(the path component of $x$)

Can you prove it is open? Can you prove its complement is open?

(Recall that open means there's a small ball around each point…)

I honestly... started learning today some stupid damn theorem and there was ONE word that made me COMPLETLY google something different the whole day and i am sitll not out of that frkn Spiral!!

One thing to the other

@MadSpaces are you a math student?

@BalarkaSen So one thing I can start thinking about is the complement of $T^2\times I$ in $D^4$… yeah?

'Cause $\partial(T^2\times I)=T^3$

so I'll hope that that complement is three slabs

but… it isn't though

Haha, yeah.

But maybe modify the idea a little bit

You mean $T^2 \times D^2$

Oh, yeah

@BalarkaSen Right, that doesn't even embed in $D^4$

But it's the correct dimension if you're thinking explicitly about 4-manifolds

@AkivaWeinberger It does though

$T^2$ embeds in $D^4$, then just take a regular neighborhood.

That's $T^2 \times D^2$.

Oh

Ah, right, of course

Or better yet, $T^2\times I$ embeds in $I^3$, and multiply both sides by $I$

($I=D$)

Still doesn't seem so helpful though

I guess I want something else that has $T^3$ as its boundary?

@BalarkaSen I'm still having trouble with this

Should I give another hint

Sure

You were trying to remove all three slabs or attach all three slabs. Do a mix, remove two slabs and attach one slab. The boundary is still the same manifold, 0-surgery on the Borromean link.

Prove that the resulting 4-manifold is $T^2 \times D^2$

Bonus hint: What is the 4-manifold given by removing a single slab from $D^4$?

@JoeShmo no

@BalarkaSen $S^1\times D^3$?

Yeah

I'm imagining a movie where it's a cube at every frame except around $t=0.5$ it's two half cubes

I suppose I can do the same but around $t=0.5$ it's three third-cubes. That gives me a solid double torus times $D^1$

@AkivaWeinberger Nice picture. I think of it as, digging a tunnel is the same as making an arch.

So digging a 2D slab is the same as adding a 1D arch

i.e. removing a $D^2 \times D^2$ is the same as attaching $D^1 \times D^3$

I see it

I feel like what I'm missing here is mainly formalism

Yeah, which is understandable. This is like a first exercise in Kirby calculus.

I knew you'd get it regardless of formalism

@BalarkaSen I didn't though

You will eventually.

"I see it" = I see the thing you just said about slabs and arches

We're doing 4-manifold topology, we can see 3+1 dimensions

@AkivaWeinberger how do you know that set is not empty

Time is immaterial

@BalarkaSen I appreciate the trust you have in me

I am guessing since a point in the first set has a ball around it.But what if the balls elements are not connectable

If something happens in future it has already happened

@MadSpaces Which set?

@AkivaWeinberger the set contains all points that have path to some point in the open set

Looking for artist mathematician to write this in an elegant and short way

ís that a matrix

what do you think?

then just 8* $I_{\text{numberofrows}}$

diagonal matrix.

@MadSpaces It contains $x$ itself

The constant path connects every point to itself

@AkivaWeinberger thats big brain, ty

Was right...

Incidentally, you can prove something stronger: if an open subset of the plane is connected, you can join every pair of points by a polygonal path (a path made up of finitely many straight line segments)

Oh because each point has a ball around it. i see, take the Radii as polygon parts

Yah. The key is, not only is the path component open, it's closed as well (which you can prove by showing that the set of points not path-connected to $x$ is also open)

("Closed in $X$", where $X$ is a subset of the plane, means it contains all limit points that are in $X$. Equivalently, it's the complement of something that's open in $X$)

@MadSpaces thanks

@CroCo $8I_8$

What are proper common notations for points may they be in sets.. i know of p and q .. what else if you used those?

$p$ is a prime and $q$ is a qrime

(jk)

In geometry they like capital letters (consider the angle AOB etc)

Yeah usually p, q, x, y, z, w, maybe u and v @MadSpaces

I mean you have freedom over it, of course

Yea, sure, i am just seeing whats common

Sometimes basepoints are denoted by $*$. Like, "let $*\in X$ be a point, consider the set of all loops starting and ending at $*$" etc

If $X$ is a topological space with base point $*$, $\Omega X$ is its loopspace, the space of all loops starting and ending at $*$

and $\Omega X$ itself can be given a base point, the constant loop at $*$

meaning you can talk about $\Omega^2X$

You can then ask things like, if $X$ is an annulus, is $\Omega X$ connected (no), is $\Omega^2X$ connected (yes), etc

or if $S^2$ means the sphere, is $\Omega X$ connected (yes), is $\Omega^2X$ connected (no), is $\Omega^3X$ connected (surprisingly, no), etc

Lots of weirdnesses if you go down that path

(Should have written $\Omega S^2$, $\Omega^2 S^2$, and $\Omega^3S^2$ there)

Are you quoting my ex girlfriend talking about dating me`?

