Isometric Embeddings of Finite Metric Trees into ( R n , d 1 ) and ( R n , d ∞ )
aa r X i v : . [ m a t h . M G ] F e b ISOMETRIC EMBEDDINGS OF FINITE METRIC TREES INTO ( R n , d ) AND ( R n , d ∞ ) ASUMAN G ¨UVEN AKSOY, ∗ MEHMET KILIC¸ , S¸AH˙IN KOC¸ AK Abstract.
We investigate isometric embeddings of finite metric trees into( R n , d ) and ( R n , d ∞ ). We prove that a finite metric tree can be isometricallyembedded into ( R n , d ) if and only if the number of its leaves is at most 2 n .We show that a finite star tree with at most 2 n leaves can be isometricallyembedded into ( R n , d ∞ ) and a finite metric tree with more than 2 n leavescannot be isometrically embedded into ( R n , d ∞ ).We conjecture that an arbitrary finite metric tree with at most 2 n leavescan be isometrically embedded into ( R n , d ∞ ). Introduction
A finite metric tree is a finite, connected and positively weighted graph withoutcycles. The distance between two vertices is given by the total weight of a simplepath (which is unique) between these vertices. For two in-between points of thetree one takes the corresponding portions of the edges carrying these points intoaccount. A leaf of a tree is a vertex with degree 1. Vertices other than the leavesare called interior vertices. A tree with a single interior vertex is called a star (orstar tree). We will not allow interior vertices of degree 2 since, from the point ofview of metric properties, they can be considered artificial.Embedding new spaces into more familiar ones is a kind of innate behaviorfor mathematicians. For metric spaces, as the first candidate for an ambientspace, the Euclidean space ( R n , d ) might come into mind; but it is a completedisappointment. No metric tree (with at least three leaves) can be embeddedisometrically into any ( R n , d ). For a star with three leaves, this is almost obvious;and any tree with more than three leaves contains a three-star as a subspace.There is no help in considering ( R n , d p ) with any 1 < p < ∞ , because these arealso uniquely geodesic spaces and they do not host any tree with at least threeleaves either.At this point we could take refuge in Kuratowski’s embedding theorem whichstates that every metric space M embeds isometrically in the Banach space L ∞ ( M ) of bounded functions on M with sup norm || f || ∞ := sup x ∈ M | f ( x ) | where f : M → R . (see [5]). But a metric tree has an uncountable number of pointsso that our ambient space would be too huge and useless. A remedy could arisefrom considering the leaves of the tree only, since they determine the whole of the Mathematics Subject Classification.
Primary 54E35; Secondary 54E45, 54E50, 05C05,47H09, 51F99.
Key words and phrases.
Metric Trees, Embeddings, Euclidean Spaces with the MaximumMetric. metric tree (as a metric space and up to isometry) as their tight span (see The-orem 8 in Dress [3]) . By this approach we get at least an isometric embeddingof the metric tree into ( R n , d ∞ ) where n is the number of leaves and d ∞ denotesthe maximum metric d ∞ ( x, y ) = max ≤ i ≤ n | x i − y i | .We will show that a metric star tree can be embedded into ( R n , d ∞ ) if andonly if it has at most 2 n leaves and an arbitrary finite metric tree with more than2 n leaves cannot be embedded into ( R n , d ∞ ). We guess that an arbitrary finitemetric tree with at most 2 n leaves can be embedded into ( R n , d ∞ ) but we are yetunable to provide a proof for this guess.The picture for ( R n , d ) as the ambient space is more clear-cut. It was alreadyshown by Evans ( [4]) that any finite metric tree can be isometrically embeddedinto l , without explicit bounds for the dimension of the target in terms of theleaf number of the given tree. We prove by other, more geometric means that afinite metric tree can be embedded isometrically into ( R n , d ) if and only if thenumber of its leaves is at most 2 n .2. Preliminaries
For the sake of clarification, we give in the following the formal definition of ametric tree and mention some of its properties. However, in this paper we willconsider a special subfamily of metric trees, namely finite connected weightedgraphs without loops. Our aim is to investigate whether we can isometricallyembed these finite metric trees into ( R n , d ) or ( R n , d ∞ ). Definition 2.1.
