Mathematics - 2018-06-23 (page 3 of 3)

The inverse sends the functional $f:V^*\to\Bbb K$ that evaluates elements of $V^*$ at a point $v\in V$ to said point $v$

hi demonic @Alessandro

it's not constructive

hi @Ted

Jo Be

@mercio :(

4:08 PM

Hi @Ted

hi Mathein

Hi ted

@JoBe the inverse sends an element $\eta$ of $V^{\star \star}$ to the unique $v \in V$ such that $\forall \xi \in V^*: \eta(\xi)=\xi(v)$

if it were constructive you would be abel to prove not(not(A)) -> A in intuitionistic logic

hi @geocalc

4:11 PM

@TedShifrin can you approximate a shape by generating finitely many geodesics

Depends on how you mean "approximate." In what sense have you approximated the plane if you draw 100 lines?

For example great circles on a sphere. The more you generate the clearer the picture of a sphere becomes

But there's still lots of space between 'em at various places and who knows what's going on on the surface there. Think about the plane.

If you know a priori what the surface is (e.g., constant curvature), then sure, you know what it is.

@geocalc: If you're interested in learning some differential geometry, with your background you should be able to read my notes (linked on my profile) and work on some of the stuff. It uses linear algebra and multivariable calc.

back when i had physics classes I was told that light always took the fastest path from A to B

is there a way to solve TSP with that ?

Least action principle, mercio.

Of course, I always wondered how the light knew what was coming up around the corner :P

4:15 PM

yes exactly

I have been befuddled ever since

hey @Ted

Hi @ÍgjøgnumMeg

Similarly, the shortest path from A to B on a surface actually depends on what's going on "in the middle," not just near A and B.

It might not even exist, right?

Light is just really good at solving the geodesic equation

@Alessandro: If the surface is complete, it exists. Otherwise, nope.

4:19 PM

If I have a punctured plane or some surface with "open" holes

Well, the punctured plane could be complete — depends on the metric.

I was thinking about the usual one! And wait, WHAT?

yay, a friend of mine is going to print my blog post and hand it to some freshmen in Bonn who are taking a rep theory seminar

now you're under pressure to produce good stuff, @Mathein.

@AlessandroCodenotti The discrete metric? :P

4:21 PM

We're doing differential geometry, @Fargle.

And hello.

@MatheinBoulomenos Another reason to be worried about the level of the grad courses in Bonn :/

@TedShifrin yeah. the stuff I have in mind is not that elementary, so I don't know how much they'll get out of the later entries

Oh. I can't read. Hey @Ted.

Hi @Fargle

Hello @Alessandro

4:22 PM

Of course, I'm not sure what freshmen are doing taking rep theory. I guess they all know linear algebra and some groups?

in Germany, you learn that in the first semester in LA, so it's not that unreasonable

Heidelberg had rep theory seminars for freshmen, too iirc, although the one I took was for sophomores

So they're sophomores? They finished the first year already?

They finished the first semester

And are in the second semester of the first year

confuzled since it's June

The summer semester I'm in right now lasts until the end of July

4:24 PM

Germans are efficient workers, there's no time for holidays!

No, we have holidays from August to the middle of October

Right ... so if it's summer semester, they've finished their first year. Stop confuzling me.

(I mean, most people don't go year-round, do they?)

They finished the first half of the first year

you start with the winter semester

What topic in math deals with

Oh, crazy ... you guys start in winter?

4:27 PM

yeah

Is there such a thing as a three dimensional group

yup, sure.

three dimensional Lie groups, sure

The three-sphere (unit quaternions) and the rotation group for 3-space.

Oh nice

4:30 PM

is there such a thing as a $\pi$ dimensional group ?

hell no

aw

Groups are homogeneous things, so they can't be fractal.

But I thought that $\pi=3$

3.2, according to a bill that almost became law in Indiana.

4:32 PM

._.

well, what do you expect from the state that elected Pence?

Oh, never mind.

#gottem

This was also in 1897, to be fair.

@TedShifrin When do you start?

