I hate $a$ and $b$ unless they're fixed constants.

Okay so quick question in Introduction to Smooth manifolds the following is said: "In any smooth local coordinates $(x^i)$ on a smooth manifold $M$, a Riemannian metric $g$ can be written as $$g = g_{ij} dx^i \otimes dx^j$$" But now $dx^i$ and $dx^j$ are covector fields, how do we take their tensor product? (Maybe $dx^i \otimes dx^j$ is just notation for a function that defines the pointwise tensor product for every $p$ in a local chart?)

ditto

i dont udnerstand why we can't stick to the x-y plane

I don't see where you got that.

my running hypothesis is so that its more difficult to google the answers

hi @loch

but that won't stop the half-intelligent googler..

I'm trying to answer 3 different things, DogAteMy, so I can't think.

hi @LeakyNun

@Perturb: You do tensor products of things pointwise, yes.

$\partial r/\partial b=0$ is what I meant

@TedShifrin parceque un intervalle coupe Q dans les points de Q

So $\partial(u^2+v^2)/\partial b=0$

That's a horse of a totally different color, DogAteMy.

@Poline: Et il y en a beaucoup!!!

@Perturbative so, there are 66 questions. Each test they decide what score having n correct answers corresponds to

Demonark: Did you finish most of it?

@TedShifrin oui tout Q y passe

$uu_b+vv_b=0$, and C–R says $u_b=-v_a$ and $u_a=v_b$

Ahh I see @Dami

imgur.com/a/0jcE07d

work^

@Poline: Si on prend un petit intervalle? Pas tout!

@TedShifrin If I wanted to think of $dx^i \otimes dx^j$ as a function between sets, what's the usual way one does so? Becuase there's two ways one can think of $dx^i$ as a function between sets

@AkivaWeinberger yes

I think I blind guessed on something like 10 questions, less than 15 for sure

It's a bilinear map on $\Bbb R^n$, @Perturb. Mapping to $\Bbb R$.

now replace a bunch of terms in the first equation from that conclusion

What's the derivative of $u/v$? @JoeShmo

I know $b$ doesn't change the modulus so I want to see if $a$ doesn't change the *angle

Mais tout de même, il n'y a rien de discrète là, @Poliine.

(which wouldn't change $u/v$)

derivative of $u/v$? im not sure

A few that I was kinda shaky on, maybe between 2 or 3. The timing was tricky to work with, and the geometric/graphical questions in particular boned me

Quotient rule, @JoeShmo.

Like what sort of geometric/graphical, Demonark. Nothing precise needed.

so then the derivative w.r.t to x?

@JoeShmo $(vu_a-uv_a)/v^2$ I think

Yeah

@TedShifrin ya toujours une infinite de nombres rationnelles entre deux reel

Subscript meaning derivative with respect to that

Evidemment ... alors, pas de topologie discrète.

right, i wasn't sure what you were asking

@TedShifrin donc c'est quoi la topologie induite?

C'est ça. C'est la topologie de la métrique. Il n'y a rien d'autre à dire.

Some Euclidean geometry type things, finding areas

OK yeah I know how to finish this

wow, like on the SAT type stuff, Demonark?

The question means $b$ doesn't change the radius of $f(z)$ (call it $r$).

We conjecture that, if $b$ doesn't affect $r$, then $a$ doesn't affect $\theta$.

ok, proof?

@TedShifrin quand on est dans un espace topologique général pas metrique comment s'appelle la topologie induite

Yeah they kinda drew some lines on a square and said to find the area of some shaded figure

Thus, we need to show that if $\dfrac{\partial(u^2+v^2)}{\partial b}=0$, then $\dfrac{\partial(v/u)}{\partial a}=0$.

@Poline: c'est ça. C'est le nom.

Some problems where you had graphs to interpret

Writing subscripts instead of $\partial$ to save time, that's, $(u^2+v^2)_b=0$ should imply $(v/u)_a=0$.

The former is $uu_b+vv_b=0$. The latter is $\dfrac{uv_a-vu_a}{u^2}=0$.

@loch so the idempotents in a ring A correspond to the clopen subsets of the compact space Spec(A)

Er, $u/v$, $~v/u$, same difference

C–R says that $u_a=v_b$ and $u_b=-v_a$.