Let x, y ∈ M , where ( M , d ) is a metric space. A geodesicsegment from x to y (or a metric segment , denoted by [ x, y ]) is the image of anisometric embedding α : [ a, b ] → M such that α ( a ) = x and α ( b ) = y . A metricspace is called geodesic if any two points can be connected by a metric segment.A metric space ( M, d ), is called a metric tree if and only if for all x, y, z ∈ M ,the following holds:(1) there exists a unique metric segment from x to y , and(2) [ x, z ] ∩ [ z, y ] = { z } ⇒ [ x, z ] ∪ [ z, y ] = [ x, y ].Next we mention some useful properties of metric segments that we will use.For the proofs of these properties we refer the reader to consult [2] and [4]. .For x, y in a metric space M , write xy = d ( x, y ). For x, y, z ∈ M , we say y is between x and z , denoted xyz , if and only if xz = xy + yz .(1) (Transitivity of betweenness [2]) Let M be a metric space and let a, b, c, d ∈ M . If abc and acd , then abd and bcd .(2) (Three point property, [4, Section 3.3.1]) Let x, y, z ∈ T ( T is a completemetric tree). There exists (necessarily unique) w ∈ T such that[ x, z ] ∩ [ y, z ] = [ w, z ] and [ x, y ] ∩ [ w, z ] = { w } Consequently,[ x, y ] = [ x, w ] ∪ [ w, y ] , [ x, z ] = [ x, w ] ∪ [ w, z ] , and [ y, z ] = [ y, w ] ∪ [ w, z ] . Here are two examples of metric trees:
SOMETRIC EMBEDDINGS OF FINITE METRIC TREES INTO ( R n , d ) AND ( R n , d ∞ ) 3 Example 2.2. (The Radial Metric)Define d : R × R → R ≥ by: d ( x, y ) = ( k x − y k if x = λ y for some λ ∈ R , k x k + k y k otherwise.It is easy to verify that d is in fact a metric and that ( R , d ) is a metric tree. Example 2.3. ( “Star Tree”)Fix k ∈ N , and a sequence of positive numbers ( a i ) ki =1 , the metric star tree isdefined as a union of k intervals of lengths a , . . . , a k , emanating from a commoncenter and equipped with the radial metric. More precisely, our tree T consistsof its center o , and the points ( i, t ), with 1 ≤ i ≤ k and 0 < t ≤ a i . The distance d is defined by setting d ( o, ( i, t )) = t , and d (( i, t ) , ( j, s )) = (cid:26) | t − s | i = jt + s i = j . Abusing the notation slightly, we often identify o with ( i, i, a i ) , i = 1 , . . . , k. In the following we will consider only a very special and simple type of metrictrees called finite metric trees or finite simplicial metric trees. They are explained,for example, in [7], p.43 and p.73 (see also [4], Example 3.16). Topologically theyare finite trees in the sense of graph theory: There is a finite set of “vertices”;between any pair of vertices there is at most one “edge” and the emerging graphis connected and has no loops. The degree of a vertex is defined in the graph-theoretical sense and a vertex with degree 1 is called a leaf.In addition to this graph-theoretical structure, non-negative weights are as-signed to the edges. Moreover, to a pair of vertices a distance is assigned byconsidering the unique simple path (in graph-theoretical sense) connecting thesevertices and adding up the weights of the edges constituting this path. This dis-tance can be naturally extended to any pair of points of the tree and convertsthe graph-theoretical tree into a metric space which is a metric tree in the senseof Definition 2.1.These special metric trees are also called finite metric trees, or finite simplicialmetric trees, though they obviously contain a continuum of points as a metricspace (except in the trivial case of a singleton). We will be concerned withembedding these finite metric trees into ( R n , d ) or ( R n , d ∞ ) where we will try tooptimize the dimension n in dependence of the number of the leaves of the tree.In general metric trees are more complicated than finite simplicial metric trees.For further discussion of non-simplicial trees and construction of metric treesrelated to the asymptotic geometry of hyperbolic metric spaces we refer the readerto [1]. 3. Embedding Star Trees
As motivational examples we consider first finite metric star trees. They canbe viewed as a union of k intervals of lengths a , a , . . . , a k , emanating from a ASUMAN G ¨UVEN AKSOY, MEHMET KILIC¸ AND SAH˙IN KOC¸ AK common center and equipped with the radial metric, as described above (seeFig 1). o b bb b b bbb (5 , a )(1 , a ) (2 , a ) (3 , a )(4 , a ) a a a (3 , t )(4 , s ) d ((3 , t ) , (4 , s )) = t + s Figure 1.
