Mathematics - 2019-06-12 (page 4 of 4)

Let $G \subset SO(n)$ be a finite subgroup. If one takes the intersection of $g\{x_n \leq 1\}$ as $g$ varies over all elements of $G$, one obtains a closed convex region $C_G$ containing $0$.

Subhasis Biswas

@RyanUnger can you please explain this a bit

Mike Miller

1) Can one algebraically determine from $G \subset SO(n)$ whether or not $C_G$ is compact? Then it is a convex polyhedron. 2) Can one determine (again, algebraically) from $G$ whether or not $C_G$ is a regular polyhedron?

Ryan Unger

@Semiclassical we were thinking of $\nabla f=\partial_rf \,e_r+r^{-1}\partial_\theta f\, e_\theta$

Mike Miller

For both questions, either sufficient or necessary conditions would be interesting.

Semiclassical

My observation should be understood as the fact that $r^{-2}\partial/\partial \theta= r^{-1}(r^{-1}\partial/\partial \theta)$

Ryan Unger

7:11 PM

I don't actually know what the original question is

@RyanUnger in polar?

@MikeMiller Oh hey!

Yeah, exactly

I thought you quit MSE

Mike Miller

I did and still am. Maybe you know the answer to the above question, though.

Semiclassical

7:12 PM

1/r is the component of that basis vector

Ryan Unger

so what was the original question

Semiclassical

o/

Akiva Weinberger

@RyanUnger Let $f(r,\theta)=\theta$ be a function, defined in polar coordinates. What is $\operatorname{grad}f$, in polar coordinates?

Ryan Unger

just compute $grad(f)u$ or whatever?

Akiva Weinberger

@MikeMiller What do you mean by $\{x_n\le1\}$?

Mike Miller

7:13 PM

subset of $\Bbb R^n$

Akiva Weinberger

Oh I see

Right, SO(n) acting on Rn

Mike Miller

if $G$ is the group of isometries of a regular polyhedron, $C_G$ is (a scaling of) that polyhedron

Whence the question for all the other choices of $G$

Subhasis Biswas

@RyanUnger are we getting $\frac{1}{r} \hat{e_\theta}$?

Ryan Unger

@SubhasisBiswas physicists have some way of computing this

Akiva Weinberger

@MikeMiller I don't understand

You mean $\{x:|x|\le1\}$?

Semiclassical

7:15 PM

g in G implies g^-1 in G

Akiva Weinberger

OH

Sorry

Half-spaces

Yes

Mike Miller

I want the last coordinate to be at most 1. This is a half-space. When you intersect over enough of them, that defines the polyhedron.

Akiva Weinberger

Right

Mike Miller

No problem, I gave a very brief description

I hope for a nice answer to (1); (2) is a pipe dream

Akiva Weinberger

So the symmetries of a cone ($\Bbb Z_n$)

give you an infinite prism?

Semiclassical

7:16 PM

I mean, you’ve got a finite number of elements in G and so finitely many half planes

Akiva Weinberger

Whose symmetry group is $D_n$?

Ryan Unger

So you just compute $g^{ij}$ and then $g^{ij}\partial_jf$

Akiva Weinberger

(I've come to understand that if we write $D_{2n}$ we should also write $S_{n!}$)

Ryan Unger

now $g^{11}=1$ and $g^{22}=r^{-2}$

Mike Miller

Yeah, exactly, and a regular polygon when it acts on $\Bbb R^2$

Semiclassical

7:17 PM

So the only issue I could see is if you don’t want unbounded polyhedra

Ryan Unger

so there you go

now for a physicist $e_\theta=r^{-1}\partial_\theta$

Akiva Weinberger

My guess is, if it lives in a copy of $SO(n-1)$

Hm wait no

OK if we write $\Bbb R^n$ as $\Bbb R^{n-1}\oplus\Bbb R$

we get a subset that's $O(n-1)\oplus O(1)$?

Mike Miller

I think you had the right statement, so long as "a copy of $O(n-1)$" means you allow the embedding to vary. There will be a family of embeddings depending on a choice of splitting, like you say.

