Convexity of Generalized Numerical Ranges
Associated with SU(n) and SO(n)
by
Dawit Gezahegn Tadesse
A thesis submitted to the Graduate Faculty of
Auburn University
in partial fulflllment of the
requirements for the Degree of
Master of Science
Auburn, Alabama
August 9, 2010
Keywords: Convexity, generalized numerical range, orthogonal group, unitary group
Copyright 2010 by Dawit Gezahegn Tadesse
Approved by:
Tin-Yau Tam, Chair, Professor of Mathematics
Geraldo S. De Souza, Professor of Mathematics
Huajun Huang, Assistant Professor of Mathematics
Abstract
We give a new proof of a result of Tam on the convexity of the generalized numerical
range associated with the classical Lie groups SO(n). We also provide a connection between
the result and the convexity of the classical numerical range.
ii
Acknowledgments
The author would like to thank Dr. Tin-Yau Tam for his excellent teaching and guidance
throughout the course of this research, and his parents for their continual support, love, and
encouragement.
iii
Table of Contents
Abstract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ii
Acknowledgments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iii
1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
2 Three simple proofs of the convexity of W(A) . . . . . . . . . . . . . . . . . . . 4
3 Basic Lie Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
4 The connection between W(A) and S(A) for small n . . . . . . . . . . . . . . . 18
5 Another proof of the convexity of S(A) . . . . . . . . . . . . . . . . . . . . . . . 23
Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
iv
Chapter 1
Introduction
A subset ?  C is said to be convex if (1 ? ?)x + ?y 2 ? whenever x;y 2 ? and
0 ? ? ? 1:
Let H be a Hilbert space overCand let B(H) denotes the algebra of all bounded linear
operators on H. The classical numerical range of T 2 B(H) is
W(T) := f(Tx;x) : x 2 H;(x;x) = 1g:
Toeplitz-Hausdorfi theorem [16, 6] asserts that W(A) is a convex set. See [11] for a simple
proof.
The flnite dimensional case may be phrased as the following. IfCn?n denotes the set of
n?n complex matrices, then
W(A) := fx?Ax : x 2Cn;kxk = 1g C
is a compact convex set, where A 2Cn?n and kxk2 = x?x. It is the image of the (compact)
unit sphere Sn?1  Cn under the nonlinear map x 7! x?Ax. It is truly remarkable since the
unit sphere is very \hallow" but its image under the quadratic map is convex.
When n = 2, W(A) is an elliptical disk (possibly degenerate) [8], known as the elliptical
range theorem.
Theorem 1.1. [8] Let A 2 C2?2 with eigenvalues ?1;?2. Then W(A) is an elliptical disk
with ?1;?2 as foci and minor axis of length ptr(A?A)?j?1j2 ?j?2j2.
1
Indeed the convexity of the numerical range of T can be reduced to the 2-dimensional
case, i.e., even the most general version of the theorem is equivalent to a statement about 2-
dimensional spaces. That is the reason why almost every approach ends up with a quadratic
computation that has no merit except correctness.
Let ?1 = x?1Ax1 and ?2 = x?2Ax2 be two points in W(A) where x1;x2 2 H are unit
vectors. We may assume that x1;x2 are linearly independent, otherwise, ?1 = ?2. Consider
the compression ^A : C ! C where C := spanfx1;x2g deflned by ^Ax = PAx, x 2 C,
where P : H ! C is the orthogonal projection onto C. Clearly ?1;?2 2 W( ^A) and thus the
line segment [?1;?2] ? W( ^A) if the theorem holds for the 2-dimensional case. Notice that
W( ^A) ? W(A) since for each unit vector x 2 C,
x? ^Ax = x?PAx = x?PAPx = (Px)?A(Px) = x?Ax
as P? = P.
The classical numerical range of A 2Cn?n may be written as
W(A) = f(U?AU)11 : U 2 U(n)g;
since each unit vector x can be extended to a U 2 U(n) of the form (x U1) where U1 2
Cn?(n?1). Here U(n) denotes the group of n?n unitary matrices. The Lie algebra u(n) of
U(n) is the set of n?n skew Hermitian matrices and gln(C) =Cn?n is the complexiflcation
of u(n).
Similar to the classical numerical range, a difierent range emerges if one replaces the
unitary group U(n) by the special orthogonal group SO(n). The Lie algebra of SO(n) is the
set of n ? n real skew symmetric matrices, whose complexiflcation is the algebra of n ? n
complex skew symmetric matrices. When n ? 3, the numerical range of a skew symmetric
2
A 2Cn?n associated with SO(n) is deflned to be the set
S(A) := f(OTAO)1;2 : O 2 SO(n)g C
and is known to be a compact convex set according to a result of Tam [14] (see [10, 15] for
related results). Indeed the result in [14] is more general and is in the context of compact
connected Lie group. The method of Tam is the usage of a lemma of Atiyah on symplectic
manifold since the adjoint orbit of a Lie algebra element has a natural symplectic structure.
From now on we denote by son(C) the algebra of n?n complex skew symmetric matrices.
Theorem 1.2. (Tam [14]) If A 2 son(C), where n ? 2, then S(A) is a compact convex set.
