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12:03 AM
@rschwieb Heh, what's the context?
Hey @anakhro everyone
Hey ultradark.
I just couldn’t solve the conjecture
Conjecture of what?
A conjecture about @anakhro; -; the: flow of information from place to place being necessary for the universe to exist. I’ve given up
And I’m never returning to it
12:10 AM
what does that even mean?
Related to the holographic principle in Physics
according to the H lophic
The universe is constructed only from the flow of info
Yeah I agree, throw this idea in the trash
Whoever came up with that idea
Is a crank
There are about 500 of them
Leaders in their field
They’re not cranks lol.
12:16 AM
Cranks are allowed to have their own subfield of crankery.
Yeah maybe they are a bit crank
Like 30 percent
Can you link me to something on the arxiv?
12:31 AM
What's a group presentation for $SO(3)$?
I have no idea.
Is it even finitely presented?
I should have asked that first
Why do you want a group presentation?
Idle curiosity
I would be shocked if it could be finitely presented. Hard to see how composing a finite number of rotations could give rotations through arbitrary angles around arbitrary axes.
12:35 AM
The orthogonal group, consisting of all proper and improper rotations, is generated by reflections. Every proper rotation is the composition of two reflections, a special case of the Cartan–Dieudonné theorem.
Yes, I know that result.
I didn't but it makes sense
Compute it in the plane first.
It's a great theorem.
But using reflections, my complaint still stands.
I actually don't know the Cartan-Dieudonné Theorem. What sayeth it?
12:37 AM
Yeah it does seem unreasonable to expect a finite presentation
Let (V, b) be an n-dimensional, non-degenerate symmetric bilinear space over a field with characteristic not equal to 2. Then, every element of the orthogonal group O(V, b) is a composition of at most n reflections.
I would hazard a guess that it's not even countably presented, but someone might show me I'm wrong.
Oh, cool.
I see: That's the generalization of the physicists' Euler angles in $\Bbb R^3$.
Oh, no, never mind. That isn't right.
That would say you can write every rotation as a composition of ($n$) rotations about the coordinate axes.
I doubt that's true for $n>3$.
These are all good questions for an industrious undergraduate to investigate.
12:54 AM
Hello nerds
Hi nerd Demonark.
@KarlKronenfeld Just filling a gap in my database , and I couldn’t believe it hadn’t been filled earlier.
How's everything going?
1:28 AM
Q: How does the Evolute of an Involute of a curve $\Gamma$ is $\Gamma$ itself?

Unknown xHow 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...

I can't even parse that title.
I got this question from a lecture video.
He is not explaining completely.
1:44 AM
Did he actually say the title anywhere
Because I still can't parse it.
yes. !7:04
evolute of the involute is the curve itself
Do you mean to ask "how is the evolute of an involute of a curve $\Gamma$ itself?"
for evolute, we have a fixed point on the curve right?
then we consider tangent to each point right of that
Involute is the curve that is perpendicular to each tangent line.
sorry, for involute we have fixed point.
then we consider tangent to each point right of that
Involute is the curve that is perpendicular to each tangent line.
For finding the Evolute of this Involute, we need to find the centre of curvature at each point.
I have never heard of an evolute or involute, sorry.
5 hours later…
6:57 AM
Thanks @TedShifrin for answering
7:34 AM
Good morning, we are trying to solve some games on infinite chessboard in one room, I will post them here also, it could be that someone finds them interesting.
Q: $6$ bishops and a knight on an infinite chessboard

GrešnikPlayer $A$ places $6$ bishops wherever he/she wants on the chessboard with infinite number of rows and columns. Player $B$ places one knight wherever he/she wants. Then $A$ makes a move, then $B$, and so on... The goal of $A$ is to checkmate $B$, that is, to attack knight of $B$ with bishop in ...

Q: A simple game on infinite chessboard

GrešnikPlayer $A$ chooses two queens and an arbitrary finite number of bishops on $\infty \times \infty$ chessboard and places them wherever he/she wants. Then player $B$ chooses one knight and places him wherever he/she wants (but of course, knight cannot be placed on the fields which are under attack ...

