Mathematics - 2015-10-15 (page 2 of 3)

theDoctor

03:00

if a vector space is of more than one dimension, then by definition it has orthogonality between it's members.

user143442

that why you are so upset

Krijn

By definition?

anon

@theDoctor do you know what orthogonal means?

Krijn

What definitions are you using, man

Forever Mozart

lol

theDoctor

03:00

i use orthogonal in two senses: 1) the sense of independent variables.

user143442

dont be like that, be the nice guy

theDoctor

and 2) the sense of completely unrelated dimensions. like geometry to whole numbers.

Forever Mozart

look

everything in math has a precise definition

you must use the definitions we all use

theDoctor

a 3d vector has 3 elements (x,y,z} that can be projected into the space R^3.

Forever Mozart

or communication will be difficult

user143442

03:02

I cant start MathJax :(

anon

that's another thing - often vector spaces are conflated with coordinate vector spaces. not every vector is a coordinate tuple.

Krijn

Well, you can use other definitions, but you should communicate that beforehand

Karim Mansour

fucken mechanics question its like 10 pages of algebra

what the hell lol

theDoctor

what other possible sense is there of a vector if not a tuple?

user143442

but every finite dimension vector space may be seen as a coordinate vector space

theDoctor

03:03

yes #user

user143442

just use the coefficients of its representation in the basis

anon

@theDoctor a vector is just any element of a vector space. you can look up "vector space"

theDoctor

@anon so you're using the term vector in an abstract, non-numerical sense of a force?

user143442

infinite dimension vector spaces are studied in functional analysis

Forever Mozart

see any abstract algebra text

Krijn

03:04

What is "the basis" @user ?

Julian Rachman

I'm FINALLY BACK! :)

Forever Mozart

woooooo

user143442

the basis you pick

Julian Rachman

How has everyone been?

user143442

now tell me how to use MathJax please

user143442

03:05

I can't install it

theDoctor

has anyone considered negative dimensional spaces? \

anon

@theDoctor vector spaces are still numerical in the sense that you multiply vectors by scalars (often numbers, specifically), but the physics interpretation of force is just a physics interpretation. like I said, look up vector space.

theDoctor

or imaginary spaces?

anon

you mean space with complex-number dimension

answer: sort-of

theDoctor

ah I see @anon. you are using non-euclidian sense of abstract vectors.

user143442

03:06

i think $\emptyset$ has dimension $-\infty$

theDoctor

@anon yes.

anon

@theDoctor non-euclidean is not the word you should be using - that refers to curved spaces in geometry.

Forever Mozart

thats what I'm working on

theDoctor

no, non-euclidean can mean more than a curved space which still implies a 2d field.

user143442

start ChatJax
^^drag this^^ to your bookmark bar or right click on it to add it as a bookmark.

theDoctor

03:07

@ForeverMozart sweet.

user143442

I can't drag it :S

user143442

don't be a drag just be a queen

Krijn

Right click it?

user143442

i also did that but it doesnt show the option add it as a bookmark

Forever Mozart

Krijn

03:09

@ForeverMozart Mentaculus?

Antonio Vargas

@user, what do you mean you can't drag it?

Forever Mozart

@Krijn yes ;)

user143442

I see the little circle of forbidden

theDoctor

which is to say non-euclidean is more than just Reinmann spaces, for exmample

anon

fractional dimensions are considered in fractal geometry, supervector spaces can have negative dimensions, and one can consider formal linear combinations of vector spaces and extend dim linearly so it can take (say) complex values (or alternatively interpret dimension as the character of a representation evaluated at id, then interpret chi(g) for different g as "twisted" dimension)

user143442

03:10

Antonio tu hablas español

user143442

dime en español

Antonio Vargas

No, sorry, only English :)

user143442

malinchista

Antonio Vargas

I know a little though from growing up in California.

Forever Mozart

non, mais je parle Francais

user143442

03:11

moi aussi

theDoctor

@anon:

user143442

mais un peu

theDoctor

i like the new idea of "twisted" dimension.

Krijn

Je ne parle pas francais bien

Antonio Vargas

@user, right click on the "start chatjax" link, then click copy link address

theDoctor

03:12

as if defining a curved line space as opposed to a curved surface.

