Too much thinking indeed.

(That is, I have been thinking too much today.)

(Also: hello chat)

Hi @Fargle @Rithaniel @Leaky @Jacksoja Demonark

hi!

Hey @Ted

Hey Ted!

@TedShifrin do you like my puzzle above =)

You could do some sort of induction, but I like Demonark's approach.

how would you do it?

:D

Evening, Ted.

I wouldn't, but I've done similar problems ... cofactors and induction.

@TedShifrin Hello Ted

@TedShifrin how does one define the map from Z---> R ( ring) ?

Also, fun problem from difftop pset: find a smooth function $f:\mathbb{R}\to \mathbb{R}$ such that the set of critical values is dense

it should be just defining where to send 1 to right?

@Jacksoja: Map? You mean ring homomorphism?

@Daminark so let $\varphi : \Bbb Z \to \Bbb Q$ be a bijection and let $(i, \varphi(i))$ be the critical values, I guess?

Demonark: That's a problem in G&P.

@TedShifrin yes ring hom

Jacksoja, so for me ring homomorphisms must send 1 to 1 ... I dunno your book.

Yeah basically Leaky, and you can accomplish that with bump functions

@TedShifrin am trying to show that if R is a field , then the kernel, ie ideal, can be only zero ideal or prime ideal

Oh, but that assumes sending 1 to 1.

Now, question that even I'm not sure about, can you have any null set as the set of critical values of some smooth function?

@TedShifrin so the map is n ---> n * unit of R?

unity of R

What's your book's definition of a ring homomorphism?

f(a+b) = f(a) +f(b) and f(ab) = f(a) f(b)

without assuming a unital element

for sure f(0) =0

then Hom(Z,R) is more complicated

That's not part of the definition. That follows immediately.

Yup.

okay we assume that f(1) =1

is it the set of idempotents?

What Leaky said. The kernel could be everything, couldn't it?

Wait ... where did that "okay we assume ..." come from?

Shouldn't a function where the set of points with 0 derivative is a dense set just be a constant function?

then the map is f(n) = n* 1_R?

It doesn't have to be, @Rithaniel

because f(n) = f (1+1...+1) = f(1) + f(1) ....+f(1) = n* 1_R

Hey everyone!

Hey @TedShifrin :)

hi @Perturb

@TedShifrin it sends 1 to 1 , i missed that part

f(1)^2=f(1) so we definitely get an injection, and as for surjection...

injection what to what?

Okay, then if we also require that it is continuous?

(an)(bn)=abn^2=abn, so it should be fine

right, so Hom(Z,R) is the set of idempotent elements of R @TedShifrin

@Rithaniel: Do you know any differentiable functions that are not continuous?

We're not mind-readers, Leaky.

Fair point.

@TedShifrin I was talking to myself until I pinged you

so can you describe all homomorpshims from Z---> R ?

Chat rooms are not for self-conversation. I hate it when that happens here.

they must have the kernel of the form nZ

I just told you @Jacksoja

Okay, in that case I don't see how such a function could exist and not be a constant function.

@Jacksoja: The only ideals in $\Bbb Z$ are what you just said, so not very informative.

okay so my proof is

It's contrary to my intuition, so, if it's possible, my intuition is wrong.

assume that the kernel is nZ for n composite, ie n =ab

f(a) f(b) = 0

but since we are in a field, we do not have zero divisors

assume f(a) = 0

so a < n , and it is an element of the kernel, so our kernel should have been aZ

contradiction

or one could put it , that n do not divide a

nZ are all multiples of n

@TedShifrin does this seem ok to you ?

Not very clearly written, but I suppose you have the idea.

thank you

Okay so I'm trying to use polar coordinates as an example to understand differentials in diff geom. So if I let $M = \{ (r, \theta) \ | \ r > 0 \text{ and } -\pi < \theta < \pi \} \subseteq (0, \infty) \times \mathbb{R}$ and define $F : M \to \mathbb{R}^2$ by $F(r, \theta) = (r\cos\theta, r\sin\theta)$, then $M$ is a smooth manifold consisting of the single chart $(M, F)$.

so $M = (0,\infty) \times (-\pi,\pi)$

If I choose $p \in M$, then $T_pM$ has basis elements: $$\frac{\partial}{\partial r}\bigg|_p := (dF_p)^{-1}\left(\frac{\partial}{\partial x^1}\bigg|_{F(p)}\right)$$ and $$\frac{\partial}{\partial \theta}\bigg|_p := (dF_p)^{-1}\left(\frac{\partial}{\partial x^2}\bigg|_{F(p)}\right)$$. Then is $M$ equipped with this coordinate chart the usual interpretation of polar coordinates in differential geometry?

why $x^1$ not $x_i$?

