Mathematics - 2015-07-26 (page 1 of 3)

Boobs are something really magic to me. But I don't put them on a shirt because I know it won't look good. Remember, you're a person, not a book. — Phaptitude Oct 23 '13 at 9:38

Soham Chowdhury

Not much. I was too busy with homework and quizzes and crap like that.

lol @r9m

Remember me

Hey @r9m How is my city?

r9m

@Rememberme nice! :D I came back to Kolkata on 23rd though .. so didn't have much time to explore around the city

Remember me

Oh.. So when will you be joining TIFR?

r9m

4:07 AM

@Rememberme :P next year possibly

well wtf

lol

@SohamChowdhury wtf

Well you know @Soham CMI entrance paper has more difficult questions than ISI's

Soham Chowdhury

Yes.

Remember me

4:09 AM

I wasn't able to solve most of 'em

Soham Chowdhury

But at least now I know that there's some place I can get into and so I won't have to roam the streets after class 12 :P

Remember me

hahahaha

But you don't know how the paper will be when we will be giving our ISI entrance examination@Soham

Well the geometry questions are a bit difficult

Remember me

4:52 AM

Hello@Eric

Deathslice

Does anyone how you can determine if a graph requires a loop or not? The subject that I'm referring to is graph theory in discrete math.

Basically the only explanation that I have is that if the sum of the lower degrees is greater than or less than the highest degree, then loops may be required.

user147690

5:09 AM

@Deathslice Learn Havel-Hakimi

Dave

can some one explain the logic to how these cancel out: sqrt(74) * (7*sqrt(74)/74) = 7

i can't visualise how it cancels out the denominator and the sqrt(74) numerator

Remember me

Hey @Alex

user147690

Hey @Rememberme, how are you?

Remember me

Good .. How about you@Alex

user147690

@Rememberme Very good! I start uni tomorrow

Remember me

5:13 AM

great...

user147690

I have 2 double clashes, and a triple clash though lmao

user147690

I.e. 3 lectures are scheduled on the same hour once, and two lectures are scheduled on the same hour twice

Remember me

Well I have a question for you if you are free

@AlexClark Seems tedious

user147690

How so?

user147690

I am not free, but you can ask anyway xD

Remember me

5:14 AM

I know Strums theorem .. Is there any other way to show that a polynomial has no real roots@Alex

user147690

@Rememberme It's more just terrible planning on their part

Remember me

True...

user147690

@Rememberme p-adics might be another good way - although I don't know enough about them

Remember me

How so ? Can you explain how they work here

user147690

I can't sorry, I don't know enough about them and I might be wrong, actually I would be interested in hearing if @Balarka could answer your question with p-adics

user147690

5:16 AM

@Rememberme I mean if you asked this in 2 weeks, I would 100% have an awesome answer

user147690

Since we will be spending 2 weeks answering this super rigorously I believe

Remember me

Okay I will wait for @Balarka then :p

user147690

I must go to uni now :)

Remember me

Well I am solving an undergrad paper so they have this question

Oh ... bye

user147690

@KajHansen how do you pronounce your first name btw, I have always read it exactly like the word 'cage'

user147690

5:18 AM

Why so many dots btw @Rememberme haha

user147690

Usually dots are annoyance in Aussieland

Remember me

Well you can say I have a paperweight kept on the full stop :p

user147690

:P okay bye for real

leo

5:43 AM

Anything wrong?

Soham Chowdhury

@Rememberme remember I said that once?

Remember me

Yes that is why I said that :p

Well do you mind looking at a question?@Soham

Soham Chowdhury

which poly has no real roots?

Remember me

0

Q: How to show that a polynomial does not have real roots

How to show generally that a polynomial does not have real roots. Well, for eg lets take the polynomial $x^8-x^7+x^2-x+15$ . Here the power($n=8$) is even so it can have real roots or it might not have real roots. Something which I thought was to find the minima and show that if the minima of $...

polynomials

^^

Hey@Huy

Soham Chowdhury

do you need that poly, specifically?