7:57 AM
Here is a room:

4 hours later…
11:43 AM
I get a 6 day ban for telling someone to shut up?
fair enough I am socially inept in the ways everyone else I've met are not, but they are also socially inept in the ways I am not
11:56 AM
The invariant formula for the exterior product, why would someone come up with something like that. I mean it looks really similar to the formula of the covariant derivative along a vector field for a tensor but otherwise I don't see why would it be something natural to come up with. The only places I have used it is deriving the poisson bracket of two one forms
@Albas What is the "invariant formula for the exterior product"
That should be exterior derivative. en.wikipedia.org/wiki/Exterior_derivative look under in terms of the invariant formula
Exterior derivative, not product!
So this is just an infinitisimal Stokes theorem
Think about what is happening for $n = 2$. $d\omega(X, Y) = X\omega(Y) - Y\omega(X) - \omega([X, Y])$ for a $1$-form $\omega$
@BalarkaSen Yea sorry I always confuse that the naming. I guess its because of the interior product
If you have two vector fields $X$ and $Y$ on your manifold, then you have the following formula relating the flows of $X$, $Y$ and $[X, Y]$:
Call $\Phi_t$ the flow of $X$ for time $t$ and $\Psi_t$ the flow of $Y$ for time $t$. Then $[X, Y] = d/dt \Phi_{\sqrt{t}} \Psi_{\sqrt{t}}\Phi_{-\sqrt{t}} \Psi_{-\sqrt{t}}$.
12:12 PM
This means starting at a point $p$, flowing along $X$ for time $\sqrt{t}$, then along $Y$ for time $\sqrt{t}$, then backwards along $X$ for the same time, backwards along $Y$ for the same time, leads you at a place different from $p$. And upto second order, flowing along $[X, Y]$ for time $t$ from $p$ will lead you to that place.
Yea measures the noncommutativity of the flows
The lie bracket that is
So there's an infinitisimal truncated square spanned by $X$ and $Y$ which is really a pentagon with an edge really small that represents $[X, Y]$
Think of evaluating $\omega$ on the edges of the truncated square and doing a signed sum of the values. You'll get value of $\omega$ on the two $X$ edges, whose difference (after taking a limit) is $Y\omega(X)$, the value of $\omega$ on the two $Y$ edges, whose difference (again after taking a limit) is $X \omega(Y)$ and on the truncation edge it's $\omega([X, Y])$
Gently taking caring of the signs, the total value is $X\omega(Y) - Y\omega(X) - \omega([X, Y])$
So value of $d\omega$ on the Lie square spanned by $X$ and $Y$ = signed sum of values of $\omega$ on the boundary of the Lie square spanned by $X$ and $Y$
Infinitisimal version of $\int_M d\omega = \int_{\partial M} \omega$
12:18 PM
Thats very nice. I have a doubt, what do you mean evaluating $\omega$ on the edge. As in $\omega$ takes in the vector field along that edge?
More or less, yeah. Think intuitively instead of too rigorously though
Cool cool
But I believe you can actually write down a proof like this, by doing $\int_{I^2} d\omega = \int_{\partial I^2} \omega$ where $I$ is the little truncated square I described and taking $\text{vol}(I) \to 0$
Didn't try it out but it should work out
Okay will try it out.
For the general case $d\omega(X_1, \cdots, X_{n+1}) = \sum_i (-1)^{i+1} X_i \omega(X_1, \cdots, \hat{X_i}, \cdots, X_{n+1}) + \sum_{i < j} (-1)^{i+j} \omega([X_i, X_j], X_1, \cdots, \hat{X_i}, \cdots, \hat{X_j}, \cdots, X_{n+1})$ says the same thing, but on a big truncated Lie cube
12:22 PM
A "multidimensional" box
@Albas Your analogy with the covariant derivative is apt. What does $X\omega(Y) - Y\omega(X) - \omega([X, Y])$ remind you of in Riemannian geometry?
Something like torsion?
Awesome man
Actually you can make this purely formal
I can tell you if you're interested
12:26 PM
Let's do bullshit generality. $E$ be a vector bundle on $M$ and $\nabla$ be a connection $E$. Remember that this means it's an $\Bbb R$-bilinear operator $\nabla : \Gamma(TM) \times \Gamma(E) \to \Gamma(E)$ denoted as $(X, s) \mapsto \nabla_X s$ which is (a) $C^\infty(M)$-linear in the first factor (b) $C^\infty(M)$-Leibniz in the second factor.
