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Topic: Symmetries in four dimensions (Read 981 times) |
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Deedlit
Senior Riddler
Posts: 476
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Symmetries in four dimensions
« on: Sep 7th, 2006, 4:14pm » |
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Something I've been wondering about.
1. What are the finite subgroups of O(4)? That is, what are the "rotations/reflections" in four dimensions? (Actually, not all origin-preserving isometries are rotations or reflections, but that's the basic idea.)
which leads into...
2. What are the "fair dice" in four dimensions? I'm calling a polytope "fair" if the symmetry group of the polytope is transitive on the facets.
Fair dice in three dimensions are known as isohedra, so I guess the four-dimensional counterparts would be called isochora.
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Michael Dagg
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Re: Symmetries in four dimensions
« Reply #1 on: Sep 8th, 2006, 7:03am » |
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The list of finite subgroups of O(n), by contrast, is complicated. It
has to be, since every finite group embeds into O(n) for sufficiently
large n. The general theorem due to Jordan is that there is a
function f : N --> N such that if G < O(n) then G contains a normal
abelian subgroup A of index at most f(n) . In particular, there are only
finitely many possibilities for G/A , which you can think of as finitely
many "families" of groups (in the same sense that all the dihedral groups
are in one "family" of subgroups of O(2) ; they all have a normal abelian
subgroup of index 2).
There's some discussion about this here http://www.math-atlas.org/95/finite_order
If your second question is sort of, "What are the cognates of the
Platonic solids in R^4", the answer is known, in fact for all R^n. (It's
actually simpler than one might want for n>4 !)
See also http://www.math-atlas.org/99/120cell and http://www.math-atlas.org/98/600.cell
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| « Last Edit: Sep 8th, 2006, 7:03am by Michael Dagg » |
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Michael Dagg
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Deedlit
Senior Riddler
Posts: 476
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Re: Symmetries in four dimensions
« Reply #2 on: Sep 8th, 2006, 2:14pm » |
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Thanks, Michael.
on Sep 8th, 2006, 7:03am, Michael_Dagg wrote:The list of finite subgroups of O(n), by contrast, is complicated. It
has to be, since every finite group embeds into O(n) for sufficiently
large n.
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I did know this, but it also occurred to me that, for instance, the collection of subgroups of finite Coxeter groups also contains every possible finite group up to isomorphism, and the finite Coxeter groups has a nice classification. (The ABCDEFGHI classification) So, I'm wondering:
1. Is every finite subgroup of O(n) (or SO(n)) a subgroup of a finite Coxeter group?
2. If the above is false, is their a similar statement we can make, i.e. "every finite subgroup of O(n) is a subgroup of some group from this collection, which has a nice classifcation."
Quote:
The general theorem due to Jordan is that there is a
function f : N --> N such that if G < O(n) then G contains a normal
abelian subgroup A of index at most f(n) . In particular, there are only
finitely many possibilities for G/A , which you can think of as finitely
many "families" of groups (in the same sense that all the dihedral groups
are in one "family" of subgroups of O(2) ; they all have a normal abelian
subgroup of index 2).
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Do you have any more information on this theorem? Is there a known upper bound for f(n)?
Quote:
If your second question is sort of, "What are the cognates of the
Platonic solids in R^4", the answer is known, in fact for all R^n. (It's
actually simpler than one might want for n>4 !)
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Yeah, they're only 3 of each dimension! What I'm actually looking for are the polytopes that are facet-transitive - that is, for any two facets, there is a symmetry of the polytope that takes one facet to the other. In three dimensions they are known as isohedra:
http://mathworld.wolfram.com/Isohedron.html
Hmmm, would the general structure be called an isotope? 
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Barukh
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Re: Symmetries in four dimensions
« Reply #3 on: Sep 10th, 2006, 5:10am » |
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This is a very interesting topic which unfortunately I know very little about.
Coxeter in his book “Regular Polytopes” considers symmetry groups of regular polytopes in any number of dimensions. He describes a method called “simplicial subdivision” and thus gives a formula for the order of the group. If I understood the method correctly, the regular polytope in dimension n is divided into (n-1)-simplexes by its symmetry hyper-planes (all of course passing through the polytope’s center). For instance, a cube is divided into 48 right tetrahedrons by its 9 planes of symmetry.
Seems like for the questions asked the material is not easily accessible (at least online).
After posting this, I found the following on the Web:
http://math.ucr.edu/home/baez/week62.html (also weeks 63 through 65)
Note the introductory paper on ADE theory of M. Hazewinkel et al. linked there.
