Some Geometry Of Symmetrized Tensor Spaces
by
Henry G. Harmon III
A dissertation submitted to the Graduate Faculty of
Auburn University
in partial ful llment of the
requirements for the Degree of
Doctor of Philosophy
Auburn, Alabama
May 4, 2014
Keywords: representation, orthogonal basis, symmetrized tensor, root system
Copyright 2014 by Henry G. Harmon III
Approved by
Randall R. Holmes, Professor of Mathematics
Michel Smith, Professor of Mathematics
Piotr Minc, Professor of Mathematics
Peter Nylen, Professor of Mathematics
Abstract
Let G be a subgroup of the symmetric group Sn for n 2 N and let V be an in-
ner product space. Orthogonality properties of the set of standard (decomposable) sym-
metrized tensors in V n corresponding to G have been studied for more than two decades
[WG91, HT92, Hol95, DP99, BPR03, Hol04, TS12]. The determination of such properties
would be facilitated by an understanding of the more general geometric properties of this set.
Building on fairly isolated insights throughout the relevant literature, we obtain some new
results, and in doing so we begin to weave a coherent narrative for the future exploration of
this geometry.
The space V n is an orthogonal direct sum of orbital subspaces, so it su ces to study
the sets of standard symmetrized tensors in these subspaces. In order to do so, we investi-
gate for each irreducible character of G and each subgroup H of G the set of standard
vectors in what we call the coset space C H. This coset space is itself a vector space and a
CG-module, and for each (H; )-pairing it corresponds to one and only one orbital subspace
of V n. Hence the coset space serves as a proxy tool for our inquiries into the geometry of
a given orbital subspace.
The structure of C H as a vector space is easily understood, but for the sake of parallelism
with orbital subspaces, we need also to understand its structure as a CG-module. The rst
result furnishes an isomorphism that renders this structure in the well-known terms of the
associated group algebra. The isomorphism may prove useful in the future, as groundwork
for a more module-theoretic approach to these matters.
ii
In order to generate meaningful conjectures, we then devote considerable e ort to the con-
struction of bases for our coset spaces. As it turns out, nding such a basis is itself a far
from trivial problem. After developing su cient machinery, we obtain bases for coset spaces
associated with the groups S3, S4, and A4, allowing the (H; )-pairings to vary. We then
compute the Gram matrices for the basis vectors in each coset space. The rst conjectures
arise from these computations. In several cases we nd that, possibly after dividing the
entries of the Gram matrix by 2, we obtain the Cartan matrix for a crystallographic root
system of type A2 or A2 A2. Thus in these cases a basis for the coset space is also a
base for a root system. By reference to the correspondence between coset spaces and or-
bital subspaces, we conclude that certain symmetrized tensor spaces possess the geometry
of crystallographic root systems. The section culminates with a theorem stating that, if a
coset space gives rise to a root system, then that root system has irreducible components of
type A1 or A2, which are simply laced.
In the next section, we examine in depth the geometric properties of C H when the group is
dihedral, of order a power of 2. It has already been proven that the standard symmetrized
tensors in V n for this choice of G have an orthogonal basis for every (H; )-pairing in the
corresponding coset space. We now deploy more explicitly geometric methods to obtain
the same conclusion. The result is a more intuitive proof, the methodology of which will
hopefully shed light on the orthogonality properties of other sets of standard symmetrized
tensors. Ultimately, one goal of future work is to provide necessary and su cient conditions
for a nite group to give rise to an orthogonal basis of symmetrized tensors.
Finally, we show how this more geometric approach con rms a result in the literature. Given
an m-dimensional inner product space V, the orbital subspace V of V n is determined by
the irreducible character of Sm and the element 2 n;m; where we may view n;m as the
set of functions : f1;:::;ng!f1;:::;mg. In [TS12], the authors prove that the standard
iii
symmetrized tensors in a certain orbital subspace of V n form a root system of type Am 1.
Again we take a more conceptual avenue to the same result. It is a tting way to close: The
proof demonstrates how, by bringing the power of representation and character theory to
bear, we can accomplish what was formerly done through combinatorial means.
iv
Acknowledgments
I owe a boundless debt of gratitude to Dr. Randall R. Holmes. He has been an intellec-
tual father to me. From teaching me how to write my rst proper mathematical string, to
helping me re ne and polish my research, he has overseen my mathematical growth. For six
years, the precision, economy, and clarity of his thought has set the standard towards which
I aspire. I am humbled and honored that he chose me as one of his graduate students. It
has been a an unrivaled privilege to learn from him.
I thank Dr. Michel Smith, Dr. Piotr Minc, and Dr. Peter Nylen for their unwavering
support and encouragement. Each of them has made a unique and irreplaceable contribu-
tion to my education and progress.
To my partner of over thirteen years, Geo rey Scott Turk, I give thanks for all his lov-
ing support over what must have seemed like an interminable time. He has always allowed
me the space and comfort I needed to pursue my education, while waiting patiently so that
we can begin a new life together. I could not have done this without him.
Finally, I thank my Mom and brother in Birmingham, AL for understanding why I had
to do this. I have missed them terribly since embarking on this journey on a summer day
seven years ago, and it has been a long winter away from them since then. I know they are
proud of me.
v
Table of Contents
Abstract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ii
Acknowledgments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . v
1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.1 Basics of Representation Theory . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.2 Selected Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
1.3 Root Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
1.4 Coset Spaces and Symmetrized Tensor Spaces . . . . . . . . . . . . . . . . . 10
2 The Module Structure of the Coset Space . . . . . . . . . . . . . . . . . . . . . 15
3 Construction of Bases for Coset Spaces and Root System Geometry . . . . . . . 18
4 The Geometry of Coset Spaces For Certain Dihedral Groups . . . . . . . . . . . 29
5 Root Systems For A Special Coset Space . . . . . . . . . . . . . . . . . . . . . . 39
6 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50
Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54
vi
Chapter 1
Preliminaries
1.1 Basics of Representation Theory
Let G be a nite group, let K be a eld, and let V be a nite-dimensional vector space
over K. Denote by GL(V) the group of invertible linear transformations of V onto itself.
We de ne a linear K-representation of G in V to be a group homomorphism : G!
GL(V). When the eld K is understood to be C, as it is throughout this work, we refer to
simply as a representation. A group G usually has multiple representations, and we may
think of the images of these representations informally as di erent \snapshots" of the group.
By studying the \snapshots," we indirectly obtain information about the group at hand.
Now, given a representation , and an element g 2 G; we have that (g) 2 GL(V), so
it follows that (g) : V !V can be viewed as a matrix [ (g)]B relative to some ordered basis
B: Recall that the trace of a matrix is the sum of its diagonal elements. Although we may
change the basis for V to some other ordered basis, the trace of (g) remains unchanged
relative to this new basis. It makes sense, then, to speak of the trace of (g). Hence the
following notion is well-de ned: The character a orded by is the function : G!K
given by (g) = Tr[ (g)]B; where B is some ordered basis for V.
Let W be a subspace of V such that (g)(W) W for all g 2 G. Then we obtain a
new map G! GL(W), called the sub-representation of a orded by W. If has no
proper, non-trivial sub-representations, then we say that is irreducible, and the character
it a ords is an irreducible character. The set of irreducible characters of a group G,
denoted Irr(G), will play a central role in the ensuing text. Before proceeding, though, we
1
must learn to see representations of G as equivalent to what we call KG-modules.
Given a nite group G and a eld K, KG is the vector space whose elements are linear
combinations of the form Pa2G aa; where a 2K: Its basis is the set G; and it gets its
additive structure in a natural way. To give KG multiplicative structure, we de ne multi-
plication by
X
a2G
aa
! X
b2G
bb
!
=
X
a;b2G
( a b)ab:
The element 1e serves as the multiplicative identity of KG; where e is the identity element
of the group G. Now, a K-algebra is a ring A with identity that is also a vector space over
K such that (ab) = ( a)b = a( b) for all 2K and all a;b2A. With the above structure,
KG is a K-algebra, and since its basis is G; we refer to KG as a group algebra.
Recall that, for an arbitrary ring R; an R-module is an abelian group M equipped with
a map R M!M such that for all r;s2R and m;n2M;
(i) r(m+n) = rm+rn
(ii) (r +s)m = rm+sm
(iii) r(sm) = (rs)m
(iv) 1m=m
Let V be a vector space over the ring KG; and again let : G!GL(V) be a representation
of G: Then V becomes a KG-module by de ning gv = (g)(v) (g2G;v2V) and extending
linearly to KG. In this way, each representation gives rise to a KG-module.
