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\title{Algebras of Smooth Functions and Holography of Traversing Flows}
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\author[1]{Gabriel Katz}
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\date{\small \em Received: 1 January 1970 Accepted: 1 January 1970 Published: 1 January 1970}
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\begin{abstract}
Let be a smooth compact manifold and a vector field on which admits a smooth function such that 0. Let be the boundary of . We denote by the algebra of smooth functions on and by the algebra of smooth functions on . With the help of ( ), we introduce two subalgebras and of and prove (under mild hypotheses) that the topological tensor product. Thus the topological algebras and , viewed as , allow for a reconstruction of . As a result, and allow for the recovery of the smooth topological type of the bulk .
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\section[{Introduction}]{Introduction}\par
It is classically known that the normed algebra C 0 (X) of continuous real-valued functions on a compact space X determines its topological type {\ref [GRS]}, {\ref [Ga]}, {\ref [Br]}. In this context, X is interpreted as the space of maximal ideals of the algebra C 0 (X). In a similar spirit, the algebra C ? (X) of smooth functions on a compact smooth manifold X (the algebra C ? (X) is considered in the Whitney topology {\ref [W3]}) determines the smooth topological type of X {\ref [KMS]}, {\ref [Na]}. Again, X may be viewed as the space of maximal ideals of the algebra C ? (X).\par
Recall that a harmonic function h on a compact connected Riemannian manifold X is uniquely determined by its restriction to the smooth boundary ?X of X. In other words, the Dirichlet boundary value problem has a unique solution in the space of harmonic functions. Therefore, the vector space H(X) of harmonic functions on X is rigidly determined by its restriction (trace) H ? (X) := H(X)| ?X to the boundary ?X. As we embark on our journey, this fact will serve us as a beacon. This paper revolves around the following question: Which algebras of smooth functions on the boundary ?X can be used to reconstruct the algebra C ? (X) and thus the smooth topological type of X?\par
Remembering the flexible nature of smooth functions (in contrast with the rigid harmonic ones), at the first glance, we should anticipate the obvious answer "None!". However, when X carries an additional geometric structure, then the question, surprisingly, may have a positive answer. The geometric structure on X that does the trick is a vector field (i.e., an ordinary differential equation), drawn from a massive class of vector fields which we will introduce below.\par
Let X be a compact connected smooth (n + 1)-dimensional manifold with boundary and v a smooth vector field admitting a Lyapunov function f : X ? R so that df (v) > 0. We call such vector fields traversing. We assume that v is in general position with respect to the boundary ?X and call such vector fields boundary generic (see {\ref [K1]} or {\ref [K3]}, Definition 5.1, for the notion of boundary generic vector fields). Temporarily, it will be sufficient to Abstract-Let be a smooth compact manifold and a vector field on which admits a smooth function such that 0. Let be the boundary of . We denote by the algebra of smooth functions on and by the algebra of smooth functions on . With the help of ( ), we introduce two subalgebras and of and prove (under mild hypotheses) that the topological tensor product. Thus the topological algebras and , viewed as , allow for a reconstruction of . As a result, and allow for the recovery of the smooth topological type of the bulk .X X X X v f :X ? R df (v) > ?X C ? (X ( X C ? (?X ( ?X v, f A(v) B(f ( C ? (?X ( C ? (X)? A(v) ?B(f ), A(v ( B(f ) C ? (X ( A(v ( B(f (\par
boundary data {\ref [Br]} Louis de Branges, L.,\par
The Stone-Weierstrass theorem, Proc.\par
Amer. Math.\par
Soc., (1959) 10 ( {\ref 5}), 822-824.
\section[{Ref}]{Ref}\par
think of the boundary generic vector fields v as having only v-trajectories that are tangent to the boundary ?X with the order of tangency less than or equal to dim(X). Section 3 contains a more accurate definition. Informally, we use the term "holography" when some residual structures on the boundary ?X are sufficient for a reconstruction of similar structures on the bulk X.\par
Given such a triple (X, v, f ), in Section 3, we will introduce two subalgebras,A(v) = C ? (?X, v) and B(f ) = (f ? ) * (C ? (R))\par
