Anyone have an opinion on this?

@BalarkaSen

what do you think?

Not sure if I have any comments about publishing.

@PedroTamaroff you have published, yes?

@ForeverMozart No, I haven't.

Have you talked to your advisor?

arxiv?

No.

my paper will appear on ArXiv Tuesday

Congratulations.

thanks, but anyone can post there

it could be a bunch of nonsense :|

It is moderated, you cannot upload rubbish.

ok, but mathematical rubbish may be hard to detect

just messing around, my paper is correct

You're panicking. Drink some water and relax.

you want the link when it is available?

Sure

Tuesday will be very exciting !!!

Indeed.

so are you still working on manifolds?

Yup.

Getting a bit distracted lately because of school.

@BalarkaSen How old are you now?

i will guess

hmm 22?

maybe 24-25

I think I am a few years older

No. Balarka is in high school.

oh, so 27-18

@PedroTamaroff 16 I think.

wow very advanced

You should be sure about your age. =)

in USA the type of math you study is very unusual in high school

Hi

@PedroTamaroff It was an attempt at a weird joke. But yes, I am 16.

Hi @Akiva

How goes camp?

Good

one day Balarka will be famous

he already knows more mathematics than most undergraduates

When I was here two years ago, camp was amazing. But I seem to have enjoyed it less last year and this year.

Yes, I plan to get a Nobel prize.

Oh wait.

@BalarkaSen It seems you're digging down the geometry hole. Are you learning some algebra, some analysis, some good ol' counting?

@BalarkaSen (Took me a minute)

@AkivaWeinberger Ah, well, you have progressed a lot during these years in my opinion, that could be a reason.

What are the hot areas of math research today?

@PedroTamaroff Hah, indeed. I spent a good few months on multivariable analysis (calculus?), and studying some some complex analysis for now. Things are going too slow though :(

I have no idea, most of what I do is "outdated"

@BalarkaSen You have plenty of time, don't be ridiculous.

For counting, not really systematically. Haven't studied algebra for a long time.

Rushing will not help you here.

hopefully he has a mentor at school

I recently set my phone to Spanish

Trying to learn it

@PedroTamaroff I agree, but I feel like I cannot give enough time into studying these. Hopefully I'll manage.

@Forever Not at school, but I do talk to a few people.

nobody has an answer to my question???

@ForeverMozart I don't know what your question is.

Hot areas of research? 4 dimensional smooth Poincare conjecture.

What are the hot areas of math research today?

so you just replace sphere with 4-dim sphere in the conjecture?

Apparently there's something called a "cluster algebra" that people are studying

@ForeverMozart I think the "smooth" bit is probably important also

Well, you need to be careful. Simply connected is not enough to specify a 4-dimensional manifold.

oh

I guess instead of curves you need surfaces

@BalarkaSen Up to homeomorphism or diffeomorpism?

The correct version is, if a smooth 4-manifold is homotopy equivalent to S^4, it's diffeomorphic to S^4.

@AkivaWeinberger Neither. $S^2 \times S^2$ is simply connected, as well as $\Bbb{CP}^2$.

Ohh.

lots of crazy things can happen when you go up 1 dimension

I mean, not homeo implies not diffeo so whatever.

it is probably false

Hey guys, I have a quick terminology question. Say you have a finite-dim'l vector space $V$ and an element $t$ of the tensor algebra $T(V)$

Se puede hablar espanol aqui?

Do you say that $t$ is a "tensor in $V$", a "tensor on $V$", or something else?

@ForeverMozart It's true if you replace "diffeomorphic" with "homeomorphic" though!

I thought the Poincaré conjecture was "If simply connected and has homology of sphere then is homeo to sphere"

That's Freedman's theorem.

plus or minus grammar

oh, was that after Perelman?

or before?

Perelman is 3 dimension.

yes but he did diffeo for 3

@AkivaWeinberger Simply connected homology sphere does imply homotopy eq to sphere and vice versa

Those are equivalent.