The main goal of this thesis is to provide an elementary proof of Theorem 1.2 and point
out some relation between S(A) and W(A) and the k-numerical range when n is small.
When n = 2, S(A) is the set ffig, where A =
0
B@ 0 fi
?fi 0
1
CA. From now on we assume
that n ? 3 to avoid the trivial case. Rewrite
S(A) = fxT1 Ax2 : x1;x2 are the two columns of some O 2 SO(n)g:
In particular,
S(A) = f(OTAO)1;2 : O 2 O(n)g
= fxT1 Ax2 : x1;x2 are the two columns of some O 2 O(n)g
= fxT1 Ax2 : x1;x2 are orthonormal in Rng:
Clearly xTAy = ?yTAx for all x;y 2Rn, because A is skew symmetric. So ? 2 S(A) if and
only if ?? 2 S(A), i.e., S(A) is symmetric about the origin. We remark in general that such
symmetry property is not present in W(A), e.g., Theorem 1.1.
3
Chapter 2
Three simple proofs of the convexity of W(A)
We flrst provide three proofs of the convexity of classical numerical range, namely,
Raghavendran?s proof [11], Gustafson?s proof [3] and the proof of Davis [1].
1. Raghavendran?s proof:
Proof. We need to consider only the case where W(A) contains at least two points. Let xk
(k = 1;2) be any two elements of H with kxkk = 1 such that x?kAxk = wk are two distinct
points of W(A): As x1 + zx2 = 0 for z 2C will imply that ?zz = 1 and then that w1 = w2,
we see that kx1 +zx2k6= 0 for all z 2C. So the theorem will be proved if we show that, for
any given real number t with 0 < t < 1, there exists at least z = x+iy 2C (with x;y real)
which satisfles the equation
(x1 +zx2)?A(x1 +zx2) = (tw1 +(1?t)w2)kx1 +zx2k2:
The equation may be rewritten in the form
pjzj2 +qz +r?z +s = 0; (2.1)
where p = t(w2 ?w1), s = (1?t)(w2 ?w1) and q;r 2C. Dividing this equation by p; and
then separating the real and imaginary parts, we get the two equations
x2 +y2 +ax+by ?(1?tt ) = 0; (2.2)
cx+dy = 0; (2.3)
4
where a;b;c;d are some well-deflned real numbers such that this pair of equations is equiv-
alent to the single equation (2.1).
Equation (2.2) represents a real circle with a positive radius having the origin in its
interior (because the constant term in this equation is negative); and when c;d are not both
zero, the straight line represented by the equation (2.3) meets this circle in two real and
distinct points. We can, therefore, always flnd (at least) two distinct complex numbers zk
such that z = zk satisfy the equation (2.1). This completes the proof.
We remark that the above proof is actually revealing that the general statement is
reduced to the n = 2 case.
2. Gustafson?s proof:
Proof. Since W(?A +  ) = ?W(A) +  ; for any scalars ?; 2C, it su?ces to consider the
situation
(Ax1;x1) = 0; (Ax2;x2) = 1; kxik = 1; xi 2 H; i = 1;2:
Let x = fix1 +flx2;fi and fl real, and require
kxk2 ? fi2 +fl2 +2fiflRe(x1;x2) = 1; (2.4)
and desire (for each 0 < ? < 1)
(Ax;x) := fl2 +fiflf(Ax1;x2)+(Ax2;x1)g = ?: (2.5)
Let B := (Ax1;x2)+(Ax2;x1). If B is real, then the system (2.4) and (2.5) describe an
ellipse (intercepts ?1;?1) and a hyperbola (intercepts??1=2) clearly possesses four solutions
since kRe(x1;x2)k < 1 by Schwartz?s inequality. But B can always be guaranteed real by
using an appropriate (scalar multiple of) x1, i.e., explicitly, use ~x1 = ?x1; where ? = a + ib
satisfles
j?j2 ? a2 +b2 = 1 (2.6)
5
and
ImB(~x1) ? aImB(x1)+bRef(Ax1;x2)?(Ax2;x1)g = 0; (2.7)
a system clearly possessing (two) solutions.
3. Davis?s idea:
Proof. Accordingly, let us assume without loss of generality that dimH = 2. Notice that
x?Ax = tr(Axx?) which the key to Davis? proof. Consider the mapping ' which takes the
arbitrary hermitian operator X on H to
'(X) = tr(AX):
It is plainly real-linear. Its domain is a real 4-dimensional space, i.e., the space M of matrices
X =
0
B@a b
c d
1
CA (a = ?a;b = ?c;d = ?d:)
The range of ' is a real 2-dimensional space, namely, the complex numbers. The conclusion
which is to be proved is that ' takes the set of one-dimensional orthoprojectors xx? onto a
convex set.
In the matrix representation of M, these orthoprojectors xx? may be parametrized as
follows. It is enough to consider
x =
0
B@ cos 
ei? sin 
1
CA;  ;? 2R
because any other unit vector is a scalar multiple of one of these. These matrices comprise a
2-sphere centered at
0
B@ 12 0
0 12
1
CA and lying in a 3- at in M (the set of H having trace 1). But
the image of a 2-sphere, under a linear map with range in real 2-space, is either an ellipse
with interior, or a segment, or a point { in any case it is convex.