A metric star tree.The following two properties give the tight embedding bounds for finite metricstar trees.
Proposition 3.1.
A metric star tree ( X, d ) with k leaves can be isometricallyembedded into ( R n , d ∞ ) if and only if k ≤ n .Proof. ( ⇐ ) First assume k ≤ n . Note that there are 2 n extreme points onthe unit ball of the space ( R n , d ∞ ), which are ǫ ∈ {− , } n . We will denotethese extreme points by v , v , . . . , v n and define the map f : X → ( R n , d ∞ ) by f ( o ) = O ∈ R n and f (( i, t )) = t · v i . The map f is then an isometric embeddingfrom X to ( R n , d ∞ ): d ∞ ( f ( i, t ) , f ( o )) = d ∞ ( t · v i , O ) = t = d (( i, t ) , o ) ,d ∞ ( f ( i, t ) , f ( i, s )) = d ∞ ( t · v i , s · v i ) = | t − s | = d (( i, t ) , ( i, s )) ,d ∞ ( f ( i, t ) , f ( j, s )) = d ∞ ( t · v i , s · v j ) = t + s = d (( i, t ) , ( j, s )) , when i = j. ( ⇒ ) Assume that k > n but there exists an isometric embedding f from X to ( R n , d ∞ ). We can assume that the embedding takes the center of X to theorigin of R n because translation is an isometry. Then, we can find two points( i, a i ) and ( j, a j ) in X with i = j such that all corresponding coordinates of theirimages have the same sign. If their images are f ( i, a i ) = A i = ( A i , A i , · · · , A ni )and f ( j, a j ) = A j = ( A j , A j , · · · , A nj ), then | A mi | ≤ a i and | A mj | ≤ a j for all m =1 , , . . . , n since d ∞ ( A i , O ) = d (( i, a i ) , o ) = a i and d ∞ ( A j , O ) = d (( j, a j ) , o ) = a j .Hence, we obtain d ∞ ( A i , A j ) = n max m =1 {| A mi − A mj |} < a i + a j = d (( i, a i ) , ( j, a j )) , SOMETRIC EMBEDDINGS OF FINITE METRIC TREES INTO ( R n , d ) AND ( R n , d ∞ ) 5 which contradicts the assumption that f is an isometry. (cid:3) Proposition 3.2.
A metric star tree X with k leaves can be embedded isometri-cally into ( R n , d ) if and only if k ≤ n .Proof. ( ⇐ ) First assume k ≤ n . Note that there are 2 n extreme point on theunit ball of the space ( R n , d ), which are e i = ( δ i,j ) nj =1 , where δ i,j is the Kroneckerdelta. We will denote these extreme points in R n by E , E , . . . , E n and definethe map f : X → ( R n , d ) by f ( o ) = O and f (( i, t )) = t · E i . We will show that f is an isometric embedding from X to ( R n , d ): d ( f ( i, t ) , f ( o )) = d ( t · E i , O ) = t = d (( i, t ) , o ) ,d ( f ( i, t ) , f ( i, s )) = d ( t · E i , s · E i ) = | t − s | = d (( i, t ) , ( i, s )) ,d ( f ( i, t ) , f ( j, s )) = d ( t · E i , s · E j ) = t + s = d (( i, t ) , ( j, s )) , where i = j. ( ⇒ ) Assume that k > n but there exists an isometric embedding f from X to ( R n , d ). We can assume that the embedding takes the center of X to theorigin because translation is an isometry. Then, we can find two points ( i, a i )and ( j, a j ) in X with i = j such that at least one common coordinate of theirimages are nonzero and have the same sign. If their images are f ( i, a i ) = A i =( A i , A i , · · · , A ni ) and f ( j, a j ) = A j = ( A j , A j , · · · , A nj ), then we obtain d (( i, a i ) , ( j, a j )) = d (( i, a i ) , o ) + d ( o, ( j, a j )) = d ( A i ,
0) + d (0 , A j )= | A i | + | A i | + · · · + | A ni | + | A j | + | A j | + · · · + | A nj | > | A i − A j | + | A i − A j | + · · · + | A ni − A nj | = d ( A i , A j ) , which is a contradiction. (cid:3) Embedding Arbitrary Metric Trees
For later use, we recall some metric preliminaries.
Definition 4.1.