, of the algebra C ? (?X), which depend only on v and f , respectively. By Theorem 3.1, A(v) and B(f ) will allow for a reconstruction of the algebra C ? (X). In fact, the boundary data, generated by these subalgebras, lead to a unique (rigid) "solution"C ? (X) ? C ? (?X, v) ? (f ? ) * (C ? (R)),\par
the topological tensor product of the two algebras. As a result, the pair A(v), B(f ), "residing on the boundary", determines the smooth topological type of the bulk X and of the 1-dimensional foliation F(v), generated by the v-flow.\par
Let X be a compact connected smooth (n + 1)-dimensional manifold with boundary ? 1 X = def ?X (we use this notation for the boundary ?X to get some consistency with similar notations below), and v a smooth traversing vector field, admitting a smooth Lyapunov function f : X ? R. We assume that v is boundary generic.\par
We denote by? + 1 X(v) the subset of ? 1 X where v is directed inwards of X or is tangent to ? 1 X. Similarly, ? ? 1 X(v) denotes the subset of ? 1 X where v is directed outwards of X or is tangent to ? 1 X.\par
Let F(v) be the 1-dimensional oriented foliation, generated by the traversing v-flow. We denote by ? x the v-trajectory through x ? X. Since v is traversing and boundary generic, each ? x is homeomorphic either a closed segment, or to a singleton {\ref [K1]}.\par
In what follows, we embed the compact manifold X in an open manifold X of the same dimension so that v extends to a smooth vector field v on X, f extends to a smooth function f on X, and d f (v) > 0 in X. We treat ( X, v, f ) as a germ in the vicinity of (X, v, f ).\par
We say that a boundary generic and traversing vector field v possesses Property A, if each v-trajectory ? is either transversal to ? 1 X at some point of the set ? ? ? 1 X, or ? ? ? 1 X is a singleton x and ? is quadratically tangent to ? 1 X at x. ?\par
A traversing vector field v on X induces a structure of a partially-ordered set (? 1 X, v ) on the boundary ? 1 X: for x, y ? ? 1 X, we write y\par
x if the two points lie on the same v-trajectory ? and y is reachable from x by moving in the v-direction.\par
We denote by T (v) the trajectory space of v and by Î?" : X ? T (v) the obvious projection. For a traversing and boundary generic v, T (v) is a compact space in the topology induced by Î?". Since any trajectory of a traversing v intersects the boundary ? 1 X, we get that T (v) is a quotient of ? 1 X modulo the partial order relation v .\par
The map Î?" : X ? T ( v ) for a traversally generic (vertical) vector field v on a disk with 4 holes. The trajectory space is a graph whose verticies are of valencies 1 and 3. The restriction of Î?" to ? 1 X is a surjective map Î?" ? with finite fibers of cardinality 3 at most; a generic fiber has cardinality 2.\par
A traversing and boundary generic v gives rise to the causality (scattering) mapC v : ? + 1 X(v) ? ? ? 1 X(v (2.1) that takes each point x ? ? + 1 X(v) to the unique consecutive point y ? ? x ? ? ? 1 X(v) that can be reached from x in the v-direction. If no such y = x is available, we put C v (x) = x.\par
We stress that typically C v is a discontinuous map (see Fig. {\ref 2}).\par
We notice that, for any smooth positive function ? : X ? R + , we have C ??v = C v ; thus the causality map depends only on the conformal class of a traversing vector field v. In fact, C v depends only on the oriented foliation F(v), generated by the v-flow.\par
In the paper, we will discuss two kinds of intimately related holography problems. The first kind amounts to the question: To what extend given boundary data are sufficient for reconstructing the unknown bulk and the traversing v-flow on it, or rather, the foliation F(v)? This question may be represented symbolically by the two diagrams:?Holographic Reconstruction Problem (? 1 X, v , ) ?? ?? (X, F(v)), (2.2) (? 1 X, v , f ? ) ?? Issue ersion I V II ( F )\par
Algebras of Smooth Functions and Holography of Traversing Flows ( Figure {\ref 1}:T(v) X v Notes x C (x) v x+ x- y C (y) v X 1 + X 1 - X 2 + X 1 + X 1 - v v An example of the causality map C v : ? + 1 X(v) ? ? ? 1 X(v). Note the essential discontinuity of C v in the vicinity of x.\par
where v denotes the partial order on boundary, defined by the causality map C v , and the symbol " ??\par
??" points to the unknown ingredients of the diagrams.\par
