C OMPOSITIO M ATHEMATICA
M ARC A. B ERGER
S HMUEL F RIEDLAND
The generalized Radon-Hurwitz numbers
Compositio Mathematica, tome 59, n
o1 (1986), p. 113-146
<http://www.numdam.org/item?id=CM_1986__59_1_113_0>
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113
THE GENERALIZED RADON-HURWITZ NUMBERS
Marc A.
Berger
and Shmuel Friedland© Martinus Nijhoff Publishers, Dordrecht - Printed in the Netherlands
Abstract
The generalized Radon-Hurwitz number, p ( m, n ), designed to characterize the dimensions for which normed bilinear maps exist, is discussed. The values of p ( m, n ) are computed
when (i) n - m 3; (ii) n - m = 4 and m is odd; (iii) m 9. Tables are provided and
many structure results for the Radon-Hurwitz matrices are developed.
Notation
Rm denotes the real m-dimensional Euclidean space with norm
and inner
product
S m -1
denotes the unit ball(i.e.
the set of unitvectors)
in IRm.R Pm-1
denotes the real
projective
space obtained fromSm-1 by identifying antipodal points. x always
denotes a vectorin Rm, y’ always
denotes avector in
Rk and y always
denotes a vectorin Rk-1.
Since we have muchmore occasion to work in
IR k -1
thanin Rk
we have chosen tolet y (without
theprime)
denote a vectorin Rk-1,
and theprime
indicates anextra dimension. Thus
The dimension of a
subspace e
is denoteddim(u).
For two spacesu, v
we let
lin(u, v)
denote the set of linearmappings
from u to V(with domain u);
and we let(u, v)
denote the set of continuousmappings
from e to Y
(with
domaino¡¡).
For a matrix A we use thefollowing symbols:
apq - the (p, q ) entry of A
At - the
transpose
of Atr( A ) -
the trace of Aker( A ) -
the kernel(i.e.
the nullspace)
of Anull( A ) -
thenullity
of A(i.e.
the dimension of itskernel)
§ 1.
IntroductionLet
Mmn
denote the vector space of all real m X nmatrices,
and letQmn
denote the subset of those matrices A E
Mmn
whichsatisfy
AAt = 03B1(A)I (1)
for some scalar
a( A). (Of
coursea( A)
must benonnegative
and it is zeroif and
only
if A = 0. It can be writtenexplicitly
asa(A) = 1 m tr(AAt).)
Ifm > n then
Qm n
istrivial,
so we willalways
take m n. When m = n wedenote these sets
by Mn
andQm, respectively.
DEFINITION I : The
generalized
Radon-Hurwitz numberp(m, n)
is themaximal dimension of a
subspace
contained inQm n .
The number
p(m) := p(m, m)
is the classicalRadon-Hurwitz
num-ber. It was
computed independently by
Radon[23]
and Hurwitz[18].
Ifwe factor m as
then
p ( m )
isgiven by
This number
03C1(m)
is of centralimportance
in many mathematicalproblems. 03C1(m)-1
is the dimension of the maximal Cliffordalgebra
inMm.
Thus this number can be traced as far back as Clifford[10].
Thefamous result of Adams
[1]
inalgebraic topology (K-theory)
asserts that03C1(m)-1
is the maximal number ofindependent
vector fields onsm-le
This number appears in the
study
ofnonsingular
bilinear maps(Lam [20]), imbeddings
of realprojective
spaces(Adem [4],
Berrick[9]),
or-thogonal designs (Geramita
andSeberry [14])
andstrictly hyperbolic partial
differentialequations (Friedland,
Robbin andSylvester [13]).
The
problem
ofdetermining
thegeneralized
Radon-Hurwitz number03C1(m, n )
was first formulatedby
Hurwitz[17],
as we shall describeshortly. (See
the remarksfollowing
Definition2.V.)
In the classical case m = n Eckmann[11]
used aspecial
group structure inMm
to prove that03C1(m)
isgiven by (2), (3)
above. When m =1= n we do not have this nice group structure. Theproofs
in§3
and§5 show, however,
that some ofthis structure can be
recovered,
and that there is a very intricatealgebraic
structure involved here. Cf. Adem
[6], [7].
In
§2
we discuss several alternate definitions of03C1(m, n).
In§3
weanalyze
the structure of matrices whichsatisfy
the Radon-Hurwitz condition andpresent
someelementary
calculations forp ( m, n ).
Here wecompute
the values of03C1(m, n ) for m
9. In§4
we describe the Adams result in theperspective
of a nonlinear Radon-Hurwitz number. In§5
we
compute
the values ofp ( m, n )
for(i) n - m 3,
and(ii) n -
m = 4with m odd. The vaues of
p(m, m
+1), p(m, m
+2)
can be obtaineddirectly
from Lam[19],
and some values ofp(m,
m +3), p(m, m
+4)
can be obtained
by applying
thetopological techniques
in Adams[2].
