Does anyone have an example of a real-world thing that is naturally viewed as being infinite-dimensional?

waves?

yeah, I know about Fourier series representations of signals and such, but I am trying to explain this to my non-mathematical friend, and I don't want to overwhelm him with these kinds of technical things. Anything lower-tech?

So, for instance, I was able to impress upon him that he thinks in high-dimensional space all the time by appealing to finance: the market can be viewed as a 1d curve sitting in R^n where n is the number of commodities under consideration

parameterized by time, of course.

indeed, dimensions and variables / degrees of freedom can be interchanged conceptually

right, but what sorts of phenomena in the real world have infinitely many such degrees of freedom?

or even, thought experiments that are easy to explain, like the infinite hotel or something

there's a distinction between hilbert space and vector space to keep in mind when it comes to infinity

in the former I think you can have infinite sums of orthonormal vectors, but not in the latter

right it's a complete metric space

but like, I'm trying to stay away from stuff that needs a degree in math, but can still be viewed has having infinitely many dimensions

perhaps that's too much to ask for?

I can think of a contrived example

I'd love to hear it

You have an infinite hallway in Hilbert's hotel. In each room there are various things like furniture etc. that is worth money. The hotel staff moves everything on one side of the hallway to the other side, thus increasing the value of the rooms in the latter. We can view this as an infinite-dimensional array of values - two of them that we are adding together.

that's pretty good

I suppose it's also not too much of a stretch to imagine a market with infinitely many commodities, once you've convinced yourself it works for huge values of n

EUREKA. I just figured out how to prove what I've been trying to prove. And it takes very little machinery!

Congratulations, anon. I think I just discovered something too. Could you confirm that in $\mathbb F[x]/(x^n)$ the units are exactly the elements $a_0+a_1x^2+\ldots+a_{n-1}x^{n-1}+(x^n)$ such that $a_0\neq 0$? I think I found a nice way of proving this.

($\mathbb F$ is a field.)

That should be correct. I think you mean $a_1x^1$ though.

(Also, turns out my eureka moment has some bugs. :/)

Yes, sorry.

@anon Oh. :( This is always unpleasant.

My proof is this. We can embed $\mathbb F[x]/(x^n)$ in $n\times n$ matrices over $\mathbb F$ like this:$$a_0+a_1x+\ldots+a_{n-1}x^{n-1}+(x^n)\mapsto \begin{pmatrix}a_0 & a_1 & \ldots & a_{n-1}\\0 & a_0 & \ldots & a_{n-2}\\\ldots & \ldots & \ldots & \ldots\\0 & 0 & \ldots & a_0\end{pmatrix}$$

If I'm not mistaken this is a ring isomorphism onto such matrices, and now it's clear which ones are invertible...

That's more or less the proof given by "Galois Theory - Chapter II Field Theory" by Artin/Milgram, freely available online.

Thanks! I'll take a look at it.

(for sufficiently loose interpretations of "more or less")

(and "proof")

@anon is that remark directed at me?

It's probably standard but that's what I plan to read to familiarize myself with field theory.

Ooh, I just got another Necromancer badge, but not for the one @Rajesh thought I would.

@DavidWallace I didn't even know you were in the room!

@anon Just a coincidence then. I got pulled up just minutes ago for being too loose with the words "more or less" in a comment.

interesting

@anon This book is standard? Or the embedding?

The proof idea.

I wouldn't have thought it was possible to be too loose with these words, but never mind. Continue your Galois Theory conversation - I'll try to grok it from the transcript.

I think it might be standard indeed, although I haven't seen it before. But I've seen something similar that involved infinite matrices.

This is where I got this from.

@anon Could you point me to a page? I can't find it...

it's the 8th page in the pdf, labelled page 28

actually, I think it's a little different

yes, similar idea, but definitely different, because x^n is obviously reducible

I think I've got the wrong pdf. Is it this one?

the idea being "expand the product and re-interpret as a linear system," I guess

that's odd, mine looks the same but is labelled differently. labelled 20 on that one.

Oh, strange. OK.

It too difficult for me now to clearly see analogies between what's on that page and what I did. I need to think about it some more.

hey guys

Hi

yo

Hey, Ben.

yo wassup

Where are all the CA guys

@BenjaminLim Trying to understand infinite matrices better and how some things sit in them.

@ymar Ah ok

@ymar I'm pretty busy this week

I have an annoying analysis assignment to do

not to mention a galois theory assignment coming up as well

Analysis is annoying indeed.

@ymar The analysis course I am doing now is really putting me off analysis

Is the lecturer bad?

