Mathematics - 2019-06-25 (page 2 of 3)

Balarka Sen

5:00 PM

eg $S_3$ is symmetry group of the triangle which gives a non-free action on $S^1$

Alessandro Codenotti

Hi @Ted

Semiclassical

Simpler question, maybe: What's the smallest finite group which is not the symmetry group of some 3D polyhedron?

Balarka Sen

Symmetry groups of 3D regular polyhedrons are either cyclic, dihedral, A_4, S_4, or A_5

Semiclassical

nice.

so just find the smallest group which isn't one of those

(for regular polyhedra, anyways)

oh hey

Polyhedral group

In geometry, the polyhedral group is any of the symmetry groups of the Platonic solids. == Groups == There are three polyhedral groups: The tetrahedral group of order 12, rotational symmetry group of the regular tetrahedron. It is isomorphic to A4. The conjugacy classes of T are: identity 4 × rotation by 120°, order 3, cw 4 × rotation by 120°, order 3, ccw 3 × rotation by 180°, order 2 The octahedral group of order 24, rotational symmetry group of the cube and the regular octahedron. It is isomorphic to S4. The conjugacy classes of O are: identity 6 × rotation by 90°, order 4 8 × rotation by...

Alessandro Codenotti

@Semiclassical $(\Bbb Z/(2))^2$ I guess?

Semiclassical

5:02 PM

this page is probably worth linking too: List of finite spherical symmetry groups

Balarka Sen

I actually really listed all the discrete subgroups of SO(3), so the first two are extra

Ted Shifrin

Yeah, polyhedra have very few possible symmetry groups.

Semiclassical

polyhedra or regular polyhedra?

Ted Shifrin

If they're not regular, the symmetry groups are even smaller!

Semiclassical

huh

neat

Ted Shifrin

5:03 PM

Even if you go to higher dimensions, you're not going to get very many groups, I believe.

I think Hilbert-Cohn Vossen have some discussion of this in their lovely Geometry and the Imagination.

Balarka Sen

I don't know what the discrete subgroups of SO(n) are

Ted Shifrin

I don't either, but I believe the standard reference is Joe Wolf's book.

Semiclassical

I guess this all ties back to point groups

Balarka Sen

@TedShifrin I don't think there's too few of them though. Remember elliptization says every finite 3-manifold group is a discrete subgroup of O(4)

That's an uh a lot

Ted Shifrin

I don't remember that, because I never knew it. :)

Is this a consequence of Thurston?

Balarka Sen

5:07 PM

Yeah it says every group acting freely on S^3 acts linearly... surprising theorem

Yup

Semiclassical

I guess another question to lump on this is: What's the simplest example of a finite group which isn't a point group for any dimension?

Balarka Sen

Oh I am dumb

Take any finite group G, and take a faithful representation G -> GL_n(R). You can do this by embedding G in S_n for some large n, and then composing with S_n -> GL_n(R)

So G is acting on R^n. Average to get a invariant inner product

You have a faithful representation G -> O(n).

Ted Shifrin

Yup.

The baby unitary trick.

Balarka Sen

Yeah

@Semiclassical So in fact every finite group is a discrete subgroup of O(n) for some n, hence a polyhedral symmetry group

Semiclassical

neat

that leaves open a practical question of how large your n needs to be for a given G

Balarka Sen

5:13 PM

I guess you construct the polyhedron by just marking a point in S^(n-1), then looking at the G-orbit of that point?

Should do the trick

@Semiclassical This says n = |G| is enough, because we used the Cayley embedding G -> S_n from the getgo

Semiclassical

nice

Ted Shifrin

Of course, that's a big n. :P

Semiclassical

I suppose sometimes you'd be able to pick a smaller n, but that presumably depends on the group

Balarka Sen

Cayley is tight for Q8 (it doesn't embed in S_n for n < 8) but I think Q8 is a subgroup of O(4)

Hm. It's a subgroup of SU(2) for sure

The usual 2x2 complex matrix representation

Ted Shifrin

Yeah, the Cayley route is not geometrically efficient.

Balarka Sen

5:18 PM

Makes sense

Sounds like a complicated representation theory question in general, but I won't put bets on it

Ted Shifrin

I guess we need Mathein or Tobias. It should just boil down to basic representation theory of the finite group.

