Mathematics - 2017-10-12 (page 7 of 9)

MatheiBoulomenos

17:00

Hi @BalarkaSen @TedShifrin

Semiclassical

@MatheiBoulomenos Here's a question I have no idea how one would answer. Suppose we take that method of generating additional solutions from a given solution as an equivalence relation, i.e. two solutions are equivalent if you can obtain one from the other.

Ted Shifrin

Hi @Mathei,@Balarka

Semiclassical

How many distinct equivalence classes will there be?

Balarka Sen

Hi @Ted

MatheiBoulomenos

@Semiclassical that's a good question. I guess someone knowledgable about elliptic curves could answer that, but unfortunately I am not knowledgable about that

Semiclassical

17:01

Nor I

user2154420

@BalarkaSen I thought a discrete set is one with the discrete topology. But for a cover, don't we only know that the preimage of an evenly covered open neighborhood $U$ of that point is a disjoint union of $V_i$ where the $V_i$ is homeomorphic to $U$. So why is this disjoint union a discrete set?

I think that is where my confusion is coming from

Balarka Sen

@user2154420 That means every point $y_i$ of $p^{-1}(x)$ has a neighborhood in the cover (namely, the slice $V_i$) separating itself from the other points

So under the subspace topology, $p^{-1}(x)$ becomes discrete

Because $V_i \cap \{y_i\} = \{y_i\}$, implying each point of $p^{-1}(x)$ is open in the subspace topology.

user2154420

Oh right :)

Semiclassical

@MatheiBoulomenos hmm, the comments to that question link this page: mathpages.com/home/kmath164.htm

user2154420

So that is what I wrote initially, but I convinced myself that was wrong because apparently we need a path-connected assumption...

Semiclassical

17:06

and I'm not sure what conclusion to draw

user2154420

but that's confusing because we only want that it is finite, not the same finite number everywhere

Which I assume is why one would also require path-connectedness

MatheiBoulomenos

@Semiclassical well, we have another curve here, so I'm not sure if the same argument applies

Balarka Sen

@user2154420 Well, you need $X$ to be connected, that is true. Because otherwise the cover might be disconnected and have a compact component somewhere and a noncompact component elsewhere

Like cover of the circle by a disjoint union of a circle and $\Bbb R$

Semiclassical

well, the key sentence is this: "As Poincare conjectured and Mordell proved, all the rational points on a curve such as x^3 + y^3 = 6 can be generated by tangent and chord constructions applied to a finite set of points."

Balarka Sen

That's pretty cute

Semiclassical

17:09

it seems to be this theorem: en.wikipedia.org/wiki/Mordell%E2%80%93Weil_theorem

Balarka Sen

Aha

MatheiBoulomenos

Well, the theorem seems to only say that the group has finite rank

not that the rank is $1$

and even if the rank is $1$, we would have to make sure that our initial solution is actually a generator

user2154420

Oh wait! Is another way of saying this that if $X$ is finitely covered at each point, each of these might be different from each other if $X$ is not connected

Semiclassical

definitely

MatheiBoulomenos

I really want to learn about elliptic curves eventually

Semiclassical

17:11

and the question about rank appears to be a pretty deep one

Lucas Henrique

@MatheiBoulomenos thanks. Isn't it an overkill?

MatheiBoulomenos

Well, you don't need the theory of elliptic curves to verify that this method generates additional solutions

Semiclassical

sure it's overkill

or, rather, it's definitely pushing the question beyond what you need

Balarka Sen

@user2154420 No, if $X$ is connected, the cardinality $\#p^{-1}(x)$ for a cover $p : Y \to X$ is always the same, irregardless of $x$.

PVAL-inactive

Think the tangent chord operation is group multiplacation.

Semiclassical

17:15

Yeah, I think that's right as well

Balarka Sen

chord operation is addition, tangent is doubling

Lucas Henrique

I mean, it's from an undergrad olympiad

PVAL-inactive

and what Mordell proved is that elliptic curves are f.g. as groups.

Balarka Sen

A + A

MatheiBoulomenos

Well, I guess from olympiads you might expect to get to these formulas out of thin air

Semiclassical

17:16

to put it a little differently: If you start from the solution (2,1,1) then you can keep doubling and get new solutions

so you'll get an infinitely family of them

MatheiBoulomenos

I'm not sure how to solve it differently

Semiclassical

question is whether or not all rational solutions can be obtained in this way

user2154420

Yeah, that's what I meant

I was saying that's why we need the connectedness assumption

PVAL-inactive

Mordell predates modern algebraic geometry so I don't know what the easiest proof is now.

user2154420

Sorry for the confusion, and thanks for the clarity

Ted Shifrin

17:17

hi @Lucas

Balarka Sen

@user2154420 Ah, I misread the "not. You are right, of course, then.

