Mathematics - 2016-10-23 (page 4 of 5)

Hi @Ted.

@TedShifrin Thank you again.

Just worrying is never productive, @Mahmoud, but I know that's easy for me to say.

G'night @MikeM.

@TedShifrin I know what you mean. There was a problem on the first quiz which was 1) identical to a homework problem, and 2) very similar to a discussion problem

6:02 PM

I should really get my ass up and to work.

and yet people still managed to foul it up

It's almost evening, @MikeM.

@Semiclassic: I'm compounding that. I told them when I did the example in class and assigned the homework to expect to see such a question on the test! rolls 6 of 8 eyes

lol

;D

There's some research computations I should work on myself

6:04 PM

Maybe they wasn't able to solve it.

@Semiclassical We're the worst.

Namely, given a Riemann surface like $(x^2+y^2-1)^2=mx+b$ and the form $\lambda=y\,dx$, consider the variation of $\lambda$ with $m,b$

@Mahmoud: As I said, I showed how to do it in class and then they had to do it for homework. They had ample opportunity to ask me questions in office hours. Nothing sneaky or surprising here.

Somehow we always talk about how we're not working.

and find a linearly combination of $m-$,$b-$derivatives of $\lambda$ that equals an exact form

6:07 PM

Same thing in my differential geometry class when I told them certain required homework problems would be appearing on the exams and final exams. Plenty of opportunity to come work on it in office hours and make sure they could ace it

@MikeMiller I won't even waste the energy to roll any eyes.

I know how to do that in some cases, by brute force

@TedShifrin I'm just stating a fact: we do.

but those are always cases with $y^2=p(x)$ and $p(x)$ having parameters

I don't know how to deal with the non-hyperelliptic case yet

You still have the standard formulas for the holomorphic differentials that we've discussed, @Semiclassical.

Danu

Working so hard right now (jk)

6:09 PM

(the above case has the additional complication that $(x^2+y^2-1)^2=mx+b$, once projectivized, corresponds to a Riemann surface with singularities)

(but I'm hoping that doesn't hurt me too much :/ )

Yeah, I should dig out the Griffiths text you had suggested

I feel crummy for not being able to work due to nonmathematical distractions. I usually invariably procrastinate when I should work, and feel crummy about it afterwards -- the order of the appearance of crummy feeling is important.

Prove that : $$(\forall n\in\mathbb N) \exists(a,b)\in\mathbb Z^2 Such that n=11a+7b$$ Why do you think about this ? I don't even know what does it represent.

For any natural numbur $n$, show that there are integers $a,b$ such that $n=11a+7b$.

Or, more succinctly: Show that every natural number can be written as an integer combination of 11 and 7.

@BalarkaSen Don't worry about it. It's part of being human.

So should I turn it into a linear equation of two variables ?

6:13 PM

One way to get the spirit of it, I suppose, is to note that if you find such an $a$ then $n-11a$ is a multiple of 7.

@Mahmoud: Start by finding $a$ and $b$ so that $11a+7b=1$.

No, this is about integers, not algebraic manipulation. If you want to cheat, look up the Euclidean algorithm.

Yeah, @ted's got the right of it. It should remain true if you replace $11,7$ by any pair of primes.

But I there is a universal quantifier there.

@Mahmoud: Figure out first why once you know how to do it for $1$, then you can do it for any integer.

@Semiclassical any pair of coprime numbers (and only pairs of coprime numbers)

6:15 PM

Right. For some reason I thought you couldn't weaken it past primes. But yeah, Bezout's identity in number theory

Hi @Tobias. Mike was asking something about representation rings of finite covers of Lie groups. You're the one to know the answer, I bet.

@MikeMiller Thanks. Trying to finish up the nonmathematical work quickly so that I can do what I want to.

@TedShifrin Hmm, I don't think I have worked much with decategorification of categories as "large" as that

LOL. Oh, OK.

Heya @Alessandro.

good evening

6:17 PM

To give a hint: Suppose you find $a,b$ such that $7a+11b=1$. Now imagine you want to find $a',b'$ such that $7a'+11b'=2$. How could I make the right-hand side of the second equation look like the right-hand side of the first?

I'd ask that last question in reverse, @Semiclassic.

I suppose you haven't heard of Bezout's lemma @Mahmoud?

@Alessandro It's not a number theory course, so probably not :/

Absolutely no @Alessandro

I have no idea how Bezout's name got on Euclid's algorithm ... except in more high-falluting settings.

Start thinking about small multiples, @Mahmoud. Explore. Experiment.

6:19 PM

You can approximate my level to Precalculus :/

@TedShifrin Was he in fact the one to originally study Bezout domains?

