Mathematics - 2017-04-28 (page 3 of 4)

Daminark

7:32 PM

Hey @Akiva!

Akiva Weinberger

@Daminark Hi! Is the decreasing intersection of triangles a (possibly degenerate) triangle?

Or, for arbitrary dimensions, "simplices"

Daminark

Presumably triangle means shaded in here?

Akiva Weinberger

I'm sure it's true, but I'm having a hard time proving it rigorously.

@Daminark Yeah.

Daminark

So basically whether the limit of triangles is also one

If our thing is decreasing

I'm pretty sure the answer is yes

Like, I don't see how triangles can tend to something curved, and there's no way to just add a side

Well, assume you have a sequence of shape "converging" a given one, can we prove that curvature and number of sides of the "limit shape" must appear after finitely many terms

Astyx

The intersection of triangles isn't necessarilly a triangle is it ?

Daminark

7:39 PM

@Astyx We're allowing degenerate cases, like collapsing into a point or line segment

And this is a decreasing sequence, so each triangle is a subset of the one before

Astyx

It's a decreasing intersection of triangles

Not an intersection of decreasing triangles

Or am I saying nonsense ?

Balarka Sen

Yes, it's a decreasing intersection.

Daminark

@Astyx What does "decreasing intersection" mean?

Balarka Sen

Otherwise obviously garbage

@Daminark $\Delta_1 \supset \Delta_2 \supset \Delta_3 \supset \cdots$ is a sequence of triangles

You look at $\bigcap \Delta_n$

Astyx

Well take a sequences of triangles $(T_n)$ and define $I_n = \bigcap_{i=0}^n T_i$

Yeah that

No not that

Daminark

7:41 PM

@Astyx What you're saying is different from what Balarka is

Balarka Sen

Opposite direction

Astyx

But still, why should the triangles be included one into the other ?

Balarka Sen

Because that's what decreasing means?

Daminark

I usually don't hear "decreasing intersection" to mean what you said, it means what Balarka said @Astyx. Plus, What Akiva said becomes immediately wrong if the triangles aren't necessarily nested, so I think we can infer that the interpretation he was going for was that the triangles formed a decreasing sequence

Astyx

Alright, fair enough

But why talk about the intersection then ? Why not just look at $\Delta_n$ ?

Balarka Sen

7:44 PM

The decreasing sequence might not stop.

It does not make sense to talk about $\Delta_n$

If it stops or stabilizes, the intersection is just some $\Delta_k$

Astyx

Oh you meant $\bigcap_{n\in \Bbb N} \Delta_n$ my bad

Balarka Sen

@AkivaWeinberger Was my suggestion about looking at the limit of the vertices and the convex hull useful?

Astyx

Well it has to be non empty and compact

And convex

So yeah, looking at the limits of the vertice would give you sufficient information I guess

Balarka Sen

I was trying to write some stuff down here

Astyx

And I guess you could prove they have to converge

Balarka Sen

7:47 PM

Yeah, that is easy.

Because you get a monotonically decreasing sequence (in terms of distance) inside a compact domain (the biggest triangle).

Astyx

so the intersection is a triangle

Balarka Sen

No, why? The limit of the vertices are three points $P, Q, R$. Why's their convex hull the intersection?

Astyx

Because you can get as near of these points as you want withthe vertice of actual triangles from the intersection

So anything out of their convex hull will be removed at some point

And anything inside has to be inside because each triangle is contained in the preceding one

Balarka Sen

I don't understand your argument. $x \in \Delta$ be an element in the convex hull of $P, Q, R$ (the limit vertices). $x$ is a limit point of $\bigcap \Delta_n$ by what I wrote in the comments... oh, because $\bigcap \Delta_n$ is compact that means $x \in \bigcap \Delta_n$.

I think?

