This paper investigates how the normal forms of complex points on real 4-manifolds embedded in complex 3-manifolds change under small perturbations, providing a detailed understanding of their quadratic structure.
Contribution
It characterizes the stability of normal forms of complex points under small perturbations and describes their quadratic normal forms in embedded real 4-manifolds.
Findings
01
Normal forms are stable under small perturbations.
02
Quadratic parts of normal forms can be explicitly described.
03
Results apply to small $ ext{C}^2$-perturbations of real 4-manifolds.
Abstract
The purpose of this paper is to give a better understanding of complex points up to quadratic terms of real codimension 2 submanifolds embedded in a complex 3-manifold. We answer the question how a normal form of a pair of one arbitrary and one symmetric 2×2 matrix with respect to a certain linear group action changes under arbitrarily small perturbations. This result is then applied to describe the quadratic part of normal forms of complex points of small C2-perturbations of real 4-manifolds embedded in a complex 3-manifold.
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TopicsGeometric Analysis and Curvature Flows · Geometric and Algebraic Topology · Advanced Combinatorial Mathematics
Full text
On normal forms of complex points of small C2-perturbations of real 4-manifolds embedded in a complex 3-manifold
Tadej Starčič
Faculty of Education, University of Ljubljana, Kardeljeva Ploščad 16, 1000 Ljubljana, Slovenia
Institute of Mathematics, Physics and Mechanics, Jadranska
19, 1000 Ljubljana, Slovenia
We answer the question how arbitrarily small perturbations of a pair of one arbitrary and one symmetric 2×2 matrix can change a normal form with respect to a certain linear group action. This result is then applied to describe the quadratic part of normal forms of complex points of small C2-perturbations of real 4-manifolds embedded in a complex 3-manifold.
Research supported by grant P1-0291 and J1-7256
from ARRS, Republic of Slovenia.
2000 Mathematics Subject Classification:
32V40,58K50,15A21
1. Introduction
The study of complex points was started in 1965 by E. Bishop with his seminal work on the problem of describing the hull of holomorphy of a submanifold near a point with one-dimensional complex tangent space [3].
This is now very well understood for surfaces (see Bishop [3], Kenig and
Webster [26], Moser and Webster [30]),
and it later initiated in many researces in geometric analysis. For instance, (formal) normal forms for
real submanifolds in Cn near complex points were considered by Burcea [6], Coffman [7],
Gong [16] and Gong and Stolovich [17], Moser and Webster [30] among others.
We add that topological structure of complex points was first considered by
Lai [27] and in the special case of surfaces by Forstnerič [14]. Up
to C0-small isotopy complex points of real codimension 2 submanifolds in complex manifolds were treated by Slapar [31, 32, 33].
In this paper we describe the behavior of
the quadratic part of normal forms of complex points of small C2-perturbations of real 4-manifolds embedded in a complex 3-manifold (see Corrolary 3.8). It is a direct consequnce of a result that clarifies how a normal form for a pair of one arbitrary and one symmetric 2×2 matrix with respect to a certain linear algebraic group action changes under small perturbations (see Theorem 3.6); by a careful analysis we also provide information how small the perturbations must be. Due to technical reasons, these results are precisely stated in Section 3 and then proved in later sections.
Let f:M2n↪Xn+1 be a C2-smooth embedding of a real smooth 2n-manifold into a complex (n+1)-manifold (X,J). A point
p∈M is CR-regular if the dimension of the complex
tangent space TpCM=df(TpM)∩Jdf(TpM)⊂Tf(p)X is n−1, while p is called complex when the complex dimension of TpCM equals n, thus TpCM=df(TpM).
By Thom’s transversality theorem [2, Section 29], for generic embeddings the intersection
is transverse and so complex points are isolated.
Using Taylor expansion M can near a complex point p∈M be seen as a graph:
[TABLE]
where (z,w)=(z1,z2,…,zn,w) are suitable local coordinates on
X, and A∈Cn×n, B,C∈CSn×n, r∈Cn.
By Cn×n we denote the group of all n×n complex matrices, and by CSn×n, GLn(C), respectively, its subgroups of symmetric and nonsingular matrices. After a simple change of coordinates (\widetilde{z},\widetilde{w})=\big{(}z,w-b^{T}z-\tfrac{1}{2}z^{T}(\overline{B}-C)z\big{)} it is achieved that df(TpM)≈{w=0}, and the normal form up to quadratic terms is:
[TABLE]
A real analytic complex point p is called flat, if local coordinates can be chosen so that the graph (1.1) lies in Czn×R⊂Czn×Cw. It is quadratically flat, if
the quadratic part of (1.1) is real valued; this happens precisely when A in (1.1) is Hermitian.
Any holomorphic change of coordinates that
preserves (0,0) and {w=0} as a set in (1.1), has the same effect on
the quadratic part as a complex-linear change
[TABLE]
Furthermore, using this linear changes of coordinates and a biholomorphic change
(\widehat{z},\widehat{w})\mapsto\left(\widetilde{z},\widehat{w}-\frac{1}{2}\widehat{z}^{T}\big{(}\frac{\overline{c}-c}{|c|^{2}}P^{T}BP\big{)}\widehat{z}\right)
transforms (1.1) into the equation that can by a slight abuse of notation be written as
[TABLE]
Studying the quadratic part of complex points thus means examining the action of a linear group C∗×GLn(C) on pairs of matrices Cn×n×CSn×n, introduced in [7]:
[TABLE]
Problems of the quadratic part thus reduce to problems in matrix theory.
When n=1 complex points are
always quadratically flat and locally given by the equations
w=zz+2γ(z2+z2)+o(∣z∣2), 0≤γ<∞ or w=z2+z2+o(∣z∣2) (Bishop
[3]). If in addition they are real analytic and elliptic (γ<1),
they are also flat (see [30]). A relatively simple
description of normal forms of the action (1.2) was obtained for n=2 (see Coffman
[7] and Izotov [25]), while in dimensions 3 and 4 a complete list of normal forms has been given only in the case of quadratically flat complex points (see Slapar and Starčič [34]). Nevertheless, if B in (1.1) is nonsingular the classification has been done even in higher dimensions by the result of Hong [20].
The problem of normal forms of matrices under perturbations was first studied by Arnold (see e.g. [2]), who considered matrices depending on parameters under similarity (miniversal deformations). The change of Jordan canical form has been then succesfully investigated also through the works of Markus and Parilis [29], Edelman, Elmroth and Kågstrom [11], among others; the software Stratigraph [12] contructs the relations between Jordan forms. However, the problem of normal forms for ∗-conjugation (or T-conjugation) under small perturbations seems to be much more involved, and has been so far inspected only in lower dimensions; check the papers Futorny, Klimenko and Sergeichuk [9], Dmytryshyn, Futorny and Sergeichuk[10] (Dmytryshyn, Futorny, Kågström, Klimenko and Sergeichuk [15]). Virtually nothing has been known until the time of this writing about simultaneous small perturbations of pairs of matrices under these actions.
In connection to these problems we mention results of Guralnick [18] and Leiterer [28], who respectively studied similarity of holomorphic maps from Riemann surfaces or Stein spaces to a set of matrices.
We shall not consider this matter here.
2. Normal forms in dimension 2
We recall the basic properties of an action of a Lie group on a manifold (check e.g. [5, Theorem IV.9.3]).
These are well known and they are not difficult to prove.
Proposition 2.1**.**
Let Φ:G×X→X be a smooth (analytic) action of a real (complex) Lie group G with a unit e, acting on a smooth (complex) manifold X, i.e.
[TABLE]
Then Φ satisfies the following properties:
(1)
For any g∈G the map Φg:X→X, x↦Φ(g,x) is an automorphism and the map Lg:G→G, h↦gh is a transitive automorphism.
2. (2)
*For any x∈X the orbit map Φx:G→X, g↦Φ(g,x) is transitive and equivariant
*(for any g∈G we have Φx(Lg(h))=Φg(Φx(h)), h∈G), and Φx is of constant rank with dgΦx=dxΦg∘deΦx∘(deLg)−1.
3. (3)
*An orbit of x∈X, denoted by
OrbΦ(x)={Φ(g,x)∣g∈G},
is an immersed (locally embedded) homogeneous submanifold of dimension equal to rank(Φx).
*
Any orbit can be endowed (globally) with the structure of a manifold, but it does not necessarily coincide with the subspace topology ([5, Theorem IV.9.6]).
We proceed with the list of representatives of orbits (normal forms) of the action (1.2) for n=2 obtained by Coffman (see [7, Sec. 7,Table 1]). In addition, we compute tangent spaces of orbits and then arrange normal forms into a table according to dimensions of their orbits (42 types); these are calculated similarily as in the case of similarity (see e.g. Arnold [2, Section 30]).
To simplify the notation, a⊕d denotes the diagonal matrix with
a, d on the main diagonal, while the 2×2 identity-matrix
and the 2×2 zero-matrix are I2
and 02, respectively.
Lemma 2.2**.**
Orbits of the action (1.2) for n=2 (represented by its normal forms (A,B)) are immersed manifolds, their dimensions are given in the first columns of the tables:
[TABLE]
[TABLE]
Here 0<τ<1, 0<θ<π, a,b,d>0, d0∈{0,d}, r≥0, ζ∈C, 0≤φ<π.
A minor change is made in comparison to the original list in [7], as
([0110],[1000]) is taken instead of
([100−1],[1111]); conjugate the later one with
21[112−2].
By Proposition 2.1 (2), (3), orbits of the action Ξ in (1.2) for n=2 are immersed manifolds. To compute the tangent space of the orbit OrbΞ(A,B) we fix (A,B)∈C2×2×CS2×2, choose a path going through (1,(I2,I2)):
[TABLE]
and then calculate
[TABLE]
Writing X=∑j,k=12(xj,k+iyj,k)Ejk, α=u+iv, where Ejk is the elementary matrix with one in the j-th row and k-th column and zeros otherwise, we deduce that
[TABLE]
and in a similar fashion we conclude that
[TABLE]
Let a 2×2 complex (symmetric) matrix be identified with a vector in a real Euclidean space R8≈C2×2 (and R6≈CS2×2), thus C2×2×CS2×2≈R8×R6=R14 with the standard basis {e1,…,e14}. The tangent space of an orbit OrbΞ(A,B) can then be seen as the linear space spanned by the vectors {w1,w2}∪{ujk,vjk}j,k∈{1,2}, where
[TABLE]
We splitt our consideration of tangent spaces according to the list of normal forms in [7, Sec. 7,Table 1] into several cases.
In each case the tangent space will be written as a direct sum of linear subspaces V1⊕V2 (with trivial intersection) such that V2⊂Span{ej}9≤j≤14 and V1 will be either trivial or of the form ∑j=114αjej with some nonvanishing αj0, j0∈{1,…,8}.
It is apparent that v22=2w1−v11. By further setting
[TABLE]
we can choose V1=Span{w1,w2,v11,v12,v21,u11,u21}, V2=Span{u12,w3} and observe that dimV1=7\mathop{\rm dim}\nolimits V_{2}=\small{\left\{\begin{array}[]{ll}0,&a=b=d=0\\
1,&a=b=0,d\neq 0\\
2,&\textrm{otherwise}\end{array}\right.}.
2. Case II.
For λ=0 we set
V1=Span{w1,w2,v12,u12}, V2=Span{v21,v22,u11,u21} and we have dimV1=4 and
\mathop{\rm dim}\nolimits V_{2}=\small{\left\{\begin{array}[]{ll}0,&a=b=d=0\\
1,&d=b=0,a>0\\
4,&a=b=0,d=1\textrm{ or }a=d=0,b=1\\
5,&0<a,d=1,b=0\end{array}\right.}.
Next, for λ=eiθ, 0<θ<π we take
V1=Span{w1,w2,v12,v21,u22,u12,u21} and V2=Span{u11,u22} with
dimV1=7,
\mathop{\rm dim}\nolimits V_{2}=\small{\left\{\begin{array}[]{ll}0,&a=b=d=0\\
1,&ad\neq 0,a+d\neq 0\textrm{ or }a,d=0,b\neq 0\\
2,&\textrm{otherwise}\end{array}\right.}.
Finally, for λ∈{−1,1} we set
[TABLE]
If we choose
V1=Span{w1,w2,v12,v22,u12}, V2=Span{u11,u22,w3,w4},
then dimV1=5,
\mathop{\rm dim}\nolimits V_{2}=\small{\left\{\begin{array}[]{ll}0,&a=b=d=0\\
3,&d>0,a\in\{0,d\},b=0\textrm{ or }a,d=0,b\neq 0\\
4,&0<a<d,b=0\end{array}\right.}.
3. Case III.
For 0<t<1 we set
V1=Span{w1,w2,v12,v21,u11,u12,u21}, V2=Span{w3,w4} and get that dimV1=7,
\mathop{\rm dim}\nolimits V_{2}=\small{\left\{\begin{array}[]{ll}0,&a=b=d=0\\
1,&a=d=0,b>0\\
2,&\textrm{otherwise}\end{array}\right.}.
Next, if t=1 we take
V1=Span{w1,w2,v12,v12,u11}, V2=Span{w3,w4,u21,u12}. Observe that dimV1=5,
\mathop{\rm dim}\nolimits V_{2}=\small{\left\{\begin{array}[]{ll}3,&d=b=0,a=1\\
4,&0\neq 0,a=1\textrm{ or }b>0,d=1\end{array}\right.}.
Finally, when t=0 we set
[TABLE]
with
V1=Span{w1,w2,v12,v22,u12,u21}, V2=Span{w3,w4,w5}.
It follows that dimV1=6 and
\mathop{\rm dim}\nolimits V_{2}=\small{\left\{\begin{array}[]{ll}0,&a=b=d=0\\
1,&a=d=0,b>0\\
2,&ad=0,b=0\\
3,&\textrm{otherwise}\end{array}\right.}.
4. Case IV.
These vectors are contained in Span{ej}7≤j≤14 and 2w1=v11+2v22, 2w1=v11+2v22. It is now easy to compute the dimension of their linear span.
This finishes the proof.
∎
Remark 2.3*.*
Sometimes it it is more informative to understand the stratification into bundles of matrices, i.e. sets of matrices having similar properties. Again, this notion was introduced first by Arnold [2, Section 30].
Given a parameter set Λ with smooth maps
λ↦Aλ,
λ↦Bλ,
one considers a bundle of pairs of matrices under the action Ξ in (1.2), i.e.
a union of orbits ⋃λ∈ΛOrbΞ(Aλ,Bλ). We set
[TABLE]
and observe that for any g∈C∗×GLn(C) we have Ξg∘ΞΛ=ΞΛ∘(Lg×idΛ),
so the rank of d(g,λ)ΞΛ depends only on λ∈Λ.
In a similar manner as we computed the tangent space of an orbit, the tangent space of a bundle can be obtained.
It follows that the generic 2×2 pairs of one arbitrary and one symmetric matrix (forming a bundle with maximal dimension 14) under the action (1.2) for n=2 are:
[TABLE]
Indeed, tangent spaces of these bundles are spanned by the tangent vectors in Case II and Case III of the proof of Lemma 2.2 (for the appropriate parameters).