6
Chapter 3
Basic Lie Theory
A (real) Lie group is a group which is also a flnite-dimensional real smooth manifold,
and in which the group operations of multiplication and inversion are smooth maps. For
example SU(n), SO(n), Sp(n) are compact Lie groups.
A real vector space L with an operation L?L ! L; denoted (x;y) 7! [x;y] and called
the bracket or commutator of x and y; is called a Lie algebra if
(L1) The bracket operation is bilinear.
(L2) [x;x] = 0 for all x 2 L:
(L3) [x;[y;z]]+[y;[z;x]]+[z;[x;y]] = 0 for all x;y;z 2 L:
The condition (L3) is called the Jacobi identity.
There is a Lie algebra g associated with G. The following is the description of g.
1. LetM be a real smooth manifold and denote byC1(M) the ring of all smooth functions
f : M ! R. A map v : C1(M) ! R is called a tangent vector at p 2 M if for all
f;g 2 C1(M), p 2 M
(i) v(f +g) = v(f)+v(g) and
(ii) v(fg) = v(f)g(p)+f(p)v(g).
Each tangent vector can be thought as a derivative. The set Tp(M) of all tangent
vectors at p is a flnite dimensional vector space.
2. Denote by T(M) = [p2MTp(M) the tangent bundle of M. A vector fleld on any
smooth manifold M, X : M ! T(M), is a smooth map such that X(p) 2 Tp(M). The
7
extension of X is the map ~X : C1(M) ! C1(M) deflned by ~X(f)(0) = X(p)(f). It
is known that ~X = ~Y if and only if X = Y. Now the bracket [X;Y] of two vector flelds
is deflned by
[X;Y](p)(f) = X(p)(~Y(f))?Y(p)( ~X(f))
3. The left translations Lg : G ! G, g 2 G is given by Lg(h) = gh, h 2 G. The left
invariant vector flelds (vector flelds satisfying dLg(X) = X?Lg for every g 2 G, where
dLg denotes the difierential of Lg) on a Lie group is a Lie algebra under the Lie bracket
of vector flelds, i.e., the bracket of two left invariant vector fleld is also left invariant.
4. The map X ! X(1) deflnes a one to one correspondence between the left invariant
vector flelds and the tangent space Te(G) at the identity e and therefore makes the
tangent space at the identity into a Lie algebra, called the Lie algebra of G, usually
denoted by g.
For example:
1. The special unitary group
SU(n) = fg 2 GLn(C) : g?g = 1;detg = 1g
has Lie algebra
su(n) = fA 2Cn?n : A? = ?A; trA = 0g
which is a real subspace of Cn?n.
2. The complex special linear group
SLn(C) = fg 2 GLn(C) : detg = 1g
has Lie algebra
sln(C) = fA 2Cn?n : trA = 0g
8
.
3. The complex orthogonal group
SOn(C) = fg 2 GLn(C) : gTg = 1;detg = 1g
has Lie algebra
son(C) = fA 2Cn?n : AT = ?Ag:
We now recall some basic notions about adjoint representation Ad : G ! Autg. Let G be
a Lie group and let g be its Lie algebra (which we identify with TeG, the tangent space to
the identity element in G). Deflne a map
? : G ! AutG
by the equation ?(g) = ?g for all g 2 G, where AutG is the automorphism group of G and
the automorphism ?g is deflned by
?g(h) = ghg?1
for all h 2 G. It follows that the derivative of ?g at the identity is an automorphism of the
Lie algebra g. We denote this map by Adg:
Adg: g ! g:
The map
Ad : G ! Autg
9
which sends g to Adg is called the adjoint representation ofG. This is indeed a representation
of G since Autg is a Lie subgroup of GL(g) and the above adjoint map is a Lie group
homomorphism. The dimension of the adjoint representation is the same as the dimension
of the group G.
One may always pass from a representation of a Lie group G to a representation of its
Lie algebra by taking the derivative at the identity. Taking the derivative of the adjoint map
Ad : G ! Autg
gives the adjoint representation of the Lie algebra g:
ad : g ! Autg:
The adjoint representation of a Lie algebra is related in a fundamental way to the structure
of that algebra. In particular, one can show that
adX(Y) = [X;Y]
for all X;Y 2 g.
The adjoint group Intg is the analytic group of adg, which is contained in GL(g). It is
known that if G is connected, then AdG = Intg [7, p.129].
A Lie algebra homomorphism ` : g ! h is a vector space isomorphism that respects
bracket: ?[X;Y] = [?X;?Y]:
Lemma 3.1. The map sl2(C) ! so3(C) given by
0
B@a b
c ?a
1
CA7!
0
BB
BB
@
0 ?2ia i(b+c)
2ia 0 c?b
?i(b+c) b?c 0
1
CC
CC
A
10
is a Lie algebra isomorphism.
Proof. One can directly verify that it is a Lie algebra isomorphism. Here we establish the
result via a double covering ? : SU(2) ! SO(3).
