12:02 AM
@Thorgott did you ever read a full proof of this theorem?

12:13 AM
no, I haven't done alg top yet, though I should

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

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.

1:04 AM
do you suggest that i do differential topology before dwelling onto theorems with just assumptions of continuity only?

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).

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

Hello.
@TedShifrin Inverse function theorem? Wouldn't just chain rule suffice?
(assuming invariance of dimension for vector spaces, which follows from rank-nullity)

1:28 AM
what does it mean that a manifold has a countable base?
what is a countable base?

@MikeMiller Oh right. I didn't use the commutativity of the diagram.
@Masterphile Do you know what is a basis for a topological space?

i forgot that! :D
tell me

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...

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

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

1:37 AM
yes, that seems to be usual practice
also that they are Hausdorff

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

@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).

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

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.

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

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/…?

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

2:40 AM
Hi @feynhat

3:40 AM
what if I integrate under the differentiation sign

Does the approach [here](https://math.stackexchange.com/a/3273950/94817) to show that $f(x,y) = \begin{cases} \frac{x^3+y^3}{x-y} &x\neq y\\ 0 &x = y \end{cases}$ is discontinuous at origin, that is going along path $y=mx^{1/3}$ really work?
It seems like $\lim _{x\to \:0}\left(\dfrac{2m^3x}{x-mx^{\frac{1}{3}}}\right)=0$

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

Neither can I

4:36 AM
High school calculus: A Geometric Approach
3

this one actually makes sense

4:51 AM
yeah, but I couldn't find any book with that title?

5:40 AM
from Wikipedia

6:06 AM
@skullpatrol I'll get you for that.

Hi @Ted!

Hi, a @Balarka!

@TedShifrin absolutely no offense intended sir :-)

Uh huh.
a @Balarka, any thoughts?

6:12 AM
@TedShifrin Question seems to be deleted

Huh?

Ted and Theodore @TedShifrin

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

Oh yeah OK let me go to the correct link

Is this better, @Balarka.

6:13 AM
Got it
Also yeah it is

Where is that, @skull?

Amazon

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

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

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.

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.

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

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

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

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

6:24 AM
Yes

Oh, I see.
Shrug. Not my problem.

It's only on some of their pages anyway.

@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 $p$.

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

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

6:35 AM
@Silent y = x + (a small term)
2 didn't work

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

Right.

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.

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

Do you want to post that, @Balarka?
Is that our fault, @Leaky?

6:37 AM
Thanks, Leaky!!

Which language are you coding in? @LeakyNun

python
in codewars.com

coolio

@TedShifrin OK I will add a comment

@LeakyNun no more chess?

6:39 AM
@LeakyNun RIP

oh wait I have a referral link
www.codewars.com/r/dM2VPA
2
if you use this link to join then I get something I think
@skullpatrol for the moment

:-)
@LeakyNun Your search - www.codewars.com/r/dM2VPA - did not match any documents.

ok
Wow! So many languages.

1 hour later…
8:19 AM
Let's try something out. Say $g_2 = e^f g_1$.
According to Koszul formula $2g_2(\nabla^2_X Y, Z) = X g_2(Y, Z) + Y g_2(Z, X) - Z g_2(X, Y) + g_2(X, [Y, Z]) + g_2([Y, [Z, X]) - g_2(Z, [X, Y])$
Ah interesting
$X g_2(Y, Z) = X(f) g_2(Y, Z) + e^f X g_1(Y, Z)$ and $g_2([X, [Y, Z]) = e^f g_1([X, [Y, Z]])$
I mean, I get $2g_2 (\nabla^{(2)}_X Y, Z) = e^f 2g_1(\nabla^{(1)}_X Y, Z) + X(f) g_2(Y, Z) + Y(f) g_2(Z, X) - Z(f) g_2(X, Y)$
I should probably use $e^{2f}$ to make that annoying factor of $2$ go away
What am I writing
Christ, OK. I do get $\nabla_X^{(2)} Y = \nabla_X^{(1)} Y + X(f) Y + Y(f) X - \mathrm{grad}(f) g_2(X, Y)$
Do I dare compute the curvature

9:26 AM
(removed)

If I have $u\in H^1(\Bbb R^d)$ (Sobolev space) with $d\ge 3$, how can I see that $u/|x|\in L^2$ ?
I know that for any function $v\in C^\infty_c(\Bbb R^d)$, $||{v\over |x|}||_{L^2}$ is bounded by $C||\nabla v||_{L^2}$

1 hour later…
10:37 AM
Is the unary negation operator separate from actual negative numbers? For example $1-10=-9$ would the $-$ in front of the $9$ be considered an operator or the actual number?

11:07 AM
What's the definition you have for $u$ being in $H^{1}(\Bbb{R}^{d})$, @Astyx?

$\int |k|^2|\hat u(k)|^2dk\le \infty$

oh no

oh no

Gotta go eat, see ya

you're supposed to say "oh yeah"

11:25 AM
oh yeah
As in, the construciton worker bashing in through a wall ?

