Want to take part in these discussions? Sign in if you have an account, or apply for one below
Vanilla 1.1.10 is a product of Lussumo. More Information: Documentation, Community Support.
I am splitting off an entry classification of finite rotation groups from ADE classification in order to collect statements and references specific to the classification of finite subgroups of $SO(3)$ and $SU(2)$.
Is there a canonical reference for the proof of the classification statement? I find lots of lecture notes that give the proof, but all of them without citing sources or original publications of proofs.
I wrote:
Is there a canonical reference for the proof of the classification statement? I find lots of lecture notes that give the proof, but all of them without citing sources or original publications of proofs.
In Burban’s lecture notes (pdf) it has this statement:
The classification of finite isometry groups of $\mathbb{R}^3$ is a classical result of F. Klein (actually of Platon)
but still no concrete pointer…
according to Lamotke’s book, it seems that the result goes back to Klein 1884
(have not checked yet, need to run now)
Thanks for this! Excellent. I have added that reference also to all the entries on the separate finite rotation groups, as well as to that of the corresponding platonic solids.
Surely the binary cyclic groups (ie the even-order cyclic groups) are also pre-images of the cyclic group of half the order?
Yes, but not every cyclic subgroup of $SU(2)$ is in the image of taking pre-images. The odd order cycile groups are clearly not.
Oh, now I see. Thanks!
For what it’s worth I think the odd-order cyclic subgroups of $SU(2)$ are the image of the section $\mathbb{Z}/(2k+1) \to \mathbb{Z}/(4k+2) \simeq \mathbb{Z}/(2k+1) \times \mathbb{Z}/2$, so the odd-order cyclic subgroups of $SU(2)$ are index-2 subgroups of inverse images of finite subgroups of $SO(3)$, and the projection map $SU(2) \to SO(3)$ restricts to an isomorphism on these odd-order cyclic subgroups.
added statement of integral group (co-)homology of the finite subgroups of $SU(2)$ (here) together with a cross-pointer to discrete torsion of the sugra C-field
I am dabbling with drawing subgroup lattice under both $2O$ and $2I$ (a beginning here)
I didn’t fully appreciate before that $2T$ sits inside both of $2O$ and $2I$, forming a kind of axis of exceptional exceptionalism with $2D_4$. The three exceptional groups are not on the same footing, $2T$ is special.
Okay, I have produced a graphics of the full subgroup lattice of $SU(2)$ under the three exceptional finite subgroups. Now here.
It’s clearly not exactly symmetric under $2T \leftrightarrow 2T$ and $2O \leftrightarrow 2I$, but it comes surprisingly close. At least I had no idea that this is what the result was going to look like before I drew this.
This appendix seems to tally.
Given that $2 I$ is rather different with respect to normal subgroups (I guess for similar reasons that $A_5$ is simple), perhaps things are less similar than they seem.
But $2 D_4$ is a subgroup of $2 I$, so that needs an arrow.
[Edit: which of course is there indirectly.]
Thanks for the pdf. Can you see which publication this is the appendix of?
I should say that I am not really concerned about the symmetry-or-not about $2T$, but the fact that $2T$ takes a special place.
For instance $\beta$ is surjective over $\mathbb{R}$ for $2T$, but only surjective over $\mathbb{R}$ onto the integral characters for $2O$ and $2I$.
This has a curious consequence: If we do away with the irrational reps, then under McKay the corresponding vertices in the Dynkin diagram disappear, so the gauge group corresponding to the singularity should change. In other words, for $2O$ and $2I$, the McKay correspondence gets slightly modified as equivariant K-theory is replaced by equivariant stable cohomotopy, but not so for $2T$.
This may be relevant. GUT theory based on $2T \leftrightarrow E_6$ works. But GUT for $2O \leftrightarrow E_7$ and $2I \leftrightarrow E_8$ doesn’t actually work (at least not without bending over backwards), since these groups don’t admit chiral fermions.
It seems Cyclic Subgroups of the Sphere Braid Groups by Daciberg Lima Goncalves, John Guaschi, Springer.
Thanks! Had just found it, too, it’s on the arXiv: arXiv:1110.6628. Will add to the entry.