Yes, @Tobias, I believe so. But we're talking $\Bbb Z$, for damn sake :P

Per Wikipedia: "French mathematician Étienne Bézout (1730–1783) proved this identity for polynomials. However, this statement for integers can be found already in the work of another French mathematician, Claude Gaspard Bachet de Méziriac (1581–1638)."

I don't know @Ted, we call it Bezout's lemma in Italian too though

Dino Lorenzini is an expert on this stuff and the history. :)

6:20 PM

What identity @Semiclassical ?

right. At least it was not referred to as Bezout's theorem as it sometimes is (which is just a shame when compared to the beauty that is really Bezout's theorem)

I understand, @Semiclassic. But Euclid's name should go on the $\Bbb Z$ case.

Can't disagree with you there @TedShifrin

Bezout's Theorem on intersection numbers of algebraic varieties is something totally different. :)

It's a lemma in number theory, Mahmoud. Basically it says what we were indicating above:

6:21 PM

@TedShifrin Right, but so much more deep also.

Just let Mahmoud follow my advice. Explore. Experiment. Don't listen to all the crap in here.

If I pick any two numbers $p,q$ such that their greatest common divisor is $d$, then there exist $a,b$ such that $ap+bq=d$

Bezout theorem in algebraic geometry looked a bit like dark magic the first time I saw it

Yes, I'm very far from number theory.

I've always understood it topologically/geometrically, not algebraically @Alessandro.

6:22 PM

Me too

If you consider that for $p=7,q=11$ then the gcd is $d=1$ and so it asserts that there exists $a,b$ such that $7a+11b=1$

But you can define local intersection numbers in terms of lengths of ideals.

So now you've stated what I asked Mahmoud to explore/experiment with 15 minutes ago. Seriously.

sigh

yeah

Sometimes too much knowledge makes people forget things can be simple and not scary if you don't make them too complicated and scary sounding.

Think like a teacher.

@TedShifrin I remember Bill Graham going through a quite slick proof of it as part of a course he gave on intersection theory while I was at UGA, reducing it to the statement that any two distinct lines intersect at precisely one point.

6:24 PM

Right, degeneration arguments, @Tobias.

Sometimes tough to prove you have a flat family, though :P

But that's the classic 19th century Italian algebraic geometry argument.

You know Bill is now head, @Tobias? :)

@TedShifrin I did not

Ok I'll explore, I can't just take the theorem and have to explain where did I get it from :P

@Mahmoud What we're rambling on about goes to the fact that the only thing special about the pair 7 and 11 is that they have no common factors.

No, seriously, ignore them and play around with small numbers for $a$ and $b$, @Mahmoud. I'm not going to say it a fifth time.

I didn't talk that much to him apart from taking that course, but he seemed like a really nice guy.

6:25 PM

so yeah, just play around with them and see what comes out.

@TedShifrin The algebraic proof of Bezout for curves inside P^2 at least is not too distinct from the cohomology proof, actually. One can "algebraically" homologue a curve of degree $k$ to a monomial to the power of $k$, in which case the intersection being product of degrees is obvious.

He is, @Tobias, and a very gifted mathematician.

Upto multiplicity.

@Balarka: The difficulty comes with singularities, but sure.

Okay then.

6:26 PM

Believe it or not, @Tobias, I was originally one of the people pushing to hire him. :)

@TedShifrin Good call :)

Yeah, I am working with smooth curves.

But it shouldn't matter for homology, @Balarka.

It's not too hard to write down a homology from a singular curve to a smooth curve algebraically either however.

Anyhow, we're far, far afield. I need to eat lunch and go play bridge for the afternoon.

@Tobias: I'm not totally without taste and judgment, despite what your mentor thinks of me :P

6:28 PM

@TedShifrin You mean Dan? (my mentor is Mazorchuk as I see it)

Yeah, I meant your UGA mentor, sorry.

Buon appetito! @Ted (actually what do you say in English before eating? There's guten Appetit in German and something similar in French but I don't know of an english expression)

I don't think I have ever heard him talk about you actually.

Just as well, @Tobias. He slammed me in some unprofessional ways.

Interesting, @Alessandro. I don't know what Americans say. I say "Bon appétit."

Happy bridge.

6:30 PM

Better finish your work before it's too late, @Balarka. I won't be here to remind you it's past your bedtime :D

I'm not making any progress :(

Yup, almost done. Have to wake up 7 in the morning tomorrow.

Remember that negative integers are allowed, @Mahmoud. You're trying to almost balance out and get $1$.

Yes, there isn't so good arithmetic going on here :P

Just concentrate, @Mahmoud. You don't need big numbers.

Part of being a mathematician is not being afraid to play around ... and NOT give up.