@LeakyNun Si. Pero no esperes que te entiendan.

@PedroTamaroff perfecto

@LeakyNun Se puede, pero pienso que estoy el único que te entenderé (y solamente un poco) EDIT Never mind

@ForeverMozart Right.

see what I mean, Balarka knows a lot more than most undergrads

@AkivaWeinberger eh, ve arriba

@AkivaWeinberger mdr

he is a prodigy of mathstack

@PedroTamaroff tienes un apellido aleman :o

@LeakyNun ?

@ForeverMozart I need to fill in stuff I don't know and which a generic undergrad does...

@AkivaWeinberger ignore it

@BalarkaSen Have you heard of the book All the Mathematics You Missed: But Need to Know for Graduate School?

(which I started reading partway)

Oh, there is such a book?

Whoa, I gotta read this, hah.

I keep watching this video youtube.com/watch?v=g3dkRsTqdDA

very good

@BalarkaSen Table of contents: assets.cambridge.org/97805217/92851/toc/9780521792851_toc.pdf

That's pretty cool. I should keep it just to remember what I need to fill in.

most of the people in my citations are dead

@AkivaWeinberger What's with those colons there?

@PedroTamaroff Separates the title and subtitle of the book

Ah, Stokes' theorem reminds me that I need to think about something.

I copied it off of Amazon

@AkivaWeinberger Reading the index, it looks like a standard undergraduate programme, at least here in Argentina.

Except some topics, which look very specific.

The title is misleading at least. I don't think people miss such topics.

So, here's a problem: What integers can be written as ratios of two Fibonacci numbers?

@AkivaWeinberger I believe every

Clearly, every Fibonacci number is itself over 1, for example.

But 6, 9, and 10, for example, seem not to be on that list.

7 is 21/3, 11 is 55/5

what integers

@AkivaWeinberger I remember somewhere that semiprimes cannot be written

but my memory is a mess

17 is 34/2, 18 is 144/8

@AkivaWeinberger never mind

What's a semiprime, again?

@AkivaWeinberger product of two primes

Ah

@LeakyNun 21=21/1

@AkivaWeinberger I said never mind

@AkivaWeinberger for 6: nothing in the first 1000 Fibonacci numbers

@AkivaWeinberger same for 9

@AkivaWeinberger same for 10

same for the first 2000

same for the first 10000

@LeakyNun Well, you only need to check finitely many, since ratios of Fibonacci numbers converge to powers of the golden ratio

I believe that's way more than is needed

^

@AkivaWeinberger then i conclude that they don't exist

and that for 6 you only need to check the first four or so

sorry no numbers for me

except infinite numbers

@AkivaWeinberger I guess you can use the theorem that a new prime is generated for each Fibonacci number beyond i-dont-remember

Also, $L_n=F_{2n}/F_n$, so all Lucas numbers

but 17=34/2 is neither Fibonacci nor Lucas

@AkivaWeinberger en.wikipedia.org/wiki/…

> Fibonacci numbers that have a prime index p do not share any common divisors greater than 1 with the preceding Fibonacci numbers

@PedroTamaroff How do I prove the fundamental theorem of algebra using max mod principle? Say $f$ is a polynom, I look at $1/f$ - that's hol. if $f$ has no zeroes. Look at a disk around $0$ - maximum of $|1/f|$ lies on the boundary of that disk. Take bigger and bigger disk, so you get a sequence of strictly decreasing local maximums $\{z_i\}$ of $1/f$, hence minumums of $|f|$, tending to infinity. This means $|f|$ is not proper. I think that cannot happen, can it?

@AkivaWeinberger Are you with me?

I guess it would be everything of the form $F_{mn}/F_n$

Sorry, what I meant is, values of $|1/f|$ at $\{z_n\}$ are strictly increasing.

Aka, values of $|f|$ at $\{z_n\}$ are strictly decreasing.