Remember me

5:50 AM

Yes I would like to find a way for that polynomial then generalize the method I used for that polynomial for other polynomials

Soham Chowdhury

heh, commenter just killed the problem

where is this from?

Huy

@Rememberme: I doubt there's a general way but approaches as presented in the answers will work often.

(an easy general way)

Soham Chowdhury

@Nemo's is nice IMO.

where is this from, @Rem?

Remember me

ISI 2012

Soham Chowdhury

hehe

Remember me

5:54 AM

Well I thought my minima idea was a streak of brilliance but then it failed :p@Soham

@Soham were you able to get a solution straight after looking at the question?

Anyways sturm polynomials are a pain to solve sometimes

phoenix

6:11 AM

has anyone got any ideas about this counting problem? math.stackexchange.com/questions/1373679/…

Balarka Sen

6:29 AM

@AlexClark The thing you do with p-adics is called the Local-Global principle, which works for a restricted class of diohantine equations over $\Bbb Z$. $p$-adics aren't devised to tell you things about $\Bbb R$.

Soham Chowdhury

Quadratic reciprocity is ridiculous.

Balarka Sen

QR is very cool.

Soham Chowdhury

Yes, I meant that in a good way.

Balarka Sen

I figured.

Soham Chowdhury

I'm reading Fearless Symmetry, actually. It's not a very big book, but I guess it gets harder as one goes further.

Balarka Sen

6:32 AM

I got to knew QR's from the algebraic number theory classes I took. Very fun.

@Soham Where did you buy it from?

Soham Chowdhury

Did I? ;)

Balarka Sen

oh. you pirated it.

Soham Chowdhury

QR seems too good to be true.

yes, as always.

Balarka Sen

send me a copy, I am too lazy to pirate it.

Soham Chowdhury

okay, wait.

Balarka Sen

6:36 AM

no, I can open djvu's. I am downloading it right now.

Soham Chowdhury

use the pdf, it's a newer ed.

Balarka Sen

oh?

Soham Chowdhury

yes.

you can get in around ~120 pages in an hour or so (where I am now). then they hit Galois theory. woo.

Balarka Sen

I'm skipping to part 2.

Soham Chowdhury

you took algebraic NT? what things have you taken, exactly? :P

@BalarkaSen of course you are.

Balarka Sen

6:37 AM

@SohamChowdhury yes, I have been in a basic algebraic number theory class.

Soham Chowdhury

Christmas thm?

Balarka Sen

I have taken algebraic number theory, commutative algebra and algebraic topology so far.

@Soham What's that? Don't know it by name.

Soham Chowdhury

oije sum of squares iff 1 mod 4

"We will call it "the Galois group" for short and denote it by "G," because we believe it is the group par excellence." :P

Balarka Sen

oh. but that's very basic.

Soham Chowdhury

I know.

But I can't wait to prove it.

Balarka Sen

6:39 AM

you just enumerate the squares mod 4.

no 3 mod 4 number appears.

Soham Chowdhury

It was the first nontrivial NT theorem I learned.

@BalarkaSen is that a proof?

Balarka Sen

ah, iff.

Soham Chowdhury

I thought you had to study ${\bf Z}[i]$.

Balarka Sen

yes, you have to, for the other direction.

Soham Chowdhury

ah.

Balarka Sen

6:40 AM

sorry, I thought it was an "only if"

Soham Chowdhury

I like \bf except for $\bf R$ and $\bf Q$.

Balarka Sen

I use \bf for Z only to denote p-adics.

Soham Chowdhury

14

Q: Blackboard bold

Why do people use blackboard bold here and elsewhere in print? I thought the whole point of the font was as a substitute for bold when one was writing out something by hand. Shouldn't we be using just bold R, Z, N, etc.?

discussion soft-question

Epic jhogra.

:P

Balarka Sen

aha, they talk about etale cohomology too.

rubs hands

Soham Chowdhury

hahaha

that Galois group is related to ell curves as well?!

Balarka Sen

6:42 AM

of course.

Soham Chowdhury

how connected is that thing?

goodness.

Balarka Sen

the absolute galois group acts on the p-torsion points of a give ell curve over Q

that's how the 2-dimensional representations appears.