Forever Mozart

i will publish that

Antonio Vargas

then create a new bookmark with the title "start chatjax" and paste what you copied as the location

user143442

ya

user143442

no se que diablos es bookmark

user143442

in spanish

user143442

03:13

es marcadores?

user143442

how do i create a new bookmark? :(

Antonio Vargas

These things are bookmarks: i.imgur.com/FVeHZ7w.png

user143442

you are from chihuahua

user143442

i catched you

Antonio Vargas

haha no, I want to buy this for a friend: themountain.com/chihuahua-face-t-shirt

Krijn

03:15

God, thats horrendous

Antonio Vargas

I know, it's great

user143442

it's horrendous

Antonio Vargas

in chrome, to create a new bookmark, right click up there, then click "add page..."

I don't know what it would be in spanish

user143442

agregar pagina

Antonio Vargas

Did it work?

user143442

03:17

nope

user143442

i dont know where to click

Antonio Vargas

are you using chrome?

user143442

my bookmark bar is different

user143442

yes

Antonio Vargas

click anywhere that is gray

right-click*

theDoctor

03:19

i'm not sure i buy into the idea of negative dimensions being related to supervectors, though.

user143442

postimg.org/image/4eboa14s5

user143442

here's my bookmark bar

Antonio Vargas

oh no

thanks, that was a good laugh ;)

user143442

?

theDoctor

anyway, i think math has gotten very confused about the notion of domains, because of how physics seems to relate to the abstraction of number itself.

Antonio Vargas

03:21

try right-clicking to the right of where it says "Importar marcadores ahora..."

in the blank gray space

theDoctor

and people use R like they do Z, even though they are completely different domains.

user143442

yess :D

user143442

thank you so much

Antonio Vargas

no problem

user143442

y de nada por la risa jaja

theDoctor

03:24

oh sry, i thought this was a math chatroom./

Krijn

It is, but the discussion is tiresom

And not fruitful

theDoctor

you telling me you've thought of dimensions beyond integer?

user143442

@theDoctor are you the one of your picture?

theDoctor

@user sure.

i can tell you practical applications of negative and imaginary dimensions.

Forever Mozart

lord

theDoctor

03:26

yes, son.

Forever Mozart

haha

theDoctor

lol

you people need some guidance.

Forever Mozart

but you speak in too much generality

theDoctor

you have been getting lost in the byways of math.

Krijn

You say that like it's a bad thing

Forever Mozart

03:28

yes that is true

theDoctor

it's not too much, you just need to understand domain theory.

Forever Mozart

wtf is domain theory??

theDoctor

but consider this, if there was not some imposed structure on math, it wouldn't exist.

Forever Mozart

most mathematicians are platonists and would disagree

theDoctor

domain theory relates different dimensions of number, like dimensions specifies different fields of geometry.

Forever Mozart

03:30

dimension theory is an area of topology

but I never heard of domain theory

Krijn

Domain theory

Domain theory is a branch of mathematics that studies special kinds of partially ordered sets (posets) commonly called domains. Consequently, domain theory can be considered as a branch of order theory. The field has major applications in computer science, where it is used to specify denotational semantics, especially for functional programming languages. Domain theory formalizes the intuitive ideas of approximation and convergence in a very general way and has close relations to topology. An alternative important approach to denotational semantics in computer science is that of metric spaces....

theDoctor

that definition is too constrained. R is not an ordered set nor a poset.

Forever Mozart

yes it is!!!

theDoctor

R gets an order by being related to Z, after that.....

Krijn

Your obsession with $\mathbb{R}$ being somehow completely misunderstood by all mathematicians here is really...

theDoctor

03:31

i know, hard to believe.

Forever Mozart

lol

theDoctor

but that's because they've been dealing with abstract albebraic expressions, never the real numbers.

functions, in otherwods

otherwords.

Forever Mozart

99% of mathematicians will disagree with you

Krijn

Yep, I'm out. Good luck @ForeverMozart!

theDoctor

get those bastards on the line.

anon

03:32

2 hours ago, by anon

you're very trolly/cranky.

theDoctor

and we'll talk some real (sic) math.

Forever Mozart

@Krijn well it was sortof entertaining

theDoctor

i'm not trolling, sometimes a premise is wrong.

Antonio Vargas

Does anyone ever feel like they're right and everyone else is wrong?

theDoctor

or, let's say, not wrong, which make shiver's down Chaitin's back, but less pleasing.