Now let V be a KG-module. Using the embedding K!KG given by 7! e; we view V
as a ( nite-dimensional) vector space over K. De ne : G!GL(V) by (g)(v) = gv: Then
the properties of a KG-module can be used to verify that is a representation of G. Hence
2
each KG-module gives rise to a representation of G; so the two objects are equivalent via
this one-to-one correspondence. In this instance, we say that is the representation of G
a orded by V. If is the character a orded by , then we may also say that is a orded
by V.
Just as a representation has subrepresentations, so too does a KG-module have submod-
ules. If a KG-module V has no non-zero, proper submodules, we call V simple. Simple
KG-modules correspond to irreducible representations of G. Like the latter, these simple
KG-modules a ord irreducible characters. We will pass freely between representations
of G and KG-modules throughout, depending on contextual convenience. As mentioned, we
will always take K = C. We now present, for the reader?s convenience, a list of results we
will need.
1.2 Selected Results
De nition 1.1. For a character of G; the positive integer (e) is the degree of . We
say is linear if (e) = 1.
Theorem 1.2. ([Isa94, p.16]) A group G is abelian if and only if every irreducible char-
acter of G is linear.
Proposition 1.3. Let be a linear character of G. Then is irreducible.
Proposition 1.4. ([Isa94, p.20]) Let be a character of G. Then for all g2G;
(i) j (g)j (e); and
(ii) (g 1) = (g):
Proposition 1.5. ([Isa94, p.16]) The cardinality of the set Irr(G) equals the number of
conjugacy classes of G.
3
Theorem 1.6. ([Isa94, p.19]) Generalized Orthogonality Relation. Let ; 2
Irr(G). Then the following holds for every h2G:
1
jGj
X
g2G
(gh) (g 1) = (h) (e);
where is the Kronecker delta.
The set of all class functions G ! C is a vector space over C: Since characters are
constant on conjugacy classes, characters are class functions, and we have the following
de nition.
De nition 1.7. Let ; be characters of G. Then
( ; ) = ( ; )G = 1jGj
X
g2G
(g) (g)
is the inner product of and .
Proposition 1.8. ([Isa94, p.21]) Let be a character of G. Then is irreducible i
( ; ) = 1.
De nition 1.9. Let be a character of G. The kernel of is the subgroup of G de ned
by ker =fg2Gj (g) = (e)g.
Theorem 1.10. [Isa94, p.81]) Let N G and let 2Irr(G) with ( ;1)N 6= 0: Then
N ker :
Proposition 1.11. [Isa94, p.24]) Let N G:
(i) If is a character of G and N ker ; then is constant on cosets of N in G, and
the function ^ on G=N de ned by ^ (gN) = (g) is a character of G=N.
4
(ii) If ^ is a character of G=N, then the function de ned by (g) = ^ (gN) is a character
of G.
(iii) In both (i) and (ii), 2Irr(G) i ^ 2Irr(G=N).
Theorem 1.12. Let H 6 G and let 2Irr(G): Then the restriction of to H, denoted
H; is a character of H:
De nition 1.13. ([Isa94, p.17]) Let i2Irr(G) for each i. If = Pki=1ni i is a character,
then those i with ni > 0 are called the irreducible constituents of .
De nition 1.14. ([Isa94, p.62]) Let H 6G and let be a class function of H. Then G;
the induced class function on G, is given by
G(g) = 1jHj
X
x2G
(xgx 1);
where is de ned by (h) = (h) if h2H and (y) = 0 if y =2H:
Proposition 1.15. ([Isa94, p.63]) Let H 6G and suppose is a character of H. Then
G is a character of G. We say that G is an induced character of G:
De nition 1.16. Let H 6 G; let be a character of H, and let g2G. The conjugate
character of , denoted g ; is the character of gH de ned by g (gh) = (h) for all h2H.
Here, gh = ghg 1; and gH =fghjh2Hg:
Proposition 1.17. [CR62, p.323]) Let H be a subgroup of G, let M be a CG-module,
and let L be a submodule of MH. De ne LG as the module corresponding to the induced
character G; where L a ords : If M = _Pa2AaL; where A is a set of representatives for
the left cosets of H in G, then M = LG.
5
De nition 1.18. [Hun80, p.135] Elements e1;:::;en in a ring R are orthogonal central
idempotents if eiej = ij, ei2Z(R) for each i; and e2i = ei for each i.
Theorem 1.19. [Isa94] The elements e ; 2Irr(G); are orthogonal central idempotents
in the ring CG, where e = (e)jGj
X
g2G
(g 1)g:
Proposition 1.20. [JL93] If is an irreducible character of G, and V is any CG-module,
then e V is equal to the sum of those CG-submodules of V that have character .
Proposition 1.21. [Suz82, (3.8),p.23] Let G be a group and let H and K be subgroups of
G. For each x2G; the double coset HxK is a union of right cosets of H and the cardinality
of the set of these cosets is jK : K\Hxj.
Theorem 1.22. Frobenius Reciprocity. Let H be a subgroup of G, let be a character
of G, and let be a character of H. Then ( G; ) = ( ; H).
Theorem 1.23. Mackey?s Subgroup Theorem. Let X and Y be subgroups of a group
G. If L is a CX-module, then
(LG)Y =
M
a2A
((aL)aX\Y )Y;
where A is a set of representatives for the (Y;X)-double cosets in G.
We will need to refer to the character table of the dihedral group
D2n =ha;bjan = e;b2 = e;bab = a 1i
[Ser77, pp. 37{38]:
6
ak bak
0 1 1
1 1 1
2 ( 1)k ( 1)k (n even)
3 ( 1)k ( 1)k+1 (n even)
j 2 cos(2 kj=n) 0 (1 j ><
>>:
1 if is even;
1 if is odd:
Proposition 1.31. If is a partition of n, then 0 = "n ; where "n is the alternating
character of the group Sn.
1.3 Root Systems
The following de nitions and results about root systems can be found in [Hum72, pp.42-
44, 47, 52, 55]. Let E be a nite-dimensional vector space over R endowed with a positive
de nite symmetric bilinear form ( ; ).
8
De nition 1.32. A re ection in E is an invertible linear transformation leaving point-wise
xed some hyperplane (subspace of co-dimension one) and sending any vector orthogonal to
that hyperplane into its negative.
De nition 1.33. The re ection determined by a nonzero vector is given by
( ) = 2( ; )( ; ) :
We abbreviate 2( ; )( ; ) by h ; i.
De nition 1.34. A subset of the Euclidean space E is called a (crystallographic) root
system in E if the following axioms are satis ed:
(R1) is nite, spans E, and does not contain 0
(R2) If 2 ; the only multiples of in are .
(R3) If 2 ; ( ) =
(R4) If ; 2 ; then h ; i2Z
We call the elements of roots.
De nition 1.35. The rank of the root system is the dimension of E.
Proposition 1.36. We have h ; ih ; i= 4cos ; where is the angle between and .
De nition 1.37. A subset of is called a base if
(B1) is a basis for E as a vector space
(B2) Each root 2 can be written as = Pk ( 2 ) with integral coe cients
k all nonnegative or nonpositive.
9
De nition 1.38. The root system is irreducible if it cannot be partitioned into the
union of two proper subsets such that each root in one set is orthogonal to each root in the
other.
De nition 1.39. Fix an ordering ( 1;:::; l) of the roots in a base of the root system .
The matrix (h i; ji) is then called the Cartan matrix of . The entries of the Cartan
matrix are called Cartan integers.
1.4 Coset Spaces and Symmetrized Tensor Spaces
Here we provide a brief account of the construction of coset spaces. We then present
some fundamental results, and discuss the nature of the correspondence between coset spaces
and orbital subspaces of symmetrized tensor spaces.
Let G be a nite group and let H be a subgroup of G. Denote by G=H the set of (left)
cosets of H in G. We denote by C(G=H) the complex vector space having G=H as basis.
Let Irr(G) denote the set of irreducible characters of G, and let 2 Irr(G). We de ne a
form B H on C(G=H) by
B H(aH;bH) = (e)jHj
X
h2H
(b 1ah);
extending linearly in the rst component and antilinearly in the second. By [Hol04, p.2],
B H is a well-de ned Hermitian form, and it is G-invariant: For all g;a;b2G; we have that
B H(gaH;gbH) = B H(aH;bH):
Now put C H := C(G=H)=kerB H; where kerB H := fx 2 C(G=H) : B H(x;y) = 0 for all
y2C(G=H)g. We callC H a coset space. The set of vectors = H =faH2C H ja2Gg
10
are called the standard vectors in the coset space. The coset space and its standard vec-
tors will play a pivotal role in the sequel. IfC H has a basis consisting of pairwise orthogonal
standard vectors, we say that C H has an o-basis. We will call G an o-basis group if for
every H 6 G and 2Irr(G) the vector space C H has a basis that is orthogonal relative to
B H (de ned below) consisting entirely of standard vectors.