Semiclassical

Seems hard. I mean, you can pick arbitrarily large cyclic groups and they’ll always be the symmetry group of the appropriate polygon

So setting bounds on what n has to be is probably tough

Balarka Sen

Yeah but our point is that n = |G| is probably not even globally tight

MatheinBoulomenos

@TedShifrin I'm needed? :)

Semiclassical

As in, there may be a better upper bound?

Ted Shifrin

5:22 PM

As if by magic ...

hey everyone

LOL

Heya

@MatheinBoulomenos Can you go through the previous few messages and comment appropriately?

Ted Shifrin

Hi, magic @Mathein, and hi, @ÍgjøgnumMeg

ÍgjøgnumMeg

5:22 PM

Hi @Ted :)

N. Maneesh

16 hours ago, by N. Maneesh

1

Q: How does the Evolute of an Involute of a curve $\Gamma$ is $\Gamma$ itself?

How does the Evolute of an Involute of a curve $\Gamma$ is $\Gamma$ itself? Definition from wiki:-The evolute of a curve is the locus of all its centres of curvature. That is to say that when the centre of curvature of each point on a curve is drawn, the resultant shape will be the evolute of th...

differential-geometry

still I am struggling :(

Ted Shifrin

Are you talking about planar curves? Otherwise, there are non-uniqueness issues.

yes

@Tobias !!!

@BalarkaSen Hi

5:25 PM

Did Tobias show up magically too?

I'm so powerful.

N. Maneesh

involute vary for different initial points.right?

Balarka Sen

@Ted Your summoning powers are fantastic

Ted Shifrin

LOL

There is no "initial point" with the standard definition, @N.Maneesh.

MatheinBoulomenos

So if I understood correctly, you're essentially asking for a given finite group $G$, what is the smallest $n$ of a faithful representation of $G$?

Balarka Sen

Globally in terms of the order of $G$, yeah. Is $n = |G|$ tight?

N. Maneesh

5:27 PM

okay

Ted Shifrin

$\beta$ is an involute of $\alpha$ if for every $t$, $\beta(t)$ lies on the tangent line to $\alpha$ at $\alpha(t)$ and if they have perpendicular tangent vectors at those points.

Tobias Kildetoft

@BalarkaSen what does that even mean?

you mean if there are arbitrarily large groups for which it is?

Balarka Sen

@TobiasKildetoft Can you have a finite group $G$ such that $G$ doesn't embed in $O(n)$ for all $n < |G|$? If not, what's the tightest $n$?

MatheinBoulomenos

I think you can always do $n-1$, the standard permutation representation has a trivial subrepresentation, if you quotient that out, you don't lose faithfulness

Ted Shifrin

At any rate, @N.Maneesh, you need to compute with some Frenet formulas and not just write words.

Tobias Kildetoft

5:29 PM

@MatheinBoulomenos and that is even independently of the field

Balarka Sen

@MatheinBoulomen Ah, OK, the diagonal is fixed.

N. Maneesh

@TedShifrin then, what happens when we take the evolute of $\beta$?

Ted Shifrin

@N.Maneesh: So if $\alpha$ is arclength parametrized, you get your locus of centers of curvature by taking $\beta(s) = \alpha(s) + \frac1{\kappa(s)}N(s)$. Now compute the involute.

N. Maneesh

@TedShifrin okay. I will try,

Ted Shifrin

You need to use Frenet formulas.

BTW, if you're playing around with differential geometry of curves, etc., you should download my diff geo text.

Lots of good exercises for you.

N. Maneesh

5:31 PM

Thank you

I will

Subhasis Biswas

I have a little problem. Prove that between any two irrational numbers, there is a rational number. (no hints now)

MatheinBoulomenos

It seems like an interesting research question, I can't think of anything more interesting than the easy upper bound of $n-1$. This is tight for $n=3$ (though this depends on $\Bbb R$ not having a third root of unity, over $\Bbb C$ it is not tight), but not for $n=4$ or $n=5$.

Balarka Sen

Interesting!

Ted Shifrin

@SubhasisBiswas: If you don't want a hint, why are you announcing it?