The second kind of problem is: Given two manifolds, X 1 and X 2 , equipped with traversing flows, and a diffeomorphism ? ? of their boundaries, respecting the relevant boundary data, is it possible to extend ? ? to a diffeomorphism/homeomorphism ? : X 1 ? X 2 that respects the corresponding flows-generated structures in the interiors of the two manifolds? This problem may be represented by the commutative diagrams:?Holographic Extension Problem (? 1 X 1 , v 1 ) inc ?? (X 1 , F(v 1 )) â??" ? ? â??" ?? ? (? 1 X 2 , v 2 ) inc ?? (X 2 , F(v 2 )) (? 1 X 1 , v 1 , f ? 1 ) inc ?? (X 1 , F(v 1 ), f 1 ) â??" ? ? â??" ?? ? (? 1 X 2 , v 2 , f ? 2 ) inc ?? (X 2 , F(v 2 ), f 2 ),\par
where inc denotes the inclusion of spaces, accompanied by the obvious restrictions of functions and foliations. The symbol "â??" ?? " indicates the unknown maps in the diagrams.\par
These two types of problems come in a big variety of flavors, depending on the more or less rich boundary data and on the anticipated quality of the transformations ? (homeomorphisms, PD-homeomorphisms, Hölder homeomorphisms with some control of the Hölder exponent, and diffeomorphisms with different degree of smoothness).\par
Let us formulate the main result of {\ref [K4]}, Theorem 4.1, which captures the philosophy of this article and puts our main result, Theorem 3.1, in the proper context. Theorem 2.1 reflects the scheme depicted in (2.4). Ref boundary generic vector fields v 1 , v 2 on X 1 and X 2 , respectively. In addition, assume that v 1 , v 2 have Property A from Definition 2.1.\par
Let a smooth orientation-preserving diffeomorphism ? ? : ? 1 X 1 ? ? 1 X 2 commute with the two causality maps:C v 2 ? ? ? = ? ? ? C v 1 Then ? ? extends to a smooth orientation-preserving diffeomorphism ? : X 1 ? X 2 such that ? maps the oriented foliation F(v 1 ) to the oriented foliation F(v 2 ).\par
Let us outline the spirit of Theorem 2.1's proof, since this will clarify the main ideas from Section 3. The reader interested in the technicalities may consult {\ref [K4]}.\par
Proof. First, using that v 2 is traversing, we construct a Lyapunov functionf 2 : X 2 ? R for v 2 . Then we pull-back, via the diffeomorphism ? ? , the restriction f ? 2 := f 2 | ? 1 X 2 to the boundary ? 1 X 2 . Since ? ? commutes with the two causality maps, the pull back f ? 1 = def (? ? ) * (f ? 2 ) has the property f ? 1 (y) > f ? 1 (x)\par
for any pair y x on the same v 1 -trajectory, the order of points being defined by the v 1 -flow. Equivalently, we getf ? 1 (C v 1 (x)) > f ? 1 (x) for any x ? ? + 1 X(v 1 ) such that C v 1 (x) = x.\par
As the key step, we prove in {\ref [K4]} that suchf ? 1 extends to a smooth function f 1 : X 1 ? R that has the property df 1 (v 1 ) > 0. Hence, f 1 is a Lyapunov function for v 1 . Recall that each causality map C v i , i = 1, 2, allows to view the v i -trajectory space T (v i ) as the quotient space (? 1 X i ) \{C v i (x) ? x\}, where x ? ? + 1 X i (v i\par
) and the topology in T (v i ) is defined as the quotient topology. Using that ? ? commutes with the causality maps C v 1 and C v 2 , we conclude that ? ? induces a homeomorphism ? T : T (v 1 ) ? T (v 2 ) of the trajectory spaces, which preserves their natural stratifications.\par
For a traversing v i , the manifold X i carries two mutually transversal foliations: the oriented 1-dimensional F(v i ), generated by the v i -flow, and the foliation G(f i ), generated by the constant level hypersurfaces of the Lyapunov function f i . To avoid dealing the singularities of F(v i ) and G(f i ), we extend f i to fi : Xi ? R and v i to vi on Xi so that d fi (v i ) > 0. This generates nonsingular foliations F(v i ) and G( fi ) on Xi . By this construction, F(v i )| Xi = F(v i ) and G( fi )| X i = G(f i ). Note that the "leaves" of G(f i ) may be disconnected, while the leaves of F(v i ), the v i -trajectories, are connected. The two smooth foliations, F(v i ) and G( fi ), will serve as a "coordinate grid" on X i : every point x ? X i belongs to a unique pair of leaves ? x ? F(v i ) and L x := f ?1 i (f i (x)) ? G( fi ). Conversely, using the traversing nature of v i , any pair (y, t), where y ? ?x ? ? 1 X i and t ? [f ? i (? x ? ? 1 X i )] ? R, where [f ? i (? x ? ? 1 X i )] denotes the minimal closed interval that contains the finite set f ? i (? x ? ? 1 X i ), determines a unique point x ? X i .\par
Note that some pairs of leaves L and ? may have an empty intersection, and some components of leaves L may have an empty intersection with the boundary ? 1 X i .\par
In fact, using that f i is a Lyapunov function, the hyprsurfaceL = f ?1 i (c) intersects with a v i -trajectory ? if and only if c ? [f ? i (? ? ? 1 X i )].\par