The form B H induces a well-de ned form B H on C H given by B H( x; y) = B H(x;y) (x;y2
C(G=H), where x denotes the coset x+ kerB H. From the natural left action of G on C(G=H)
we get a well-de ned action of G on C H, and B H inherits G-invariance. It is a theorem in
[Hol04, p.2] that, since B H is positive semide nite, the form B H is positive de nite. Thus
C H is an inner product space. Given vectors v1;:::;vn 2C H; the matrix of inner products
given by (B H(vi;vj))i;j is the Gram matrix of the vectors v1;:::;vn. Also from [Hol04, p.2],
we have the following formula for the dimension of the complex vector space C H:
dimCC H = (e)( ;1)H = (e)jHj
X
h2H
(h):
Before discussing the correspondence between coset spaces and orbital subspaces, we note
the following proposition, which we use in a later chapter. Let N G; let 2 Irr(G),
and assume that N ker . Put bG := G=N and denote by ^a the image of a2G under
the canonical epimorphism G ! bG: The function b : bG ! C given by b (^a) = (a) is a
well-de ned irreducible character of bG [Isa94, p.24]. Let H 6G.
Proposition 1.40. [Hol04, p.4] The map ? :C H !C^ ^H given by ?(aH) = ^abH is a well-
de ned linear isometry. In particular, C H has an o-basis if and only if C^ ^H has an o-basis.
Let us now explore the notion of a symmetrized tensor space. Fix positive integers n
and m and set n;m = f 2Znj1 i mg. Fix a subgroup G of the symmetric group
Sn. A right action of G on the set n;m is given by = ( (1);:::; (n)) ( 2 n;m; 2G).
The stabilizer of 2 n;m is the set G =f 2Gj = g.
11
Let V be an inner product space of dimension m and letfeij1 i mgbe an orthonormal
basis for V. The inner product on V induces an inner product on V n (the nth tensor power
of V) and, with respect to this inner product, the setfe j 2 n;mgis an orthonormal basis
for V n, where e = e 1 e n.
The space V n is a (left) CG-module with action given by e = e 1 ( 2G; 2 n;m),
extended linearly. The inner product on V n is G-invariant, which is to say ( v; w) = (v;w)
for all 2G and all v;w2V n.
Let 2 Irr(G). The symmetrizer corresponding to is
s = (e)jGj
X
2G
( 1) 2CG;
where e denotes the identity element of G. This element s is the central idempotent of CG
corresponding to [CR62, 33.8].
Let 2 n;m. The standard (decomposable) symmetrized tensor corresponding to
and is e = s e . The orbital subspace of V n corresponding to and , denoted V ,
is the span of the set = =fe j 2Gg: The space V n is an orthogonal direct sum
of orbital subspaces.
To understand the nature of the correspondence between coset spaces and orbital subspaces,
we rely on the following de nition, theorem, and corollary.
12
De nition 1.41. Let V and V0 be inner product spaces, let ? : V !V0 be a linear map,
and let r be a positive real number. Then ? is a similarity transformation of ratio r if
k?(v)k= rkvk for all v2V: In this case, we write V V0.
A similarity transformation of ratio 1 is commonly known as an isometry. All similarity
transformations, however, preserve angles as well as relative lengths (i.e., k?(v)k=k?(w)k=
kvk=kwk).
Theorem 1.42. ([HH13]) For 2 n;m, we have V C G .
Proof. Let 2 n;m and put H = G . The map G=H ! V , given by H 7! e 1, is
well-de ned and it induces a surjective linear map ? : C(G=H)!V . For ; 2G, we have
(?( H);?( H)) = (e 1;e 1) = (e)jGj
X
2H
( 1 ) (1.1)
= rB H( H; H);
where r =jG : Hj 1 and where the second equality is from [Fre73, p. 339] (with in place
of ). Using linearity we get (?(x);?(y)) = rB H(x;y) for all x;y2C(G=H), and it follows
that the induced map ? :C H !V given by ?( x) = ?(x) is a well-de ned bijective similarity
transformation of ratio r.
According to Theorem 1.42 and its proof, every orbital subspace can be identi ed with
a coset space in such a way that the standard symmetrized tensors in the orbital subspace
identify, in an angle-preserving and relative length-preserving manner, with the standard
vectors in the coset space. The following result says that, conversely, every coset space can
be similarly identi ed with an orbital subspace. The statement requires some explanation:
Let G = fg1;:::;gng be a nite group. The Cayley embedding of G in the symmetric
group SjGj is the monomorphism ? : G ! SjGj given by ?(g) = g, with g : G ! G
de ned by g(a) = ga. Here, we regard g as an element of SjGj by using the identi cation
f1;:::;ng$G, i$gi. Using this same identi cation, we write gi to mean i for 2 jGj;m.
13
Corollary 1.43. ([HH13]) Assume that m = dimV 2. Let G be a nite group, let
2Irr(G), and let H 6G. Identifying G as a subgroup of SjGj via the Cayley embedding,
we have C H V , where 2 jGj;m is de ned by putting g equal to 1 or 2 according as
g2H or g =2H.
Proof. We have H = G , so the claim follows from Theorem 1.42.
We will reference this correspondence frequently in the fourth and fth chapters. Now,
with preliminaries behind us, we present new work.
14
Chapter 2
The Module Structure of the Coset Space
The coset space is the fundamental object enabling us to explore the geometry of sym-
metrized tensor spaces. Accordingly, we should familiarize ourselves with its structure as a
CG-module. Although we do not rely heavily on the module-theoretic properties of the coset
space throughout, we might glean future insight from its decomposition as a direct sum of
simple CG-modules. Let G be a nite group, let H 6G, and let 2Irr(G).
Theorem 2.1. C H =
( ;1)HM
j=1
L ; where L is the simple CG-module a ording .
Proof. First, we claim that C(G=H) = (CH)G: De ne : CG CH ! C(G=H) by
(r;s) = rsH. Letting r1 =
X
a2G
aa2CG, r2 =
X
a2G
aa2CG, and s2CH, we have that
(r1 +r2;s) =
X
a
( a + a)saH =
X
a
asaH +
X
a
asaH = (r1;s) + (r2;s):
Another straightforward argument shows that is linear in the second slot as well. Now let
r2CG; h2CH; and let s2CH. Then
(rh;s) = (rh)sH = s(rh)H = sr(hH) = srH = rsH = r(hs)H = (r;hs);
where we have used the fact that s2Z(CG) and the fact that H acts trivially on CH. Thus
is a middle-linear map, and so there exists a unique homomorphism : CG CH CH !
C(G=H) such that (r s) = (r;s): In fact, is a CG-homomorphism: Let a2G: Then
(a(r s)) = (ar s) = (ar)sH = a(rs)H = a (r s):
15
Now de ne : C(G=H) !CG CH CH by (aH) = a 1: Let aH;bH 2G=H; and
suppose aH = bH. Then a 1b2H; so a 1b = h 1 for some h2H. Thus
(aH) = a 1 = bh 1 = b h 1 = b 1 = (bH);
so is well-de ned. We claim is the inverse map of . For a2G; and r =
X
g2G
ggH 2
C(G=H); extending linearly gives
(ar) =
X
g2G
gagH
!
=
X
g2G
gag
!
1 = a
X
g2G
gg
!
1 = a (r);
so is a CG-homomorphism. Letting r be as above, we then have that
( )(r) = ( (r)) =
X
g
gg
!
1
!
=
X
g
gg
!
H =
X
g
ggH = r:
Now let m n2CG CH CH; where m2G and n2CH: Then
( )(m n) = ( (m n)) = (mnH) = mn 1 = m n:
Therefore, C(G=H) = CG CH CH = (CH)G; establishing the claim.
By virtue of the above isomorphism, C(G=H) a ords the character (1H)G: Thus the number
of times the simple CG-module L appears in the direct sum decomposition of C(G=H) is
((1H)G; ): By Frobenius Reciprocity, we have ((1H)G; ) = (1H; jH) = ( ;1)H:
Thus
( ;1)HM
j=1
L is the direct sum of the copies of L appearing in the decomposition of
C(G=H). Therefore, 1.20 gives
( ;1)HM
j=1
L = e C(G=H):
We now prove the theorem. De ne : C(G=H) !e C(G=H) in the obvious way. Since
16
is an epimorphism, we need only show that ker = kerB H:Lets =
X
a2G
aaH2C(G=H);
and let X be a set of left coset representatives of H in G. Then
s2ker () (e)jGj
X
2G
( 1)
X
a2G
aaH
!