MatheinBoulomenos

if we restrict ourselves to finite simple groups (a big restriction, I know), then we can use the fact that any non-trivial representation is faithful and the fact that the squares of the degrees of the irreps sum to $n$, to get an upper bound of $d^2+1 \leq n$, so $d \leq \sqrt{n-1}$

Ted Shifrin

5:34 PM

I knew Mathein would have some representation theory magic.

MatheinBoulomenos

$d$ being the degree of the smallest faithful rep, which is necessarily irreducible

Balarka Sen

Nice, I didn't know that fact

Tobias Kildetoft

yeah, if it is to be tight, the group probably has to be far from simple but still also not abelian obviously

MatheinBoulomenos

@BalarkaSen I prove it in my blog using the center of the group algebra and Artin-Wedderburn :)

Ryan Unger

@MatheinBoulomenos are you doing a PhD yet

MatheinBoulomenos

5:35 PM

@RyanUnger nope, still undergrad

Balarka Sen

@MatheinBoulomenos Hahah, nice advertisement. Now I have to read it

Tobias Kildetoft

I completely avoid the group algebra in my notes

MatheinBoulomenos

no, wait we don't need the center of the group algebra for that that's for number of irreps= number of conjugacy classes

Tobias Kildetoft

yeah

Ted Shifrin

pops popcorn for representation theory fight

Balarka Sen

5:37 PM

This is in Simplicity and Representations part 1 and 2, right?

LOL @Ted

MatheinBoulomenos

Yes, this is in Semisimplicity and Representations part 1

the thing about conjugacy classes is in part 2

Ryan Unger

I have vague memories of this from Thistlethwaite’s class

Tobias Kildetoft

so if we take two primes $p$ and $q$ that allow a nonabelian group of order $pq$, what is the bound for that group?

MatheinBoulomenos

no wait, actually considering real representations, we get a worse upper bound

I think it should be $2 \sqrt{n-1}$

due to the presence of non-trivial division algebras over $\Bbb R$, the sum-of-squares formula doesn't work out

but since their degree is bounded by $4$ (due to Frobenius' classification), we still get an inequality

Subhasis Biswas

Let $a$ and $b$ two positive irrational numbers with $a<b$. Both of the numbers must have infinite non-recurring decimal expansion. Let $a=c.r_1r_2r_3...$ and $b=d.s_1s_2s_3...$. If $c<d$, then $(c+d)/2$ is such a number. Now, we consider the case when $c=d$. Now, since the two numbers are different, then they must differ at some position of their decimal expansion. Let the point of "mismatch" be $ r_n$ and $s_n$. It must be $r_n<s_n$.

MatheinBoulomenos

5:40 PM

@TobiasKildetoft I appreciate that, but I personally really like noncommutative algebra

Subhasis Biswas

Now, we consider the number $c.r_1r_2r_3...t_n$, where $t_n=(r_n+s_n)/2$.

Ted Shifrin

No, no, no decimals, @SubhasisBiswas.

MatheinBoulomenos

@TedShifrin lol, I don't fight over multiple viewpoints

Ted Shifrin

That doesn't even make sense a lot of the time.

MatheinBoulomenos

not if I like both of them, anyway

Tobias Kildetoft

5:41 PM

@MatheinBoulomenos So do I. But the students were not properly prepared for it.

Subhasis Biswas

@TedShifrin did I make any mistake?

Tobias Kildetoft

seeing as they had only just seen commutative rings at that point, and nothing about modules at all

MatheinBoulomenos

yeah, reducing it to character theory seems like the correct choice for that audience

Tobias Kildetoft

not that I think they truly appreciated characters (there was just not enough time for that)

Ted Shifrin

@SubhasisBiswas: You haven't written enough for it to make sense. But please don't use decimals. There are better ways.

@SubhasisBiswas: So here's a hint. Suppose the distance between the two irrationals is at least 1. What must be true?

Subhasis Biswas

5:48 PM

@TedShifrin I am thinking. Will let you know when I have an answer

Thorgott

Even with decimals, there're easier ways (though I agree that not using decimals is the best option)

Ted Shifrin

I think it's better just to think about two arbitrary real numbers, honestly.

@SubhasisBiswas: In your proof, are you actually using the hypothesis that they're both irrational? I don't think you are.

Subhasis Biswas

@TedShifrin yes. I was considering the case when both are irrational

@TedShifrin there is an integer between them?

Ted Shifrin

How did you use that in what you wrote?