Geez it's half past 3

I'm sick so I didn't notice any difference. I've been tired all day.

He is worse than bad

I had pretty good lecturers in analysis. And I hate it with all my heart anyway.

@ymar Stick with commutative algebra

@ymar Can I ask you something?

@BenjaminLim Well, if I want to stick to algebra I need to pass some exams in other subjects first.

@BenjaminLim sure

I find in my life now

I don't have time to do anything else

except just maths

while people are planning to go skiing

or to go like on holiday or something

I don't have any kind of plans like that

the only kind of plans I have are like

like?

what to study next

Like I am planning to study commutative algebra for at least two semesters

Do you enjoy that?

well it is pretty intense

but @ymar I seem less able to relate to people

because like I have no plans nothing

my life seems very isolated from the rest

It is difficult to have conversations with people

@ymar Yeah so I'm just telling you because you do maths

and probably understand what I'm saying

I definitely do.

Problem is, when it comes to communication with other people I'm like the worst person to ask advice of.

why?

Perhaps I can help you out

you know @ymar what's the oddest thing ever

like kanna and I have never met in person

but then because of maths

I frequently speak to him on skype

it's crazy man

I think it's quite normal that people talk to people who share their interests.

yeah

@BenjaminLim I just like being alone. I'm not a misanthrope but I don't like deeper relationships. I feel uncomfortable with them.

ah ok

@ymar But certainly maths

it's crazy man it goes beyond all boundaries!!!

It does. :) I've never been on chats much, and here I am at 3:45 am

talking to a guy from Australia, while I rarely talk to my brothers. :)

hahahahahahaha!!!!!!!!!!

@ymar math.stackexchange.com/questions/137876/…

Now can anyone help me with a Galois Theory problem?

@DavidK We just learned about automorphisms of fields extensions, so prob I can't help you out yet!

Ok. :(

my galois theory is not so advanced :D :D :D

I'm probably no good either, sorry.

Ok, well how about this, what do you think "the squarefree part of the polynomial $f_{1}(x)f_{2}(x)$" means?

What's left after you divide out all repetitions of irreducible polynomials in its factorization?

@BenjaminLim Nice question. Do you know the answer?

I'd better be going. I hate going to sleep after sunrise.

See you guys!

No

@ymar I am going to think about it using localisation

@ymar Not quite. Just divide out even numbers of repetitions.

In this context $f_{1}$ and $f_{2}$ are separable with splitting fields $K_{1}$ and $K_{2}$ so, the compositum $K_{1}K_{2}$ is the splitting field of the squarefree part of $f_{1}(x)f_{2}(x)$

OK, I have no idea what any of that means. Apart from the word squarefree.

Oh.

Just divide out any polynomial that's the square of some other polynomial; until you can't any more.

guys

quick question

Yes, I think that's right now. In this context that makes perfect sense.

suppose you have a ring $A$

consider the multiplicative set $S = A - \{0\}$

What is $S^{-1}A$?

So the "squarefree part" of $(x+1)^2(x+2)^3(x+3)$ is $(x+2)(x+3)$.

Wait, why isn't it $(x+1)(x+2)(x+3)$ ?

Divide out squares. There's no $(x+1)$ left.

Hmm.... But that would change the splitting field....

Umm, OK, I'll just consult a few references to see if the words are used differently in different fields (no pun intended).

I mean, what you're saying makes terminological sense, but in this context I think we should be left with $(x+1)(x+2)(x+3)$.

Umm, whoops. It looks like the a different definition is more commonly used than the one I gave you. Sorry to put you crook.

You're quite right - in my example it's $(x+1)(x+2)(x+3)$.

Ok. Great.

In other words, you divide away individual factors until you're left with one that is squarefree.

mathworld.wolfram.com/SquarefreeFactorization.html

Yes, that sounds right. Do you know anything about the alternating group?

Just to prove that I'm not talking through my arse this time.

The alternating group is the even permutations of a finite set.

Right. So suppose that $A_{3}$ is the set of even permutations on 3-cycles. What is $A_{3}\times A_{3}$ ?

What do you specifically need to know about it?

Well, $A_3 \times A_3$ would have nine elements; each would be a pair of permutations.

Ok. Well at least I've got that much right.

I have no idea what I'm doing.

What do you mean "even permutations on 3-cycles"? $A_3$ is just the even permutations on any set of 3 elements.

@DavidWallace Right, same difference.

Also, 8 of the elements of $A_3 \times A_3$ have order 3.