Our
techniques
here areentirely
different from those of Adams or Lam.§2.
Alternate def initions f orp ( m, n )
For any
subspace *C Qmn
we can choose a basisEl, .... Ek ~ u
suchthat
where
Equivalent
to(1)
Thus we arrive at the
following
alternate definition for03C1(m, n ).
DEFINITION I :
(Alternate): 03C1(m, n )
is the maximal k for which there exists a mapA ~ lin(Rk, Qmn) satisfying (1). Equivalently
it is the maximalnumber, k,
of real m X n matricesEl=(e(l)pq), i = l, ... , k,
which can
satisfy (3).
For A E
Mm n
let us writeA = [B, C] (4)
in block
partitioned form,
where B EMm
and C eM,nl ( l
= n -m ).
LetQ*mn
be the subsetIf u ~
Qmn
is anysubspace
and V is any n X northogonal matrix,
thenu03A6 is a
subspace
ofQmn
of the same dimension as u. Inparticular by
such a transformation we may assume without loss of
generality
in(3)
that
It then follows from
(3)
thatEl’...’ Ek-1 ~ Q*mn,
and thesubspace
vspanned by {E1,...,Ek-1}
is a(k - 1)-dimensional subspace
ofQmn.
Conversely,
if v is any(k - l)-dimensional subspace
ofQmn
then thesubspace 4Y spanned by v
and[1, 0]
is a k-dimensionalsubspace
ofQmn.
Thus we also have thefollowing
definition for03C1(m, n ).
DEFINITION II:
(Alternate): p(m, n ) -1
is the maximal dimension of asubspace
ofQ:n.
The condition on
A(y) = [B(y), C(y)]
thatis
If
then
equivalent
to(8), (9)
in terms of the basis elementsE,
=[Bl, Cl ]
for i, j = 1,..., k - 1
withi ~ j.
This then leads us to thefollowing
définition.
DEFINITION III:
(Alternate): 03C1(m, n) -1
is the maximal k’ for which there exists a map A Elin(lRk’, Q*mn) satisfying (7). Equivalently
it is the maximalnumber,
k’ of real m X m and m X lmatrices Bl
andCl,
i =
1,..., k’, respectively,
which cansatisfy (11)-(13).
The
following
result is immediate.PROPOSITION IV:
The following
areequivalent.
( a )
Condition(3)
holds.(b) Et1x,..., Etkx
are orthonormal whenever x ESm-1.
We note that
(d)
amounts to the bilinear mapD:Rm Rk~Rn
definedby
satisfying
Such a map
(bilinear, satisfying (15))
is called a normed map. Thus we arrive now at the next alternate definition.DEFINITION V:
(Alternate): p(m, n)
is the maximal k for which there exists a normed mapRm Rk~Rn.
Hurwitz
[17]
first formulated theproblem
offinding
the minimal n forwhich there exists a normed map
Rm Rk~Rn, given k
and m. Ofcourse this is
equivalent
to the determination ofp(m, n ),
since what isreally
underinvestigation
is the set oftriples (m, k, n)
for which such anormed map exists. Some results about normed maps are
given
in Adem[3]-[7].
Lam[22]
contains recent new results. For ageneral
survey of results andtechniques
see Lam[21]
andShapiro [24].
The k m matrices
D1,...,Dn
inProposition IV(d)
are said to bedual to the m n matrices
El, .... Ek.
Thisduality, d(q)ip
=e(l)pq,
is bestillustrated
by
the three-dimensional box inFigure
1 in theAppendix.
Namely,
the D’s arethe " horizontally
stacked" k X mmatrices,
and theE ’s are the "
vertically
stacked" m X n matrices. Thisfigure
also il-lustrates the
symmetry
in thek, m variables, corresponding
toProposi-
tion
3.I(a)
below.§3. Preliminary
resultsPROPOSITION I:
PROOF:
(a)
It follows at once from Definition 2.V.(b)
It is a direct consequence of(a)
above that03C1(m, n )
decreases inm. To see that it increases in n note that
[A, 0]
satisfies(1.1)
whenever A does.
(c)
The lower bound follows from(b) above,
and the upper bound follows fromProposition 2.IV(b).
(d) Let 0?/,
be asubspace
ofQmn ,
i =l, ... ,
t. Thenis a
subspace
ofdimension of 4Y is
t
(e)
The direct sum~ Al(y’)
satisfies(2.1)
whenever eachAl(y’)
does.