= 0
() 1jGj
X
a2G
X
2G
a (e) ( 1) aH = 0
() 1jGj
X
x2X
X
a2G
X
h2H
a (e) (ah 1x 1)xha 1aH = 0
() 1jGj
X
x2X
X
a2G
X
h2H
( a (e) (ahx 1) )xH = 0
()
X
a2G
X
h2H
a (e) (ahx 1) = 0 8x2X
() 1jHj
X
a2G
X
h2H
a (e) (x 1ah) = 0 8x2X
()
X
a2G
a (e)jHj
X
h2H
(x 1ah) = 0 8x2X
()
X
a2G
aB H(aH;xH) = 0 8x2X
()B H
X
a
aaH;xH
!
= 0 8x2X;
()s2kerB H
The proof is complete.
17
Chapter 3
Construction of Bases for Coset Spaces and Root System Geometry
We fuel our conjectures by rst computing the Gram matrices forC H in the case where
G is a symmetric group of small order, using various choices for H and :
Example 3.1. Let G = S3, let H = feg, and let 2 Irr(G) of degree 2. We have that
dimC H = 4, and we obtain the basis fH;(12)H;(13)H;(123)Hg by trial and error. After
dividing all entries of the resulting Gram matrix by 2, we get
2
66
66
66
64
2 1 0 0
1 2 0 0
0 0 2 1
0 0 1 2
3
77
77
77
75
This is the Cartan matrix for the rank 4 root system A2 A2:
The construction of bases for coset spaces by trial and error is time-consuming. To
expedite the next example, we use the following.
Proposition 3.2. Let H 6K 6G; and letfkiHgni=1 be a complete set of distinct left cosets
of H in K. Let 2Irr(G) and let a;b2G. Then B K(aK;bK) = jHjjKj
X
i
B H(aH;bkiH):
Proof. Let 2 Irr(G), and let a;b2G. For each k2K; we have that k = kih for some i
and some h2H, and this expression for k is unique, so
18
B K(aK;bK) = (e)jKj
X
k2K
(a 1bk)
= (e)jKj
X
i
X
h2H
(a 1bkih)
= (e)jKj
X
i
jHj
(e)
(e)
jHj
X
h2H
(a 1bkih)
!
= jHjjKj
X
i
B H(aH;bkiH):
Now we vary only the subgroup, but in doing so, we obtain a result that we later prove
holds generally for a symmetric group of arbitrarily large degree, given a speci c subgroup
and a speci c degree 2 irreducible character.
Example 3.3. Let G = S3; let K = h(12)i; and let be as in the rst example. Then
dimC K = 2, and using 3.2, we quickly obtain a basis for C K. The Gram matrix corresponds
to the Cartan matrix for the root system A2:
2
64 2 1
1 2
3
75
Already, with two cases, a pattern has begun to emerge.
As we construct bases for coset spaces of higher dimension, we require more results in
order to make calculations manageable. The next proposition is helpful in this regard.
Proposition 3.4. Let n = dimC H; let a1;:::;an 2G; and let 1;::: n 2C: The following
are equivalent:
(i) B=faiHgni=1 is a basis for C H
19
(ii) If B H
nX
j=1
jajH;aiH
!
= 0 for all 1 i n; then 1 = = n = 0:
Proof. Assume (i) holds, and suppose that B H
nX
j=1
jajH;aiH
!
= 0 for all 1 i n:
Fix y2G. Since B is a basis for C H, we have that yH = 1a1H + + nanH for some
i2C; with 1 i n: Then
yH + kerB H = ( 1a1H + + nanH ) + kerB H
=)yH ( 1a1H + + nanH )2kerB H
=)yH ( 1a1H + + nanH ) = k;
for some k2kerB H
=)yH = ( 1a1H + + nanH ) +k:
Now, we claim that B H
nX
i=1
iaiH;yH
!
= 0, and this will be true if and only if
B H( 1a1H + + nanH; 1a1H + + nanH +k) = 0
if and only if
nX
i=1
nX
l=1
i lB H(aiH;alH) +
nX
i=1
iB H(aiH;k) = 0
if and only if
(1)
nX
i=1
nX
l=1
i lB H(aiH;alH) = 0:
By assumption, we have the following system of equations:
20
(1) 1B H(a1H;a1H) + + iB H(aiH;a1H) + + nB H(anH;a1H) = 0
(j) 1B H(a1H;ajH) + + iB H(aiH;ajH) + + nB H(anH;ajH) = 0
(n) 1B H(a1H;anH) + + iB H(aiH;anH) + + nB H(anH;anH) = 0
Now, multiply (1) by 1, (2) by 2, and continue in this fashion until nally multiply-
ing (n) by n. Note that each row remains equal to zero. Summing both sides of the newly
obtained system, we have (1) in the chain of double implications, which establishes the claim.
Since y 2 G was arbitrary, we have B H
nX
i=1
iaiH;yH
!
= 0 for all y 2 G. Hence
nX
i=1
iaiH
!
2 kerB H; implying that
nX
i=1
( iaiH + kerB H) = kerB H, which implies that
nX
i=1
iaiH = 0: Since B is a basis for C H; we must have 1 = = n = 0:
Assume (ii) holds. Since the fact that B H
nX
j=1
jajH;aiH
!
= 0 for all i is enough to
imply that all the i are zero, it certainly follows that if B H
nX
i=1
iaiH;yH
!
= 0 for all
y2G; then all the i are zero. By an argument above, this is tantamount to saying that
21
1a1H + + nanH = 0 implies that all the i are zero. Hence faiHgni=1 is a linearly
independent set of n vectors, which proves the set constitutes a basis for C H.
Example 3.5. Let G = S4, let H = feg, and let be of degree 2. Then dimC H = 4, and
using 3.4 we have that the Gram matrix corresponds to the root system A2 A2. This result
is in line with our previous ndings.
We would like to calculate the Gram matrix for a coset space of higher dimension in
order to see what patterns emerge. Since S4 has two degree 3 irreducible characters, spaces
with dimension as high as 9 are available for H = feg. In constructing a basis for such a
space, however, the number of candidates for basis vectors is daunting. A more manageable
task is to nd a basis for a 9-dimensional space associated with the subgroup A4. Even for
this, we require two more results. Yet a third result will then be needed to draw a conclusion
for S4. We present those results now.
Proposition 3.6. Let H 6 K 6 G. Let 2Irr(G); and let a1;:::;an2G: If faiKgni=1 is
linearly independent in C K, then faiHgni=1 is linearly independent in C H.
Proof. Assume that faiKgni=1 is linearly independent in C K, and suppose 1a1H + +
nanH = 0, where i2C for all i. Then
1B H(a1H;yH) + + nB H(anH;yH) = 0 8y2G:
Let fkjHgmj=1 be a set of distinct left cosets of H in K, and let x2G: Then we have
22
1B H(a1H;xk1H) + + nB H(anH;xk1H) = 0
1B H(a1H;xk2H) + + nB H(anH;xk2H) = 0
1B H(a1H;xkmH) + + nB H(anH;xkmH) = 0:
Thus by summing the above system of equations, we have
1
X
j
B H(a1H;xkjH) + + n
X
j
B H(anH;xkjH) = 0;
implying by 3.2 that
1jKjjHjB K(a1K;xK) + + njKjjHjB K(anK;xK) = 0:
Factoring jKjjHj from the left-hand side, we now have that
1B K(a1K;xK) + + nB K(anK;xK) = 0;
which in turn implies that
B K( 1a1K + + nanK;xK) = 0 8x2G:
Hence
1a1K + + nanK = 0:
23
By assumption, the vectors faiKgni=1 are linearly independent, so 1 = = n = 0: We
conclude that faiHgni=1 is linearly independent in C H.
Theorem 3.7. Let H 6 K 6 G; let 2 Irr(G); and assume vanishes o K. Let
fkjHgtj=1 (kj 2 K) be linearly independent in C H, and let fgiKgsi=1 be a complete set of
distinct left cosets of K in G. Then fgikjH j 1 i s; 1 j tg is linearly independent
in C H.
Proof. First, assume aH and bH are left cosets of H in K such that aH cK and bH dK,
where cK and dK are distinct left cosets of K in G. Let h2H: We claim that a 1bh =2K:
Suppose otherwise. Since a2cK we get aK = cK: Similarly, bK = dK: Thus aK 6= bK:
Then a 1bh2K would give a 1b2Kh 1 = K; a contradiction. Thus a 1bh =2K:
Now, since fkjHgtj=1 is linearly independent in C H, so is the set of translates Ti :=fgik1H;
:::;giktHg for each 1 i s. For each i, hTii is a subspace of C H.
Let 1 m;n s with m 6= n; and let 1 p;q t: We have that kpH K and
kqH K, so it follows that gmkpH gmK and gnkqH gnK: Since gmK 6= gnK; we
get that B H(gmkpH;gnkqH) = (e)jHj
X
h2H
((gmkp) 1(gnkq)h) = 0 by the rst claim of the
proof and the fact that vanishes o K.