So what would the subgroups of $A_{3}\times A_{3}$ ?

Can we make an answer community wiki, get votes on it, then change it back to normal and get the rep? I have a partial answer on someone's question and plan to ask my own question in order to finish it..

So I guess it has 4 subgroups; each isomorphic to $A_3$. Plus the trivial subgroup and the whole group, of course.

@anon No idea sorry.

Ok well, that seems right. But it means I'm pretty well stuck on this problem then.

So if we write $A_3$ as $\{0,a,b\}$, then the four subgroups of order 3 of $A_3\times A_3$ would be $\{(0,0),(a,0),(b,0)\}, \{(0,0),(a,a),(b,b)\}, \{(0,0),(a,b),(b,a)\}$ and $\{(0,0),(0,a),(0,b)\}$ with all the usual operations.

So, what's the problem?

Determine the Galois group of $(x^{3}-2)(x^{3}-3)$ over $\Bbb{Q}$. Determine all the subfields which contain $\Bbb{Q}(\rho)$ where $\rho$ is a primitive $3^{\text{rd}}$ root of unity.

No idea, sorry. It was 1991 that I studied all of this. The grey matter has gone decidedly rusty since then.

No worries. I think I'm on the right track, but it's the second part that has me stuck. Finding the subfields.

So you've got the Galois group part sorted then?

I'm beginning to think there aren't any. Yeah I think so.

I'm pretty sure that the Galois group is isomorphic to $A_{3}\times A_{3}$.

Then I don't understand the question. What do you have to find the subfields of?

But that implies that $|\operatorname{Gal}(K/\Bbb{Q})|=9$ where $K$ is the splitting field of the above polynomial.

I need to find the subfields of the splitting field that contain $\Bbb{Q}(\rho)$.

OK, I don't remember the term "splitting field". Let me go and look it up.

The splitting field is $\Bbb{Q}(\sqrt[3]{2},\sqrt[3]{3},\rho)$.

Its the field in which a polynomial factors into linear terms

Oh, OK, I understand.

But if the Galois group is isomorphic to $A_{3}\times A_{3}$, then its degree is 9, but then this thread implies that $\rho$ is in $\Bbb{Q}(\sqrt[3]{2},\sqrt[3]{3})$.

So surely $Q(\rho)$, $Q(\sqrt[3]{2},\rho)$, $Q(\sqrt[3]{3},\rho)$ and the whole splitting field all qualify?

This follows because $[K:\Bbb{Q}]=[K:\Bbb{Q}(\sqrt[3]{2},\sqrt[3]{3})][\Bbb{Q}(\sqrt[3]{2},\sqrt[3]{3‌​}):\Bbb{Q}]$

And that thread shows that $[\Bbb{Q}(\sqrt[3]{2},\sqrt[3]{3}):\Bbb{Q}]=9$.

I don't think that $\rho$ is in $Q(\sqrt[3]{2},\sqrt[3]{3})$.

I know. That seems to be the problem I'm having.

Because everything in $Q(\sqrt[3]2, \sqrt[3]3)$ is real, isn't it?

Yes.

But the splitting field of your original polynomial has to contain $\rho$.

Yes.

So what's wrong with my answer above? The line that starts "So surely".

So either, $\Bbb{Q}(\sqrt[3]{2},\sqrt[3]{3})$ is not a subfield of $\Bbb{Q}(\sqrt[3]{2},\sqrt[3]{3},\rho)$ or the Galois group is not $A_{3}\times A_{3}$.

How could $\Bbb{Q}(\sqrt[3]{2},\sqrt[3]{3})$ not be a subfield of $\Bbb{Q}(\sqrt[3]{2},\sqrt[3]{3},\rho)$ ?

I don't know. It must be right? But the tower law requires that $[\Bbb{Q}(\sqrt[3]{2},\sqrt[3]{3‌​},\rho):\Bbb{Q}]=[\Bbb{Q}(\sqrt[3]{2},\sqrt[3]‌​{3‌​},\rho):\Bbb{Q}(\sqrt[3]{2},\sqrt[3]{3})][\Bbb{Q}(\sqrt[3]{2},\sqrt[3]{3‌​}): \Bbb{ Q }]$.

So I must've gone wrong somewhere else.

Maybe, but I don't understand why you think something's wrong.

Well, I'm trying to determine the fields that contain $\Bbb{Q}(\rho)$ as a subfield. But $[\Bbb{Q}(\rho):\Bbb{Q}]=2$ and $2\nmid9$.

Right, so it would seem that the Galois group isn't $A_3 \times A_3$.