Note that using the list of normal forms in dimension 2, recently a result on holomorphical flattenability of CR-nonminimal codimension 2 real analytic submanifold near a complex point in Cn, n≥2, was obtained through the works of Huang and Yin [23, 24], Fang and
Huang [13].
3. Change of the normal form under small perturbations
In this section we study how small deformations of a pair of one arbitrary and one symmetric matrix can change its orbit under the action (1.2) for n=2.
First recall that (A,B), (A′,B′) are in the same orbit with respect to the action (1.2) if and only if there exist P∈GLn(Cn), c∈C∗ such that (A′,B′)=(cP∗AP,cPTBP). By real scaling P we can assume that ∣c∣=1, and after additional scaling P by c1 we eliminate the constant c.
Thus the orbits of the action (1.2) are precisely the orbits of the action of S1×GLn(C) acting on Cn×n×CSn×n by:
[TABLE]
The projections are smooth actions as well (the second one is even holomorphic):
[TABLE]
Next, let (A,B), (A′,B′) be in the same orbit under the action (3.1) (A′=cP∗AP, B′=PTBP for some P∈GLn(C), c∈S1) and let (E,F) be a perturbation of (A,B):
[TABLE]
A suitable perturbation of (A,B) is in the orbit of an arbitrarily choosen perturbation of (A′,B′).
It is thus sufficient to consider perturbations of normal forms.
Observe further that an arbitrarily small perturbation of (A,B) is contained in OrbΨ(A,B) if and only if (A,B) (and hence the whole orbit OrbΨ(A,B)) is contained in the closure of OrbΨ(A,B). The same conclusion also holds for actions Ψ1, Ψ2.
For the sake of clarity the notion of a closure graph for an action has been introduced. Given an action Φ, the vertices in a closure graph for Φ are the orbits under Φ, and there is an edge (a path) from a vertex (an orbit) V to a vertex (an orbit) V precisely when V lies in the closure of V. The path from V to V is denoted briefly by V→V. There are few evident properties of closure graphs:
•
For every vertex V, there exists V→V (a trivial path),
•
Paths V→V and V→V imply the path V→V (usually not drawn).
•
If there is no path from V to V (denoted by V→V), then for any vertex W it follows that either V→W or W→V (or both).
To simplify the notation V→V, we usually write V→V, where V∈V, V∈V.
In case V→V it is useful to know the distance of V from the orbit of V.
We shall use the standard max norm ∥X∥=maxj,k∈{1,…n}∣xj,k∣, X=[xj,k]j,k=1n∈Cn×n To measure the distance between two matrices. This norm is not submultiplicative, but ∥XY∥≤n∥X∥∥Y∥ (see [21, p. 342]).
Proceed with basic properties of closure graph for the actions
(3.1), (3.2), (3.3).
Lemma 3.1**.**
Suppose A,A,E∈Cn×n, B,B,F∈CSn×n, and p=∣detAdetB∣−∣detBdetA∣.
(1)
There exists a path A→A (a path B→B) in the closure graph for the action (3.2) (for the action (3.3)) if and only if there exist sequences Pj∈GLn(C), cj∈S1 (a sequence Qj∈GLn(C)), such that
[TABLE]
(a)
The existence of a path A→A (a path B→B) implies the following:
(i)
If detA=0 (or detB=0), then detA=0 (or detB=0). Apparently, A=0 (or B=0), then A=0 (or B=0).
2. (ii)
If detA=0, detA=0, ∥E∥<∥A−1∥−1 (detB=0, detB=0, ∥F∥<∥B−1∥−1), then A+F∈OrbΨ1(A) (B+F∈OrbΨ2(B)). Trivially, if A=0, ∥E∥<∥A∥ (and B=0, ∥F∥<∥B∥), then A+E=0 (B+F=0).
2. (2)
There exists a path (A,B)→(A,B) in the closure graph for an action (3.1) precisely when there exist sequences Pj∈GLn(C), cj∈S1 such that
[TABLE]
Morover, if A and A (B* and B) are in the same orbit and sufficiently close to each other, then it may be assumed in (3.5) that cjPj∗BPj=A (PjTBPj=B).*
(a)
When (A,B)→(A,B), it follows that
(i)
A→A, B→B, and p=0.
2. (ii)
dimOrbΨ(A,B)>dimOrbΨ(A,B).
2. (b)
If A,B,A,B∈GL2(C) are such that p=0 and \|E\|<\max\big{\{}1,\frac{|p|}{4|\det B|(2\|\widetilde{A}\|+1)}\big{\}}, \|F\|<\max\big{\{}1,\frac{|p|}{4|\det A|(2\|\widetilde{B}\|+1)}\big{\}},
then (A+E,B+F)∈OrbΨ(A,B).
Proof.
By definition A→A (B→B) if and only if A+Fj∈Orbψ1(A) (B+Gj∈Orbψ2(B)) for some Gj→0 (Fj→0). It is equivalent to (3.4) (see (3.2), (3.3)), so the first part of (1) is proved.
Apparently, (A,B)→(A,B)
is then equivalent to (3.5).
Since the orbit map of the action Ψ1 (the action Ψ2) is by Lemma 2.1 (2) of constant rank and hence locally a submersion (see e.g. [5, Theorem II.7.1]), this action has the so-called local Lipschitz property, i.e. if A, A′ (B, B′) are sufficiently close and cP∗AP=A′ (PTBP=B′), then P can be chosen near to identity and c near 1.
For any sufficiently small E (or F) such that A, A+E (B, B+F) are in the same orbit, then there exists some P close to the identity-matrix and c close to 1, so that (A+E,B+F) is equal to (cP∗AP,B+F)=(A,P−T(B+F)P−1) (equal to (A+E,PTBP)=(P−∗(A+E)P−1,B)).
As the inverse map X→X−1
is continuous, P−1 is close to identity-matrix, too.
Hence (A+E,B+F) is in the orbit of (A,B+F′) (or (A+E′,B)), where F′=P−∗FP (E′=P−TEP) is arbitrarily close to the zero-matrix. This concludes the proof of the first part of (2).
for some {Pj}j⊂GLn(C), {cj}j⊂S1 ({Qj}j⊂GLn(C)).
This implies (1) (a) (1(a)i).
A necessary condition for (A,B) to be in the closure of OrbΨ(A,B) is that A and B are in the closures of OrbΨ1(A) and OrbΨ2(B), respectively. Further, by multiplying the limits in (3.6) for Pj=Qj by detB or detA, and by comparing the absolute values of the expressions, we deduce (2) (a) (2(a)i).
It is well known that the distance from a nonsingular matrix X to the nearest singular matrix with respect to the norm ∥⋅∥ is equal to ∥X−1∥−1 (see e.g. [21, Problem 5.6.P47]). Thus (1) (b) follows.
Next, applying the triangle inequality, estimating the absolute values of the entries of the matrices by the max norm of the matrices, and by slightly simplifying, we obtain for X,D∈C2×2:
[TABLE]
Let A, B, B, A be nonsingular matrices,
p=∣detAdetB∣−∣detBdetA∣=0.
Using (3.7) for X=A, D=E and X=B, D=F with ∥E∥,∥F∥≤1, respectively,
we get
[TABLE]
[TABLE]
To estimate the left-hand sides of the above inequalities from above by 2∣p∣ it suffices to take ∥E∥<4∣detB∣(2∥A∥+1)∣p∣ and ∥F∥<4∣detA∣(2∥B∥+1)∣p∣. By combining (3.8), (3.9) and applying the triangle inequality we then conclude
[TABLE]
By (2) (a) (2(a)i) (allready proved) we have (A+E,B+F)→(A,B) and this gives (2) (b).
It is left to show the orbit-dimension inequalities (1) (a) (1(a)ii) and (2) (a) (2(a)ii).
Since orbits of Ψ, Ψ1, Ψ2 can be seen as nonsingular algebraic subsets in Euclidean space (zero loci of polynomials),
these facts can be deduced by using a few classical results in real (complex) algebraic geometry [4, Propositions 2.8.13,2.8.14] (or [35, Propositions 21.4.3, 21.4.5], [19, Exercise 14.1.]). Indeed, orbits OrbΨ(A,B), OrbΨ1(A), OrbΨ2(B) are contained in the closures (also with respect to a coarser Zariski topology) of orbits OrbΨ(A,B), OrbΨ1(A), OrbΨ2(B), respectively. Hence algebraic dimensions of orbits mentioned first are strictly smaller than algebraic dimensions of the later orbits.
Finally, orbits are locally regular submanifolds and their manifold dimensions agree with their algebraic dimensions.
∎
Remark 3.2*.*
Lemma 3.1 provides a quantitative information on the distance from an element to another orbit. We observe that it suffices to consider only normal forms.
Given any Q∈GLn(C) the induced operator norms of maps Cn×n→Cn×n, X↦Q∗XQ and CSn×n→CSn×n, X↦QTXQ are bounded from above by n2∥Q∗∥∥Q∥ and n2∥QT∥∥Q∥, respectively.
If (A,B), (A,B) are in the same orbit (i.e. A=cQ∗AQ, B=QTBQ with Q∈GLn(C), c∈S1), then for any P∈GLn(C), c∈S1 we get:
[TABLE]
When inspecting (A,B)→(A,B) where either A, A or B, B are in the same orbit and sufficiently close, it is by (2) enough to analyse perturbations of the matrix A (the matrix B). Unfortunately, we do not know how close the matrices should be since the constant rank theorem (even the quantitative version [22, Theorem 2.9.4]) does not provide the size of the local charts which define the orbits.
By Autonne-Takagi factorization (see e.g. [21, Corolarry
4.4.4]), any complex symmetric matrix is
unitary T-congruent to a diagonal matrix with non-negative diagonal entries, hence T-congruent to a diagonal matrix with ones and zeros on the diagonal.
Therefore 2×2 symmetric matrices with respect to T-conjugacy consist of three orbits, each containing matrices of the same rank. Their closure graph is thus very simple. (For closure graphs of all 2×2 or 3×3 matrices see [9].)
Lemma 3.3**.**
The closure graph for the action (3.3) (T-conjugacy on CS2×2) is
[TABLE]
where dimOrb(1⊕0)=2, dimOrb(I2)=3.
Moreover, if B, B are vertices in the above graph, and such B→B with PTBP=B+F for some P∈GL2(C), then ∥F∥≥1.
The (non)existence of most paths in the closure graph for the action Ψ1 in (3.2) follows immediately from the (non)existence of paths in the closure graph for ∗-conjugacy ([15, Theorem 2.2]). The remaining paths are treated by a slight adaptation of the ∗-cojugacy case. By a careful analysis we provide neccesarry (sufficient) conditions for the existing paths; see Lemma 3.4 (its proof is given in Sec. 4). These turn out to be essential in the proof of Theorem 3.6. Furthermore, if A→A we find a lower bound for the distance from A to OrbΨ1(A). Note that normal forms for Ψ1 were first observed by Coffman [8, Theorem 4.3], and by calculating their stabilizers eventually normal forms for the action (3.1) were obtained ([7, Subsection 2.4]).
Lemma 3.4**.**
Closure graph for the action (3.2) is the following (0<θ<π, 0<τ<1):
[TABLE]
The closure graph contains an infinite set of vertices corresponding to the orbits with normal forms
1⊕eiθ,
[0τ10],
indexed by the parameters θ, τ, respectively.
Orbits at the same horizontal level have the same dimension and these are indicated on the right.
Furthermore, let A, A be normal forms in (3.10), and let E=cP∗AP−A for some c∈S1, P=[xuyv]∈GL2(C), E∈C2×2. Then one of the following statements holds:
(1)
If A→A, then there exists a positive constant μ such that ∥E∥≥μ.
2. (2)
If A→A, then there is a positive constant ν such that the moduli of the expressions (depending on c, P) listed in the fourth column (and in the line corresponding to A, A) of the table below are bounded from above by ν∥E∥. (If A,A∈GL2(C) then also ∥E∥≤8∥A∥+4∣detA∣ is assumed.)
Conversely, if A, A correspond to any of the lines (C3.4), (C3.4), (C3.4), (C3.4), then there exists a positive constant ρ such that: if the moduli of expressions listed in the fourth column of this line
are bounded from above by s∈(0,1], then ∥E∥≤ρs.
[TABLE]
Remark 3.5*.*
Constants μ or ν in Lemma 3.4 are calculated for any given pair A,A (see Lemma 3.1 and the proof of Lemma 3.4). The existence of the constant ρ (computable as well) in the converse in Lemma 3.4 (2) is showed only in those cases where it turns to be useful in the proof of Theorem 3.6, though it is expected to be proved (possibly by a slight modification) for most of the cases.
We are ready to state the main results of the paper. The following theorem describes the closure graph for the action (3.1). Its proof is given in Sec. 5.
It is expected that by adapting the proof a similar result should hold for the restriction of the action (3.1) with c=1; in this case there are few more types of orbits.
Theorem 3.6**.**
Orbits with normal forms from Lemma 2.2 are vertices in the closure graph for the action (3.1). The graph has the following properties:
(1)
If (A,B)→(A,B), it is possible to provide some lower bound for the distance from (A,B) to the orbit of (A,B).
2. (2)
There is a path from (02,02) to any orbit. There exist paths from (1⊕0,02) to all orbits, except to \big{(}1\oplus 0,\begin{bmatrix}0&1\\
1&0\end{bmatrix}\big{)},
\big{(}\begin{bmatrix}0&1\\
\tau&0\end{bmatrix},\begin{bmatrix}0&b\\
b&0\end{bmatrix}\big{)} or b>0, and to (02,B), B∈CS2×2.
3. (3)
There exist paths from (1⊕0,a⊕0) with a>0 to all orbits, except to \big{(}1\oplus 0,\begin{bmatrix}0&1\\
1&0\end{bmatrix}\big{)},
\big{(}\begin{bmatrix}0&1\\
1&i\end{bmatrix},0\oplus d\big{)} for d>a, (02,B) for B∈CS2×2,
\big{(}\begin{bmatrix}0&1\\
\tau&0\end{bmatrix},\begin{bmatrix}0&b\\
b&0\end{bmatrix}\big{)} for b>0 such that a∈[1+τ2b,1−τ2b], and to
\big{(}1\oplus e^{i\theta},\begin{bmatrix}a&b\\
b&d\end{bmatrix}\big{)} for 0≤θ<π and such that a>M, where M is the maximum of the function given with a constraint
[TABLE]
4. (4)
All nontrivial paths (A,B)→(A,B) with nontrivial (A,B)=(1⊕0,a⊕0) for a≥0 (not mentioned in (2), (3)) are noted in the following two diagrams. Orbits at the same horizontal level have equal dimension (indicated on the right).
[TABLE]
[TABLE]
Remark 3.7*.*
The non-existence of some paths in the closure graph for the action (3.1) follow immediately from Lemma 3.3, Lemma 3.4 and Lemma 3.1 (a) (2) (2(a)i), (2(a)ii). However, to get some lower bound on the distance from a certain normal form to another orbit, many intrigueing and sometimes tedious estimates need to be done (see the proof of Theorem 3.6 in Sec. 5).
