Date: Thu, 9 Mar 95 11:46:42 CST
From: rusin (Dave Rusin)
To: columbus@osf.org
Subject: Re: Thanks
I'm glad my remarks were helpful. They pay me "to do mathematics",
and communicating what I know is how I do that.
>You mention an alternate definition of induced representation I'm not
>familiar with, F[G/H]\tensor V. I assume this is equivalent to the
>(G/H)xV version? I can see how F[G/H]\tensor V is a G-module--- it's
>enough to define the action g.(aH\tensor v) for all g in G, aH in G/H,
>and v in V, and this obviously should be gaH\tensor v. But I'm
>puzzled about the action of H on this. H already has an action on V,
>so if h is in H, we have two possible meanings for h.(aH\tensor v),
>namely haH\tensor v and also aH\tensor(h.v). Do you need to mod out
>by the submodule generated by all the differences of these two
>expressions?
No. Let me try to clarify the nature of the induced representation.
Your notation "(G/H)xV" is really a sort of bastard mix which would be
repudiated by either of two respectable parent concepts.
A representation of a group means a homomorphism of G to a "known"
group. Typically the phrase means that the known group is GL_n(F) where
F is a field like C (or perhaps a field of characteristic p), but
one also speaks of "permutation representations", meaning that one
takes the "known" group to be a symmetric group. Depending on the context,
the "induced representation" has to be defined a little differently.
In the case of permutation representations, given H --> Sym(S) (that is,
H is viewed as permuting a (usu. finite) set of elements S), we make an
induced representation for G by letting G act on a larger discrete set.
Fix a collection of coset representatives for the cosets of H in G and
take S' to be the cartesian product of the sets S and G/H (the set
of cosets). Then the action of g in G on an ordered pair (g'H, s)
is to send it to the pair (g''H, hs), where g'' is the preselected coset
representative of the coset (gg')H and h in H is defined by
gg'=g''h. There are a few things to observe about this induced representation.
First of all, we have enlarged the set on which G is to act. This is in
general necessary since there is typically _no_ action of G on the original
set S which extends the action of H. Secondly we note that the specific
action of G on S' depends on the choice of coset representatives in G/H;
however, replacing each representative g' with g'h' will give an
equivalent representation (that is, there exists a permutation f: S'-->S'
such that the two actions of G on S' are related by
phi1(g) (s) = f^-1 ( phi2(g) ( f (s) ) ).) Finally, we note that H, as a
subgroup of G, acts on this new enlarged set S'. Taking 1 as the
representative of the coset H in G/H, we see that the action of H on
the subset S x {1} of S' is essentially that of H on S itself.
There is no telling how H acts on the remainder of S', although in the
special case that H is normal in G, the set S' will decompose into
G/H orbits under the action of H, each of them conjugate to the first, i.e.,
for each phi there is a g in G so that phi(h) (s) = phi1( g^-1 h g ) (s)
for all h and s).
Now, in the case of linear representations, the "induced representation" is
defined in a similar way, but the goal is to find another _vector space_
on which G can act. You can't take G/H x V this time (where V is the
vector space on which V is acting). Instead, you can create a vector space
F[G/H] over F whose basis elements are labelled by the cosets of H in G,
and then take the tensor product of this with V. This gives you a new
vector space of dimension [G:H].dim(V), and G acts on it using essentially
the same formula as above (H is, after all, permuting the vectors of V,
so this is a permutation representation, too). I think you had this in mind
in your example: you let G be the group of the Rubik's cube and H the
stabilizer of one corner vertex, then [G:H]=8, and you viewed H as acting
on V = R^3. Then the induced representation makes G act on a vector space
of dimension 24. You might want to check out some details of this
induced representation. For example, in the induced representation, H itself
acts on R^24, but the representation breaks naturally into 4 parts:
there is the 3-dimensional representation space (R.1)\tensor V, another
3-D space (R.a)\tensor V (where aH is the coset in G/H which contains
a 180 rotation sending this vertex to its antipode; H union aH is the
normalizer of H in G), and two 9-D spaces (R.b + R.c + R.d)\tensor V
and (R.ab + R.ac + R.ad)\tensor V, where b, c, and d are rotations which
carry this vertex to the three adjacent vertices. Each of these 4 subspaces,
incidentally, may be further decomposed.
>Anyway, you're quite right, my question really is
>
> does a homomorphism H --> Aut(V) extend to a map G --> Aut(V)?
>
>for various examples.
Fine, that's an important general question (with no easy answer) but that
does not explicitly refer to the induced representation. What is true is
that if such an extension exists, then it will be a summand of the
induced representation, but it is not easy to decompose the induced
representation in general.
Your discussion of the Rubik's cube group seems right on target. Ignoring
the edge cubies (that is, dealing with a 2x2x2 cube) the group of Rubik
motions maps to Sym(8) with a kernel which maps to (Z/3)^8. A group theorist
would observe that this makes the Rubik's group a subgroup of the
wreath product (Z/3) wr Sym(8). As I recall, the Rubik group is all of this
last group, or maybe a subgroup of index 2 or 3 or so. In particular, a
representation of it involves a representation of Sym(8) which I think is of
dimension at least 7 if not linear.
(These observations are at the heart of the solution of the Rubik group.
A simple combination of 4 90-degree rotations yields a permutation of
three vertices of the cube. Then combinations of these yield all 3-cycles
among the 8 vertices. But the set of 3-cycles generates the alternating
group, and since the Rubik group also includes odd permutations, the
homomorphism (Rubik) --> Sym(8) is onto. To get more information we look
at preimages in (Rubik) of nice elements in Sym(8), such as the 3-cycles.
The cubes of these group elements map to the identity in Sym(8), and so
provide elements in the kernel of the homomorphism. One then studies the
map of this kernel onto (Z/3)^8, and again finds enough elements to show
the map is (essentially?) onto. Then you can back up to the kernel of
_that_ homomorphism and start looking at actions of the edge cubies.
And so on. As a practical matter this is not a fast way to solve the
cube, but to a group theorist it's very pretty and easy.)
dave