Astyx

Yup

Balarka Sen

7:52 PM

@Akiva ^^^

Evinda

Hello!!! How do we deduce from the fact that the period of cos and sinx is $2 \pi$, that the functions $\sin{\alpha x}$ and $\cos{\alpha x}$ have period $\frac{2 \pi}{\alpha}$ ?

Astyx

Look what $\sin(\alpha (x+{2\pi\over \alpha}))$ is

I think you can make it easier by just worrying about the points on the boundaries @Balarka

Evinda

It is equal to $\sin{\alpha x+ 2 \pi}$ @Astyx

Balarka Sen

I am a little angry that @Akiva didn't look at my argument, which was more or less literally the proof, thanks to @Astyx :P /rant.

Astyx

And proving that these are in the intersection (otherwise the vertice of the intersection are some positive distance away from the triangle, which is absurd)

Not much really, I nodded at what Balarka said :p

Akiva Weinberger

7:57 PM

Oh, you mean what you're writing right now?

I didn't see that until just now

Balarka Sen

It's the same argument provided here. The last statement is easily justified by just saying $x$ is a limit point of $\bigcap \Delta_n$, hence by compactness is in $\bigcap \Delta_n$.

Akiva Weinberger

@BalarkaSen I don't understand the argument

Astyx

Neither do I

Balarka Sen

Which bit?

Astyx

Why it's a limit point

Daminark

8:00 PM

That bit

Over there

:P

Akiva Weinberger

Why would $x$ be a linear combination of the $P/Q/R_n$? Besides, that would show that $\Delta\subseteq\bigcap\Delta_n$, but we need to show the other direction as well.

@BalarkaSen

Astyx

I think dealing with double inclusion is more appropriate

You know the vertice of you triangle are in the set because it's a decreasing sequence of compacts

You elso know it's convex because it's compact

Now take a point outside the convex hull of your three vertice

Balarka Sen

@AkivaWeinberger $x$ is in the convex hull of $P, Q, R$. So that's a convex combination of $P, Q, R$. Agree?

Akiva Weinberger

Sure.

Astyx

Get an actual triangle close enough to your convex hull

the point isn't inside so it's not in the limit

Thus the limit is the convex hull of your three vertices, ie a triangle

Akiva Weinberger

8:03 PM

And I guess you could construct a sequence of $x_n$ with the same barycentric coordinates in $\Delta_n$

Balarka Sen

Let $x_n$ be the convex combination of $P_n, Q_n, R_n$ with the same coefficients.

As $P_n \to P, Q_n \to Q, R_n \to R$, $x_n \to x$. Is that right?

Akiva Weinberger

@Astyx Isn't inside what?

@BalarkaSen OK

OH!

Astyx

the triangle of which the vertice are close to your limit vertices

Akiva Weinberger

@BalarkaSen Choose an $N$. Then $x_{N+n}$, $n\ge0$ is a sequence in $\Delta_N$ that goes to $x$, so by closedness $x\in\Delta_N$. Repeat for all $N$ to see that it's in the intersection. Is that what you wanted to do?

@Astyx Why isn't the point inside it?

Balarka Sen

@Akiva Yeah, exactly.

Should have written it like that.

Astyx

8:08 PM

Cause it's not in the convex hull of the vertice of the triangle, since the limit set is compact, the point has to be some distance away from it, and the limit set is also convex

Balarka Sen

@Akiva In fact, by your logic, $P, Q, R$ are all in $\bigcap \Delta_n$ (because it's a limit of $P_n, Q_n, R_n$, which, for $n > N$, are all in $\Delta_N$ - for each $N$.) EDIT of course that's a special case of Delta being in it.