Just for the record, I found the association of reps to quiver vertices here (around p. 10 )
Turns out that
removing the two irrational nodes from the Dynkin diagram for 2O makes it become that of $2D_4$ and two disconnected nodes.
removing the four irrational nodes from the Dynkin diagram for 2I makes it become that of $A_4$ and one disconnected node.
now $A_4$ corresponds to $SU(5)$. So this case happens to match my little speculation… Not sure what to make of the other case, yet.
It’s KR-theory that’s at stake, isn’t it, rather than KO-theory?
A coincidence that it has been connected by Matthew Young with the representations of finite 2-groups (“This strongly suggests a role for Real 2-representation theory in M-theory”), as in the Platonic 2-groups for which Epa and Ganter remark: “The fact that there are canonical categorical extensions of all these groups suggests a categorical aspect of McKay correspondence that seems worth exploring” ?
We expect KR in general, but it reduces to KO at the orientifold fixed point.
Thanks for the pointer to Young, had not seen that.
From the maths ingredients it seems compelling to bring in the Platonic 2-groups. But despite some trying, I don’t see yet what’s really going on with them, in the physics story.
added pointer to
for classification of the finite subgroups of $O(5)$.
for classification of finite subgroups of $O(4)$ I have added pointer to
Patrick du Val, Homographies, Quaternions and Rotations, Oxford Mathematical Monographs, Clarendon Press (1964)
also(?): Journal of the London Mathematical Society, Volume s1-40, Issue 1 (1965) (doi:10.1112/jlms/s1-40.1.569b)
for finite subgroups of $O(4)$ also:
John Conway, D. A. Smith, On quaternions and octonions: their geometry, arithmetic and symmetry A K Peters Ltd., Natick, MA, 2003
Paul de Medeiros, José Figueroa-O’Farrill, appendix B of Half-BPS M2-brane orbifolds, Adv. Theor. Math. Phys. Volume 16, Number 5 (2012), 1349-1408. (arXiv:1007.4761, Euclid)
Which of these finite subgroups is the symmetry of the 120-cell?
Thanks. Can you figure out which group that is in terms of the list of all subgroups in tables 16, 17 and 18 in arXiv:1007.4761 ? (Beware that these tables are in the middle of the References, scattered over the very last pages).
This Google books page claims that the orientation preserving symmetries of the 120-cell (and so 600-cell) is $Y' \times \Y'/\mathbb{Z}_2$, where $Y'$ is the binary icosohedral group. So that makes it the final entry in one of those tables.
So $\pm [I\times I]$.
So the full group is the Coxeter group $F_4$ with 1152 elements. Then the orientation-preserving subgroup of $SO(4)$ is of order 576. But how does it break down relative to the double cover $Spin(4) \equiv Sp(1) \times Sp(1) \to SO(4)$?
What’s this telling us:
There is an analogy in four dimensions, where the 24-cell and its dual can be composed to yield an object with a rotational symmetry group of order 1152 instead of 576. The double cover of this, with order 2304, is obtained by choosing any ordered pair (l,r) of quaternions from the binary octahedral group 2O.
Recall that 2O has 2T as a normal subgroup, allowing us to refer meaningfully to ‘odd’ and ‘even’ rotations. If we add the additional constraint that l and r have the same parity, we get down to a group of order 1152, the double cover of the symmetry group of the 24-cell.
Does that make the right subgroup of $SO(4)$ isomorphic to $\pm \frac{1}{2}[O \times O]$?
Thanks again! Much appreciated. Sounds plausible. Would need to think about this.
The phrase “24-cell and its dual” sounds a little surprising, given that the 24-cell is supposed to be self-dual. But I guess the group action may depend on whether we think of it one way or the other?
Presumably that would be like taking a tetrahedron and then superimposing it on its (isomorphic) dual as in Compound of two tetrahedra.
