Date: Wed, 1 Mar 95 16:05:51 CST
From: rusin (Dave Rusin)
To: Michael Weiss
Subject: Re: Question on group reps
>Briefly, let H be a subgroup of G; let H act on a vector space V.
>How does one tell if the induced representation on G is trivializable,
>i.e. equivalent to an action of G on GxV that extends the action of H?
Well, I'll confess I'm not really sure what trivializable means. To me
the induced representation of G is an action of G on either the set
(G/H)xV or the vector space F[G/H]\tensor V, where F[G/H] is the
vector space spanned by G/H. I'm not sure how you intend these to be
equivalent to an action of G on GxV. Under the induced action of G,
H (now a subgroup of G) acts on the newly constructed G-set or G-vectorspace,
and includes the original vector space V as an invariant subset
(resp. subspace). I take it from your example #2 (G=Alt(4), H = C(4),
V= rotation space) that your question is
does a homomorphism H --> Aut(V) extend to a map G --> Aut(V)?
As far as I can tell this does not permit an easy answer.
Incidentally, the connection with the induced representation is that
you want to know if the induced representation (V\uparrow G) has a
summand W (i.e. a G-invariant subspace) with (W\downarrow H)= V. For
finite groups there is an inner product on the representation ring given
for irreducible reps by (f, g) = 1 if f=g, 0 otherwise. This may be
useful for you because of a duality theorem: (Ind(f), g) = (f, Res(g)).
(an irreducible rep g of G is a component of the induced rep Ind(f)
of f precisely as often as f shows up as a component of the restriction
Res(g) of g.) [PS: here and below I deal with _complex_ representations.]
Forgive me if I've misunderstood the question; since you were
describing in your examples groups which are already groups of
symmetries of something, it is a little unclear how you intended those
actions to blend with the sought-after ones. I will below treat the
groups just as abstract sets with binary operations on them; perhaps
some of what follows can be useful even if I've misinterpreted you.
You already speak of some of what I know about when
you start talking about H as the isotropy subgroups of certain sets
(like faces) and then start looking at G-conjugate sets. These other
sets will have isotropy groups which are G-conjugate to H. Then if
an extension of H--> Aut(V) exists to G, the actions of H^g will
have to be conjugate _within Aut(V)_ to that of H. This gives you
some restrictions due to cases when H^g = H (the restrictions will
have to be conjugate) and when H^g meets H in a proper subgroup of H.
Here's how I would state these group-theoretically. You have first the
invariance under the normalizer. If H --> Aut(V) extends to G--> Aut(V),
it must in particular extend to N(H) --> Aut(V). Well, this situation is
easier to analyze. If you take any representation of N(H)
and restrict down to H, you will get a sum of conjugate representations.
The representations conjugate to f: H --> Aut(H) are f o g where
g: H --> H is conjugation by an element of N(H). For example, in the
dihedral group G of order 8, the cyclic subgroup H = C4 is normal, and
there is a conjugation by an element of G sending each h in H to
its inverse. This H has 4 irreducible complex representations, all
1-dimensional: the trivial and +-1 reps are self-conjugate, and the
2 faithful representations are conjugates of each other under G. So
neither of these latter two can separately be the restriction of a
representation of G, although their direct sum can, as could either of
the other 2. Indeed, these two self-conjugate reps, as well as the sum of
the other two reps, are indeed the restrictions of reps of G (of
dimensions 1, 1, and 2 respectively, of course.)
The bad news is that there is more involved. In general there is a
"degree of inertia" or something, call it e, associated to an N(H)-invariant
representation f such that the direct sum of e copies of f is the
restriction of a rep of N(H). You'll have to look up "Clifford theory"
for the relations of the reps of a normal subgroup to those of the big group
because I don't quite remember how it works, but I think if you had a
rep of N(H), then you find a couple of intermediate subgroups containing H
such that among the three steps of the restriction you find one case of
an irreducible rep. staying irreducible, one case of an irreducible rep
restricting to a sum of identical reps (all N(H)-invariant of course), and
one case of a split into pairwise distinct (but conjugate) reps. I just
can't remember which comes first. Example: The center C2 of the
dihedral group G of order 8 has a non-trivial linear rep. Since C2 is
in the center, this rep is N(H)-invariant, but it's not the restriction of
any linear rep of G (since C2 lies in the commutator subgroup of G).
You could take the sum of two copies of this rep, and then you do have
something which extends to G.
At least there is some compensation: the Clifford theory allows some
duality between induction and restriction. Again, you'll have to check
for details elsewhere.