Krijn

03:33

You are either trolling, or you should take a class in Real Analysis

theDoctor

haha, real analysis is a joke.

it's like a subject taken by Cthulhlu.

to try to understand the world above water.

Forever Mozart

Chaitin is way out on the fringe, and does not actually contribute to mathematics.

theDoctor

after only ever knowing the world within water.

Forever Mozart

he is messing up your mind I think

theDoctor

haha, what counts as a contribution to you?

no, he's is awesome. if you want to understand yourself some philosophy of math, check it out.

if you just want to get lost in the byways of math, well, there's SE.

Forever Mozart

03:35

he talks a lot about philosophy

but he doesn't actually do math

theDoctor

EXACTLY! where number has been for quite awhile.

you are partially right, depending on whether you consider compsci math.

anon

out of curiosity, what is your native language @theDoctor ?

theDoctor

computational math, math.

native language is English.

i know, i don't talk with the semantics of most math, but that's because i've gotten a new religion, so to speak.

it's like

Pythagoras.

Antonio Vargas

On a different note, would anyone be willing to try out a small userscript for Math.SE I've been working on? I'm trying to add live previews for when you post a comment.

anon

@theDoctor I think you should collaborate with John Gabriel. I would be interested in the result.

theDoctor

03:39

hmm, will look into that....

Karim Mansour

Hahaha I am tempted in a question that involves physics to say by maple and some algebra we get the following

Forever Mozart

go ahead

Karim Mansour

its like 1 page equation

I am not gonna go through hrs of algebra to get solution lol

Forever Mozart

there has to be more to the problem than just algebra

Karim Mansour

yeah ofcoz

setuping it up and stuff

it involves Lagrangian mechanics

but I am just lazy to do the algebra i already did the steps that goes through the algebra

if your interested the question reads as follows

theDoctor

03:43

@anon: I am both intrigued and repelled by John Gabriel page you gave me.

Karim Mansour

A point particles moves in space under the influence of a force derivable from a generalized potential of the form $U(r,v) = V(r) + \sigma . L$ Find components of the force in both cartesian coordinates and polar and show that the components are related to each other

lol wtf is that john gabriel @anon

Mike Miller

@anon: I suppose you'd need to define volume for me to agree. But either way the lack of a group "SSp(2) = S^7" is annoying in a number of ways; first that S^7 isn't actually a group; second that if there was one, then we (should) have H^*(BSSp(2)) = Z[x] where x is in dimension 8, but this is entirely impossible

(analagous to H^*(BSU(2)) = Z[x], x degree 4; H^*(BSO(2)) = Z[x], x degree 2)

anon

dunno cohomology or classifying spaces, so, I have no idea what's annoying about that

but I believe you

Mike Miller

it annoys me personally i suppose

anon

@KarimMansour someone who tried evangelizing on m.se a while back but to no avail.

Mike Miller

03:48

@anon: it's just annoying in the same way that any pattern breaking is

Karim Mansour

oh

Galois in the Field

Do $\Bbb F$-automorphisms of $\Bbb F(\alpha)$ take roots of $\alpha$ to other roots of the minimal polynomial of $\alpha$ and fix everything in $\Bbb F$?

anon

yes

try to prove it

Galois in the Field

Which means that any $\Bbb F$-automorphism of $\Bbb F(\alpha)$ couldn't possibably take an element to the root of the minimal polynomial of $\alpha$ unless it was already a root of the minmial polynomial of $\alpha$

anon

(well, being an F-auto by definition entails fixing F pointwise)

right

Galois in the Field

03:53

Good good good, yes I will prove this thanks

Karim Mansour

04:25

Hey just @anon is it true that intersection of all maximal subgroups of Z is actually isomorphic to $Z_2$ right ?

just making sure

anon

the maximal subgroups of Z are pZ for primes p, and their intersection is {0} (under +, that's the trivial group)

Karim Mansour

yeah pZ

is maximal subgroup of Z

oh yeah

anon

Z doesn't have any (nontrivial) finite subgroups, so in particular none iso to Z2

Karim Mansour

I see

I have a question can we have uncountable group and a countable maximal subgroup of it ?

anon

no

If G is uncountable and H<G is countable, then picking any g in G\H we have that <H,g> is countable and bigger than H.