As we know SU(2) is connected. Thus the image of SU(2) by ? is connected as well,
hence contained in the connected component SO(3) of O(3): Let d?e : su(2) ! so(3) be the
difierential of ? at the identity e. Since ? = Ad, and d?e = ad, the kernel of d?e consists
of the X 2 su(2) satisfying XY ?YX = 0 for all skew Hermitian Y 2 C2?2 of zero trace,
hence for all skew Hermitian Y 2 C2?2, hence for all Y 2 C2?2 (as one sees by writing
Y = iY1 + Y2 with Y1 and Y2 skew Hermitian). Thus X is a scalar and therefore equals 0,
because trX = 0: Since the kernel of d?e is f0g; d?e is an isomorphism. It follows that ? is
a covering. Its kernel consists of all a 2 SU(2) satisfying aYa?1 = Y for all skew Hermitian
Y; hence are scalar and thus equal to ?1:
To compute the Lie algebra isomorphism explicitly let
X =
0
B@ ix3 ?x1 +ix2
x1 +ix2 ?ix3
1
CA; Y =
0
B@ iy3 ?y1 +iy2
y1 +iy2 ?iy3
1
CA2 su(2):
Then
ad(X)Y
= XY ?YX
=
0
B@ 2i(x2y1 ?x1y2) ?2(x3y2 ?x2y3)+2i(?x3y1 +x1y3)
2(x3y2 ?x2y3)+2i(?x3y1 +x1y3) ?2i(x2y1 ?x1y2)
1
CA
$
0
BB
BB
@
2(x3y2 ?x2y3)
2(?x3y1 +x1y3)
2(x2y1 ?x1y2)
1
CC
CC
A
=
0
BB
BB
@
0 2x3 ?2x2
?2x3 0 2x1
2x2 ?2x1 0
1
CC
CC
A
0
BB
BB
@
y1
y2
y3
1
CC
CC
A
11
which gives us the desired isomorphism
0
B@ ix3 ?x1 +ix2
x1 +ix2 ?ix3
1
CA7!
0
BB
BB
@
0 2x3 ?2x2
?2x3 0 2x1
2x2 ?2x1 0
1
CC
CC
A
:
Then extend it to sl2(C) ! so3(C).
Lemma 3.2. The map sl2(C)'sl2(C) ! so4(C)
0
B@a b
c ?a
1
CA+
0
B@d e
f ?d
1
CA
7!
0
BB
BB
BB
B@
0 i(a?d) 12(c?b?f +e) 12i(b+c?e?f)
?i(a?d) 0 12i(b+c+e+f) ?12 (c?b+f ?e)
?1
2 (c?b?f +e)
?1
2 i(b+c+e+f) 0 i(a+d)
?1
2 i(b+c?e?f)
1
2(c?b+f ?e) ?i(a+d) 0
1
CC
CC
CC
CA
is a Lie algebra isomorphism.
Proof. One can directly verify that it is a Lie algebra isomorphism. However it is good to
see how one can get it from the double covering ? : SU(2)?SU(2) ! SO(4) [12, p.42]. The
difierential at the identity d?e : su2 ' su2 ! so(4) is the desired Lie algebra isomorphism.
Let ? : SU(2)?SU(2) ! SO(4) be the map acting on the quaternions
H= fQ =
0
B@?1 +i?2 ??3 ?i?4
?3 ?i?4 ?1 ?i?2
1
CA : ?
1;?2;?3;?4 2Rg?!f
?!Q =
0
BB
BB
BB
B@
?1
?2
?3
?4
1
CC
CC
CC
CA
: ?1;?2;?3;?4 2Rg
as follows:
12
?((U;V))?!Q = ?????!UQV ?1 where Q 2H and (U;V) 2 SU(2)?SU(2). The corresponding
Lie map is
d?e : su2(C)'su2(C) ! so(4) deflned by d?e((X;Y))?!Q = ????????!?XQ+QY:
Using the same idea we used above it can be verifled that d?e is a Lie algebra isomor-
phism. To compute this Lie algebra isomorphism explicitly. Let
X =
0
B@ ix3 ?x1 +ix2
x1 +ix2 ?ix3
1
CA; Y =
0
B@ iy3 ?y1 +iy2
y1 +iy2 ?iy3
1
CA2 su(2)
and
Q =
0
B@?1 +i?2 ??3 ?i?4
?3 ?i?4 ?1 ?i?2
1
CA2H:
Direct computation yields
d?e((X;Y))?!Q = ????????!?XQ+QY =
0
BB
BB
BB
B@
0 x3 ?y3 x1 ?y1 ?x2 +y2
?x3 +y3 0 ?x2 ?y2 ?x1 ?y1
?x1 +y1 x2 +y2 0 x3 +y3
x2 ?y2 x1 +y1 ?x3 ?y3 0
1
CC
CC
CC
CA
0
BB
BB
BB
B@
?1
?2
?3
?4
1
CC
CC
CC
CA
which gives us the Lie algebra isomorphism
0
B@ ix3 ?x1 +ix2
x1 +ix2 ?ix3
1
CA+
0
B@ iy3 ?y1 +iy2
y1 +iy2 ?iy3
1
CA
7!
0
BB
BB
BB
B@
0 x3 ?y3 x1 ?y1 ?x2 +y2
?x3 +y3 0 ?x2 ?y2 ?x1 ?y1
?x1 +y1 x2 +y2 0 x3 +y3
x2 ?y2 x1 +y1 ?x3 ?y3 0
1
CC
CC
CC
CA
Then extend it to sl2(C)'sl2(C) ! so4(C) in order to get the desired isomorphism.