No, polar coordinates usually are Gauss normal coordinates, based on following geodesics (instead of straight lines).

Upper indices get used to make consistent upper/lower symbols, Leaky.

I thought vectors are lower and coefficients are upper

But that's wrong, @Perturb. Sit down and write out carefully what you said.

Well, $\partial/\partial x^i$ puts the index "down," Leaky, so that's consistent

But $dx^i$ puts it up.

hmm

@TedShifrin Wait was one of my statements incorrect?

Your equations for $\partial/\partial r$ and $\partial/\partial\theta$ seem quite wrong to me, @Perturb. $\partial/\partial r$ is a unit length radial vector field, so it should be $(x/r)\partial/\partial x+ (y/r)\partial/\partial y$, for example.

Is that what you've written down?

when do we have a canonical isomorphism between $TM$ and $T^\ast M$?

When you have a symplectic structure, Leaky.

Don't know what "canonical" means ....

@Ted Ohh okay I was just taking a guess that this was the interpretation of polar coordinates, what I've written down must be wrong based on what you said earlier

maybe delete that word

You have the coordinates living in the wrong place, @Perturb. You should be writing coordinates on the usual $\Bbb R^2$, so your $F$ is backwards.

oh, there's demonic Alessandro !

Hi @Ted

So you're saying I should take coordinate charts on $\mathbb{R}^2$ and then take $F : \mathbb{R}^2 \to M$?

I was hoping to find Mathei but it looks like I'm out of luck

Writing $M$ is totally confusing here, @Perturb. We're talking about polar coordinates on an open subset of $\Bbb R^2$. That is your $M$.

He hasn't been here whilst I have, @Alessandro.

There's a course being offered here next semester by a lecturer who did the same course at his uni this semester so I was curious if he knows something about it, it's nothing urgent

Don't you have email to reach him, anyway?

and by "polar coordinates" we're really taking $\Bbb R^2 \setminus \{0\}$ to be our manifold, right?

Actually, deleting a closed ray from the origin?

why?

You tell me ... We're talking charts on manifolds ...

@TedShifrin True, I should have his phone number too somewhere, but I didn't want to bug him

I'm sure he'll reappear soon, Alessandro.

@TedShifrin Okay yeah, but if polar coordinates aren't thought of in this way then there isn't much reason to carry on thinking about what I said since it's probably nonsense

Sure sure, I have no hurry. Italians are never in a hurry :P

As I said, in general, there's geodesic polar coordinates on a Riemannian manifold.

I'll take a look at that once I get that far in diff geom

what role does the funny $\mathrm dx$ play as an element of $\Omega^1(S^1)$, $S^1 \subseteq \Bbb R^2$?

What are you asking, Leaky?

a broad question, I guess

I can define the form "dx" right?

You mean with $(x,y)$ coordinates on $\Bbb R^2$?

right

So you can restrict (pull back) to any submanifold.

ah

I have much to learn

Of course, it may vanish places (or everywhere, conceivably) when you do so.

do you prefer writing a 1-form as fdx+gdy instead of fdθ?

Depends if you're doing a line integral coming from something ambient or if you're being intrinsic.

actually how to differentiate fdθ lol

wait no that's rubbish

I mean how to express df in terms of gdθ

You're starting with $f(\theta)$?

right

Then I think you should be able to answer.

aha, if I start with f(x,y) then I get gdx+hdy easily; if I start with f(θ) then I get gdθ easily; but not if I swap them

Pull back by the parametrization. You need to learn basics.

See my lectures :P

I see

Gram-Schmidt is a continuous map $GL_n \to O_n$ right

"is"?

But the mapping is smooth, in fact.

So $O_n$ is a deformation retract of $GL_n$?

Yup.

I guess the mapping wouldn't be algebraic then

wait it should be

Um, no ... $\|x\|$ is hardly algebraic.

oh

Let's say you have a function $f : U \subseteq \mathbb{R}^n \to \mathbb{R}^m$ where $U$ is an open subset of $\mathbb{R}^n$. And the only definition that you have of a partial derivative of $f$ is the following, the $j$-th partial derivative of $f$ at $a$ is the directional derivative of $f$ at $a$ with respect to the basis vector $e_j$, that is $$\frac{\partial f}{\partial x^j}(a) = \lim_{t \to 0} \frac{f(a+te_j)-f(a)}{t}$$.

Then someone comes along and gives you the function $f : \mathbb{R}^2 \to \mathbb{R}^2$ defined by $f(r, \theta) = (r\cos\theta, r\sin\theta)$ and asks you to find $\frac{\partial f}{\partial r}$ how do you do that using your definition of partial derivative given above?