Explicitly, (b) is $\nabla_X (fs) = X(f)s + f\nabla_X s$
Makes sense
You can verify that this in particular means it's a pointwise defined on the first factor. This means to evaluate $\nabla_X s(p)$ you only need $X(p) \in T_p M$ not the full vector field. That makes sense, right? You can take directional derivative of a function at a point in the direction of a single vector at that point
Yeah that's how a connection when defined for a riemannian manifold behaves. So I guess a generality should also follow that
Suppose that $G$ is a group acting freely on a tree $T$ via graph automorphisms; let $T'$ be the associated spanning tree. Call an edge $e = \{u,v\}$ in $T$ essential if $e$ doesn't belong to $T'$. Note: it is easy to prove that if $u \in T'$, then $v \notin T"$ (this follows from uniqueness of paths between vertices).
Now, let $e = \{u,v\}$ be an essential edge with $u \in T'$. I am reading through a proof and the author claims that there is a $g \in G$ such that $g \cdot v \in T'$? My thought was to try to show that $orb(u) \neq orb (v)$ and then use the fact that the spanning tree contains exactly vertex from each orbit. But I can't seem to prove that orb(u) \neq orb(v)...
12:38 PM
@Albas Right, more or less. So it defines an operator $d^\nabla : \Gamma(E) \to \Gamma(E \otimes T^*M)$, which takes a section $s$ of $E$ and spits out $d^\nabla(s)$ which is a section of $E \otimes T^*M$, which is the same as a bundle homomorphism $TM \to E$ ($V \otimes W^* \cong \text{Hom}(W, V)$ for vector spaces). So what is this homomorphism $d^\nabla(s) : TM \to E$? Just $d^\nabla(s)(X) = \nabla_X s$.
This might be complicated to grok first but basically think of it as currying. Making a billinear map a linear one, like in linear algebra.
You can replace $E \otimes T^*M$ just by the Hom-bundle $\text{Hom}(TM, E)$ in your head if you want. Nothing is lost.
I'll use the latter notation consistently if that's what you're comfortable with
Yea kind of makes sense. So you're fixing your s so that the map now varies with $X$. This is somewhat similar like saying that $\nabla Y$ is $(1,1)$ tensor
Yeah precisely
$\nabla s$ is a bundle homomorphism $TM \to E$, sending $X$ to $\nabla_X s$. That's the (1, 1) tensor in your context.
And then I also "contract $s$" to just make a map $d^\nabla : \Gamma(E) \to \Gamma(\text{Hom}(TM, E))$, defined by $d^\nabla(s) = \nabla s$
(Technical point: Note how contracting $X$ in $\nabla_X s$ made a bundle-homomorphsm $TM \to E$ but contracting $s$ in $\nabla s$ only gave as a map $\Gamma(E) \to \Gamma(\text{Hom}(TM, E))$ at the level of space of sections, not a bundle-homomorphism $E \to \text{Hom}(TM, E)$. This is because $\nabla_X s$ is pointwise defined on $X$ and not $s$)
Ah yes.
12:46 PM
@Albas So this fella is called the exterior covariant derivative. Denote $\Omega^0(M; E) = \Gamma(E)$ ($0$-forms with values on $E$ aka functions on $M$ with values on $E$ aka sections of $E \to M$), denote $\Omega^1(M; E) = \Gamma(\text{Hom}(TM, E))$ ($1$-forms with values on $E$ aka bundle-homs $TM \to E$)
Then this is the 0 level exterior derivative $d : \Omega^0(M; E) \to \Omega^1(M; E)$. That's what a connection is, a 0-level exterior derivative of a bundle-valued theory of differential forms
So what's $\Omega^k(M; E)$ for higher $k$? Define it as $\Omega^k(M; E) = \Gamma(\text{Hom}(TM^{\wedge k}, E))$. Just space of alternating multilinear bundle homomorphisms $TM \times \cdots \times TM \to E$.