15

6:33 PM

$a=7$ and $b=-11$ ?

That gives you 0?

I said small numbers.

Oh crap

small a,b

I'm not going to give up this time >:) Just bear with me please.

This isn't absurd like that other question, @Mahmoud.

@Balarka: Did you just star that?

6:35 PM

Nope

LOL, ok. :)

$a=2$ and $b=-3$ ?

But I agree with you

Yippee.

:D

6:35 PM

See, @Mahmoud. Just jump in and play.

Now do the general $n$.

nah, that was me. I liked it.

Using Induction ?

No.

You don't need anything that fancy.

to keep with the theme of exploration, try doing it for $n=2$.

Back to Semiclassic's question. Knowing the equation $1=2\cdot 11 - 3\cdot 7$, how do you write $2=?$. Yup.

6:36 PM

Okay sounds good

Multiply by two ?

So you get ... ?

Woop woop.

Heya @Krijn

hi

Hi @Ted and @ Bala

6:39 PM

$$n=22n-21n$$ Doesn't make much sense. :I

Better to write it as $n=(2n)\cdot 11 + (-3n)\cdot 7$.

Sometimes groupings change meanings and understandings.

$a=2n$ and $b=-3n$ :D

Got it?

$\forall n\in\mathbb N$

$\forall n\in\Bbb Z$, in fact.

OK, I'm heading out. You all have a great day/evening.

(g'night @Balarka)

6:41 PM

One question @Ted

Night

Yes, @Krijn?

G'night @Ted!

$\exists (a=2n,b=-3n)$

Would you as an author of math books like to receive a list of small typo's of one of your books? @Ted

6:42 PM

@Krijn: If they're not already on the list(s) that are posted on my webpage, absitively.

Because the book I am reading is fuuuuuuull of them

I'll see if Rosen has one of those then

Oh, you didn't mean me. Damn you.

Most authors keep a webpage for that purpose.

:D

My algebra book has a bunch because the asshole publishers never did a second edition.

Is that the proof @TedShifrin ?

Studentmath

6:43 PM

I just realised what a mistake it was to take this PDE course. Not that I had a choice for the M.Sc, but I just know absolutely nothing..

You're done, @Mahmoud. I would say for every $n\in\Bbb Z$, let $a= ...$ and let $b=...$. I'm not one for formal logic symbols.

how do I find $\tan(\pi/8)$ without double/half angle formula or complex numbers?

Okay that gave me a bit of confidence

Using the double-angle formula, @Huy.

without

6:44 PM

I doubt you can, @Huy

F*** that, @Huy.

lol

yes, I doubt it too

one of my former students has an exercise where $\tan(\pi/8)$ comes up

Oh, maybe using my favorite angle-bisector theorem from geometry? I doubt it works, but ...

but they've neither seen the angle formulae or complex numbers

6:45 PM

draw an octagon and do something fancy?

Do they know the law of sines/cosines, @Huy?

I know them !

Oh, don't need it. Just need my favorite angle-bisector theorem.

yes, but I doubt the idea of the exercise (it's an integral to be solved with substitution, since that's their current topic) is to derive some more identities

which is that?

@TedShifrin What's your favorite theorem?

6:46 PM

that doesn't ring a bell

One thing worth noting for a test scenario: You could just as easily see the same problem but with something other than 7,11

An angle bisector divides the opposite side in a the same proportion as the adjacent sides.

I love the proof(s).

any pair of numbers that lack a common divisor would work

Ah, yes

That's all you need here, @Huy, plus simple algebra.

6:46 PM

ah

victory

:D

But any pair of numbers with a common divisor wouldn't work

$$c^2=a^2+b^2-2ab*cos(\theta)$$

So you can't use 9 and 12 to write every integer.

6:47 PM

You get it, @Huy? Oh, and hi, btw.

good, @Mahmoud. :) But we don't need it this time.

And that shouldn't entirely shock you. What's a number that divides any integer combination of 9 and 12?

hi, too. I need to get a paper and draw and think, but I'll figure it out

Cool. I like it, @Huy. Make sure your former student knows how to prove that theorem, too :) Beautiful geometric proof.

OK, bubye, all.

I wanted to test my knowldge

G'bye @TedShifrin

But one more Q

ask us, @ted's gone :)

6:49 PM

Are you sure he's 63 ?

I mean he is very cool

The two statements you made are not mutually exclusive.

@Semiclassical

he's old and awesome

but, back to what I was saying before

try to figure out why the same statement you gave initially would be false if you replaced 7,11 by 8,12 @Mahmoud

Maks

Hey guys, can someone give me a hand with homework ?? The topic is vector spaces & subspaces

(i switched 9 to 8 because why not)

6:53 PM

Is it because 8/12 is not a supermarket?