@Ted I can't see why $dx^i \otimes dx^j$ would be defined on $\mathbb{R}^n \times \mathbb{R}^n$. The reason I say that is if I choose some local chart $(U, \varphi)$ in the atlas of $M$ where $\varphi$ has component functions $(x^i)$, then $dx^i : TU \to \mathbb{R}$ (or I could view it as $dx^i : U \to T^*U$) where $TU$ is the tangent bundle on $U$. So I would say that $dx^i \otimes dx^j$ must be defined at least on (a cartesian product) of $TU$ or $U$ depending on which way you view $dx^i$

Making both substitutions, we get $-uv_a+vu_a=0$, and thus our conjecture is proven.

(Since that's $-(uv_a-vu_a)$; compare the numerator of the thing we wanted to show is zero)

(If I did $u/v$ instead of $v/u$ I wouldn't have gotten that minus sign, but whatever)

so what did we prove again? (how is it useful?)

The usual use of the notation, @Perturb, is to pull everything back by a chart so that we're working in an open set of $\Bbb R^n$ once you write this.

Oh, wait, crap, the final step isn't working like I thought it would

So points in the manifold correspond to points in $\Bbb R^n$ and tangent vectors to the manifold correspond to tangent vectors in $\Bbb R^n$.

Let me think

@JoeShmo We showed that $r$ is entirely a function of $a$ and that $\theta$ is entirely a function of $b$

Or, equivalently, $u^2+v^2$ is entirely a function of $a$, and $u/v$ (or $v/u$) is entirely a function of $b$

@TedShifrin par exemple la topologie induite sur N est la topologie discrète, et pour la metrique induite?

ok

Oui, c'est vrai, @Poline.

Okay I sort of see what you're saying @TedShifrin

I was hoping we could then take the derivative of the former with respect to $a$ and the derivative of the latter with respect to $b$, and find some relation between them that limits what they can be

but that doesn't look like it works

The notation gets confusing, because the pullback gets suppressed, @Perturb. When we write $\partial f/\partial x^i$ we really mean $\partial (f\circ x^{-1})/\partial x^i$.

Where in the second expression I'm using standard coordinates in $\Bbb R^n$.

@TedShifrin et la distance induite est la discrète?

Il faut vérifier ça, Poline.

Il faut d'abord une définition.

I mean, $(u^2+v^2)_a=u^2(v/u)_b$, as you can check, but I don't see how that helps

je n'ai pas de definition

Alors il faut ne pas employer ce terme-là.

mais j'ai trouvé ca dans un exercise ya pas de definition dans le livre

Il faut demander au prof, alors.

AHA @JoeShmo

OK so I just wrote $(u^2+v^2)_a=u^2\cdot(v/u)_b$, and also we know that $u^2+v^2$ is purely a function of $a$ and that $v/u$ is purely a function of $b$

Divide both sides by $u^2+v^2$

How can a function of $a$ equal a function of $b$?

They must both be ...

Not that I'm paying much attention.

@TedShifrin Notice the $u^2$ on the right though

That messes us up, since $u^2$ depends on them both

Oh, right.

But if we divide by $u^2+v^2$, we get

OK, I need to get going ... Bye, all.

$\dfrac{(u^2+v^2)_a}{u^2+v^2}=\dfrac{u^2}{u^2+v^2}(v/u)_b$

$\dfrac{(u^2+v^2)_a}{u^2+v^2}=\dfrac{1}{1+(v/u)^2}(v/u)_b$

@TedShifrin OK just look at that^ before you go

And now we know that both sides are constants

(Because the only way a function of $a$ can equal a function of $b$ is if they're both constant functions)

('Cause the LHS doesn't depend on $b$, and the RHS doesn't depend on $a$, and since they're equal, they can't depend on either variable)

In particular, $\dfrac{(u^2+v^2)_a}{u^2+v^2}$ is a constant

@TedShifrin Okay I see what you're saying

or, $(r^2)_a/r^2$ is a constant

and $r=K/\cosh(a)$

I need to learn more about pullbacks

but $(r^2)_a=2rr_a$, meaning $(r^2)_a/r^2=2r_a/r$

Show that a finite commutative ring with unity in which $x^2=x$ (i.e. a Boolean ring) is isomorphic to $(\Bbb Z/2\Bbb Z)^n$ as rings for some $n \in \Bbb N$

And you can check that that's most definitely not a constant if $r=K/\cosh(a)$ @JoeShmo

^ this question generates a lot of thoughts and interesting results

Any ideas about the question I posted ?