Soham Chowdhury

I don't understand "p-torsion points" yet, as you know. But, still, interesting.

Group actions.

Balarka Sen

similar representations come from modular forms -- that's the beginning of the connection between elliptic curve and modular forms.

Soham Chowdhury

all this is Weil conjectures / Langlands stuff, no?

Balarka Sen

6:44 AM

@SohamChowdhury $p$-torsion points of a group $G$ is just collection of elts $x \in G$ such that $x^p = 1$.

Soham Chowdhury

oh.

I know the group law, before you start.

interesting, indeed.

Balarka Sen

good.

Soham Chowdhury

ICM2014 VideoSeries PL9: Manjul Bhargava (Laudation by Benedict Grossi) on Aug16Sat

►

the part with Bhargava himself is pretty understandable.

(unlike the "introductory" speaker's)

Balarka Sen

yes, that's an excellent lecture.

Soham Chowdhury

isn't most of arithmetic top about finding analogs of higher rec. laws in linked-up knots?

Balarka Sen

6:47 AM

no, it's just about making the list of analogies longer and formal.

Soham Chowdhury

oh.

anyway. are you going to read the book now?

Balarka Sen

higher reciprocity laws is Langlands. you're mixing two things up.

effect of knowing too much words.

:P

Soham Chowdhury

because I know nothing about either

Balarka Sen

@SohamChowdhury no, I am trying to (re)discover the analogies by myself.

Soham Chowdhury

"(note that the knots corresponding to the three primes are not the unknot but more complicated). Here's where the story gets interesting : in number-theory one would like to discover 'higher reciprocity laws' (for collections of n prime numbers) by imitating higher-link invariants in knot-theory."

oije.

Balarka Sen

6:48 AM

ok.

Soham Chowdhury

from the MO question with those pretty slides. :P

13

Q: What is the knot associated to a prime?

I can't help but ask this question, having found out about arithmetic topology here on MO. There is a concise description of the MKR dictionary central to this philosophy here. This dictionary is used to translate statements in three-manifold topology into statements in number theory, and vice ve...

knot-theory nt.number-theory

Balarka Sen

I am not looking there.

Richard

@SohamChowdhury - Please don't post links to copyrighted material unless you're the rights holder. Doing so could result in a chat ban.

Soham Chowdhury

Didn't know that, sorry.

are you reading the book, @Balarka?

Balarka Sen

yes.

Soham Chowdhury

6:51 AM

I wonder how they get to cohomology in the span of 100 pages . . .

Richard

@SohamChowdhury - No worries. I know how frustrating it can be when you're referencing something and want to pass it over. Resist that urge :-)

Soham Chowdhury

:)

Balarka Sen

bleh, they don't really define etale cohomology.

Soham Chowdhury

displeased noises emanate from Balarka

what were you expecting?

Balarka Sen

a sketch of the definition.

The Substitute

6:55 AM

off topic algebra. On pg 176 of Dummit/Foote Abstract Algebra 3rd ed, regarding semidirect products, they prove in Theorem 10 (e) that $\phi(k)(h)=khk^-1$, that is, that the map $\phi$ from $K$ into $Aut(H)$ is conjugation by $k$. Doesn't this imply that there is only one $\phi$ for each pair $H,K$? I was under the assumption that $phi$ could be ANY homomorphism from $K$ into $Aut(H)$

Soham Chowdhury

isn't it supposed to be some kind of black box even when you learn AG for the first time?

Balarka Sen

no.

Soham Chowdhury

is $\bf Q^{alg}$ what you call $\bar{\bf Q}$?

Balarka Sen

yes.

Soham Chowdhury

well, a bunch of people on MSE and MO say it should be.

anyway.

Balarka Sen

6:56 AM

etale cohomology is a precise, defined thing, as far as I know.

Soham Chowdhury

so you lost interest?

Balarka Sen

yeah.

anon

@TheSubstitute you can build a semidirect product using any homomorphism phi. conversely, given a group which is an internal semidirect product, there is a corresponding homomorphism phi.