Aka, not continuous at infinity. Aka, not proper.

I mean, $z \mapsto |z|$ is proper. A polynomial as a map $\Bbb C \to \Bbb C$ is proper. Composition is obviously proper, so that seems right.

Can you rewrite all that in the form of a proof?

To get intuition about properness: what can you say about an holomorphic proper function $f:\mathbb C \longrightarrow \mathbb C$?

@PedroTamaroff Not sure what you mean by 'what can you say'. I think of proper as being extendable to one point compactifications.

So it extends to a holomorphic function from $\Bbb P^1$ to $\Bbb P^1$, I guess.

Properness is a topological property.

A map is proper if the preimage of compact sets is compact.

Sure, preimage of compact set is compact.

What can you say about an entire proper function?

Well, it has to be finite-to-one, for one.

Because preimage of a point is a discrete set since so is a zero set of any hol. function, which is moreover compact.

Aha.

That's one thing.

So $\exp,\sin,\cos$ are not proper.

Agreed.

Polynomials so far are passing the properness test.

What else?

Hmm.

You are sitting inside a metric space.

Properness can be stated in a different way.

You're telling me to look at the continuous-near-infinity definition?

Yes.

It is that.

So $f$ is a function such that $|z|\to\infty$ implies $|f(z)|\to\infty$.

Mhm.

In particular, polynomials are proper entire.

I agree.

Can you show those are all of them? =)

Oh. That'd be interesting. I'll think.

Dang it.

(Because of the last remark.)

Can you do anything with that?

Um. I have an idea but not sure if it's going to work.

I'll think a bit more.

@BalarkaSen It is not that trivial. I just looked up some proofs and they use some little theorems.

Ah, ok.

I think I knew a simple proof, but cannot remember it now.

I was thinking of doing that bigger and bigger circles argument. $z_k$ has a local max at each $k$-th boundary circle, and $|f(z_k)|$ is an increasing sequence escaping to infinity. If one can show that this is dominated by $R_k^n$ for large enough $n$ ($R_k$ = radius of the k-th circle), one would be done as then the coefficients of the power series would eventually vanish. I am not sure how to do that.

I cannot recall any result on how fast holomorphic functions are allowed to grow.

Well, as fast as you wish.

By the way, returning to your proof of FTA.

Do you know say Rouché?

That does it, because a polynomial is dominated by its principal term in big circles.

That's more or less equivalent to the homotopy proof :)

That is the homotopy proof.

Rouché is a homotopical fact.

Sure, agreed.

Another proof is indeed what you're suggesting, but you need not take an infinite sequence, I think.

You mean the Liouville proof?

No, no.

You can show your polynomial has no roots outside a big circle.

Sure, the zero set is discrete.

Ugh, cannot recall the details. Spivak has this proof in his book, and it is really nice.

Let me try to recall.

Hmm, I'll have look.

OK.

I mean his Calculus book.

Ah, Balarka.

It is essentially a use of the max-min principle.

But you can do it manually on polynomials.

Max min principle as in you always have a max/min on a compact set?

Well, yes and no.

You use that you have a minimum, and proceed to show this minimum must be a root.

To do so, you essentially show that if your minimum is not zero, you can obtain a smaller value.

Ohh.

There is some big $R$ where $|p(z)|$ is positive.

In fact say, bigger than some constant $C$.

For example, look at $z=1$ and take $C$ to depend on the coefficients of $p$ in a clever way.

Now you are in a compact ball $B(0;R)$ where $|p(z)|$ attains its global minimum.

Call it $\xi$.

OK.

The claim is that $|p(\xi)|=0$.

Yes. I am worried about $\xi$ being on the boundary.

Why?

if one can show it's not on the boundary, we're done by minimum modulus, is what I am thinking.

Well, this argument will work with $z^n$, so you shouldn't go that way.

Hmm, alright. Go on, then.