Yeah, I'm starting to think that must be the case. But then I'm not sure where I went wrong in applying the theorem I used to get that conclusion. On another note. What do you think $\Bbb{Q}(\sqrt[3]{2},\rho)\cap\Bbb{Q}(\sqrt[3]{3},\rho)=$ ?

Well,it's obviously $Q(\rho)$.

Yeah.

Sigh

Can I ask a stupid question?

I always do when I'm here.

Where does $[Q(\rho):Q]=2$ come from?

Oh, sorry, I get it now.

Well $\rho$ is a $3^{\text{rd}}$ root of unity, so the degree of it's minimal polynomial is $\varphi(3)$ where $\varphi$ is the Euler $\varphi$ function

So I know that the Galois group is isomorphic to the direct product $\operatorname{Gal}(\Bbb{Q}(\sqrt[3]{2},\rho)/\Bbb{ Q }(\rho))\times\operatorname{Gal}(\Bbb{ Q }(\sqrt[3]{3},\rho)/\Bbb{Q}(\rho))$

$\rho^2+\rho+1$

No, that's OK, I just had a temporary lack of faith, for a moment.

And $\operatorname{Gal}(\Bbb{Q}(\sqrt[3]{2},\rho)/\Bbb{ Q }(\rho))\times\operatorname{Gal}(\Bbb{ Q }(\sqrt[3]{3},\rho)/\Bbb{Q}(\rho))\cong A_{3}\times A_{3}$.

Right, but [Q(\sqrt[3]{2},\sqrt[3]{3},\rho):Q] = 18, right?

@DavidWallace How do you get this? I was just about to ask that.

Well, how would you normally calculate this?

Well I have a theorem that states if $K_{1}$ and $K_{2}$ are Galois over $F$ then $$[K_{1}K_{2}:F]=\frac{[K_{1}:F][K_{2}:F]}{[K_{1}\cap K_{2}:F]}$$

Consider $[Q(\sqrt[3]{2}):Q]$. This is 3, right?

So in this case $[K_{j}:\Bbb{Q}]=6$ and $[K_{1}\cap K_{2}:\Bbb{Q}]=2$.

Yes.

Now $[Q(\sqrt[3]{2},\sqrt[3]{3}):Q(\sqrt[3]{2})]$ is also 3, right?

Yes.

Right. Yes. I see where you're going with this.

So you're happy that it's 18? Or do I need to type more?

Well yes I'm happy that it is 18. However, I'm not really sure what the Galois group is now.

I'm starting to think that it's $A_3\times A_3\times S_2$.

Hmm...

but like I said, I'm really rusty, not having done any of this for over 20 years.

Actually I think that's right. I need to work at it a bit to confirm, but I'll have to do that tomorrow.

@DavidWallace Well I'm impressed. I hope to remember as much as you do 20 years from now.

Thanks.

Thank you too!

I have to go now and do something else for a while. See you later, David K.

Me too . Thanks.

Hmm. When posting a question, the system seems to automatically permute the order of the tags you entered into something different. How does that work?

It's probably undefined. They'll go via some kind of Java HashSet or something like it.

hashes to permute a set of tags?

Well, to store them on their way from one part of the system to another.

maybe the hash isn't dependent on the order of the tags, so when they are reconstructed they come out in a predefined order?

@anon Just out of curiosity--what is your mathematical background?

I never know how to answer that question, because I haven't taken classes outside of high school (vector calc, lin alg and diff equ). I've learned a little bit of algebra, number theory etc. reading for fun. (I have a collection of notes and texts on my hard drive.)

Are you still in high school?

No.

So, you just do it for fun?

I plan to study it in uni at some point. But yes, that is the reason I do it.

That's good stuff man. What interests you the most?

Number theory.

Oh yeah? Any particular branch/topic?

I'm really unbalanced - I know a lot more from the analytic side than the algebraic side. But I want to work up to Langland's program. (Also another interest might be representation theory, but I haven't really looked at it deeply.)

I guess that's not really something too "particular." :D

Haha, I think everyone would like to learn a little about the Langland's program. Yeah, if you're interested in that you will HAVE to learn a LOT of representation theory. My (infinitesimal) understanding of Langland's is that it's a correspondence between "algebraic" and "analytic" representations.

Haha, yeah, but that's cool stuff :P

Yes, the unification between the analysis and the algebra of number theory is what attracts me to it.

Haha, yeah. The best subjects are the ones that use forty-seven-thousand branches of mathematics in one proof.

There are a couple like that.