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| « Last Edit: Sep 10th, 2006, 5:20am by Barukh » |
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Michael Dagg
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Re: Symmetries in four dimensions
« Reply #4 on: Sep 10th, 2006, 8:02am » |
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> Is every finite subgroup of O(n) (or SO(n)) a subgroup of a finite Coxeter group?
That's a very good question. It may be known but not by me. On the
other hand, your question reminds me of a broader question that I thought
about years ago without reaching a conclusion. Let me describe this to
you in a little detail, because I think this would make a very good
project for someone.
The context in which I was thinking about this kind of question once
before was the cohomology of groups. Let's be specific here and suppose we
consider only the mod-2 cohomology, i.e. we will study H^*(G, Z/2Z)
for finite 2-groups G. For each G this will be a finitely-generated,
graded, Z/2Z - algebra. For example, when G is an elementary-abelian 2-group
G = (Z/2Z)^k, then H^*(G,Z/2Z) is a polynomial ring in k generators,
each in H^1(G,Z/2Z). I don't know what you know about cohomology in
general; if you've seen cohomology of topological spaces, that's good
enough because for each group G there is a topological space BG
whose cohomology is exactly the cohomology of the group G. (BG is
characterized by the property that its fundamental group is G and its universal
cover (called EG) is contractible; in particular, all the higher homotopy
groups of BG are zero.) But it's not necessary to think topologically.
Anyway, there were some interesting papers around the 1970s and 1980s
to the effect that the cohomology of G is pretty well captured by its
elementary-abelian subgroups, in the sense that for each such subgroup
E we have a "restriction" map H^*(G) --> H^*(E), which is a homomorphism
of rings, and so if we put together these maps for all elementary-abelian
subgroups E we have a homomorphism of rings from H*(G) to the
direct sum \oplus H*(E) of a bunch of polynomial rings. Now the main
theorem is that this last map is pretty close to an isomorphism: the kernel
(which must contain all nilpotent elements of the ring H*(G) ) consists of
ONLY the set of nilpotent elements; a similar statement is true of the
cokernel, once you replace this direct sum with a related but more complicated
inverse limit kind of thing (which takes into account that some of the
E's are contained in other ones). For my purposes, that cokernel part
is kind of irrelevant.
Let me give you some examples. When G = Z/(2^k)Z is a cyclic group
of order greater than 2, the cohomology ring H*(G) may be described
as (Z/2Z)[x,y]/(x^2) where x is a nonzero element of H^1(G) and
y is a nonzero element of H^2(G). As a consequence of this description,
H^n(G) is isomorphic to Z/2Z for every n -- it's either y^k or x y^k
depending on whether n is even or odd. The only elementary-abelian
subgroup is the lone element of order 2 in G, and the restriction map
H*(G) --> H*(E) sends x to zero and sends y to the square of the
generator of H*(E). So "up to the extraction of roots" this restriction
map is an isomorphism, though there definitely is a kernel.
When G is a dihedral group of order divisible by 8, then it turns
out H*(G) = (Z/2Z)[x,y,z]/(xy) where x and y generate the group
H^1(G) = (Z/2Z)^2 and where z is a certain element in H^2(G). There
are two maximal elementary-abelian subgroups of G, each one being a Klein
4-group.
So the restriction map this time is a ring homomorphism of the form
(Z/2Z)[x,y,z]/(xy) --> (Z/2Z)[x1,y1] + (Z/2Z)[x2,y2]
where x1,y1,x2,y2 are all in dimension 1. (remember, these polynomial
rings are all supposed to be _graded_ rings, and the homomorphism
preserves the grading.) OK, well I happen to know exactly what this homomorphism
does: it sends x to (x1, 0) and y to (0, x2) and z to (y1(y1+x1), y2(y2+x2)).
You can see that xy does indeed map to 0. The key thing about this example
is that H*(G) has NO nilpotent elements, so that in fact this map of
rings is an injection: the elementary-abelian subgroups "detect" all the
cohomology there is in H*(G).
One more example: if G is dihedral and H is its cyclic subgroup of
index 2, then the restriction map H*(G) --> H*(H) turns out to be
(Z/2Z)[x,y,z]/(xy) --> (Z/2Z)[x1,y1]/(x1^2)
given by x --> x1, y --> x1, z --> y1. In particular we know xy is
zero in the first ring, and sure enough this presciption will send xy
to x1^2, which is zero in the second ring.