It makes the proof of the theorem much longer and more involved. Moreover, one must also work with inequalities instead of inspecting the convergence of sequences (see Lemma 3.1).
It would be interesting to have the closure graph for bundles of matrices with respect to the action (3.1) (see (2.3)) or its restriction. One would need to consider the same equations as in our case, but possibly with less constraints.
Let M be a compact codimension 2 submanifold in a complex manifold X. Since the complex dimension of the complex tangent spaces near a regular point of M is preserved under small perturbations, complex points of a small deformation M′ of M can arise only near complex points of M.
Recall that (A,B)∈C2×2×CS2×2 (a normal form up to quadratic terms) can be associated to a complex point p∈M, so that in a neighborhood of p the submanifold M is of the form (1.1) for A=A, B=B. If M′ is a C2-small deformation of M, then it is seen near p as a graph:
[TABLE]
where (z,w) are local coordinates with w0∈C, r,s∈C2, C∈C2×2 small, and A∈C2×2, B∈CS2×2 close to A, B, respectively. Similarily as in the exposition in Sec. 1 a complex point on this graph are put into the standard position (1.1) for A=A, B=B. (Translate the complex point to (0,0), use a complex-linear transformation close to identity to insure the tangent space at (0,0) to be {w=0}, and finally eliminate z-terms.) The next result is hence a direct consequence of Theorem 3.6.
Corollary 3.8**.**
Let M be a compact real 4-manifold embedded C2-smoothly in a complex 3-manifold X and let p1,…,pk∈M be its isolated complex points with the corresponding normal forms up to quadratic terms (A1,B1),…,(Ak,Bk)∈Cn×n×CSn×n. Asumme further that M′ is a deformation of M obtained by a smooth isotopy of M, and let p∈M′ be a complex point with the corresponding normal form (A,B). If the isotopy is sufficiently C2-small then p is arbitrarily close to some pj0, j0∈{1,…,k}, and there is a path (Aj0,Bj0)→(A,B) in the closure graph for the action (3.1).
Remark 3.9*.*
In the proof of Theorem 3.6 the lower estimates for the distances from normal forms to other orbits are provided, therefore it can be told how small the isotopy M′ (of M) in the assumption of Corollary 3.8 needs to be.
In this section we prove Lemma 3.4. We start with the following technical lemma related to actions (\refactionpsi1) and (\refactionpsi2).
Lemma 4.1**.**
(1)
Suppose A,A,P∈GL2(C), E∈C2×2, c∈S1 and such that cP∗AP=A+E,
∥E∥≤8∥A∥+4∣detA∣. Denote further \Delta=\arg\bigl{(}\tfrac{\det\widetilde{A}}{\det A}\bigr{)}. It then follows that
[TABLE]
2. (2)
If F∈C2×2, B,B,P∈GL2(C) and such that PTAP=B+F,
∥F∥≤8∥B∥+4∣detB∣, then
[TABLE]
Proof.
First, we observe the following simple fact. For ξ,ζ,h∈C:
[TABLE]
Indeed, we have ξζ−1=1+ζh=∣1+ζh∣eiψ with ∣ζh∣≤21, hence ψ∈(−2π,2π) and
|\sin\psi|=\bigl{|}\mathop{\rm Im}\nolimits\big{(}\tfrac{1+\frac{h}{\zeta}}{|1+\frac{h}{\zeta}|}\big{)}\bigr{|}\leq\tfrac{|\mathop{\rm Im}\nolimits\frac{h}{\zeta}|}{|1+\frac{h}{\zeta}|}\leq\tfrac{|\frac{h}{\zeta}|}{1-|\frac{h}{\zeta}|}\leq\tfrac{2|h|}{|\zeta|}.
By assuming ∥D∥≤8∥X∥+4∣detX∣ and applying (4.4) to (4.5) we obtain
[TABLE]
We apply det to cP∗AP=A+E, QTBQ=B+F and after simplifying we get
[TABLE]
From (4.6) for X=A, D=E and (4.7) it follows that
c=(−1)kei(2Δ+2ψ), k∈Z, \Delta=\arg\bigl{(}\frac{\det\widetilde{A}}{\det A}\bigr{)} with ψ as in (4.6) with X=A, D=E.
Using well known facts ei2ψ=1+2isin(4ψ)ei4ψ and
2∣sin4ψ∣≤∣2ψ∣≤∣sinψ∣ for ψ∈(−2π,2π), we deduce (4.1).
Furthermore, (4.5) for X=A, D=E and (4.7) for X=B, D=F give
[TABLE]
respectively. To conclude the proof we observe another simple fact. If ∣s∣≤1 then
[TABLE]
To see this, we take ρ=1+s′ to be the square root of 1+s (thus ρ2=1+s) with Re(ρ)≥0 and Re(s′)≥−1. It yields ∣s′∣=∣ρ−1∣=∣ρ+1ρ2−1∣≤1∣s∣.
For ∥E∥≤8∥A∥+4∣detA∣ and
∥F∥≤8∥B∥+4∣detB∣ (∣p∣,∣q∣≤21≤1 in (4.8)),
we apply (4.9) to (4.8) for s=p and s=q, respectively. It implies (4.2) and (2).
∎
For actions Ψ, Ψ1 (see (3.1) and (3.2)), it follows that (A′,B′)∈OrbΨ(A,0) if and only if B′=0 and A′∈OrbΨ1(A). Hence \mathop{\rm dim}\nolimits\bigl{(}\mathop{\rm Orb}\nolimits_{\Psi_{1}}(A)\bigr{)}=\mathop{\rm dim}\nolimits\bigl{(}\mathop{\rm Orb}\nolimits_{\Psi}(A,0)\bigr{)}, where dimensions of orbits of Ψ1 are obtained from Lemma 2.2.
To prove A→A it is sufficient to find c(s)∈S1,
P(s)∈GL2(C) such that
[TABLE]
It is straightforward to see that
P(s)=1⊕s
with c(s)=1 in (4.10) yields 1⊕0→1⊕λ, while
to prove
1⊕0→[0τ10],
0≤τ≤1, and
1⊕−1→[011i],
we take
P(s)=1+τ1[110s]
and
P(s)=21[s1ss1−s], respectively, and in both cases again with c(s)=1.
(Compositions of these paths represent paths as well.)
It is then left to find necessary (sufficient) conditions for the existence of these paths, i.e. given A, A, E satisfying
[TABLE]
we must find out how c, P depend on E (how E depends on c, P).
On the other hand, if (4.11) fails for every sufficiently small E, it gives A→A. To prove (1),
upper estimates for ∥E∥ will be provided in such cases.
This has allready been done for A=0, A=0 and detA=0, detA=0 (see Lemma 3.1 (1) (1(a)i)).
Throughout the rest of the proof we denote
[TABLE]
and splitt our consideration of the remaining paths A→A
into several cases.
Case I.
A=[011i]
It is straightforward to compute
[TABLE]
Multipling the equation (4.11) by c−1 and then writing them componentwise yields
[TABLE]
The real and the imaginary parts of the first and the last equation of (4.13) give:
[TABLE]
while by adding (subtracting) the second and the complex-conjugated third equation of (4.13) for β,γ∈R we deduce
[TABLE]
(a)
A=[0τ10], 0≤τ<1
For α=ω=0 the second and the last equation of (4.14) give ∣u∣2≤∥E∥ and ∣v∣2≤∥E∥ (hence ∣uv∣≤∥E∥). Further, the first equation of (4.15) for β=1, γ=τ yields
[TABLE]
If ∥E∥<21−τ we get a contradiction (remember 1>τ≥0).
2. (b)
A=α⊕0, α∈{0,1}
The first and the last equation of (4.13) for ω=0 imply
[TABLE]
respectively. Next, from (4.15) for β=γ=0 we obtain
[TABLE]
Thus the first part of (2) for (C3.4) follows; the converse is immediate by (4.13).
3. (c)
detA=eiϑ, 0≤ϑ≤π, ∥A∥=1
By (4.1) in Lemma 4.1 we have c−1=i(−1)ke−i2ϑ+g with k∈Z, ∣g∣≤12∥E∥, provided that ∥E∥≤121. Using this and rearranging the terms in (4.14), (4.15) we deduce:
[TABLE]
Observe that for β=γ the first equation in (4.17) yields
[TABLE]
while for ω=0 the last equation of (4.13) implies that 2\big{|}\mathop{\rm Re}\nolimits(\overline{y}u)\big{|}\leq\|E\|, ∣v∣2≤∥E∥.
(i)
A=1⊕eiϑ, 0≤ϑ<π
After multiplying the third and the fifth equation of (4.17) for α=1, ω=eiθ, β=γ=0, estimating the imaginary parts by its moduli, applying the triangle inequality and using ∣g∣≤12∥E∥, we obtain
[TABLE]
Combining it with (4.16) and ∥E∥≥∥E∥2 leads to a contradiction for ∥E∥<392∣1+cosϑ∣.
2. (ii)
A is either equal to 1⊕−1, [0110] or [011i] (ϑ=π)
The statement (2) for (C3.4) follows immediately from (4.17), (4.18) for ϑ=π and either α=1, ω=−1, β=0 or α=ω=0, β=1 or α=0, β=1, ω=i.
2. Case II.
A=1⊕λ, ∣λ∣∈{1,0}
We have
[TABLE]
Thus (4.11) multiplied by c−1 and written componentwise (also rearranged) yields:
[TABLE]
Subtracting the second complex-conjugated equation (and multiplied by λ) from the third equation (and multiplied by λ) for β,γ∈R further gives
[TABLE]
(a)
λ=eiθ, 0≤θ≤π
By taking the imaginary and the real parts ob the first and the last equation of (4.19) for λ=eiθ we obtain
[TABLE]
(i)
A=[0τ10],
0≤τ≤1
If θ=0 then (4.21) for θ=α=ω=0 implies that ∣x∣2,∣u∣2,∣y∣2,∣v∣2≤∥E∥, which contradicts the second equation of (4.19) for β=λ=1 (θ=0) with ∥E∥<31.
Next, let θ=π (λ=−1). From (4.20) for Im(λ)=0, β=1, γ=τ it then follows ∣1−τ∣≤2∥E∥, which fails if τ=1, ∥E∥<21−τ. Further, when τ=1 we deduce from (4.1) (Lemma 4.1) that c−1=(−1)k+g, ∣g∣≤12∥E∥, so the second equation of (4.19) for λ=−1, β=1 yields \big{|}\overline{x}y-\overline{u}v-(-1)^{k}\big{|}\leq 13\|E\|.
By combining it with the first and the last equation of (4.19) for λ=−1, α=ω=0 we get (2) with (C3.4).
It is left to consider 0<θ<π.
From (4.21) for α=ω=0 it follows that
[TABLE]
Applying the triangle inequality to the second equation of (4.19), and using the estimates (4.22) leads to an inequality which fails for ∥E∥<(2+cotθ+sinθ1)−1:
[TABLE]
2. (ii)
A=[011i]
By (4.1) in Lemma 4.1 we have c−1=−i(−1)kei2θ+g, ∣g∣≤12∥E∥, provided that ∥E∥≤121.
The third equation of (4.21) for θ=π, ω=i yields 0=(−1)k+Im(ig+c−1ϵ4), which failes for ∥E∥<131.
If θ=0, then the second and the last equation of (4.21) for α=0, ω=i imply ∣x∣2,∣u∣2≤∥E∥ and ∣y∣2,∣v∣2≤1+∥E∥, respectively. From the second equation of (4.19) for β=1, λ=eiθ we then conclude 1−∥E∥≤2∥E∥(1+∥E∥), so we have a contradiction for any ∥E∥ small enough.
Finally, for 0<θ<π we use (4.21) for α=0, ω=i, c−1=−i(−1)kei2θ+g to get ∣u∣2≤sinθ∥E∥, ∣x∣2≤∥E∥(1+cotθ) (see (4.22)) and
[TABLE]
Applying these estimates to the second equation of (4.19) for β=1, λ=eiθ we get an inequality, which fails for every sufficiently small ∥E∥:
[TABLE]
3. (iii)
A=α⊕0, α∈{0,1}
If 0≤θ<π (θ=π) the equations (4.19) for ω=β=γ=0, λ=eiθ give the first part of (the complete statement) (2) with (C3.4) (with (C3.4) for ω=0, σ=−1). To see the converse for (C3.4), we fix s>0 and assume ∣y∣2,∣v∣2≤s and \big{|}|x|^{2}+e^{i\theta}|u|^{2}-c^{-1}\alpha\big{|}\leq s. By observing the imaginary and the real part of ∣x∣2+eiθ∣u∣2−c−1α for θ=0 we deduce that ∣x∣2,∣u∣2≤∣α∣+s and
[TABLE]
The second (third) equation of (4.19) for β=0 (γ=0), λ=eiθ then yields ∣ϵ2∣,∣ϵ3∣≤(∣α∣+∣cotθ∣)(∣α∣+s)s, so (2) with (C3.4) is proved. The converse for (C3.4) is trivial.
4. (iv)
A=1⊕θ, 0≤θ≤π.
By (4.1) (Lemma 4.1) we have
c−1=(−1)kei2θ−θ+g, ∣g∣≤12∥E∥, assuming that ∥E∥≤121.
Thus the first and the last equation of (4.19) for α=1, ω=eiθ are of the form:
[TABLE]
We now take the
imaginary parts of equations (4.23), slightly rearrange the terms and use the triangle inequality:
[TABLE]
In particular we have
[TABLE]
Multiplying these inequalities and using the triangle inequality we deduce that
[TABLE]
From (4.20) for β=γ=0, Im(λ)=sinθ we get
\big{|}(\sin\theta)\overline{v}u\big{|}\leq\|E\|.
Combining this with (4.25) we obtain that
[TABLE]
If θ=θ∈{0,π} then equations (4.23) and the second equation of (4.19) for λ=eiθ allready give us the statement (2) for (C3.4) in case ω=σ∈{1,−1}. (Note that if θ=θ=0, then k is even.)
Further, when θ=θ∈{0,π} the first equation of (4.24) fails for \|E\|\leq\frac{1}{13}\big{|}\sin(\tfrac{\theta-\widetilde{\theta}}{2})\big{|}.
Next, let 0<θ<π. For θ=θ we have 21(θ+θ),21(θ−θ)=lπ, l∈Z, hence it is easy to choose ∥E∥ so small that(4.26) fails.
If θ=θ, then the first equation of (4.24) leads to ∣u∣2≤sinθ∥E∥, and by comparing the real parts of the first equation of (4.23), slightly rearranging the terms, we furter get:
[TABLE]
The second equation of (4.24) also yields \big{|}|v|^{2}-(-1)^{k}\big{|}\leq\frac{13}{\sin\theta}\|E\|, thus ∥E∥<13sinθ implies that k is even. This concludes the proof of (2) about (C3.4).
2. (b)
λ=0 (It suffices to consider the case when detA=0.)
