Mathematics - 2020-05-06 (page 3 of 3)

Jasper

18:00

@CaptainAmerica16 I am not even sure what it treats that Spivak does not.

CaptainAmerica16

idk either

Akiva Weinberger

@BalarkaSen I thought Jordan–Schoenflies fails in higher than two dimensions

Horned spheres and all that

Alessandro Codenotti

Let 🗺️ and 🗾 be charts

Balarka Sen

That's not locally flat though

Akiva Weinberger

What's locally flat?

Is that like having a collar?

Balarka Sen

18:02

Bicollar yeah

Akiva Weinberger

Ah

Sounds good to me then

Jasper

@AkivaWeinberger There is the generalised Schonflies theorem. Covered in Bredon's Topology and Geometry, together with generalised Jordan curve theorem. In n dimensions. Look at the exact statement there if interested.

Balarka Sen

generalized schoenflies theorem is usually the statement that spheres separate

locally flat implies connected components are balls is the schoenflies conjecture, proved by Morse-Mazur and Brown independently

Jasper

Also, there is a graph theoretic proof of the Jordan Schonflies theorem. Look at the paper by Carsten Thomassen: The Jordan Curve Theorem and the Classification of Surfaces. There he also proves that every surface is triangulable.

This proof is also in the book Curves and Surfaces by Marco Abate and Francesca Tovena.

Akiva Weinberger

Is it actually true that $M\#N\cong S^n$ implies $M,N\cong S^n$?

That proof seems convincing

Balarka Sen

18:06

Topologically yeah

There is also a swindle proof

$M \# N \# M \# N \# \cdots \cong M \# \Bbb R^n$

Alessandro Codenotti

Time for black magic

geocalc33

Roger Heath-Brown

amWhy

@robjohn Given the times, perhaps you could encourage safety by sporting a face-mask. (Note @AlexanderGruber's clever avatar!)

Akiva Weinberger

Where does your proof fail for diffeomorphisms

Balarka Sen

If $D^n$ denotes the standard $n$-disk and $S^n$ denotes the standard $n$-sphere and $\varphi$ is a self-diffeomorphism of $\partial D^n$, it is not true that $D^n \cup_{\varphi} D^n$ is diffeomorphic to $S^n$

Jasper

18:09

@amWhy I am not so sure about the efficacy of face masks. It is a controversial subject. Hello by the way!

Balarka Sen

It may be a smooth manifold homeomorphic to $S^n$ which is not diffeomorphic to it. There are a couple of those out there

amWhy

@Jasper Hello!! Glad to see you!

Akiva Weinberger

Ah

Jasper

@amWhy If you do wear a mask, like I am forced to do now, make sure you don't touch your face unnecessarily, and make sure the mask is washed and dried properly if it is a cloth one. That is what I know. =)

Akiva Weinberger

I think I've successfully trained myself not to touch my face when outside (and wearing a mask)

Balarka Sen

18:11

What is "outside"?

Akiva Weinberger

You can adjust your glasses by dabbing

Balarka Sen

You mean the open world environment I am trying explore?

Thorgott

I've successfully ~~trained myself~~ just continued to not go outside

Jasper

Of course, the most important factors are social distancing and hand hygiene, just to add on.

Thorgott

wait, that's not how strikethrough works here

amWhy

18:12

@Jasper It's hard not to do when you have allergies (itchy eyes, in particular), but I'm learning. I gave up biting my fingernails!

Akiva Weinberger

~~test~~

Three hyphens

Alessandro Codenotti

@BalarkaSen they wasted all the budget on graphics so the gameplay kinda sucks, but it was a good effort

Thorgott

ah

Jasper

@AkivaWeinberger One for italics, two for bold, three for strike through. QED.

Balarka Sen

justgamerthings

Jasper

18:17

With the integrated graphics in the latest Intel and AMD chips, light gaming can be done without using a discrete graphics card.

The Ice Lake chips have better integrated GPU than the Comet Lake chips, but both are tenth generation Intel.

As for the 7nm AMD chips they will be out in the market only later this year.

Balarka Sen

@AkivaWeinberger Did you see the new trailer for Hyperbolica

Seems like a good game

Hyperbolica: A Non-Euclidean Adventure [Official Trailer]

►

Akiva Weinberger

I did not

feynhat

Why does every polygon admit a triangulation?

For polygons without holes, we can prove this by induction.

Balarka Sen

What is a polygon with holes man

Oh like nonconvex

feynhat

Think of $1 \le |x| + |y| \le 2$.

Like a square inside a square, but the inner square is deleted.

Akiva Weinberger

18:21

Looks cool

@feynhat Take two adjacent edges

Draw the segment between them

If they intersect the polygon, find the vertex in the resulting triangle furthest from the edge you just drew

Draw the segment between that vertex and the one between the two adjacent edges

Induct

Oh - you're including holes

Hm, does this still work?