So, assume that $|p(\xi)|=c>0$. We want to show there is some other $\xi_0$ such that $|p(\xi_0)|< c$.

Yup.

Now I think you need to do some trickery with roots, and this is essential: you are using, again, that you know what the roots of $z^n=c$ are (and every polynomial looks like such a map, locally).

@BalarkaSen

Um, the strategy there is not clear to me.

It is not clear at all, no.

At this point I have to look at Spivak, however.

So, by translating we can assume the minimum occurs at $0$.

Say your polynomial is $g(z) = a+b z^m +\cdots + a_n z^n$.

So that $g(0)=a$ is nonzero.

Uh-huh.

Find $c$ so that $c^m = -a/b$, and let $c_k= a_k c^{k}$.

(Here we use we can solve for roots.)

What's $d_k$? Gotcha.

Fixed.

I mean, just compute.

Sure, ok.

Now, look at the last expression in parenthesis.

You can always find, by continuity (that's a polynomial that is zero at zero), a small enough nonzero $z$ so that the modulus of such expression is at most $1$.

In fact, you can always rotate $z$ so that $z$ is real and positive.

The last expression as in $z^m(...)$?

The parenthesis, no $z^m$.

In such case, $|z^m (c_{k+1}a^{-1}z+\cdots)|<|z^m|=z^m$.

Yes, I used that $z^m$ before to refer to the parenthesis (too many of them).

Alright, why not.

Yes, it appears both times.

Now, you can also assume you chose $0<z<1$.

Yeah.

In such a case, the triangle inequality on $\left | a(1-z^m+z^m(c_{k+1}a^{-1} z+\cdots))\right|$ gives $\leqslant |a||(1-z^m|+z^m (\text{something less than 1}))$

And because $0<z<1$ you get $|1-z^m|=1-z^m$, so all in all you get $<|a|$.

Mhm.

Odd proof.

But this is absurd, because it gives $|g(cz)|<|a|$.

Help! I could talk in stack overflow chat yesterday but today it says that I need 20 reutation.

It's not at all clear to me what goes behind the inequality manipulations.

@BalarkaSen What do you mean?

I mean to say, I follow the strategy of coming up with a new $p_0$ such that $|g(p_0)|$ is less that the minumum per se, but I probably won't ever come up with the proof of existence of such a $p_0$.

1 more question- Yesterday in my math book there was a question saying- a pole which is 6m ling casts a shadow of 4m and at the same time , a building casts a shadow of 28m. Find the height of the building

we had to find it using similarity of triangles

@BalarkaSen I guess the strategy to "clear up" $a$ and $b$ is clear, though.

so my teacher said that the upper angles of the triangles were same due to "sun's elivation"

how does sun's elevation make the angles equal?

like it forms 2 right triangles

Because the sun is far far far far away.

but how does the sun make the 2 angles equal?

@Adi The sun doesn't make anything.

the upper angles of the triangles

did u inderstand the question in my math book?

Did you get to draw your problem?

Can you make a drawing to exemplify what the situation is here?

how do i post the drawing here?

i have made a drawing on paper

Describe it.

What does it look like.

there are 2 right triangles. The base of both of them are the shadows. The heights are the pole and the building. Now my teacher says the the upper angles of both the triangles will be equal.

You have a triangle I suppose. Label its vertices A,B,C.

@Adi Are the triangles one inside the other?

They have to share sides.

the upper vetice is A. then B is the right angle vertice

no

My teacher says we can prove them similar by

the A.A. property

by taking the right angles in both of them and the angles A and angle B

Do you agree that is your scenario?

more like this vt-s3-files.s3.amazonaws.com/uploads/problem_question_image/…

Where are the other numbers coming from...?

@Adi The ray from the sun cannot be two different things.

lets take the upper vertice of the bigger triangle as A

ignore the other numbers, coulsnt find a better image

It has to be one, single ray, coming from one, single light source, aka, the sun.