Lucas Henrique

@Semiclassical but when you double everything you're just playing with equivalence classes

Semiclassical

sure.

Lucas Henrique

hey @Ted :)

N. Maneesh

2

Q: uniform continuity and which of the following statements are true??(NBHM-$2014$)

Let $g_n(x)=n[f(x+\frac{1}{n})-f(x)]$, where $f: R\to R$ is a continuous function . Which of the following are true? a. If $f(x)=x^3$, then $g_n\to f'$ uniformly on $R$ as $n\to \infty$. b. If $f(x)=x^2$, then $g_n\to f'$ uniformly on $R$ as $n\to \infty$. c. If f is differentiable and if $f'$...

analysis

please help me

PVAL-inactive

17:18

Mordell–Weil theorem

In mathematics, the Mordell–Weil theorem states that for an abelian variety A over a number field K, the group A(K) of K-rational points of A is a finitely-generated abelian group, called the Mordell-Weil group. The case with A an elliptic curve E and K the rational number field Q is Mordell's theorem, answering a question apparently posed by Poincaré around 1908; it was proved by Louis Mordell in 1922. == History == The tangent-chord process (one form of addition theorem on a cubic curve) had been known as far back as the seventeenth century. The process of infinite descent of Fermat was well...

Lucas Henrique

but $\frac{a}{b}$ is, in fact, equal to $\frac{ka}{kb}, \forall k \in \Bbb Z$

so you won't get new solutions

Semiclassical

it's not that kind of doubling

PVAL-inactive

This probably is done in a modern efficient way in Silverman's book.

Lucas Henrique

I'm not getting the point

Semiclassical

it's not (a,b,c) is a solution -> (2a,2b,2c) is a solution

Lucas Henrique

17:19

oh

I'm sorry

how is the transform like?

Kasmir Khaan

@anon Hello !

anon

hello

Kasmir Khaan

@TedShifrin Ted did you get my message ? =p

Lucas Henrique

hey there @anon

Kasmir Khaan

Anon !

Semiclassical

17:20

it uses the group operation on elliptic curves

Kasmir Khaan

I got a hard question for you

Semiclassical

but it amounts to the tangent-chord construction

Lucas Henrique

Gee

Kasmir Khaan

Let me post a picture of it here :D

Lucas Henrique

I know nothing about group theory

There's probably an easier solution

Semiclassical

17:21

definitely

Lucas Henrique

I'm almost sure it's a proof by construction or something that uses elementary number theory

Like, 'pidgeonholing'

MatheiBoulomenos

$(a,b,c) \to (-ab^3-9ac^3,a^3b+9bc^3,-a^3c+b^3c)$

Kasmir Khaan

Let p be a prime , and suppose m is a number not divisible by p. prove that there can be only fintely many numbers k for which there is a simple group of order mp^k @anon

Semiclassical

that's the doubling?

MatheiBoulomenos

yes

Semiclassical

17:23

hooray

Lucas Henrique

@MatheiBoulomenos heck, this looks awful

MatheiBoulomenos

@LucasHenrique I know. But you don't need group theory to verify that it works

Semiclassical

main point is that this (apparently) has a nice geometric interpretation in terms of the elliptic curve $x^3+y^3=9z^3$

Lucas Henrique

in fact.

Ted Shifrin

@Kasmir: Use your own judgment. I do not need to be such a babysitter at this point.

Kasmir Khaan

17:24

@TedShifrin omg :(

@TedShifrin I only asked because i dont know if you written that book for the same course as we doing or not

-.-

Ted Shifrin

Well, seriously, the answer to that is clearly NO.

Kasmir Khaan

hmm

Then i dont see the point

Anyway ill figure it out , did not think this question would bother you , sorry again

Semiclassical

@LucasHenrique okay, here we go

the blue curve is x^3+y^3=9z^3

the point on the right is (a/c,b/c) for (a,b,c)=(1,2,1)

Alessandro Codenotti

Hi chat

Tobias Kildetoft

Hi

I realized today that I don't actually know any good applications of the Jordan normal form

Semiclassical

17:33

if you plug that (a,b,c) into @MatheiBoulomenos's formula, you'll get the point (-17,20,7) which corresponds to the point (-17/7,20/7). that's also a solution of x^3+y^3=9 and it's point on the left

Tobias Kildetoft

Which felt like a fairly big lack when lecturing on it

Ted Shifrin

Hi @Alessandro

Semiclassical

if we now draw a line through these two points, we see that the left point is obtained as the intersection of the cubic with the tangent line through (2,1)

MatheiBoulomenos

Hi @AlessandroCodenotti

Ted Shifrin

@Tobias: There are lots in terms of matrix exponentials and dynamical systems. ... But there are more "straight algebra" applications as well. See, for example, exercises in Herstein's chapter on linear algebra.