In fact, if $2r_a/r$ is a constant, that means $r_a$ is a constant multiple of $r$

The derivative of $r$ (wrt $a$) is a constant multiple of $r$

That means $r$ has to be $e^{Ca}$ for some $C$!

(You can write it as $2r_a/r=C$, or $2\frac{\partial r}{\partial a}/r=C$, and solve it like a differential equation)

interestingly, for large $|a|$ you would have $r\approx 2K e^{-|a|}$

Thus, the only way we can have a function $f(z)$ whose modulus depends only on ${\rm Re}(z)$ is if that modulus is exponential

In fact, if you go through with this analysis, I think you can show that $f$ has to equal $e^{Cz}$

(i.e. the angle has to be ${\rm Im}(z)$)

one sec

Yeah, just look at $\dfrac{1}{1+(v/u)^2}(v/u)_b=\dfrac1{1+\tan^2\theta}(\tan\theta)_b$, which we also said was a constant

stepped out for a sec i need to catch up with what you were writing

Right OK sorry

Where was the last point you read

gotcha

i wouldn't have thought all of this up by myself

but i was getting somewhere similar, although i didn't switch to polar form of the function

and i ended up with the differential equation -- wolframalpha.com/input/?i=(d(u(x,y))%2Fdy)%2Fv(x,y)+%3D+tanhx

@JoeShmo I was kinda cheating a little bit, in that I "knew" that $f(z)$ would end up looking like $e^z$ and I worked backwards a little

($e^z$ has radius $e^a$ and angle $b$)

i know, but thats the thing. is that the solution to my differential equation would have an $e^z$ form anyway

Oh, um, I skipped a factor of 2 somewhere

$(u^2+v^2)_a$ is $2(uu_a+vv_a)$, not $uu_a+vv_a$

It doesn't really change anything though

The rest of the argument still works

can you take a look at this link -- wolframalpha.com/input/?i=(d(u(x,y))%2Fdy)%2Fv(x,y)+%3D+tanhx

and tell me if anything pops up into your head?

Notice that you're only differentiatin wrt y

and tanh(x) is essentially a constant, wrt y

yes

but v(x,y) isn't

so you just want $u'$ to be a constant multiple of $v$, essentially

yes

which is way too broad

right

so this is a dead end right

Like, if you let $u$ be anything, you can find a $v$ that works (by solving)

Yeah

just so i can sleep tight tonight :)

@JoeShmo Here's a more ordered approach to the problem

exaclty^

i wasn't getting a contradiction

Yeah

C–R says $u_a=v_b$ and $u_b=-v_a$, yeah?

But we want to know about the moduli

so we should compute $r_a$, $r_b$, $\theta_a$, and $\theta_b$ and see if we can see patterns

And, if successful, you'll derive a "polar C–R"

well that was my approach

for clarification, what's the actual problem?

but for instance, trying to see if the modulus is purely a function of $u$ and the angle is purely a function of $v$ is knowing what to look for

$r=\sqrt{u^2+v^2}$, so $r_a=\dfrac{uu_a+vv_b}{2\sqrt{u^2+v^2}}=\dfrac{uu_a+vv_b}{2r}$, and similarly for $b$

(I imagine it's somewhere up there in the transcript)

@semi yeah

Ah.

$\theta=\arctan(v/u)$, so $\theta_b=\dfrac{uv_b-vu_b}{u^2}\cdot\dfrac1{1+(v/u)^2} =\dfrac{uv_b-vu_b}{u^2+v^2}$

I guess what I notice is that, if you pick any rectangular region in the complex plane, then the modulus achieves its maximum along one of the vertical sides

which seems incongruous with the maximum modulus principle

${}=\dfrac{uv_b-vu_b}{r^2}$

I suppose that can be reduced to the observation that you'd have $|f(z)|=|f(z+i y)|$ for any $y$

Using the regular C–R and combining those, we get $\dfrac{r_a}r=\theta_b$

I forgot the $2$ in the numerator

$r_a=\dfrac{2uu_a+2vv_b}{2\sqrt{u^2+v^2}}=\dfrac{uu_a+vv_b}r$

@Semiclassical ah! $|f(z)| = |f(\Re(z))|$

ya

OK, whatever, point is, $\dfrac{r_a}r=\theta_b$, and I bet that $\dfrac{r_b}r=-\theta_a$

and it's a very similar calculation so I won't do it again

But whatever, we now have polar C–R, and now we can attack the problem

and it's done very easily: $r_b=0$, therefore the $\theta_a=0$ from the second equation

Thus $r$ is a function of $a$ and $\theta$ is a function of $b$. Thus, the first equation implies that $\dfrac{r_a}r$ and $\theta_b$ are both constants

Plugging in $r=1/\cosh(a)$ demonstrates that this is not the case.