When
A=α⊕0, α∈{0,1}
the statement (2) with (C3.4) follows immediately from the first and the third equation of (4.19) for ω=λ=0. Applying (4.4) for ∥E∥≤21 to the first equation of (4.19) for α=1, λ=0 (multiplied by c), yields ψ=arg(c)∈(−2π,2π), ∣sinψ∣≤2∥E∥. Therefore c−1=eiψ−1=2isin(2ψ)ei2ψ with ∣sin(2ψ)∣≤∣sinψ∣≤2∥E∥.
If A=[0010], the first and the last equation of (4.19) for λ=α=ω=0 yield ∣x∣2,∣y∣2≤∥E∥, thus the third equation of (4.19) for λ=0, γ=0
fails for ∥E∥<21.
3. Case III.
A=[0τ10], 0≤τ≤1
We calculate
[TABLE]
Thus (4.11) multiplied by c−1 and rearranged is equivalent to
[TABLE]
Rearranging the terms of the first and the last equation immediately yields
[TABLE]
while multiplying the third (second) complex-conjugated equation with τ, subtracting it from the second (third) equation, and rearranging the terms, give
[TABLE]
(a)
τ=1
Since
A=[0110]
is ∗-congruent to 1⊕−1, the existence of paths in this case was allready analysed in Case II. It is only left to check (2) with (C3.4). If either
A=[0110] or A=1⊕−1, then by (4.1) we have c−1=(−1)k+g, k∈{0,1}, ∣g∣≤12∥E∥, provided that ∥E∥≤121.
The second (third) equation of (4.27) and (4.28) for τ=1 then imply:
[TABLE]
The inequalities (4.30) are valid also if we consider (4.27) (and (4.28)) for α∈{0,1}, β=γ=ω=0 (A=α⊕0).
Note that for α=1 the first equation of (4.28) for τ=1, and multiplied by c, is of the form
2cRe(xu)=1+ϵ1. Therefore, by applying (4.4) for ∥E∥≤21 we get that c=(−1)keiψ, k∈Z, ψ∈(−2π,2π), ∣sinψ∣≤2ϵ. Moreover, c−(−1)k=(−1)k2isin(2ψ)ei2ψ with ∣sin(2ψ)∣≤∣sinψ∣≤2∥E∥.
Conversely, we assume that the expressions (C3.4) for c=(−1)k (precisely the left-hand sides in (4.27)) are bounded from above by some s>0. Thus the right-hand sides of (4.27) (and hence ∥E∥) are bounded from above by s as well.
2. (b)
By combining these inequalities and making some trivial estimates we obtain
[TABLE]
Together with (4.33) for α=0 and using ∥E∥≥∥E∥2 we get
[TABLE]
which fails for ∥E∥<(1+τ)2(γ+∣ω∣+3)∣τγ−1∣∣γ−τ∣; remember γ=τ, 0≤γ≤1, 0≤τ<1.
2. (ii)
A=[0τ10]
By (4.1) in Lemma 4.1 for τ=0, ∥E∥≤12τ≤121 we have c−1=(−1)k+g, k∈Z, ∣g∣≤τ12∥E∥, thus (4.29) for β=1, γ=τ yields
[TABLE]
It further implies
[TABLE]
Moreover, from (4.29) for τ=γ=0, β=1 we deduce ∣yu∣≤∥E∥, ∣xv−c−1∣≤∥E∥, and (4.31) for α=ω=0 concludes the proof of (2) for (C3.4). (The converse is apparent.)
3. (iii)
Combining this with (4.32), rearranging the terms we get
[TABLE]
If α,ω=0 then by choosing ∥E∥<∣α∣+∣ω∣+2∣αω∣ we contradict the above inequality.
Furthermore, (4.28), (4.31), (4.34) give the first part of (2) for (C3.4).
Conversely, assuming that the expressions of (C3.4) for α=1 are bounded from above by s>0, then (4.27) (and 4.28) imply that ∥E∥≤3s.
Recall that the existence of a path (A,B)→(A,B) in the closure graph for the action (3.1),
immediately implies (see Lemma 3.1):
[TABLE]
When any of the conditions (5.1) is not fullfiled, then (A,B)→(A,B) and we allready have a lower estimate on the distance from (A,B) to the orbit of (A,B) (see Lemma 3.1, Lemma 3.3, Lemma 3.4). Further, (A,02)→(A,02) if and only if A→A, and trivially (A,B)→(A,B) for any A,B.
From now on we suppose (A,B)=(A,B), B=0, and such that (5.1) is valid. Let
[TABLE]
Due to Lemma 3.4, Lemma 4.1 the first equation of (5.2) yields the restrictions on P, c imposesed by ∥E∥. Using these we then analyse the second equation of (5.2).
When it implies an inequality that fails for any sufficiently small E, F, it proves (A,B)→(A,B). The inequality just mentioned also provides the estimates how small E, F should be; this calculation is very straightforward but is often omitted.
On the other hand, if given matrices A, B, A, B we can choose E and F in (5.2) to be arbitrarily small, this will yield (A,B)→(A,B).
In most cases we find c(s)∈S1, P(s)∈GL2(C) such that
[TABLE]
However, to confirm the existence of a path we can also prove the existence of suitable solutions of (5.2) by using the last part of Lemma 3.4 (2).
Throughout the rest the proof we denote δ=ν∥E∥ (the constant ν>0 is provided by Lemma 3.4), ϵ=∥F∥,
[TABLE]
where sometimes polar coordinates for x,y,u,v in P might be prefered:
[TABLE]
The second matrix equation of (5.2) can thus be written componentwise as:
[TABLE]
For the sake of simplicity some estimates in the proof are crude, and it is allways assumed ϵ,δ≤21. When appying Lemma 4.1 with A,A or B,B nonsingular we in addition take νδ=∥E∥≤8∥B∥+4∣detB∣ or ϵ=∥F∥≤8∥B∥+4∣detB∣, respectively.
Furthermore, we use the notation (A,B)⇢(A,B) when the existence of a path is yet to be considered.
We splitt our analysis
into several cases. (For normal forms recall Lemma 2.2.)
Case I.
(1⊕eiθ,B)⇢(1⊕eiθ,B), 0<θ<π, B=B
Denoting u2=δ2, y2=δ1 and slightly rearranging the terms in (5) yields
[TABLE]
From Lemma 3.4 (2) for (C3.4) we get ∣δ1∣,∣δ2∣≤δ and \bigl{|}|v|^{2}-1\bigr{|},\bigl{|}|x|^{2}-1\bigr{|}\leq\delta (therefore 1−δ≤∣x∣,∣v∣≤1+δ, \bigl{|}|vx|-1\bigr{|}\leq\delta).
By applying the triangle inequality we conclude from the first equation of (Case I) that
[TABLE]
and similarly the last two equations of (Case I) yield
[TABLE]
Since B=B, a comparison of the left-hand and the right-hand sides of the above inequalities implies that at least one of them fails
for
ϵ,δ such that
[TABLE]
2. Case II.
(1⊕σ,B)⇢(1⊕σ,B), σ∈{1,−1}
From Lemma 3.4 (2) for (C3.4) for α=1, ω=σ∈{1,−1} we have
From the second equation of ((a)) and the second inequality (5.7), it follows that \big{|}|axy|-|duv|\big{|}\leq\epsilon and \big{|}|dxy|-|duv|\big{|}\leq d\delta, respectively, and thus
[TABLE]
(i)
a=a=0, d=d, d>0
Applying the triangle inequality to ((a)) for a=a=0 yields
[TABLE]
respectively. Using (5.7) and (5.9) for a=d we further obtain
∣y∣2≤1−δ−dϵ(δ+dϵ)2 and
[TABLE]
These inequalities and the last inequality in (5.10) eventually lead to
[TABLE]
which fails if ϵ,δ are such that Ω=dϵ+δ is so small that
Ω+1−ΩΩ2<∣dd−1∣.
2. (ii)
0<a≤d,
0<a≤d, a=a,a=d (remember ad=ad, B=B)
Using the notation (5.4) the following calculation is validated trivially:
[TABLE]
Furthermore, one easily computes:
[TABLE]
Using the second equation of ((a)) and the second inequality of (5.7) we conclude
[TABLE]
The above implies that at least one of the moduli of the second terms on the right-hand sides of equations (5.11) is bounded from above by aϵ+δ,
while from the first and the last inequality in (5.7) it follows that the moduli of the first terms on the right-hand sides of (5.11) are bounded from above by 1+δ and from below by 1−δ.
For ϵ=δ<a+2min{a−a,d−a} thus the first or the last equation of ((a)) fails (a=a,d).
2. (b)
B=[0bb0], b>0, B=[a00d], 0<a≤d, σ=−1 (see Lemma 2.2)
By using Lemma 4.1 (2) and 1=∣detAdetA∣=∣detBdetB∣=∣−b2ad∣ (see (5.1)) we deduce that
detP=xv−yu=i(−1)l+δ′ with ∣δ′∣≤b2ϵ(4d+2), l∈Z.
Combining it further with the second equation of ((b)) we get vx=\frac{1}{2}\big{(}i(-1)^{k}+\delta^{\prime}+\frac{\epsilon_{2}}{b}\big{)} with ∣vx∣≥21(1−b2ϵ(4d+2+b)).
By taking ϵ≤4b(1−b2ϵ(4d+2+b)) we guarantie ∣ϵ2∣≤2∣bvx∣, hence
by applying (4.4) to the second equation of ((b)) we conclude (see (5.4)):
[TABLE]
Multplying the first and the last equation of ((b)) and using the triangle inequality gives
4b2∣uvxy∣≥ad−(a+d)ϵ−ϵ2, so ∣xy∣2 or ∣uv∣2 (or both) is at least equal to 2b1(ad−(a+d)ϵ−ϵ2).
Using (4.4) the second inequality of (5.7) for σ=−1 implies
[TABLE]
Adding it to (5.14) yields ψ1+ψ2=π+(2l1+2l2+1)π. Applying sin finally gives
[TABLE]
It is now easy to see that for any appropriately small ϵ and δ we get a contradiction.
3. (c)
B=[0bb0], b>0, B=[a00d], 0<a≤d, σ=−1 (see Lemma 2.2)
It is easy to check that
P(s)=[s0s2s−1] and
P(s)=[s−1s20s] with c(s)=1 in (5.3) prove
\big{(}\begin{bmatrix}0&1\\
\tau&0\end{bmatrix},\begin{bmatrix}0&b\\
b&0\end{bmatrix}\big{)}\to\big{(}\begin{bmatrix}0&1\\
\tau&0\end{bmatrix},\begin{bmatrix}a&b\\
b&0\end{bmatrix}\big{)}
and
\big{(}\begin{bmatrix}0&1\\
\tau&0\end{bmatrix},\begin{bmatrix}0&b\\
b&0\end{bmatrix}\big{)}\to\big{(}\begin{bmatrix}0&1\\
\tau&0\end{bmatrix},\begin{bmatrix}0&b\\
b&d\end{bmatrix}\big{)}, respectively.
Set ux=δ1, uy=δ2, vy=δ4 and after rearranging the terms we write (5) as
[TABLE]
∣δ1∣,∣δ2∣,∣δ3∣≤δ.
Applying the triangle inequality to the second equation gives
[TABLE]
Next, multiplying ∣xu∣≤δ and ∣vy∣≤δ with ∣vx∣≤1+δ (see (5.17)) gives ∣x2uv∣≤δ(1+δ)≤2δ and ∣v2xy∣≤δ(1+δ)≤2δ, respectively (recall δ≤21≤1). Thus either ∣x2∣ or ∣v2∣ or ∣uv∣, ∣xy∣ (or more of them) are bounded by 2δ.
(a)
∣x2∣≤2δ
From (5.17) we get ∣vx∣≥1−δ≥21, ∣yv∣≤δ and it further yields ∣xy∣=∣xv∣∣yv∣∣x∣2≤2δ2δ.
Applying the triangle inequality to the last equation of (Case III) multiplied by x2 we obtain an inequality that fails for d=0 and any sufficiently small ϵ,δ:
[TABLE]
Further, if d=0 the last and the first equality of (Case III) yield
[TABLE]
respectively. For a=0 (if d=0) the second (the first) inequality fails for ϵ=δ<2∣a∣+2b+1∣d∣ (for ϵ2=δ2<2∣a∣+2b+1∣a∣).
Since ∣x∣2≤2δ, ∣y∣2≤2δ we have ∣axy∣2≤∣a∣28δ3, so (5.19) for d=0 gives ∣a∣44δ3+bδ+ϵ≥∣b−b∣−bδ.
When b=b it is not too difficult to choose δ,ϵ such that the above inequality fails.
Remember that the case d=a=d=0, b=b has allready been considered.
2. (b)
∣v2∣≤2δ
We deal with this case in the same manner as in Case III(a), we only replace x,y,v,a,d,a,d by v,u,x,d,a,d,a, respectively; we get similar estimates for δ,ϵ.
3. (c)
∣uv∣,∣xy∣≤2δ
Here the inequality (5.19) fails for b=b, ϵ=δ<(∣a∣+∣d∣)2+2b∣b−b∣. Moreover, if b=b:
[TABLE]
Next, using (5.17) we obtain \delta\sqrt{2\delta}\geq\big{|}(xu)(vu)\big{|}=|u^{2}xv|\geq|u|^{2}(1-\delta) and \delta\sqrt{2\delta}\geq\big{|}(xy)(vy)\big{|}=|y^{2}xv|\geq|y|^{2}(1-\delta). Hence ∣y∣2,∣u∣2≤1−δδ2δ≤2δ (recall 21≥δ), so the first and the last equation of (Case III) give
[TABLE]
We set x,y,u,v as in (5.4) and let d=∣d∣eiϑ, d=∣d∣eiϑ, a=eiι, a=eiι. Applying (4.4) to the inequality (5.20), to the estimates (5.21) and to \big{|}x\overline{v}-(-1)^{k}\big{|}\leq\delta, k∈Z (see Lemma 3.4 (2) for (C3.4) with 0<τ<1), then leads to:
[TABLE]
respectively. Subtracting (adding) the first and the last equation from the second-one (the third-one), rearranging the terms, and applying sin, gives for 1>τ>0:
[TABLE]
[TABLE]
(i)
adad=0
First, let a=a=0 (or d=d=0). Since B=B, we have d=d (or a=a). The case d=0 (or a=0) has allready been considered, thus it is left to check the case 0<τ<1 with d=eiϑ, d=eiϑ, ϑ,ϑ∈[0,π) (from a=eiι, a=eiι, ι,ι∈[0,π)); see Lemma 2.2.
By (5.23), (5.22) it is easy to choose suitable ϵ,δ to get a contradiction.
For a=0, a=0 (for d=0, d=0) the first (the second) inequality of (5.21) fails for ϵ=δ<1+2b+2∣d∣∣a∣ (for ϵ=δ<1+2b+2∣a∣∣d∣).
Next, let a,d=0.
Multiplying the estimates for ∣ax2∣, ∣dv2∣ given by (5.21) yields
[TABLE]
From (5.20) we further deduce |vx|\geq 1-\tfrac{1}{b}\big{(}\sqrt{2\delta}(|a|+|d|)+b\delta+\epsilon\big{)}, so by combining it with the above inequality we obtain an inequality that fails to hold,
provided that at least one of a,d vanishes and ϵ,δ are small enough.