Yeah

though probably oplus is bad notation for groups

Akiva Weinberger

Right

But yeah so in 3D I'm basically describing the cyclic and dihedral groups

and everything else gives us a solid

Hm

Mike Miller

To be clear I'm not saying I think your conjecture is correct, I just think that would be a natural statement to think about. I have no idea if that's sufficient.

Rithaniel

7:20 PM

Heya Mike

Semiclassical

Which convention for D_n are we using ?

Akiva Weinberger

Depends on the orientation of this subgroup relative to the axes

Mike Miller

Hi

Ted Shifrin

OMG ... who is that?

Akiva Weinberger

@Semiclassical If $|D_n|=n$ then $|S_n|=n$

We don't write $S_{120}$, right?

I think that's just the most consistent way to do it

Rithaniel

7:22 PM

Haven't seen you around in a little while. What've you been up to?

Semiclassical

Well there’s D_n vs D_2n

But D_n is fine

Akiva Weinberger

But yeah the polyhedron isn't determined by the isomorphism class of the subgroup

Like take the cube/octahedron group

We can get a cube, or an octahedron

or things in between

Semiclassical

As a check, for the D_n subgroup of SO(2), we’d get an n-gon $\times \mathbb{R}$?

Akiva Weinberger

Not necessarily

Semiclassical

Oh. Yeah

Akiva Weinberger

7:26 PM

Is the half-space perpendicular to then plane the n-gon lives in?

If it's parallel we get $\Bbb R^2\times[-1,1]$

Semiclassical

Right

Akiva Weinberger

If it's kinda skewed we get a… bipyramid?

Semiclassical

@AkivaWeinberger That is a convex unbounded polyhedron, though

Subhasis Biswas

@AkivaWeinberger with infinite base?

Akiva Weinberger

No @SubhasisBiswas

Semiclassical

7:28 PM

And so will any rotation thereof

Ryan Unger

@BalarkaSen the TA for the CMC course is like a ten year old indian kid

Akiva Weinberger

But I think both bipyramids and trapezohedra are possible

(which are both bounded)

Subhasis Biswas

@AkivaWeinberger ummm...can you draw me a sketch? I can't get an idea what would that look like

Semiclassical

So it’d seem like we’re intersecting a finite list of convex unbounded polyhedra, which should yield a convex polyhedron

Akiva Weinberger

Semiclassical

7:30 PM

Only question would seem to be under what conditions will it be bounded

Rithaniel

Akiva's art skills are amazing. He even colored it.

Subhasis Biswas

@Rithaniel yeah.. really beautiful. Someone 3d print this

Akiva Weinberger

A trapezohedron is this

And you can imagine a half-way polyhedron where the faces are like "lopsided" kites

Ryan Unger

wow tht was fast

Semiclassical

I should make a Mathematica attempt on this, showing how the polyhedron varies with the symmetry axis

Akiva Weinberger

7:32 PM

Semiclassical

Noice

Akiva Weinberger

The proportions on that look wonky

Subhasis Biswas

what are you using to draw these?

Semiclassical

Has dihedral symmetry though, as it should?

Akiva Weinberger

@SubhasisBiswas …Google images

Did you really think I was drawing these

Semiclassical

7:33 PM

Loool

Subhasis Biswas

@AkivaWeinberger bamboozled

Akiva Weinberger

Also they make both bipyramid- and trapezohedron-shaped dice, I think

Rithaniel

Akiva is a speed painter

Akiva Weinberger

Like 10-sided dice are usually trapezohedra I think

Look I can do shading and reflections

Semiclassical

amaaazing

simply amaaaaaazing

Akiva Weinberger

7:37 PM

6

Q: Why are $10$-sided dice not bipyramids?

Commonly used $10$-sided dice are pentagonal trapezohedrons, as opposed to pentagonal bipyramids. Given that bipyramids are a more "obvious" shape for a fair die with an even number of faces, it's curious to me that the trapezohedrons are the more commonly used shape. So, what are the advantages...

Oh that's why

Rithaniel

Unfortunately, Akiva can only draw images of trapezohedra.