So I think to show the other inclusion one should be able to do some convex hull is unique argument... thinking

Adeek

hey

Akiva Weinberger

@Astyx I guess you're kind of implying that if $d(P',P),d(Q,Q'),d(R,R')<\epsilon$, then $\Delta'\subseteq B_\epsilon(\Delta)$

Astyx

If I get what these notations mean, yeah I am

Except not really the same $\epsilon$ each time

Akiva Weinberger

$d$ is distance, $B_\epsilon$ is the set of things within $\epsilon$ of the set, and $\Delta'$ is the triangle with vertices $P',Q',R'$

Astyx

8:14 PM

Oh yeah I guess it's also true when it's the same $\epsilon$

Akiva Weinberger

Ah, yeah, 'cause $B_\epsilon(\Delta)$ is convex

(In fact, the $\epsilon$-neighborhood of any convex set is convex, right?)

Ran Wang

(Good question.... I want to think about this for a while... Don't mind me, I am just passing by)

Akiva Weinberger

And clearly $P',Q',R'\in B_\epsilon(\Delta)$.

Astyx

Oh actually you could probably see it only considering the halfplanes delimited by the edges of your final triangle

Maybe, and then it becomes a problem of continuity of scalar product and closeness of $\Bbb R_+$

Akiva Weinberger

So their convex hull, $\Delta'$, is in $B_\epsilon(\Delta)$.

Sophie

8:18 PM

Is there an easy way to describe what number are of the form $a^4+4b^4$ in the same way there is for numbers of the form $a^2+b^2$?

Akiva Weinberger

@Astyx I'm not sure I get what you're saying

@Sophie Hint: there's an identity named after you!

Sophie

I tried that, it didn't help a lot

Akiva Weinberger

Oh :(

I dunno, then

Sophie

I want to know for which primes $p$ there is an integer $q>1$ so that $pq^2=a^4+4b^4$

Akiva Weinberger

@Astyx @BalarkaSen But, in any case, that lemma finishes it. If any point $x$ has distance $\epsilon>0$ away from $\Delta$, consider the $\Delta_n$ whose vertices are less than $\epsilon$ away from the vertices of $\Delta$; the lemma says that $x\notin\Delta_n$.

Technically, there's one more piece left, which is assuring that $P_n,Q_n,R_n$ actually converge to something in the limit. But that's fine by the fact that, in $(\Bbb R^2)^3$, the set $(\Delta_1)^3$ is compact (and thus sequentially compact).

Astyx

8:23 PM

If a point $x$ is not in $\Delta$, if you take $u, v, w$ unitary vectors perpendicular to each edge of the convex hull of $P, Q, R$ pointing inwards the triangle from the edges (which can be obtained as a continuous function of $P, Q, R$) and any point $a, b, c$ of the edge, then being in the triangle is satisfying $\langle u, x-a \rangle \ge 0$, $\langle v, x-b \rangle \ge 0$ and $\langle w, x-c \rangle \ge 0$

Balarka Sen

Ok, I see it.

@AkivaWeinberger Yeah that's easy.

Astyx

And everything is continuous

Akiva Weinberger

(And by the fact that you can delete co-infinitely many elements of a decreasing sequence of sets without changing the intersection)

Balarka Sen

@AkivaWeinberger How did you conclude this fact, by the way? I haven't been following the discussion.

Akiva Weinberger

The lemma is:

12 mins ago, by Akiva Weinberger

@Astyx I guess you're kind of implying that if $d(P',P),d(Q,Q'),d(R,R')<\epsilon$, then $\Delta'\subseteq B_\epsilon(\Delta)$

Astyx

8:25 PM

Actually with that approach it feels quite easy ?

Akiva Weinberger

The proof is that $B_\epsilon(\Delta)$ is convex, so

$P',Q',R'\in B_\epsilon(\Delta)$ by the condition, and thus their convex hull is in it as well.

Balarka Sen

Ah ok, I see.

Astyx

If you take $u, v, w$ unitary vectors perpendicular to each edge of the convex hull of $P, Q, R$ pointing inwards the triangle from the edges (which can be obtained as a continuous function of $P, Q, R$) and any point $a, b, c$ of the edge, then being in the triangle is satisfying $\langle u, x-a \rangle \ge 0$, $\langle v, x-b \rangle \ge 0$ and $\langle w, x-c \rangle \ge 0$. @Akiva @Balarka What d'you think ?