@BalarkaSen: what u up to, sorry I've been pretty busy :(

@krijin I'm not Ted, so I probably can't smack you

doesn't mean I don't want to, though

@Semi But you can spell my name wrong, which is sort of the same, really

true.

@Huy still doing bits of topology and bits of geometry :) but right now doing physics

6:54 PM

What physics are you on now?

U WAT M8

no, what kind of physics?

hydrostatics. easy stuff

gotcha

Still no number theory and algebra :(

yeah, that's pretty simple

6:55 PM

yup

abenthy

Topology question: Is there a smooth manifold $M$ which does not admit any non-trivial periodic self-diffeomorphism, but a non-trivial periodic self-homeomorphism?

I'm gonna cook some food brb

@Krijn I'll promise to read some if you finish watching the film I recommended :P

hydrostatics would basically just be Bernoulli's equation and the principle of continuity?

Or would it just be Archimedes' principle?

Easier than that. Archimedes, yes.

6:58 PM

mmkay

Here's a fun Archimedes' principle problem

Suppose I have a glass of water. If I put an ice cube in it, the water level goes up.

If I now wait for it to melt, what happens to the water level---does it go up, down, or stay the same?

Guys what do you think of 3Blue1Brown ?

I'll warn you that I haven't yet encountered Archimedes though: just started. We'll be introduced tomorrow.

Okay

Use that as a practice problem, then, when you get a chance

interesting question.

ice weighs less than water, yup?

right. so it's less dense.

also, you can assume that the melted ice has the same density as the rest of the water

7:01 PM

3Blue1Brown ? youtube.com/channel/UCYO_jab_esuFRV4b17AJtAw

mhm

Please ?

(you could have a version of the problem where that's not true, e.g. a freshwater iceberg melting into a salty ocean)

I've heard his linear algebra series is well done, but I haven't seen them

@Semiclassical I'll keep that in mind when I learn Archimedes. Thanks.

7:05 PM

np

brb getting some ice cubes

Speaking of, I confused myself for a while on the force exerted by a cone filled with water on the ground

@Huy :P

that's how experimental physics works!

not string theory tho

or should I say, conformal field theory on an (infty, n)-topoi, as nlab would say

lololol

7:08 PM

Why is the general theory of relativity so easy to learn about ?

@Semiclassical It turns out the force exerted by the cone towards the bottom is the same as the case with cones replaced by a cube of the same height and base.

hmm, nice

that's really counterintuitive at a glance because the total masses are clearly different

@Mahmoud It's not really that easy to learn about.

You need a pretty solid amount of math to actually learn it.

so the weight is more for the cube version, of course.

7:10 PM

I mean it is in course of Mechanics, which isn't very far

I'd be skeptical if it's covered in any depth in such a course.

@BalarkaSen You should look at their motivation for K-theory.

this page?

How do you name links ?

just like you name kids. give them one.

7:13 PM

Command ?

press "help" bottom right corner

(if you want help)

Click Me !

Done.

Thanks @Huy

Q: Necessary and Sufficient Condition for cyclic $H$ to be a homomorphic image of cyclic $G$

no problem my dear friend

Jessy Cat

1

Let $G$ and $H$ be cyclic groups. I need to find and prove a necessary and sufficient condition in order for $H$ to be homomorphic to $G$. By the Fundamental Theorem on Homomorphisms, the possible homomorphic images of $G$ are isomorphic to $G/N$, where $N$ is a normal subgroup of $G$. So, is a...

@MikeMiller I searched the transcript; get what page you're referring to.

Jessy Cat

7:20 PM

I am still REALLY confused about this. I have a bounty for 100 points up and the only answer I've gotten kind of sucks.

money doesn't guarantee quality

Their main motivation seems to be "we cannot deloop, so we categorify it, make it an infinity-stack and take it's geometric realization"

arctic tern

@JessyCat what do you think the answer is?

actually, Berci gave a complete answer to your question

that's nice, only that nlab's sense of motivation seems to $\infty$-differ from mine.

7:39 PM

Back.

Unfortunately when I need the abstract stuff the nLab rarely actually suffices.

Vrouvrou

Hello, for a set $E=\{a,b,c\}$ i can made 29 topology on $E$ , but in found only one which is Hausdorff that is $(E,\mathcal{P}(E))$ can i find other Hausdorff space ?

If a finite topological space is T1 (in particular, if it is Hausdorff) then it must, in fact, be discrete. This is because the complement of a point is a finite union of closed points and therefore closed. It follows that each point must be open.

Vrouvrou

so the only Hausdorff space is the discrete space

right ?

@Krijn