Since the two smooth leaves, ?y and f ?1 i (f i (z)), depend smoothly on the points y, z ? ? 1 X i and are transversal, their intersection point ?y? f ?1 i (f i (z)) ? Xi depends smoothly on (y, z) ? (? 1 X i ) × (? 1 X i ), as long as f ? i (z) ? [f ? i (? y ?? 1 X i )]. Note that pairs (y, z), where y, z ? ? 1 X i , with the property f ? i (z) ? f ? i (? y ? ? 1 X i ) give rise to the intersections ?y ? f ?1 i (f i (z)) that belong to ? 1 X i . Now we are ready to extend the diffeomorphism ? ? to a homeomorphism ? : X 1 ? X 2 .\par
In the process, following the scheme in (2.4), we assume the the foliations F(v i ) and of the Lyapunov functions f i on X i (i = 1, 2) do exist and are "knowable", although we have access only to their traces on the boundaries.\par
Take any x ? X 1 . It belongs to a unique pair of leaves L x ? G(f 1 ) and ? x ? F(v 1 ). We define ?(x) = x ? X 2 , where x is the unique point that belongs to the intersection off ?1 2 (f 1 (x)) ? G(f 2 ) and the v 2 -trajectory ? = Î?" ?1 2 (? T (? x )). By its construction, ?| ? 1 X 1 = ? ? . Therefore, ? induces the same homeomorphism ? T : T (v 1 ) ? T (v 2 ) as ? ? does.\par
The leaf-hypersurface f ?1 2 (f 1 (x)) depends smoothly on x, but the leaf-trajectory ? = Î?" ?1 2 (? T (? x )) may not! Although the homeomorphism ? is a diffeomorphism along the v 1 -trajectories, it is not clear that it is a diffeomorphism on X 1 (a priori, ? is just a Hölder map with a Hölder exponent ? = 1/m, where m is the maximal tangency order of ?'s to ? 1 X). Presently, for proving that ? is a diffeomorphism, we need Property A from Definition 2.1. Assuming its validity, we use the transversality of ? x somewhere to ? 1 X to claim the smooth dependence of Î?" ?1 2 (? T (? x )) on x. Now, since the smooth foliations F(v i ) and G( fi ) are transversal, it follows that x = ?(x) depends smoothly on x. Conjecturally, Property A is unnecessary for establishing that ? is a diffeomorphism.\par
Note that this construction of the extension ? is quite explicit, but not canonic. For example, it depends on the choice of extension off ? 1 := (? ? ) * (f ? 2 ) to a smooth function f 1 : X 1 ? R,\par
which is strictly monotone along the v 1 -trajectories. The uniqueness (topological rigidity) of the extension ? may be achieved, if one assumes knowing fully the manifolds X i , equipped with the foliation grids F(v i ), G(f i ) and the Lyapunov function f i . In Theorem 3.1, we will reflect on this issue.\par
The next theorem (see {\ref [K4]}, Corollary 4.3) fits the scheme in (2.2). It claims that the smooth topological type of the triple \{X, F(v), G(f )\} may be reconstructed from the appropriate boundary-confined data, provided that Property A is valid.\par
(Holography of Traversing Flows) Let X be a compact connected smooth (n + 1)-dimensional manifold with boundary, and let v be a traversing boundary generic vector field, which possesses Property A.\par
Then the following boundary-confined data:? the causality map C v : ? + 1 X(v) ? ? ? 1 X(v), ? the restriction f ? : ? 1 X ? R of the Lyapunov function f ,\par
are sufficient for reconstructing the triple (X, F(v), f ), up to diffeomorphisms ? : X ? X which are the identity on the boundary ? 1 X.\par
Proof. We claim that, in the presence of Property A, the data \{C v , f ? \} on the boundary ? 1 X allow for a reconstruction of the triple (X, F(v), f ), up to a diffeomorphism that is the identity on ? 1 X.\par
Assume that there exist two traversing flows (X 1 , F(v 1 ), f 1 ) and(X 2 , F(v 2 ), f 2 ) such that ? 1 X 1 = ? 1 X 2 = ? 1 X, \{C v 1 , f ? 1 \} = \{C v 2 , f ? 2 \} = \{C v , f ? \}.\par
Applying Theorem 2.1 to the identity diffeomorphism ? ? = id ? 1 X , we conclude that it extends to a diffeomorphism ? :X 1 ? X 2 that takes \{F(v 1 ) ? ? 1 X 1 , f ? 1 \} to \{F(v 2 ) ? ? 1 X 2 , f ? 2 \}.\par
Unfortunately, Corollary 2.1 and its proof are not very constructive. They are just claims of existence: at the moment, it is not clear how to build the triple (X, F(v), f ) only from the boundary data (? 1 X, C v , f ? ).
\section[{Notes}]{Notes}(v) (v) R X Î?" Embedding ? : X ? T (v) × R, produced by Î?" : X ? T (v) and f : X ? R.\par
Fortunately, the following simple construction ([K4], Lemma 3.4), shown in Fig. {\ref 3}, produces an explicit recipe for recovering the triple (X, F(v), f ) from the triple (? 1 X, C v , f ? ), but only up to a homeomorphism.\par
As we have seen in the proof of Theorem 2.1, the causality map C v determines the quotient trajectory space T (v) canonically. Let f : X ? R be a Lyapunov function for v.\par
The pair (F (v), f ) gives rise to an embedding ? : X ? T (v) × R, defined by the formula ?(x) = ([? x ], f (x)), where x ? X and [? x ] ? T (v) denotes the point-trajectory through x.