(f) According
toHopf [16]
a necessary condition for the existence of a normedmap Rm Rk~Rn is
that(n)
be even for n - k r m.(Stiefel [25]
first arrived at this condition for bilinear maps. Behrend[8]
extended it to oddhomogeneous polynomials and, finally, Hopf
extended it to biskew maps. This work
pioneered
thedevelopment
of the
theory
ofcohomology rings.)
0Using Proposition I, along
with the formula(1.3)
forp(m),
we cannow fill in Table 1 of the
Appendix.
TheSupplement
which follows that table containssteps
which can be used as instructions.By carefully going
over thesteps
in thatSupplement
one discovers twointeresting periodicities
PROPOSITION II:
PROPOSITION III : Let A =
[ B, C ] ~ Q*mn.
(a) If
m is odd then thespectral
radiusof cct
andCtC
is a, wherea =
a(A)
is the scalarappearing
in(1.1),
and any other nonzeroeigenvalues
come inpairs.
(b) If
m is even then any nonzeroeigenvalues of cct
andCtC
come inpairs.
(c) If
A =1= 0 then m -rank(C)
is even.PROOF :
The
Proposition
follows from thefollowing
sequence of observations.(i)
Since A satisfies(1.1)
and B is skewsymmetric
(ii)
Since B is skewsymmetric - B 2 is positive
semi-definite.Thus, by (1) r(CCt)
a.Furthermore,
if m is odd then B must besingu- lar,
and sor(CCt)
= a.(iii)
Since B is skewsymmetric
the nonzeroeigenvalues
ofB 2 come
inpairs.
(iv)
The nonzeroeigenvalues
and theirrespective multiplicities
coin-cide for
CC and C tC.
(v) a
is aneigenvalue
ofaI - CC
t ofmultiplicity
m -rank(CCt),
and
rank(C)
=rank( CC t ).
DPROPOSITION IV:
Then k 03C1(m + q, n - p).
PROOF :
(a)
Sinceker(C’C)
=ker(C)
the firstequality
is immediate. To estab-k-1
lish the second
equality
let v en ker( I - CtlCl).
Then it follows1=1
from
(2.11), (2.12)
thatNext we
compute, using (2.11), (2.12),
and for i
~ j, using (2.11), (2.13)
and(2) above,
Thus
and so
(b) (Part I).
Ifp > 0
choose an orthonormalbasis ( vl, .... vp 1
forn ker(Ct(y)C(y)),
and thencomplete
this to an orthonormal basis y~Rk-1{u1,...,ul-p, v1,...,up}
for all ofR 1,
where l = n - m. Let (D be theorthogonal
1 X 1 matrixIf Yc
Q*mn
is thesubspace
and if we set
then
is a
subspace
ofQmn
of the same dimension(namely, k - 1).
On theother
hand,
sinceit is clear that the
last p
columns of each matrixin
are zero. Thusk p ( m, n - p ). Furthermore,
sinceit is also clear that
and
thus q
remainsunchanged
under the transition fromC( y ) to C( y ).
(Part II).
Ifq > 0
choose an orthonormalbasis {w1,...,wq}
forSince
it follows from
(2.8), (2.9)
thatSet
Then it follows from
(2.8), (2.9)
and(7), (8)
above thatA(y’)
satisfies(2.1).
nPROPOSITION V:
Let A(y) = [B(y), C(y)] satisfy (2.8)-(2.10).
Let vitand fi be the
subspaces of Rk-1
(c) If C(~)
is notinjective
then 11 E~.
Inparticular,
UN.(d) If B(~)
issingular
then 11 EJÍÎ 1- .
N.B. u is a
subspace
sincePROOF:
(a)
Forany 03BE, ~ ~ Rk-1
It follows from
(2.8), (2.9)
that forany 03BE ~ u
Thus
by multiplying (10)
on the leftby Ct(03BE)
and on theright by C(03BE),
andusing
the definition of vit we have(for
any~)
(b)
This follows at once from(10).
(c)
IfC(~)
is notinjective
choose w ~ 0 such thatC(~)w=0.
Itfollows from
(11)
that forany 03BE~ u
By taking
the innerproduct
of both sides withrespect
to w we arrive atHence 17 1
03BE.
(d)
IfB(~)
issingular
choose w ~ 0 such thatB(,q)w = 0.
It followsfrom
(10)
that forany 03BE ~
JÍÎBy taking
the innerproduct
of both sides withrespect
to w andusing
the skewsymmetry
ofB(~)
we arrive atCOROLLARY VI: Let
A(y) = [B(y), C(y)] satisfy (2.8)-(2.10).
Let 17 E
Rk-1
be such thatB(~)
issingular,
and setThen
where X is the
subspace
inProposition
V.PROOF: It follows from
Proposition V(b)
that =ker(B(~))
is invariantunder
B(03BE),
forany 03BE ~ . By restricting B(03BE to 5°, then,
we obtain afamily
of skewsymmetric
matrices whichsatisfy
Thus, by
Definition2.IV,
the desiredinequality
follows. DThe next
Proposition
is from Friedland[12,
Thm.8.11].