Right, that must be it.
in making up for my sins in another thread, I have added more detail on how exactly all the finite subgroups of $SU(2)$ sit inside and how they map to $SO(3)$. Now it reads like so:
(…)
Here under the double cover projection (using the exceptional isomorphism $SU(2) \simeq Spin(3)$)
$SU(2) \simeq Spin(3) \overset{\pi}{\longrightarrow} SO(3)$all the finite subgroups of $SU(2)$ except the odd-order cyclic groups are the preimages of the corresponding finite subgroups of $SO(3)$, in that we have pullback diagrams
$\array{ \left\langle \exp \left( \pi \mathrm{i} \tfrac{1}{n} \right) \right\rangle & = & \mathbb{Z}/(2n) &\overset{\phantom{AA}}{\hookrightarrow}& Spin(2) &\overset{\phantom{AA}}{\hookrightarrow}& Spin(3) \\ && \big\downarrow &{}^{(pb)}& \big\downarrow &{}^{(pb)}& \big\downarrow^{ \mathrlap{\pi} } \\ \left\langle Ad_{ \exp \left( \pi \mathrm{i} \tfrac{1}{n} \right) } \right\rangle & = & \mathbb{Z}/n &\overset{\phantom{AA}}{\hookrightarrow}& SO(2) &\overset{\phantom{AA}}{\hookrightarrow}& SO(3) }$exhibiting the even order cyclic groups as subgroups of Spin(2), including the the minimal case of the group of order 2
$\array{ \left\langle \exp \left( \pi \mathrm{i} \right) = -1 \right\rangle & = & \mathbb{Z}/2 &\overset{\phantom{AA}}{\hookrightarrow}& Spin(2) &\overset{\phantom{AA}}{\hookrightarrow}& Spin(3) \\ && \big\downarrow &{}^{(pb)}& \big\downarrow &{}^{(pb)}& \big\downarrow^{ \mathrlap{\pi} } \\ \left\langle Ad_{ \exp \left( \pi \mathrm{i} \right) } = e \right\rangle & = & 1 &\overset{\phantom{AA}}{\hookrightarrow}& SO(2) &\overset{\phantom{AA}}{\hookrightarrow}& SO(3) }$as well as
$\array{ \left\langle \exp\left( \pi \mathrm{i} \tfrac{1}{n} \right), \, \mathrm{j} \right\rangle &=& 2 D_{2n} &\overset{\phantom{AA}}{\hookrightarrow}& Pin_-(2) &\overset{\phantom{AA}}{\hookrightarrow}& Spin(3) \\ && \big\downarrow &{}^{(pb)}& \big\downarrow &{}^{(pb)}& \big\downarrow^{ \mathrlap{\pi} } \\ \left\langle Ad_{\exp\left( \pi \mathrm{i} \tfrac{1}{n} \right) }, \, Ad_{\mathrm{j}} \right\rangle && D_{2n} &\overset{\phantom{AA}}{\hookrightarrow}& O(2) &\overset{\phantom{AA}}{\hookrightarrow}& SO(3) }$exhibiting the binary dihedral groups as sitting inside the Pin(2)-subgroup of Spin(3),
but only commuting diagrams
$\array{ \left\langle \exp \left( 2 \pi \mathrm{i} \tfrac{1}{{2n+1}} \right) \right\rangle & = & \mathbb{Z}/(2n+1) &&\overset{\phantom{AA}}{\hookrightarrow}&& Spin(3) \\ && \big\downarrow && && \big\downarrow^{ \mathrlap{\pi} } \\ \left\langle Ad_{ \exp \left( 2 \pi \mathrm{i} \tfrac{1}{2n+1} \right) } \right\rangle & = & \mathbb{Z}/(2n+1) &\overset{\phantom{AA}}{\hookrightarrow}& SO(2) &\overset{\phantom{AA}}{\hookrightarrow}& SO(3) }$[ removed, sorry for the noise ]
The entry states the group cohomology of finite subgroups of $SU(2)$ as an abelian group:
$H^n_{grp}(G_{ADE}, \mathbb{Z}) \;\simeq\; \left\{ \array{ \mathbb{Z} &\vert& n = 0 \\ G_{ADE}^{ab} &\vert& n = 2 \, mod \, 4 \\ \mathbb{Z}/{\vert G_{ADE}\vert} &\vert& n \, \text{positive multiple of} \, 4 \\ 0 &\vert& \text{otherwise} } \right.$Question: What’s the ring structure under cup product?
In particular, what’s
$H^2 \otimes H^2 \overset{\cup}{\longrightarrow} H^4$??
Partial answers:
For cyclic groups it’s clear.
For generalized quaternion groups ( in fact for all binary dihedral groups) I see it’s Theorem 3 in
For the ordinary quaternion group and for the binary icosahedral group it’s in
1 to 40 of 40