I think it should

Instead of dividing the polygon into more pieces, sometimes that new edge just decreases the genus

Balarka Sen

Better to do that from the start IMO. Draw edges joining boundary of holes with the outermost boundary (all of this makes sense)

If you delete some polygonal nbhd of those edges you remain with something which is bounded by a polygonal simple closed curve

Polygonal simple closed curves bound triangulable regions - this is easy

Akiva Weinberger

Even in that case it's not 100% trivial but I outlined a way to do it

Balarka Sen

Yeah true

feynhat

@AkivaWeinberger "the vertex in the resulting triangle furthest from the edge you just drew". This will just be the common vertex of the two adjacent edges that I started with, right?

Akiva Weinberger

Oh sorry other than that

Furthest among the vertices in the interior of the triangle

feynhat

18:31

I see.

Balarka Sen

Imagine filling the region inside the polygonal simple closed curve by water

Choose some vertex and choose some direction in which the water level grows such that the water level is never horizontal to an edge at any stage

I claim the triangulation is clear :)

Akiva Weinberger

I guess you join the first two vertices it hits?

feynhat

For polygonal simple closed curve, its obvious no? You draw any diagonal. This will split the interior into two polygonal region with strictly fewer number of vertices. Apply induction.

Érico Melo Silva

@BalarkaSen nice

Akiva Weinberger

@feynhat The question is, how do you know there exists a diagonal that stays inside the polygon?

You can find concave polygons with diagonals that poke outside.

How do you know they don't all do that?

Balarka Sen

18:34

@Akiva I'd rather draw an edge along the water level whenever it hits a vertex. The region between two such "water-level edges" is always either a convex geometric quadrilateral or a triangle.

So you can draw an extra diagonal to triangulate

Akiva Weinberger

It can hit a vertex before it hits either of its edges

Balarka Sen

That's true

In which case you just draw a degenerate edge at that vertex

Those are what gives triangles

Mike Miller

@Thorgott I appreciate the commitment to the bit

Balarka Sen

(FWIW, what I am doing has a name and is called Morse theory)

feynhat

@BalarkaSen I don't get this. When water-level hits a vertex, the other end of the level may not be vertex, no? How can this be a valid diagonal?

Mike Miller

18:41

I think Feynhat's approach is cleanest

Rotate so that no arcs are horizontal

And no two vertices have the same y value

Balarka Sen

That doesn't actually matter but worth doing ^

Mike Miller

Now at every vertex, draw a horizontal line through it. This subdivides your polygon into a bunch of simpler things

Eh there's too much finickiness here

Balarka Sen

Just do Morse theory damn it

Mike Miller

I'll let you do this

Balarka Sen

@feynhat The water level is a line in $\Bbb R^2$. Intersect it with your polygon whenever it passes a vertex - this will give a bunch of points (corresponding to vertices of the kind you speak of) and a bunch of lines inside the polygon.

I call all of them water-level edges. The vertices are degenerate edges. They occur whenever, as Akiva said, the water level hits the vertex before it hits the edges adjacent to it

Akiva Weinberger

18:46

I need you to draw a picture

for the dart quadrilateral

robjohn

@amWhy Better?

Balarka Sen

This one?

Akiva Weinberger

Sure. Or upside-down

Balarka Sen

Start with the topmost vertex and fill water from there, translating downwards. There are three water-level lines of relevance.

Akiva Weinberger

Of the two diagonals, only one works, so that's the one your algorithm needs to draw

amWhy

18:48

@robjohn Thanks! I feel safer already! ;D

Balarka Sen

@AkivaWeinberger Idk what to draw. There are three horizontal lines, one passing through the top, one through the weird vertex inside, and one through both the bottommost vertices

These give rise to (1) degenerate edge (2) 2 edges and (3) 2 degenerate edges

That's the tringulation

robjohn

@BalarkaSen Is tringulation how you kill a triangle?

Akiva Weinberger

Wait

You're subdividing the edge?

Balarka Sen

Yes

Akiva Weinberger

Oh I wasn't thinking that's allowed

Balarka Sen

18:53

@feynhat Of course you're allowing that, right?

Akiva Weinberger

feynhat

Triangles can only have diagonals or the edges of the polygons as their sides.

Akiva Weinberger

I was thinking like this

Yeah

feynhat

You can't introduce new vertices.