13
Lemma 3.3. ([9, p.162]) The map sl4(C) ! so6(C)
0
BB
BB
BB
B@
a11 a12 a13 a14
a21 a22 a23 a24
a31 a32 a33 a34
a41 a42 a43 ?a11 ?a22 ?a33
1
CC
CC
CC
CA
7!
0
BB
BB
B@
0 (a23+a14?a41?a32)2 (a24?a13+a31?a42)2 i(a11+a22) i(a23?a14?a41+a32)2 i(a24+a13+a31+a42)2
?(a23+a14?a41?a32)2 0 a34+a12?a21?a432 i(a23+a14+a41+a32)2 i(a11+a33) i(a34?a12?a21+a43)2
?(a24?a13+a31?a42)2 ?(a34+a12?a21?a43)2 0 i(a24?a13?a31+a42)2 i(a34+a12+a21+a43)2 ?i(a22+a33)
?i(a11+a22) ?i(a23+a14+a41+a32)2 ?i(a24?a13?a31+a42)2 0 (a23?a14+a41?a32)2 (a24+a13?a31?a42)2
?i(a23?a14?a41+a32)2 ?i(a11+a33) ?i(a34+a12+a21+a43)2 ?(a23?a14+a41?a32)2 0 (a34?a12+a21?a43)2
?i(a24+a13+a31+a42)2 ?i(a34?a12?a21+a43)2 i(a22+a33) ?(a24+a13?a31?a42)2 ?(a34?a12+a21?a43)2 0
1
CC
CC
CA
is a Lie algebra isomorphism.
Proof. Let I3;3 be the 6-by-6 diagonal matrix given by I3;3 = diag(1;?1;1;?1;1;?1) and
deflne g = fX 2 gl6(C) : XTI3;3 + I3;3X = 0g: Let S = diag(i;i;i;1;1;1): For X 2 g; let
Y = SXS?1: One easily sees that the map X 7! Y is an isomorphism of g onto so6(C): Any
member of sl4(C) acts on the 6-dimensional complex vector space of alternating tensors of
rank 2 by
M(ei ^ej) = Mei ^ej +ei ^Mej;
where feig4i=1 is the standard basis of C4:
Using the following ordered basis:
e1^e2+e3^e4;e1^e3?e2^e4;e1^e4+e2^e3;e1^e2?e3^e4;e1^e3+e2^e4;e1^e4?e2^e3
of the exterior space ^2C4:
14
Let
A =
0
BB
BB
BB
B@
a11 a12 a13 a14
a21 a22 a23 a24
a31 a32 a33 a34
a41 a42 a43 a44
1
CC
CC
CC
CA
2 sl4(C)
such that a44 = ?(a11 + a22 + a33): Consider the map A(ei ^ej) = Aei ^ej + ei ^Aej. We
now compute the matrix representation of A with respect the above basis.
A(e1 ^e2 +e3 ^e4)
= A(e1 ^e2)+A(e3 ^e4)
= 0?(e1 ^e2 +e3 ^e4)+ 12(a32 ?a14 ?a23 +a41)(e1 ^e3 ?e2 ^e4)
+12(a42 +a13 ?a31 ?a24)(e1 ^e4 +e2 ^e3)+(a11 +a22)(e1 ^e2 ?e3 ^e4)
+12(a32 ?a14 +a23 ?a14)(e1 ^e3 +e2 ^e4)+ 12(a42 +a13 +a31 +a24)(e1 ^e4 ?e2 ^e3)
A(e1 ^e3 ?e2 ^e4)
= A(e1 ^e3)?A(e2 ^e4)
= 12(a23 +a14 ?a41 ?a32)(e1 ^e2 +e3 ^e4)+0?(e1 ^e3 ?e2 ^e4)
+12(a43 ?a12 +a21 ?a34)(e1 ^e4 +e2 ^e3)+ 12(a23 +a14 +a41 +a32)(e1 ^e2 ?e3 ^e4)
+(a11 +a33)(e1 ^e3 +e2 ^e4)+ 12(a43 ?a12 ?a21 +a34)(e1 ^e4 ?e2 ^e3)
15
A(e1 ^e4 +e2 ^e3)
= A(e1 ^e4)+A(e2 ^e3)
= 12(a24 ?a13 +a31 ?a42)(e1 ^e2 +e3 ^e4)+ 12(a34 +a12 ?a21 ?a43)e1 ^e3 ?e2 ^e4)
+0?(e1 ^e4 +e2 ^e3)+ 12(a24 ?a13 ?a31 +a42)(a23 +a14 +a41 +a32)(e1 ^e2 ?e3 ^e4)
+12(a34 +a12 +a21 +a43)(e1 ^e3 +e2 ^e4)?(a22 +a33)(e1 ^e4 ?e2 ^e3)
A(e2 ^e3 ?e3 ^e4)
= A(e1 ^e3)?A(e2 ^e4)
= (a11 +a22)(e1 ^e2 +e3 ^e4)+ 12(a32 +a14 +a23 +a41)(e1 ^e3 ?e2 ^e4)
+12(a42 ?a13 ?a31 +a24)(e1 ^e4 +e2 ^e3)+0?(e1 ^e2 ?e3 ^e4)
+12(a32 +a14 ?a23 ?a14))(e1 ^e3 +e2 ^e4)+ 12(a42 ?a13 +a31 ?a24)(e1 ^e4 ?e2 ^e3)
A(e1 ^e3 +e2 ^e4)
= A(e1 ^e3)+A(e2 ^e4)
= 12(a23 ?a14 ?a41 +a32)(e1 ^e2 +e3 ^e4)+(a11 +a33)(e1 ^e3 ?e2 ^e4)