I think Alabama or Mississippi tried to pass a law that $\pi = 22/7$ about 20 years ago. Maybe they did pass it.

@Alessandro: August/September.

But bad math is wrong in any time period. Just like bigotry or thinking the Star Wars prequels are good.

4:34 PM

"And he made a molten sea, ten cubits from the one rim to the other it was round all about, and...a line of thirty cubits did compass it round about....And it was an hand breadth thick...." — First Kings, chapter 7, verses 23 and 26
The bible says $\pi=3$

That's early, we start in mid September/early October

I think quarter system schools in the US start late September, @Alessandro.

Hey everyone!

And yeah this year we're gonna start October 1st, for example

Hi @Dami

hi Demonark

4:40 PM

@TedShifrin is there any motivation for three dimensional families of functions

Probably ...

quallenjäger

Hi chat

This isn't exactly what you're talking about, but in geometry you can try to fill up 3-space with three families of surfaces that intersect orthogonally at each point.

Hi @Daminark

Hi @quallenjäger

Secret

4:42 PM

Hmm... I don't know if there is a name for an object where the gridlines are all geodesics

Heya @Daminark

Hi @quallenjäger

@Secret: The issue is the gridlines, not the object itself.

Hi @quallenjäger.

locally you can always force "half" of the gridlines to be geodesics

If you don't force constant angle, you can certainly have both families being geodesics.

heya @MikeM

4:45 PM

hi

Fair, I was thinking about a constant (square) angle because all the formulas become way nicer

Yeah, that happens iff the surface is flat, Alessandro. Good exercise.

Wait, what exactly happens iff flat?

if f is a function on R^3 then grad f is a 3-dimensional family of functions

Geodesics making a constant angle with one another.

4:48 PM

hmm, wait I'm missing something, at any point I can find geodesics intersecting at every angle I think?

I meant the same angle at every point.

A coordinate system $(u,v)$ where the $u$-curves and $v$-curves are all geodesics and they meet at a constant angle.

Ahhh, of course, I forgot we were only thinking about the geodesics which are also coordinate lines

Well, we wanted two families of geodesics ... so yeah.

Mary Star

Hello!! Is someone of you familiar with the multigrid method? Especially with the 2-grid and the V-cycle.

not I.

4:59 PM

should mobius transformations be taught in terms of 2x2 matrices?

maybe

@TedShifrin you're the expert here

in pedagogy

@LeakyNun I'm taking a course on modular forms right now. Thinking about them as 2x2 matrices is really important. At least I don't think you can reason e.g. about congruence subgroups of the modular group if you just think of them asprojective linear transforms

yes, but I'm not asking you because you're an algebraist like me :P

this is more from a modular forms PoV, which is the area that actually heavily uses Möbius transforms

5:03 PM

maybe teach them in terms of linear relation between $1,z,w$ and $zw$

@LeakyNun I was just saying that my impression is that people who use Möbius transforms a lot think about them as an action of matrices

I know that you despise that description for some reason

there is always the annoying part about $-I_2$ acting trivially

you can just mod that out and have equivalence class of matrices, but it's usually not an issue

@MatheinBoulomenos do I?

I like eigenvalues

5:14 PM

@LeakyNun I had the impression when you talked about it the last time

interesting

I don't remember that at all

May 21 at 1:54, by Leaky Nun

I just feel like mobius transformations are being taught the wrong way everywhere

Also hi @MikeMiller

Hello

@MatheinBoulomenos I did say that

I was referring to the fact that they were taught as special functions $\Bbb C \to \Bbb C$

the horror

5:18 PM

@LeakyNun no, you talked about linear transformations of $\Bbb C^2$

I think they're defined on the Riemann sphere for the most part, what does that have to do with matrices or not matrices?

@MatheinBoulomenos maybe if you would like to view the conversation following it?

complex analysis course notes

Are you arguing against not introducing the Riemann sphere or against not introducing projective linear transforms or are you arguing against using matrices?

@LeakyNun I agree that's not the way to do it

@MatheinBoulomenos I think matrices are great. Riemann sphere is great. Projective linear transform is great.

