5:00 PM
eg $S_3$ is symmetry group of the triangle which gives a non-free action on $S^1$

Hi @Ted

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

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

nice.
so just find the smallest group which isn't one of those
(for regular polyhedra, anyways)
oh hey
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...

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

5:02 PM

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

Yeah, polyhedra have very few possible symmetry groups.

polyhedra or regular polyhedra?

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

huh
neat

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.

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

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

I guess this all ties back to point groups

@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

I don't remember that, because I never knew it. :)
Is this a consequence of Thurston?

5:07 PM
Yeah it says every group acting freely on S^3 acts linearly... surprising theorem
Yup

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?

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).

Yup.
The baby unitary trick.

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

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

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

nice

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

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

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

Yeah, the Cayley route is not geometrically efficient.

5:18 PM
Makes sense
Sounds like a complicated representation theory question in general, but I won't put bets on it

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

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

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

@TedShifrin I'm needed? :)

As in, there may be a better upper bound?

5:22 PM
As if by magic ...

hey everyone

LOL

Heya

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

Hi, magic @Mathein, and hi, @ÍgjøgnumMeg

5:22 PM
Hi @Ted :)

16 hours ago, by N. Maneesh
1

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...

still I am struggling :(

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.

involute vary for different initial points.right?

@Ted Your summoning powers are fantastic

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

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 $G$?

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

5:27 PM
okay

$\beta$ 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.

@BalarkaSen what does that even mean?
you mean if there are arbitrarily large groups for which it is?

@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 $n$?

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

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

5:29 PM
@MatheinBoulomenos and that is even independently of the field

@MatheinBoulomen Ah, OK, the diagonal is fixed.

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

@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.

@TedShifrin okay. I will try,

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.

5:31 PM
Thank you
I will

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

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 $n=5$.

Interesting!

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

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 $d \leq \sqrt{n-1}$

5:34 PM
I knew Mathein would have some representation theory magic.

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

Nice, I didn't know that fact

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

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

@MatheinBoulomenos are you doing a PhD yet

5:35 PM

I completely avoid the group algebra in my notes

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

yeah

pops popcorn for representation theory fight

5:37 PM
This is in Simplicity and Representations part 1 and 2, right?
LOL @Ted

Yes, this is in Semisimplicity and Representations part 1
the thing about conjugacy classes is in part 2

I have vague memories of this from Thistlethwaite’s class

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

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

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 $r_n<s_n$.

5:40 PM
@TobiasKildetoft I appreciate that, but I personally really like noncommutative algebra

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

No, no, no decimals, @SubhasisBiswas.

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

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

not if I like both of them, anyway

5:41 PM
@MatheinBoulomenos So do I. But the students were not properly prepared for it.

@TedShifrin did I make any mistake?

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

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

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

@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?

5:48 PM
@TedShifrin I am thinking. Will let you know when I have an answer

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

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.

@TedShifrin yes. I was considering the case when both are irrational
@TedShifrin there is an integer between them?

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?

@TedShifrin by "shrinking " the distance accordingly

5:54 PM
Can you elaborate?

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

I am trying to formulate it

that was nonsense

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

5:54 PM

We live in an unfaithful world.

Has there ever been a world that was fully faithful?

Certainly not a free one.

this is so brilliant

5:57 PM
Ayy my man jamming to Bach

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, $ma < t < mb \implies a<t/m<b$

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

Yes!
They're beyond brilliant

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

@SubhasisBiswas: There you go.

5:58 PM

@BalarkaSen I've always liked Bach

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

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

@TedShifrin yes yes...

@Tobias Haha.

5:59 PM
anyone likes little fugue ?

me!

@LeakyNun <3

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

@BalarkaSen Can you really stand my algebraic musings?

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

6:00 PM
youtube.com/watch?v=5ZdDvB6eMDc this is so brilliant too

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

@LeakyNun What other classical music do you like

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

@BalarkaSen Mozart
but Bach is always my favourite :P

6:02 PM

baroque music is true music
3

Oof we got a baroque elitist bois

I like Rachmaninoff

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

Oh, Satie is awesome
I listened to the Gnossiennes so many times

6:04 PM
Gnossiennes > Gymnopedie agree or disagree
YES

anyone likes vivaldi?