+12(a43 +a12 +a21 +a34)(e1 ^e4 +e2 ^e3)+ 12(a23 ?a14 +a41 ?a32)(e1 ^e2 ?e3 ^e4)
+0?(e1 ^e3 +e2 ^e4)+ 12(a43 +a12 +a21 ?a34)(e1 ^e4 ?e2 ^e3)
16
A(e1 ^e4 ?e2 ^e3)
= A(e1 ^e4)?A(e2 ^e3)
= 12(a24 +a13 +a31 +a42)(e1 ^e2 +e3 ^e4)+ 12(a34 ?a12 ?a21 +a43)(e1 ^e3 ?e2 ^e4)
?(a22 +a33)(e1 ^e4 +e2 ^e3)+ 12(a24 +a13 ?a31 ?a42)(e1 ^e2 ?e3 ^e4)
+12(a34 ?a12 +a21 ?a43)(e1 ^e3 +e2 ^e4)+0?(e1 ^e4 ?e2 ^e3)
So we get the following 6?6 matrix
0
BB
BB
@
0 a23+a14?a41?a322 a24?a13+a31?a422 a11+a22 a23?a14?a41+a322 a24+a13+a31+a422
a23+a14?a41?a32
2 0
a34+a12?a21?a43
2
a23+a14+a41+a32
2 a11+a33
a34?a12?a21+a43
2
a24?a13+a31?a42
2
a34+a12?a21?a43
2 0
a24?a13?a31+a42
2
a34+a12+a21+a43
2 ?(a22+a33)
a11+a22 a23+a14+a41+a322 a24?a13?a31+a422 0 a23?a14+a41?a322 a24+a13?a31?a422
a23?a14?a41+a32
2 a11+a33
a34+a12+a21+a43
2
a23?a14+a41?a32
2 0
a34?a12+a21?a43
2
a24+a13+a31+a42
2
a34?a12?a21+a43
2 ?(a22+a33)
a24+a13?a31?a42
2
a34?a12+a21?a43
2 0
1
CC
CC
A
in g which gives us the desired Lie algebra isomorphism after conjugation A 7! SAS?1 where
S = diag(i;i;i;1;1;1):
Theorem 3.4. ([17, p.101])
1. Let G and H be Lie groups with Lie algebras g and h and with G simply connected.
Let ? : g ! h be a homomorphism. Then there is a unique homomorphism ? : G ! H
such that d?e = ?.
2. For each Lie algebra g, there is a simply connected Lie group G with Lie algebra g.
17
Chapter 4
The connection between W(A) and S(A) for small n
We want to study S(A) and its relation to the classical numerical range when the
dimension is small, i.e., 2 ? n ? 6. The k-numerical range of B 2Cn?n, 1 ? k ? n, is
Wk(B) := fx?1Bx1 +???+x?kBxk : x1;??? ;xk 2Cn are orthonormalg:
Halmos [4] asked if Wk(B) is convex and Berger [5, p.110-111] provided an a?rmative answer
for any B 2Cn?n.
Theorem 4.1. (Berger) The k-numerical range of B 2Cn?n is convex.
Lemma 4.2. Let ?1 and ?2 be convex subsets of C. The sum of of ?1 and ?2, denoted by
?1 +?2 := fa+b : a 2 ?1;b 2 ?2g;
is a convex subset of C:
Proof. Let a1 + b1;a2 + b2 2 ?1 + ?2. Then (1 ? ?)(a1 + b1) + ?(a2 + b2) = ((1 ? ?)a1 +
?a2)+((1??)b1 +?b2) 2 ?1 +?2
Lemma 4.3. Let ? be a convex subset of C; and fi be a scalar (either real number or
complex number), then fi? := ffia : a 2 ?g is a convex subset of C:
Proof. Let a1;a2 2 ?. Then fia1;fia2 2 fi?, and (1??)(fia1)+?(fia2) = fi((1??)a1+?a2)) 2
fi?:
18
Theorem 4.4. 1. If A 2 so3(C), then S(A) is equal to W(B) for some B 2 sl2(C) and
thus is an elliptical disk (possibly degenerate) centered at the origin.
2. If A 2 so4(C), then S(A) is the sum of W(B) and W(C), where B;C 2 sl2(C). Thus
S(A) is the sum of two elliptical disks (possibly degenerate) centered at the origin.
3. If A 2 so5(C), then S(A) = iW2(B) for some B 2 sp2(C) ? C4?4, i.e., B =0
B@B1 B2
B3 ?BT1
1
CA where B
1;B2;B3 2C2?2 and B2;B3 are symmetric.