Karim Mansour

04:30

oh I see

why is it bigger than H ?

Ben Lim

@anon hello.

Karim Mansour

1 sec thinking

anon

@KarimMansour I mean H is proper in <H,g>

@BenLim hey

Karim Mansour

ok yeah

Ben Lim

@anon wassup?

Karim Mansour

04:32

yeah makes sense

anon

@BenLim thinking about accidental spin groups

Ben Lim

I see. I've been busy with grad school crap lately so haven't been on here or MO

anon

no worries

Karim Mansour

which grad school you go to @BenLim

Ben Lim

@KarimMansour Stanford.

Karim Mansour

04:34

nice

Anthony

Ayo @anon

You in the mood for some Dedekind domains?

anon

meh

Anthony

You in the mood to help me?

Forever Mozart

by grad school crap you mean teaching incompetent calculus students?

:)

Ben Lim

@ForeverMozart I mean handling three weekly assignments and being a CA

Anthony

04:39

I'm wondering why every prime ideal being maximal in a dedekind domain implies that the CRT holds on powers of prime ideals

Forever Mozart

what is a CA?

Ben Lim

@ForeverMozart course assistant

Anthony

@anon In particular, I don't see why powers of prime ideals are pairwise coprime.

Ben Lim

@Anthony Ok. you agree that if p,q are prime ideals, then p +q = (1) yea?

Anthony

I actually don't know the proof of this either.

But I know it's true.

Ben Lim

04:40

@Anthony So we just need to show the following. Suppose, I,J are such that I + J = (1). Then for all $n$, $I^n + J^n = (1)$.

Anthony

I mean couldn't the powers be different?

Ben Lim

Ok sure if you'd like

@Anthony This is easy enough. Suppose $I^n + J^m \neq A$. Then we can find a maximal ideal $m$ with $I^n + J^m$. Then in particular $I^n \subseteq m$ and $J^m \subseteq m$. Hence $I \subseteq m$ and $J \subseteq m$, so that $I+ J \subseteq m$, contradicting $I+ J = (1)$.

Anthony

@BenLim Thanks. Give me a moment will I try to read this. I guess there's a bigger problem though, because I can't think of why the sum of two prime ideals should be (1).

Ben Lim

@Anthony Well you know in this case that your prime ideals are maximal.

anon

If $P$ is prime ($\Rightarrow$ maximal in dedekind domains) $P\subseteq I$ implies $I=P$ or $I=(1)$. Apply this with $P,Q\subseteq P+Q$ with $P\ne Q$ prime ideals and you get $P+Q=(1)$.

Anthony

04:46

Oh, d'oh.

Thanks, ya'll.

@BenLim Why does $I^n\subseteq m$ imply $I\subseteq m$?

Ben Lim

Choose $x \in I$, Then $x^n \in I \subseteq m$, so $x^n \in m$. Now $m$ is maximal, hence prime so that $x \in m$.

Anthony

lol

Thanks.

Julian Rachman

#lovealuffi

Huy

ok

Julian Rachman

ok

sorry

Huy

04:52

Julian you are surely interested in studying functional analysis with me

Julian Rachman

nice to see you again @Huy

??

Huy

I'm looking for someone to study with so I have more motivation

Julian Rachman

ok sure

Huy

are you working on a lot of other things currently?

Julian Rachman

not really

Huy

04:53

you have enough time?

Julian Rachman

you mean now?

Huy

in general

Julian Rachman

I mean ya

i guess

Huy

what's your background? sorry I don't quite remember @Julian

Julian Rachman

Real Analysis, Topology, Category Theory (current), Alg. Top. (trying :/), Abstract Alg

Huy

04:58

well if it doesn't interfere with your studies I'd be happy if you'd join me

Julian Rachman

I would love to join

Galois in the Field

What does studying functional analysis entail here?

Huy

@JulianRachman: dl.dropboxusercontent.com/u/2098511/FAnotes.pdf I'm using my professor's draft of his textbook. I'm currently revising some measure theory in the appendix and going through chapters 2-4. I want to focus on chapters 10-11 eventually, so after 4 I'll check what else I need to start with 10. You can diverge of course.

@GaloisintheField: What do you mean by that?

