Happy New Year to all who celebrate it!

Happy New Year! 🎆🎇

Lies. The year started months ago.

Happy New Year everyone.

happy new year!

Happy every year

I'm going to read classification of smooth connected 1-manifolds with boundary tomorrow

:D

friedl.app.uni-regensburg.de/papers/… Huge book on topology.

That’s a proof I’ve never bothered with, I confess.

HNY to all.

I just realized that a flat earther is just one point compactification away from living on the sphere.

The moral is: Geometry is a richer structure than topology.:

Happy new year and may the odds be ever in your favor

Happy new year everyone!

this year I hope to learn maths proper! woohoo!

@Jakobian Uhm not sure, I always thought it is kind of interesting by itself

Having the FPP is an obstruction to being a topological group so that for example the Hilbert cube has no group structure despite being homogeneous but I wouldn't really call this an application

there's one important paper I want to read but it's in French. And I'm semilingual.

Are analytic varieties also called analytic sets?

four conformal compactifcations of something isometric to Minkowski (1+1) spacetime in the light cone metric on top of each other

i just did it because sYmeTrY

actually i missed counted there

and the diagram is missing a copy

@SoumikMukherjee analytic sets are a different thing, at least in descriptive set theory

Ohh okay

We're not in old ages anymore thankfully

1. A manifold with fixed point property is compact.

2. There is a non-compact convex subset of locally convex space with fixed point property

@AlessandroCodenotti Every locally connected, locally compact metrizable space with FPP is a Peano continuum

Do you know of any non-compact subsets of $\mathbb{R}^n$ with fixed point property?

Found one in $\mathbb{R}^2$

@Jakobian I guess any space with a single cut-point will do

Two tangent disks without borders except for the shared point where they touch for example

Maybe it's more interesting to look for an example $X$ with the FPP but such that $\Homeo(X)\curvearrowright X$ has only infinite orbits. Or maybe even homogeneous examples

@AlessandroCodenotti This space contains a ray though

What do you mean with a ray?

Closed copy of $[0, 1)$

Any such set won't have a fixed point property

The set $[(0, 0), (1, 0)]\cup \bigcup_n [(1/n, 0), (1/n, 1)]$ of the plane is non-compact with fixed point property

Here $[a, b]$ is the segment from $a$ to $b$

Oh wait the FPP says "continuous map" not "homeomorphism"

I forgot

Yep

My bad, I always think about homeomorphisms

Its alright, help appreciated

So what is your example in the plane?

The one above

@Jakobian here

Ah I see

A while back I was wondering about spaces $X$ such that any nonconstant continuous map $X\to X$ is universal, but the only example I could find is boring (the cook continuum, because the only nonconstant continuous self map is the identity)

A much weaker version is spaces such thay any continuous self map is either constant or surjective

I've discussed some of my investigations into the fixed point property in general topology chat as well. It seemed to me like spaces with fixed point property are mostly compact, found a counter-example of non-compact convex set with fixed point property in $C(\omega_1+1)^*$ with weak$*$ topology, but I was still wondering about Euclidean space case

For a long time there was a famous open question asling whether every tree like continuum has the FPP, but it is not known to be false

It is still open for planar continua (maybe with some extra adjectives) iirc

I found the same problem on some site about continua

web.mst.edu/~continua/Number1.html

Yes it was a very famous problem

At least among continuum theory people

I don't consider myself a continuum theory person

Me neither

:0

That's a surprise

I would say I'm a topological dynamics and descriptive set theory person. But I do look at actions on continua mostly

@AlessandroCodenotti Above example is path-connected. Any ideas on how to get rid of this property?

HAPPY MONDAY!

Happy monday Xander

Let $\varphi$ be defined by $\varphi(x)=\frac{15}{16}(x^2-1)^2$ for $|x|<1$ and $\varphi(x)=0$ otherwise. Let $f$ be a function with a continuous derivative. Find the limit $$\lim_{n\to\infty}\int_{-1}^1n^2\varphi '(nx)f(x)dx.$$

Integrating the integral by parts, I get \begin{equation}\big(n^2\varphi(n)f(1)-n^2\varphi(-n)f(-1)\big)-\int_{-1}^1 n^2\varphi(nx)f'(x)dx.\end{equation}

This is a problem from a section on positive summability kernels, but I have been unable to verify what the kernel is in this exercise, see here (i.e. $n^2\varphi(nx)$ doesn't integrate to $1$ over $[-1,1]$). Moreover, I'm unsure how to approach this problem any further. Appreciate any help.