In fact, solving $\dfrac{r_a}r=C$ gives $r=e^{Ca}$.

quora.com/What-is-an-intuitive-explanation-of-Ricci-flows

Cool reading^

Ricci flow distributes curvature like heat

let me try to put it all together on paper. there are a few nice insights in this proof

@AkivaWeinberger, very neat! Thanks for your help

yes, converting to polar coordinates is the big insight. neat

@JoeShmo Incidentally, $r'/r=(\ln r)'$

so, what we've just proven is that if $u$ and $v$ satisfy C–R, then so do $\ln r$ and $\theta$

why do we care?

Which makes sense, since if $f=u+iv$ is analytic, then $\ln f=\ln r+i\theta$ is analytic too

(because $\ln$ is an analytic function)

So you can prove it more easily from that

how did you get lnf=lnr+iθ ?

u is a function of r, v of theta

What? $r=\sqrt{u^2+v^2}$

I got it from $f=re^{i\theta}$, which comes from Euler's formula

how do you know what $\ln f$ is

$\ln f=\ln(re^{i\theta})=\ln(r )+\ln(e^{i\theta})$

Hello!

oh. durp

so how did you get $\frac{r'}{r} = (\ln r)'$

Chain rule

(As a corollary, we get that $\dfrac{f'}f+\dfrac{g'}g=\dfrac{(fg)'}{fg}$, which you can also check directly)

oh. dur

i thought you were doing something deep. my brain is bust

i thought you meant that within the scope of the problem the equality holds

@AkivaWeinberger. (1) Your posts have inspired quite a few of my posts... Thanks for contributing to this site. (2) Are you currently "In gap year between high school and college" because I suspect that was true when you set up your profile but no longer true... but maybe I am wrong.

(Proof: This is $(\ln f)'+(\ln g)'=(\ln fg)'$, which follows from $\ln f+\ln g=\ln fg$ and taking the derivative. Alternate proof: Simply use the product rule.)

but its just the derivative

@Mason I am one month into my gap year

Oh very good.

Thanks for the compliment

what are you up to now?

Gap year in Israel

israel is my birth place

Jerusalem

Ah, really? Nice

oh youre one of them...

Ani lo mevin

I did that! I hung out in Tzuba. And then worked on Sde Eliyahu

i spent a few months in jerusalem last year

the israel freshmen

Nu? Eich haya

You asking me?

Joe

Well, both I guess

it was wonderful. im glad i got to do that

i was back in ny within 3 months :)

where do you live? we lived on yaffo

Where in New York are you from?

Right now I'm in Katamonim (southwesty bit)

monsey

but I was up on Mount Scopus for a little bit earlier

ish. monsey area

Never been but I hear it's nice

I'm from Brooklyn

never been to where

Monsey

@Mason Nice profile pic

oh. you havent missed anything

Life was great. I picked anavim b'kerem col boker and also I picked up some Hebrew.

so wait so you learned all of this by reading?

hehe

@AkivaWeinberger if youre not in college, how did you pick all of that up?

I forget what anavim are

Books, internetz

Grapes.

Ah yeah

Right, thanks

stones*

Kerem is a vineyard.

anavim**

sorry. im dyslexic

anavim hayínu lefar'o bemitzráyim

very impressive, @AkivaWeinberger

Wait, no

hhhhh

We were grapes!

That seems correct.

I like to tell my students that my profile pic is an old photo... my hair isn't that long right now. Your profile image is also cool. Did you make it?

Ever notice how the Hebrew for "this book" is literally "the book the this"?

haséfer haze

@Mason Asked someone online to make it

lh3.googleusercontent.com/…

Basically, it's a bit of a joke - it's impossible to have a polyhedron with only hexagons.

(If you did, it would be another Platonic solid)

Hmmm

(You can prove that it's impossible from V-E+F=2. Note that I'm assuming there are three hexagons per vertex)

So, yeah, got someone to animate it rotating

Nice. Interesting.