Julian Rachman

Ok

I will take a look and start reading

Galois in the Field

I mean what functional analysis are you studying, but I guess your link answers that

Huy

05:00

@JulianRachman: I think it's quite nicely written and has lots of motivation, but maybe that's just me.

Julian Rachman

Ok. So what do you what me to go over so that we can discuss?

Huy

idk, I've never done this before, online :D

Galois in the Field

Should you be posting his textbook draft online?

Huy

@GaloisintheField: The link is available from his website.

Galois in the Field

I didn't mean to offend(if I did)

Julian Rachman

05:01

What to email me?

Galois in the Field

Why go straight to 10?

Julian Rachman

@Huy

Huy

@JulianRachman: Maybe you can go through (parts of) chapter 2 and next time we meet you tell me where you're at? I'll have a bit of a head start but I'm sure you'll catch up quickly.

Julian Rachman

I think I am going to start with motivation so that I get a feel

Ok sounds good

Huy

@GaloisintheField: Because I'm mostly interested in the spectral theory of self-adj unbounded ops

Galois in the Field

05:03

@Huy Fair enough, and you already know chapter 5 and 6?

Huy

I've seen most of the results already during my studies, but it was a while ago, which is why I will go through it again, just not as much in detail as with the other things.

Julian Rachman

TBH, I have not learned anything about fundamental analysis so it is good for me to start from some motivation then go from there

Huy

@JulianRachman: I really like his explanations on ODEs and PDEs, made me understand those two topics much more in a very short time even though I've been dealing with such equations for years.

Julian Rachman

Ok Ill take a look and see if I will be on the same page with you

did you print this pdf or did you just go digital

Huy

I always read digitally

but that's just my personal preference

Julian Rachman

05:06

Oh ok

Galois in the Field

Maybe you should make another chat room for this? I think I would be interested in following your progress

Julian Rachman

lol I like to write on stuff

Huy

@JulianRachman: I usually just write it down and then have hundreds of pages of notes :P

Julian Rachman

Cool. @Huy can you make the room/

Galois in the Field

(and maybe participate when I have time

Julian Rachman

05:08

Do you go notebook or loose leaf paper?

^ the debate :/

Galois in the Field

I throw out all of my notes

Julian Rachman

@GaloisintheField lol

Krijn

05:45

Hey @GaloisintheField, I'm sorry I couldn't answer last time

Did you manage to solve what you were doing?

Galois in the Field

08:15

@Krijn I don't know, I am mainly just working through understanding stuff

Balarka Sen

08:28

@GaloisintheField Cool username.

Chris's sis the artist

@robjohn here is an amazing question I created last night. Proving that (in a very easy way)

$$\int _0^1\int _0^1\int _0^1\log \left(x^2+y^2+z^2-x y-x z-y z\right) \ dx \ dy \ dz=\frac{1}{3} \left(\sqrt{3} \pi -11\right).$$

Sure, here is also a basic thing to prove for beginners: Prove that $$x^2+y^2+z^2-x y-x z-y z\ge0$$

robjohn

08:43

@Chris'ssistheartist $\frac92\left(\frac{x^2+y^2+z^2}3\right)-\frac92\left(\frac{x+y+z}3\right)^2$ and Jensen's Inequality

Chris's sis the artist

@robjohn Yeah. However, the left side, after multiplying by $2$, can be viewed as a sum of squares.

Tobias Kildetoft

@BalarkaSen So, what are you up to currently?

Balarka Sen

@Tobias I was mostly preparing for exams (you know, high-school stuff) the whole week. Finally, I'm all free.

Tobias Kildetoft

@BalarkaSen Cool. How did it go?

Balarka Sen

Pretty good.

Now that I am free, I am planning to do some multivariable calculus, relearn commutative algebra and revise algebraic topology, putting a stress on calculus since I have ignored it for so long and now I can't study the cool stuffs (differential topology) without it.

@TobiasKildetoft What are you upto?

Tobias Kildetoft

08:53

@BalarkaSen Updating/rewriting my research statement. I need to start applying for jobs again soon

Balarka Sen

Ah. Must be a boring work, updating research statements.

Tobias Kildetoft

@BalarkaSen Yeah, though it was interesting to read the old one again and see what I needed to change

I have had to remove several things that I have not actually thought about at all since writing the old one (making them perhaps not the best inclusions back then)

Vrouvrou

09:45

please in $(\mathbb{R},|.|)$ how to prove that $\overset{\circ}{Q}=\emptyset$ ?