\section[{The dependece x}]{The dependece x}\par
[? x ] is continuous by the definition of the quotient topology in T (v). Note that ? maps each v-trajectory ? to the line [?] × R, and, for any c ? R, each (possibly disconnected) leaf G c := f ?1 (c ) to the slice T (v) × c of T (v ) × R of the embedding ?, each trajectory ? ? F(v) may be identified with the closed interval [f ? (? ? ? 1 X)] ? R, and the vector field v| ? with the constant vector field ? u on R.\par
Consider now the restriction ? ? of the embedding ? to the boundary ? 1 X. Evidently, the image of ?? : ? 1 X ? T (v) × R bounds the image ?(X) = [?]?T (v) [f ? (? ? ? 1 X)].\par
Therefore, using the product structure in T (v) × R, ? ? (? 1 X) determines ?(X) canonically. Hence, ?(X) depends on C v and f ? 1 only! Note that ? is a continuous 1-to-1 map on a compact space, and thus, a homeomorphism onto its image. Moreover, the topological type of X depends only on C v : the apparent dependence of ?(X) on f ? is not crucial, since, for a given v, the space Lyap(v) of Lyapunov functions for v is convex.\par
The standing issue is: How to make sense of the claim "? is a diffeomorphism"? Section 3 descibes our attempt to address this question (see Lemma 3.3 and Theorem 3.1).\par
In what follows, we are inspired by the following classical property of functional algebras: for any compact smooth manifolds X, Y , we have an algebra isomorphismC ? (X × Y ) ? C ? (X) ?C ? (Y )\par
, where ? denotes an appropriate completion of the algebraic tensor product C ? (X) ? C ? (Y ) {\ref [Grot]}.\par
The trajectory space T (v), although a singular space, carries a surrogate smooth structure {\ref [K3]}. By definition, a function h : In the same spirit, we may talk about diffeomorphisms ? T : T (v) ? T (v) of the trajectory spaces, as maps that induce isomorphisms of the algebra C ? (T (v)).T (v) ? R is smooth if its pull-back Î?" * (h) : X ? R is a smooth function on X. As a subspace of C ? (X), the C ? (T (v)) is formed exactly by the smooth functions g : X ? R, whose directional derivatives L v g vanish in X. If L v (g) = 0 and L v (h) = 0, then L v (g ? h) = L v (g) ? h + g ? L v (h) = 0. Thus, C ? (T (v)) is indeed a subalgebra of C ? (X).\par
If two (v-invariant) functions from C ? (T (v)) take different values at a point [?] ? T (v), then they must take different values on the finite set ? ?? 1 X ? ? 1 X. Therefore, the obvious restriction homomorphism res ? T :C ? (T (v)) ? C ? (? 1 X)\par
, induced by the inclusion ? 1 X ? X, is a monomorphism. We denote its image by C ? (? 1 X, v). Thus, we get an isomorphismres ? T : C ? (T (v)) ? C ? (? 1 X, v). We think of the subalgebra C ? (? 1 X, v) ? C ? (? 1 X\par
) as an integral part of the boundary data for the holography problems we are tackling.\par
Let ? k : J k (X, R) ? X be the vector bundle of k-jets of smooth maps from X to R. We choose a continuous family semi-norms | ? | k in the fibers of the jet bundle ? k and use it to define a sup-norm ? k for the sections of ? k . We denote by jet k the obvious map C ? (X, R) ? J k (X, R) that takes each function h to the collection of its k-jets \{jet k\par
x (h)\} x?X . The Whitney topology {\ref [W3]} in the space C ? (X) = \{h : X ? R\} is defined in terms of the countable family of the norms \{ jet k (h) k \} k?N of such sections jet k (h) of ? k . This topology insures the uniform convergence, on the compact subsets of X, of functions and their partial derivatives of an arbitrary order. Note also that jetk (h 1 ?h 2 ) k ? jet k (h 1 ) k ? jet k (h 2 ) k for any h 1 , h 2 ? C ? (X).\par
Any subalgebra A ? C ? (X) inherits a topology from the Whitney topology in C ? (X). In particular, the subalgebraC ? (T (v)) ? C ? (X, v) does.\par
As a locally convex vector spaces, C ? (T (v)) and C ? (R) are then nuclear ([DS], {\ref [Ga]}) so that the topological tensor product C ? (T (v)) ? C ? (R) (over R) is uniquely defined as the completion of the algebraic tensor product C ? (T (v)) ? C ? (R) {\ref [Grot]}.\par
We interpret C ? (T (v)) ? C ? (R) as the algebra of "smooth" functions on the productT (v) × R and denote it by C ? (T (v) × R). The intersection C ? (T (v)) ? (f ) * (C ? (R)) = R, the space of constant func- tions on X.\par
Proof. If a smooth function h : X ? R is constant on each v-trajectory ? and belongs to (f ) * (C ? (R)), then it must be constant on each connected leaf of G(f ) that intersects ?. Thus, such h is constant on the maximal closed connected subset A ? ? f ?1 (f (?)) that contains ?. Each trajectory ?, homeomorphic to a closed interval, has an open neighborhood such that, for any trajectory ? from that neighborhood, we have A ? ? A ? = ?. Since X is connected, any pair ?, ? of trajectories may be connected by a path ? ? X. Using the compactness of ?, we conclude that the function h must be a constant along ?. Therefore, h is a constant globally.
\section[{Let us consider two subalgebras}]{Let us consider two subalgebras}, f * (C ? (R)) ? C ? (X) and (f ? ) * (C ? (R)) ? C ? (? 1 X),\par
the second one is assumed to be a "known" part of the boundary data. Proof. The restriction operator H ? f is an algebra epimorphism, since any composite function
\section[{The restriction operator H}]{The restriction operator H}? f : f * (C ? (R)) ? (f ? ) * (C ? (R)) to the boundary ? 1 X is an epimorphism of algebras. If the range f ? (? 1 X) of f ? is a connected closed interval of R (which is the case for a connected ? 1 X), then H ? f is an isomorphism. © 2023 Global Journals? ? f ? , where ? ? C ? (R), is the restriction to ? 1 X of the function ? ? f . Lemma 3.1. Lemma 3.2.\par
[Grot] Grothendieck, A., Produits tensoriels topologiques et espaces nucléaires (French), Providence: American Mathematical Society, (1966), ISBN 0 -8218-1216-5.