PROPOSITION VII : Let
A(y)
be a real m X n matrixpolynomial
in thevariable y E
Rk-1.
Assume thatA( y)
is rank one, so that it is notidentically
zero and all 2 X 2 minors vanish. Then there exist vectorpolynomials u(y)
ERm, v(y)
E Rn withrelatively prime coordinates,
anda
(scalar) polynomial a(y)
such thatu(y)
andv(y)
areunique
up to scalarmultiplication.
Furthermoreif A(y)
is
symmetric
then we can chooseu(y)
=v(y).
PROOF: Choose y E
Rk-1
, a ERn, 03B2
E Rm such thatSince the
polynomial ring R[y] is
aunique
factorization domain we can setwhere
b(y), c(y)
are(scalar) polynomials,
andu(y), v(y)
have rela-tively prime
coordinates. Thus in aneighbourhood of y
we have anequality
Since
A(y), u(y), vt( y )
arepolynomials, a(y)
must be a rationalfunction of y.
Say, then,
where
p(y)
andq( y )
arerelatively prime polynomials.
We claim thatq( y)
must be constant. Otherwise there would be an irreducible factorr(y)
inq(y).
Since the coordinates ofu(y)
arerelatively prime
theremust be a coordinate
ui(y)
which isrelatively prime
tor y).
Since the(i, j)-entry
ofA(y)
is apolynomial
it must be thatr(y)
divides each coordinatevj(y) -
but this contradicts our choice ofv(y).
Note that once y,
a, /3
are chosen thenu(y)
andv(y)
areunique
up to scalarmultiplication.
IfA( y)
issymmetric
we can take03B1 = 03B2,
andthus
u(y)
andv(y)
can be chosen the same. 0PROPOSITION VIII:
Let A (y) = [B(y), C(y)] satisfy (2.8)-(2.10).
(a) If
then
(b) If
then
PROOF:
(a)
If m is even it follows fromProposition III(c)
thatC( y ) = 0,
andthus k p(m). Suppose, then,
that m is odd.According
toProposition III(a)
theeigenvalues
ofCt(y )C(y)
must be( |y| 2, 0, ...,0). According
to
Proposition
VII we writewhere
u(y)
is a vectorpolynomial
in Rn-m anda(y)
is a(scalar) polynomial.
SinceC( y )
is linearonly
one of a, u can vary with y. If u is constant thenand
thus, by Proposition IV, k p ( m,
m +1)
=03C1(m
+1). Otherwise,
ifa is constant we can write
where
Then
and ( u 1, ... , uk-1}
must be an orthonormalsystem.
Hence k -1 n -
m.(b)
If n is even it follows fromProposition
III thatCt(y)C(y) = |y|2I,
and
thus k 03C1(n). Suppose, then,
that n is odd.According
toProposi-
tion III the
eigenvalues
ofCt( y ) C( y )
must be(1 |y|2, |y|2,...,|y|2, 0).
According
toProposition
VII we writeExactly
asabove,
if u is constant thenand
(by Proposition IV) k 03C1(n - 1, n) = 03C1(n). Otherwise,
if a is con-stant then k -
1 n -
m. D§4.
Nonlineartheory
Let
Rmn
denote the subset of those matrices AE Mm n
with rank m.Conventionally
when m = n this set is denotedGLm .
LetR0mn, GLm
denote the sets
Rmn U (0), GLm ~ {0}, respectively,
with the zero matrixappended.
p ( m, n )
is the maximal k for whichlin(Rk, Qmn )
contains aninjective
map. It is this
perpective
that reveals howstriking
the theorem of Adams[1] is,
for a consequence of it is thefollowing. (See Friedland,
Robbinand
Sylvester [13]).
THEOREM I :
If (Sk-1, GLm )
contains an odd map thenk 03C1(m).
Inparticular 03C1(m)
is the maximal dimensionof
asubspace of GLm .
This theorem tells us that there exists an
injective
linear mapRk ~ Qm
if and
only
if there exists a continuous odd mapSk-1 ~ GLm.
COROLLARY II:
If (Sk-2, GLm)
contains an odd map which is skewsymmetric
thenk
p( m ).
PROOF:
If 03C8: Sk-2 - GLm
is an odd map which is skewsymmetric
thenthe map
is an odd map
in (Sk-1, GLm).
0COROLLARY III : Let
A(y)
=[B(y), C(y)]
be an odd map in(Sk-2, Q*mn).
(a) If B(y)
isnonsingular for
all y ES k - 2
thenk 03C1(m).