I don't know what you're talking about.

$\Bbb C \setminus \{-d/c\}$ is not great

@LeakyNun oh I thought you found the matrix stuff unmotivated. I think matrices and Riemann sphere are really useful and projective linear transform, while great, doesn't help you much in doing complex analysis (depending on what you do, of course, if you prove that Riemann surfaces are projective, then have fun), so it's better as a side remark or maybe in a bonus exercise or something

that would be my opinion on that

5:28 PM

but then my professor says that this course is for literally everyone (it isn't optional) studying maths and not everyone likes algebra

and i kinda agree with my professor

What's algebraic about matrices? As I said, we're using that in modular forms all the time. The prof and TA who do the modular forms course are officially complex analysts or analytic number theorists

I don't really get what's algebraic about the Riemann sphere either, that's more of a geometric thing (stereographic projection etc.)

õ..o

oh it wasn't my professor, but still there's that opinion

I thought modular form is quite algebraic

Really? We spent two lectures computing integrals this week

they are

what integrals did you compute ?

5:33 PM

We're doing a proof of the Eichler–Selberg trace formula

That's in Lang "introduction modular forms", for example, there are really a lot of integrals in that proof, though Lang just skips over the details

also you have to compute Fourier series, worry about convergence of series and products etc. lots of analysis

@Leaky: Although I don't teach the matrices particularly in a complex analysis setting, I'm totally in favor of $\Bbb PGL(2,k)$. I even do that in my algebra book. And geometers use matrix Lie groups all over the place.

Well, we can make the sphere more algebraic, thinking of it as a symmetric space. There are plenty of places I'm in favor of that.

Perhaps I haven't told you guys that my first coauthor (a French mathematician) called me an algebraist as we were writing a paper together because I used differential forms, and to him those are algebraic. Sigh.

you don't even need to tell people what a group action is, just state and prove the properties of the map $\operatorname{GL}_2(\Bbb C) \times \overline{\Bbb{C}} \to \overline{\Bbb{C}}$ directly. If you're doing stuff like inverse and compose of Möbius transforms are again Möbius transforms, then you'll do the same computations anyway

Sure. The only thing that's less natural with matrices is the game of figuring out a Möbius transformation given where three points go.

Let $V$ be a topological vector space with respect to some topology $T$. If $T'$ is a topology on $V$ that is coarser than $T$ (i.e., $T' \subseteq T$), does $T'$ also make $V$ a TVS?

I mean, if they're $[0,1]$, $[1,0]$, and $[1,1]$, then matrices are easy.

Well, @user193319, what's the definition of a TVS?

5:41 PM

@TedShifrin yeah that's true

@Mathein: I actually ended up teaching this stuff in my algebra course a bit differently the last time, as compared to the way I did it in my book. But now I've forgotten the difference.

And I didn't keep a scan of those notes.

@TedShifrin A topology on $V$ makes it a topological vector space provided vector addition and scalar multiplication are continuous with respect to it.

OK, so can you do that with a coarser/finer topology?

I believe you can do it with a coarser topology, but not a finer topology (e.g., discrete topology on $\Bbb{R}$).

Oh, I put a note in my book, @Mathein, so I do remember.

@user193319: Well, you need open sets in both domain and range, though.

5:44 PM

@TedShifrin I think the statement that $\operatorname{Aut}_k(k(x)) \cong \operatorname{Aut}(\Bbb{P}^1(k)) \cong \operatorname{PGL}_2(k)$ is a nice thing to teach or have as an exercise in an algebra course

But doesn't restrictions to coarser topologies preserve continuity?

Not if the space is domain and range.

If you do it in the range only, sure.

Oh wait. That's backwards.

Coarser.

Yeah, it's right.

"Coarser" and "finer" always confuzle me.

Well, that was basically my definition (not using polynomials anywhere), Mathein.

