Mathematics

Masterphile

12:02 AM

@Thorgott did you ever read a full proof of this theorem?

Thorgott

12:13 AM

no, I haven't done alg top yet, though I should

Masterphile

yes, it seems that you are familiar much with classical analysis and some abstract algebra, so it seems as a natural extension of your current knowledge

Ted Shifrin

12:54 AM

Note: If you do differential topology (so diffeomorphisms instead of homeomorphisms), then this result is an easy consequence of the Inverse Function Theorem — no homology theory needed.

Masterphile

1:04 AM

do you suggest that i do differential topology before dwelling onto theorems with just assumptions of continuity only?

Ted Shifrin

Well, it's a matter of taste. I personally think differential topology (Guillemin & Pollack) is the best course in the undergraduate curriculum, but you do need solid linear algebra and some multivariable calculus/analysis (in particular, the derivative as a linear map and the inverse function theorem).

Masterphile

yes, that´s why, because i am not skilled much in Taylor series for many variables and definition of the derivative in those cases, and because of some other obstacles, i have chosen to do first theorems that only assume continuity, let´s be fair, differentiability is a rather rare property and the theory is less general than only when assumptions of continuity are supposed,

but despite being less general it is equipped with some special theorems that sometimes do not have existant only-continuous-but-not-everywhere-differentiable counterpart

feynhat

Hello.

@TedShifrin Inverse function theorem? Wouldn't just chain rule suffice?

(assuming invariance of dimension for vector spaces, which follows from rank-nullity)

Masterphile

1:28 AM

what does it mean that a manifold has a countable base?

what is a countable base?

feynhat

@MikeMiller Oh right. I didn't use the commutativity of the diagram.

@Masterphile Do you know what is a basis for a topological space?

Masterphile

i forgot that! :D

tell me

feynhat

Base (topology)

In mathematics, a base (or basis) B for a topological space X with topology T is a collection of subsets of X such that every open set in X can be written as a union of elements of B. We say that the base generates the topology T. Bases are useful because many properties of topologies can be reduced to statements about a base generating that topology, and because many topologies are most easily defined in terms of a base which generates them. == Definition and basic properties == A base is a collection B of subsets of X satisfying the following properties: The base elements cover X. Let B1, B2...

Masterphile

ok i think i understand, then how some manifolds do not have countable base?

feynhat

Most people would include existence of countable basis in the definition of manifolds.

Masterphile

1:37 AM

yes, that seems to be usual practice

also that they are Hausdorff

feynhat

It really depends on how you define a manifold. Some texts don't require second countability axiom (I'm looking at Hatcher, Milnor & Stasheff etc.). In their world manifolds can have no countable basis.

Hi @BalarkaSen

Ted Shifrin

@feynhat: No, to prove invariance of domain you need the inverse function theorem to prove that a smooth map with invertible derivative is an open map.

@Masterphile: Unless you want to fret over the long line as a manifold, we all assume countable basis (or metrizable).

Thorgott

If all you want to know is that open subsets of $\mathbb{R}^n$ and $\mathbb{R}^m$ can only be diffeomorphic if $m=n$, then the chain rule suffices

schn

What property is it that allows one to move the del symbol under the integral sign (in the second equation)? This is from Griffiths Introduction to Electrodynamics.

feynhat

@TedShifrin Yes, I was wrong. I had a different statement of invariance in mind (what @Thorgott wrote).

user76284

2:08 AM

Is the cumulative hierarchy stage $V_\alpha$ corresponding to the ordinal $\alpha$ always the intersection of all supertransitive, powertransitive supersets of $\alpha$?

where powertransitive is as defined in math.stackexchange.com/questions/3611428/…?

Ted Shifrin

@schn: It's called differentiation under the integral sign. Super important and powerful.

Balarka Sen

2:40 AM

Hi @feynhat

Leaky Nun

3:40 AM

what if I integrate under the differentiation sign

Silent

Does the approach [here](https://math.stackexchange.com/a/3273950/94817) to show that $\begin{equation}
f(x,y) =
\begin{cases}
\frac{x^3+y^3}{x-y} &x\neq y\\
0 &x = y
\end{cases}
\end{equation}$ is discontinuous at origin, that is going along path $y=mx^{1/3}$ really work?