(b) If C(y)
isof
rank n -m for
all y ESk - 2
thenk 03C1(n).
PROOF:
(a)
IfB(y)
isnonsingular
then it must be an odd map in(Sk-2, GLm )
which is skew
symmetric.
Henceby Corollary II k p ( m ).
(b)
For asingle
matrix A =[B, C] ~ Q*mn
supposeC tC
is invertible.Let
Since
where
a = a(A)
is the scalar in(1.1)
it follows thatCtC
andCtBC
commute. Thus since B is skew
symmetric
so is D. Furthermoreby using
the
commutativity
ofC tC
andCtBC,
thecommutativity
ofCC and B,
and(2)
it follows thatThus
lies in
Q*,
and satisfiesIn terms
of y e Sk-2
if we now setthen
(y)
will be an odd mapin (Sk-1, Q*n).
Since(y)
is never zero( C( y )
is of maximalrank),
it follows as abovethat k 03C1(n).
DGitler and Lam
[15] have
demonstratedthat (S27, R1332)
containsan odd map,
yet lin(R28, R01332)
does not contain anyinjective
maps.Thus we cannot
hope
togeneralize
Theorem 1 in itsstrongest
form. Wecan obtain the
following result, though.
PROPOSITION IV:
If (Sk-1, Rmm+1)
con tains an odd map thenk
max( p ( m ), p ( m
+1)).
Inparticular (a) p(m, m
+1)
=max( p(m), p(m
+1));
( b ) p ( m, m
+1) is
the maximal dimensionof
asubspace of R0mm+1.
PROOF:
Define 03C8: Rmm+1 ~ GLm+1 by setting 03C8m+1q(A) equal
to(-1)q
multiplied by
the determinant of the m X m submatrix obtained from Aby striking
theq th column, and 03C8pq(A) = apq for p
m.Any
odd map~ E
(Sk-1, Rmm+1)
can be liftedto ) = 03C8 o
cr E(Sk-1, GLm+l).
Ifm is odd
then (p
is odd andthus, by
TheoremI, k p (m
+1).
If m iseven, then for any y E
Sk-1
the rows of(y’)
form a basis forRm+1,
thelast one an even function
of y’
and the rest odd functions.Hence, according
toFriedland,
Robbin andSylvester [13,
Thm.A] k p(m).
~
The result
03C1(m, m+1) = max(03C1(m), 03C1(m+1))
can beproved by elementary
means, without resort to TheoremI,
as will beapparent
in§5
below.
§5.
Basic resultsIn this section we
compute p ( m, m
+l)
for 13,
and03C1(2m’ + 1, 2m’ + 5).
THEOREM 1 : For odd m
PROOF: It follows from Table 1 and
Proposition 3.11(a)
that03C1(3, m ) m - 2 (for
allm). Hence,
fromProposition 3.I(a)
follows03C1(m,m+2)3 (for
allm ).
It follows fromProposition I(c)
thatp ( m, m
+
2) p ( m
+1) (for
allm ).
This establishes the lower boundp ( m, m
+2)
max(03C1(m
+1), 3) (for
allm ).
Let
A(y) = [B(y), C(y)] satisfy (2.8)-(2.10). According
toProposi-
tion
3.III(c) rank(C(y)) = 1
fory ~ 0. Thus, according
toProposition 3.VIII(a), k max(p(m, m
+1), 3).
Since m is oddp ( m, m + 1) = 03C1(m + 1).
This establishes the upper bound03C1(m,
m +2) max(03C1(m
+1), 3).
DTHEOREM II: For even m
PROOF: From
Proposition 3.I(b)
follows the lower bound03C1(m, m + 2) max(03C1(m), 03C1(m+2)).
LetA(y) = [B(y), C(y)] satisfy
(2.8)-(2.10). According
toProposition 3.III(b)
theeigenvalues
ofCt(y)C(y)
are(03B3(y), 03B3(y)),
where03B3(y)
is aquadratic
formsatisfying
Thus
Let JV be the
subspace
For y e
J1I’ we can defineas in
(4.7),
andwill be an odd continuous function
of y satisfying (2.7).
Sinceit follows that
where
Suppose
now that k >p(m). (Otherwise
we arethrough.) According
to
Corollary
4.111 there must be a non-zero 11 for whichB(~)
issingular.
Since
null(B(~))
= 2 it follows then fromCorollary
3.VI thatdim(N)
1.Thus
w ( y )
in(3)
is defined on all ofSk- 2 except
for at most twoantipodal points,
where03B3(y) =
0. Sincew ( y )
canonly approach ± 1
atthese
points
and sinceSk-1
issimply
connectedfor k à 4,
it follows that03C9(y),
and hence(y),
has an odd continuous extension to all ofSk-2,
as
long
ask >
4. Furthermore we maysuppose k >
4 sincemax(03C1(m), 03C1(m+2)) 4.