Akiva Weinberger

Makes sense. (And those inner products are clearly independent of the $a,b,c$, which is nice.)

Balarka Sen

That's a cool proof of the opposite inclusion, @Akiva, @Astyx.

Astyx

8:27 PM

I think what I said proves both inclusions

Semiclassical

hi chat.

Balarka Sen

What you just wrote now?

Astyx

Yes

Balarka Sen

I'm still reading it.

Astyx

Hi @Semi

Balarka Sen

8:28 PM

Ah, yes, that is true.

Astyx

In short, just consider the triangle as the intersection of three half planes

And see the limit of a half plane is a half plane

Balarka Sen

Yeah. Excellent.

Ran Wang

hmmmmm guys, can I ask suggestions? I am teaching very basic general topologies to students from probability and statistics, and want to introduce the uniform space. I have decided to start the thing by introducing the metric space first. However, now I am not entirely sure what will be an appropriate way to motivate uniform spaces.

Astyx

Right cool :) time to sleep now, See ya !

Ran Wang

(ignored....)

Astyx

8:33 PM

(don't know much about unfirom spaces sorry :/ @RanWang)

Ran Wang

(No problem~ Just kidding~ have a nice sleep~)

Balarka Sen

Good night.

Ted Shifrin

@Balarka: Any luck with that strange topology?

Daminark

Uniform spaces as in, like a space of functions equipped with the sup norm? @Ran

Ran Wang

To add a few context, I want to introduce the entourage definition. From the definition it is clear a generalization of the metric, but what exactly is that it is trying to overcome? I can think of weak convergence as examples, but how do I "derive" from the definition of weak convergence the definition of uniform space? Or maybe I should introduce the file:///C:/Users/Gebruiker/Downloads/9781461450573-c1.pdf?

Balarka Sen

8:38 PM

@Ted I stored it on my thinking list. Was busy figuring out a problem by Akiva with Astyx. Let me get back to thinking about it.

Ran Wang

Thank you very much~

Ted Shifrin

No problem. I don't think I have any clue. My guess is it doesn't happen.

Ran Wang

Not really the sup norm. I think by sup norm you mean continuous function on compact. Then the space is actually metrizable.

What I want to introduce at the end, is the topology is convergence on compacts

PVAL-inactive

Hi @Ted

Hi @Balarka

Ted Shifrin

heya @PVAL

Ran Wang

8:40 PM

Hi @PVAL-inactive

Balarka Sen

Hi, @PVAL

Daminark

Ah, well in this case I shall probably defer to people who know what they're talking about. I only know topology in the context of analysis, so metric spaces and uniform convergence

Ran Wang

Hmmmm how did I see a PVAL-inactive??

Daminark

Hey @PVAL!

PVAL-inactive

@Daminark Hiya

Ran Wang

8:41 PM

No problem, thanks @Daminark

Daminark

How's it going?

Ted Shifrin

In my 45+ years of doing mathematics, I've never done anything with uniform spaces (other than the usual metric ones).

PVAL-inactive

Well convergence on compact subsets is like a uniform topology right.

Pretty sure you need to undestand that to do a lot of even basic complex analysis.

Ted Shifrin

well, that's all coming from metric ...

Daminark

@PVAL Rang was asking about more general uniform spaces

PVAL-inactive

8:43 PM

Yeah but convergence on compact subsets shouldn't be metrizable.

I think

Like the topology for that convergence should be uniform but not metrizable

Ran Wang

Nope, it is not metrizable in general

PVAL-inactive

or am i missremeberig.

Ted Shifrin

Yeah, OK, I've taught function spaces a few times. In my research life I've never cared about this degree of generality.

Ran Wang

The problem is that there is this book by Le Cam, which basically built on the language of uniform space...