OK, so assume you've been able to decide that a representation
f: H --> Aut(V) extends at least to f: N(H) --> V. Now, how do you
decide if it extends all the way to G? Well, I don't think you can
answer this as clearly as in the normal case, but you can give necessary
conditions for an extension to exist. Here you'll need to look up
Mackey's name in your index. I guess the best way to check for these
conditions is to say, if f extends all the way to G, then it also
extends to other subgroups including N(K) where K is any subgroup
of H. (I think you were alluding to this in your examples: a representation
of G would restrict to give representations of _all_ of the face
groups; but these data would be inconsistent on the intersections of some
of those groups. Well, those intersections are examples of the K
I have in mind).
As a practical matter I will say it is possible to determine the
conjugates and restrictions of reps of H using the character tables,
so at least some information can be deduced there about the possibility of
extending to G. Of course, as you noted in your example 2, if you
know the whole character table for G itself, that can make the
job a lot easier!
Your last example kind of bugged me:
>Example 3: Let G = the group of Rubik's cube, let H = isotropy group
>of one corner cubie. Let V=3-space. Now this is interesting because
>there is an obvious way to set up isomorphisms between the fibers at
>each corner cubie. However, G does not act the same way on all
>fibers, i.e., the action doesn't commute with the isomorphisms. So my
>guess is that the induced action is *not* trivializable. But how to
>prove it? G is a pretty big group--- computing all the 3-dim reps of
>it is a daunting task. Surely there's a way to show that the action
>isn't trivializable without that?
Your statement "let V=3-space" implies you already have a representation
of H on R^3 in mind. I don't think you do. You have an action of H
on a certain subset in R^3, but the actions you use to permute parts of
the Rubik's cube are _not_ linear maps. (That is, you want to do things
like rotate a face without rotating the rest of the cube.) The expression
of G as the group of Rubik's cube, and H as an isotropy group thereof,
give _permutation_ representations G (or H) --> Sym (n) for some n,
not _linear_ representations G (or H) --> Aut(V).
Indeed, there is a theorem that a finite group in GL_n(R) (or O_n(R) or...)
must have an abelian subgroup of some relatively small index which
depends on n. This is the reason there are only finitely many platonic
solids: a finite subgroup of GL_3(R) has an abelian subgroup of index
60 (?) or less. Well, the Rubik's cube group maps onto Sym(8)xSym(12)xSym(6)
that is, you can put the cubies wherever you want, except of course you
have to send corners to corners, etc. (Hmm, maybe it's only of index two in
this product -- I forget). The upshot is that it does not have an
abelian subgroup of any small index. If you want a faithful representation,
I imagine you'll have to go into dimension 7x11x5 at least, and even
representations with kernels probably have dimension 5 or more, except for
linear representations (some of which may have dimension 2 if you try to
express them as real representations.).
You are right, though, the Rubik group is cool.
dave
==============================================================================
To: rusin@math.niu.edu (Dave Rusin)
Subject: Thanks
Date: Wed, 08 Mar 1995 10:36:53 -0500
From: Michael Weiss
Thanks for taking the time to reply. I still need to mull over a few
things you wrote, but I thought I should let you know how helpful your
message has been.
The definition I meant to write for trivializable was equivalent to an
action of G on MxV extending the action of H on V. Here M is the base
space of the induced bundle, i.e., essentially G/H. Sorry for the
confusion.
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?
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.
Before getting your reply, I'd found a simple proof that it *doesn't*
for my third example, Rubik's cube group. You message helped me
understand my argument better--- it's really just a special case of
your remark
if an extension of H--> Aut(V) exists to G, the actions of H^g will
have to be conjugate _within Aut(V)_ to that of H.
I didn't explain my last example very well. Let me try again. Let G
be the group of Rubik's cube--- actually I care only about the corner
cubies and cubicles (*), so let's say we have a "twobik" without edge
cubies/cubicles. When I said "let V=3-space", I mean for V to be the
fiber--- we don't have an action of G on V, at least not an obvious
one. We attach a copy of V to each point in the base space, i.e., to
each cubicle (sort of like a tangent space). The union of all these
copies gives us our bundle E, and G acts on E. (Passing to the
24-dimensional vector space of sections, we get a linear
representation of G.)
Now let H be the isotropy group of one cubicle (call it b), i.e., the
set of operations that keep b's cubie in b, twisting it perhaps.