That's why I've set it as a "lifetime goal." One of my few..

What do you study in uni?

That is a life goal I can definitely get behind.

Math, of course!

Yes, yes - but the areas!

Right now? I'm taking a course in homological algebra/group cohomology and a course in "advanced" galois theory (not that advanced). Next term I'm going to be doing algebraic number theory and a very exciting course on the algebraic side of Riemann surfaces.

analytic side*

Galois theory and AlgNT are definitely things I need to learn.

Yeah, they are definitely very cool stuff! I'm sure once you get to college (I think it's weird to call it university! :) ) you'll learn it quickly.

I assume you're familiar with Milne's notes (and KConrad and Pete L. Clark's as well)?

Not KC's, but Milne's and PLC's are part of my collection. It's just a matter of getting off my butt and learning the math.

Also, Stein.

I would take a look when you get a chance:

math.uconn.edu/~kconrad/blurbs

What is Stein?

@AlexYoucis Where I live, "college" means secondary school.

modular.math.washington.edu

@DavidWallace Where I live uni sounds really, really weird.

Potato potato.

I just read that as "potato potato", both with the same inflection.

I did the same.

No, no, you have to say "po-tay-to po-tah-to".

Haha, yes, I know.

@anon Also, I would download everything from this man--they are very good

people.brandeis.edu/~igusa

Hm! I've never heard of Igusa.

that guy is really cool

Well there goes Igusa's hipster status.

the nicest guy in the universe

Oh yeah? Do tell.

really. Met him a few times in conferences.

Good guy and good notes. Hell of a catch he is.

well, he is married...

Damn. And to think I was half-way through composing a angsty "I know you don't know who I am, but..." love letter.

@MarianoSuárezAlvarez You're from Argentinia, right?

Well, from rgentina

Argentinia* at least

what is wrong with me? Argentina.

yup

This will probably just sound really stupid, but I'm too curious not to ask. We have like four guys in our department from Argentina, none are from the same region/school, etc. They all do the same kind of PDE theoretic functional analysis. Is that big in Argentina?

heh

where are you?

there are quite a few of people doing that sort of things here, yes

University of Maryland, College Park.

do you remember the names of the guys?

Oh god, I don't know. I don't know if you'd know them, they are all Ph.D. students right now. I think their advisor is Nochetto?

Nocetto?

Ah. Nochetto.

Do you actually know him?

Ricardo H. Nochetto

I don't know him but I know who he is

Oh, I see he's from Buenos Aries, which is where you are from, correct?

yes

Haha, small world, huh?

(I've been in Buenos Aires for some 10 years now, but I am really from Río Gallegos, down south)

Well, math is a small world :)

Did you do your Ph.D. there?

(yes, indeed :) )

yes

Under anyone I might know?

half here and half in Paris

Andrea Solotar here and Max Karoubi in Paris were my advisors

I actually may have heard of Karoubi.

Have you ever heard of Jonathan Rosenberg?

he's pretty well known :)

sure

He's teaching a homotopy theory course here this term

and I remember hearing that name

Hey, isn't Argentina hosting the International Mathematical Olympiad again this year?

Is that the right subject?

@DavidWallace, I think so, yes

@AlexYoucis, yes

well, his main topic is K-theory

I was coaching the New Zealand team in 1997, which is the last time it was in Argentina.

Same as Rosenberg! Cool, cool Are you a homotopy theorist?

he wrote the book everyone learns to hate on the subject :)

Haha, do you actually dislike it? I've only read the first chapter on rings, and it seemed pretty good.

Max has a writing style which is not appreciated by all

I work on homological algebra

Oh wow. Very, very cool. I'm taking my first homological algebra course this term.

Good :)

beware: it is addictive :P

Haha, I can definitely see that!

I just gave a presentation on H^i(G,A) i=1,2 :P

that's how they get you hooked

Until you ask an expert what H^4 means and they say "Nothing!"

heh

Got any insight on that?

well, we all learn to wave our hands very significantly when we get asked that!

Ok, here's my real question for you

I have not been able to find ANY intuition for H_1

What the hell does that mean?

H^3 can be interpreted in terms of non-central extensions of groups

(you can find that in MacLane's book on Homology)

Yeah, I got 1,2,3 for cohomology. I got nothing for homology.

and (a part of) H^4 has a pretty wierd interpretation which I do not remember

but after that what you want is understand the whole cohomoogy, not the individual groups

for example, H^*(G,Z/p) is a ring with Krull dimension equal to the p-rank of G

that's the sort of things you want to interpret