Note how if $E$ is the trivial bundle $M \times \Bbb R$ of rank 1, then $\Omega^k(M; E) = \Omega^k(M)$, the usual space of differential forms.
Makes sense so far?
Yeah. It will take me some time to take all this in. There's a lot of notation to swift through
Let me know when you're ok so we can continue :)
It seems that $orb(u) \neq orb(v)$ is equivalent to there existing a $g \in G$ such that $g v \in T'$. Freeness of the action needs to be used somehow, but i don't see it.
My next plan is to define higher exterior derivatives $d : \Omega^k(M; E) \to \Omega^{k+1}(M; E)$ for $k > 0$, but only after this formalism is clear.
12:51 PM
@user193319 if the orbits were equal then $gu=v$ and $hv=u$ for some $g,h\in G$. Then $hgu=u$ and $ghv=v$, since the action is free $h=g^{-1}$. Set $w=gv$. Then $hw=u$, thus $w=v$, a contradiction
@AlessandroCodenotti Ah, very nice. Thanks!
Okay make sense let's continue @BalarkaSen. There's this itch in me which makes me think that can we define some crazy generalization of De Rham cohomology here.
@Albas Alas, we won't have any de Rham cohomology. But not having one would precisely be the point of interest.
Oh no what is happening in here
Ah I guess $(d^{\nabla})^2$ won't be $0$
1:00 PM
You get a curvature term
It would be the Riemann curvature tensor, yep
@AlessandroCodenotti Wait, why does $hw = u$?
@Albas Okay! So here is a definition of the higher exterior derivatives $d^\nabla : \Omega^k(M, E) \to \Omega^{k+1}(M, E)$ takes a $E$-valued $k$-form $\omega \in \text{Hom}(TM^{\wedge k}, E)$ to $d^\nabla \omega \in \text{Hom}(EM^{\wedge k+1}, E)$ defined by $d^\nabla \omega(X_1, \cdots, X_{k+1}) = \sum_i (-1)^{i+1} \nabla_{X_i} \omega(X_1, \cdots, \hat{X_i}, \cdots, X_n) + \sum_{i < j}(-1)^{i+j} \omega([X_i, X_j], X_1, \cdots, \hat{X_i}, \cdots, \hat{X_j}, \cdots, X_n)$
Broke the chat but you get my point
This is exactly the invariant formula for higher exterior derivatives, modified to fit the scenario
@user193319 Ah I got confused $hw=v$
Whatever, assume the orbits intersect and play around with the action until you get an edge inversion of $e$ I think
1:06 PM
What is an edge inversion?
@Albas Note how $\omega(X_1, \cdots, \hat{X_i}, \cdots, X_n)$ is a section of $E$, so I am doing $\nabla_{X_i}$ of that
An elelement swapping $u$ and $v$, thus fixing an edge
@BalarkaSen Yea its well defined like that
That's what taking derivative of a section of $E$ wrt a vector field on $M$ means, taking the connection
Alright so to verify that $d^2 \neq 0$ indeed, let's just do the computation: $(d^2s)(X, Y) = d(ds)(X, Y) = \nabla_X ds(Y) - \nabla_Y ds(X) - ds([X, Y]) = \nabla_X \nabla_Y s - \nabla_Y \nabla_X s - \nabla_{[X, Y]} s$
Voila, Riemann curvature tensor
Well, that's what it is called when $E = TM$ so that $s = Z$ is some vector field on $M$. This is the bundle curvature
This is super neat
1:11 PM
Here's a point. What is $d\omega$ for $\omega \in \Omega^k(M; E)$ "really"? What would, for example, having $d\omega = 0$ mean?
Well, the point is, $d : \Omega^k(M; E) \to \Omega^{k+1}(M; E)$ is a connection operator on $E$-valued $k$-forms on $E$. So $d\omega = 0$ would mean that the form $\omega$ is parallel with respect to the connection $\nabla$.
