mind is asking if you want the space to be a metric space since you mention boundedness and normally that is defined only for metric spaces (I actually don't know a def that does not use a metric)

A Beautiful Mind

What is the definition of boundedness in non-metric space?

ᴇʏᴇs

Oh okay

So by definition boundedness implies metric

Paul Plummer

If you do want a metric space then every compact space is bounded,

A Beautiful Mind

@PaulPlummer And also closed right?

Paul Plummer

I think so but not 100%, pretty sure if a space was not closed you could construct an infinite cover

A Beautiful Mind

02:48

I have forgotten this stuff, but I am sure it is pretty easy to prove.

Ah yes, a compact metric space is totally bounded and complete, so it is bounded and closed, lol.

Paul Plummer

yah

in fact compact subsets of Hausdorff spaces are closed.

Dave

Any of you guys know how to do the underline style shown on second like here: math4ged.com/wp-content/uploads/2011/07/… in microsoft word? i don't know the name of the underline style

ᴇʏᴇs

You can do that in Word?

Dave

03:03

no idea but its written like that in my book and it looks neater than ordinary underline i like the up flicks to seperate the terms better

easier to read

A Beautiful Mind

@ᴇʏᴇs 4 more hours to my appointment.

ᴇʏᴇs

I don't know about Word, but in LaTeX I think it's like \underbracket{} or something

Or not

Dave

i've never used it

i've written most of my assignment in word now :P

i guess using ordinary underline will be sufficient

i'm just ocd :P

Semiclassical

take a look at this question from tex SE:

20

Q: \overbrace with square bracket

I'm trying to put a nice square bracket over a complex text, it should look like an \overline but with ending like \ulcorner and \urcorner. The presence of \overrightarrow and \overleftarrow makes me believe that is possible to decorate the end of the \overline bar. I know the existence of \ove...

Dave

thats for LaTeX no ?

ᴇʏᴇs

03:08

Yea you can't do that in Word

Dave

is it not a standard ascii symbol ?

Semiclassical

ah, i misread your query

David Wheeler

here is an example: the interval $[0,1]$ is closed and bounded in the $K$-topology, but not compact.

ᴇʏᴇs

@DavidWheeler I was looking for an example of compact but not closed and not bounded

But it's not possible as some people have said

infinitesimal

Thanks for sharing that answer @Semiclassical :-)

David Wheeler

03:10

$\Bbb R$ in the indiscrete topology is compact, but not bounded.

Dave

does this make sense to you guys :

5dt + 2td = 5dt + 2dt = 7dt

ᴇʏᴇs

Oh what really

Dave

this example literally makes no sense in my book

A Beautiful Mind

@DavidWheeler What is bounded in a topological space?

Semiclassical

context?

ᴇʏᴇs

03:11

Is this a definition for totally bounded for a non-metric space?

Dave

recognizing like terms

David Wheeler

well, it's a funny thing-bounded can have different meanings

Semiclassical

oh, so it's algebra? yeah, that's something you should be able to recognize

Dave

why does it swap 2td to a dt ?

Semiclassical

because they want the two terms to be of the same form before combining them

Dave

03:12

it doesn't explain it just shows an example with no explaination for why it does it

David Wheeler

@Dave What are t and d?

Dave

@DavidWheeler they have no context its just example for simplifying expression using like terms

David Wheeler

If they are numbers (field elements) we have commutativity of multiplication

Semiclassical

let me write it a little differently: 5dt+2td=5dt+2dt = (5+2)dt=7dt

in the first equality, i use the fact that (assuming we're working in a simple context) the order in which variables are multiplied doesn't matter

Paul Plummer

$\mathbb{R}$ with the indiscete topology is bounded since $d(x,y)=0$ for all elements in $\mathbb{R}$

Dave

03:14

oh so its read as 2 * (t *d) ? meaning t * d can be either way

Semiclassical

right, in the same way that 10 is both 2*5 and 5*2

Dave

okay i get it now. so expressions never use more than one letter to represent something

David Wheeler

@PaulPlummer Not really, there's no metric.