11:56 AM
@Astyx Do objects in contact always exert a normal and a frictional force on each other?

@Astyx yeah

That video always cracks me up for some reason
@Knight Depends what you mean by that

@Astyx I’m thinking of the source of normal force (when an object rests on ground it is obvious but what about if I take two glass slabs parallel to each other vertically)

The normal force is only here to describe the fact that if you push two objects against each other, they don't go through each other
Or at least, there is some resistance to that

Imagine I’m making a “namaste” with my palms
Are my palms pushing each other if I just make them to contact.

12:02 PM
As long as you exert a force with one on the other, since your hands don't move, Newton's laws tell you there's a force compensating the one you exert

I got a confusion in the drawer problem

@Astyx I feel like the integral of a non-negative function will always be $\le\infty$

Ooops, I meant $<\infty$

> Suppose that we have a drawer with two handles $a$ distance apart so that it can be pulled and pushed. The maximum distance the drawer can come or go inside is $L$. Finding the minimum coefficient of friction $K$ So that our drawer starts to move if we pull it just by one handle, whatever the force magnitude.
My confusion is why the sides of the drawer (the sides must be the part of cupboard) would exert a normal force on the sides of the drawer
@Astyx

$\hat{u}$ being the Fourier transform?

12:14 PM
0

How do I prove the expressions in the red square? I know $\Delta A _{\perp}= |\Delta A|\sin \Delta \theta \approx |\Delta A|\Delta \theta$ $\Delta A _{||}= |\Delta A|\cos \Delta \theta \approx |\Delta A|$(For small $\Delta \theta$) I don't know how to derive

@Thorgott yes
Basically it tells you that the weak derivative of first order of the function is $L^2$
@Knight I don't understand your question

12:44 PM
@Astyx A table has an internal movable surface used for slicing bread. In order to be able to be pulled to the outside, there are two small handles on its front face with distance aaa between them put in a symmetric way with respect to the middle of the surface. The length of the surface is L. (Fig.1)
Find the minimum friction coefficient K between the sides of the movable surface and the internal surface of the respective sides of the table, so that we can pull the movable surface using only the one handle whatever force magnitude (i.e. big force) we exert
My doubt is whether there will be a normal force on the drawer by the sides?

12:56 PM
Yes
Because the force makes the drawer turn because it's not colinear with the center of mass (badly phrased)

@Astyx Okay, means putting the force on the handles actually causes a turn not a translation but due to the restrictions from sides the drawer takes on translational move. Am I right?

yes

Thanks @Astyx

You're welcome

@Astyx ok, maybe I'm being stupid, but isn't $x\mapsto e^{-x^2}$ in $H^1$, yet $x\mapsto e^{-x^2}/|x|$ not in $L^2$?

1:17 PM
That's in $\Bbb R$, here I'm in $\Bbb R^d$ with $d\ge 3$

Is the unary negation operator separate from actual negative numbers? For example $1−10=−9$
would the "$−$" in front of the $9$ be considered an operator or the actual number?

$$\int_{\Bbb R^2}e^{-x^2-y^2}/\sqrt{x^2 + y^2}dxdy = \pi^{3/2}$$ according to Mathematica
@northerner the operator is a function $\Bbb R\to \Bbb R$ whereas negative numbers are a subset of $\Bbb R$, if that's what you're asking

I guess I'm asking, is it really an operator or do just people call it that?
Or is it two different things?

It probably depends how you describe things
the thing is, if $op$ is the negation operator, then $op(n) = -n$ so it's not ambiguous when you write $-n$

@Astyx Is that a Gaussian?

1:25 PM
A Gaussian multiplied by $1/r$

@Astyx but it could just happen to be a negative number, for example $-5$ has no operator.

what about taking a bump function that's constant around the origin?

Am I looking at things correctly?

that should diverge around the origin after division by $|x|$ and be in $H^1$ cause it's smooth and compactly supported

@Thorgott $1/r$ is $L^2$ in $\Bbb R^d$ when $d\ge2$ if that's what you're asking

1:26 PM
@Astyx That's risky... can't even use By parts

turns out my intuition in higher dimensions is ass, I should shut up

manifold equation?

@AbhasKumarSinha I'm not trying to compute it

k....

@Thorgott In the end I got my answer

1:28 PM
I also have a question, will ask after this.

If you look at $\int |\nabla u + \alpha u(x)/x|^2$ and develop it, you get a polynomial in $\alpha$
maximising it gives a majoration of $\int |u(x)/x|^2$

Oh I've actually seen a question that asked the same thing before

It was me probably
yesterday

Okay...
I wanted to ask if $$S \stackrel \Delta = \int \limits_{\text{4D Manifold }\, \mathcal D} R \sqrt{-g} d^4x$$
Why minimization of that ^ integral using variation doesn't yields a flat manifold?