It seems like $\lim _{x\to \:0}\left(\dfrac{2m^3x}{x-mx^{\frac{1}{3}}}\right)=0$

Leaky Nun

4:11 AM

@Silent not only does it not work, in fact no exponent works

here's a sort of graph: desmos.com/calculator/bwwicoabzh

you need something "close to" $y=x$

looks like $y=x+x^3$ works: desmos.com/calculator/fkztkntnr4

@BalarkaSen I can't do high school calculus

Balarka Sen

Neither can I

skullpatrol

4:36 AM

High school calculus: A Geometric Approach

3

Leaky Nun

this one actually makes sense

skullpatrol

4:51 AM

yeah, but I couldn't find any book with that title?

skullpatrol

5:40 AM

from Wikipedia

Ted Shifrin

6:06 AM

@skullpatrol I'll get you for that.

Balarka Sen

Hi @Ted!

Ted Shifrin

Hi, a @Balarka!

skullpatrol

@TedShifrin absolutely no offense intended sir :-)

Ted Shifrin

Uh huh.

a @Balarka, any thoughts?

skullpatrol

Why do they list your name twice on your book? @TedShifrin

Balarka Sen

6:12 AM

@TedShifrin Question seems to be deleted

Ted Shifrin

Huh?

skullpatrol

Ted and Theodore @TedShifrin

Ted Shifrin

I just commented on it. Maybe my link is screwed up.

Balarka Sen

Oh yeah OK let me go to the correct link

Ted Shifrin

Is this better, @Balarka.

Balarka Sen

6:13 AM

Got it

Also yeah it is

Ted Shifrin

Where is that, @skull?

skullpatrol

Amazon

Ted Shifrin

I don't remember ever seeing that, @skull. Don't know.

Balarka Sen

So we wants a pointwise defined pairing on $\Omega^1(M)$ such that $\langle \omega, \eta \rangle = 0$ whenever $\omega$ is an exact form? Aka the pairing is degenerate on the subspace of exact forms

Weird

Ted Shifrin

So if $H^1 = 0$, I believe I have a proof that only $0$ works.

And for something like the circle or the torus, I can write down a nonzero thing, I think.

user76284

6:18 AM

If I understand correctly (en.wikipedia.org/wiki/…), mnk games (Tic-tac-toe- connect6, gomoku) are PSPACE-complete in their size parameter. What if we remove diagonals from the win condition?

Unfortunately the Stefan Reisch paper is in German.

Ted Shifrin

I guess I need the cohomology to generate $\Omega^1$ over smooth functions.

Balarka Sen

The pairing I mentioned descends to a pairing on $H^1$ I suppose.

Let me think

skullpatrol

@TedShifrin amazon.ca/Linear-Algebra-Geometric-Ted-Shifrin/dp/071674337X

user76284

Looks like cs.ru.nl/bachelors-theses/2017/… answers my question.

Ted Shifrin

Oh, you just mean that the book says one name and the page says the other.

skullpatrol

6:24 AM

Yes

Ted Shifrin

No need to link twice.

Oh, I see.

Shrug. Not my problem.

skullpatrol

It's only on some of their pages anyway.

Balarka Sen

@Ted This can't be right. Take any two forms, $\omega, \omega'$ which agree on a neighborhood of $p$. Let $f$ be a bump functions around $p$. $\psi(f \omega) = \psi(f \omega')$ so $f \psi(\omega) = f \psi(\omega')$ so $\psi(\omega)$ and $\psi(\omega')$ agree on a neighborhood of $p$ as well. Let $\eta$ be an arbitrary form; for any $p \in M$ produce an exact form which matches $\eta$ in a neighborhood of $p$. Then above tells you $\psi(\eta)$ is zero in a neighborhood of $p$.

Ted Shifrin

Yeah, I'm thinking it's always 0. I'm about to add a comment showing why on $S^1$.

Silent

@LeakyNun Thanks a lot! How did you come up with $y=x+x^3$?

Leaky Nun

6:35 AM

@Silent y = x + (a small term)

2 didn't work

Balarka Sen

The point is his $C^\infty$-linearity condition implies $\psi : \Omega^1(M) \to X(M)$ is pointwise defined; the value of $\psi(\omega)_p$ only depends on $\omega_p$ - this is the "tensor characterization lemma". But germinally exact forms are the same as any other form

Ted Shifrin

Right.

user76284

Interesting that a game as simple as axis-parallel 5-in-a-row is PSPACE complete. Looks like the question for 4-in-a-row is open.

Leaky Nun

@BalarkaSen help I have been coding all day for days and didn't learn any maths

Ted Shifrin

Do you want to post that, @Balarka?

Is that our fault, @Leaky?

Silent

6:37 AM

Thanks, Leaky!!