There is a linear action of H on the fiber V_b at b. Namely, see what
h in H does to the selected cubie--- it subjects it to a rigid
rotation in R^3, and so we have a mapping of H into the orthogonal
group O(3). This action is well-defined only with respect to a
particular isotropy group, of course--- if H_a and H_b are two
isotropy groups, and h belongs to their intersection, then we have one
action of h on V_a and another of h on V_b. Even though V_a and V_b
are both canonically isomorphic R^3, we *don't* get a well-defined
action of h on R^3.
And this is the key that tells us that E is not trivializable.
Namely, let H_a, H_b, and h be as just mentioned, and let g in G move
a to b, so g(H_a)g^-1 = H_b. So h and ghg^-1 both belong to H_b, and
would have to be conjugate in Aut(V_b) by your remark if the bundle
were trivializable. Choose h so it acts trivially at b and
non-trivially at a. Contradiction, QED.
(*) Do you know about this convenient terminology? The cubicles are
positions at the eight corners of the cube; the cubies are the little
pieces of plastic that move from one cubicle to another.
Thanks again.
==============================================================================
To: rusin@math.niu.edu (Dave Rusin)
Subject: Medley
Date: Fri, 10 Mar 1995 12:48:11 -0500
From: Michael Weiss
Table of contents:
1. solution to my second exercise
2. OK, now I understand F[G/H]\tensor V
3. Some questions
My second exercise asked if the vector bundle we get by attaching a
copy of R^2 to each face of the cube, and acting on these with the
rotation group of the cube, was trivial.
I now have a nice simple argument without using character tables. (I
got it from mulling over some of what you said in your messages,
though I won't bore you with the genesis of the argument.)
Dramatis Personae
O, the rotation group of the cube
C4, the isotropy subgroup of a face
C2, the subgroup of C4 of order 2
G, any old group
Theorem: if f:C4 --> G is a monomorphism, and f(C2) lies in the center
of G, then f does NOT extend to O. (So this solves my problem, and not
only for vector bundles!)
Proof: Consider three elements of O: the 90 degree counter-clockwise
rotations about the three orthogonal axes. Then you can label them
a,b,c so that the following all hold:
(1) abc = b
(2) a,b,c are all conjugate
(3) In particular, b^-1 c b = a
(4) b generates C4 (and so b^2 is in C2)
Putting (1) and (3) together, we get
(5) b^-1 c b b c = b
Now assume f extends to O, and let f(a)=A, f(b)=B, f(c)=C. Applying f
to (5) we get
(6) B^-1 C B B C = B
Since BB is in f(C2) and hence in the center of G, we get
(7) BCC = B
(8) CC = 1
But since B is conjugate to C, this implies
(9) BB = 1
This contradicts the assumption that f is faithful on C4. QED
This may seem like a totally unmotivated argument, but I got it by
working backwards from geometric intuition, plus your remark about
"inconsistent data".
F[G/H]\tensor V: thanks for explaining that. Chewing on it, I came
up with another way of thinking about it I like better. What say you?
Given: H acts on V, and G contains H. We want a vector space E and an
action of G on E which "extends H acting on V"; in particular, this
means that we must have some canonical way of injecting V into E. So
we must have a binary function G x V --> E, where we first inject V
into E and then act on it with G.
It seems reasonable to demand that the image of G x V should generate
E, otherwise we have extra baggage we didn't need to accomplish our
purpose. Anyway, let's try it. We want a vector space, and we want G
to act on it, so let's form F[G]\tensor V, where F[G] is the
group-algebra. Now we have an obvious action, obtained by linearly
extending g_1.(g_2\tensor v) = (g_1 g_2)\tensor v.
But we also wanted this action to extend the action of H on V. Now we
can inject V into F[G]\tensor V via v--> 1\tensor v. The two
actions are now:
h.(1\tensor v) = h\tensor v our newly defined action
h.(1\tensor v) = 1\tensor hv the old action, lifted up
So we want (but don't have)
h\tensor v = 1\tensor hv
So just use the standard trick to make relations true: mod out by the
G-module generated by them! Calling this A, the induced bundle
becomes (F[G]\tensor V)/A.
This doesn't involve the arbitrary choice of a system of
representatives for G/H. Also the same idea works for permutation
reps, but now you have to mod out by an equivalence relation.
(Sternberg also does a modding out, but he doesn't use tensor product.
I think his passage from the vector bundle to the space of sections
corresponds to tensor product, plus forming F[G].)
Questions: the induction idea seems pretty similar for perm reps and
linear reps. Do you have anything analogous to Frobenius duality for
perm reps?