(and a lot of other weird things)

Ted Shifrin

Oh, LeCam from Berkeley?

Ran Wang

8:44 PM

yep

PVAL-inactive

I don't even remember the actual names of the A-A theorem let alone the statement, the idea of the proof or the proof.

Ted Shifrin

Yeah, in the stat dept. I actually went to elementary and junior high school with his son. ...

PVAL-inactive

Maybe if I saw it in more generality I would have.

Ran Wang

!!!!!!!!!!!!!!!!!!!!

Akiva Weinberger

@TedShifrin Did my proof of the Rising Sun lemma make sense?

Ted Shifrin

8:45 PM

Arzela-Ascoli, @PVAL?

Daminark

Arzela-Ascoli?

Ted Shifrin

I haven't looked, DogAteMy. I will soon.

Daminark

@Ted ninja'd

PVAL-inactive

@TedShifrin That's the one.

Ran Wang

But this is a valid concern. I might just skip the part on it and continue to general topology directly

Akiva Weinberger

8:46 PM

From here

Daminark

I saw only what I've heard is the noob version of it, which is that an equicontinuous and pointwise bounded sequence of functions has a subsequence converging uniformly. Basically, you use a diagonal argument to get that you can do this on any countable set, and then use equicontinuity and separability of compact sets to get that it works on the whole

Ted Shifrin

Of course, now I need to think about it all again, DogAteMy. Haven't since the last time I taught that class 6 or 7 years ago.

Daminark

Or at least that's how I think it goes

PVAL-inactive

@Daminark That sounds like the version needed to do the constructions in complex analysis I've seen it used for.

Ran Wang

Anyway, thank everybody for the suggestions. I will think about this.

Daminark

8:49 PM

Ah, neat. That's the version done in Rudin

Which constructions use it?

Eric

hi again chat

Balarka Sen

@Ted Actually, remind me what your question was. A leaf space with a dense point? That's easy enough, eg irrational slope foliation on T^2.

Daminark

The only time I remember using it was to show that an operator was compact if and only if its dual was. Schlag only asked us to do Hilbert spaces but mentioned that the general version for Banach spaces used that, so that's how a few of us did it

Balarka Sen

Or was it something else

Eric

@Daminark the most general version of Arzela-Ascoli I've seen is actually not really more general than that version

Daminark

8:52 PM

Which did you see @Eric?

Ted Shifrin

Hi again, @Eric

Eric

it was like sequences of equicontinuous functions from separable metric spaces to compact metric spaces converge uniformly on compact subsets or something

Ted Shifrin

Um, @Balarka, I wanted a topology on an interval with a point $P$ so that $P$ is in every neighborhood of all other $Q$.

PVAL-inactive

It comes up in en.wikipedia.org/wiki/Montel%27s_theorem which I think comes up in the proof of classification of holomorphic singularities.

(in particular that around an esssential singularity the image is dense).

Eric

I remember needing it do something with geodesics or something @Daminark

Daminark

8:54 PM

Whaaaaaaa?

Balarka Sen

@TedShifrin That also sounds not possible to me.

Ted Shifrin

I think you're still reversing the question with your dense point.

At any rate, unless you can give me an example, it shows you can't get every topological 1-pseudo-manifold :P

Balarka Sen

Wait, why?

PVAL-inactive

@Ted Pretty sure the leaf space gotten from an irrational slope on the torus/cyllinder works for that.

Balarka Sen

I am really confused by how Ted is parsing his question.

Danu

8:58 PM

o/

PVAL-inactive

in that case the space has the indiscrete topology I believe.

Balarka Sen

Yes, the leaf space of irrational slopes is indiscrete.

Ted Shifrin

I don't think so, @PVAL.

Danu

Anyone have any neat lectures on mathematics that I can watch tonight? I'm not capable of doing any work of my own right now

Balarka Sen

That's why I was giving it as an example of dense point before.