"lots of wierdness going down that path"

Can we say, that for a point in a boundary of a set, applies that for each epsillon ball, the ball is not contained in the set, but it intersects not empty with it?

A boundary point is one for which every epsilon ball intersects the set but none are contained in it, yes

Every epsilon ball intersects both the set and the set's complement

Yes. i wasnt bothered to try to proof, looked obvious. thanks for confirmation

For example, if $X=\Bbb Q\subset\Bbb R$, the set of rationals, what is the boundary set of $X$?

@AkivaWeinberger I guess a corollary of the above is Borromean link is not an unlink. If it was, then there would be a diffeomorphism between $T^3$ and some surgery (possibly not the 0-surgery) on an unlink with three connected components in $S^3$. That would be a decomposable 3-manifold, but $T^3$ is prime/irreducible.

@Akiva Sorry didnt see your message. Q is thicccc in R, so Fr would be R

Yeah (Fr stands for frontier?)

Yes

The frontier is everywhere - Carl Sagan

@BalarkaSen did you see arxiv.org/abs/2205.15283

Nope

It made me think of you for some reason.

Haha, thank you for sending me the paper. I will have a look.

Finally pushed through the first chapter of this 4-manifold topology book

Which one?

4-Manifolds and Kirby Calculus by Gompf and Stipsicz?

Does that one trump Kirby's original?

Freedman and Donaldson also had books, too. I would assume both are actually decent.

Speaking of Freedman, there was also the disk embedding theorem book which is outstanding, but also intimidating.

@BalarkaSen Was my guess correct

@anak @AkivaWeinberger Akbulut's "4-Manifolds"

Ah

Although I look at G-S from time to time too, and Kirby's original

Don't ask if you can do 4-manifolds

Ask what manifolds can do 4-you

Lol

G-S is too thick mostly

@anak Dude I tried to read some smooth 4-manifold topology and got screwed

I will return to it at a later date

Way too much analysis

@BalarkaSen Ah, my university's library has it

Cool, join me

Being the summer, though, I think I will have to wait till the semester starts again

I am actually on campus again for the weekend, but I don't think I have access to the buildings

I was making some notes on smooth 4-dimensional topology techniques but stopped abruptly. Here and subsequent posts.

or if the libraries are open

You have a Wordpress called Withered Stumps?

yeah lol

Akiva where are you in school now?

@JoeShmo I took a gap semester for COVID so I'm about to start the second half of my junior year

(undergrad)

Three semesters left

where

you should acquaint yourself with Sylvain Cappell, if youre seriously into topology

at NYU

@JoeShmo Yale

oh nice

@JoeShmo I live in Brooklyn, so that might be nice

(assuming he is also in NY over the summer)

right I thought you were somewhere in NY

Yeah if you could introduce us via email that'd be really nice, I'm at akiva.weinberger@yale.edu

you got it

Thanks!

@JoeShmo What about you?

What do you do with your time (if you're comfortable sharing)?

I went to not Yale

I did my masters at NYU in math

my undergrad was not in math

how gentlemanly of them to give you firstname.lastname@school.edu

i guess yale is kind of a small school, which might help

that reminds me, when more people started getting email in the 90s a lot of universities had pretty chaotic rules around who got what address

you'd have peoples initials, numbers, sometimes student ID numbers, sometimes weird alphanumeric prefixes indicating student vs. faculty vs. staff

I'm surprised he got first.last@yale

Yeah that's how Yale does it

My brother, who did NYU (film school), got initials###@nyu I think

Like, initials plus some ID number

yes

Im ds3837@nyu.edu

it also seemed more common then for cs/math/etc departments to have their own email/IT infrastructure. they have phased this out at a lot of schools

Ted is last@uni I think

I think NYU still has that

causing collisions in the name space

I hope that isn't too private information

so blahblah@cims.nyu.edu

(Courant Institute of Mathematical Sciences)

some people have 3 different NYU emails addresses

one from their student days

another from theprfessorships at nyu

and a third one from being members of courant

when i was at iowa they began taking our own IT away bit by bit. we still had our own email server but we'd get these silly warning emails about how they weren't our official emails and to check the new ones because that's where all the important stuff would arrive

I had a high school email once upon a time but they deleted it

I think they Yale emails will be for life though

then on their new centralized email system some person somehow managed to send an email to all faculty and staff without anyone having intended them to have the access to do that, and all the addresses were in the To: line so there was a multi-day cascade of people saying TAKE ME OFF OF THIS

i think i only have an alumni address for undergrad now, although the address i actually used as an undergrad stayed active far longer than they intended it to

like 10 years longer

I assume emailing all@yale.edu has no reason to work, but I've technically never tried it…

i would cry myself to sleep the day my uni removes my email account, but once again, that would be my usual sleeping routine.

On that note, good night

Night

Take care

DogAteMy, if you wish, you can also mention me. Cappell has known me for 40 years or so (although it's been ages ...).