Mike Miller

pseudometric spaces are silly

David Wheeler

The question: are compact sets closed and bounded, hides some necessary information about context.

Semiclassical

03:17

@MikeMiller: the spacetime interval between two spacetime points on the path of a light ray is zero. hence, that interval had best be a pseudometric :P

Paul Plummer

To be bounded or not bounded there needs to be a metric, at least any defintion I have seen for those

David Wheeler

"closed" is clear-bounded implies some way of distinguishing

ᴇʏᴇs

I know an example of a compact set that is not closed, but I was curious if there's an example of a compact set that is not closed and also not bounded or just not bounded

David Wheeler

Unfortunately, inclusion is only a partial order

Mike Miller

still silly, @Semiclassical :)

Semiclassical

03:18

pfff

Mike Miller

@ᴇʏᴇs to summarize what's been pointed out: there's no notion of boundedness in a topological space, so one has to talk about metric spaces. there, compactness is equivalent to complete + totally bounded; complete implies closed, totally bounded implies bounded

if you want a noncompact set in a metric space that's bounded, pick the closed unit ball in an infinite-dimensional Banach space

Semiclassical

i'd make some joke regarding the twin paradox at this point, but uh

no can do

Mike Miller

appreciate it

Semiclassical

mostly b/c nothing is coming to mind, alas

ᴇʏᴇs

Is boundedness not useful in general topological spaces?

Semiclassical

03:23

pet peeve of the day: finding a useful review paper that covers something strongly related to what you want, but explicitly sidelines the thing you're actually interested in :P

Mike Miller

it doesn't make sense in general topological spaces

you can't define it

Semiclassical

"Our focus is always on geometric rates of convergence associated with
analytic functions, not on the next-order algebraic estimates that depend on behavior at the edge of analyticity." but it's the next-order stuff i care about now, sigh

ᴇʏᴇs

Oh

David Wheeler

My example before was a bit tongue-in-cheek-it is natural to associate the "bounded" condition in the real numbers with the natural field order. But $\Bbb R$ can have many topologies on it.

A logician would call this a "type-mismatch"

Mike Miller

point being: (0,1) is homeomorphic to the real line

David Wheeler

03:37

$\arctan\left(\pi\left(x - \dfrac{1}{2}\right)\right)$?

Mike Miller

something like that probably works. the one i had in mind was $$\log\left(\frac{x}{1-x}\right).$$

David Wheeler

lol, i get confused with trig functions

Semiclassical

i usually think in terms of stereographically projecting the real line to a semicircle, and then projecting that down to the interval (0,1)

though i usually do that with (-1,1), since that's the least tedious version of that

Mike Miller

if we all trade our favorite homeomorphisms it will take a while, since there are quite a few self-homeomorphisms of $\Bbb R$

Semiclassical

heh, yes

ᴇʏᴇs

03:41

I don't know any :(

Mike Miller

a self-homeomorphism of $\Bbb R$ is the same as an increasing surjective function

Semiclassical

another simple one is just (x-1/2)/(x-x^2), i.e. choosing the most obvious rational function to do the job

David Wheeler

@Semiclassical Yeah, I was working on that-had the denominator, but needed a "sign-switching" function, I suppose some shifted trig function on top would do, too

ᴇʏᴇs

Why can't it be decreasing

Semiclassical

right. the stereographic way generically involves a tangent function (though i forget which direction)

Mike Miller

03:48

sorry. decreasing works fine

Semiclassical

arctanh is another obvious one

David Wheeler

Normally, if you find a continuous bijection $(0,1) \to \Bbb R$, the inverse function will be continuous, too, unless it's really bizarre

Semiclassical

is there a canonical example of 'really bizarre' in this context?

ᴇʏᴇs

Oh it has to be surjective from $(0,1) \to \mathbb{R}$?