Daminark

8:58 PM

@Danu See if you can find lectures by Federer on Youtube

PVAL-inactive

@Ted Every point $P$ is in every other neighborhood of point $Q$.

Danu

On what?

Daminark

Geometric measure theory

Danu

Doesn't sound like the kind of thing I'm interested in hearing about right now... Any other suggestions?

(sorry if I come across as rude)

Daminark

I kid, that's probably a bad idea, he's apparently not the best expositor.

Ted Shifrin

8:59 PM

Oh, maybe I just want $P$ in some nbhd of every $Q$, sorry.

Danu

I can watch some Federer though...

Daminark

In reality, I mean, what types of things are you interested in?

Balarka Sen

But if you have $P$ in every nbhd of every $Q$, you automatically have it in some nbhd :P

Danu

Roger Federer vs Rafael Nadal - Australian Open 2017 Final (highlights HD)

►

Daminark

Kek

Danu

9:00 PM

@Daminark Topology/geometry of smooth manifolds and complex geometry mostly

PVAL-inactive

@Ted I agree with @Balarka that statement is vacuous.

Danu

Here's a question

Ted Shifrin

I meant some but not all, guys.

Danu

Any $S^2=\Bbb P^1$ bundle is the sphere bundle of a rank 3 real vector bundle. Under what conditions it lifts to a rank 2 complex vector bundle is not so hard to find out (sheaf cohomology). But what can you say about the holomorphic case? I'm asking under which conditions a holomorphic $\Bbb P^1$ bundle lifts to a holomorphic rank 2 bundle

PVAL-inactive

You can watch the first lecture of the Freedman Bonn seminar

Balarka Sen

9:02 PM

Oh. Hm.

PVAL-inactive

That's one of the few things I have watched online

Danu

@PVAL-inactive I'll try to find that

PVAL-inactive

that I really enjoyed

I've probably watched very few lectures online.

Danu

I can't find it by searching for those words. Any guidance?

Ted Shifrin

DogAteMy: I need to read your argument more carefully when I'm not distracted, but I doubt looking at one value will do it.

PVAL-inactive

9:04 PM

ugh

maybe it wasn't bonn

mpim-bonn.mpg.de/node/4436

There's the lectures

Danu

By the way, does Freedman still do any mathematics? I have a friend who is doing an internship at this qubit research group that he's affiliated with

@PVAL-inactive Great. Thank you so much!

PVAL-inactive

they maybe go off the deep end a few lectures in (its not clear how much he went through the technical details of proofs he and Bob Edwards wrote 30 years ago).

Danu

Haha I love how he adresses the camera :D

Balarka Sen

@Ted If you want $P$ in some nbhd of every $Q$, that should be trivial. Just take the trivial foliation on $T^2$ by parallel circles, for any two parallel circles you have an nbhd containing them both. I don't get it.

PVAL-inactive

@danu he wrote a very very accessible paper (joint with M.Scharleman ) and posted it a week or so ago.

Danu

9:07 PM

@PVAL-inactive Huh, weird that I didn't see it (was it not in AT,GT?)

PVAL-inactive

It was so accessible that I essentially looked at it and read it.

Danu

hahaha

PVAL-inactive

It was in GT

Danu

There has to be a good joke in there somewhere

Weird, I must've missed it :\ Thanks for the heads up

PVAL-inactive

arxiv.org/abs/1704.05507

Ted Shifrin

9:08 PM

I dunno, Balarka. I quit.

Danu

Thanks

So he still does math

I wonder why he stopped being an "official mathematician" at all, since he seems to have had such great success

Balarka Sen

I think your initial question was to have a leaf $L$ such that for every leaf $L'$ any nbhd of $L'$ contains $L$ (in the leaf space). In which case, irrational slopes is an example.

Ted Shifrin

A number of great mathematicians (e.g., Simons) have moved on to other more lucrative things. I know others, too.

