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• CommentRowNumber1.
• CommentAuthorUrs
• CommentTimeApr 15th 2018
• (edited Apr 15th 2018)

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.

• CommentRowNumber2.
• CommentAuthorUrs
• CommentTimeApr 15th 2018

found a good textbook account: Rees 05

but again there is no pointer in there to the origin of the statement.

• CommentRowNumber3.
• CommentAuthorUrs
• CommentTimeApr 15th 2018

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…

• CommentRowNumber4.
• CommentAuthorUrs
• CommentTimeApr 15th 2018

according to Lamotke’s book, it seems that the result goes back to Klein 1884

(have not checked yet, need to run now)

• CommentRowNumber5.
• CommentAuthorDavid_Corfield
• CommentTimeApr 15th 2018

From here

In his Vorlesungen über das Ikosaeder [Klein,1993], published in 1884, Felix Klein gives the classification of finite subgroups of $SL(2,\mathbb{C})$ up to conjugacy

There’s an online translation here.

• CommentRowNumber6.
• CommentAuthorDavid_Corfield
• CommentTimeApr 15th 2018

• CommentRowNumber7.
• CommentAuthorUrs
• CommentTimeApr 16th 2018

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.

• CommentRowNumber8.
• CommentAuthorDavidRoberts
• CommentTimeApr 16th 2018

Surely the binary cyclic groups (ie the even-order cyclic groups) are also pre-images of the cyclic group of half the order?

• CommentRowNumber9.
• CommentAuthorUrs
• CommentTimeApr 17th 2018

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.

• CommentRowNumber10.
• CommentAuthorDavidRoberts
• CommentTimeApr 17th 2018

Tweaked description of which finite subgroups are preimages under the double cover projection.

• CommentRowNumber11.
• CommentAuthorUrs
• CommentTimeApr 17th 2018

Oh, now I see. Thanks!

• CommentRowNumber12.
• CommentAuthorDavidRoberts
• CommentTimeApr 17th 2018

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.

• CommentRowNumber13.
• CommentAuthorUrs
• CommentTimeSep 9th 2018

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

• CommentRowNumber14.
• CommentAuthorUrs
• CommentTimeSep 9th 2018

• CommentRowNumber15.
• CommentAuthorUrs
• CommentTimeOct 7th 2018
• (edited Oct 7th 2018)

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.

• CommentRowNumber16.
• CommentAuthorUrs
• CommentTimeOct 8th 2018
• (edited Oct 8th 2018)

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.

• CommentRowNumber17.
• CommentAuthorDavid_Corfield
• CommentTimeOct 8th 2018

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.

• CommentRowNumber18.
• CommentAuthorDavid_Corfield
• CommentTimeOct 8th 2018
• (edited Oct 8th 2018)

But $2 D_4$ is a subgroup of $2 I$, so that needs an arrow.

[Edit: which of course is there indirectly.]

• CommentRowNumber19.
• CommentAuthorUrs
• CommentTimeOct 8th 2018

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.

• CommentRowNumber20.
• CommentAuthorDavid_Corfield
• CommentTimeOct 8th 2018

It seems Cyclic Subgroups of the Sphere Braid Groups by Daciberg Lima Goncalves, John Guaschi, Springer.

• CommentRowNumber21.
• CommentAuthorUrs
• CommentTimeOct 8th 2018

Thanks! Had just found it, too, it’s on the arXiv: arXiv:1110.6628. Will add to the entry.

• CommentRowNumber22.
• CommentAuthorUrs
• CommentTimeOct 8th 2018
• (edited Oct 8th 2018)

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.

• CommentRowNumber23.
• CommentAuthorDavid_Corfield
• CommentTimeOct 8th 2018

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” ?

• CommentRowNumber24.
• CommentAuthorUrs
• CommentTimeOct 9th 2018

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.

• CommentRowNumber25.
• CommentAuthorUrs
• CommentTimeDec 7th 2018

• Mattia Mecchia, Bruno Zimmermann, On finite groups acting on homology 4-spheres and finite subgroups of $SO(5)$, Topology and its Applications 158.6 (2011): 741-747 (arXiv:1001.3976)

for classification of the finite subgroups of $O(5)$.

• CommentRowNumber26.
• CommentAuthorUrs
• CommentTimeDec 7th 2018

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)

• CommentRowNumber27.
• CommentAuthorUrs
• CommentTimeDec 7th 2018
• (edited Dec 7th 2018)

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?

• CommentRowNumber28.
• CommentAuthorDavid_Corfield
• CommentTimeDec 7th 2018
• (edited Dec 7th 2018)

The exceptional Coxeter group $H_4$ according to Dechant here, and Wikipedia agrees.

• CommentRowNumber29.
• CommentAuthorUrs
• CommentTimeDec 7th 2018

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).

• CommentRowNumber30.
• CommentAuthorDavid_Corfield
• CommentTimeDec 7th 2018
• (edited Dec 7th 2018)

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.

• CommentRowNumber31.
• CommentAuthorDavid_Corfield
• CommentTimeDec 7th 2018

So $\pm [I\times I]$.

• CommentRowNumber32.
• CommentAuthorUrs
• CommentTimeDec 7th 2018

Thanks a lot! Have added it here.

Might you also know the corresponding identification for the symmetry group of the 24-cell?

• CommentRowNumber33.
• CommentAuthorDavid_Corfield
• CommentTimeDec 7th 2018

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 $Spin(4) \equiv Sp(1) \times Sp(1) \to SO(4)$?

• CommentRowNumber34.
• CommentAuthorDavid_Corfield
• CommentTimeDec 7th 2018

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 $\pm \frac{1}{2}[O \times O]$?

• CommentRowNumber35.
• CommentAuthorUrs
• CommentTimeDec 7th 2018

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?

• CommentRowNumber36.
• CommentAuthorDavid_Corfield
• CommentTimeDec 7th 2018

Presumably that would be like taking a tetrahedron and then superimposing it on its (isomorphic) dual as in Compound of two tetrahedra.

• CommentRowNumber37.
• CommentAuthorUrs
• CommentTimeDec 7th 2018

Right, that must be it.

• CommentRowNumber38.
• CommentAuthorUrs
• CommentTimeFeb 18th 2019
• (edited Feb 18th 2019)

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

$\array{ \left\langle \exp \left( 2 \pi \mathrm{i} \tfrac{1}{{2n+1}} \right) \right\rangle & = & \mathbb{Z}/(2n+1) &&\overset{\phantom{AA}}{\hookrightarrow}&& Spin(3) \\ && \big\downarrow && && \big\downarrow^{ \mathrlap{\pi} } \\ \left\langle Ad_{ \exp \left( 2 \pi \mathrm{i} \tfrac{1}{2n+1} \right) } \right\rangle & = & \mathbb{Z}/(2n+1) &\overset{\phantom{AA}}{\hookrightarrow}& SO(2) &\overset{\phantom{AA}}{\hookrightarrow}& SO(3) }$

for the odd order cyclic subgroups.

• CommentRowNumber39.
• CommentAuthorUrs
• CommentTimeApr 11th 2019
• (edited Apr 11th 2019)

[ removed, sorry for the noise ]