Ah, true. I was thinking about doing projective geometry synthetically, but I'm not sure how to characterize linear then. Semilinear automorphisms also induce projectivities

I'm confused as to why $\varprojlim \Bbb Z/p^n \Bbb Z$ is uncountable; isn't it just indexed by $n \in \Bbb N$...? Or is this incredible naive

@ÍgjøgnumMeg $\mathcal{P}(\Bbb N) \cong \prod_{n \in \Bbb N} \Bbb Z/2\Bbb Z$ as a set

(actually as boolean algebras)

5:55 PM

F a i r

@ÍgjøgnumMeg maybe the confusion comes from this: consider a $p$-regular tree with infinitely many layers. Index the first layer with $p$ elements by $\Bbb Z/p\Bbb Z$, the second layer by elements in $\Bbb Z/p^2\Bbb Z$ etc. such that for each vertex in the layer corresponding to $\overline{k} \in \Bbb Z/p^n\Bbb Z$, all the children, which are index by elements in $\Bbb Z/p^{n+1}\Bbb Z$ are congruent to $\overline{k}$ mod $p^n$ (there are precisely $p$ of such elements.)
Then an element in $\varprojlim \Bbb Z/p^n\Bbb Z$ can be represented as an infinite path from the parent vertex that goes …

h e r e s y, everything is countable

lol

saying everything is countable is heresy, it get's you banned from the paradise that Cantor created for us

I suppose constructivists don't believe in uncountable.

heather

6:27 PM

@TedShifrin thanks so much for your answer =)

i'm sorry i had to pop out, my dad needed me to do some stuff.

Hi

I think if you're going to make a small bound like that you should say all sets are finite

7:03 PM

I can maybe buy that all groups are finite because loss of cardinality arguments is sad but...

@Daminark even if finite groups are all that matters (which is plausible), you sometimes use infinite stuff to study them

Dwadelfri

7:18 PM

Hi does anyone know a example of a f(x,y,t) => (x,y) continious and random equation.

7:46 PM

@TedShifrin I don't think that's right

Imago

8:30 PM

Given, that X takes only the discrete values 0 and 1, with equal probability between 0 and 1. How does one get its cumulative distribution function?
According to wikipedia it's 1/2 between 0 and 1 and 1, if x $\ge$ 1

Xander Henderson

Constructivists are crazy people.

Personally, I am three-ist. I don't believe that there is any number larger than 3.

4

@MatheinBoulomenos oh I'm not endorsing ultrafinitism because of group theory, that's where the "but..." came from

@Daminark makes sense

@Daminark second post in the series is done, if you're interested

@XanderHenderson are you sure they are crazy or just maybe not not crazy?

Xander Henderson

9:01 PM

@MatheinBoulomenos No sane person would deny the law of the excluded middle, thus they are definitely crazy (as that is equivalent to not not crazy).

Nice, I'll check it out

"Or can we?" Love it

9:23 PM

@Daminark yeah I'm a fan of motivation things before the defintion

me too

Don't tell Balarka but I find Hatcher hard to read because sometimes the formal stuff is buried between intuitive explanations and motivations

@AlessandroCodenotti It's not my favourite book either, but I accept that this just a matter of taste

@AlessandroCodenotti but the layout is so beautiful

Xander Henderson

@AlessandroCodenotti So very true.

I like Hatcher's exposition, but you need to already be familiar with the material before reading it.

His book is a good supplement (to some other book that I have never found), but it was hard as hell to learn out of. :\

Oh, and I hate the font he chose. His $\varphi$s look silly. :(

heather

9:34 PM

@TedShifrin I was looking more at your answer and attempting to derive $a, b, c$ for myself, and realized, it doesn't work unless you assume that $y_2 = c$. why can you assume that? (or rather, what am I doing wrong?)

9:47 PM

Yeah I'm not too happy with it either

Profs here are kinda divided

(it = Hatcher)

I think it's a decent book but I do have trouble learning from it.