You're gonna need to specify what those letters are

1:33 PM
@Astyx I'm doing wait.... :-)
Where $R = g^{\mu \nu} R_{\mu \nu}$ and $g = \det g_{\mu \nu}$ where $g_{\mu \nu}, R_{\mu \nu}$ are tensors such that $R_{\mu \nu}$ is Ricci Tensor and $d^4 x$ is infinitesimal volume element
$g_{\mu \nu}$ is a metric tensor
Also, I'm using standard signatures $-+++$ for metric tensor
or alternatively, why $R_{\mu \nu} - \frac 12 R g_{\mu \nu} = 0$ isn't flat spacetime?

1:53 PM
When proving that $C \sum_{k=1}^n a_{ik} = \sum_{k=1}^n Ca_{ik}$. Can you use the distributive property of multiplication over addition as justification?

yes

:)
Thanks, guys

distributive property + induction

1:59 PM
@Thorgott Use an inductive proof?

yes, distributivity gives you, say, $C(a_1+a_2)=Ca_1+Ca_2$, but what about $C(a_1+a_2+a_3)$?. Well, $C(a_1+a_2+a_3)=C((a_1+a_2)+a_3)=C(a_1+a_2)+Ca_3=Ca_1+Ca_2+Ca_3$, etc.

Ah, ok.

2:29 PM
@northerner There is an operation M: Z -> Z, which has M(n) = -n. There is an operation P: Z x Z -> Z, where P(m,n) = m+n. Putting these together you get an operation S: Z x Z -> Z which deserves to be called subtraction, given as S(m,n) = P(m,M(n)) = m-n. Even if you define S in a different way (perhaps you know that P gives a group law, and you describe S as being "the unique S(m,n) such that P(m,S(m,n)) = n", then it still happens to be the case that S(m,n) = P(m,M(n)).
In other words: it doesn't matter whether you view that subtraction as being "m plus the negative of n", or some other definition.
You can view it as being a binary operation in its own right (S), or not.

2:58 PM
I've been pulling together a draft of an idea and am looking for feedback; a TeX PDF contains four total pages of set builder notation concerning number system construction and generalizations of inner product spaces over those systems. benblohowiak.com/2020-03-28MultipolarEquations.pdf

2 hours later…
4:57 PM
Not a lot of activity tonight

indeed

Hi @Edward

Hiya @Alessandro

PSA: There's a bunch of seminars in various fields being moved to an online format and usually anyone interested can join so check out the list

that's cool af

5:01 PM
If you star it more people might see it

1 hour later…
6:27 PM
@AlessandroCodenotti Thanks for sharing this.

You're welcome! It's a great opportunity

6:58 PM
0

Let $S_n = X_1 + X_2 + \cdots + X_n$ be a non-symmetric simple random walk with $S_0 = 0$. For $y>0$ define $$T_y = \min \lbrace n \geq 1 \ : \ S_n = y \rbrace$$ How can I compute in a closed form the value of $\mathbb{E}(T_y \ | \ T_y < \infty)$?

For this question, I don't understand the suggestion solution mean

Hi, demonic @Alessandro.

Hi Ted

Done with your PDE exam yet?

7:15 PM
Hiya @TedShifrin @Alessandro

@TedShifrin Nope, it'll be next Friday

but from Tao so nise

lol
There is a collatz conjecture course this term in Bonn

7:30 PM
that's cool haha

@AlessandroCodenotti "it's open", end of course, thanks for attending

Thanks for coming to my TED talk

It's not the kind of conjecture that tickles my fancy tbh

7:58 PM
Fair enough
What have you been working on lately?

Just going through some Minkowski theory, really interesting stuff
Minkowski theory is actually very cool, it considers elements of a number field $K$ as points in an n-dimensional $\Bbb C$-vector space indexed by the embeddings of $K$ into $\Bbb C$
and then combines this with some lattice theory to draw conclusions about ideals when considered as lattices in the ring of integers of a number field

@EdwardEvans I'm studying PDEs for that exam, while trying to find some time for set theory as well
I guess that the Minkowski bound for the class number is like the baby version of that then

well "proving" the Minkowski bound is basically the goal of the theory in first ANT courses

8:14 PM
I see
I wish I had done more ANT, the first course I took was very interesting
Hi @Mike

There's a theorem of Minkowski that given a certain bound on a subset $X$ of a euclidean vector space $V$ and a lattice $\Gamma$ in $V$, you can guarantee that there's at least one non-zero point of $\Gamma$ in $X$

Do you want to hear more semigroups stories?
@EdwardEvans Right that's the result you need to get the bound on the class number

exactly hehe

I was asking Mike actually, I didn't think you'd be much interested in PDEs :P

8:20 PM
fair one hahaha

8:37 PM
Sure

8:57 PM
@MikeMiller Do you know if Notre Dame is one of the stronger places in Geometry and Topology in the US at the moment?

2 hours later…
10:50 PM
Not the first place I think of for geometric topology in particular but they have some good people
Nicolaescu though I don't know that he does much topology anymore; Putman for specifically topology of surfaces
Stolz does some high-dimensional manifolds stuff