Umm let me get this wait
Let $V$ be a finite dimensional real vector space, $q$ a quadratic form on $V$ and $Cl(V,q)$ the associated Clifford algebra, with the $\Bbb Z/2\Bbb Z$-grading $Cl(V,q)=Cl(V,q)^0\oplus Cl(V,q)^1$. We define $P(V,q)$ as the group of elements of $Cl(V,q)$ with $q(v)\neq 0$ (under the identification $V\hookrightarrow Cl(V,q)$) and $\mathrm{Pin}(V)$ as the subgroup of $P(V,q)$ with $q(v)=\pm 1$. We define $\mathrm{Spin}(V)$ as $\mathrm{Pin}(V)\cap Cl(V,q)^0$.
Is $\mathrm{Spin}(V)$ the set of elements with $q(v)=1$?
Just like $ds = 0$ for a section $s \in \Gamma(E)$ would mean $\nabla_X s = 0$ for all vector fields $X$ on $M$ would mean $s$ is a "parallel section". Think in Riemannian geometry, for example.
Oh okay, yeah
If $ds(X) = 0$ then $\nabla_X s = 0$ i.e., $s$ doesn't change if you parallel transport along $X$.
1:15 PM
OK, let's wrap this up and interpret the torsion tensor in this formalism
Yeah, this is a lot to think about
Torsion only makes sense on the tangent bundle, so take $E = TM$ from the start. Consider the identity bundle homomorphism $TM \to TM$... you can think of this as an element of $\Omega^1(M; TM)$. This is called the "soldering form", comes tautologically when you work with the tangent bundle.
You'll also see this thing appearing in symplectic geometry. I think they call it the tautological 1-form
Yeah on the cotangent bundle
(The cotangent bundle is naturally a symplectic manifold)
So let's give this guy a name, $\theta \in \Omega^1(M; TM)$. It's exterior covariant derivative $d\theta$ is a $TM$-valued $2$-form on $M$, explicitly $d\theta(X, Y) = \nabla_X \theta(Y) - \nabla_Y \theta(X) - \theta([X, Y])$.
But $\theta$ is the identity operator, so this is $\nabla_X Y - \nabla_Y X - [X, Y]$. Torsion tensor!!
1:21 PM
Where did you learn all this
$d\theta = 0$, or vanishing of torsion, is exactly the condition that the soldering form is parallel with respect to the connection
Like torsion free connection
Which is natural, I mean: you'd want the identity homomorphism $TM \to TM$ to be parallel in whatever geometry you're working on. Roughly speaking it means the space doesn't "twist in itself"
@Albas I dunno, I saw it somewhere on MO once. Picked up bits and pieces of bundle-valued differential forms by working it out by hand. This is all language game at the end of the day
But a powerful one I think
Yea kind of enforces Cartan's thing that differential forms are supreme
They are everywhere
Lol well I am very bad at differential forms to be honest
This is just something cool without having to do any local computations by hand. That's what bundle geometry helps you to do, avoid local computations as long as possible
1:25 PM
Yea, computations especially here will kill you
I remember trying to find killing vector fields for various kinds of metrics. It was horrendous. But Killing Vector fields are nice
I don't know any explicit calculations man
They are nice man. If you find the Killing vector fields for the Minkowski space, I think they give you the generators of the Poincare algebra.
I really have to thank my physics background for providing me motivation to do calculations
I have 0 clue about semiRiemannian geometry. Killing fields are the ones which integrate to geodesics, right? So, what, the flows are isometries and for R^{3, 1} the pointwise isometry group is O(3,1)
The global isometry group is that semidirect product R^4 or whatever
You'd get the Lie algebra of R^4 $\rtimes$ O(3, 1)
So I was reading this thing called the Poisson bracket. With the poisson bracket you can give the space of all smooth functions on a symplectic manifold a Lie algebra structure. And then you can show that a symplectomorphism must also preserve the Poisson structure. I would like to calculate the Poisson Lie algebra for something like $S^2$. Something cool might pop up
The space of Killing fields on $M$ is just the tangent space at identity (or the Lie algebra) of $\text{Isom}(M)$ (which is a Lie group by a famous theorem)
at least when $M$ is Riemannian
1:33 PM
@BalarkaSen Yup exactly, thats why when you find the vector fields you see that they are the generators of the Poincare algebra which is just the Lie algebra of O(3,1) the lorentz group
@BalarkaSen Oh did not know that
Gtg then, thanks for this @BalarkaSen
I don't know much about the Poisson bracket, seems interesting
@Albas $S^2$ is not a symplectic manifold!