Now put T = [iTi: From the above it follows that, for all 1 j s; hTji and
X
i6=j
hTii
are orthogonal, whence hTi = _
X
i
hTii; and thus T = fgikjH j 1 i s; 1 j tg is a
linearly independent set.
Theorem 3.8. Let H 6K 6G; and let 2Irr(G); with = jK 2Irr(K): Then C H and
C H are isometric.
24
Proof. Let faiHgni=1 be a basis for C H. Since and agree on H, the de nition of coset
space dimension gives that
dimCC H = (e)jHj
X
h2H
(h) = (e)jHj
X
h2H
(h) = dimCC H:
For each 1 i n; aiH 2K=H G=H; so aiH 2C H: Thus there exists a unique and
well-de ned linear map ? :C H !C H such that ?(aiH) = aiH for each i: We now show
that the inner products on the two spaces agree. Let aH;bH2C H: Then
B H(aH;bH) = (e)jHj
X
h2H
(b 1ah) = (e)jHj
X
h2H
(b 1ah) = B H(aH;bH) = B H(?(aH);?(bH));
so by linearity of ? and the inner product, the formula holds when replacing standard
vectors with arbitrary vectors. Now letting v 2 ker?; we have that ?(v) = 0; so that
B H(?(v);?(v)) = 0: Hence B H(v;v) = 0; implying that v = 0: Thus ? is injective, which
proves that the two spaces are isometric.
Example 3.9. Let G = A4, let H = feg, and let be the degree 3 irreducible character
of A4. Then dimC H = 9. We outline the construction of a basis for C H. First put K =
h(12)(34)i6A4, so that dimC K = 3. Using 3.4, we obtain the basisfK;(12)(34)K;(13)(24)Kg
forC K. Now, by 3.6, replaceK withH to get the linearly independent setfH;(12)(34)H;(13)(24)Hg
inC H. We have that vanishes o V =f(1);(12)(34);(13)(24);(14)(23)g: Using a set of left
cosets of V in A4, 3.7 gives for C H the linearly independent set
fH;(12)(34)H;(13)(24)H;(123)H;(134)H;(243)H;(132)H;(234)H;(124)Hg:
These are 9 vectors, so they constitute a basis for C H. We nd, however, that the Gram
matrix corresponds to no Cartan matrix for a crystallographic root system.
25
Example 3.10. We have that feg6 A4 6 S4; and also that the restriction of one of the
degree 3 irreducible characters of S4 is the degree 3 irreducible character of A4. By 3.8, it
follows that the basis acquired in the last example is also a basis for C H, where G = S4;
H = feg; and where is the character formerly restricted. Hence we have found a coset
space associated with a symmetric group such that the geometry of this coset space is
not that of a crystallographic root system. The entries of the Gram matrix for the basis
vectors inC H cannot be adjusted to become the entries for a Cartan matrix. This raises the
question, however, of precisely which of the Cartan integers we can obtain from the inner
product on a coset space. The answer, as we will see shortly, points the way towards a more
general connection between coset spaces and root systems. The next results make clear this
connection.
Proposition 3.11. Let C H be a coset space and let aH;bH 2 : Then jB H(aH;bH)j
dimCC H:
Proof. Let aH;bH 2 : By the Cauchy-Schwarz Inequality, we have that jB H(aH;bH)j
kaHkkbHk: Note that equality occurs if and only if one vector is a scalar multiple of the
other [Axl97]. Thus we infer that
jB H(aH;bH)j
q
B H(aH;aH)
q
B H(bH;bH)
=
s
(e)
jHj
X
h2H
(h)
s
(e)
jHj
X
h2H
(h)
=
q
dimCC H
q
dimCC H
= dimCC H:
Lemma 3.12. Let H 6 G and let 2Irr(G): Assume that C H has a basis consisting of
standard vectors. If this basis forms a base for a crystallographic root system, then for all
distinct basis vectors aH and bH, either B H(aH;bH) = 0 or jB H(aH;bH)j= 12 dimCC H:
26
Proof. Assume thatC H has a basis consisting of standard vectors, and that this basis forms
a base for a crystallographic root system. Let aH and bH be distinct basis vectors for C H.
By 3.11, we have the strict inequality
jB H(aH;bH)j< dimCC H:
Since the basis vectors forC H form a base for a root system, we must have thathaH;bHi2Z
for all aH;bH in the basis. Thus, with (aH;bH) := B H(aH;bH), we have
2(aH;bH)
(bH;bH) 2Z:
By 3.11, this gives
2(aH;bH)
dimCC H 2Z:
Now put j(aH;bH)j= c and dimCC H = d; so that 2cd = k for some k2Z+[f0g: If k = 0;
then B H(aH;bH) = c = 0, so suppose k> 0.
By the inequality in 3.11, c i2: Left multiplying each of these two basis vectors by a i2bl;
we obtain ai1 i2feg and feg. Put k = i1 i2: Then feg and akfeg are a pair of orthogonal
elements of E. By the isometry in 4.3, we have a pair of orthogonal elements in Bn0 as well.
35
Thus there exists in G a rotation through an angle of =2, implying that 4jn and hence that
n is even.
We are now ready to prove the main theorem.
Theorem 4.5. The following are equivalent:
(i) D2n has order a power of 2,
(ii) for each H 6D2n; 2Irr(D2n); C H has an o-basis.
(i ) ii) Assume G is a dihedral group with order a power of 2. Fix H 6 G and
2Irr(G): If ( ;1)H = 0, then dimC H = 0; so the conclusion follows. Suppose ( ;1)H 6= 0:
If has degree 1, then
dimCC H = (e)jHj
X
h2H
(h) = 1jHj
X
h2H
(h) 1jHj
X
h2H
(e) = 1jHj jHj= 1
so again the conclusion follows.
Now put = j for some 1 j < n=2 and assume ( ;1)H 6= 0: Since n0 is a power of
2 and n0> 2; we have that 4jn0: Hence Bn0 has a pair of orthogonal elements, and it follows
from 4.3 that E does as well. Assume that H *Cn: Put A = H\Cn =fai1;:::;airg; where
1 ij n 1; and put B = H\Cnc =fai1b;:::;airbg; where b is the generating re ection
for the dihedral group. Noting that jAj=jBj, and that H = AtB; we have that
dimCC H = (e)jHj
X
h2H
(h) = (e)jHj
X
h2A
(h) = (e)jHj jHj2 (e) = 2;
again using that Cn ker : Hence the two orthogonal elements of E form an o-basis for
C H:
36
Assume now that H Cn: Let cH and dH be the pair of orthogonal elements guaran-
teed by 4.3. ThenfcH;dHgis linearly independent inC H. A complete set of distinct cosets
of Cn in G is fCn;bCng, where b is the generating re ection. By 3.7, fcH;dH;bcH;bdHg is
linearly independent in C H: Since cH and dH are orthogonal, the G-invariance of B H gives
us that bcH and bdH are orthogonal as well. Consider cH and bcH. We have that
B H(cH;bcH) = (e)jHj
X
h2H
(c(bc) 1h) = (e)jHj
X
h2H
(cc 1b 1h) = (e)jHj
X
h2H
(bh) = 0;
since b is a re ection. Now consider cH and bdH. We have
B H(cH;bdH) = (e)jHj
X
h2H
(c(bd) 1h) = (e)jHj
X
h2H
(cd 1bh) = (e)jHj
X
h2H
(cd 1h 1b) = 0;
since cd 1h 1 is a rotation and hence cd 1h 1b is a re ection. Similar arguments show that
dH is orthogonal to both bcH and bdH; whence the set of all four vectors are pairwise
orthogonal. Finally, since H ker ; we have
dimCC H = (e)jHj
X
h2H
(h) = (e)jHj jHj (e) = 4
Therefore, C H has an o-basis.
(ii ) i) Let G = D2n. Assume C H has an o-basis for each H 6 G and 2 Irr(G).
Let = 1 and let H = feg. Then n = n0: Hence for each cH 2E, there exists c2Cn
such that f(cv0) = cH; so that f in Theorem 4.3 is surjective and thus bijective. Thus there
exists a pair of orthogonal elements in E and hence in Bn, implying by Lemma 4.4 that n is
even. Thus, if n = 3; then C H does not have an o-basis for each (H; )-pair. If n = 4, then
jGj= 8 = 23:
We now proceed by induction on n. Fix n > 4. Since n is even, we have that an=2 2Cn:
37
Let : G !G=han=2i be the canonical epimorphism, and put bG = G=han=2i: Since the
quotient of a dihedral group is dihedral, we have that bG is a dihedral group of order n. Now
x bH 6 bG and b 2 Irr(bG): Then bH = (H) for some H 6 G and b = for some
2Irr(G): We have thathan=2i ker ; so it follows by 1.40 thatC H andCb bH are isometric.