For p = ( p , p , . . . , p n ) ∈ ( R n , d ∞ ), we define S + i ( p ) = { q = ( q , q , . . . , q n ) ∈ R n | d ∞ ( p, q ) = q i − p i } ,S − i ( p ) = { q = ( q , q , . . . , q n ) ∈ R n | d ∞ ( p, q ) = p i − q i } for i = 1 , , . . . , n and call them the sectors at the point p as shown in the followingFig. 2 and Fig. 3. Notice that if q belongs to S εi ( p ), S εi ( q ) ⊆ S εi ( p ) holds, where ε ∈ { + , −} .The following theorem gives a characterization of geodesics in ( R n , d ∞ ). Proposition 4.2 (Theorem 2.2 of [6]) . Let p = ( p , p , . . . , p n ) , q = ( q , q , . . . , q n ) ∈ R n be two points, q ∈ S εi ( p ) and α : [0 , d ( p, q ) ] → R n be a path such that α (0) = p and α ( d ( p, q )) = q . Then α is a geodesic in R n ∞ if and only if α ( t ′ ) ∈ S εi ( α ( t )) for all t, t ′ ∈ [0 , d ( p, q )] such that t < t ′ . (See Fig 4) We will also need the following “Shortening Lemma”:
ASUMAN G ¨UVEN AKSOY, MEHMET KILIC¸ AND SAH˙IN KOC¸ AK xy b p S +1 ( p ) S +2 ( p ) S − ( p ) S − ( p ) Figure 2.
Sectors of a point p in ( R , d ∞ ). b O S +2 ( O ) x yz Figure 3.
The sector S +2 ( O ) of the origin in ( R , d ∞ ). xy b p b q xy b p b q Figure 4.
Two paths between p and q in R ∞ one of which (onthe left) is a geodesic but the other is not. SOMETRIC EMBEDDINGS OF FINITE METRIC TREES INTO ( R n , d ) AND ( R n , d ∞ ) 7 Lemma 4.3.
Let α : [0 , b ] → ( R n , d ∞ ) be a geodesic and < c < d < b . Then, ˜ α : [0 , b − d + c ] → ( R n , d ∞ ) , ˜ α ( t ) = (cid:26) α ( t ) when t ∈ [0 , c ] α ( t − c + d ) − α ( d ) + α ( c ) when t ∈ [ c, b − d + c ] is a geodesic.Proof. Assume that α ( b ) ∈ S εi ( α (0)) for some i ∈ { , , . . . , n } and ε ∈ { + , −} .Since α is a geodesic, for all t, t ′ ∈ [0 , b ] such that t < t ′ we get α ( t ′ ) ∈ S εi ( α ( t )).Given t, t ′ ∈ [0 , b − d + c ] with t < t ′ . We consider three possibilities: if t, t ′ ∈ [0 , c ], then ˜ α ( t ′ ) ∈ S εi ( ˜ α ( t )) because ˜ α = α on [0 , c ]. If t, t ′ ∈ [ c, b − d + c ],then ˜ α ( t ′ ) ∈ S εi ( ˜ α ( t )) because α ( t ′ − c + d ) ∈ S εi ( α ( t − c + d )) and this implies α ( t ′ − c + d ) − α ( d ) + α ( c ) ∈ S εi ( α ( t − c + d ) − α ( d ) + α ( c )) . If t ∈ [0 , c ] and t ′ ∈ [ c, b − d + c ], we know that ˜ α ( t ′ ) ∈ S εi ( ˜ α ( c )) and ˜ α ( c ) ∈ S εi ( ˜ α ( t )). Since˜ α ( c ) ∈ S εi ( ˜ α ( t )), S εi ( ˜ α ( c )) ⊆ S εi ( ˜ α ( t )); hence, we get ˜ α ( t ′ ) ∈ S εi ( ˜ α ( t )). Thus,previous proposition implies that ˜ α is a geodesic. (cid:3) Proposition 4.4.