Also the other day I tried computing simplicial homology of S^2 and even finding the kernel of the boundary map amounted to solving a 6x4 matrix equation, like ugh

don't use simplicial, cellular is much easier

because CW-decompositions are usually much smaller than the decomposition as a simplicial complex

I see, at some point I should learn learn CW complexes

just use singular man

9:54 PM

You know what I'm just gonna git gud at computing homotopy groups and symmetric power s

@LeakyNun that's mostly useful for theory. You can't compute most things with that definition

@Daminark that's the best approach of course

just MV it all the way man

If you can't compute something with homotopy theory, you shouldn't be computing it

then you might just use only Eilenberg-Steenrod axioms as well, if you only use formal properties like MV @LeakyNun

If you can't compute something with your phone's calculator, you shouldn't be computing it*

9:55 PM

@Daminark the first paragraph of this defines CW complex

GRE d e b u n k e d

@Daminark other than loving the "or can we?" part, what do you think?

@AlessandroCodenotti damn I need to get a phone that can do homological algebra

ok I read 2 pages of Vakil and his style is brilliant

It was good!

@LeakyNun Vakil's notes are absolutely amazing

9:59 PM

Smartphones can do a lot of stuff now so I guess there's hope

@Daminark don't know if the linear algebra part in the beginning was really necessary, but I thought it might be helpful with the motivation. If you do this whole stuff for moniods, then that's actually a special case: a choice of an endomorphism is a representation of the monoid $\Bbb N$ and $K[\Bbb N]$ is the polynomial ring

Isn't it obvious from the fact that $\varprojlim \Bbb Z / p^n \Bbb Z \subseteq \prod_{n=1}^\infty \Bbb Z/p^n \Bbb Z$ what the units are in $\Bbb Z_p$?

err

@ÍgjøgnumMeg you can use that, yeah. You could also use p-adic logarithims to compute the units of $\Bbb Z/p^n\Bbb Z$ :)

but the logarithm approach generalizes to finite extensions of $\Bbb Q_p$

@Mathein I don#t know anything about p-adic logarithms yet so I guess I'll just look at the product lol

@ÍgjøgnumMeg by abstract nonsense, taking inverse limits of rings commutes with taking the unit group

10:09 PM

or just look at the godforsaken valuation people

that's for the necessary condition

for the sufficiency condition, hensel it up

@LeakyNun just saying elements of valuation $0$ doesn't really describe the structure of the unit group

@Leaky imagine if the answer wasn't immediately obvious

what?

@Mathein thanks

@MatheinBoulomenos oh, structure

then logarithm is good i suppose

10:12 PM

yeah, a Hensel argument as you said gives you a splitting of the sequence $1 \to 1+p\Bbb Z_p \to (\Bbb Z_p)^\times \to \Bbb{F}_p^\times \to 1$ and then you study $1+p\Bbb Z_p$ by the p-adic logarithm

you can also construct this splitting (which is really just taking Teichmüller representative) without Hensel if you want

fair enough

@MatheinBoulomenos do you have questions for finite galois theory?

hmm

things are starting to sink in for me

they are starting to become like high-school arithmetic: you get used to it and it becomes your second nature

high-school arithmetic definitely isn't second nature for my painfully poor mental arithmetic

lol

Arithmetic is extremely second nature for me

10:15 PM

high-school arithmetic is harder than finite galois theory

I just whip out my phone

hi @loch

ha

hi @LeakyNun

@loch you should study galois theory :P

loch

10:19 PM

@LeakyNun i did

well not a lot about the infinite case

if any

I see

@LeakyNun Let $G$ be a finite abelian group, then show there exists a finite Galois extension $K/\Bbb{Q}$ with $\operatorname{Gal}(K/\Bbb{Q}) \cong G$

@Mathein so each $a \in \Bbb Z_p$ is a tuple $(a_1 \bmod p, a_2 \bmod p^2, \dots)$ with connecting homomorphisms $\pi_n$ s.t. $\pi_n(x_{n+1}) = x_n$ for all $n \in \Bbb N$. If $a_1 = 0$ then $a$ is clearly not a unit (since $a_1$ is not invertible in $\Bbb Z/p\Bbb Z$). Conversely, if $a$ is not a unit then $(p^i, a_i) \neq 1$ for some $a_i$ so $p \mid a_i$, and then use the connecting maps to conclude that $p \mid a_1$ so $a_1 = 0$?

