LHF - National University of Singapore
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Transcript LHF - National University of Singapore
MATRIX COMPLETION PROBLEMS IN
MULTIDIMENSIONAL SYSTEMS
Wayne M. Lawton
Dept. of Mathematics, National University of Singapore
Block S14, Lower Kent Ridge Road, Singapore 119260
[email protected]
Zhiping Lin
School of Electrical and Electronic Engineering
Block S2, Nanyang Avenue
Nanyang Technological University, Singapore 639798
[email protected]
OUTLINE
1. Introduction
2. Continuous functions
3. Trigonometric polynomials
4. Stable rational functions
INTRODUCTION
R is a commutative ring with identity 1
m is unimodular if there exists
PR
m
QR
MR
R
such that
PQ 1
m m
is unimodular if
det M 1
is a Hermite ring if every unimodular row vector is
the first row of a unimodular matrix (completion)
INTRODUCTION
HERMITE RINGS INCLUDE
1. Polynomials over any field (Quillen-Suslin)
2. Laurent polynomials over any field (Swan)
3. Rings of formal power series over any field
(Lindel and Lutkebohment)
4. Complex Banach algebras with contractible
maximal ideal spaces (V. Ya Lin)
6. Principal ideal domains eg rational integers, stable
rational functions of one variable (Smith)
DEGREE OF MAP OF SPHERE
THE DEGREE
D(f) OF CONTINUOUS f : S S
n
is an integer that measures the direction and number
of times the function winds the sphere onto itself.
S {(cos , sin )}
2
S {(cos cos , cos sin , sin )}
D( k ) k
D(constant ) 0 D(identity ) 1
n 1
D(antipoda l) (1)
D(f h) D(f)D(h)
EXAMPLES
1
n
HOMOTOPY
f1 , f 2 : X Y are homotopic
(f1 f 2 ) if F : [0,1] X Y
DEFINITION
F(0,) f1 , F(1,) f 2
f1 , f 2 : S S
then f f D(f ) D(f )
1
2
1
2
COROLLARY x, f ( x) f (x) f f
1
2
1
2
Proof. Consider (1 t)f tf
1
2 || (1 t)f 1 tf 2 ||
HOPF’S THEOREM If
n
n
J. Dugundji, Topology, Allyn and Bacon, Boston, 1966.
CONTINUOUS FUNCTIONS
Define
n
R R(S )
is unimodular
For unimodular
Then
n
1
PR
x, P(x) 0
P
define
D(P) D( P || P ||)
n
1
Theorem 1. For n even, a unimodular P R
admits a matrix completion D(P) 0, hence
n
is
not
Hermite
since
the
identity
function
on
S
R
has degree 1 and thus cannot admit a matrix completion.
Proof
CONTINUOUS FUNCTIONS
Let Q be the second row of a matrix
completion
P
M of
n
1
PR
.Since det M 1
and Q linearly independent at each point, hence
D(P) D(Q). Multiply the second and third rows of
M by - 1 to obtain D(P) D(-Q) -D(Q) 0.
Hopf’s theorem implies there exists a homotopy
n
n
n
g : [0,1] S S , g(0,) P, g(1,) c S
, x)
0 t 0 t L 1 x, g(t , x) g(t
k
k-1
Construct M M L M1 where x, M k ( x ) SO(n 1) satisfies
g(t , x)M (x) g(t
, x) and yM (x) y if
k
k
k
k -1
yg(t , x) yg(t
, x) 0.M is continuous and completes P.
k
k -1
Choose
TRIGONOMETRIC POLYNOMIALS
Let
n
n
Prs ( R ), Trs ( R )
be the
Z
n
periodic symmetric
continuous real-valued functions, trigonometric polynomials.
Isomorphic to rings of functions on the space
identifying
xT
n
and
X n obtained by
- x.
[-1,1]
under the map x cos 2x
2
X2 homeomorphic to sphere S
2
under a map : T X
2
X1
homeomorphic to interval
that is 2-1 except at
{(0,0),(0,.5),(.5,0),(.5,.5)}
RESULT
Lemma
Prs ( R )
is a Hermite ring.
Proof This ring is isomorphic to the ring of real-valued
functions on the interval
Choose a unimodular
And approximate
m
F R([-1,1])
F/ | F |
by a continuously differentiable map
H : [ 1,1] S
m-1
And use parallel transporting to extend to a map
M : [1,1] SO(m)
WEIRSTRASS p-FUNCTION
Define
p : C C {}
by
p( z ) z [( z ) ]
2
2
2
\{0}
where
{m ni | m, n Z}
Lemma
p(z) p(w) z w or z w
Proof. z ( p(z), p(z)) maps the elliptic curve
curve in projective
C/ isomorphically onto the cubic
2
3
space defined by the equation p 4p - g 2 p - g 3 .
J. P. Serre, A Course in Arithmetic, Springer, New York,
1973, page 84.
WEIRSTRASS p-FUNCTION
~
2
2
Define p : R S
2
1
1
2
p
2
2
R C
C {} S
where
is stereographic projection
2
1
2 ( z ) w [2u , 2v, w 2]
w u v 1, z u iv
and (x , x ) x ix
1 1 2
1
2
~ is 2 periodic and defines
2
2
:T S
Z
WEIERSTRASS p-FUNCTION
LAURENT EXPANSION
p(z) z k2 (2k 1)G k z
2
where
Gk
2k1
2k
\{0}
is the Eisentein series of index k for the lattice
This provides an efficient computational algorithm.
RESULTS
Theorem 2.
2
Prs ( R )
is isomorphic to the ring
2
R(S )
And therefore is not a Hermite ring. Furthermore the ring
2
Trs ( R )
is not a Hermite ring.
: R(S ) Prs (R )
2
by (f ) f , f R(S ).
Results for p imply that is a surjective isomorphism.
Proof Define the map
2
2
The second statement follows by perturbing a row having
degree not equal to zero to obtain a unimodular row.
EXAMPLE
EXAMPLE OF A UNIMODULAR ROW IN
2 3
Trs ( R )
THAT DOES NOT ADMIT A MATRIX COMPLETION
F1 ( X1 , X 2 ) 2.84(G 2 - G1 )
F2 ( X1 , X 2 ) 2.51(G 3- G 4 )
F3 ( X1 , X 2 ) 10 3.55 (G 1 G 2 ) - 2.56(G 3 G 4 )
G1 cos X1 , G 2 cos X 2
G 3 cos ( X1 X 2 ), G 4 cos ( X1 X 2 )
Proof Compute
F 0.5 | F |
maps are never antipodal, hence
so these
D(F) D() D((identity)) 1
OPEN PROBLEMS
PROBLEM 1 If
F Trs ( R )
2 3
is unimodular and has degree zero does it admit a matrix extension ?
PROBLEM 2
Is the ring
2
Ts ( R )
of symmetric trigonometric polynomials a Hermite ring ?
PROBLEM 3
Is the ring
2
Tr ( R )
of real-valued trigonometric polynomials a Hermite ring ?