2. (ii)
By choosing ϵ,δ such that \frac{|a|}{|\widetilde{a}|}\big{(}\epsilon+2\delta(b+|d|)\big{)}\leq 1 we assure that ∣δ5′∣,∣δ6′∣≤1 and so ∣δ5′δ6′∣≤∣δ6′∣. Further, using (5.20) and ∣ad∣∣ad∣=1 yields:
[TABLE]
If τ=0, then d=d=1, a=a (see by Lemma 2.2 for B=B), so we easily find ϵ,δ to contradict this inequality. Similarly we treat the cases τ∈(0,1) with B=1⊕d, B=1⊕d, d=d or B=[eiιbbd],
B=[eiιbbd], where either ∣d∣=∣d∣ or ι=ι, d=−d.
Finally, let τ∈(0,1), a=eiι, a=eiι, ι,ι∈[0,π), ∣d∣=∣d∣, and let d=−d (so ∣ϑ−ϑ∣=π) precisely when ι=ι.
From (5.23), (5.22) we get a contradiction for any small ϵ,δ.
4. Case IV.
Lemma 3.4 (2) with (C3.4) for α=0, β=1, ω=i, k=0 (since δ≤21<1) gives
[TABLE]
(a)
B=[0bb0], b>0, B=a⊕d, a>0, d∈C∗
From (5) we obtain the same equations as in ((c)) and the inequality (5.16) but with d∈C∗. Applying the triangle inequality to these (in)equalities, and
using (5.24) gives ∣ax2∣≤∣d∣δ+ϵ, ∣ay2∣≤∣dv2∣+ϵ≤∣d∣(1+δ)+ϵ and finally
[TABLE]
It is straightforward to find δ,ϵ such that that the last inequality fails.
2. (b)
B=a⊕d, a>0, d∈C∗, B=[0bb0], b>0
We have the same equations as are ((b)), but with d∈C∗.
From the first and the last of these equations we obtain the estimates 2b∣xu∣≥a−ϵ and 2b∣y∣(1−δ)≤∣d∣−ϵ (∣v∣2≥1−δ by (5.24)), respectively.
Combining them and ∣u∣2≤δ, ∣v∣2≥1−δ with the second equation of ((b)) multiplied by u, yields
[TABLE]
Clearly, for any small enough ϵ,δ we get a contradiction.
3. (c)
By slightly rearranging the terms in (5) for b=b=0, d,d∈C (see also ((a))), using the last two estimates in (5.24) and applying the triangle inequality, we get
[TABLE]
On the other hand the third equation of ((a)) gives a∣y∣2=∣dv2−d+ϵ∣≤∣d∣(1+δ)+∣d∣+ϵ.
Using this and the first two estimates of (5.24) we conclude that
[TABLE]
Note that according to the list in Lemma 2.2, a=0 (a=0) implies d≥0 (d≥0).
(i)
a=0 or a=0
If a=0, a=0 the first inequality of (5.25) fails for ϵ=δ<1+∣d∣a, while for a=a=0 (hence d,d≥0, d=d) with ϵ=δ<1+∣d∣∣d−d∣ the third inequality of (5.25) fails. If a=0, a=0, then the first inequality of (5.25) yields ∣x2∣≤a∣d∣δ+ϵ. Since ∣v∣2≤1−δ, it is now not to difficult to contradict (5.26) for any sufficiently small ϵ,δ.
2. (ii)
a,a,d,d=0
Inequalities (5.25) yield ∣ax2∣≥a−ϵ−∣d∣δ, |ay|^{2}\leq\tfrac{\big{(}|d|\sqrt{\delta(1+\delta)}+\epsilon\big{)}^{2}}{\widetilde{a}-\epsilon-|d|\delta} and
[TABLE]
If ∣d∣=∣d∣ it is straightforward to obtain a contradiction for any ϵ,δ small enough.
If ∣d∣=∣d∣, then a∣d∣=a∣d∣ yields a=a, d=∣d∣eiϑ, d=∣d∣eiϑ with ϑ−ϑ=2lπ, l∈Z.
The right-hand (left-hand) side of the first (third) inequality of (5.25) thus gives
[TABLE]
If we choose ϵ,δ such that ∣δ5′∣,∣δ6′∣≤1, then using (4.9) it follows that
[TABLE]
where l1,l2∈Z. Hence
vx=(−1)l1+l2ei2ϑ−ϑ(1+δ5′)(1+δ6′)
and further
[TABLE]
Since 2ϑ−ϑ=l′π, l′∈Z, then combining the above inequality with (5.26) and using the triangle inequality yields a contradiction for every small ϵ,δ.
3. (iii)
d=d=0, a,a=0, a=a (since B=B)
The first equation of ((a)) and the last inequality of (5.24) give ∣x∣2=aa(1+δ1), ∣δ1∣≤aaϵ and ∣v∣2=1+δ2, ∣δ2∣≤δ, respectively. Further,
(4.9) leads to ∣x∣=aa(1+δ1′), ∣δ1′∣≤aaϵ and ∣v∣=1+δ2′, ∣δ2′∣≤δ≤21. Therefore we obtain \big{|}|xv|-\sqrt{\frac{\widetilde{a}}{a}}\big{|}\leq\sqrt{\frac{\widetilde{a}}{a}}\delta+\frac{3a}{2\widetilde{a}}\epsilon+\delta. From (5.26) for d=d=0 and applying the triangle inequality we deduce
From Lemma 3.4 (2) with (C3.4) for −ω=α=1, β=0 we obtain
[TABLE]
(a)
B=a⊕d, a≥0
(i)
B=a⊕d, a,d≥0
For c(s)=1, P(s)=[21s−1s−21s−1s] in (5.3) we obtain
(1\oplus-1,0_{2})\to\big{(}\begin{bmatrix}0&1\\
1&i\end{bmatrix},0\oplus d\big{)}, d>0.
We have the same equations as in ((a)).
Since ∣u∣2,∣v∣2≤δ (see (5.27)) we deduce that the first (third) of these equation fails for a=0, a=0, ϵ=δ<∣d∣+1a (a=0, d=0, ϵ=δ<∣d∣+1d).
Next, d=0, a=0 yields x2=aa+ϵ1, hence (1−δ)2≤∣2Re(xu)∣2≤∣2xu∣2≤4aa+ϵδ, which fails for any δ,ϵ chosen sufficiently small. Finally, for a,d=0 (hence a,d=0) the first and the last equation of ((a)) (with ∣u∣2,∣v∣2≤δ) give ∣ax2∣≥a−∣d∣δ−ϵ, ∣ay2∣≥d−∣d∣δ−ϵ, respectively. Provided that a,d≥ϵ+∣d∣δ we multiply these inequalities and a comparison to the second equation of ((a)) implies:
[TABLE]
which fails to hold for ϵ=δ<(∣d∣+1)(a+d)ad.
2. (ii)
B=[0bb0], b>0
We have equations ((c)). Its first and its last equation yield ∣ax2∣,∣ay2∣≤∣d∣δ+ϵ, what further with the second equation and ∣u∣2,∣v∣2≤δ (see (5.27)) gives
[TABLE]
Since b>0 this inequality fails for ϵ=δ<2(∣d∣+1)b.
2. (b)
B=[0bb0], b>0
(i)
B=[0bb0], b>0
The first equation of (5) for a=d=a=0 is 2bux=ϵ1, which implies ∣2Rexu∣≤bϵ.
For ϵ=δ<b+1b the first inequality of (5.27) then fails to hold.
2. (ii)
The first and the last equation of ((b)) give ∣2xu∣≤ba+ϵ, ∣2yv∣≤bd+ϵ.
Combined with the first two estimates in (5.27), it leads to
[TABLE]
Since ad=b2 then for a=d we either have ba<1 or bd<1. In both cases one of the above inequalities fails for ϵ=δ<b+1b−min{a,d}.
Moreover, when a=d=b, then c(s)=−1, P(s)=21[is−1−iss−1s] in (5.3) proves
(1\oplus-1,bI_{2})\to\big{(}\begin{bmatrix}0&1\\
1&i\end{bmatrix},\begin{bmatrix}0&b\\
b&0\end{bmatrix}\big{)}.
6. Case VI.
Setting u2=δ1, v2=δ2 (with ∣δ1∣,∣δ2∣≤δ), and rearranging the terms
in the first and the last equation of (5) for b=0, gives
[TABLE]
Since B=0 constants a,d do not both vanish (see Lemma 2.2). For a=0
at least one of the above equations fails, provided that ϵ=δ<∣d∣+1max{∣a∣,∣d∣}.
Further, if a>0 we obtain
|x|^{2}\leq\frac{1}{a}\big{(}|\widetilde{a}|+\epsilon+d\delta\big{)} and |y|^{2}\leq\frac{1}{a}\big{(}|\widetilde{d}|+\epsilon+d\delta\big{)},
and using (5.28) we get
[TABLE]
It is not difficult to contradict this inequality for any sufficiently small ϵ,δ.
2. (b)
B=[0bb0], b>0
(i)
B=[0bb1], b>0
Since 1=∣detA∣∣detA∣=∣detB∣∣detB∣=b2b2 (see (5.1)), we have (5) for a=d=a=0, d=1, b=b:
[TABLE]
Furthermore, Lemma 4.1 (2) gives detP=vx−uy=(−1)l+δ′, l∈Z, with ∣δ′∣≤b2ϵ(4max{1,b}+2). By combining it with the second equation of (Case VI((b))(i)) we obtain:
[TABLE]
If ϵ,δ are small, then for l even (odd), uy is close to [math] (close to 1) and vx is close to 1 (close to [math]), and hence xv (or uy) is close to (−1)k, k∈Z by (5.28).
Using the notation (5.4) for x,y,u,v we apply (4.4) to the third equation of (Case VI((b))(i)), to (5.30) and to the last estimate of (5.28), provided that bϵ+∣δ′∣+δ≤21. We deduce that
[TABLE]
(so φ−κ=(−1)l(kπ+ψl+ψl′−ψ)), where
∣sinψ∣≤2ϵ,
∣sinψl∣≤b22ϵ(b+4max{1,b}+2), ∣sinψl′∣≤b22ϵ(b+4max{1,b}+2)+2δ, l∈{0,1}.
Therefore using
∣sin(φ−κ)∣≤∣sinψl′∣+∣sinψl∣+∣sinψ∣ together with the squared second estimate of (5.28)
and the third equation of (Case VI((b))(i)) we get
[TABLE]
It is straightforward to choose ϵ,δ small enough so that this inequality fails.
2. (ii)
As 1=∣detA∣∣detA∣=∣detB∣∣detB∣=b2∣d∣, Lemma 4.1 gives detP=vx−uy=(−1)liei2ϑ+δ′, l∈Z, ∣δ′∣≤b∣d∣ϵ(4max{1,∣d∣}+2). Adding (subtracting) it from second equation of (Case VI((b))(ii)) yields
[TABLE]
We combine this with the last estimate in (5.28) and write it in the notation (5.4):
[TABLE]
where ∣δ1∣≤δ.
Using (4.4) we deduce
ψ=2ϑ−η−ϕ+lπ−kπ+2πm, m∈Z, |\sin\psi|\leq 2\big{(}\frac{2\epsilon}{b}+\delta+|\delta^{\prime}|\big{)},
while the first equation of (Case VI((b))(ii)) gives ψ′=ϕ+φ+2πm′, ∣sinψ′∣≤2ϵ. Thus ψ′+ψ=2ϑ+nπ, n∈Z and by applying sin we conclude
[TABLE]
which fails for \epsilon=\delta<|\sin\frac{\widetilde{\vartheta}}{2}|\big{(}4+\frac{4}{b}+\tfrac{2(4\max\{1,|\widetilde{d}|\}+2)}{b\sqrt{|\widetilde{d}|}}\big{)}^{-1} (remember 0<ϑ<π).
7. Case VII.
First, consider
(1\oplus-1,0\oplus\widetilde{d})\dashrightarrow\big{(}\begin{bmatrix}0&1\\
1&0\end{bmatrix},1\oplus 0\big{)}, d>0.
By applying (4.9) to the first and the last equation of
(5) for a=1, b=d=a=0 we get
[TABLE]
respectively.
Next, by manipulation of the third expression of (5.32) we deduce
[TABLE]
By multiply it with x, rearranging the terms and slightly simplify leads to
[TABLE]
To conclude, Lemma 4.1 (1) gives
∣detP∣=1+δ′, ∣δ′∣≤6∥E∥≤ν6δ with ν>0,
and combining it with the first equation of (5.32) and (5.33) eventually contradicts the above equation for any sufficiently small ϵ,δ.
For c(s)=1, P(s)=21[21+s−21+1+s2−21+s1+1+s] in (5.3)
we get
(1\oplus-1,0_{2})\to\big{(}\begin{bmatrix}0&1\\
1&0\end{bmatrix},1\oplus 0\big{)}.
2. (b)
Multiplying detP=vx−uy with b and then adding and subtracting it from the second equation of ((b)) yields
[TABLE]
respectively.
Multiplying the first (the second) equation by u (by v) and comparing it with the first (the last) equation of ((b)), multiplied by v (u), gives
[TABLE]
For v,u=0 it follows that
[TABLE]
(i)
B=[0bb0], b>0
The first and the last equation of ((b)) for a=d=0 immediately imply that \big{|}2b|ux|-|u|^{2}\big{|}\leq\epsilon and \big{|}2b|vy|-|v|^{2}\big{|}\leq\epsilon.
Further, (5.32) gives that 2∣xu∣,2∣yv∣≥1−δ and hence ∣u∣2,∣v∣2≥b(1−δ)−ϵ. For ϵ<2b we have u,v=0 (since δ≤21), so the first and the last equation of ((b)) yield 2bx=−u+uϵ1 and
2by=−v+vϵ1.
Therefore
[TABLE]
Adding the real parts of these equalities, applying the triangle inequality and using the first two equations of (5.32) with the lower estimates on ∣u∣2, ∣v∣2 gives
[TABLE]
which fails for ϵ=δ<4b+42b.
2. (ii)
B=a⊕d, 0<a≤d
If uv=0, then ((b)) fails for ϵ<a.
Next, for v=0
we easily validate the following calculation
[TABLE]
Combining this with the second and the third equation of (5.32), and comparing it with (5.35) for b=0, leads to
[TABLE]
Since 1=∣detAdetA∣=∣detBdetB∣=∣−b2ad∣ (see (5.1)), then b2=ad and Lemma 4.1 (2) yields detP=vx−uy=(−1)li+δ′, l∈Z, ∣δ′∣≤b2ϵ(4d+2). Thus either a=d=b=1 or we obtain a contradiction for any suitably small ϵ,δ.
For c(s)=1, P(s)=21[2s1s−2siis] in (5.3)
it follows that
(1\oplus-1,I_{2})\to\big{(}\begin{bmatrix}0&1\\
1&0\end{bmatrix},\begin{bmatrix}0&1\\
1&1\end{bmatrix}\big{)}.
3. (c)
The first (the second) equation of (5.32) yields that at least one of ∣x∣2,∣u∣2 (and ∣y∣2,∣v∣2) is greater or equal to 41 (remember δ≤21).