4. If A 2 so6(C), then S(A) = iW2(B) for some B 2 sl4(C).
Proof. Let K be a connected Lie group with Lie algebra k = so(n). Given A 2 g := k'ik =
son(C), consider the orbit of A under the adjoint action of SO(n)
AdK ?A := fAd(k)A : k 2 Kg
So
S(A) = ftrCY : Y 2 AdK ?Ag
where C = 12
0
B@0 ?1
1 0
1
CA'0
n?2. The orbit AdK?A depends on Ad K which is the analytic
subgroup of the adjoint group Int k ? Autk corresponding to adk [7, p.126, p.129]. Thus
AdK ?A is independent of the choice of K. In particular we can pick the simply connected
fSO(n).
(1) By Lemma 3.1 the following is a Lie algebra isomorphism sl2(C) ! so3(C)
0
B@a b
c ?a
1
CA7!
0
BB
BB
@
0 ?2ia i(b+c)
2ia 0 c?b
?i(b+c) b?c 0
1
CC
CC
A
19
and thus its restriction ? : su(2) ! so(3) is a Lie algebra isomorphism. By Theorem 3.4
there is a Lie group isomorphism ? : fSU(2) ! SO(3) so that d?e = ? which naturally
extends to sl2(C) ! so3(C). (indeed there is a double covering SU(2) ! SO(3)). So we have
the relation
d?efUAU?1 : U 2 SU(2)g
= d?efUAU?1 : U 2 fSU(2)g
= f?(U)(d?e(A))?(U)?1 : U 2 fSU(2)g
= fO(d?e(A))O?1 : O 2 SO(3)g:
where the second equality is due to the fact that
d?e(Ad(g)A) = Ad(?(g))d?e(A); A 2 sl2(C)
and that the adjoint action is conjugation for matrix group. The quadratic map x 7!
x?Ax, x 2 S1, amounts to U?1AU 7! (U?1AU)11, U 2 SU(2) and thus corresponds to
?12i(??1(U)d?e(A)?(U))12, i.e., (x;y) 7! xTd?e(A)y where x;y 2R3 are orthonormal. Thus
W
0
B@a b
c ?a
1
CA = ? 1
2iS
0
BB
BB
@
0 ?2ia i(b+c)
2ia 0 c?b
?i(b+c) b?c 0
1
CC
CC
A
:
20
(2) By Lemma 3.2 the following is a Lie algebra isomorphism ? : sl2(C) ' sl2(C) !
so4(C):
0
B@a b
c ?a
1
CA+
0
B@d e
f ?d
1
CA
7!
0
BB
BB
BB
B@
0 i(a?d) 12(c?b?f +e) 12i(b+c?e?f)
?i(a?d) 0 12i(b+c+e+f) ?12 (c?b+f ?e)
?1
2 (c?b?f +e)
?1
2 i(b+c+e+f) 0 i(a+d)
?1
2 i(b+c?e?f)
1
2(c?b+f ?e) ?i(a+d) 0
1
CC
CC
CC
CA
and thus its restriction ? : su(2)'su(2) ! so(4) is a Lie algebra isomorphism. By Theorem
3.4 there is a Lie group isomorphism ? : fSU(2)? fSU(2) ! SO(4) so that d?e = ? (indeed
there is a double covering map SU(2)?SU(2) ! SO(4) [12, p.42]). So we have the relation
d?efUAU?1 : U 2 SU(2)?SU(2)g
= d?efUAU?1 : U 2 fSU(2)? fSU(2)g
= f?(U)(d?e(A))?(U)?1 : U 2 fSU(2)? fSU(2)g
= fO(d?e(A))O?1 : O 2 SO(4)g:
where the second equality is due to the fact that
d?e(Ad(g)A) = Ad(?(g))d?e(A); A 2 sl2(C)'sl2(C)
andthattheadjointactionisconjugationformatrixgroup. ThemapU?1AU 7! (U?1AU)11+
(U?1AU)44, U 2 SU(2) ? SU(2) corresponds to i(??1(U)(d?e(A))?(U))12, i.e., (x;y) 7!
21
xTd?e(A)y where x;y 2R4 are orthonormal. So
W
0
B@a b
c ?a
1
CA+W
0
B@d e
f ?d
1
CA
= 1iS
0
BB
BB
BB
B@
0 i(a?d) 12(c?b?f +e) 12i(b+c?e?f)
?i(a?d) 0 12i(b+c+e+f) ?12 (c?b+f ?e)
?1
2 (c?b?f +e)
?1
2 i(b+c+e+f) 0 i(a+d)
?1
2 i(b+c?e?f)
1
2(c?b+f ?e) ?i(a+d) 0
1
CC
CC
CC
CA
:
Notice that W
0
B@a b
c ?a
1
CA+W
0
B@d e
f ?d
1
CA is simply the sum of two elliptical disks centered
at the origin. By Lemma 4.2 it is convex.
(4) By Lemma 3.3
0
BB
BB
BB
B@
a11 a12 a13 a14
a21 a22 a23 a24
a31 a32 a33 a34
a41 a42 a43 ?a11 ?a22 ?a33
1
CC
CC
CC
CA
7!