Balarka Sen

That's not what a triangulation means to me. Totally confusing

Akiva Weinberger

18:55

I mean I guess "subdividing an edge" is equivalent to "introducing new vertices" since you're essentially drawing an infinitely thin triangle next to that edge

which you can thicken a bit if you want

By the way

Question about volume in 3D

In 2D, the area of polygons is determined by the following:

Balarka Sen

Anyway what I showed you actually proves Jordan-Scheonflies theorem for polygonal simple closed curves. In fact, with a little more work, you get Jordan-Schoenflies for smooth simple closed curves as well.

Same for spheres in R^3

But that requires significant more work

Akiva Weinberger

(a) formula for the area of rectangles (b) congruent things have the same area (c) area is additive (you can break things into pieces)

In 3D that's not enough

basically because in 2D a right triangle is half a rectangle but you can't fit congruent tetrahedrons together into a cube

But

if we add (d) scaling things up by a factor of 2 scales the volume up by a factor of 8

Is that now enough?

For example:

To find the area of a tetrahedron, it would be enough to show that a double-size tetrahedron minus two regular tetrahedrons can be cut and rearranged into a cube (which I think it can)

because now you have $8T=2T+\text{known thing}$

feynhat

@BalarkaSen So here, you weren't joining vertices of the holes with vertices on the outer boundary?

Balarka Sen

I was, I just took the vertices as a reference point than absolutes.

Akiva Weinberger

Also: can a lattice polyhedron be broken up into lattice tetrahedra

Balarka Sen

19:01

My final thing will be a triangulation with vertex set containing the geometric vertices of the polygon

I don't see why one would care about triangulating with the exact same vertices as the geometric vertices :P

Akiva Weinberger

Hm

Schönhardt polyhedron

In geometry, the Schönhardt polyhedron is the simplest non-convex polyhedron that cannot be triangulated into tetrahedra without adding new vertices. It is named after German mathematician Erich Schönhardt, who first described it in 1928. == Construction == The Schönhardt polyhedron can be formed by two congruent equilateral triangles in two parallel planes, such that the line through the centers of the triangles is perpendicular to the planes. The two triangles should be twisted with respect to each other, so that they are neither translates of each other nor 180-degree reflections of each other...

Balarka Sen

That's why ^

The Csazar tetrahedron is also like this I think

Akiva Weinberger

If I understand this right

It's like taking a figure 8 quadrilateral (shifted a bit into the 3rd dimension so it doesn't intersect itself)

and then forming a pyramid on it on either side

feynhat

@BalarkaSen I don't see why that would be possible. Like if we had a square hole inside one of the 'legs' of the dart, how can we draw edges from vertices of the hole to the vertices on the outer boundary?

Akiva Weinberger

Hm… is that it?

Balarka Sen

19:06

@feynhat I don't particularly care, just draw some edge

Mike Miller

I see now that my proof suggestion was actually fine

It divides the region into triangles and quadrilaterals

All of which are convex

Akiva Weinberger

The 3D model in the Wiki page is very clarifying

Balarka Sen

Oh the Csaszar is worse

The only diagonals are the edges

Oh that's just because it's exactly like the tetrahedron but funky

Akiva Weinberger

Posted a question

0

Q: Can every lattice polyhedron be subdivided into lattice tetrahedra?

In 2D, every polygon can be triangulated without introducing new vertices. In particular, lattice polygons can be divided into lattice triangles. In 3D, the Schönhardt polyhedron cannot be triangulated into tetrahedra without adding new vertices. What about the following weaker question?: Can ev...

("Cha Sar polyhedron")

Balarka Sen

Good question

+1

Akiva Weinberger

19:15

Thanks

feynhat

@BalarkaSen I am sorry but I don't see why you can do that. Like imagine a polygon with one hole and draw an edge from each vertex of the hole to every vertex on the outer boundary. If it is a valid edge (by valid I mean it lies in the interior), then you add the new hole so that the edge is no longer valid. In this new polygon with extra holes, you can't draw an edge between atleast one hole to the outer boundary.

Balarka Sen

This is fixable, you just have to choose a different scheme for drawing edges in a way that reduced number of holes successively

I am not interested in writing an algorithm admittedly

@AkivaWeinberger Yo if I take the usual triangulation of $S^2$ as a tetrahedron and $S^1$ as a triangle I get a 3*4 = 12 vertex triangulation of $S^3$ and a 4-vertex triangulation of $S^2$ and a simplicial map $S^3 \to S^2$ which realizes the Hopf map

Do you think we can write down the map on the level of finite topological spaces? :P

By collapsing each clopen face to a point

(I think that works? I'd normally collapse open faces but I think that makes it unnecessarily larger)

feynhat

19:31

@BalarkaSen Ah... okay.

@BalarkaSen @AkivaWeinberger Thanks for the discussion.

I gotta sleep now.

Later.