@SubhasisBiswas Precisely. So in that case you have a rational. Now how do you fix it in general?

Subhasis Biswas

@TedShifrin by "shrinking " the distance accordingly

Ted Shifrin

5:54 PM

Can you elaborate?

Tobias Kildetoft

@MatheinBoulomenos What do you mean by reducing to the cyclic case?

Subhasis Biswas

I am trying to formulate it

MatheinBoulomenos

@TobiasKildetoft lol

that was nonsense

Tobias Kildetoft

Clearly not all $2$-groups have faithful reps of dimension $1$

MatheinBoulomenos

I forgot about faithful

Tobias Kildetoft

5:54 PM

you are correct about irreps

MatheinBoulomenos

yeah

Ted Shifrin

We live in an unfaithful world.

MatheinBoulomenos

Has there ever been a world that was fully faithful?

Ted Shifrin

Certainly not a free one.

this is so brilliant

5:57 PM

Ayy my man jamming to Bach

Subhasis Biswas

Let $b>a$ and $|b-a| > \epsilon>1/m$ for some $m \in \mathbb{N}$

There is an integer b/w $ma $ and $mb$, say $t$ . Therefore, $ma < t < mb \implies a<t/m<b$

Tobias Kildetoft

@BalarkaSen Have you ever listened to the band called Heilung?

Balarka Sen

Yes!

They're beyond brilliant

Tobias Kildetoft

Indeed. I read a review of their performance at Copenhell and had to check them out

Ted Shifrin

@SubhasisBiswas: There you go.

Subhasis Biswas

5:58 PM

@LeakyNun trademark Bach

Leaky Nun

@BalarkaSen I've always liked Bach

Ted Shifrin

(@SubhasisBiswas: You don't need the absolute value, of course, since you assumed $b>a$.)

Tobias Kildetoft

Awesome to code to as well, as you can hardly ever understand the lyrics, so they don't disturb

Subhasis Biswas

@TedShifrin yes yes...

Balarka Sen

@Tobias Haha.

Subhasis Biswas

5:59 PM

anyone likes little fugue ?

Leaky Nun

me!

Subhasis Biswas

@LeakyNun <3

Balarka Sen

I listen to King Gizzard and The Lizard Wizard a lot when writing down math sometimes

MatheinBoulomenos

@BalarkaSen Can you really stand my algebraic musings?

Leaky Nun

@TobiasKildetoft I revised for my maths final this year to WTC Book II

Adam

6:00 PM

youtube.com/watch?v=5ZdDvB6eMDc this is so brilliant too

Balarka Sen

@Tobias I have been into this Finnish(?) Krautrock band called Circle for some time

Alessandro Codenotti

@TobiasKildetoft They're great

Balarka Sen

@LeakyNun What other classical music do you like

Alessandro Codenotti

I found them listening to Aurora, which brought me to Wardruna which brought me to Heilung

Leaky Nun

@BalarkaSen Mozart

but Bach is always my favourite :P

Balarka Sen

6:02 PM

What about non-baroque

Leaky Nun

baroque music is true music

3

Balarka Sen

Oof we got a baroque elitist bois

MatheinBoulomenos

I like Rachmaninoff

Balarka Sen

I have been into French Romantic stuff lately. Satie, Chopin, Debussy, ...

MatheinBoulomenos

Oh, Satie is awesome

I listened to the Gnossiennes so many times

Balarka Sen

6:04 PM

Gnossiennes > Gymnopedie agree or disagree

YES

Subhasis Biswas

anyone likes vivaldi?

Hildegard von Bingen - Voices of Angels - Voices of Ascension

►

Adam

This is the biggest social media network on the Internet, both in terms of total number of users and name recognition. Founded on February 4, 2004, Facebook has within 12 years managed to accumulate more than 1.59 billion monthly active users and this automatically makes it one of the best mediums for connecting people from all over the world with your business. It is estimated that more than 1 million small and medium-sized businesses use the platform to advertise their business.

I love classical music, but there is just as good modern music and I prefer it because of it's relevance

a lot of people will disagree, but it's because they are inept in the regard that allows appreciation of art from an unbias unindoctrinated perspective

but each to their own opinion

Balarka Sen

6:20 PM

Lots of good modern music but modern pop sucks tho

I'm actually centrally a metal fan but I like all sorts of good music

Semiclassical

main thing about music nowadays is that there's some much of it

or, rather, that there's so much of it accessible

Balarka Sen

@Semiclassical I heard The Mountain Goats released an underwhelming album

Semiclassical

which?