[continued]
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| « Last Edit: Sep 10th, 2006, 8:06am by Michael Dagg » |
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Michael Dagg
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Michael Dagg
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Re: Symmetries in four dimensions
« Reply #5 on: Sep 10th, 2006, 8:05am » |
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All right, now I can finally tell you how your question fits in to my
deliberations on cohomology. The cyclic groups have cohomlogy (the
stuff in odd dimensions) which is not "detected" by its elementary-abelian
subgroup. On the other hand, that cohomology is in the image of the
restriction from the dihedral group, and the cohomology of the dihedral
group IS "detected" by its elementary-abelian subgroup. In some sense
you can argue that the elementary-abelian subgroups of the cyclic group
are too "thin" to detect all its cohomology, but what we have done is
to "fatten up" the cyclic group by embedding it into a dihedral group,
whose elementary-abelian groups are "fatter" .
From this one example, I was led to ask whether a similar construction
was possible for every group in the place of the cyclic groups. Given
a group G we want to know what its cohomology ring looks like. We
get a partial picture from the cohomology of the elementary-abelian
subgroups of G and how they are included inside each other. But we
know that this picture will miss some of the cohomology of G, namely,
all the nilpotent elements of H*(G). But maybe for each group G we
can construct a larger group X which has the properties that
(1) H*(X) --> H*(G) is a surjection
(2) H*(X) --> \oplus H*(E) is an injection
so that all the cohomology of G will be found from that of X, which
in turn is accounted for by the elementray subgroups of X.
Well, I once computed the cohomology of all the groups of order 32, in
part just to check this conjecture! At the very least I could see which
X had property (2), and so these became the candidates for the X's
that we could use to resolve the conjecture for groups G of order 16
or less. I never found a counterexample, but I had nothing like a
proof either, in part because the method of construction the X's from the
G's was so ad hoc (try all X's and see which ones work...)
Now here's the specific connection to Coxeter groups. The whole game
of restriction to elementary-abelian subgroups has this one obvious
bottleneck: if all the E's for a given G lie in a proper subgroup
H of G, then the restrictions H*(G) --> H*(E) all factor through a
single homomorphism H*(G) --> H*(H). This map can NEVER be an
injection if [G:H] is a multiple of 2, so there will ALWAYS be a nontrivial
kernel H*(G) --> H*(E) unless there is no such H, which is
equivalent to saying G is generated by its involutions. I know by example that
there ARE groups G generated by involutions which still have nilpotent
elements in their cohomology ring, so that's not the end of the story,
but it points towards what we'd have to do to make X : somehow, given
a group G, we have to find a way to embed it into a group generated
by involutions (satisfying (1) and (2)).
OK, well one kind of group generated by involutions is a group generated
by REFLECTIONS. I believe it's true that all these groups satisfy (2).
(I can't remember right now because there are multiple definitions of
"Coxeter group", depending on whether the reflections have to be in a
Euclidean space, and on whether there are any additional relations
besides the orders of the reflections and their pair-wise products.
It's been a looong time since I worked with this!) So the Coxeter groups
(meaning the ones for which property (2) has been proven) provide a
possible resolution of my general conjecture: is it true that for
every 2-group G there is a Coxeter group X containing G such that
H*(X) --> H*(G) is a surjection? If so, then these X's serve as a
way of capturing all cohomology with elementary-abelian subgroups !
(Of X, not G.)
As you can see, I have now woven your question into this more high-falutin'
one: you asked whether, for each G, there exists any Coxeter group X
at all which contains G. My question demands more of X but includes
your question as a sidebar.
I don't know whether your "easier" question is already solved, but you
could set your sights higher and try to resolve the harder question I
asked. That would be a really great contribution to understanding where group
cohomology "comes from" !
IIRC Jordan's theorem can be found in Finite Reflection Groups, by Grove
and Benson.
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| « Last Edit: Sep 10th, 2006, 8:05am by Michael Dagg » |
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Regards,
Michael Dagg
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Deedlit
Senior Riddler
Posts: 476
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Re: Symmetries in four dimensions
« Reply #6 on: Sep 13th, 2006, 4:25pm » |
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Thank you for the interesting ideas, Michael. Wow, you figured out the cohomologies of all the groups of order 32? You were pretty serious!
I admit I'm not that comortable with cohomology. I may try to address your questions later, but for now I'll still be looking at O(4) from a more elementary angle.
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Michael Dagg
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Re: Symmetries in four dimensions
« Reply #7 on: Sep 13th, 2006, 5:04pm » |
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Many people have computed them and will likey continue
just to see what emerges form it.
But computing them is different than really figuring
them all out;you start seeing patterns as you go
along and the picture that starts to emerge depends
on what you can recognize.
If you come up with anything about your Coxeter
group question please let me know about it.
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Michael Dagg
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