By assumption, C H has an o-basis for each (H; )-pair. Thus Cb bH has an o-basis for each
(bH;b )-pair. By the induction hypothesis, jbGj= n is a power of 2, whence jGj= 2jbGj= 2n
is a power of 2 as well. The proof is complete.
38
Chapter 5
Root Systems For A Special Coset Space
We now extend our results on root systems to include a special case. Earlier, we looked
at symmetric groups and their associated coset spaces, and we drew conclusions about the
type of rank 1 and 2 crystallographic root systems embedded in the geometry of these spaces.
If we consider symmetric groups Sm with m arbitrarily large, it turns out that we need not
restrict ourselves to rank 1 and 2. By xing a particular subgroup and irreducible character
of Sm, we always obtain the root system Am 1. This result agrees with a conclusion found
in [TS12, Theorem 14]. In that paper, however, the authors take a combinatorial approach,
drawing upon substantial graph-theoretic preliminaries. Our proof is much more geometrical.
Put G = Sm. The inner product space Cm =fa = (a1;a2;:::;am)jai2Cg is a CG-module
with action given by a = (a 1(1);a 1(2);:::;a 1(m)). This CG-module is an internal direct
sum Cm = T _+V, where T = C(1;1;:::;1) and V =fa2CmjPiai = 0g.
The CG-module T a ords the trivial character 1. Let and denote the characters of
G a orded by Cm and V, respectively. By the preceding paragraph, we have = + 1.
Denote by C" the vector space C on which G acts according to the formula x = "( )x,
where " is the sign character of G. Then C" is a CG-module a ording the character ".
The CG-module V" = C" V a ords the character = " . We identify the vector space V"
with V using the map x v7!xv. With this identi cation we de ne the action of G on V"
by the formula a = "( )(a 1(1);a 1(2);:::;a 1(m)).
39
Given a partition of m; recall that there exists a bijective correspondence 7! from
the set of partitions of m to the set Irr(Sm) of irreducible characters of Sm.
For each 1 i m, let ei 2 Cm be the m-tuple with jth entry ij. The R-span of
the set = fei ejji 6= jg V" is a root system of type Am 1 for 2 R. Put
a = (1; 1;0;:::;0)2 .
Fix H 6G and 2Irr(G). Again, let be the standard vectors of C H.
Theorem 5.1. Assume that H =h(12)i; and = [2;1m 2]. There exists a CG-isomorphism
? :C H !V" satisfying the following:
(i) ? is an isometry,
(ii) ?( H) = a for all 2G,
(iii) =
8>
><
>>:
?( _[ ) if m = 3;
?( ) if m 4:
The proof will require several lemmas. After inducing the map ? in the statement of
the theorem, we will give the proof of the theorem.
Consider vi2Cm 1; the (m 1)-tuple with j-th entry ij. We identify vi with ei2Cm, the
m-tuple with the same entries but with a xed m-th entry of zero. Put v = v1 + +vm 1:
Putting L = Cv; we have that L is a submodule of CmjH; where we de ne H as the subgroup
of Sm isomorphic to Sm 1 obtained by xing m. Throughout, we will identify H and Sm 1
for the sake of simplicity.
Lemma 5.2. Cm = LSm
40
Proof. Using 5.2, we must show that Cm = _
X
a2A
aL; where A is a set of representatives for
the left cosets of H in G. The CG-isomorphism then follows.
We claim that a complete set of left coset representatives forH inGis A =f(1;m);(2;m);:::;(m
1;m);(m;m) = eg: Suppose (i;m)H = (j;m)H for some i 6= j; with 1 i;j < m:
Note that m =2 for any 2 H, and also that (i;j) 2 H: If (j;m)(i;m)H = H, then
(j;m)(i;m)(i;j)2H, in which case (i;m)2H; a contradiction. Since jG : Hj= m; A is a
set of distinct coset representatives of H in G:
We have that aL = Cav for each a 2 A: Label the elements of A as a1 = (1;m); a2 =
(2;m);:::;am = (m;m): We must show that aiLT( Pj6=iajL ) = f0g for each 1 i m:
To show this, it su ces to prove that the statement holds for i = 1; since the other cases
are handled similarly.
Let w2a1L
\
(
X
i>1
aiL ): Since w2a1L;
(1) w = 1a1v = 1v2 + 1v3 + + 1vm
for some 1 2C. On the other hand, w2
X
i>1
aiL; so w = 2a2v + 3a3v + + mamv
where i 2 C for 2 i m. Now, v1 occurs as a summand in the vectors a2v through
amv; so the combined coe cient of v1 as a summand of w is 2 + + m: Equating this
coe cient of v1 with its coe cient in (1), we have
(2) 2 + + m = 0:
For each vi with i> 1;vi fails to occur as a summand in iaiv: Hence the combined coe cient
of vi in the expression of w2
X
i>1
aiL is 2 + + i 1 + i+1 + + m: Equating this
41
coe cient with the coe cient of vi in (1) yields
(3) 2 + + i 1 + i+1 + + m = 1:
Now, taking (2) with each equation of the form (3) corresponding to an i > 1, we have a
system of m equations on whose left side each i occurs m 1 times. Summing both sides
of this system, we have (m 1)
X
i>1
i = (m 1) 1, so that
X
i>1
i = 1: But by (2), this
left-hand sum is zero, implying that 1 = 0; and hence that w = 0:
Thus aiL
\
(
X
j6=i
ajL ) = f0g; as claimed. Since
X
i
aiL is the sum of m submodules,
we must have Cm = _
X
i
aiL: The isomorphism follows.
We continue to view Sm 1 as a subgroup of Sm.
Lemma 5.3. = (1Sm 1)Sm
Proof. The module L = Cv is one-dimensional, and clearly stable under the action of Sm 1,
so it a ords 1Sm 1. Since Cm a ords ; the above isomorphism gives the result.
Lemma 5.4. The character is irreducible.
Proof. We have that and are the characters ofG = Sm a orded by Cm andV;respectively,
and that = + 1: Now is irreducible if and only if ( ; ) = ( 1; 1) = 1: Put
= 1H; where again H = Sm 1; so that = G: Then by using Frobenius Reciprocity twice,
we have
42
( G 1G; G 1G) = ( G; G) 2( G; 1G) + (1G; 1G)
= ( ; ( G)H) 2( ; (1G)H) + 1
= ( ; ( G)H) 2( ; ) + 1
= ( ; ( G)H) 1
It remains to determine that the value of ( ; ( G)H) = 2: By Mackey?s Subgroup Theorem,
( G)H =
M
a2A
( (a )aH\H )H;
where A is a set of representatives for the (H;H)-double cosets in G. Let A =fe;(m 1;m)g:
Then HeH and H(m 1;m)H are (H;H)-double cosets of H in G, and since (m 1;m)2
H(m 1;m)H but (m 1;m) =2H; these cosets are distinct.
Let K 6 H = Sm 1; with K = Sm 2; and now identify K with the subgroup of Sm 1
xing m 1: With a = (m 1;m); we claim that aH\H = K: Let let k2K: Then
k = ek = (m 1;m)(m 1;m)k = (m 1;m)k(m 1;m)2aH;
and we have aH\H K: Now let aha 12aH\H: Since aH\H H; we have m =2aha 1:
Assume m 1 2aha 1: Since conjugation applies a to each entry of h, this implies that
m2H; a contradiction. Hence aH\H K; and the claim follows.
Since K = Sm 2; we havejaH\Hj=jKj= (m 2)!; and thusjH : aH\Hj= m 1: Using
1.21, and incorporating the case a = e; we now have that
jSmj= m! = (m 1)! + (m 1)!(m 1) =jHj+jHjjH : aH\Hj=jHj+jH(m 1;m)Hj:
43
By reasons of order, we must in fact have a complete set of (H;H)-double cosets in G. Thus
Mackey?s Subgroup Theorem gives
( G)H = ( H)H + (a K)H = + (a K)H:
Then
( ;( G)H) = ( ; + (a K)H) = ( ; ) + ( ;(a K)H):
By Frobenius Reciprocity, we have
( ;(a K)H) = ( K;a K)
Since K is trivial, the de nition of the conjugate character yields that a K(ak) = 1 for all
ak2aK; so that a K is trivial as well. Thus ( K;a K) = 1; and we conclude that
( ; ) = ( 1; 1)
= ( G 1G; G 1G)
= ( ; ( G)H) 1
= ( ; ) + ( ;(a K)H) 1
= ( ; ) + ( K;a K) 1
= 2 1 = 1;
whence is irreducible.