With thisextension, then, (y)
is an oddmap in
(Sk-2, GLm+2)
which is skewsymmetric.
Thusby Corollary
4.11
k 03C1(m
+2).
This establishes the upper boundp(m,
m +2) max( p(m), p(m
+2)).
0THEOREM III: For even m
PROOF: From
Proposition 3.I(b)
follows the lower bound03C1(m,m+3)03C1(m,m+2) (for all m ).
LetA(y) = [B(y), C(y)] satisfy (2.8)-(2.10). According
toProposition 3.III(b)
theeigenvalues
ofCt(y)C(y)
are(03B3(y), 03B3(y), 0),
where03B3(y)
is aquadratic
formsatisfying
Accordingly
is a
symmetric
rank onepolynomial
matrix. IfF( y )
isidentically
zerothen
03B3(y),
and henceC( y ),
is alsoidentically
zero, in which casek p ( m ). Otherwise,
ifF( y )
is notidentically
zero, we can useProposi-
tion 3.VII to write
for a vector
polynomial u(y) ~ R3 and
a(scalar) polynomial a( y).
If uis a constant vector we can
let |u| = 1,
in which caseThen
and it follows from
Proposition 3.IV(b)
thatk 03C1(m,
m +2).
Otherwise,
ifu(y)
varieswith y
it must be linear in y, anda(y)
mustbe constant. We can then write
Furthermore,
Condition
(2.7)
remains valid under anorthogonal
transformation of the yvariable, y = 03A6y
and thus it may be assumed without loss of gener-ality
thatwhere
0 p
k - 1 and each Y, ispositive.
From(7), (8)
follows that infact p
3. Let N be thesubspace
Exactly
as in theproof
of Theorem IIabove,
we conclude that if k >p ( m )
thendim(N)
1. Sinceand since
p(m, m+2) 4,
it suffices now, in order to conclude theproof,
to demonstrate that ifk > 03C1(m)
and if fi is non-trivial thenp 2.
Suppose
thenthat p
=3, C4
= 0. We show how this leads to acontradiction. Observe that from
(4), (6), (8)
followsExpanding
this we obtainFrom
(7), (8), (9)
followsFrom
(4), (6), (7)
followsand thus
using (13)
we further conclude thatFrom
(12), (13), (14)
follows thatC1u2, CIU3, C2u3
aremutually orthogo-
nal nonzero vectors, and
hence, using (11), (12), (13)
we conclude thatFrom
(6), (10)
follows thatC(y)F(y) = 0,
and so from(2.9), (4)
weconclude that
In
particular, for ~~N~,
q ~ 0 it follows that the two-dimensionaleigenspace of -B2(~) corresponding
to theeigenvalue |~|2 - 03B3(~)
isprecisely range(C(~)).
On the otherhand, according
toProposition 3.IV(b),
Thus this
eigenspace, range(C(~)),
is invariant underB4.
Their unionU range(C(~))
is then also invariant underB4. According
to(15),
11 E JV’
though,
this union is a three-dimensionalsubspace
of Rm. SinceB4
isboth skew
symmetric
andnonsingular (in
factorthogonal)
it cannot havean odd dimensional invariant
subspace.
This is our desired contradiction.D
THEOREM IV: For odd m
PROOF: From
Proposition 3.I(b)
follows the lower bound03C1(m,m+3)03C1(m+1,m+3) (for
allm).
LetA(y)=[B(y), C(y)]
satisfy (2.8)-(2.10). According
toProposition 3.III(a)
theeigenvalues
ofCt(y)C(y)
are(|y|2, 03B3(y), 03B3(y)),
where03B3(y)
is aquadratic
formsatisfying
Accordingly
is a
symmetric
rank onepolynomial
matrix. IfF( y )
isidentically
zerothen
rank( C( y )) = 3, y * 0,
andthus k p ( m
+3) (Corollary 4.III(b)).
Otherwise,
ifF( y)
is notidentically
zero, we can useProposition
3.VIIto write
for a vector
polynomial u(y) ~ R3
and a(scalar) polynomial a(y).
If uis a constant vector we can
let |u| = 1,
in which caseThen
and it follows from
Proposition 3.IV(b) that k 03C1(m
+1,
m +3).
Otherwise,
ifu(y)
varies with y it must be linear in y, anda ( y )
mustbe constant. We can then write
Furthermore,
Whenever
03B3(y) ~
0 we can defineas in
(2.7),
and thenwill
satisfy (2.7).