What your teacher is probably wanting you to do is use that in that picture, $6/4=x/28$.

lets take the vertice with the right angle as B

yes thats what she want me to do

That is, if two triangles are similar (as in the picture) then the ratios of their sides are equal.

but for that we have to prove them to be simlar

exactly

The point is that you don't have to prove it here.

You are supposed to understand how to "set up the scenario".

so she said that we can prove em by the A.A. property

Oh, you're asking why the two angles of the topmost vertex of both the triangles are same?

That's because the pole and the building are parallel, @Adi.

I don't know what that is. But you have two triangles with two equal angles.

yes

Do you understand why I drew that picture, though?

but how does the pole and building being parallel make the upper angles equal?

@PedroTamaroff which picture?

The picture I uploaded here.

no

@Adi If two lines are parallel, then any line intersecting them forms equal angles on the same side.

oh! its a tranversal

If that's what it is called, yes.

We have to start calling the Greeks here.

They will get offended. =)

:P

corrsponding angles

:P

yea, they are called transversals

and its corrsponding angles

ok i got it ty

now i need to go. Jusr curious.. how old are you guys?

81.

are you no kidding?

No, completely serious.

are you rly 81?

nice

so should i call you grandpa?

xd

ok i need to go now

im 14

bye!!

@Adi He's not 81

@BalarkaSen: thanks, that makes more sense, yet I'm confused why Pete added a comment just to point out (apparently wrongly) that it is a Riemann surface

What is the comment we are referring to?

@BalarkaSen: see here

Hmm. I don't think H^2/SL2(Z) is biholomorphic to C.

Odd.

@BalarkaSen: on the other hand, how would I express H^2/SL2(Z) as a genus g surface with n punctures and b boundary components in a way compatible that it should be hyperbolic?

actually idk after some pondering...how can the triangles be smilar?\

Well it's topologically S^2 minus puncture. Are you asking me what's the explicit metric on it which makes it isometric to H^2/SL2(Z)? I have no earthly idea, @Huy.

@BalarkaSen: topologically, yes, but shouldn't it be a hyperbolic surface if anything?

isometrically

brb

Hyperbolic surface means a surface with a hyperbolic metric. You already know what the surface is: S^2 - point. You need to specify the hyperbolic metric on it which makes it H^2/SL2(Z). That I do not know.

ok

so what are you up to @BalarkaSen

I am going to fix my sleep schedule today.

11 in the morning, fully awake. Not sure how far I can manage.

I can probably push through 6 PM, maybe 7.

Any suggestions?

ok, just don't give up if you feel terrible after a day or two

it usually takes several days for the body to adjust

and don't give in to the desire to sleep in the afternoon

@Huy hmm. i hope not, because i have to do a bunch of things tomorrow.

also, if you end up waking up in the middle of the night and feel awake, just stay in bed and try to get back to sleep until say 6/7 am

@BalarkaSen: you should have started earlier then :P

@BalarkaSen: but of course there's no general rule, some people can adapt quicker, some slower

I have been sleeping 5PM - 10PM for like 3 days straight.

yeah, that's not right

try something like midnight to 5am at first, if you can only sleep for so long

something like 11pm to 6am would be perfect, imo

Yeah, will try that.

I'm usually doing something like midnight/1am to 7-8am, because I never have to get up early

Me neither, really. But my current sleep schedule is more dangerous because I have to go somewhere at about 2PM tomorrow, come back home and go somewhere else at about 5PM.

I'll stay there for probably till 8PM and have to get some studying done.

There's no way out than to fix it immediately.

indeed

I just uploaded 501 images to Google photos with the intention of making an album with them. but only 500 images can be added to an album at a time. so I just added 500, but now I have forgotten which was the last image.

uh oh

sometimes I think those "limits" change specifically to piss me off

hehe

that's the purpose of it

@BalarkaSen: do you drink coffee?

no, only tea

black tea?

Sorry, got disconnected. Yes.