Semiclassical

17:35

and moreover we can repeat this: the tangent line of a rational point through a cubic intersects with the cubic to give another rational point

so that's a geometric construction of an infinite family of rational pointst

Tobias Kildetoft

@TedShifrin In which of the books?

Topics or Abstract?

Kasmir Khaan

guys should one take homologic algebra and algabraic topology after abstract algebra?

Alessandro Codenotti

My linear algebra professor used the real variant of the JNF to answer my question on which matrices have a square root, but I've never seen a real application

Semiclassical

i'd suggest no

Tobias Kildetoft

@KasmirKhaan Yes

Ted Shifrin

17:36

The "real" book, Topics.

Kasmir Khaan

@TobiasKildetoft thanks tobias :D

MatheiBoulomenos

@TobiasKildetoft: the Jordan canonical form let's you show that any linear endomorphism of a finitely-dimensional vector space over an algebraically closed field may be written as a sum of a nilpotent and a diagonizable endomorphism

Ted Shifrin

@Alessandro: In geometry and dynamical systems, the Jordan form is used all over the place.

Semiclassical

or, at least, i'd say those topics are probably not what you should be doing directly after abstract algebra

Ted Shifrin

@Mathei: That's circular. That's merely restating the canonical form.

Tobias Kildetoft

17:37

@MatheiBoulomenos That is much more trivial to show that JNF

Kasmir Khaan

@TobiasKildetoft like what does some need to know before studying homologic algebra and algebraic topology ( exept abstract algebra ) ?

Tobias Kildetoft

as that is just bringing the matrix to upper triangular form

MatheiBoulomenos

apparently it is used in the theory of Lie algebras as well, but I don't know any details

Ted Shifrin

You want them to commute, @Tobias.

MatheiBoulomenos

yeah, I missed that

Tobias Kildetoft

17:37

@TedShifrin Ahh, right

MatheiBoulomenos

I mean, I prefer that formulation over the matrix one

Tobias Kildetoft

@MatheiBoulomenos Hmm, it may just have been hidden somewhere when I did the basics of Lie algebras

Ted Shifrin

That's because you're a head-in-the-clouds theory person who doesn't believe in using math :P

Alessandro Codenotti

@TedShifrin I trust you, I know nothing about those

Balarka Sen

What is it that we are talking about?

Ted Shifrin

17:38

uses of Jordan canonical form

Balarka Sen

Classifying solutions to linear ODEs

MatheiBoulomenos

@TedShifrin I won't even try to deny that

Ted Shifrin

I said matrix exponentials and dynamical systems, @Balarka.

Tobias Kildetoft

@MatheiBoulomenos Hmm, I suppose it is basically the same as the Jordan decomposition of a matrix. I just never really gave that much thought

Balarka Sen

@TedShifrin Oh well, you're sniping me

MatheiBoulomenos

17:40

matrix exponentials and ODEs is really the same thing when it comes to the Jordan form

Balarka Sen

@Mathei I mean, columns of the matrix exponential span the solution space of the corresponding matrix ODE

Ted Shifrin

@Tobias: Here's a reasonable one. Prove that $A$ and $A^\top$ are similar.

I don't think that's very obvious without something powerful.

Balarka Sen

That bit doesn't have much to do with the Jordan form

Tobias Kildetoft

@MatheiBoulomenos Ahh, just found it in Humphreys. He indeed recalls the JNF (except he calls it the Jordan canonical form instead) in order to do stuff

MatheiBoulomenos

@TobiasKildetoft so my memory was right

Tobias Kildetoft

17:41

@TedShifrin Ahh, that is indeed a neat one

If only I had had more time to prepare before lecturing on it

Ted Shifrin

Well, don't bitch at us about that!

Balarka Sen

The commutative algebra version of the JNF is the primary decomposition I believe

MatheiBoulomenos

Pretty much

Ted Shifrin

Hmm, @Balarka, I'd never thought of it that way.

Tobias Kildetoft

@TedShifrin I won't. I will have to consider bitching to the guy who was supposed to lecture :)

MatheiBoulomenos

17:43

It's an easy application of the classification of finitely generated modules over a PID

Balarka Sen

Right.

MatheiBoulomenos

(which is a special case of the classification of finitely generated modules over a Dedekind domain)

that's how I think about it

Balarka Sen

@Tobias I'm pretty much learning about the Jordan form on the go while learning how to use them in ODEs

That's why I love this course so much, it gets the blood running and the linear algebra pumping

Ted Shifrin

Hirsch-Smale derive all the linear algebra from scratch, Balarka.

Balarka Sen

Yeah I saw

MatheiBoulomenos

17:44

But only linear algebra over $\mathbb R$ and $\mathbb C$, I always found that restrictive

Balarka Sen

They have a very clean exposition

Ted Shifrin

Blah @Mathei.