We can get your type of "illusion" whenever we can tile the plane with the polygon?

Explain?

What other shapes could we use to make your "illusion" of a platonic solid?

let me riddle yous a riddle -- a library has $n$ books, and $n+1$ subscribers. Prove there exist two (disjoint) sets of readers who read the exact same books. Hint: basic linear algebra

imgur.com/download/mBKypPo/What%20shape%20is%20this

Well... if we don't have to show how things connect in the back... we only need to show part of this solid: Even when you have it rotate you are only showing part of it.

Does that link work?^

So I think we can manage this illusion whenever we have a shape that tiles the plane.

Yes.

The link works.

the link works

Yeah so that's the same thing but with two adjacent faces making a 120 degree angle

Other link

@Mason I believe you're correct

@JoeShmo I know this, but I didn't figure it out the first time I saw it

the riddle?

Oh, wait, that's slightly different than the version I remember

Not sure if it makes a difference

Ah, for your version, we can argue like this:

I wonder if the condition is sufficient but not necessary...

There are $2^{n+1}$ different ways of selecting a set of subscribers

and $2^n$ sets of books

I can't tile the plane with a regular pentagon... but maybe one can still make your illusion? @AkivaWeinberger

By the pigeonhole principle, there exist two sets of subscribers who read the same books

@Mason Why not just grab a dodecahedron

No need for illusions...

So then it's clearly sufficient but NOT necessary

What are the necessary conditions?

not following, @AkivaWeinberger. there are twice as many sets of subscribers than there are books. so what?

So there exists two sets of subscribers who read the same books

To make them disjoint get rid of the people who are in both sets

@Mason The thing in the animation is essentially this:

The angle is hiding infinitely many hexagons in the back

And it's this infinite thing which makes this an illusion.

oops. i misphrased the puzzle

@JoeShmo Oh, wait, that doesn't work because by getting rid of the people in both sets you might break it

yeah, exactly

But yeah you wanted something about odds and evens in the puzzle description I think?

nope

there, verbatim

Oh so you were right the first time?

Or what changed

My proof was wrong

I see how to do it now though

I'm curious if there's a simpler proof

@AkivaWeinberger. I think you can probably create the same illusion with a heptagon. I am just guessing based on this image

the union of books, etc. is an important point

go for it. whats your proof

Associate to each reader a vector

If he reads books 1, 2, and 4 he gets $(1,1,0,1,0,0)$ for example

(if there are six books)

There are more vectors than the dimension of the space

Thus, there's a linear dependence

Move all of the vector is with negative coefficients to the other side of the equation

there you go

The vectors on each side form the two sets

convinced

I feel like there should be a simpler solution, though

this is pretty simple

One that doesn't require knowing linear algebra

perhaps induction

Could be

but yeah, the problem with brute force is that you lose track of who went where

This has the benefit of, each book was read it the same number of times by each set

Never mind, this is false

because a "set" can't contain more than one of the same element

so we lose the coefficients when we go from the linear dependence to the two sets

@Mason Here's a puzzle. Does$$(x-\sqrt2-\sqrt3)(x-\sqrt2+\sqrt3)(x+\sqrt2-\sqrt3)(x+\sqrt2+\sqrt3)$$have integer coefficients when you expand it out?

In other words, multiplying together all four choices of $(x\pm\sqrt2\pm\sqrt3)$

I mean, theoretically you can brute force it by just multiplying it out, but there are nicer ways

I am confident it has an integer coefficient associated with x^4

@JoeShmo Here's a puzzle for you. Show that there is some cubic polynomial $ax^3+bx^2+cx+d$ with rational coefficients such that, when you plug in $x=\sqrt2+\sqrt3$ into it, it equals $\sqrt2$. (Hint: linear algebra)

I am heading back to working on my post...

We get this the same way we would with having i=sqrt(-1) right: each root's conjugate is also a root. This is because sqrt(2), sqrt(3) are independent over the rationals?

Apparently I am not leaving very quickly... :p. No. I am leaving to work on my post!

That's an idea, but not a proof

Although, fun fact, those are called Galois conjugates

That makes some sense.

On the main Questions page... what does it mean when some questions are highlighted yellow?

How can we get from the recursive definition of sequence: $$a_1=0 \\ a_{n+1}=(-1)^{n+1}\cdot (a_n+2\cdot n)$$ to the explicit one?