\section[{Ref}]{Ref}\par
On the other hand, when f? (? 1 X) is a connected subset of R, we claim that H ? f is a monomorphism. Indeed, take a function ? ? C ? (R), such that ? ? f ? ? 0, but ? ? f is not identically zero on X.\par
Then there is x ? X such that ? ? f (x) = 0. On the other hand, by the hypothesis, f (x) = f ? (y) for some y ? ? 1 X. By the assumption, f ? ? ? ? 0, which implies that ?(f ? (y)) = 0. This contradiction validates the claim about H ? f being a monomorphism. Therefore, when f? (? 1 X) is a connected interval, H ? f is an isomorphism of algebras.\par
Consider the homomorphism of algebrasP : C ? (T (v)) ? f * (C ? (R)) ? C ? (X) that takes every finite sum i h i ? (f ? g i ), where h i ? C ? (T (v)) ? C ? (X) and g i ? C ? (R), to the finite sum i h i ? (g i ? f ) ? C ? (X). Recall that, by Lemma 3.1, C ? (T (v)) ? (f ) * (C ? (R)) = R, the constants. For any linearly independent \{h i \} i , this lemma implies that if i h i ? (g i ? f ) ? 0, then \{g i ? f ? 0\} i ; therefore, P is a monomorphism.\par
Let us compare the, so called, projective crossnorms \{ ? k \} k?Z + (see (3.1)) of an element? = i h i ? (f ? g i )\par
and the norms of the elementP(?) = i h i ? (f ? g i ).\par
By comparing the Taylor polynomial of the product of two smooth functions with the product of their Taylor polynomials, we get that, for all k ? Z + ,? k = def inf i h i k ? (f ? g i ) k ? P(?) k , (3.1)\par
where inf is taken over all the representations of the element ?? C ? (T (v)) ? f * (C ? (R)) as a sum i h i ? (f ? g i ).\par
Here we may assume that all \{h i \} i are linearly independent elements and so are all \{f ? g i \} i ; otherwise, a simpler representation of ? is available.\par
By the inequality in (3.1), P is a bounded (continuous) operator. As a result, by continuity, P extends to an algebra homomorphismP : C ? (T (v)) ? f * (C ? (R)) ? C ? (X)\par
whose source is the completion of the algebraic tensor productC ? (T (v)) ? f * (C ? (R)).\par
The embedding ? : X ? T (v) × R (introduced in the end of Section 2 and depicted in Fig. {\ref 3}) induces an algebra epimorphism? * : C ? (T (v)) ? C ? (R) id ? f * ?? C ? (T (v)) ? f * (C ? (R)) P ?? C ? (X). (3.2)\par
Moreover, the map P is an isomorphism.\par
Proof. First, we claim that the subalgebraP C ? (T (v)) ? f * (C ? (R)) ? C ? (X)\par
satisfies the three hypotheses of Nachbin's Theorem {\ref [Na]}. Therefore, by {\ref [Na]}, the P-image of C ? (T (v)) ? f * (C ? (R)) is dense in C ? (X)
\section[{Ref}]{Ref}\par
(1) For each x ? X, there is a function q ? C ? (T (v))?f * (C ? (R)) such that q(x) = 0. Just take q = f ? (t + c), where c > min X f and t : R ? R is the identity.\par
(2) For each x, y ? X, there is a function q? C ? (T (v)) ? f * (C ? (R)) such that q(x) = q(y) (i.e., the algebra C ? (T (v)) ? f * (C ? (R))) separates the points of X) If f (x) = f (y), q = f will do. If f (x) = f (y), but [? x ] = [? y ], then there is a v-invariant function h ? C ? (T (v)\par
) such that h(x) = 1 and h(y) = 0. To construct this h, we take a transversal section S x ? X of the v-flow in the vicinity of x such that all the v-trajectories through S x are distinct from the trajectory ? y . We pick a smooth function h : S x ? R such that h is supported in int(S x ), vanishes with all its derivatives along the boundary ?S x , and h(x) = 1. Let S denote the set of v-trajectories through S x . Of course, h extends to a smooth function h ? : S ? R so that h ? is constant along each trajectory from S. We denote by h ? the obvious extension of h ? by the zero function. Finally, restriction h of h ? to X separates x and y.\par
(3) For each x ? X and w ? T x X, there is a function q? C ? (T (v)) ? f * (C ? (R))\par
such that dq x (w) = 0. Let us decompose w = av + bw ? , where a, b ? R and the vector w ? is tangent to the hypersurfce S x = f ?1 (f (x)). Then, if a = 0, then df (w) = 0. If a = 0, then the there is a function h : S x ? R which, with all its derivatives, is compactly supported in the vicinity of x in S x and such that d hx (w ? ) = 0. As in the case (2), this function extends to a desired function h ? C ? (T (v)). Now put q = h ? 1.\par
As a result, the image of P :C ? (T (v)) ? f * (C ? (R)) ?? C ? (X) is dense. Therefore, P and, thus, (?) * : C ? (T (v)) ? C ? (R) ?? C ? (X) are epimorphisms.\par
Let us show that P is also a monomorphism. Take a typical element? = ? i=1 h i ? (f ? g i ) ? C ? (X, v) ? f * (C ? (R)),\par
