From: rusin@vesuvius.math.niu.edu (Dave Rusin)
Newsgroups: sci.math
Subject: Re: Q: homotopy groups of SO(3)/(Z_2 x Z_2)
Date: 26 Nov 1996 16:09:07 GMT
Keywords: homotopy, topology
In article <573qqi$ceo@sparcserver.lrz-muenchen.de>,
wrote:
>
>Can anybody tell me the homotpy groups pi_1, pi_2, p_3
>of the homogenous space SO(3)/(Z_2 x Z_2) (which is the
>order parameter for a fixed frame of directors) ?
>Or is there a theorem saying that these homotopy groups
>are equal to those of SO(3) ?
>pi_1(SO(3))=Z_2
>pi_2(SO(3))=0
>pi_3(SO(3))=Z
If D is a discrete subgroup of a Lie group G, then the quotient map
G -> G/D is a covering space (with D as the fibre).
Every covering is also a fibration (with discrete fibres).
If F -> E -> B is a fibration then there is a long exact sequence
...-> pi_n(F) -> pi_n(E) -> pi_n(B) -> pi_(n-1)(F) -> ...
... -> pi_1(F) -> pi_1(E) -> pi_1(B) -> pi_0(F) -> 0
(where the last maps are in general simply maps between _sets_ of course).
Here I'm assuming E is connected.
Combining the last two paragraphs we see that if E -> B is a covering
space, then pi_n(E)=pi_n(B) for all n > 1, pi_1(E) is a subgroup of
pi_1(B), and the cosets pi_1(B)/pi_1(E) are naturally in correspondence
with the fibre of the covering.
Combining with the first paragraph we see that pi_n(G) = pi_n(G/D) for
all n>1, and there is a short exact sequence
1-> pi_1(G) -> pi_1(G/D) -> D -> 1
of groups (Yes, I know those last two words are not immediate from the
previous statements, but they are still correct.)
In your example, we should have (I'll trust your statements about pi_n(SO)
since I keep forgetting those groups)
pi_3( SO(3)/(Z_2 x Z_2) ) = Z
pi_2( SO(3)/(Z_2 x Z_2) ) = 0
pi_1( SO(3)/(Z_2 x Z_2) ) = an extension of Z_2 by Z_2^2, therefore one of
Z_2^3, Z_2 x Z_4, Dihedral(8), or Quaternion(8).
There is a little voice at the back of my head telling me that I've seen
this before and that the correct choice among these options for pi_1 is
the last, but if you put down as a reference that this result is known to
be true by voices inside someone's head, you'll get us both in trouble.
This might work instead but I'll confess I'm not really very familiar with
this group so I can't provide details. Every map f: G-> G induces a
homomorphism f*: pi_1(G)-> pi_1(G). Such a map on G=SO(3) which
preserves cosets of D also gives a map f2: G/D -> G/D and hence
f2*: pi_1(G/D) -> pi_1(G/D). If you can find such a map f such that
f2* is of order 3, it will follow that pi_1(SO(3)/Z_2^2) has an
automorphism of order 3 fixing the normal subgroup of order 2, which
can only happen (among the four possible groups) if pi_1 is the
quaternion group. I would guess you can find such a map f which is
in fact a homomorphism, but it needn't be.
dave
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From: rusin@moriarty.math.niu.edu (Dave Rusin)
Newsgroups: sci.math
Subject: Re: Q: homotopy groups of SO(3)/(Z_2 x Z_2)
Date: 30 Nov 1996 18:55:44 GMT
In article ,
Pertti Lounesto wrote:
>I can understand what is SO(3)/Z_2, where Z_2={I,-I} is a normal subgroup.
>But what is SO(3)/(Z_2 x Z_2)?
This is the coset space of the subgroup. Since the subgroup is discrete,
the topology is locally the same as SO(3), that is, it's another manifold.
Since the subgroup is _not_ normal, the new space is not a group. (It's
also true, therefore, that the covering map G -> G/D is not a
normal covering, i.e., D isn't the group of covering autmorphisms).
If you were asking what the embedding of D = Z_2 x Z_2 into SO(3) is,
that's easy: this D has 4 irreducible representations, all linear
and all real. The minimal ones whose determinants are trivial are:
the trivial representation, the double of any of the other three, and
the direct sum of those other three. Only the last occurs in an injection
into SO(3). The subgroup intended is then the image of this representation
(then unique, up to conjugacy).
Incidentally, this makes it clear that there is an automorphism f of
order 3 of SO(3) which restricts to a nontrivial automorphism of D.
(We can take f to be conjugation by a matrix of order 3).
Then we have a commutative diagram
1 -> pi_1(SO(3)) -> pi_1(SO(3)/D) -> D -> 1
| | | | |
|= |f* |f* |f* |=
v v v v v
| | | | |
1 -> pi_1(SO(3)) -> pi_1(SO(3)/D) -> D -> 1
which shows that the group in the middle has an automorphism of order 3
and so must be the quaternion group of order 8 (as explained in a
previous post).
Note that since G/D has a nonabelian fundamental group, it's not
itself a Lie group (or even an h-space).
dave
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From: rusin@moriarty.math.niu.edu (Dave Rusin)
Newsgroups: sci.math
Subject: Re: Q: homotopy groups of SO(3)/(Z_2 x Z_2)
Date: 2 Dec 1996 14:12:53 GMT
In article ,
Pertti Lounesto wrote:
>rusin@moriarty.math.niu.edu (Dave Rusin) writes:
>> Incidentally, this makes it clear that there is an automorphism f of
>> order 3 of SO(3) which restricts to a nontrivial automorphism of D.
>
>Is this automorphism f of order 3 of SO(3) a linear automorphism, that is,
>is it U->f(U)=MUM^{-1}, where M is a 3x3-matrix? And what is M?
Yes.
D = { diag(a,b,c); a*b*c=1, a^2=b^2=c^2=1 }, and M=
( 0 0 1 )
( 1 0 0 )
( 0 1 0 )
[I wish I had had your questions before I posted the first time. Each
time I have to stare at the situation for a few more minutes it becomes
more concrete.]
dave
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