Balarka Sen

10:11

er, @robjohn. I think you have a dinosaur on your keyboard.

Loong

@robjohn :

111

Q: How can I keep my cat off my keyboard?

This is a common scenario when typing: When the family assembled for Sunday dinner, With their minds made up that they wouldn't get thinner On Argentine joint, potato^DR&FTGYB`kuhadrggoy867rt98wouth4bfgdhjlkhdsfghhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhf This happens beca...

Александър Гьорев

10:24

Has anyone in here studied transformational plane geometry ?

Vrouvrou

@robjohn can you help me please

Chris's sis the artist

@Hippalectryon

Hippalectryon

@Chris'ssistheartist o/

Chris's sis the artist

o/

$$\int _0^1\int _0^1\int _0^1\log \left(x^2+y^2+z^2-x y-x z-y z\right) \ dx \ dy \ dz=\frac{1}{3} \left(\sqrt{3} \pi -11\right).$$

@Hippalectryon ^^^

Hippalectryon

:D

Huy

11:44

@DanielFischer: My prof asked me whether I knew the "invariant measure" and I hesitated and he said "it's just $1/y^2$". Of course I know the Riemannian metric, but that's not the same, is it? He then told me that what's interesting is that $1/y^2$ isn't invariant on tangent vectors, but the standard one is. Do you know what he means by this "invariant measure on $\mathbb{H}^2$"? Invariant under what? Möbius transformations?

Daniel Fischer

12:09

@Huy Invariant under automorphisms, typically. In this case, those would be Möbius transformations (and maybe compositions with reflection in the $y$-axis, depends on whether you are also interested in orientation-reversing maps or not), since the orientation-preserving isometries are Möbius transformations.

I assume you're doing this in differential geometry, since you speak of a Riemannian metric, so the pertinent concept of automorphism is isometry, and you're using the standard constant-curvature metric. In the context of (locally compact) topological groups, "invariant measure" would refer to "the" Haar measure.

Huy

@DanielFischer: but does that rule out Haar measure, since hyperbolic plane isn't a topological group?

I'm a bit puzzled because usually he is very correct with his wording but "measure" didn't quite make any sense for me

(he asked specifically about "the invariant measure on H^2")

Daniel Fischer

12:32

@Huy Well, $H^2$ is homeomorphic to $\mathbb{R}^2$, so you could make it a topological group. I just mentioned Haar measures to have another example of what "invariant measure" can mean. It's not so different, since in both cases you have a ~~bunch~~ group of maps, and you want the measure invariant under those maps.

Huy

@DanielFischer: and then would the Haar measure look anything at all like $1/y^2$?

Daniel Fischer

@Huy I don't think so. Under the obvious homeomorphism, you'd get $\frac{1}{y}\,dx\,dy$. I don't see how you could make it a topological group in a way that $1/y^2$ becomes the density of a Haar measure. And by the way, saying $1/y^2$ is the measure is an abuse of terminology. It's the density of the invariant measure with respect to the Lebesgue measure on $\mathbb{R}\times (0,+\infty)$.

Huy

@DanielFischer: and can you make anything out of his statement that "$1/y^2$ isn't invariant for tangent vectors, but the standard one is"?

Daniel Fischer

@Huy Not really. What "standard one"?

Huy

@DanielFischer: I think he said "x+y" which is why I thought he meant the Euclidean metric (and then the 1/y^2 also made more sense to me, but then again I didn't understand why he'd say measure)

Daniel Fischer

12:43

Can't make heads nor tails from it. I can't even fathom what "invariant for tangent vectors" could mean in connection with a measure.

UserX

13:14

Can we do operations on matrices that have elements from different fields? How much can we extend that?(my terminology is a bad attemt to translate greek math to english math)

I mean we can multiply the matrices $A\in\Bbb Q^{m\times n},\; B\in\Bbb R^{n\times r}$

Balarka Sen

@UserX How do you define multiplication of two elements of $F$ and $F'$, two different fields?

It works for $\Bbb R$ and $\Bbb Q$ because $\Bbb Q$ is a subfield of $\Bbb R$

UserX

Do the corresponding fields have to be connected with one being a subfield to the other and the matrix we get is in the largest field? That's what makes sense for me.