viewed as a sum that converges in all the norms ? k from (3.1). We aim to prove that if P(?) = ? i=1 h i ? (f ? g i ) vanishes on X, then ? = 0. For each point x ? int(X), there is a small closed cylindrical solid H x ? int(X) that contains x and consists of segments of trajectories through a small n-ball D n ? f ?1 (f (x)), transversal to the flow. Thus, the product structure D 1 × D n of the solid H x is given by the v-flow and the Lyapunov function f : X ? R.\par
We localize the problem to the cylinder H x . Consider the commutative diagram C ? (X, v) ? f * (C ? (R)) P ?? C ? (X). â??" res ?res â??" res C ? (D n ) ? C ? (D 1 ) ? Q ?? C ? (H x ), (3.3) where res : C ? (X) ? C ? (H x ) is the natural homomorphism, (res ?res ) ? i=1 h i ? (f ? g i ) = def ? i=1 h i | D n ? (f ? g i )| D 1 , and Q ? i=1 hi ? gi = def ? i=1 hi ? gi for hi ? C ? (D n ), gi ? C ? (D 1 ). ©\textbf{2023}
\section[{Notes}]{Notes}\par
Since Q is an isomorphism {\ref [Grot]} and P(?) = 0, it follows from (3.5) that ? ? ker(res ?res ) for any cylinder H x . After reshuffling terms in the sum, one may assume that all the functions \{h i | D n \} i are linearly independent. Using that the functions h i | D n and (f ? g i )| D 1 depend of the complementary groups of coordinates in H x , we conclude that these functions must vanish for any H x ? int(X). As a result, ? = 0 globally in int(X) and, by continuity, ? vanishes on X.\par
Consider now the "known" homomorphism of algebras(? ? ) * : C ? (T (v)) ? C ? (R) ?res ? T ? (f ? ) * ?? ?? C ? (? 1 X, v) ? (f ? ) * (C ? (R)) R? ?? C ? (? 1 X),\textbf{(3.4)}\par
utilizing the boundary data. Here, by the definition of C ? (? 1 X, v), res ? T : C ? (T (v)) ? C ? (? 1 X, v) is an isomorphism, and R? denotes the completion of the bounded homomorphism R ? that takes each element i h i ?(f ? ?g i ), where h i ? C ? (? 1 X, v) and g i ? C ? (R), to the sum i h i ? (g i ? f ? ).\par
The next lemma shows that the hypotheses of Theorem 3.1 are not restrictive, even when ? 1 X has many connected components.\par
Any traversing vector field v on a connected compact manifold X admits a Lyapunov function f : X ? R such that f (X) = f (? 1 X).\par
Proof. Note that, for any Lyapunov function f , the image f (? 1 X) is a disjoint union of finitely many closed intervals \{I k = [a k , b k ]\} k , where the index k reflects the natural order of intervals in R. We will show how to decrease, step by step, the number of these intervals by deforming the original function f . Note that the local extrema of any Lyapunov function on X occur on its boundary ? 1 X and away from the locus ? 2 X(v) where v is tangent to ? 1 X. Consider a pair of points A k+1 , B k ? ? 1 X \textbackslash ? 2 X(v) such that f (A k+1 ) = a k+1 and f (B k ) = b k , where a k+1 > b k . Then we can increase f in the vicinity of its local maximum B k so that the B k -localized deformation f of f has the property f (B k ) > f (A k+1 ) and f is a Lyapunov function for v. This construction decreases the number of intervals in f (? 1 X) in conparison to f (? 1 X) at least by one.\par
We are ready to state the main result of this paper.\par
Assuming that the range f ? (? 1 X) is a connected interval of R, 1 the algebra C ? (X) is isomorphic to the subalgebraC ? (? 1 X, v) ? (f ? ) * (C ? (R)) ? C ? (? 1 X) ?C ? (? 1 X).\par
Moreover, by combining (3.2) with (3.4), we get a commutative diagram {\ref (3.5)} C ? (T (v)) ? f * (C ? (R)) R ?? C ? (X) â??" id ? H ? f â??" res C ? (? 1 X, v) ? (f ? ) * (C ? (R)) R? ?? C ? (? 1 X),\begin{figure}[htbp]
\noindent\textbf{}\includegraphics[]{image-2.png}
\caption{\label{fig_0}(}\end{figure}
\begin{figure}[htbp]
\noindent\textbf{}\includegraphics[]{image-3.png}
\caption{\label{fig_2}}\end{figure}
\begin{figure}[htbp]
\noindent\textbf{}\includegraphics[]{image-4.png}
\caption{\label{fig_3}}\end{figure}
\begin{figure}[htbp]
\noindent\textbf{}\includegraphics[]{image-5.png}
\caption{\label{fig_5}}\end{figure}
\begin{figure}[htbp]
\noindent\textbf{}\includegraphics[]{image-6.png}
\caption{\label{fig_6}}\end{figure}
\backmatter
\subsection[{Notes}]{Notes}\par
whose vertical homomorphism id ? H ? f and the horizontal homomorphism R are isomorphisms, and the vertical epimorphism res is the obvious restriction operator.\par
As a result, inverting id ? H ? f , we get an algebra isomorphism\par