A finite metric tree with more than n leaves can not be em-bedded isometrically into ( R n , d ∞ ) .Proof. Let us assume that X has k leaves, k > n and f : X → ( R n , d ∞ ) bean isometric embedding. Denote the leaves a , a , . . . , a k , and their images under f by A , A , . . . , A k , i.e. f ( a i ) = A i . Let us denote the vertex points on X by b i for i = 1 , , . . . , k such that there exists an edge between a i and b i andassume f ( b i ) = B i . Note that the vector A i B i can not be equal to t · ( A j B j )for i = j and t >
0. Because if we assume A i B i = t · ( A j B j ) or equivalently B i − A i = t · ( B j − A j ), we get || A i − A j || = || ( A i − B i ) + ( B i − B j ) + ( B j − A j ) || = || (1 − t )( B j − A j ) + ( B i − B j ) ||≤ | − t | · || B j − A j || + || B i − B j || On the other hand, since the geodesic from a i to a j passes through b i and b j respectively, we have || A i − A j || = || A i − B i || + || B i − B j || + || B j − A j ) || = t || B j − A j || + || B i − B j || + || B j − A j ) || = (1 + t ) || B j − A j || + || B i − B j || and this contradicts previous inequality.Now consider the set { t. ( A i − B i ) | ≤ t ≤ , i = 1 , , . . . , k } . According toLemma 4.3, this set is a star tree with k > n leaves. But this contradicts theProposition 3.1. (cid:3) Proposition 4.5.
A finite metric tree with more than n leaves can not be em-bedded isometrically into ( R n , d ) . Proof of this claim is very similar to the proof above. In fact, let us assume that X has k leaves with k > n and f : X → ( R n , d ) be an isometric embedding. ASUMAN G ¨UVEN AKSOY, MEHMET KILIC¸ AND SAH˙IN KOC¸ AK
Denote those leaves by a , a , . . . , a k , and their images by A , A , . . . , A k , i.e. f ( a i ) = A i . Let us denote the vertices on X by b i for i = 1 , , . . . , k such thatthere is an edge between a i and b i and assume f ( b i ) = B i . Now consider the set { t · ( A i − B i ) | ≤ t ≤ , i = 1 , , . . . , k } . This set is a star tree with k > n leaves. But this contradicts the Proposition 3.2. Theorem 4.6.
Let ( X, d ) be a metric tree. If X contains at most n leaves, itcan be embedded isometrically into ( R n , d ) .Proof. We will prove this theorem by induction on n . If n = 1, the statement isobviously true. Assume that we can embed isomerically any metric tree whichhas 2 n leaves into ( R n , d ). Let X be any metric tree which has 2( n + 1) leaves.We will show that X can be embedded isometrically into ( R n +1 , d ). We canchoose two leaves in X such that after deleting them with the adjacent edges(and discarding possibly emerging vertices with degree 2), the rest of the tree has2 n leaves (see Figure 5). Let us call these leaves as A and A and their adjacentedges as B A and B A . According to our assumption, there is an isometricembedding f from rest of the tree Y = X − (( B A ] ∪ ( B A ]) to ( R n , d ). Definethe map F : X → ( R n +1 , d ), F ( x ) = ( f ( x ) , when x ∈ Y ( f ( B ) , − d ( x, B )) when x ∈ [ B A ]( f ( B ) , d ( x, B )) when x ∈ [ B A ] b A b B B A bb bb b bb b b b Figure 5.
If we delete ( B A ] and ( B A ] and discarding the ver-tex B , we get a tree which has two fewer leaves.We will show that F is an isometric embedding. Let x, y ∈ X be two arbitrarypoints. Since f is an isometric embedding, If x, y ∈ Y , we get d ( F ( x ) , F ( y )) = d ( f ( x ) , f ( y )) = d ( x, y ) . If x ∈ ( B A ] and y ∈ Y , d ( F ( x ) , F ( y )) = d ( f ( B ) , f ( y )) + d ( x, B )= d ( B , y ) + d ( x, B )= d ( x, y ) . SOMETRIC EMBEDDINGS OF FINITE METRIC TREES INTO ( R n , d ) AND ( R n , d ∞ ) 9 If x ∈ ( B A ] and y ∈ Y , d ( F ( x ) , F ( y )) = d ( f ( B ) , f ( y )) + d ( x, B )= d ( B , y ) + d ( x, B )= d ( x, y ) . If x ∈ ( B A ] and y ∈ ( B A ], d ( F ( x ) , F ( y )) = d ( f ( B ) , f ( B )) + d ( x, B ) + d ( y, B )= d ( x, B ) + d ( B , B ) + d ( B , y )= d ( x, y ) . (cid:3) References
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E-mail address : [email protected] Department of Mathematics, Anadolu University, 26470, Eskisehir Turkey.
E-mail address : [email protected] Department of Mathematics, Anadolu University, 26470, Eskisehir Turkey.
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