Balarka Sen

hi @anon

anon

hi

Soham Chowdhury

6:58 AM

is the set of all numbers which are expressible in radicals given a symbol or name?

Q^ab

^

name?

abelian closure of Q.

anon

or "maximal abelian extension of Q"

The Substitute

6:59 AM

@anon isn't it a bit strange that for any choice of $\phi$ that we have $\phi(k)(h)=khk^-1$? I understand the proof, but it's proved in a roundabout way by identifying $H,K$ with subgroups of the direct product. They identify $h$ with $(h,1)$ and $k$ with $(1,k)$

anon

@TheSubstitute if you construct a semidirect product out of H, K and phi, then the relation holds by construction. we literally defined the semidirect product by imposing that relation on the elements.

so of course it holds!

we made it hold.

Balarka Sen

@anon $\Bbb Q$ are $S^3$ are similar in the sense that there is no unramified extension of $\Bbb Q$ and on the other hand there are no unbranched covering of $S^3$. cool, eh?

The Substitute

@Anon I thought we just wanted $khk^-1$ to be in $H$, not to necessarily fix $h$. So why does it matter how I define $\phi$ if the associated group action will always be exactly the same? Doesn't this mean that the semidirect product doesn't depend on $\phi$?

anon

@TheSubstitute what do you mean by "the associated group action will always be exactly the same"?

Balarka Sen

@anon I like to think of semidirect products as the analogs for "split extensions" in Grp, and thinking about the aut definition as a "nice presentation" for such groups. much more satisfactory.

anon

7:03 AM

different phis yield different group operations on the set HxK. some of those operations make isomorphic groups ... some may not!

@BalarkaSen not sure, really

The Substitute

@Anon The associated group action of $\phi$ is $k*h=\phi(k)(h)$, where $\phi(k)$ is an automorphism of $H$. By the theorem I referred to, this action is always $khk^-1$, regardless of $\phi.$

anon

@TheSubstitute yeah, because we define phi(k)(h) to equal khk^-1

Balarka Sen

@anon many other analogies appear in arithmetic topology which are seemingly not just coincidence.

anon

just like how in Z/3Z we define 1+1+1 to be 0

The Substitute

@Anon, where exactly does the choice of $\phi$ affect the isomorphism type?

anon

7:10 AM

@TheSubstitute consider $(H\times K,*_\phi)$ to be the the group structure on the set $H\times K$ afforded by the semidirect product construction. suppose $\phi,\psi$ are different maps $K\to{\rm Aut}(H)$. Thus, $\phi(k)\ne\psi(k)$ as automorphisms of $H$ for some $k\in K$. Thus, for that $k$, there is an $h\in H$ for which $\phi(k)(h)\ne \psi(k)(h)$.

Then in $(H\times K,*_\phi)$ we know that $k*_\phi h*_\phi k^{-1}$ equals $\phi(k)(h)$, whereas in $(H\times K,*_\psi)$ we know that $k*_\psi h*_\psi k^{-1}$ equals $\psi(k)(h)$. So the binary operation on the set $H\times K$ (remember, a binary operation on a set $G$ is a subset of $G\times G$) depends on choice of $K\to{\rm Aut}(H)$.

Just because the group operation is different doesn't mean we didn't create an isomorphic group - but in general, we can get nonisomorphic groups with different choices of homomorphisms. Really, you should be asking yourself why you believe your unfounded assumption that the isomorphism type of $(H\times K,*_\phi)$ doesn't depend on $\phi$. Where does that assumption come from?

Soham Chowdhury

so the abs. Gal. grp. of $Q^{alg}$ is just its aut?

anon

@SohamChowdhury heh. careful how you phrase things. the absolute galois group of Q^alg is trivial. the absolute galois group of Q, on the other hand, consists of all field automorphisms of Q^alg.

Soham Chowdhury

oops.

anon

"the absolute galois group of K" means Gal(K^alg/K) (or K^sep if you need that instead). if K is already algebraically closed then K^alg=K, so G is trivial

Soham Chowdhury

oh, okay.

The Substitute

7:15 AM

@Anon, but aren't $\phi(k)(h)$ and $\psi(k)(h)$ both equal to $khk^-1$ by the theorem?