I wasn't that knocked out by "Goths" (except for the first track, which was goddamn brilliant)

Balarka Sen

"In League With Dragons", the one after Goths

Very recent, released in April 26th

Semiclassical

ahh

haven't checked it out yet tbh, i'm not great at following these things

Ted Shifrin

6:26 PM

Hmm, we need to return to our cooking chat :P

Balarka Sen

I haven't either, theneedledrop says it's not great

Semiclassical

I mean, they're allowed some misses

Balarka Sen

but I am not that big a believer of Fantano

Yeah haha massive discography

Semiclassical

Beat The Champ was a great album, for instance

Balarka Sen

mostly great shit

Semiclassical

6:27 PM

(the one that preceded Goths)

Balarka Sen

Haven't listened to 70% of it tbh

Semiclassical

i've listened to a lot of it

both old and new

Subhasis Biswas

what is a nonconstructive proof?

Balarka Sen

I went through The Sunset Tree sometime ago

Liked it a lot

Semiclassical

yeah

MatheinBoulomenos

6:29 PM

@TedShifrin I made Vietnamese summer rolls with a friend, they were delicious :)

Ted Shifrin

Nice, @Mathein. Did you remember to send me one?

Ryan Unger

@SubhasisBiswas an incorrect one

Ted Shifrin

blah @Ryan

MatheinBoulomenos

I'm not sure how they would take the travel

Ryan Unger

@TedShifrin I only accept explicit constants in my inequalities

Semiclassical

6:31 PM

Trying to figure out how to explain something simple without being laborious about it

Ted Shifrin

You'll wise up eventually, @Ryan :P

Ryan Unger

@TedShifrin actually though I’ve run into this problem lately

Ted Shifrin

So you're going to find the maximum value of every continuous function on every compact set. I'm impressed.

Ryan Unger

I need some “algebraic” inequalities for matrices

I can only prove them using compactness :P

But there should be a way to explicitly find the constant

Ted Shifrin

Finding the operator norm is doable in dimension 2, but almost impossible in general.

Ryan Unger

6:32 PM

These are symmetric matrices

Everything can be assumed diagonal

I’m sure one just has to be clever

Semiclassical

Suppose I take the convex hull of some point set, and then project that convex hull onto a subspace. The image of the convex hull is nothing other than the convex hull of the image of the point set.

Ryan Unger

My compactness proof is nice too

Ted Shifrin

In my multivariable math class, I made them compute explicit operator norms for some $2\times 2$ matrices. But the Lagrange multipliers gets challenging if dimensions go up ...

Ryan Unger

I’m using the Hilbert Schmidt norm of course

The basic thing is to bound |A|/tr(A) from above

Ted Shifrin

I've never worked with that.

Ryan Unger

6:35 PM

It’s important in mean curvature flow

Where A is the second fundamental form

Semiclassical

Hilbert-Schmidt is $\langle A,B\rangle = \text{tr}(A^\top B)$ right

Ryan Unger

Yeah

Semiclassical

(guess it should be a dagger depending on the context but w/e)

Ryan Unger

I don’t use complex numbers ;)

Semiclassical

hah, fair enough

I did come across a case where thinking in terms of H-S was useful for deriving the CHSH inequality in QM

Ryan Unger

6:37 PM

Which inequality?

Ted Shifrin

Yes, sure, I know.

Subhasis Biswas

@TedShifrin, if I start constructing the Cantor set with $[a,b]$ with both $a,b$ irrational , will every point in the set be irrational?

Semiclassical

Clauser-Horne-Shimony-Holte if I remember right, let's see if I do

Ted Shifrin

@Semiclassic: It's just the $\ell^2$ norm.

Semiclassical

ah, just Holt

Ted Shifrin

6:38 PM

@SubhasisBiswas: I have no idea.

Semiclassical

anyways, it's like the Bell inequality but with slightly looser conditions

Subhasis Biswas

@TedShifrin impossible.