Since equations ((c)) for a=d=0 are the same as in ((c)) for a=1, d∈C, also inequalities (5.16) for a=1 are valid; if ∣x∣2≥41, then ∣du∣2≥41−ϵ (if ∣u∣2≥41 we have ∣x∣2≥4∣d∣−ϵ). Similarly holds for y,v as well, so:
[TABLE]
Applying (4.4) with \epsilon\leq\frac{1}{2}\big{(}\frac{1}{4}\min\{1,|d|\}-\epsilon\big{)} to the first and to the last equation of ((c)) for a=d=0, d=∣d∣eiϑ and x,y,u,v as in (5.4), leads to
[TABLE]
The last equation of (5.32) (using (4.4)) similarly gives
[TABLE]
Collecting everything together yields 21(ψ1+ψ2)+ψ=−2π−ϑ and
[TABLE]
Since Imd>0 we can easily choose ϵ, δ to contradict this inequality.
2. (ii)
B=a⊕d, 0<a≤d
The first two equations of (5.32) imply ∣xu∣,∣vy∣≥21(1−δ).
Hence ∣xvuy∣≥41(1−δ)2, so the second equation of ((c)) for b=0 (the last equation of (5.32)) yields that either ∣duv∣ or ∣xy∣ (either ∣xv∣ or ∣uy∣) or both are at least 21∣d∣(1−δ) (at least 21(1−δ)).
By setting x,y,u,v as in (5.4), d=∣d∣eiϑ and using (4.4) we also deduce:
[TABLE]
with ∣sinψ1∣≤∣d∣(1−δ)4ϵ, ∣sinψ2∣≤1−δ4δ, l1,l2∈Z. Hence ψ1+ψ2=ϑ+(2l1+2l2−1)π,
[TABLE]
which fails for ϵ=δ<4+5∣d∣+∣d∣∣sinϑ∣∣d∣∣sinϑ∣ (remember Imd>0).
8. Case VIII.
We have equations ((a)) for a=1, Imd>0.
By combining calculations (5.11), (5.12) for σ=−1 with the first two equations of ((a)) for a=1 and the first two inequalities of (5.41) we deduce
[TABLE]
By rearranging the terms and appying the triangle inequality we further get
[TABLE]
It is immediate that \big{|}a|u|^{2}-|uv|\big{|}\,|e^{2i(\phi-\eta)}+\tfrac{d}{a}|\leq(a+1)(\tfrac{\epsilon}{a}+\delta). Moreover, for a=d:
[TABLE]
It follows (see also (5.41)) that for sufficiently small ϵ,δ we have ∣x∣,∣u∣ (and ∣y∣,∣v∣) arbitrarily close and bounded (for a=d), so it is straightforward to get a contradiction with the second equation of ((a)).
Next, let a=d. By combining (5.11), (5.12) for σ=−1, a=d with equations (5.41), ((a)) for a=1, and slightly simplifying, we obtain
[TABLE]
By applying (4.4) to these equations with d=∣d∣eiϑ
we then deduce
[TABLE]
where l1,l2,l3,l4∈Z. It implies further that
ψ1+ψ3−ψ2−ψ4=2π(l1−l2+l3−l4)+ϑ, which yields an inequality that fails for ϵ=δ=8+(1+a)(1+∣d∣)∣Imd∣ (recall Imd>0):
Combining the first and the last equality of (5.41) with the first and the last equation of (5) for d=1, a=b=0, and applying the triangle inequality, we deduce
[TABLE]
These inequalities and (5.41) imply the upper estimates for ∣x∣,∣y∣,∣u∣,∣v∣ in case a=d, which further gives an inequality that fails for any ϵ,δ sufficiently small:
[TABLE]
Next, when a=d (hence b=d) it is convenient to conjugate the first pair of matrices with
21[212−1]:
[TABLE]
Using (5.7) for σ=−1 and (5) for a=41(4d+1), d=41(−4d+1), b=41, a=d, b=0, we can write the identities (5.11), (5.12) for a=d, σ=−1 in the form
[TABLE]
where ∣δ1′∣,∣δ2′∣,∣δ3′∣≤δ, k∈Z. By combining the first two and the last two equations (rearranging the terms and then multiplying the equations) we obtain
[TABLE]
respectively. By comparing the right-hand sides of the equations and rearranging the terms we eventually conclude
[TABLE]
where ∣ϵ′∣≤2(4d1+4d+dϵ+δ)(dϵ+δ)+2(4d1+dϵ+δ)(dϵ+δ). Thus \tfrac{1}{2d}\big{|}\sin(\phi-\varphi)\big{|}\leq\epsilon^{\prime} and either 2\big{|}\sin(\phi+\tfrac{k\pi}{2})\big{|}\leq\sqrt{\epsilon^{\prime}} or 2\big{|}\sin(\varphi+\tfrac{k\pi}{2})\big{|}\leq\sqrt{\epsilon^{\prime}} (or both). Observe also that if one of the expressions 2∣sin(ϕ+2kπ)∣ or 2∣sin(φ+2kπ)∣ is bounded by ϵ′, then by using the angle diffence formula for sine and applying the triangular inequality we deduce that the other expression is bounded by 1−ϵ′ϵ′+ϵ′.
In the same manner, but with x,u,y,v,ϕ,η,φ,κ,δ1′,δ2′,δ3′,k replaced by u,x,v,y,η, ϕ,κ,φ,−δ1′,−δ2′,−δ3′,k+1, respectively, in (5.12), (5.11) for a=d=b, σ=−1 and (5.42), we conclude that
2\big{|}\sin(\eta+\tfrac{(k+1)\pi}{2})\big{|},2\big{|}\sin(\kappa+\tfrac{(k+1)\pi}{2})\big{|}\leq\tfrac{\sqrt{\epsilon^{\prime}}+\epsilon^{\prime}}{\sqrt{1-\epsilon^{\prime}}}. It follows that
\big{|}\cos(\phi+\kappa)\big{|}=\big{|}\sin(\phi+\frac{k\pi}{2}+\kappa+\frac{(k+1)\pi}{2})\big{|}\leq 2\tfrac{\sqrt{\epsilon^{\prime}}+\epsilon^{\prime}}{\sqrt{1-\epsilon^{\prime}}}.
On the other hand, the third equation of (5.42) with its terms rearranged and simplified can be written as
[TABLE]
Provided that 8d1≥δ+dϵ, then applying (4.4) yields ∣sin(ϕ+κ)∣≤8d(δ+dϵ). We have a contradiction for any small enough ϵ,δ.
2. (b)
B=[0bb0], b>0
(i)
B=[0bb1], b>0
The first equation of (Case VI((b))(i)) and (5.41) yield that at least one of ∣x∣2 or ∣u∣2 is not larger that 2bϵ and the other is not larger than 2bϵ+δ. Similarly, the last equation of (Case VI((b))(i)) and the last equation of (5.41) give ∣y∣2,∣v∣2≤2b1+ϵ+δ. The second estimate of (5.41) finally implies a contradiction for any sufficiently small ϵ,δ:
We consider
\big{(}\begin{bmatrix}0&1\\
1&0\end{bmatrix},1\oplus 0\big{)}\approx\big{(}1\oplus-1,\begin{bmatrix},1&1\\
1&1\end{bmatrix}\big{)}\dashrightarrow(1\oplus-1,0\oplus d), d>0. The first and the last equation of (5) for a=b=0, a=b=d=1 yield that u2=d1+ϵ1 and v2=d1+ϵ4. Hence the first and the last estimate of (5.7) for σ=−1 give
[TABLE]
respectively. For d≥1 at least one of these inequalities fails (provided that δ+dϵ<1−d1), while for d<1
by combining the inequalities with (\frac{1-\epsilon}{d}-\delta)^{2}\leq\big{(}|\overline{u}v|-\delta\big{)}^{2}\leq|\overline{x}y|^{2} (see the second estimate of (5.7) for σ=−1) and simplifying, we get
[TABLE]
It is not difficult to choose ϵ,δ small enough to contradict this inequality.
9. Case IX.
\bigl{(}\begin{bmatrix}0&1\\
1&0\end{bmatrix},\widetilde{B}\bigr{)}\dashrightarrow\bigl{(}\begin{bmatrix}0&1\\
1&0\end{bmatrix},B\bigr{)}, detB=detB=0 (see Lemma 2.2 and (5.1))
Lemma 3.4 (2) with (C3.4) for α=ω=0, β=1 and with some ∣δ1∣,∣δ2∣,∣δ4∣≤δ gives
Due to the second equation of ((c)) for b=0 it follows that given a positive constant s≤161 (to be choosen) it suffices to consider the following two cases:
(i)
∣xy∣,∣duv∣≤s
It is clear that either ∣x∣2≤s or ∣y∣2≤s. We only consider ∣x∣2≤s (the case ∣y∣2≤s is treated similarly, we just replace x,y,u,v,a,d,ϑ,arga with y,x,v,u,d,a,arga,ϑ). The first equation of ((c)) yields ∣du2−a∣≤ϵ+s and (4.4) gives further ψ1=2η+ϑ−arga∈(−2π,2π) with ∣sinψ1∣≤∣a∣2(ϵ+s), provided that s+ϵ≤2∣a∣1. Since ∣duv∣≤s we also get ∣dv2∣≤∣a∣−ϵ−ss2≤∣a∣2s2. Applying (4.4) to the third equation of ((c)) and to the second equation of (5.43) implies
[TABLE]
By choosing s<min{4∣a∣1,8∣ad∣} then for any sufficiently small ϵ,δ we provide ∣sinψ1∣,∣sinψ2∣,∣sin2ψ∣≤21 (2ψ,ψ1,ψ2∈(−3π,3π)).
Thus 2ψ+ψ1−ψ2=ϑ+ϑ−arga−2kπ∈(−2π,2π)∖{0} (recall 0<ϑ,ϑ<π, ϑ=ϑ, a=1) and if in addition s<\min\big{\{}\big{|}\frac{\sin(\widetilde{\vartheta}+\vartheta-\arg\widetilde{a})}{4\widetilde{a}}\big{|},\sqrt{\frac{|\widetilde{a}\widetilde{d}\sin(\widetilde{\vartheta}+\vartheta-\arg\widetilde{a})|}{16}}\big{\}} the next inequality fails for any small ϵ,δ:
[TABLE]
2. (ii)
∣xy∣,∣duv∣≥s−δ
We write the first and the last equality of (5.43) in the form:
[TABLE]
hence either 8∣cos(η−ϕ)∣3,8∣cos(κ−φ)∣3≤δ or ∣v∣3≤δ or ∣y∣3≤δ or ∣u∣3≤δ or ∣x∣3≤δ (or more at the same time).
First, let 8∣cos(η−ϕ)∣3,8∣cos(κ−φ)∣3≤δ. Applying (4.4) to the second equation of ((c)) for b=0 yields
[TABLE]
Using the angle sum formula for sine and the triangle inequality we obtain an inequlity that fails for any small ϵ,δ:
[TABLE]
Next, suppose ∣v∣3≤δ. The first and the third equation of ((c)) lead to
\big{|}|x|-|\sqrt{d}u|\big{|}^{2}\leq\big{|}|x|^{2}-|du^{2}|\big{|}\leq|\widetilde{a}|+\epsilon and
\big{|}|y|^{2}-|\widetilde{d}|\big{|}\leq|y^{2}-\widetilde{d}|\leq\epsilon+|d|\sqrt[3]{\delta^{2}}.
Combining these facts with the second equation of ((c)) for b=0 gives
[TABLE]
For v=0 (hence ∣u∣=∣dv∣∣duv∣≥∣d∣4δs−δ) we get a contradiction for any small ϵ,δ.
If v=0, then ((c)) for b=0, yields
y2=d+ϵ4, xy=ϵ2 (hence x2=d+ϵ4ϵ22), du2=a+ϵ1−d+ϵ4ϵ22.
Combining this with the squared second equation of (5.43) gives the equality which fails to hold for any ϵ,δ sufficiently small (ϑ=ϑ, ∣d∣=∣d∣):
[TABLE]
The case ∣y∣3≤δ is due to a symmetry treated similarly, with x,y,du,dv,d,a replaced by du,dv,x,y,a,d; the cases when ∣u∣3≤δ or ∣x∣3≤δ are dealth likewise.
2. (b)
B=1⊕d, Imd>0, B=[0bb1], b>0
In this case it is convenient to conjugate (A,B)
with
21[212−1] and consider
[TABLE]
By setting y,v as in (5.4) we conclude from the first equation of (5) that
[TABLE]
Observe that the first summand on the right-hand side is smaller than δ (as in Case VIII we have (5.41)), thus we get the estimate ∣4+d∣216∣Im(d)∣∣xu∣≤∣4+d∣4ϵ+δ. Combining it with the first equality of (5.41) we deduce that ∣x∣2,∣u∣2≤16∣Im(d)∣∣4+d∣2(∣4+d∣4ϵ+δ)+δ.
Similarly as in ((b)) we write the third equation of (5) in the form
[TABLE]
and using the first and the last equality of (5.41) we deduce
∣4+d∣216∣Im(d)∣∣yv∣≤∣4+d∣4+4ϵ+δ, ∣y∣2,∣v∣2≤16∣Im(d)∣∣4+d∣2(∣4+d∣4+4ϵ+δ)+δ. Applying the triangle inequality to the squared second equality of (5.41) now leads to an inequality that fails for any small ϵ,δ:
[TABLE]
3. (c)
B=[0bb1], b>0, B=1⊕d, d=∣d∣eiϑ, π>ϑ>0, ∣d∣=b2 (see (5.1))
We have equations ((b)) for a=1, b=0 and (5.35) is valid for b=0, a=1, u,v=0, thus uv=1+ϵ4bdetP+ϵ2. (If u=0 (v=0) the first (the last) of equations ((b)) fails for ϵ<1 (ϵ<∣d∣).)
Further, from (Case VII((b))(ii)), (5.43) we obtain (−1)k+δ2=e−2iκdetP+δ4vu. Similarly, with x,y,u,v,φ,κ replaced by y,x,v,u,ϕ,η in (Case VII((b))(ii)) we get (−1)k+δ2=e−2iηdetP+δ1uv. Since ∣detBdetB∣=∣−b2ad∣=1, then Lemma 4.1 (2) yields detP=(−1)liei2ϑ+δ′, l∈Z, ∣δ′∣≤b2ϵ(4∣d∣+2).
We now collect everything together to deduce:
[TABLE]
∣ϵ′∣,∣ϵ′′∣,∣ϵ′∣,∣ϵ′′∣≤Kmax{ϵ,δ}, where a constant K>0 can be obtained by a straightforward computation. If ϵ,δ are small enough it contradicts π>ϑ>0.
10. Case X.
(02,1⊕σ)⇢(A,B)
(a)
σ=0
(i)
A=1⊕eiθ, 0≤θ≤π
For θ=π we have ∥P∥2≤δ (Lemma 3.4 (2) for (C3.4) with α=0), hence 1=∥B∥=∥PTBP−F∥≤4δ∥B∥+ϵ fails for ϵ=δ<4∥B∥+11.