If03B3(y)
isidentically
zero thenrank(C(y)) 1,
y~ Rk-1
and we conclude from
Proposition
3.VIII thatOtherwise
D(y)
definedby (23)
is a rational function of y. We show now that in fact it islocally
linear in y, henceglobally
linear - in which caseÂ
aboveactually
lies inlin(Rk-1, Q*m+3).
ThenAccordingly,
we devote the remainder of thisproof
toestablishing
thatD(y)
islocally
linear in y. From(18), (20), (22)
followsand thus from
(2.8), (2.9)
followsUsing (23)
we see then thatFrom
(4.3)
followsChose 11
such that0 03B3(~) lq 1 2.
This can be done since we areassuming
thaty( y )
is neitheridentically
0or |y| 2.
Thenu(~) ~
0. SinceD(~)
is askew-symmetric
3 X 3matrix,
the conditions(24), (25) require
that
where X is the vector cross
product.
SinceD(y)
is continuous the choice of + must be fixed in aneighbourhood
of q, and we arethrough.
0THEOREM V: For odd m
p(m, m
+4)
=max(p(m, m
+3), 5).
PROOF: It follows from Table 1 and
Proposition 3.II(a)
thatp(5, m) m -
4(for
allm ). Hence,
fromProposition 3.I(a)
followsp ( m, m
+4)
5(for
allm ).
It follows fromProposition 3.I(b)
thatp ( m, m + 4) 03C1(m, m + 3) (for
allm ).
This establishes the lower boundp ( m, m
+4) max( p ( m,
m +3), 5) (for
allm ).
Let
A(y) = [B(y), C(y)] satisfy (2.8)-(2.10). According
toProposi-
tion
3.II I(a)
theeigenvalues
ofCt(y)C(y)
are(|y|2, 03B3(y), 03B3(y), 0),
where
03B3(y)
is aquadratic
formsatisfying
Accordingly
is a
symmetric
rank onepolynomial
matrix. IfF( y)
isidentically
zerothen
y( y ) ~ |y|2
and it follows fromProposition 3.VIII(b)
thatk max(03C1(m
+3), 5).
Otherwise we useProposition
3.VII to writewhere
u(y)
is a vectorpolynomial
inR4
anda(y)
is a(scalar) poly-
nomial.
Similarly,
is also a
symmetric
rank onepolynomial
matrix. IfG(y)
isidentically
zero then
03B3(y) ~ 0
and it follows fromProposition 3.VIII(a)
thatk max( p ( m
+1), 5).
Otherwise weagain
useProposition
3.VII to writewhere
v(y)
is a vectorpolynomial
inR4 and b(y)
is a(scalar) poly-
nomial. Observe from
(27)-(30)
thatCombining (27)-(30)
we obtainFurthermore
u(y)
andv(y)
must beorthogonal
sinceF(y)G(y) = 0.
Thus we find from
(33)
thatSince
F( y )
andG(y)
are fourth order in y, there are several cases to considerregarding
thedegrees of a, b,
u, v in(28), (30).
All but one ofthe cases,
however,
are immediate.Indeed,
ifu(y)
is constant(i.e.
independent
ofy )
then we use(34)
to conclude fromProposition
3.IVthat
k p ( m
+1,
m +4). Similarly
ifv(y)
is constantthen k p ( m, m + 3).
Ifa(y)
is constant then we may choose it to be one, and(33)
becomes
Thus
Since
(for k - 1 2) |y|2 is
irreducible and the coordinates ofu( y)
arerelatively prime (recall Proposition 3.VII)
it follows thatThis means that for
v(y)
to be linearin y
we must havek - 1 4. (Cf.
the
proof
ofProposition 3.VIII(a) -
inparticular
the remarkfollowing (12), (13).) Similarly
ifb(y)
is constant andu(y)
is linearthen, again,
wemust have
k - 1 4. Suppose
thatu(y), v(y)
are both linear in y. Thenusing
theirreducibility of lyl2@
it follows from(31), (32)
that one of thealternatives
holds. If the first or second alternative holds
then, as just
mentionedabove, k -
1 4. For the third alternative observe thatsince
u(y)
andv ( y )
areorthogonal. Thus,
under this alternative aswell,
k -
1
4. Theonly remaining
case to consider then is whereu(y), v(y)
are both
quadratic.
Accordingly,
we devote the remainder of thisproof
to the caseu(y), v(y) quadratic.
We choosea(y), b ( y )
to be one. Observe that from(31), (32)
and theorthogonality
ofu(y), v(y)
followsSince
(from (33))
and since the coordinates of
u(y)
are assumed to berelatively prime,
itfollows that one of the alternatives
holds,
wherew ~ R4
is a fixed unit vector.(That
it is a unit vectorfollows from
(35).)