First, you didn't antidifferentiate correctly. Second, what is $\varphi(n)$?

Ok, so $\int \varphi'(nx)dx=\frac1{n}\varphi(nx)$, is that correct?

You can check easily for yourself.

ok, as for your second question, $\varphi(n)=0$, right? Since $n\geq 1$.

Right.

👍

hmm, so I get, by integration by parts, $$\begin{equation}\big[n\varphi(xn)f(x)\big]_{-1}^1-\int_{-1}^1 n\varphi(nx)f'(x)dx=0-\int_{-1}^1 n\varphi(nx)f'(x)dx.\end{equation}$$ I'm still unsure what the remaining integral evaluates to in the limit $n\to\infty$. If $n\varphi(nx)$ is a kernel, then I have maybe more of a clue.

Pretend you’re a beginning calculus student.

do integration by parts again?

@TedShifrin compact manifold has fixed point property?

ah sorry, no

circle is the obvious example

Ted, was pretending to be a beginning calculus student a hint? :)

the question which compact manifolds have the FPP seems quite complicated

@Thorgott It'd surely be less complex than classification of connected $n$-manifolds

also you don't need to specify compact

every manifold with FPP is compact

For dimension $0$ and $1$ its clear that $\{0\}$ and $[0, 1]$ are the only examples

Warning. To most of us, manifold means without boundary.

Oh, right. Then there's no such $1$-dimensional examples

the only positive examples I know are certain Grassmannians

idk, I think this might be an approachable problem

ah no, I have some more examples

its "approachable" in the sense that you can say things about it

Lefschetz FP Thm might help.

it's not approachable in the sense that there's a complete answer

All examples people list are projective spaces of even dimension, or some variation of that. That's curious

Homology.

uh... nice, someone is serially downvoting me now

it was reversed, never mind

yeah, homology projective spaces in those dimensions work too

if you allow boundaries, anything that's acyclic over some field works

it was reversed but the downvotes are still visible... so...

eh

do the downvotes disappear after a while? Or do I just have to be downvoted

oh okay it just didn't update, I had to click it for some reason

was it a secret sabotage by Ted? I guess I'll never know (I'm joking)

Ted has no idea what’s going on.

I also don't know, I didn't make any enemies recently

I think someone was trying to be malicious is all

I never experienced a serial downvote so I wasn't sure how it works

I have definitely acquired some personal enemies here over the years.

he knows what he did.

It must be time for Munchkin to wreak havoc before descending upon her innocent teachers.

@psie integrating by parts again, I just get that $\int_{-1}^1n^2\varphi '(nx)f(x)dx=\int_{-1}^1n^2\varphi '(nx)f(x)dx$, which is nothing new, so this can't be the right approach...

If $\varphi(s)\geq0$ and $\int_{\mathbb R}\varphi (s) ds=1$, then putting $K_n(s)=n\varphi(ns)$, we get a positive summability kernel, which I believe is the case here.

I think the person retracted downvotes from my questions but kept the downvotes on my answers

Its weird, I think it might be someone here actually

or maybe its multiple people and the other account started downvoting me

@psie so if $\varphi(x)=\frac{15}{16}(x^2-1)^2$ for $|x|<1$ and $\varphi(x)=0$ otherwise, and $f$ is continuously differentiable, then $\lim_{n\to\infty}\int_{-1}^1n^2\varphi '(nx)f(x)dx$ should evaluate to $-f'(0)$.

@psie So what's the basic beginning calculus fact you need to use here?

I'm still unsure about that beginner's calculus fact :) I used a theorem in my book, which is quite involved I think

integration by substitution?

yeah, maybe that, but then you need the dominated convergence theorem, don't you?

Not maybe that. Come on.

If you know blah works for $\phi(x)$, then it works for $n\phi(nx)$.

ah, makes sense

it does?

positive :)

@psie can't you just substitute in this and use Lebesue's dominated convergence theorem?

I think you can, yeah

$\int_{-1}^1 n^2\varphi'(nx)f(x)dx = - \int_{-1}^1 n\varphi(nx)f'(x)dx = -\int_{-n}^n \varphi(u) f'(u/n)du = -\int_{-1}^1 \varphi(u)f'(u/n)du$

and now just take the limit

I'm cross-posting this from the Sand Trap, but:

I don't actually intend to replace math terms, but this is just so fun.