Akiva Weinberger

Say you and your neighbor each own a field

and you want to build a wall such that neither of you can see into the other's field

but you can each still walk to the other's field

I think that shape^ is the most efficient way to do it

Semiclassical

the area in the semicircle belongs to neither of them?

Akiva Weinberger

Yeah

And the line extends horizontally infinitely. You each own a half-plane (minus the semicircle)

If only one of you is comfortable with ceding territory

then you can do this

…You and your neighbor are shaped like discs of radius 1

You're spherical cows in a vacuum

I discussed this with Secret a few months ago

I don't remember the end of the discussion but I don't think we ended up with a proof that these are optimal

imgur.com/a/E6e0GOz

If the walls had thickness, though, like if they were made of Minecraft blocks of sidelength 1, You could do this

Stan Shunpike

8:06 PM

hello

I am reading from

cds.caltech.edu/~murray/courses/cds101/fa02/caltech/…

I am confused what they mean by "standard conditions" for the existence and uniqueness of solutions

Is the list they provide in the next sentence sufficient?

Or is that an incomplete list and there are more conditions

Semiclassical

@AkivaWeinberger hmm, doing it in mathematica I'm only finding bipyramids and no trapezohedra

Akiva Weinberger

I dunno but it kinda feels like the author was saying "I'm too lazy to go check a differential equations textbook"

Stan Shunpike

@AkivaWeinberger yeah exactly lolol

Akiva Weinberger

@Semiclassical Is the cutting plane parallel to the sides of the polygons?

What if you give the cutting plane random coordinates?

Semiclassical

No, I'm allowing it to vary. That is, I'm keeping the dihedral axis fixed as z but allowing the cutting plane to take arbitrary direction (ranging from +z to +x)

If I start with an angle between those two extremes, I get a bipyramid

Akiva Weinberger

8:16 PM

What are the "flip" elements of the group?

Uh

Semiclassical

if I let the cutting plane to go towards the z direction, then the top/bottom vertices pull apart until I get an infinite prism

Akiva Weinberger

The order-2 ones

Semiclassical

if I let the cutting plane go towards the x-direction, then the vertices in the plane move away from the origin until I get an infinite plane width

Thorgott

@StanShunpike I believe so. See en.wikipedia.org/wiki/Picard%E2%80%93Lindel%C3%B6f_theorem

Akiva Weinberger

@Semiclassical Yeah but you have two dimensions of freedom

not just one

for positioning a plane in space

Semiclassical

8:19 PM

hmm, I'd convinced myself that that other degree of freedom wouldn't matter. but you may be right, and it's easy enough for me to modify my code to find out

Ryan Unger

You’re still doing this?

Akiva Weinberger

We have yet to generate the pretty pictures

What n are you doing by the way

Semiclassical

6 right now

Ryan Unger

Lol

Semiclassical

yep, trapezohedra emerge when I allow that other degree of freedom to vary

Akiva Weinberger

8:22 PM

Yay

Stan Shunpike

@Thorgott super! thanks

Semiclassical

Now to figure out how to make the pictures not s***

(mathematica is surprisingly bad at plotting polyhedra generated by half-plane inequalities)

May I see it?

Cool

8:24 PM

It's not great at figuring out where the edges are supposed to be

Akiva Weinberger

and also wonky rendering, yeah

Semiclassical

Yeah

The smart way to do this would be to use my sagemath expertise

Akiva Weinberger

And the half-way shape?

Semiclassical

with the way I'm doing this, this is the halfway shape. But I think I know what you mean:

(I upped the PlotPoints for this one to make it render nicer)

Akiva Weinberger

I meant halfway between bipyramid and trapezohedron

Semiclassical

8:27 PM

ohhh

Akiva Weinberger

"Quarter-way shape" then?