SetIt then follows from
(31), (34), (36)
and theorthogonality
ofu(y), v(y)
that
Thus from
(2.8), (2.9)
we deduce thatBy rotating
if necessary let us assume thatThis
corresponds
tow ( y ) being
the fourth column ofC( y ),
and we writewhere the dimensions of
E(y)
are m X 3. Thenand
since, by (38), 03C9(y)
is aneigenvector
ofC( y ) Ct( y )
witheigenvalue
|y|2,
it follows from(37)
that theeigenvalues
ofE(y)Et(y)
are(03B3(y), 03B3(y), 03B3(y), 0,...,0). Thus
Observe that
since, by (2.9), C(y)Ct(y)
andB(y)
commute, it follows from(39), (40)
thatE(y)Et(y)
andB(y)
likewise commute.Wherever
03B3(y)
> 0 setThis matrix
D(y)
is skewsymmetric,
sinceB(y) is,
and sinceE(y)Et(y)
and
B(y)
commute it satisfiesThus the matrix
satisfies
Clearly
this isnonsingular
for03B3(y)
> 0. Thus if03B3(y)
> 0 forall y ~
0then
Â
constitutes an odd mapin (Sk-2, GLm+3)
which iseverywhere
skew
symmetric. According
toCorollary 4.11, then, k p ( m
+3).
Thus we must concentrate on the case where
is a nontrivial
subspace,
for the remainder of thisproof.
Sincerank(C(y))
3 it follows thatThen, using
thelinearity
of03C9(y)
andE(y)
in y, we obtain thatSince, by (37), w 1 foi
is anisometry
it follows thatFurthermore,
we know on account of(41)
thatfor y tE JV, range(E(y))
isprecisely
theeigenspace
ofE(y)Et(y) corresponding
to theeigenvalue y( y ).
Thus we conclude from(42)
thatThis shows that if %=1= 0 then
Y( 17)
can be written as the sum of three terms, each aproduct
of two linear forms in q. SinceY( 17)
ispositive
definite we must have
Since
we conclude the
proof by establishing
and
invoking (43).
Tosimplify
the rest of this discussion we assume witheut loss ofgenerality,
as in theproof
of Theorem III(see
the remarkpreceding (9)),
thatwhere p =
dim(X )
and each yl ispositive. Then, analogous
to(2.3),
the condition
(41)
isP
where
E(y) = 03A3 ylEl.
To establish(45)
suppose p 2 and that in factrange(E1) = range(E2).
Then there is a 3 X 3orthogonal matrix Q
suchthat
E2
=EIQ.
From(47), though,
it followsthat Q
must also be skewsymmetric,
which isimpossible.
Thusrange(E1) ~ range(E2) and,
sinceeach range is three-dimensional
(45)
follows.To establish
(46)
suppose p 3 and let ai =Ela,
w2 =E103B2
be any twolinearly independent
vectors. Choose ~2, Yl3 not both zero such thatThen it follows from
(47)
thatwhere M =
span( a, 03B2).
Sincerange(~2E2 + ~3E3)
is three-dimensional it isimpossible
that it should containE1,
and thus(46)
follows. DRegarding
the next result in line we conclude this Section with thefollowing
CONJECTURE VI: For m even
For a summary of the
preceding
results see Tables 2 and 3 in theAppendix.
Appendix: Figures
and tablesTABLE 1. p(m, n)
TABLE 2. Values of 03C1(m, n )
Figure 1. Duality condition. The horizontally stacked k X m matrices are dual to the vertically stacked m X n matrices.
d(q)lp
=e p q
m 1
Supplement Steps for filling
in Table 1(i)
Use(1.3)
to fill in the 15 entries(ii)
Use the fact thatp(m, n) n m03C1(m)
whenevermin (Proposition
m
3.I(d))
and the lower bound03C1(m, n)
n fromProposition 3.I(c)
toconclude that
for the 23 entries
Then use the
monotonicity
condition(Proposition 3.I(b))
to concludethat
(1)
holds for the 6 additional entries.(iii)
Use the fact that03C1(m, n) n - m + 1
whenevern
is odd(Proposition 3.I(f))
and themonotonicity
condition(Proposition 3.I(b))
to conclude that
for the 27 entries
Then use the
monotonicity
condition(Proposition 3.I(b))
to concludethat
for the 32 entries
(iv)
Use the fact thatp(m, n) n - m + 1
whenevern
is odd(Proposition 3.I(f))
and the lower bound inProposition 3.I(e) (with t
= 2and
( m, n )
=(4, 4)
or(8, 8)
or(12, 12))
to conclude that(2)
holds for the 7 entriesThen use the lower bound from the
symmetry
condition(Proposition 3.I(d))
to conclude that(1)
holds for the 6 additional entriesRepeat
thisstep
once more to add the 2 additional entriesto the list.
(The
calculation for03C1(11, 15) depends
onprior knowledge
that