Mike Miller

@Semiclassical no, because every such continuous bijection is a homeomorphism

Semiclassical

03:50

hmm

Mike Miller

just installed texlive for the first time on an old computer, trying to compile a document. click the start button for pdflatex (in TeXworks) and nothing happens.

ugh

David Wheeler

@MikeMiller oh yeah, that's right, because we can use compactness arguments (darn it, $\Bbb R$ is too special)

Mike Miller

same is true on a manifold in general by invariance of domain (where by 'same' i mean 'every continuous self-bijection is a homeomorphism')

ᴇʏᴇs

I use miktex and texniccenter

Mike Miller

the miktex download page didn't work for me :(

i have web-based solutions i've used until now but this is too large a file

ᴇʏᴇs

03:56

Oh

I use Overleaf a lot now because I'm getting lazy transferring files from my USB drive to the school computer to print it out and such

(formerly writelatex)

Mike Miller

yeah, i use that and LaTeXLab, a google docs based latex editor

ᴇʏᴇs

Oh wow I didn't know Google had that

Might switch to that instead so I can use it in conjunction with Google Drive

Mike Miller

it doesn't, it's a webpage that you go to that enhances google docs

i find it very valuable

the only problem is that it takes way too long to compile

Stan Shunpike

I don't understand why rings are important.

In what situation might I use a ring where a field wouldn't suffice.

Paul Plummer

The ring of matrices

Stan Shunpike

04:10

That's a good example, I have never heard of that.

Hmm....you are right. Interesting.

Awesome.

Semiclassical

the ring of integers: multiplication and addition make sense, but you don't have multiplicative inverses

Kaj Hansen

Integers mod $n$ if $n$ isn't prime

Stan Shunpike

Wow, that's cool too.

@Semiclassical because you don't have things like 1/2 to serve as an inverse to 2 , is that what you mean?

Semiclassical

right

Kaj Hansen

Oh, an important one: The set of polynomials over a field forms a ring

Stan Shunpike

04:12

How does that form a ring?

David Wheeler

You can add them and multiply them

Semiclassical

multiplication and addition of polynomials are well-defined, but not all polynomials have multiplicative inverses

Stan Shunpike

Example?

Dave

I need help with an explaination

David Wheeler

In fact the invertible polynomials are non-zero constant polynomials

Dave

04:15

I'm confused why this: 3 + 64a^2 – 32a^3 when simplified is -(32a^3-64a^2-3)

How is that simplified it just looks re-arranged?

David Wheeler

the power set of a set forms a ring under symmetric difference as + and intersection as *

Semiclassical

depends on the definition of simplified. i really wouldn't count that as a simplification either

Kaj Hansen

Think about the polynomial "x" in $\mathbb{Q}[x]$

Dave

wolframalpha considers it the simplified answer =/

Kaj Hansen

The way multiplication works with polynomials, degrees can only go up.

Semiclassical

04:16

yes, and wolframalpha is a computer program

Kaj Hansen

So there cannot be an $f(x)$ such that $f(x) \cdot x = 1$.

David Wheeler

@Dave wolfram alpha can go suck a lollipop

Semiclassical

just b/c it spits something back at you doesn't mean it's holy writ

Dave

oh :P

Stan Shunpike

Is Q[x] the set of rational polynomials?

Semiclassical

04:16

to be fair, wolfram alpha is useful

Dave

it makes me doubt myself :P

David Wheeler

@Semiclassical W|A is just a glorified calculator

Semiclassical

but it's giving you a simplified answer, not -the- simplified answer

Dave

I'm not sure i can simplify it further

Semiclassical

true, though a calculator that can give you integrals in terms of special functions is nice :)

Stan Shunpike

04:17

@infinitesimal hola amico

@KajHansen Is Q[x] the set of rational polynomials?