Semiclassical

nah, it's halfway between those two extremes

So that's fine

unfortunately, it seems to be way slower to do the cases other than the bipyramid

so waiting on that now

Akiva Weinberger

Hm

Strange

Semiclassical

oh, maybe I was making it slower than needed. trying again

ok, yeah

so yeah, tilted

Akiva Weinberger

Weird-looking thing

I think you get to name it

Semiclassical

8:31 PM

lol

well, I think it's just the same thing you were showing before

Akiva Weinberger

True

So OK for Mike's question

Semiclassical

so still just trapezohedron

Akiva Weinberger

I think for any subgroup, almost any cutting plane is gonna give us a bounded thing

Well he phrased it so that the cutting plane was fixed and the subgroup varies

but the point is it's bounded almost always

Semiclassical

meh, I don't think that's an issue

i mean, i see why it's the clearer formulation

Akiva Weinberger

Now I'm curious about other groups

like the symmetries of a tetrahedron

Theoretically you'd be able to smoothly vary between a tetrahedron and its (self-)dual

Seems like it would be hard to code though

Semiclassical

8:35 PM

Yeah, the coding is the tricky part

Akiva Weinberger

Or the cube group

Semiclassical

Basically I'd need a good way to generate all of the relevant rotation matrices

Akiva Weinberger

Yeah

Semiclassical

That's easy enough to do for the dihedral group, since it's just rotations and reflections

Akiva Weinberger

With the tetrahedron with vertices at (-1,-1,-1), (-1,1,1), (1,-1,1), and (1,1,-1)

Semiclassical

8:37 PM

Probably the smart way to do it is to exploit a useful presentation

Akiva Weinberger

you only have three matrices to worry about

No wait

Oh

Eleven?? Whoops

Semiclassical

(accidental recursion)

oops

Tetrahedral symmetry

A regular tetrahedron has 12 rotational (or orientation-preserving) symmetries, and a symmetry order of 24 including transformations that combine a reflection and a rotation. The group of all symmetries is isomorphic to the group S4, the symmetric group of permutations of four objects, since there is exactly one such symmetry for each permutation of the vertices of the tetrahedron. The set of orientation-preserving symmetries forms a group referred to as the alternating subgroup A4 of S4. == Details == Chiral and full (or achiral tetrahedral symmetry and pyritohedral symmetry) are discrete point...

Akiva Weinberger

I was skipping the identity

@AkivaWeinberger Hm

Semiclassical

we're working with SO(2), so I guess just the 12?

Akiva Weinberger

Orientation-preserving, yeah

Hm

What if we went back to the dihedral example

but instead of orientation-preserving symmetries of a polygon, we had all symmetries

What group is that

Semiclassical

8:40 PM

keep in mind that, in the dihedral case, you can think of the 'reflections' as rotations w/r/t lines in the plane

and I was treating it as such

Akiva Weinberger

Yeah but if you also added in reflection in the plane

Semiclassical

If you only use the rotation symmetries, you won't get something bounded

Akiva Weinberger

You can already tell that if you took the trapezohedron and reflect it in the plane, the two aren't equal

Not sure what their intersection would be

Semiclassical

( e.g. x+z < 1, y+z<1,-x+z<1,-y+z<1 is unbounded)

Akiva Weinberger

A bipyramid with twice as many sides?

Semiclassical

8:42 PM

gross

Akiva Weinberger

@Semiclassical Those aren't the same in 3D

is the point

Semiclassical

hmm

that's how i was implementing them, anyways

it does go to O(n) vs. SO(n), though

Akiva Weinberger

What if you did the group with both

The subgroup of O(n)

Semiclassical

dunno

Akiva Weinberger

I'm guessing we get a bipyramid, with potentially twice as many sides

Semiclassical

8:51 PM

I think I see the right way to do the code better, at any rate. Write out symbolically all the group elements in terms of the generators, then replace the generators with their representations as rotation matrices

the former task is a bit tedious but I can probably look it up

Akiva Weinberger

You can probably do a speedup

where like instead of intersecting $S\cap aS\cap bS\cap abS$

you first compute $T:=S\cap aS$ and then do $T\cap bT$

That does $ba$ rather than $ab$ but whatever

> "Absolutely no one on this team is good at baseball," said Tom, ruthlessly

Semiclassical

well, I'm just collecting all the inequalities and having mathematica plot the resulting region

So I'm not sure how I'd implement that

Akiva Weinberger

Hm