Soham Chowdhury

khk^{-1}

The Substitute

Is there something I can download to convert your text into TeX?

anon

@TheSubstitute khk^-1 equals phi(k)(h) in the first group, and equals psi(k)(h) in the second group.

did you read what I wrote?

The Substitute

Yes.

Soham Chowdhury

yesterday, by anon

Chat guidelines | $\LaTeX$ in chat

anon

7:16 AM

you are likely confusing internal semidirect products with the semidirect product construction, or your text is confusing them, or both

why are you quoting my thing Soham?

Soham Chowdhury

to @TheSubstitute.

anon

oh, didn't see his/her question

Given two groups $H$ and $K$ and given any choice of map $\phi:K\to{\rm Aut}(H)$, one may construct the semidirect product $H\rtimes_\phi K$, in which $khk^{-1}$ is defined to be $\phi_k(h)$. Conversely, suppose we have a group $G$, with $G=HK$ with $H$ normal and $H\cap K=1$. Then there is a map $\phi:K\to{\rm Aut}(H)$ where $\phi_k$ is defined by the automorphisn $\phi_k(h)=khk^{-1}$. Note that $\phi$ is determined by $G$; we don't get to pick it here!

It then so happens that for this $G$, we have $H\rtimes_\phi K\cong G$ via $(h,k)\mapsto hk$.

The Substitute

@SohamChowdhury thank you!

Soham Chowdhury

:)

The Substitute

@anon I will review this and my text. Thank you!

@anon it makes sense. My confusion was due to some identification made by Dummit & Foote. Your remarks make much more sense.

anon

7:28 AM

@TheSubstitute here's an example you might be able to appreciate. If $\phi$ is trivial then $H\rtimes_\phi K$ is just $H\times K$, because $khk^{-1}=h$ for all $h,k$. But if $\phi$ is not trivial, then it isn't $H\times K$ because $khk^{-1}\ne h$ for some $h,k$. More concrete instance: $(C_2\times C_2)\rtimes_{\rm triv}C_3$ equals $C_2\times C_2\times C_3$ is abelian, whereas $(C_2\times C_2)\rtimes_{\rm nontriv}C_3\cong A_4$ is nonabelian.

Soham Chowdhury

algebra is fast turning more and more interesting.

anon

actually, I like $(C_2\times C_2)\rtimes_{\alpha}S_3\cong \color{Blue}{C_2\times C_2\times S_3}$ ($\alpha$ trivial) versus $(C_2\times C_2)\rtimes_{\beta}S_3\cong \color{Red}{S_4}$ ($\beta$ nontrivial) better

Soham Chowdhury

"One of the great mysteries of Galois theory is that there are certain polynomials with the property that, given a list of their roots ar, a2, . . ., an, we can find an element g that permutes these roots any way we like."

e.g.?

The Substitute

@anon I got it. That's also an example in my text. It wasn't until I reread the section I started to confuse myself. But it makes more sense now.

anon

@SohamChowdhury have you ever rationalized a denominator Soham?

Soham Chowdhury

7:33 AM

yes.

anon

what's the conjugate of, say, $a+b\sqrt{2}$? ($a,b\in\Bbb Q$)

Soham Chowdhury

$\frac{a - b \sqrt 2}{a^2 - 2b^2}$

anon

Indeed, the map $\sigma:(a+b\sqrt{2})\mapsto (a-b\sqrt{2})$ is an automorphism of $\Bbb Q(\sqrt{2})$. it restricts to a permutation of the roots $\{-\sqrt{2},\sqrt{2}\}$ of the polynomial $x^2-2$.

El'endia Starman

New blog post! Exploring and Understanding Hyperbolic Geometry Part 3: Correct Movement and Circles. Pretty pictures at the bottom!

2

anon

@SohamChowdhury no, the conjugate of $a+b\sqrt{2}$ is $a-b\sqrt{2}$

you did more than I was asking you about

Balarka Sen

7:35 AM

@SohamChowdhury that's the inverse.