Kaj Hansen

Indeed @StanShunpike

David Wheeler

$3 + 64a^2 - 32a^3$ (which is a polynomial expression in $a$) is already "simplified", unless there is a non-trivial factorization

infinitesimal

Hi pal @stan

:-)

David Wheeler

factorization is a "complicated" subject, though-what you allow the factors to be has a lot to do with what is possible

Stan Shunpike

@infinitesimal I just learned why rings are cool! Awesomeness

Semiclassical

04:19

and there probably isn't, since the roots are just terrible

not over the complex numbers, at any rate

infinitesimal

cool

Dave

can't the 64a^2 - 32^3 be subtracted from each other?

David Wheeler

@StanShunpike Factorization is actually one of the reasons we study rings

Dave

although i would get 32a^-1

Stan Shunpike

Really? @DavidWheeler I'd like to learn more about polynomials. I feel like I know nothing. What's a good book to read about them? My uncle gave me this book Topics in Algebra, but I don't really think they cover polynomials extensively in there

David Wheeler

04:21

@StanShunpike That has a couple of chapters on polynomials

Stan Shunpike

Does it? Good, I'll start with that then.

It's a nice book.

@infinitesimal I've been confused for like a year about them. definitely a eureka moment

David Wheeler

Before studying polynomials, however, you want to know some "ring basics"

Semiclassical

an example where the commutative ring v. fields distinction is practical is that of block matrices

suppose i've got a big nm-by-nm matrix whose elements are from some field F

i can take the determinant of that, no problem

infinitesimal

@StanShunpike now you know how Archimedes felt :P

David Wheeler

Well, it's also important in systems of simultaneous linear equations, in general

Stan Shunpike

04:24

@DavidWheeler Rings are?

David Wheeler

often, one wants solutions in the system that are integer solutions

Semiclassical

now, suppose i partition that matrix into m-by-m blocks (for a total of n^2 blocks)

Stan Shunpike

@Semiclassical okay. i follow i think

Semiclassical

and let me further assume that, for whatever reason, my matrix is of such a form that all of the m-by-m blocks commute

in that case, those blocks must all belong to some commutative ring, and i'll be able to think of the determinant of that n-by-n matrix

whose output will be a member of the ring, i.e. an m-by-m matrix. i can then take the determinant of that, and get the same determinant as i got from starting with nm-by-nm

(i have this on the brain for practical reasons. for a nice summary, see this paper)

Stan Shunpike

nice! that's awesome. I'm going to have to study this. I first saw block matrices when I was reading about Lorentz transformations.

Semiclassical

04:29

i actually don't much like the proof in that note, since it seems like there should be some more elegant way to think about it

might make a main site question out of that

Stan Shunpike

I give my +1 in advance lol

David Wheeler

The trouble, when generalizing from field elements, to matrix elements, is the loss of commutativity.

For example, with matrices, people often try to factor $A^2 - B^2 = (A + B)(A - B)$

But that doesn't work if $AB \neq BA$

Stan Shunpike

Yeah, exactly

Semiclassical

right. the lack of commutativity in general is what makes block matrices difficult

Stan Shunpike

Matrices like stunned me. They are the first time I ever noticed something wasn't commutative and I was like 8o

Semiclassical

04:44

though i wish i knew a nice article on block matrices from a more formal perspective

David Wheeler

Fortunately, there is a ring of matrices that is commutative-one can form the ring $R[A]$ for a commutative ring $R$ and a specific matrix $A$.

Semiclassical

most of the stuff i know seems rather piecemeal, and i imagine there is some higher-level POV

Stan Shunpike

@DavidWheeler really? That's amazing. I didn't know there were such examples.

Does any know a good book on complex analysis? I haven't studied it at all and it seems needed for QFT.

David Wheeler

$R[A]$ is a ring of "matrix polynomials", commutative since $AA = AA$

and by induction $A^kA = AA^k$

Stan Shunpike

That's amazing too. I have never really considered these properties.

I didn't realize people even studied these things.

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$Mathematics$ Mathematics