==============================================================================
Date: Sat, 11 Mar 95 00:20:00 CST
From: rusin (Dave Rusin)
To: Michael Weiss
Subject: Re: Medley
>Theorem: if f:C4 --> G is a monomorphism, and f(C2) lies in the center
>of G, then f does NOT extend to O. (So this solves my problem, and not
>only for vector bundles!)
Right. You have a very geometric understanding of groups, which is great,
although in some instances (like this) you can speed things up by
working literally (='with letters'). See, I would observe that if
f: G1 --> G is any homomorphism of groups, and H < G1, then
f([G1,H]) <= [f(G1),f(H)] <= [G,f(H)], where [ , ] denotes the group of
commutators. In particular, if f(H) lies in the center of G, then
the last term is just {1}, so f has [G1,H] in its kernel. In particular,
if f is supposed to be an injection on H, then [G1,H] had better not
include any elements of H other than {1}. This is not the case in
the cube group -- [G1,H] includes the commutator of a 90-degree rotation in
H and a 180-degree rotation outside of H; this commutator generates your C2.
(Note incidentally that it's usually easier to start with a purported
homomorphism of the big group and consider its restriction to H, than
to start with a homom. on H and try to extend it to G1.)
>Given: H acts on V, and G contains H. We want a vector space E and an
>action of G on E which "extends H acting on V"; in particular, this
>means that we must have some canonical way of injecting V into E. So
>we must have a binary function G x V --> E, where we first inject V
>into E and then act on it with G.
>
>It seems reasonable to demand that the image of G x V should generate
>E, otherwise we have extra baggage we didn't need to accomplish our
>purpose. Anyway, let's try it. We want a vector space, and we want G
>to act on it, so let's form F[G]\tensor V, where F[G] is the
>group-algebra. Now we have an obvious action, obtained by linearly
>extending g_1.(g_2\tensor v) = (g_1 g_2)\tensor v.
>
>But we also wanted this action to extend the action of H on V. Now we
>can inject V into F[G]\tensor V via v--> 1\tensor v. The two
>actions are now:
>
> h.(1\tensor v) = h\tensor v our newly defined action
> h.(1\tensor v) = 1\tensor hv the old action, lifted up
>
>So we want (but don't have)
>
> h\tensor v = 1\tensor hv
>
>So just use the standard trick to make relations true: mod out by the
>G-module generated by them! Calling this A, the induced bundle
>becomes (F[G]\tensor V)/A.
This is very good! Indeed, this is an example of what a topologist would
call a 'universal construction' (as in 'universal covering space'). What
you have constructed is a homomorphism G --> Aut(E) lying over the
original H --> Aut(V) which is universal in the sense that given _any_
map G --> Aut(W) with W a vector space including V such that
the action of H on W includes the original action on V, there will
be a unique map E --> W consistent with all the other data. (There's
reallly no need to belabor details. I just wanted to congratulate you on
discovering the category-theoretic approach.)
>Questions: the induction idea seems pretty similar for perm reps and
>linear reps. Do you have anything analogous to Frobenius duality for
>perm reps?
Hmm, I never thought about that. I guess I would note that every permutation
rep. gives a linear rep in a natural way (permuting basis vectors), indeed,
this correspondence pairs induced representations and restrictions. It
also pairs a sum of permutation reps with a sum of linear reps; unfortunately,
not every linear rep comes from a permutation rep, so a splitting of a
linear rep into constituent parts need not come from a splitting of the
corresponding permutation rep.
So what would we expect? Given g: G --> Sym(n) and h: H --> Sym(k),
we'd like to know how many times g is a constituent of Ind(h).
Write L(g) : G --> GL(n) for the corresponding linear rep; likewise
for L(h). Then L(Ind(h)) = Ind(L(h)) and L(Res(g)) = Res(L(g)). By
reciprocity, = , so
we get [ Ind(h), g ] = [ h, Res(g) ] where we define the inner product
for premutation representations by [ f, g ] = .
So, sure, we get a reciprocity theorem for permutation reps by leaning
on the one for linear reps. The difficulty will come in interpretation. It's
no longer true that for irreducible permutation representations f and g
we have [f,g] = 0 or 1. For example, if f : Sym(2) --> Sym(2) is the
identity map, then L(f) is the linear representation sending a generator
of Sym(2) to the matrix [ [0,1], [1,0] ], which is the direct sum of the
+1 and the -1 representation. Hence = 2, i.e., [f, f]=2, not 1.
The collection of permutation representations is of course quite important.
You might want to look up the Burnside ring which collects these into a
ring (a la Grothendieck); lots is known about the idempotents, induction
theorems (e.g. Conlon induction theorem), and so on.
[deletia]
dave