Balarka Sen

If one of the field is subfield of the other, it is clear how to multiply elements of the two fields, thus you can define multiplication of the matrices with values from those fields too.

UserX

That's what I thought @BalarkaSen. Are there any other circumstances where it may be possible?

Huy

@DanielFischer: I'll just ask him asap. :P

Balarka Sen

13:23

@UserX Depends on what you mean. If $F$ is a field, $F'$ is another field, there is generally no natural way to define a multiplication between elements of $F$ and elements of $F'$ unless you're inside some bigger field already. That is, $F$ and $F'$ must be subfields of some bigger field $F''$

Otherwise, you have to define for me what multiplication even means.

robjohn

@Vrouvrou note that $(\sqrt2+\mathbb{Q})\cap\mathbb{Q}=\emptyset$ and since $\mathbb{Q}$ is dense, so is $\sqrt2+\mathbb{Q}$. What can you say about $\overline{\mathbb{Q}^C}$?

Galois in the Field

Dumb question:

If I take $f(x) + \langle g(x)\rangle$ and $g(x)$ is degree $n$, why do I always get a degree $\leq n-1$ remainder?

Huy

@DanielFischer: I got an answer by email (German): "Das invariante Mass ist 1/y^2 dxdy (glauben sie mir nichts und überprüfen Sie dies). Eigentlich ist die obere Halbebene keine Lie Gruppe sondern bloss ein homogener Raum (genauer G/K=SL_2(R)/SO_(R)). Aber andererseits kann man die Gruppe der obere Dreicksmatrizen (fast) mit der oberen Halbebene identifizieren, denn diese wirken einfach transitiv (wenn man entweder nur positive Zahlen auf der Diagonale zulässt oder durch -I dividiert).
Doch bei dieser Identifizierung muss man mit dem y-Parameter aufpassen, vielleicht liegt hier die Verwirr…

Balarka Sen

13:42

lol!

3

(it was originally posted in the homotopy theory chat. it's hilarious, so i thought i might link it here too)

Daniel Fischer

@Huy I think they mean that when you identify the half-plane with the group of triangular matrices (modulo $\{\pm I\}$) then $1/y^2\,dx\,dy$ becomes the Haar measure. Sounds credible. I still have no idea what the tangent vectors have to do with anything.

Huy

@DanielFischer: I'll try to work it out in detail until tomorrow and see if I can understand what is meant with the tangent vectors.

robjohn

14:05

@GaloisintheField polynomial division

anon

14:48

@Huy SL(2,R) acts transitively by mobius transformations on the upper half plane. with the poincare metric dxdy/y^2, they are the orientation-preserving isometries (and we get the poincare upper half plane model of hyperbolic geometry). the stabilizer of the point i is SO(2,R), so we can identify SL(2,R)/SO(2,R) with it. Integrating over a region in SL(2,R)/SO(2,R) amounts to pulling the region and function back to SL(2,R) and integrating with Haar measure I think.

Also SL(2,R) acts (sharply) transitively on the bundle of unit tangent vectors, making H=SL(2,R)/SO(2,R) an isotropic manifold (the space "looks the same" while standing at any point looking in any direction).

(Disclaimer: I can't read a lick of german, I'm just guessing what's being said based on the symbols, and Daniel's response.)

You didn't say what question the email was an answer to.

anon

15:10

ah, there is discussion between you and Daniel further up. I see.

err, I meant (dx^2+dy^2)/y^2 is the Poincare metric, which induces the volume form dxdy/y^2, which agrees with the Haar measure pushed from SL(2,R) down to SL(2,R)/SO(2,R)=H^2.

also not sure what is meant by "invariant measure for tangent vectors"

Danu

15:29

Does anyone have any idea what this quote (from V.I. Arnol'd) is referring to?

Sir Michael Atiyah once told me that he was always delighted by the way Petrovskii dealt with algebraic geometry in his works on PDEs. One of these, the paper on the lacunas of hyperbolic PDEs, was later rewritten by Atiyah, Bott, and Gårding in modern terminology in two long papers in Acta Mathematica. It is a far-reaching generalization of the well-known fact of the impossibility of acoustic communication in the even-dimensional spaces (for instance, in the “plane” world), while in our three-

(when talking about the work of Petrovskii)

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$Mathematics$ Mathematics