Balarka Sen

But yeah I dunno

Akiva Weinberger

By the way, interesting construction: Take a triangle, and for each vertex $v_i$ draw a segment to a point $p_i$ other side. (The three segments need not intersect at a point.)

Then create the triangle with the $p_i$ as the vertices, and repeat (using the same lines as you started with).

PVAL-inactive

9:12 PM

@Danu The kind of math he did (Bing-style topology) has fallen out of favor.

Danu

@TedShifrin Heh, yeah Simons is rich^rich. I saw some talk he gave recently. I must admit that he didn't seem the most likeable kind of guy. Kind of... business-y...

Akiva Weinberger

Pretty sure: If the segments concur, the limit (intersection) of the sequence of triangles is the point where they concur. Otherwise, limit is the triangle bounded by the three original line segments.

Danu

@PVAL-inactive I'm not very aware of the kinds of topology that exist. What is Bing-style?

Balarka Sen

Bing shrinking, etc

Danu

I don't know what that is.

Akiva Weinberger

9:13 PM

Bing shrinking

In geometric topology, a branch of mathematics, the Bing shrinking criterion, introduced by Bing (1952), is a method for showing that a quotient of a space is homeomorphic to the space. == References == Bing, R. H. (1952), "A homeomorphism between the 3-sphere and the sum of two solid horned spheres", Annals of Mathematics. Second Series, 56: 354–362, ISSN 0003-486X, JSTOR 1969804, MR 0049549...

Thanks, Wiki

PVAL-inactive

Some people blame some of this falling out on the inaccessibility of some of the proofs written by Edwards, Freedman and Quinn.

Studentmath

Prof @Ted, @Balarka and all \o

Balarka Sen

Me neither! I think there are some notes by Freedman flying around somewhere

Akiva Weinberger

Well, the reference is probably good

Ted Shifrin

heya @Studentmath

PVAL-inactive

9:14 PM

But I won't state my opinion on this.

Danu

@PVAL-inactive ha-ha.

Studentmath

Wie geht's?

Balarka Sen

Hi @Studentmath

Akiva Weinberger

Oh, @MikeMiller mentioned that sort of thing a while ago I think

PVAL-inactive

@Danu If you watch that lecture you'll get a good introduction to Bing style topology.

Danu

9:14 PM

So 4-manifold topology is kinda crazy huh

Hi Mike

Akiva Weinberger

A homeomorphism between the 3-sphere and the sum of two solid horned spheres, I mean.

Balarka Sen

The double of the gored ball is S^3, yeah

pretty cool result

I have 0 idea why that is even a manifold

PVAL-inactive

I went through the proof of that in a talk.

Danu

@PVAL-inactive I'll definitely give it a go

PVAL-inactive

Essentially the idea is if you look at a sequence of closed subsets R_i of a product, and the diameter of the preimages {x|(x,y) \in R_i} for fixed y and images {y|(x,y) \in R_i} are both getting small when you take i large and the projections are surjective onto each coordinate.

Then the only possiblity for $\cap R_i$ is the graph of a homeomorphism.

Balarka Sen

9:19 PM

@Ted How does one prove that if $f : M \to M$ is an isometry, then the fixed point set of $f$ is totally geodesic?

PVAL-inactive

So what Bing did was show that the map from S^3 to the double of the Alexander horned ball which just quotients out the right thing can actually be modified to have arbitraily small preimages and images , and then applied the previous result to neighborhoods of these graphs.

Ted Shifrin

what's your defn of totally geodesic, @Balarka?

PVAL-inactive

(The sequences should be nested).

Balarka Sen

@Ted If every geodesic on the submanifold is also a geodesic on the ambient manifold

PVAL-inactive

for my above comments everything is happening inside some compact metric space.

Balarka Sen

9:25 PM

Interesting, @PVAL. The idea does not sound natural at all to me though!

PVAL-inactive

Well it becomes very geometric when you are actually trying to modify these preimage sets.

Also Bing was a very smart guy.