yeah that's true, the units are the elements not divisible by $p$

Cool :) Thanks, gonna stop for the day now I think

lol

10:30 PM

math.stackexchange.com/questions/2828059/… does anyone know how to answer this?

im stuck

@LeakyNun this is not an easy exercise, but it's doable with some basic Galois theory. You'll need a theorem from number theory that you probably heard of, though

kronecker-weber?

no

CRT?

not algebraic number theory actually

10:32 PM

division theorem?

kronecker-weber might give you a hint in the right direction ;)

sure, I get that I need to construct a surjection from $\widehat{\Bbb Z}^\times$ to our abelian group

note that you're only using the obvious part of K-W here: that cyclotomic extensions are abelian is not difficult to see

...

so I need to find a large enough $N$ such that $(\Bbb Z/n\Bbb Z)^\times$ surjects onto $G$

if $n = \prod p_i^{a_i}$ then that thing becomes $\prod (\Bbb Z/(p_i^{a_i-1}(p_i-1))\Bbb Z)$

which obviously surjects onto $\prod (\Bbb Z/(p_i^{a_i-1})\Bbb Z)$

10:37 PM

you need to be slightly more careful with $p_i=2$, but yeah

done?

the abelian group can have two cyclic factors with the same prime

you're right

ok

$\Bbb Z/(p_i^{a_i-1}(p_i-1))\Bbb Z$ surjects onto $\Bbb Z/(p_i-1)\Bbb Z$

so for each cyclic factor, go through the multiples until you're one less than a prime

why does that work?

does this work?

some theorem about arithmetic progression

10:41 PM

it does, but it's not obvious

if a and b are coprime, then an+b contains infinitely many primes

here b=1

right, that's the number theory theorem I was thinking of

nice

If you want to do more Galois theory, I recommend learning some ANT, there the Galois groups or subgroups thereof basically act on everything and everyone and everyone's dog

confirmed

10:43 PM

well when I said finite galois theory, I meant finite fields

lol

I just thought without Krull topology

so would you have a problem about finite fields?

$\mathcal{O}_K/\mathfrak{p}$ is a finite field! pokes in Galois

@MatheinBoulomenos so in $\Bbb F_9$, $X^9-X = X(X-1)(X+1)(X^2+1)(X^2+X-1)(X^2-X-1)$

which one is the minimal polynomial of the primitive root of unity?

@LeakyNun you only need to look at the $8$th cyclotomic polynomial which is $x^4+1$ which has precisely the last two factors

10:54 PM

right, so which one is it?

you will have $4$ $8$th roots of unity, but not all with be roots of the same polynomial

the question is not well-defined

:(

it's not like over $\Bbb Q$ where all primitive roots of unity are conjugate

so if all I tell you is that $\zeta^8=1\ne\zeta^4 \in \Bbb F_9$, then you can't tell me the min-poly of $\zeta$?

@MatheinBoulomenos :'( I hate maths

indeed

10:56 PM

maths sucks

@LeakyNun woah! watch your mouth!

dont make me wash it out with soap

:(

lol, that happens already over extensions of $\Bbb Q$. Note that $x^4+1=x^4+2x^2+1-2x^2=(x^2+1)^2-(\sqrt{2}x)^2=(x^2+\sqrt{2}x+1)(x^2-\sqrt{2}x+1‌)$, so you can't answer the same question over $\Bbb{Q}(\sqrt{2})$ either

but $\sqrt2$ and $-\sqrt2$ are conjugates

$1$ and $-1$ aren't

you still can't give a minimal polynomial

10:58 PM

:(

You should think of the fact that all primitive roots of unity are conjugate over $\Bbb Q$ as a very special property of $\Bbb Q$

@LeakyNun finite field field question: let $p$ be a prime and $\Bbb{F}$ be a finite field, how many irreducible polynomials are there over $\Bbb{F}$ of degree $p$ in terms of $p$ and the order of $\Bbb{F}$?

But I'll go to sleep now

See you @everyone

@Mathein schlof guat :P