Proof. Consider the commutative diagram (3.5). Its upper-right conner is "unknown", while the lower row is "known" and represents the boundary data, and res is obviously an epimorphism. By Lemma 3.2, the left vertical arrow id ? H ? f is an isomorphism. Since, by Lemma 3.3, R is an isomorphism, it follows that R ? (id ? H ? f ) ?1 must be an isomorphism as well. In particular, R? is an epimorphism, whose kernel is isomorphic to the smooth functions on X whose restrictions to ? 1 X vanish. If z ? C ? (X) is a smooth function such that zero is its regular value, z ?1 (0) = ? 1 X, and z > 0 in int(X), then the kernel of res is the principle ideal m(z), generated by z. Therefore, by the commutativity of (3.5), the kernel of the homomorphism R? must be also a principle ideal M ? , generated by an element ( R\par
X) determine, up to an isomorphism, the algebra C ? (X), and thus determine the smooth topological type of the manifold X.\par
Proof. We call a maximal ideal of an algebra A nontrivial if it is different from A.\par
By Theorem 3.1, the algebra C ? (X) is determined by the two algebras on ? 1 X, up to an isomorphism. In turn, the algebra C ? (X) determines the smooth topological type of X, viewed as a ringed space. This fact is based on interpreting X as the space M(C ? (X)) of nontrivial maximal ideals of the algebra C ? (X) {\ref [KMS]}.\par
Therefore, with the help of the isomorphism H(v, f ) from (3.6), the nontrivial maximal ideals of C ? (X) (which by {\ref [KMS]} \par
Thus, an action of any group G of such isomorphisms ? ? extends canonically to a Gaction on the algebra C ? (X) and, via it, to a G-action on X by smooth diffeomorphisms.\par
Proof. By {\ref [Mr]}, any algebra isomorphism ? : C ? (X 1 ) ? C ? (X 2 ) is induced by a unique smooth diffeomorphism ? : X 1 ? X 2 . With this fact in hand, by Theorem 2.1 and Theorem 3.1, the proof is on the level of definitions.\par
It remains to address the following crucial question: how to characterize intrinsically the trace C\par
Evidently, functions from C ? (? 1 X, v) are constant along each C v -"trajectory" ? ? := ? ? ? 1 X of the causality map. Furthermore, any smooth function ? : ? 1 X ? R that is constant on each finite set ? ? gives rise to a unique continuous function ? on X that is constant along each v-trajectory ?. However, such functions ? may not be automatically smooth on X (a priory, they are just Hölderian with some control of the Hölder exponent that depends on the dimension of X only)! This potential complication leads to the following question.\par
To get some feel for a possible answer, we need the notion of the Morse stratification of the boundary ? 1 X that a vector field v generates {\ref [Mo]}.\par
Let dim(X) = n + 1 and v be a boundary generic traversing vector field on X. {\ref (v)}. This construction self-replicates until we reach finite sets ? ± n+1 X(v). By definition, the boundary generic vector fields {\ref [K1]} are the ones that satisfy certain nested transversality of v with respect to the boundary ? 1 X, the transversality that guaranties that all the Morse strata ? j X(v) are regular closed submanifolds and all the strata ? ± j X(v) are compact submanifolds. For a traversing boundary generic v, the map {\ref ]}). Let us describe now a good candidate for the subalgebra C ? (? 1 X, v) in the algebra C ? (?X).
\subsection[{Let us recall the definition of the Morse stratification \{?}]{Let us recall the definition of the Morse stratification \{?}\par
We denote by L Let v be a traversing and boundary generic vector field on a smooth compact (n + 1)-manifold X. Then the algebra C ? (? 1 X, v) coincides with the subalgebra M(v) Cv ? C ? (? 1 X).\par
In particular, C ? (? 1 X, v) can be determined by the causality map C v and the restriction of v to ? 2 X(v). ?\par
It is easy to check that C ? (? 1 X, v) ? M(v) Cv ; the challenge is to show that the two algebras coincide. \begin{bibitemlist}{1}
\bibitem[Katz (2017)]{b9}\label{b9} ‘Causal Holography in Application to the Inverse Scattering Problem’. G Katz . arXiv:1703.08874v1. \textit{Inverse Problems and Imaging J} June 2019. Mar 2017. 13 (3) p. 27. (Math.GT)
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\bibitem[Indeed, the subalgebras C ? (? 1 X, v) and (f ? ) * (C ? (R)) would acquire a description in terms of C v and f ? . This would deliver an independent proof of a natural generalization of Corollary 2 drop Property A from the hypotheses of the Holography Theorem]{b0}\label{b0} ‘Indeed, the subalgebras C ? (? 1 X, v) and (f ? ) * (C ? (R)) would acquire a description in terms of C v and f ? . This would deliver an independent proof of a natural generalization of Corollary 2’. \textit{drop Property A from the hypotheses of the Holography Theorem}, (.1. Acknowledgments: Notes)
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