Balarka Sen

I see. How do you see the map S^3 to the double of the ball as a quotient map?

lol, true

Ted Shifrin

@Balarka: Isn't it just that if not you'd violate uniqueness of geodesics in the big manifold?

Balarka Sen

@Ted AH! Yeah, isometry sends geodesics to geodesics.

Ted Shifrin

nods

Balarka Sen

9:27 PM

Cool, thanks!

PVAL-inactive

The Alexander Horned ball is B^3 modulo some intersections of nested links of cylinders.

If you think about it you can see this though its probably not immediately obvious from the intersection definiton.

If you look at figure 1 here ams.org/journals/tran/1965-116-00/S0002-9947-1965-0185585-9/…

Mike Miller

There are still people (even young people) who understand the Freedman constructions. I don't at all though.

PVAL-inactive

I understand all the constructions of surfaces (casson towers and capped gropes and the like), I haven't met anyone who knows how to turn a capped grope into a topological disk.

at least I certainly know people who completely understand how to deal with the combinatorial machinery.

I don't know if many people understand the Bing topology aspect.

Mike Miller

I don't know people's work or understanding well enough to say.

PVAL-inactive

Lots of people working on topological concordance theory understand how to find towers of gropes and casson handles.

Balarka Sen

9:53 PM

@PVAL-inactive Sorry, I was away. I see the picture now.

PVAL-inactive

Well the thing getting modded out is the components of intersections of those nested cyllinders

in other words if two points are in the same sequence of infinite nested cyllinders they are identified.

Balarka Sen

That makes sense

PVAL-inactive

I remember spending a few hours in front of a chalkboard trying to convince myself this was equivalent to the usual embedding definition.

I don't know a very obvious way to see it.

Balarka Sen

Yeah it's not very clear to me, but I'm believing it

ALannister

In $\mathbb{Z}[\sqrt{-5}]$, how do I show that $2+\sqrt{-5}$ is not divisible by $3$?

PVAL-inactive

10:03 PM

It should be sort of obvious that you can't modify this quotient map from B^3 \to in any reasonable way(isotopy preserving the boundary for instance) which make all the inverse images small as at least one cyllinder at any stage basically reaches at least the center of the first cyllinder.

So the inverse sets are going to be "long" and can't be made to have small diameter.

ALannister

I've looked at so many examples of showing that $\mathbb{Z}[\sqrt{-5}]$ is not a PID and something like this is always mentioned, but it's just glazed over and never worked out in detail.

PVAL-inactive

But when you double these cyllinders will turn into tori and in that case Bing came up with a really nice way of making these inverse images small, by making the diameter of the nested tori get smaller and smaller as you get to a very deep stage.

@BalarkaSen The pictures of that are in figures 5a-5d here jstor.org/stable/pdf/1969804.pdf

(wrong paste)

ALannister

10:19 PM

0

Q: Showing that $2+\sqrt{-5}$ is not divisible by $3$ in $\mathbb{Z}$

I'm trying to show that $\mathbb{Z}[\sqrt{-5}]$ is not a principal ideal domain by showing that $3$, which I've already shown is irreducible in $\mathbb{Z}[\sqrt{-5}]$, is not prime there. To that effect, My goal is to show that although $3$ divides $(2+\sqrt{-5})(2-\sqrt{-5})=3^{2}$, neither of...

LegionMammal978

Hmm

Not sure how I'd apply the DFT to retrieve non-linearly-spaced frequencies from a waveform

Akiva Weinberger

10:57 PM

@BalarkaSen @Astyx By the way, there's a generalization of the triangle thing. I'm sure you could solve it trivially by modifying the proof or something, but I haven't thought about it.

But, in any case, the decreasing intersection of polygons of at most $n$ sides (not necessarily convex) should be a polygon of at most $n$ sides.

Note that you need to bound the number of sides, or else you can easily get circles and curves and whatnot. In any case, something to think about.

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