If θ=π then P(s)=a+d+2b1[110s], c(s)=1 in (5.3) proves
(02,1⊕0)→(1⊕−1,B) for B=[0bb0], b>0 and B=a⊕d,d>0.
2. (ii)
A=[011i]
From Lemma 3.4 (2) with (C3.4) for α=0 we get ∣u∣2≤δ. Further, the first equation of (5) for a=b=0, a=1 is du2=1+ϵ1 and fails for δ=ϵ<1+∣d∣1.
Taking P(s)=2b1+i[s−1−iss2s2] and
P(s)=a1⊕s with c(s)=1 in (5.3), we prove
(0_{2},1\oplus 0)\to\big{(}\begin{bmatrix}0&1\\
1&i\end{bmatrix},\begin{bmatrix}a&b\\
b&d\end{bmatrix}\big{)} for a=d=0,b>0 and a>0, b=0, d∈C, respectively.
3. (iii)
A=[0τ10], 0≤τ≤1
When a=0 and d=0 we can take
P(s)=a1⊕s
and
P(s)=[0d1s0] with c(s)=1 in (5.3), respectively, to prove
(0_{2},1\oplus 0)\to\big{(}\begin{bmatrix}0&1\\
\tau&0\end{bmatrix},\begin{bmatrix}a&b\\
b&d\end{bmatrix}\big{)}.
Next, let a=d=0 (hence 0≤τ<1 by Lemma 2.2). From Lemma 3.4 (2) for (C3.4) with α=0 we get ∣ux∣≤δ, so the first equation of (5) for a=d=0, a=1
fails for ϵ=δ≤1+2∣b∣1.
4. (iv)
A=1⊕0
To prove (02,1⊕0)→(1⊕0,a⊕1), a≥0, and
(0_{2},1\oplus 0)\to\big{(}1\oplus 0,\begin{bmatrix}0&1\\
1&0\end{bmatrix}\big{)}
we can take
P(ϵ)=[s1ss]
and
P(s)=21[ss−1s2s2], both with c(s)=1 in (5.3), respectively.
From Lemma 3.4 (2) with (C3.4) for α=0 we have ∣x∣2≤δ. When B=a⊕0, a>0 the first equation of (5) for a=b=d=0 (i.e. 0=a+ϵ2) fails for ϵ<a.
2. (b)
σ=1
(i)
A=1⊕0
We prove (0_{2},I_{2})\to\big{(}1\oplus 0,\begin{bmatrix}0&1\\
1&0\end{bmatrix}\big{)} by taking
P(ϵ)=21[ss−1is−is−1] with c(s)=1 in (5.3).
For B=a⊕d, a,d≥0 we have ∣x∣2,∣y∣2≤δ (Lemma 3.4 (2) with (C3.4) for α=0). Combining it with ((a)) for a=d=1 yields ∣du∣2,∣dv∣2≥1−ϵ−aδ and ∣duv∣≤ϵ+aδ. Thus we obtain a contradiction for ϵ=δ<2(a+1)1.
2. (ii)
A=[0010]
By Lemma 3.4 (2) for (C3.4) with τ=0 we have ∣ux∣,∣uy∣,∣vx∣,∣vy∣≤δ.
It implies that either ∣x∣2,∣y∣2≤δ or ∣u∣2,∣v∣2≤δ (or both). If ∣x∣2,∣y∣2≤δ then the first and the last equation of (5) for b=0, a=d=1 give ∣d∣∣u∣2,∣d∣∣v∣2≥1−2bδ−∣a∣δ−ϵ, and the the application of the triangle inequality to second equation further gives
[TABLE]
which fails for ϵ=δ<2+4b+2∣a∣1.
The case ∣u∣2,∣v∣2≤δ is for the sake of symmetry treated in a similar fashion, with u,v,x,y,a,d replaced by x,y,u,v,d,a, respectively.
11. Case XI.
(1⊕0,B)→(1⊕0,B)
From Lemma 3.4 (2) with (C3.4) for α=1 we get that
[TABLE]
(a)
B=[0110] (hence
B=a⊕d,
a≥0, d∈{0,1} by Lemma 2.2 we have)
The first equation of ((b)) for b=1 and ∣x∣2≥1−δ≥21 (see (5.45)) yield ∣u∣≤2a+ϵ,
while the second equation (multiplied with y) and the third equation give 21x(d+ϵ4)+buy2=bxvy+buy2=yϵ2.
Using the upper (the lower) estimates on ∣u∣, ∣y∣ (on ∣x∣) we get an inequality that fails for d=0 and small enough ϵ,δ:
From (5) for b=0 and by slightly rearranging the terms we obtain
[TABLE]
Thus using (5.45) and applying the triangle inequality to ((b)) we deduce
[TABLE]
(i)
B=a⊕d, a≥0, d∈{0,1}
Combining all the estimates in ((b)) for b=0, d=1 yields
[TABLE]
which fails for a=a and appropriately small ϵ,δ. (Note that d=1 implies a=a (see (5.1), Lemma 2.2) for B=B.)
Next, when B=a⊕0, B=a⊕0, a=a, the first inequality of ((b)) for d=0 fails for ϵ=δ<1+a∣a−a∣.
Finally,
c(s)=1,
P(s)=[1a−a0s] in (5.3)
proves
(1⊕0,a⊕0)→(1⊕0,a⊕1).
2. (ii)
B=[0110]
We have B=a⊕1, a>0. The inequalities in ((b)) for a=d=0, b=1 then give
[TABLE]
which fails for ϵ,δ small enough.
12. Case XII.
(1⊕0,a⊕d)⇢(1⊕eiθ,B), 0≤θ≤π, d∈{0,1}, a≥0, ad=0 (see (5.1))
(a)
0≤θ<π
By Lemma 3.4 (2) for (C3.4) we have
∣y∣2,∣v∣2≤δ.
If d=1, then the third equation of (5) fails for \epsilon=\delta<\big{(}|a|+2|b|+|d|+1\big{)}^{-1}.
Next, let d=0. To prove (disprove) the existence of a path we need to solve (to see that there are no solutions) the equations (5.2) for arbitrarily small E,F. Given any small s>0
we must (not) find x,y,u,v,c and ϵ1,ϵ2,ϵ4 with
∣ϵ1∣,∣ϵ2∣,∣ϵ4∣≤s satisfying (5) and such that the expressions in Lemma 3.4 for (C3.4) are bounded from above by s. Observe that by choosing v,y sufficiently small we achieve that the last two equations in (5) for b=d=0 are fulfilled trivially for some small ϵ2,ϵ4, and the first two expressions in Lemma 3.4 (C3.4) are arbitrarily small. It is thus left to consider the remaining first equation of (5) and the third expression of Lemma 3.4 (C3.4); note that they do not depend on y,v:
[TABLE]
The second equation can be rewritten as
\big{(}|x|^{2}+\cos\theta|u|^{2}\big{)}+i\sin\theta|u|^{2}=c^{-1}(1+\delta_{1})
and it is clearly equivalent to compare only the (squared) moduli of both sides
[TABLE]
By (4.9) for ∣δ1∣≤31 it further implies ∣x∣4+2cosθ∣ux∣2+∣u∣4=1+δ2 for some δ2∈R, ∣δ2∣≤3∣δ1∣≤3s.
By writting x,u in view of (5.4) we can see (5.48) as
[TABLE]
Clearly, (5.49) for δ2=0 is equivalent to (5.48) for δ1=0.
Next, we observe the range of the function f(r,t,β)=∣ar2eiβ+2brt+dt2e−iβ∣ given with a constraint r4+2r2t2cosθ+t4=1 (see (3.11)).
Provided that (R,T)=(r2, t2) lies on on an ellipse R2+2RTcosθ+T2=1, we can further assume that either tr or rt is any real nonegative number. With a suitable choice of β we achieve finally that the following expression (and hence f) vanishes:
[TABLE]
If the maximum of f given with a constraint is M, its range is thus [0,M].
Provided that ϕ−η, ∣x∣, ∣u∣ (corresponding to β, r, t in (3.11)) are chosen appropriately, the modulus of the left-hand side of the first equation in (5.49) (hence ∣ax2+2bux+du2∣) can be any number from the interval [0,M] and the second equation of (5.49) for δ2=0 (thus ∣x∣2+eiθ∣u∣2−c−1=0) is valid, simultaneously.
By a suitable choice of ϕ+η (see (5.49)) we arrange ax2+2bux+du2∈R≥0 and so (1⊕0,a⊕0)→(1⊕eiθ,B) for
a∈[0,M] is proved. In particular, for a=b=0 we get M=∣d∣, for d=b=0 we obtain M=∣a∣, and for a=d=0 we have M=2∣b∣max{rt∣r4+2(cosθ)r2t2+t4=1}. Moreover, if θ=0, 0≤a≤d, d>0 we get M=d, and to prove the existence of a path we take
P(s)=a+d1[a+dia−a0s] for 0≤a≤a≤d
and
P(s)=d−a1[d−aa−a0s] for 0≤a<a≤d, both with c(s)=1 in (5.3).
Furthermore
[TABLE]
Applying (4.9) to 1+δ2′ and its root yields 41+δ2′=1+δ3, ∣δ3∣≤∣δ2′∣. For r′=r41+δ2′, t′=t41+δ2′, δ2′∈R, ∣δ2′∣≤∣δ2∣≤3s with ∣r∣,∣t∣≤1 (note that r4+2cosθr2t2+t4=1), we then deduce
[TABLE]
Since M is the maximum of f with respect to a constraint (3.11),
it follows that the maximum of f on a compact domain given by ∣r4+2cosθr2t2+t4−1∣≤∣δ2∣≤3s
is at most M_{1}=M+3s\big{(}|a|+2|b|+|d|\big{)}. Assuming that the second equation of (5.49) is valid, the modulus of the left-hand side of the first equation of (5.49) is then at most M1; if a>M, this equation fails for s<1+3∣a∣+6∣b∣+3∣d∣)a−M.
2. (b)
From Lemma 3.4 (2) for (C3.4) for σ=−1, α=1, ω=0 we get
[TABLE]
As in Case II(a) we obtain ∣a−d∣∣xy∣≤dδ+ϵ (see (5.9)).
Further, from the first equation of ((a)) for a=0 we get a∣x∣2+d∣u∣2≥∣ax2+du2∣≥ϵ, and by combining it with the first estimate of (5.50) we deduce
∣a+d∣∣x∣2≥d(1−δ)+ϵ.
Moreover,
[TABLE]
When a=d we have
∣y∣2,∣v∣2≤∣a−d∣2(d(1−δ)+ϵ)(a+d)(dδ+ϵ)2+δ (see the last estimate in (5.50)), and if further d=1 the third equation of ((a)) gives a contradiction for any ϵ,δ small enough.
Next, for a=d and d=1 (hence a=0) the first equation of ((a)) and the first estimate in (5.50) yield
[TABLE]
which fails for ϵ=δ<1+dd. The existence of a path for d=0 follows from (5.3) for
[TABLE]
2. (ii)
B=[0bb0], b>0
We take u2=2b−b+b2+a2, x2=2bb+b2+a2, y=v=s, c(s)=1 to satisfy (5.3).
13. Case XIII.
(1\oplus 0,\widetilde{a}\oplus\widetilde{d})\to\big{(}\begin{bmatrix}0&1\\
1&0\end{bmatrix},B\big{)}, d∈{0,1}, ad=0 (see Lemma 2.2 and (5.1))
From the first estimate in (5.51) and the first equation of ((c)) for a=0 we deduce that ∣du∣2,∣x∣2≥4min{1,∣d∣}−ϵ (see Case VII(c)Case VII((c))(i) (5.38) for the same conclusion).
The last equation of ((c)) for d=1 further yields that either ∣y∣2≥21−ϵ≥41 or ∣v∣2≥2∣d∣1−ϵ≥4∣d∣1. By combining these facts we get that either ∣xv∣2 or ∣uy∣2 is at least \frac{1}{4|d|}\big{(}\frac{\min\{1,|d|\}}{4}-\epsilon\big{)}.
Similarly, either ∣xy∣2 or ∣duv∣2 is greater or equal to 41(4min{1,∣d∣}−ϵ).
As in Case VII(c)Case VII((c))(ii) we obtain precisely (5.39), (5.40), but now with estimates
[TABLE]
which give a contradiction for any suitably small ϵ,δ as well.
2. (ii)
B=a⊕0
To prove the existence of a path for any a≥0, then given an arbitrary s>0 it is sufficient to solve the system of equations ax2+du2=a+ϵ1 and 2Re(xu)=−1+δ1, where ∣ϵ1∣,∣δ1∣≤s. Indeed, provided that y,v are chosen small enough, the last two equations in ((c)) for b=0 are satisfied for some ∣ϵ2∣,∣ϵ4∣≤s, and the absolute values of expressions in Lemma 3.4 (C3.4) for k=0 are bounded by s (it implies ∥E∥≤ρs for some constant ρ>0 independent of s; see also (5.2)).
In view of the notation (5.4) with d=∣d∣eiϑ, 0<ϑ<π we obtain
[TABLE]
where ∣ϵ1∣,∣δ1∣≤s. We now observe the range of the following functions:
[TABLE]
Since g→∞ as r→∞ and g(r)→0 as r→sin2Δ4∣d∣, d=0 or r→0, d=0, the range of g (and hence f with a constraint) is (0,∞). Thus the modulus of the left-hand side of the first equation in (5.52) can be any number from the interval (0,∞) and the second equation in (5.52) is valid at the same time, provided that ϕ−η, ∣u∣, ∣x∣ (corresponding to β,t,r in (5.53)) are chosen appropriately.
Finally, a suitable choice of ϕ+η guarantees that the left-hand side of (5.52) is real and positive. Thus (5.52) is solvable fo any a≥0 (and any d, Imd>0).
2. (b)
B=[0bb1], b>0
(i)
B=0⊕1
Let x,u be as in (5.4). Since ∣ux∣≥∣Re(xu)∣≥1−δ≥21 (see (5.51)), then assuming ϵ≤2b and applying (4.4) to the first equation of ((b)) for a=0 yields
[TABLE]
We multiply the first equation of ((b)) for a=0 with uv to get
2bvx=(u2ϵ1−1)uv.
Together with the last estimate in (5.51) and providing ϵ≤b we obtain:
[TABLE]
Further, we have
[TABLE]
Combining it with the second equation of ((b)) for b=0 (estimate of (5.55)) gives
Multiplying the third equation of ((b)) for d=1 by u2, and rearranging the terms gives 2b(uv)(uy)+(uv)2=(1+ϵ4)u2. Combining this with (5.55) and with the upper estimate on ∣uv∣ finally yields an inequality that fails for any ϵ,δ small enough.
2. (ii)
B=a⊕0, a≥0
To prove the existence of a path we use the same argument as in Case XIII(a)Case XIII((a))(ii). For a given s>0 it suffices to solve the equations 2bux+u2=a+ϵ1, ∣ϵ1∣≤s (see ((b))) and −Re(xu)=1+δ1, ∣δ1∣≤s (k=1 in Lemma 3.4 (C3.4)).