Akiva Weinberger

$S^2$ can certainly be replaced by a finite topological space. Not sure about $S^1$ or $S^3$ in there

I dunno

Balarka Sen

Any simplicial complex can be replaced by one, right? Collapse each open simplex to a point, that's a weak homotopy equivalence

i think

Akiva Weinberger

If you replace $S^2$ by a finite topological space (call it $s^2$?), then the preimages of the map $S^3\to s^2$ are two linked circles, the two twisted halves of the torus between them, and the inside and outside of the torus

Balarka Sen

Lol nice notation

Akiva Weinberger

Not thinking of $S^2$ as a tetrahedron

Thinking of it as that thing^

Balarka Sen

19:38

Got it. I wanted the whole map $f : S^3 \to S^2$ to be realized as a mapping of finite topological guys

So thats why I took a larger realization

Akiva Weinberger

It's probably doable, but you can't use just any representation of $S^3$ - it has to be "big enough" to fit all the preimage pieces

You can probably start with what I described and subdivide

Balarka Sen

I am guessing the 12 vertex thing works, because over each of those four 2-simplices the Hopf fibration trivializes to a product

But I guess I have to check the circle to circle maps preserve the triangle

Someone must have done it, let's see

emis.impa.br/EMIS/journals/AG/2-2/2_99.pdf

Apparently it's minimal as well

Jasper

@robjohn Your square looks lovely now, with the blue mask.

genescuba

Let's say I have 3 groups of data A, B, C.
A has two datapoints
B has three datapoints
C has five datapoints

I want to assign some weight to each datapoint so that each datapoint is used down the line for some analysis and is not biased by the size of the group it belongs to.

Hence the criteria I wish to satisfy here is the sum of the weights in each group is equal to 10 / 3 (total number of data points / total num groups). Hence my weights for datapoints in A: 10 / 6, B: 10 / 9, C: 10 / 15

Balarka Sen

Actually kssarkaria.org/docs/Hopf.pdf

Figure 6 sort of makes it apparent

I might try to write down the map on finite spaces for the lulz

Alessandro Codenotti

19:58

@Balarka Are you going to sleep anytime soon?

Balarka Sen

Yeah I think I'll hit the sack

Alessandro Codenotti

Unfortunate

I wanted to ask whether you were up to go through the Grigorchuk group paper

Balarka Sen

Unfortunately not any time soon I think. I've been doing too many things at once already

Alessandro Codenotti

Fair enough

I'll just read it myself sometimes soon I think

robjohn

@BalarkaSen poor, bruised sack

Balarka Sen

20:04

Holy shit

sciencedirect.com/science/article/pii/S0012365X97002732

This is written by a (currently retired) faculty in my university

Mike Miller

Wiw

Wow

Balarka Sen

Amazing result

I will now go to sleep with the crushing question of how people are so good

2

Ciao

Alessandro Codenotti

Good night

genescuba

Anybody have an idea on how to take into account pos/neg imbalance?

Leaky Nun

@AlessandroCodenotti ironic

I just woke up

Alessandro Codenotti

20:14

Good morning

Wait where are you now?

Leaky Nun

Hong Kong, 04:13 (GMT+8)

Alessandro Codenotti

I see

Ted Shifrin

21:04

Howdy, demonic.

Good morning, Leaky.

Alessandro Codenotti

Hi @Ted

genescuba

21:21

I'm wondering if this is even possible now it seems simple enough but haven't been able to come up w/ a way to satisfy the criteria? math.stackexchange.com/questions/3662322/…

Stupid Kid

23:25

Hello

Please quickly say me

Why removing a cube n^3 from sub cube (n-1)^3 gives only hexagon

This is counter intutive

my book was deriving series for 1^2 +2^2+3^2...

sht now I can't use phone

robjohn

23:42

@StupidKid what does that mean? I don't understand.

Akiva Weinberger

The nth hexagonal number is $n^3-(n-1)^3$

robjohn

$n^3-(n-1)^3=3n^2-3n+1$

Akiva Weinberger

You basically get three square faces, bounded by six edges

You can "flatten" it to form a hexagonal shape

These things, to be clear

robjohn

@AkivaWeinberger That's half a cube: $6$ square faces and $12$ edges.

Akiva Weinberger

@robjohn Take an nxnxn cube and remove an (n-1)x(n-1)x(n-1) section from it

That has $n^3-(n-1)^3$ blocks in it, and has only three faces

The claim is $n^3-(n-1)^3=H(n)$

and the idea is to just kinda smush the blocks

Compare the 19 cubies (the technical term for a 1x1x1 piece of a Rubik's cube) visible in that image, to the picture accompanying H(3)=19 in the last image

(not to be confused with the 27 cubies total*, or the 27 faces visible)

(*a real Rubik's cube will actually only have 26 cubies because there isn't one in the center, but whatever)

@StupidKid

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$Mathematics$ Mathematics