Soham Chowdhury

oops, yes.

duh

Balarka Sen

Galois group of a polynomial is like symmetry group of a figure.

except that the figure is replaced by roots of a polynomial.

Soham Chowdhury

right.

Balarka Sen

it's like an "arithmetic symmetry group"

anon

Given a polynomial $f(x)\in\Bbb Q[x]$, if we adjoin all of its roots to $\Bbb Q$ to obtain the splitting field $K$, then every field automorphism in ${\rm Gal}(K/\Bbb Q)$ restricts to a permutation of the roots of $f$.

Soham Chowdhury

7:37 AM

interesting.

anon

@BalarkaSen I like to think of it as a symmetry group of an entire number system, actually. The feature its symmetries preserve is the truth of all arithmetic equations involving its elements (a necessary and sufficient condition for a self-bijection to do that is to be a unital ring automorphism).

Balarka Sen

makes sense

anon

Indeed, we have thus defined a group action ${\rm Gal}(K/\Bbb Q)\to{\rm Perm}(f^{-1}(0))$. This action is transitive, but not every permutation of the roots generally comes from a Galois symmetry however.

Balarka Sen

yep

otherwise it'd just be a symmetry group of a set. boring.

Soham Chowdhury

but, well, every non-boring group is a subgroup of the boring symmetry group of some set. :P

Balarka Sen

7:43 AM

one should think of galois group of a field extension as automorphism of the fields preserving not only the base field point-wise, but also preserving the permutations of the intermediate fields.

indeed, that's how the "big galois group" becomes inverse limit of "intermediate galois groups"

Soham Chowdhury

I once read something about how nobody seems to have made use of the fact that every group is a subgroup of a symmetric group in proving anything.

Balarka Sen

it'a a good fact, though.

anon

@BalarkaSen "preserving the permutations of the intermediate fields"?

Soham Chowdhury

> Serge Lang began his book Elliptic Curves: Diophantine Analysis by writing: "It is possible to write endlessly about elliptic curves. (This is not a threat.)"

anon

perhaps balarka means that ${\rm Gal}(K^{\rm alg}/K)$ induces automorphisms of the lattice of algebraic extensions of $K$

Soham Chowdhury

7:58 AM

it turns the subgroup lattice upside down in some way, I seem to remember from somewhere

anon

that's the fundamental theorem, which says (the lattice of subgroups of Gal) and (the lattice of intermediate extensions) are lattice anti-isomorphic

Remember me

Hello@BalarkaSen

Balarka Sen

@anon sorry, I got disconnected. for any $\sigma \in \mathsf{Gal}(K/k)$, and an intermediate $K/L/k$, there is a naturally induced aut $\sigma|_L$ of $\mathsf{Gal}(L/k)$. so an aut of the top field in your lattice of subfields induces an aut on every field in the lattice.

so the galois auts of the top field "preserves" the galois auts of the bottom fields

anon

if L/k is also normal

Balarka Sen

this is precisely why the galois group of the top field is inverse limit of galois group of the bottom fields.

yeah, sorry.

I am assuming every extension is Galois.

@SohamChowdhury $\mathsf{Gal}(K/-)$ is a contravariant functor, yes.

hi @Remember

Remember me

8:09 AM

@Balarka I had a question to ask you about solvability and finding solutions to polynomials but since you are not free I will ask later

Bye for now

Balarka Sen

you were asking if a polynomial has a real root

that's not solvability.

Remember me

No about p adics .... They help me in getting solutions ?

Balarka Sen

it has nothing to do with p-adics

Remember me

@AlexClark did say about that

Balarka Sen

yes, what he said was wrong

2 hours ago, by Balarka Sen

@AlexClark The thing you do with p-adics is called the Local-Global principle, which works for a restricted class of diohantine equations over $\Bbb Z$. $p$-adics aren't devised to tell you things about $\Bbb R$.

Remember me

8:12 AM

Oh. nvm then

Balarka Sen

should have said equations over $\Bbb Q$, but whatever

anon

same diff

Soham Chowdhury

8:32 AM

@BalarkaSen ooh

@Rem: p-adics will not help you in univ entrance exams. plain and simple. :P

Balarka Sen

galois theory will not help you in university entrace exams either, @Soham

Soham Chowdhury

yes, I know.