Try torus, maybe, if you want model examples
The standard symplectic structure on $\Bbb R^{2n}$ descends to $T^{2n}$
@BalarkaSen $S^2$ is, its the only even dimensional sphere to satisfy the cohomology condition
Oh yeah sorry rip the volume form of course lol
Mentally replaced that with $S^{2n}$, smh
1:44 PM
Lol. S^2 even has the almost complex structure which originally made me think about symplectic manifolds without Kahler structure.
It has a complex structure. It's CP^1
CP^n are the prime examples of Kahler manifolds
In particular it means any smooth projective algebraic variety are also Kahler, because the Fubini-Study metric on CP^n restricts
@Albas You should ask that question to Ted, he would know an example
Oh cool, I will. The question has actually peaked my interest, let me see if I can construct some example on my own. Most probably won't, I don't even know the theory of complex manifolds properly
Idk any of this
just barely understand out of folkloric knowledge
You know more than I do man. I mean you have the language, now you just have to sit with the material.
@ÉricoMeloSilva its not possible to ask Alice Chang a question
The lecture must go on uninterrupted
1:55 PM
And the point is not to compare how much one knows. Just have to learn I guess
sure its whatever just learn because its fun
ill have to read symplectic geometry seriously at some point
You at TIFR now?
just need to read these 50 things before that
@Albas not yet
i leave on 30th
Hmm for a month?
a little less than that
3 weeks more or less
1:58 PM
@BalarkaSen come to Princeton. Pardon is teaching it next semester
o shit
I have a couple of friends who are doing something with elliptic curves there. Went through VSRP.
aha nice
Yo what’s up chat. Anyone know an inter sing result in symplectic geometry? Like geodesic and the metric on the manifold maybe torus or that possibly convex thing?
I actually have no idea what it’s about
2:09 PM
Oh :( that’s okay though
You know a lot lol
Not enough
Spoken like a true champ
@RyanUnger y
2:27 PM
@BalarkaSen lol @ bullshit generality.
Say hwat’ Adam @Adam
Say something
Something modern.
> There exists an infinite geodesic that is geodesically complete
That is indeed something modern
you modernist
@Secret how do you know so much math if you’re a PhD chemistry student
I’m perplexed. I really am. My brother barely knows basic cell biology and his IQ is nearly 106!
You just learn along the way. Passion makes a difference
I think I’m going to purchase a text book about that geodesic thing you said. Where did you read that fact?
Yes you are right.
That’s awesome though
Not many people are that adventurous
2:46 PM
@ÉricoMeloSilva there’s a good chance she’ll just ignore you
Or she’ll say something like “did I make a mistake? Sorry” and just carry on
Paul Yang didn’t bring the notes so I guess I just won’t understand anything
3:38 PM
Hey everyone!
Nerd alert
There ^
4:03 PM
If someone has the time to quickly check my result, I would appreciate.
Let $X_{i},....,X_{n} \sim \Gamma(2,\,\frac{2}{\lambda})$
Is $\mathbb{E}([\frac{(\frac{X_{1}+...+X_{n}}{n})^2}{2}] = \frac{1}{n^2\lambda^2}+\frac{2}{\lambda^2}$ ?
$\frac{1}{n\lambda^2}+\frac{2}{\lambda^2}$ *
Uh apparenty there are metrizable Baire spaces $X$ such that $X^2$ not only is not Baire, but it has a countable family $D_\alpha$ of dense open sets such that $\bigcap_{\alpha<\omega}D_\alpha$ is empty
@Ultradark I don't know what you mean, but you seem down in the dumps champ. Remember, girls are not as significant as you might think, design an attachment for a cordless drill and a flesh light that oscillates perpendicular to the drill's rotation and your done. Even better than the natural method
4:24 PM
@Ultradark the non-squeezing theorem is one of the defining problems of symplectic geometry: en.wikipedia.org/wiki/Non-squeezing_theorem
It's one I always like to tell people who don't know anything about math who wonder about symplectic geometry and what it means.