0
BB
BB
B@
0 (a23+a14?a41?a32)2 (a24?a13+a31?a42)2 i(a11+a22) i(a23?a14?a41+a32)2 i(a24+a13+a31+a42)2
?(a23+a14?a41?a32)2 0 a34+a12?a21?a432 i(a23+a14+a41+a32)2 i(a11+a33) i(a34?a12?a21+a43)2
?(a24?a13+a31?a42)2 ?(a34+a12?a21?a43)2 0 i(a24?a13?a31+a42)2 i(a34+a12+a21+a43)2 ?i(a22+a33)
?i(a11+a22) ?i(a23+a14+a41+a32)2 ?i(a24?a13?a31+a42)2 0 (a23?a14+a41?a32)2 (a24+a13?a31?a42)2
?i(a23?a14?a41+a32)2 ?i(a11+a33) ?i(a34+a12+a21+a43)2 ?(a23?a14+a41?a32)2 0 (a34?a12+a21?a43)2
?i(a24+a13+a31+a42)2 ?i(a34?a12?a21+a43)2 i(a22+a33) ?(a24+a13?a31?a42)2 ?(a34?a12+a21?a43)2 0
1
CC
CC
CA
Then apply the same argument in (1) to have the desired result.
(3) When A 2 sp4(C), the flfth row and column are zero so that the map in (4) yields
an isomorphism of sp4(C) and so5(C). Then apply the argument in (1).
22
Chapter 5
Another proof of the convexity of S(A)
Remark 5.1. Alike Davis? treatment [1] for W(A) when A 2C2?2, there is a simpler and
more geometric way to see that S(A) is an elliptical disk if
A =
0
BB
BB
@
0 a b
?a 0 c
?b ?c 0
1
CC
CC
A
2 so3(C):
Direct computation yields
xTAy = a(x1y2 ?x2y1)+b(x1y3 ?y1x3)+c(x2y3 ?x3y2) = (c;?b;a)?(x?y):
Since fx ? y : x;y 2 R3 are orthonormalg = S2, S(A) is the image under the linear map
z 2S2 7! (c;?b;a)?z 2C. Thus S(A) ?C is an elliptical disk using Davis? idea. We just
establish the convexity of S(A) when n = 3.
Lemma 5.2. Let A 2 son(C) and n ? 3.
1. Suppose x1;x2 2 Rn and y1;y2 2 Rn are orthonormal pairs, and spanfx1;x2g =
spanfy1;y2g. Then yT2 Ay1 = ?xT2 Ax1.
2. If ? 2 S(A), then t? 2 S(A) if jtj? 1 and t 2R.
Proof. (1) Notice that [y1 y2] = [x1 x2]O where O is a 2?2 orthogonal matrix, that is,
[y1 y2] = [x1 x2]
0
B@ cos sin 
?sin cos 
1
CA
23
or
[y1 y2] = [x1 x2]
0
B@?cos sin 
sin cos 
1
CA
Then for the flrst case
yT2 Ay1 = (sin x1 +cos x2)TA(cos x1 ?sin x2)
= cos2  xT2 Ax1 ?sin2  xT1 Ax2
= xT2 Ax1
and for the second case yT2 Ay1 = ?xT2 Ax1.
(2) Because of Remark 5.1 we may assume that n ? 4. Suppose that ? = xTAy where
x;y 2Rn areorthonormal. Chooseaunitvectorz thatisorthogonaltoy, y1 := (ReA)y 2Rn
and y2 := (ImA)y 2 Rn since n ? 4. Then choose ? 2 R so that w := tx + ?z is a unit
vector. Hence y and w are orthonormal and t? = txTAy = wTAy 2 S(A).
A proof of the convexity of S(A):
We now provide a proof of the convexity of S(A) which is difierent from [14] and is
based on Theorem 4.4. The cases n = 3;4 are proved in Theorem 4.4. Consider n >
4. Let w1 = xT1 Ay1 and w2 = xT2 Ay2 be two distinct points in S(A), where x1;y1 and
x2;y2 are orthonormal pairs in Rn. Let ^A : C ! C be the compression of A onto C :=
spanfx1;x2;y1;y2g. It is easy to see that the matrix of ^A is complex skew symmetric and
S( ^A) contains w1 and w2. Since w1;w2 are distinct, 2 ? dimC ? 4.
Case 1: 3 ? dimC ? 4. By Theorem 4.4, S( ^A) is convex and hence contains the line segment
[w1;w2]. So does S(A) since S( ^A) ? S(A).
Case 2: dimC = 2. Then spanfx1;y1g = spanfx2;y2g so that by Lemma 5.2(1) w1 =
?w2 6= 0. Pick x3 2 Rn such that x3 62 spanfx1;y1g since n ? 4. Apply the previous
argument on C0 := spanfx1;x3;y1g to have the desired result.
24
Finally we remark that S(A) may not be convex if A 2Cn?n, even though S(A) is well
deflned for all A 2Cn?n.
Example 5.3. If A =
0
B@1 i
0 0
1
CA, then
S(A) = f?cos (sin ?icos ) :  2Rg
by direct computation. So S(A) contains the points ?12 + i2 but not their midpoint i2.
25
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26
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27