Lemma 5.5. The CG-module V" a ords the character = [2;1m 2]:
Proof. We have that + 1 = = (1S[m 1;1])Sm; after identifying the partition [m 1;1] of m
with the partition [m 1] of m 1. The only partitions of m that majorize [m 1;m] must
have m 1 or m as the rst entry, giving [m 1;1] and [m]. Thus by 1.29, we have that
(1S[m 1;1])Sm = c [m 1;1] + d [m] for some positive integers c and d. Since = (1S[m 1;1])Sm;
44
and 1Sm = [m]; we must have = [m 1;1]:
Now, the conjugate partition of = [m 1;1] is the partition 0 with i-th component
0i equal to the number of indices j for which j i: Hence 0 = [2;1m 2]: By 1.31, we have
that [2;1m 2] = "m [m 1;1]; where "m is the sign character of Sm. Since C" a ords "m and V
a ords [m 1;1]; we infer that [2;1m 2] is the character a orded by V" = C" V:
Since H xes a ; we get a well-de ned linear map ? : C(G=H)!V" satisfying ?( H) =
a ( 2 G). This map is a CG-homomorphism. For the next result, we prove that ?
preserves the bilinear forms on C(G=H) and V". To this end, we put =
p
(e)=jHj =
p
(m 1)=2 and we use the character = [2;1m 2] of degree m 1 from 5.5. This character
can be de ned by
( ) =
8
>><
>>:
jFix( )j 1 if is even;
1 jFix( )j if is odd:
where Fix( ) denotes the subset of elements off1;:::;mg xed by . For 1 i m; we use
the notation i2 ; i =2 to indicate that moves or xes the integer i, respectively.
Lemma 5.6. The map ? : C(G=H)!V" preserves the bilinear forms.
Proof. Due to the G-invariance of B H, it su ces to consider products of elements of the
form H; H: Throughout, k denotes the number of xed points of ; and we calculate the
value of on an element of G according to the formula above. We proceed by considering
cases.
Case 1: Suppose 1;2 =2 :
B H( H;H) = m 12 [ ( ) + ( (1;2))] = m 12 [(k 1) + (3 k)] = m 12 2 = m 1 =
2h(1; 1;0;:::;0);(1; 1;0;:::;0)i=h?( H);?(H)i
Case 2: Suppose 12 ; 2 =2 . (The reverse case is similar)
We have (1;2) = (1;:::) (1;2) = (1;2;:::) (:::).
45
B H( H;H) = m 12 [(k 1)+(2 k)] = m 12 = 2h(1; 1;0;:::;0);(0; 1;:::;1;:::;0)i=
h?( H);?(H)i
Case 3: Suppose 1;22 ; with 1 and 2 in di erent cycles.
We have (1;2) = (1;:::)(2;:::) (1;2) = (1;:::;2;:::) (:::).
B H( H;H) = m 12 [(k 1)+(1 k)] = 0 = 2h(1; 1;0;:::;0);(0;0;:::; 1;:::;1;:::;0)i=
h?( H);?(H)i
Case 4: Suppose 1;22 ; with 1;2 in same cycle, (1;2) not a cycle.
Subcase i: 1,2 nonadjacent
We have (1;2) = (1;:::;2;:::) (1;2) = (1;:::)(2;:::) (:::).
B H( H;H) = m 12 [(k 1)+(1 k)] = 0 = 2h(1; 1;0;:::;0);(0;0;:::; 1;:::;1;:::;0)i=
h?( H);?(H)i
Subcase ii: 1,2 adjacent
We have (1;2) = (1;2;:::) (1;2) = (1;:::) (:::).
B H( H;H) = m 12 [(k 1)+( k)] = (m 12 ) = 2h(1; 1;0;:::;0);(0;1;:::; 1;:::;0)i=
h?( H);?(H)i
Case 5: Suppose 1;2 =2 ; and (1;2) a cycle.
We have (1;2) = (1;2)(:::) (:::)(1;2) = (:::) (:::):
B H( H;H) = m 12 [(k 1)+( k 1)] = (m 1) = 2h(1; 1;0;:::;0);( 1;1;0;:::;0)i=
h?( H);?(H)i
Thus ? preserves the bilinear map.
46
The last step in getting the induced map ? is to realize the kernel of B H in terms of the
orthogonal idempotent e = (e)jGj
X
2G
( 1) associated with G and .
Lemma 5.7. We have ker B H = (1 e ) R(G=H):
Proof. Let
X
a
aaH2 kerB H: It is enough to prove that e
X
a
aaH = 0: We have
e
X
a
aaH = (e)jGj
X
( 1)
X
a
aaH
= (e)jGj
X
X
a
a ( 1) aH
= (e)jGj
X
g
X
a
a (ag 1)gH
= (e)jGj
X
b
X
a
X
h
a (ah 1b 1)bH
= (e)jGj
X
b
X
a
X
h
a (hb 1a)bH
= (e)jGj
X
b
X
a
a
X
h
(b 1ah)bH
= (e)jGj
X
b
X
a
a(
X
h
(b 1ah) )bH
= (e)jGj
X
b
X
a
a( B H(aH;bH) )bH
= (e)jGj
X
b
B H(
X
a
aaH;bH )bH
= 0
Now letr =
X
6=
e
X
a
aaH2(1 e ) R(G=H): It su ces to show that (e)jGj
X
( 1)
X
a
aaH2
kerB H for a single summand 6= ; for then r2B H as well. Then for all b in G; we have
47
B H
(e)
jGj
X
( 1)
X
a
aaH;bH
!
= (e)jGj
X
X
a
a ( 1)B H( aH;bH)
= (e)jGj
X
X
a
a ( 1)
X
h
(b 1 ah)
= (e)1
X
a
a
X
h
1
jGj
X
(b 1 ah) ( 1)
= (e)1
X
a
a
X
h
1
jGj
X
( ahb 1) ( 1)
= 0
by the Generalized Orthogonality Relation.
We now prove Theorem 5.1.
Proof. First, we claim that kerB H ker?. Using Lemma 5.7, and the fact that ? is a
CG-homomorphism, we have that ?((1 e )
X
a2G
aaH) = (1 e )?(
X
a2G
aaH ): Now
?( Pa2G aaH )2V"; and since V" a ords by Lemma 5.5, e acts as the identity map on
V". Thus
(1 e )?(
X
a2G
aaH ) = ?(
X
a2G
aaH ) e ?(
X
a2G
aaH )
= ?(
X
a2G
aaH ) ?(
X
a2G
aaH )
= 0;
and the claim follows.
We now have an induced CG-isomorphism ? : C H ! V" with ?( H) = ?( H) = a ;
proving (ii).
48
Since B H is an inner product, and since B H( v; w) = B H(v;w) = h?(v);?(w)i for all
v; w2C H; the induced map preserves the inner product. Therefore ? is an isometry, yielding
(i).
To establish (iii), rst let m = 3. Since (132)H = (23)H and (123)H = (13)H; we have
that = f(132)H;(123)H;Hg; after choosing representatives for the left cosets of H in
G. Hence ?( ) = f (1; 1;0); (0;1; 1); (1;0; 1)g: To map onto the remaining roots
( 1;1;0);(0; 1;1); and ( 1;0;1); we must include =f (132)H; (123)H; Hgin the
domain of ?: After doing so, we get ?( _[ ) = :
Now let m 4: We have that ?(H) = a = "(e)a = (1; 1;0;:::;0) 2 : Since
m 4; Sm contains the element = (12)(34); and so ?( H) = a = "( )a = a =
( 1;1;0;:::;0): By [Hum72, p.53, Lemma C], we have that Sm acts transitively on the set
of those roots having a xed length, so that the orbit of a under Sm is precisely . That
is, ?( ) = : The proof is complete.
49
Chapter 6
Conclusions
Originally, the coset space was developed as a tool for deciding whether an orbital
subspace of a symmetrized tensor space had a basis consisting of pairwise orthogonal standard
symmetrized tensors (an o-basis). As we have seen, when G is a dihedral group of order a
power of 2, these orbital subspaces V have such a basis for each 2 and 2 Irr(G).
Dihedral groups are distinctive in the sense that all of their irreducible characters have degree
at most 2. Hence for any (H; )-pairing, we have that
dimCC H = (e)jHj
X
h2H
(h)1(h) (e)jHj
X
h2H
(e) = (e)jHj jHj (e) = 2(e) 4:
Their relatively small dimension makes the coset spaces for the dihedral group fairly easy
to investigate. With the symmetric group, however, the maximum degree of its irreducible
characters increases with the cardinality of the group. Consequently the maximum dimen-
sions of coset spaces for these groups grow as well. The approach of constructing bases and
computing Gram matrices becomes unwieldy for Sn when n is high, a fact which became
apparent with the concluding example of S4 in Chapter 3.