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

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

main thing about music nowadays is that there's some much of it
or, rather, that there's so much of it accessible

@Semiclassical I heard The Mountain Goats released an underwhelming album

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

"In League With Dragons", the one after Goths
Very recent, released in April 26th

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

6:26 PM

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

I mean, they're allowed some misses

but I am not that big a believer of Fantano
Yeah haha massive discography

Beat The Champ was a great album, for instance

mostly great shit

6:27 PM
(the one that preceded Goths)

Haven't listened to 70% of it tbh

i've listened to a lot of it
both old and new

what is a nonconstructive proof?

I went through The Sunset Tree sometime ago
Liked it a lot

yeah

6:29 PM
@TedShifrin I made Vietnamese summer rolls with a friend, they were delicious :)

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

@SubhasisBiswas an incorrect one

blah @Ryan

I'm not sure how they would take the travel

@TedShifrin I only accept explicit constants in my inequalities

6:31 PM
Trying to figure out how to explain something simple without being laborious about it

You'll wise up eventually, @Ryan :P

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

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

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

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

6:32 PM
These are symmetric matrices
Everything can be assumed diagonal
I’m sure one just has to be clever

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.

My compactness proof is nice too

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 ...

I’m using the Hilbert Schmidt norm of course
The basic thing is to bound |A|/tr(A) from above

I've never worked with that.

6:35 PM
It’s important in mean curvature flow
Where A is the second fundamental form

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

Yeah

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

I don’t use complex numbers ;)

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

6:37 PM
Which inequality?

Yes, sure, I know.

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

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

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

ah, just Holt

6:38 PM
@SubhasisBiswas: I have no idea.

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

@TedShifrin impossible.

@SubhasisBiswas Seems unlikely

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

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

6:41 PM
Isn't there a problem like that in Rudin?

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.

@TedShifrin Yes!

OK, I solved that once years ago.

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

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

6:41 PM
with CHSH you let Alice and Bob use two different settings each

@SubhasisBiswas Sure there's many ways to do that

Yeah, I remember that problem in Rudin.

i.e. there's no common setting b

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

anyways

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

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

ROFL @Balarka

@BalarkaSen what is $dH^2$?

I learnt this from Mahan. Typical hyperbolic geometer

@BalarkaSen did I tell you I randomly met him

6:45 PM
@BalarkaSen Mahan Maharaj?

@RyanUnger What? lmao
How

I sat next to him at a talk

LOL

@BalarkaSen LOL what's wrong with BCT

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

6:46 PM
Nah, Balarka is imaginary.

@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

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

That's also a cool approach

Brought to you by cardinality arguments
Oh yours is the measure theoretic thing?

That's also one way to argue it

6:48 PM
Pick an appropriate translate of the standard Cantor set

Otherwise you can invoke BCT

Hm how does that one go

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

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

@BalarkaSen he confirmed you exist

6:49 PM
@Alessandro Ah

It was in Rio

I feel uncool for only knowing the standard way

@RyanUnger Oh I see

@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

@Thorgott Poor Thorgott.

6:51 PM
Nine Inch Nails is pretty good sometimes
I was never into Metallica, I should look into their older albums some day

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?

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

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

@TedShifrin I messed it up

@BalarkaSen im not kicking you only because I don't want to spam the moderators with notifications

6:53 PM
LOL
Lulu is a masterpiece believe it or not

You could put him in time out, @Alessandro.
I haven't used my authority yet.

How would that help you in concluding the perfect set contains no rational numbers?

Hint: how do you construct a perfect set in $[e,\pi]$ not containing 3?

Nice endpoints

But $[3,3]=\{3\}$

6:57 PM
chosen to minimize the typing

@AlessandroCodenotti I will try to come up with an answer, but I can use any irrational, right?

\sqrt{2} is way too long
@SubhasisBiswas sure

ok. It will take me some time.

@AlessandroCodenotti \sqrt2 will work just fine.

It's longer to type is what he meant
oh you shortened it ok

6:59 PM
I saved several keystrokes!
Blah!

Lol

People overuse curly braces.

"Well now that you had to explain it, no you didn't!"