Alessandro Codenotti

@SubhasisBiswas Seems unlikely

Ted Shifrin

Hi, demonic Alessandro. That was my intuition, too, but I'm often misintuited.

I guess we just stretch and translate the usual ternary expansion, but shrug

Subhasis Biswas

@AlessandroCodenotti I am trying to construct my own perfect set without any rational in it. Nonempty

Ted Shifrin

6:41 PM

Isn't there a problem like that in Rudin?

Semiclassical

with the typical Bell inequality, you've got observers Alice and Bob. You first consider experiments where Alice/Bob look at measurement types a/b respectively. You then do the same for types a/c and b/c.

Subhasis Biswas

@TedShifrin Yes!

Ted Shifrin

OK, I solved that once years ago.

Semiclassical

main point is that alice can measure using settings a,b and bob using settings b,c

Ted Shifrin

When I taught a reading course to two students, I assigned it.

Semiclassical

6:41 PM

with CHSH you let Alice and Bob use two different settings each

Alessandro Codenotti

@SubhasisBiswas Sure there's many ways to do that

Thorgott

Yeah, I remember that problem in Rudin.

Semiclassical

i.e. there's no common setting b

Thorgott

Mimicking the Cantor set construction is a good idea, but you need to be a bit more careful.

Semiclassical

anyways

Subhasis Biswas

6:43 PM

@Thorgott I was trying to follow the route, but my intuition (very likely a flawed one) says that some rational numbers will sneak into it

Balarka Sen

There's a completely outrageous way to do it, by cooking up a discrete subgroup of PSL2(R) whose limit set on dH^2 doesn't contain rationals

Ted Shifrin

ROFL @Balarka

Subhasis Biswas

@BalarkaSen what is $dH^2$?

Balarka Sen

I learnt this from Mahan. Typical hyperbolic geometer

Ryan Unger

@BalarkaSen did I tell you I randomly met him

Subhasis Biswas

6:45 PM

@BalarkaSen Mahan Maharaj?

Balarka Sen

@RyanUnger What? lmao

How

Ryan Unger

I sat next to him at a talk

Balarka Sen

LOL

Alessandro Codenotti

@BalarkaSen LOL what's wrong with BCT

Ryan Unger

He’s the only reason I know you’re real

Ted Shifrin

6:46 PM

Nah, Balarka is imaginary.

Balarka Sen

@Alessandro Oh my personal favorite way to do it is to set up a homeo C -> C x C and then picking out a nice vertical

Thorgott

If you just do the regular construction with irrational endpoints, then you will probably get some rationals. You can carefully exclude them though.

Alessandro Codenotti

That's also a cool approach

Balarka Sen

Brought to you by cardinality arguments

Oh yours is the measure theoretic thing?

Alessandro Codenotti

That's also one way to argue it

Balarka Sen

6:48 PM

Pick an appropriate translate of the standard Cantor set

Alessandro Codenotti

Otherwise you can invoke BCT

Balarka Sen

Hm how does that one go

Alessandro Codenotti

Same thing, but you need to justify why the translates don't cover everything

And that can be done either by measure theory or BCT

Balarka Sen

@RyanUnger I mean you don't necessarily know I'm real because he's real do you

Which talk was this

Ryan Unger

@BalarkaSen he confirmed you exist

Balarka Sen

6:49 PM

@Alessandro Ah

Ryan Unger

It was in Rio

Thorgott

I feel uncool for only knowing the standard way

Balarka Sen

@RyanUnger Oh I see

Adam

@BalarkaSen yeah I used to be a die hard metallica fan when I was 18-22, but in terms of the substance of the lyrical content they are pretty garbage, which is difficult to ignore considering how awesome their instrumental component is. But Trent Rez (Nine inch Nails) and his wife and her band (how to destroy angels) are amongst my favourites for modern

Ted Shifrin

@Thorgott Poor Thorgott.

Balarka Sen

6:51 PM

Nine Inch Nails is pretty good sometimes

I was never into Metallica, I should look into their older albums some day

Subhasis Biswas

Every closed set in a separable metric space is the union of a perfect set and an at most countable set . (applying it with $R^k$). Does that help?

Balarka Sen

My favorite Metallica album is Lulu by Lou Reed and Metallica :P

Ted Shifrin

Huh, @SubhasisBiswas? How did the separable metric space turn into $\Bbb R^k$?