It works for any algebraically closed field. Nothing different that I can recall.

MatheiBoulomenos

I mean the ODE course

Balarka Sen

I think there's a non-algebraically closed version, the rational normal form or something

Tobias Kildetoft

@BalarkaSen Well, there is the Smith normal form for integral matrices

Balarka Sen

17:46

True

I have forgotten these various things

Tobias Kildetoft

And one can replace the Jordan blocks by larger blocks based on the degree of the extension of the field which contains the eigenvalues

MatheiBoulomenos

@BalarkaSen that form is named after Frobenius I think?

Ted Shifrin

Rational canonical form, @Balarka, but it's a very different decomposition.

Tobias Kildetoft

similarly to how one can do it with those $(a\ b\ -b\ a)$ matrices in place of complex numbers

Balarka Sen

@MatheiBoulomenos Sometimes it's good to be concrete, even at the cost of being restrictive.

Ted Shifrin

17:47

@Mathei: You're being ridiculous. People do not generally do dynamical systems over arbitrary fields.

MatheiBoulomenos

Nah, abstract is the way to go

Balarka Sen

I hope you mean that ironically.

Ted Shifrin

I'm gonna seriously end up putting you on ignore.

No, he's worse than Demonark.

MatheiBoulomenos

I'm just joking

Balarka Sen

@Ted Nah he's joking

Tobias Kildetoft

17:48

I don't mind intro linear algebra restricting to the reals and complex numbers, though I prefer if a note is made that the choice is not important (except where it is of course)

Balarka Sen

Just like we joke about pictures being the end of math :)

He's just less obvious

MatheiBoulomenos

linear algebra over finite fields is important in applications

Ted Shifrin

Pedagogy is a serious issue, folks. Most students are not going to end up being abstract algebraists, or even mathematicians.

Balarka Sen

@MatheiBoulomenos That's true.

Ted Shifrin

That's true, @Mathei, but that doesn't mean that a first course needs to do finite fields.

Tobias Kildetoft

17:49

I mean, are there ever any exercises that needs more than rationals plus square roots?

(until one gets to exponentials that is)

Ted Shifrin

What are you talking about, @Tobias?

Exercises in what?

Tobias Kildetoft

@TedShifrin I mean in intro linear algebra

MatheiBoulomenos

Well, I'm glad I was taught linear algebra by an abstract algebraist how did everything over general fields :P

Tobias Kildetoft

They work over the reals usually, but practically all exercises are really over the rationals

Leaky Nun

@MatheiBoulomenos someone ending up dealing with images do not need to know how matrices work with finite fields

Ted Shifrin

17:50

Right.

And I have no problems with that.

Tobias Kildetoft

@TedShifrin Neither do I, I just find it curious

MatheiBoulomenos

@LeakyNun finite fields are pretty important in image processing

Ted Shifrin

Indeed, in my books I've gone out of my way to be extra artificial and try to make arithmetic not the point of the course. But I'm not opposed to some use of computers/calculators to do RREF, etc. Just why make the computations horrendous unless it's a truly meaningful application?

Balarka Sen

@Mathei The thing is, historically most of the algebraic abstractness are truly motivated by concrete, analytical/geometric happenings over C or R

MatheiBoulomenos

Honestly, I find it easier to work in a small finite fields than over rationals where sometimes the denominators get huge

Balarka Sen

17:53

@Ted I saw the Lypapunov function yesterday, in dynamic classification of linear ODEs

Tobias Kildetoft

@TedShifrin I agree with not making computations the main point. Thought there can be a good opportunity to teach the students to keep a $\pi$ around in calculations rather than replacing it with a decimal.

Semiclassical

Eh, just do pi^2 =10. :P

(That’s a surprisingly good approximation)

Balarka Sen

just do pi = 3

engineering intensifies

Ted Shifrin

Other than matrix exponentials, I can't recall $\pi$'s making an urgent appearance, @Tobias.

Tobias Kildetoft

@TedShifrin I meant in general, by introducing it in some linear equations in such a way that the answer comes out nicely if one just keeps it around

Ted Shifrin

17:57

Talk about contrived.

MatheiBoulomenos

@TedShifrin maybe when you derive canonical forms for orthogonal endomorphisms in terms of trigonometric functions?

Semiclassical

pi = 3 is off by 4.7 percent, whereas pi^2 =10 is only 1.3 percent off

Silent

Please someone answer this

Balarka Sen

@MatheiBoulomenos Realification?

Ted Shifrin

But then you're not typically doing row operations or solving linear equations when you do a normal form for a rotation matrix, etc.

MatheiBoulomenos

17:59

@BalarkaSen you mean restriction of scalars?

Semiclassical

Of course, pi = 22/7 does a lot better than both of them

Balarka Sen

lol, yes

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$Mathematics$ Mathematics