In view of (5.4) they can be written in the form
[TABLE]
As in the case mentioned before we define the function given with constraint:
[TABLE]
Since g(r)=f(r,r1,π)=∣2b−r2∣ with g(2b)=0 and g(r)→∞ as r→∞, the range of g (and hence f) is [0,∞). Thus the left-hand side of the first equation in (5.56) can be any number from [0,∞) and the second equation in (5.56) stays valid for δ1=0, simultaneously, provided that ϕ−η, ∣u∣, ∣x∣ (corresponding to β,t,r above), ϕ+η (makes the left-hand side of (5.56) real and positive) are chosen appropriately.
3. (c)
B=1⊕0
It is equivalent to consider
\big{(}1\oplus 0,\widetilde{B}\big{)}\dashrightarrow\big{(}1\oplus-1,\begin{bmatrix}1&1\\
1&1\end{bmatrix}\big{)}\approx\big{(}\begin{bmatrix}0&1\\
1&0\end{bmatrix},1\oplus 0\big{)}.
In case B=a⊕0, a≥0 we apply (5.3) for c(s)=1 and
P(s)=[p2+1−p0s2],
p=\left\{\begin{array}[]{ll}s^{-1},&\widetilde{a}=0\\
\frac{|1-\widetilde{a}|}{2\sqrt{\widetilde{a}}},&\widetilde{a}>0\end{array}\right..
When B=0⊕1 (d=1, a=0) and a=b=d=1 we get from (5):
[TABLE]
Hence u+x=(−1)l1′ϵ1 and v+y=(−1)l2′(1+ϵ4′) with ∣ϵ4′∣≤∣ϵ4∣≤ϵ, l1′,l2′∈Z (see (4.9)).
From the first estimate of (5.50)
it follows that either ∣x∣2≥41 or ∣u∣2≥41 (note δ≤21), and furthermore u+x=(−1)l1′ϵ1 yields ∣x∣,∣u∣≥21−ϵ.
Next, on one hand we have ∣y∣+∣v∣≥∣y+v∣≥1−ϵ≥21 and on the other hand (see (5.50)):
[TABLE]
hence ∣y∣,∣v∣≥41−δ. Using the lower estimates on x, y, u, v and
applying (4.4) to u+x=(−1)l1′ϵ1 and to ∣xy−uv∣≤δ (see (5.50)) and symplifying, gives
[TABLE]
Further we denote y,v as in (5.4) and manipulate the following expression:
From (5.59) it follows that either ∣x∣2,∣y∣2≤δ or ∣v∣2,∣y∣2≤δ or ∣u∣2,∣v∣2≤δ.
If ∣v∣2,∣y∣2≤δ the last equation of (Case XIV) with d=1 fails for ϵ=δ<1+∣a∣+∣d∣+2b1.
Next, if ∣x∣2,∣y∣2≤δ the first equation of (Case XIV) for a=0 and the last estimate of (5.59) give
[TABLE]
For d=0, ϵ=δ<1+∣a∣+2b(1+τ)−12b(1+τ)−1 we immediately get a contradiction. The last equation of (Case XIV) for d=1 and the second equation of (Case XIV) for b=0 yield
[TABLE]
(Note that for v=0 the last equation of (Case XIV) with d=1 fails for ϵ=δ<1+∣a∣+2b1.)
Using this and (5.61) we conclude that
[TABLE]
It is straightforward to see that this fails for any sufficiently small ϵ,δ.
The case ∣u∣2,∣v∣2≤δ is done in a similar manner, with a,x,u replaced by d,y,v, respectively.
2. (b)
As in Case XII(a) we argue that (5.2) has solutions for arbitrarily small E, F precisely when given s∈(0,1] we can solve the system of equations
[TABLE]
where ∣ϵ1∣,∣δ1∣≤s.
By choosing y,v small enough the remaining two equations of (Case XIV) are satisfied for some ∣δ2∣,∣δ4∣≤s and the moduli of the first three expressions in Lemma 3.4 (C3.4) for α=1 are bounded by s.
(i)
[0bb0], b>0
Observe that since ∣c∣=1 the second equation of (5.62) is equivalent to the equation which compares the moduli (squared) of both sides of the equation:
[TABLE]
In particular it follows that 1−τ1+s≥∣xu∣≥1+τ1−s. Using the first equation of (5.62) for a=d=0 we further get \big{|}\widetilde{a}-2b|ux|\big{|}\leq\epsilon, and thus
[TABLE]
Clearly, when a∈[1+τ2b,1−τ2b] we see that the above inequality fails for every sufficiently small s>0. For a∈[1+τ2b,1−τ2b] we choose u,x such that 2bxu=a and so that R+iT=xu=2baxx lies on an ellipse (1+τ)2R2+(1−τ)2T2=1 (see (5.63)). Therefore (5.62) for a=d=ϵ1=δ1=0 is satisfied and it yields the existence of the path.
2. (ii)
B=[abbd], ∣a∣+∣d∣=0, b>0
Set x0=∣x0∣eϕ0, u0=∣u0∣eiη0 and let ux=u0x0=R0eiβ0 (with β0=ϕ0−φ0) solve the equation 0=ax^{2}+2bux+du^{2}=u^{2}\big{(}a(\frac{x}{u})^{2}+2b\frac{x}{u}+d\big{)} (see (5.62) for a=0).
A direct computation shows that by choosing |x_{0}|^{2}=R_{0}\big{|}(1+\tau)\cos(-\beta_{0})+(1-\tau)\sin(-\beta_{0})\big{|}^{-1} we assure that x0u0 fullfils the second equation of (5.62) for δ1=0 and some ∣c∣=1.
Denoting further x,y,u,v as in (5.4) with ∣x∣=r, ∣u∣=t, a=∣a∣eiι, d=∣d∣eiϑ the equations (5.62) for ϵ1=δ1=0
are seen as
[TABLE]
Define a function with constraint
[TABLE]
where r,t≥0,β∈R.
Fot t=rΓ(β), \Gamma(\beta)=\big{|}(1+\tau)\cos(\beta)+i(1-\tau)\sin(\beta)\big{|}^{-1} we set
[TABLE]
Since Γ is bounded we get g(r)→∞ as r→∞. As also g(∣x0∣)=0 (see the above observation) and g is continuous for r∈[∣x0∣,∞), the range of g (and hence f with constraint) is [0,∞).
Thus the left-hand side of the first equation in (5.64) can be any number from the interval [0,∞) and the second equation in (5.64) stays valid, provided that r,t, ϕ+η (ϕ−η=β0) are chosen appropriately.
Thus (5.64) and hence (5.62) are solvable for any a≥0, which proves the existence of a path.
3. (iii)
B=a⊕d, a,d∈{0,1}, ad=0
If a=0, d=0 (or d=0, a=0) we take y=v=s, x=aa+s (or u=aa+s) and u (or x) such that xu=1+τ1 in (5.3) to prove (1\oplus 0,\widetilde{a}\oplus 0)\to\big{(}\begin{bmatrix}0&1\\
\tau&0\end{bmatrix},a\oplus d\big{)}.
3. (c)
Since ∣vy∣≤δ by (5.59), then either ∣y∣2≤δ or ∣v∣2≤δ (or both). Let us consider ∣y∣2≤δ. Then the last equation of (Case XIV) for d=1 yields that ∣d∣∣v∣2≥1−2bδ−∣a∣δ. If d=0, we get a contradiction for δ<∣a∣+2b1, while for d=0 we get
∣u∣≥∣x∣1−δ≥δ1−δ∣d∣1−2bδ−∣a∣δ (by (5.59) for τ=0 we have ∣vx∣≤δ, \big{|}|\overline{x}u|-1\big{|}\leq\delta). Note that x=0 would yield 1≤δ, and δ=0 would imply x=y=0, both is not possible. Applying the triangle inequality to the right-hand side of the second equation of (Case XIV) for b=0, and using the lower estimates for ∣u∣,∣v∣ give:
[TABLE]
By taking ϵ,δ sufficiently small this inequality fails.
The case ∣v∣2≤δ is for the sake of symmetry done in a similar manner, but with a,x,u replaced by d,y,v.
4. (d)
If a=d=0 the second equation of (Case XIV) for a=d=0, b=1 fails for ϵ=δ<1+2b1.
Next, suppose ∣a∣+∣d∣=0.
As in Case XIV(a) we get that either ∣x∣2,∣y∣2≤δ or ∣v∣2,∣y∣2≤δ or ∣u∣2,∣v∣2≤δ.
If ∣x∣2,∣y∣2≤δ the first (the last) equation of (Case XIV) for a=0 (or d=0) and ∣xu∣≤1+δ (see (5.59) for τ=0) yield ∣du2∣≤∣a∣δ+2b(1+δ)+ϵ (or ∣dv2∣≤ϵ+2bδ+∣a∣δ). Combining this with the second equation of (Case XIV) for b=1 gives
[TABLE]
which failes for any sufficiently small ϵ,δ.
In case ∣u∣2,∣v∣2≤δ we get allmost the same estimates, but with a,x,y replaced by d,u,v.
Finally, let ∣v∣2,∣y∣2≤δ.
Since ∣ux∣≤1+δ, then either ∣x∣2≤1+δ or ∣u∣2≤1+δ (or both). If ∣ax2∣≤∣a∣(1+δ) (or ∣du2∣≤∣d∣(1+δ)) the first equation of (Case XIV) for a=0 yields ∣du∣2≤ϵ+2bδ+∣a∣(1+δ) (or ∣ax∣2≤ϵ+2bδ+∣d∣(1+δ)). Thus we deduce that
[TABLE]
From the second equations of (Case XIV) for b=1 (using the triangle inequality) we get
[TABLE]
which fails for any sufficiently small ϵ,δ.
15. Case XV.
(1\oplus 0,\widetilde{a}\oplus\widetilde{d})\dashrightarrow\big{(}\begin{bmatrix}0&1\\
1&i\end{bmatrix},B\big{)}, d∈{0,1}, a≥0, ad=0 (see Lemma 2.2, (5.1))
(a)
B=[0bb0], b>0
If d=0 then P(s)=21[2ba(1−i)1+iss], c(s)=−i in (5.3) yields
(1\oplus 0,\widetilde{a}\oplus 0)\to\big{(}\begin{bmatrix}0&1\\
1&i\end{bmatrix},\begin{bmatrix}0&b\\
b&0\end{bmatrix}\big{)}.
Next, let a=0, d=1.
From Lemma 3.4 (2) for (C3.4) with α=1, c−1=eiΓ we get
[TABLE]
We have equations ((b)); the first one for a=0 (the third one for d=1) gives 2bϵ≥∣ux∣ (and 2b1−ϵ≤∣vy∣). The last estimate in (5.65) further implies
\frac{\epsilon}{2b}\geq|ux|\geq\big{|}\mathop{\rm Re}\nolimits(\overline{x}u)\big{|}\geq\frac{|\cos\Gamma|-\delta}{2} and
[TABLE]
Multiplying the second (rearranged) equation of ((b)) with bv, and using the estimates on ∣x∣, ∣u∣, ∣v∣, ∣vy∣ we get an inequality that fails to hold for ϵ,δ small enough:
[TABLE]
2. (b)
B=a⊕d, a≥0
(i)
B=0⊕1 (d=1, a=0)
We have equations ((a)) for a=0, d=1 and it is apparent that the last of these equations fails to hold for ϵ=δ<ϵ+∣d∣δ1. Next, we set set uv=δ1, v2=δ2, ∣δ1∣,∣δ2∣≤δ. The last equation of ((a)) for d=1 gives ay2=1+ϵ4−dδ2. Combining it with the second and the first equation of ((a)) for a=0 then yields ax2=ay2(axy)2=1+ϵ4−dδ2(ϵ2−dδ1)2 and du2=ϵ1−ax2=ϵ1−1+ϵ4−dδ2(ϵ2−dδ1)2, respectively. From the last estimate in (5.65) we finally get an inequality that fails for a,d=0 and ϵ,δ small enough:
[TABLE]
When d=a=0, d=1, a=0 the first and the last equation of ((a)) for a=0 give x2=aϵ1, ∣y∣2=∣a1+ϵ4∣≤2a3. Further, the last estimate in (5.65) first leads to the upper estimate ∣u∣2≤1+δ≤23 (hence 2a3ϵ≥∣xu∣2≥∣Re(xu)∣) and then the lower estimate ∣u∣2≥21−a3ϵ≥21−a3ϵ. Combining these facts with the first estimate in (5.65) implies the inequality that fails for \epsilon=\delta<\frac{3}{4}\big{(}(1+\sqrt{a})^{2}+\frac{9}{2a}\big{)}^{-1}:
[TABLE]
2. (ii)
B=a⊕0, a≥0
In case d=0, a=0 the path is guaranteed by
P(s)=[aais0], c(s)=−i in (5.3).
Next, let a=0 (hence d>0). The first equation of ((a)) for a=0 and the last estimate in (5.65) imply 1+δ≥∣u∣2≥da−ϵ, a contradiction for a>d, ϵ=δ<d+1a−d. When a≤d the existence of a path for a=0 and a>0 is proved with (5.3) for c(s)=1,
P(s)=21[s−1ss0] and c(s)=d1(d2−a2−a),
P(s)=2ad1[d2−a22as0], respectively.
It is left to consider the case a>0, d=0.
Using a similar argument as in Case XIII(a)Case XIII((a))(ii) we prove the existence of a path by solving the system of equations
ax2+du2=a and \big{|}2\mathop{\rm Re}\nolimits(\overline{x}u)+i|u|^{2}\big{|}=1; by choosing v,y sufficiently small we assure that the remaining expressions in Lemma 3.4 for (C3.4) are arbitrarily small (then ∥E∥ is arbitrarily small) and the last two equations in ((a)) are satisfied.
By setting x=reiϕ, u=teiη with d=∣d∣eiϑ we can write the above equations as:
[TABLE]
We show that the range of the following function given with a constraint is [0,∞):
[TABLE]
If ϑ=2lπ, l∈Z then β=2π, t=1 satisfy the constraint and
f(r,1,\frac{\pi}{2})=\big{|}ar^{2}-|d|\big{|}, thus the range of f is [0,∞).
When ϑ=2lπ, l∈Z then by taking β=2ϑ+π the constraint can be rewritten as r^{2}\big{(}4t^{2}\sin^{2}(\frac{\vartheta}{2})\big{)}=1-t^{4}. Letting t→0 implies r→∞ and hence
f(r,t,\frac{\vartheta+\pi}{2})=\big{|}ar^{2}-|d|\big{|}\to\infty. For r0=a∣d∣ and by setting t0 to be the positive root of the equation t4+4a∣d∣sin2(2ϑ)t2−1=0
we have f(r0,t0,2ϑ+π)=0. The range of f is again [0,∞). Therefore (5.66) has a solution for any a≥0, provided that ϕ−η, ∣u∣, ∣x∣ (corresponding to β, t, r), ϕ+η are chosen appropriately.
Thic completes the proof of the theorem.
∎
*Acknowledgement.
*The research was supported by Slovenian Research Agency (grant no. P1-0291).
The author wishes to thank M. Slapar for helpful discussions considering the topic of the paper.
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