I'm not studying it for that, am I?

but it seems like a boatload of fun.

I can understand why you want to connect it to everything you know.

Balarka Sen

I don't want to connect it to everything I know.

it is connected to everything I know.

Soham Chowdhury

I'm talking about your analogies.

Balarka Sen

which analogies?

Soham Chowdhury

8:35 AM

all those covering space thingies.

are you reading the book?

Balarka Sen

@SohamChowdhury i rediscovered them, sure. but all of those has been done in far more greater generality by Grothendieck, as I found after talking with about it in the homotopy theory chat

Soham Chowdhury

okay

Balarka Sen

and covering space really are related to galois theory

i'm not trying to connect it. the connection is natural and obvious.

@SohamChowdhury nope.

Soham Chowdhury

ah. no etale cohomology turned you off. :P

Balarka Sen

yes, kind of

i guess i roughly know whatever's in there. i'd rather read all of them thoroughly from a textbook rather than reading them from a pop-math book

you should read it. Ash and Gross are good at sketching the general picture out of nothing.

Soham Chowdhury

8:42 AM

yeah, I can see that.

didn't you read their ell curves book?

Balarka Sen

yes, i did

Soham Chowdhury

ok, so they say that you can prove that the ell curve group law is assoc "elegantly" using AG.

any clue how?

Balarka Sen

i guess there's a theorem about cubic curves passing through 9 points in general position.

it's a generalization of the statement "given 5 points in general position, a conic can be drawn through them"

Soham Chowdhury

oh.

how do you do $P+P$ on an ell curve?

Balarka Sen

tangent

Soham Chowdhury

8:45 AM

oh ok

Balarka Sen

ok, it's called the Cayley Bacharach theorem. any two cubics passing through nine points in general position over P^2 are the same.

hmm. that doesn't seem like the right thing.

oh, I see. there's another version which says if a cubic passes through a given set of eight points, it must also pass through the ninth.

right, there you go.

@Soham an $n$-torsion point on the elliptic curve is a point $P$ such that $P + P + \cdots + P = O$, addition is $n$ times

Soham Chowdhury

yes, I read that.

Balarka Sen

the $p$-torsion points of $E(\Bbb Q)$ form the subgroup $\Bbb Z/p\Bbb Z \times \Bbb Z/p\Bbb Z$.

Soham Chowdhury

wow.

Balarka Sen

$p$ is a prime

Soham Chowdhury

8:54 AM

nvm

columbus8myhw

$\mathcal C\cap\mathbb Q$

Balarka Sen

@Soham now there's a natural action of $\mathsf{Gal}(\Bbb Q^{alg}/\Bbb Q)$ over the $P$-torsion points. if the book doesn't explain what the action is, read it up from somewhere.

Soham Chowdhury

it's good I learned a little about varieties from here.

Balarka Sen

such an action gives a hom $\mathsf{Gal}(\Bbb Q^{alg}/\Bbb Q) \to \mathsf{Aut}(\Bbb Z/p\Bbb Z \times \Bbb Z/p\Bbb Z)$

Soham Chowdhury

@BalarkaSen I will.

Balarka Sen

8:56 AM

that aut group is just $GL_2(\Bbb F_p)$ (prove this)

so we have a representation $\mathsf{Gal}(\Bbb Q^{alg}/\Bbb Q) \to GL_2(\Bbb F_p)$

this is how the 2-dimensional representation theory of Gal(Q^alg/Q) is related to elliptic curves

columbus8myhw

Hey, is there a name for a variant of the Cantor set where, instead of removing $\left(\dfrac13,\dfrac23\right)$ at the first step, you remove $\left(\dfrac13,\dfrac12\right)\cup\left(\dfrac12,\dfrac23\right)$ (and similarly for the rest of the moves)?

Like, you remove the middle third, but you leave in the middle point.

Balarka Sen

it'll just be homeomorphic to the cantor set

You sure?

yes

Pretty sure it's not

8:58 AM

why d'you think it's not?