Oh cool this is nice
This is very nice lmao. Gromov is life
How the heck did he think of proving something like this
The linear one is a little bit easier to fathom.
Yeah I believe that
From Mcduff & Salamon, by the way.
I need to read that someday
4:33 PM
I need to finish it some day. :(
looks at his bookcase sadly.
I am trying to show that if $d$ divides $24$, then $S_4$ has a subgroup of order $d$. The only proof I could come up with is a brute force proof. It actually was too bad. E.g., orders $2$, $3$, and $4$ are easy (just take the subgroup generated by a 2 cycle, 3 cycle, and 4 cycle, respectively); $d=8$ is Sylow's theorem; $d=12$, take $d=24$, take $S_4$. The only case that presented a semblance of trouble was $d=6$. But the group generated by $(1,2)$ and $(1,2,3)$ does the job.
My only quibble with this solution is that it doesn't seen very elegant. Is there a better way?
$d = 6$ shouldn't be hard, it's an embedded $S_3$ in $S_4$!
Maybe it's easier if you think about $S_4$ as the orientation preserving symmetry group of the cube.
which is isomorphic to the subgroup generated by $(1,2)$ and $(1,2,3)$
That's why I said "semblance of trouble"
Yeah but I mean you shouldn't have to think more than 2 seconds or even write an explicit generating set
It's just a stabilizer subgroup of the usual action of $S_4$ on a set of size $4$ by permutation
"orientation preserving symmetry group of the cube" that doesn't sound much easier.
4:44 PM
It's immediate from that what the $d = 8$ guy is. Take the equatorial cube, it's symmetry group is a $D_8$ embedded in $S_4$.
No Sylow theory required
The $d = 12$ guy is the symmetry group of an embedded tetrahedra in the cube, which is an $A_4$ in $S_4$
You can list down all the groups purely geometrically from this
Well, the only reason why I used sylow is because this problem is part 3 of a problem, where in part 2 I showed that S4 has a sylow 2 subgroup of order 8 (three actually).
What is an equatorial cube?
Equatorial square I meant, whoops
Think of the cube as $[-1, 1]^3$ and slice by $z = 0$
Ah, okay.
4:48 PM
You can also slice by $x = 0$ and $y = 0$, which gives three squares inside the cube, which correspond to those three nonconjugate 2-Sylow $D_8$'s in $S_4$ that you speak of!
equatorial square is a nice phrase, i may have to steal it for my own purposes
as an example of that kind of usage: the elliptope, defined by $x^2+y^2+z^2\leq 2xyz+1$ in the [-1,1]^3 cube, looks like this
and, amusingly, the three equatorial cross-sections are all circles
e.g. $z=0\implies x^2+y^2 \leq 1$
In fact, the action of $S_4$ on these three 2-Sylows by conjugation gives a surjective homomorphism $S_4 \to S_3$ whose kernel is a $V_4$. This $V_4$ can be thought as the sub-symmetries of the cube which act on the three pairs of faces {{top, bottom}, {right, left}, {front, back}}.
(the non-equatorial cross-sections are all ellipses, hence why it's called theh elliptope)
Clearly these are 180 degree rotations along the $x$, $y$ and the $z$-axis. But composing the 180 rotation along the $x$ with a 180 rotation along the $y$ gives you a 180 rotation along the $z$, indicative of the $ab = c$ relation in Klein's 4-group
Everything about $S_4$ is encoded in the cube, in a way
The same can be said of $A_5$ and the dodecahedron, say
Hi, a @Balarka. You're channeling Artin :)
4:54 PM
and A_4 for the tetrahedron, etc
Hi @Ted! Yup
This was one of my favorite parts of my algebra course when I used to teach it :P
Haha I love it a lot
hmm. Now I'm wondering if every finite group is the symmetry group of some polyhedron (of appropriate dimension)
certainly not regular
4:57 PM
Not seeing why that restriction is obvious
if $G$ is symmetry group of a regular polyhedron in $\Bbb R^{n+1}$ say then $G$ acts on $S^n$ freely I think
Because embed the regular polyhedron in a sphere
but groups acting freely on spheres have nontrivial cohomological restrictions
Hm no reason for the action to be free I suppose

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