Still, for the xed (H; )-pairing in Chapter 5, we were able to obtain a geometric description
of C H for Sn with n arbitrarily large. In this case, the standard vectors C H formed a
root system of type An 1: Considering this alongside our prior results, it is notable that, for
each of the root systems we obtained, all of its irreducible components were of type Am for
some m2N. Thus all of the root systems we obtained have roots of equal length: They are
\simply laced." Since we arrived at our results, work has been done [HH13] indicating that
50
the following is a theorem:
There exists an orbital subspace such that the standard symmetrized tensors in the sub-
space form a root system isomorphic to a given irreducible root system if and only if the
irreducible root system is simply laced.
We saw a glimmer of this already with our concluding result of Chapter 3, which placed
tight constraints on the rank 1 and 2 root systems which could be realized with standard
vectors in a coset space. As we established in 3.12, if C H has a basis of standard vectors
that forms a base for a root system, then the value of B H(aH;bH) is either 0 or 12 dimCC H
for distinct basis vectors aH;bH in the coset space. Naturally, we revisit the question that
arose in the context of research on o-basis groups: Given a nite group G; what conditions
on the group insure that for every H 6 G and 2Irr(G); the vector space C H has a basis
that is orthogonal relative to B H consisting entirely of standard vectors? This is a question
about the structure of a nite group, and about its irreducible characters. Having posed
the question, the formula for the inner product of two vectors in a coset space is itself very
suggestive. We have
B H(aH;bH) = (e)jHj
X
h2H
(b 1ah):
If we could determine the conditions under which an irreducible character vanishes on a
subset of G, this could shed light on the values we obtain from the summation on the right.
Speci cally, note that the right-hand side of the equation above is a sum of the elements of
a set of the form (cH) =f (ch) jh2Hg, where cH2G=H: Thus, if we knew when irre-
ducible characters vanish on entire cosets, we could know one way in which this summation
becomes zero.
Much literature on zeros of irreducible characters already exists. For instance, we have
the following theorem in [Nav01]:
51
Theorem 6.1. Let G be a nite group and let N C G: Let 2 Irr(G). Then N is not
irreducible if and only if vanishes on some coset Nx of N in G.
To illustrate an easy consequence of this theorem, let N CG and let 2 Irr(G). Sup-
pose N is not irreducible. By the theorem, vanishes on some coset of N in G. Call this
coset xN. Then xN and N are a pair of orthogonal vectors inC N: Extension of results such
as this theorem might prove very fruitful in generating more orthogonal cosets, and hence
in generating an entire o-basis for a coset space.
We nd other results of interest scattered throughout the literature. In [DRB07], the authors
prove that if G is a nite solvable group which has an irreducible character which vanishes
on exactly one conjugacy class, then G has a homomorphic image which is a nontrivial 2-
transitive permutation group. Now, Holmes proved in [Hol95] that any 2-transitive subgroup
of Sn (n 3) is not o-basis. In [Hol04] he proved that the class of o-basis groups is closed
under taking homomorphic images. The above results, taken in concert, imply that if G is
a nite solvable group which has an irreducible character which vanishes on exactly one
conjugacy class, then G cannot be o-basis.
Another possible line of inquiry lies in the related notions of a Camina group and the
vanishing-o subgroup. Given a nite group G and the set of nonlinear irreducible char-
acters nl(G) Irr(G), we de ne the vanishing-o subgroup V(G) = hg2Gj there exists
2 nl(G) such that (g) 6= 0i [Isa94]. It turns out that Camina groups can be de ned
by the condition that V(G) = G0 [Lew09]. When restricting attention to Camina groups of
prime power order, we nd in [DS96] that such groups have nilpotency class at most 3. This
suggests a possible connection with results in [Erv07], wherein the author conjectures that
having nilpotency class no more than 3 may be a necessary condition for a nilpotent group
to be o-basis.
52
In working towards a proof that a nite group is o-basis, we will always take cognizance
of the geometric proof methods we have employed throughout this work. These methods
hold promise in furnishing intuitive ways to explore future conjectures in this area.
53
Bibliography
[Axl97] Sheldon Axler, Linear algebra done right, second ed., Undergraduate Texts in
Mathematics, Springer-Verlag, New York, 1997. MR 1482226 (98i:15001)
[BPR03] C. Bessenrodt, M. R. Pournaki, and A. Reifegerste, A note on the orthogonal
basis of a certain full symmetry class of tensors, Linear Algebra Appl. 370 (2003),
369{374. MR 1994340 (2004f:15052)
[CR62] Charles W. Curtis and Irving Reiner, Representation theory of nite groups and
associative algebras, Pure and Applied Mathematics, Vol. XI, Interscience Publish-
ers, a division of John Wiley & Sons, New York-London, 1962. MR 0144979 (26
#2519)
[DP99] M. R. Darafsheh and N. S. Poursalavati, Orthogonal basis of the symmetry classes
of tensors associated with the direct product of permutation groups, Pure Math.
Appl. 10 (1999), no. 3, 241{248. MR 1752864 (2001f:15030)
[DRB07] John D. Dixon and A. Rahnamai Barghi, Irreducible characters which are zero
on only one conjugacy class, Proc. Amer. Math. Soc. 135 (2007), no. 1, 41{45
(electronic). MR 2280172 (2008j:20016)
[DS96] Rex Dark and Carlo M. Scoppola, On Camina groups of prime power order, J.
Algebra 181 (1996), no. 3, 787{802. MR 1386579 (97b:20022)
[Erv07] Jason Ervin, On o-basis groups and generalizations, Ph.D. Dissertation.
[Fre73] Ralph Freese, Inequalities for generalized matrix functions based on arbitrary char-
acters, Linear Algebra and Appl. 7 (1973), 337{345. MR 0340273 (49 #5028)
[HH13] Hank G. Harmon and Randall R. Holmes, Geometry of standard symmetrized ten-
sors, submitted.
[Hog07] Leslie Hogben (ed.), Handbook of linear algebra, Discrete Mathematics and its
Applications (Boca Raton), Chapman & Hall/CRC, Boca Raton, FL, 2007, Asso-
ciate editors: Richard Brualdi, Anne Greenbaum and Roy Mathias. MR 2279160
(2007j:15001)
[Hol95] Randall R. Holmes, Orthogonal bases of symmetrized tensor spaces, Linear and
Multilinear Algebra 39 (1995), no. 3, 241{243. MR 1365442 (96k:15015)
[Hol04] , Orthogonality of cosets relative to irreducible characters of nite groups,
Linear Multilinear Algebra 52 (2004), no. 2, 133{143. MR 2033134 (2004j:20011)
54
[HT92] Randall R. Holmes and Tin Yau Tam, Symmetry classes of tensors associated
with certain groups, Linear and Multilinear Algebra 32 (1992), no. 1, 21{31. MR
1198818 (93j:20019)
[Hum72] James E. Humphreys, Introduction to Lie algebras and representation theory,
Springer-Verlag, New York, 1972, Graduate Texts in Mathematics, Vol. 9. MR
0323842 (48 #2197)
[Hun80] Thomas W. Hungerford, Algebra, Graduate Texts in Mathematics, vol. 73,
Springer-Verlag, New York, 1980, Reprint of the 1974 original. MR 600654
(82a:00006)
[Isa94] I. Martin Isaacs, Character theory of nite groups, Dover Publications Inc., New
York, 1994, Corrected reprint of the 1976 original [Academic Press, New York;
MR0460423 (57 #417)]. MR 1280461
[JL93] Gordon James and Martin Liebeck, Representations and characters of groups,
Cambridge Mathematical Textbooks, Cambridge University Press, Cambridge,
1993. MR 1237401 (94h:20007)
[Lew09] Mark L. Lewis, The vanishing-o subgroup, J. Algebra 321 (2009), no. 4, 1313{
1325. MR 2489902 (2010a:20020)
[Nav01] Gabriel Navarro, Irreducible restriction and zeros of characters, Proc. Amer. Math.
Soc. 129 (2001), no. 6, 1643{1645 (electronic). MR 1814092 (2001k:20012)
[Ser77] Jean-Pierre Serre, Linear representations of nite groups, Springer-Verlag, New
York, 1977, Translated from the second French edition by Leonard L. Scott, Grad-
uate Texts in Mathematics, Vol. 42. MR 0450380 (56 #8675)
[Suz82] Michio Suzuki, Group theory. I, Grundlehren der Mathematischen Wissenschaften
[Fundamental Principles of Mathematical Sciences], vol. 247, Springer-Verlag,
Berlin, 1982, Translated from the Japanese by the author. MR 648772 (82k:20001c)
[TS12] Maria M. Torres and Pedro C. Silva, Tensors, matchings and codes, Linear Algebra
Appl. 436 (2012), no. 6, 1606{1622. MR 2890942
[WG91] Bo Ying Wang and Ming Peng Gong, A high symmetry class of tensors with an
orthogonal basis of decomposable symmetrized tensors, Linear and Multilinear Al-
gebra 30 (1991), no. 1-2, 61{64. MR 1119469 (93f:15028)
55