CONICAL GEODESIC BICOMBINGS ON SUBSETS OF NORMED ...
arXiv:1604.04163v1 [math.MG] 14 Apr 2016
CONICAL GEODESIC BICOMBINGS ON SUBSETS OF NORMED VECTOR SPACES GIULIANO BASSO AND BENJAMIN MIESCH
Abstract. In this paper we establish existence and uniqueness results for conical geodesic bicombings on subsets of normed vector spaces. Concerning existence, we give a first example of a convex geodesic bicombing that is not consistent. Furthermore, we show that under a mild geometric assumption on the norm a conical geodesic bicombing on an open subset of a normed vector space locally consists of linear geodesics. As an application, we obtain by the use of a Cartan-Hadamard type result that if a closed convex subset of a Banach space has non-empty interior, then it admits a unique consistent conical geodesic bicombing.
1
Introduction
In [DL15], D. Descombes and U. Lang initiated the systematic study of metric spaces that obey a weak form of non-positive curvature which is encapsulated in the existence of a conical geodesic bicombing. Essentially, a conical geodesic bicombing on a metric space is a selection of geodesics such that the distance function between each two selected geodesics is bounded from above by the linear interpolation of its values at the endpoints. Conical geodesic bicombings arise naturally in the context of injective metric spaces, cf. [Lan13], and the class of metric spaces that admit a conical geodesic bicombing is closed under 1-Lipschitz retractions and ultralimits. The term bicombing originates from W. Thurston and is prominently used by J. Alonso and M. Bridson to define semihyperbolic metric spaces, cf. [AB95]. Notions related to conical geodesic bicombings also occur in metric fixed point theory, most notably W-convexity mappings, cf. [Tak70], and hyperbolic spaces in the sense of S. Reich and I. Shafrir, cf. [RS90]. Recently, classical results from the theory of CAT(0) spaces have been transferred to metric spaces that admit conical geodesic bicombings, cf. [Bas15, Des15, DL16, Kel16] and [Mie15]. In this article we proceed with the systematic study of metric spaces that admit conical geodesic bicombings and we enlarge the collection of examples from [DL15]. Let (X, d) denote a metric space. We abbreviate D(X) := X × X × [0, 1]. A map σ : D(X) → X is said to be a geodesic bicombing if the path σpq (·) := σ(p, q, ·) is a constant speed geodesic from p to q for all points p, q in X, that is, we have σpq (0) = p, σpq (1) = q and d(σpq (t), σpq (s)) = |t − s| d(p, q) for all real numbers s, t ∈ [0, 1]. Furthermore, a geodesic bicombing σ : D(X) → X is called conical if it satisfies d(σpq (t), σp0 q0 (t)) ≤ (1 − t)d(p, p0 ) + td(q, q 0 ) Date: April 15, 2016. 1
(1.1)
CONICAL GEODESIC BICOMBINGS ON SUBSETS OF NORMED VECTOR SPACES
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for all points p, q, p0 , q 0 in X and all real numbers t ∈ [0, 1]. Examples of metric spaces that admit conical geodesic bicombings include convex subsets of normed vector spaces, Busemann spaces and injective metric spaces. Our first result deals with convex geodesic bicombings. A geodesic bicombing σ : D(X) → X is convex if the map t 7→ d(σpq (t), σp0 q0 (t)) is convex on [0, 1] for all points p, q, p0 , q 0 in X. Note that if the underlying metric space is not uniquely geodesic, then a conical geodesic bicombing is not necessarily convex. Examples of conical geodesic bicombings that are not convex are ubiquitous; for instance, nonconvex conical geodesics bicombings may be obtained via 1-Lipschitz retractions of linear geodesics, see [DL15, Example 2.2] or Lemma 3.1. In [DL15], it is shown that metric spaces of finite combinatorial dimension in the sense of Dress, cf. [Dre84], possess at most one convex geodesic bicombing. If it exists, this unique convex geodesic bicombing, say σ : D(X) → X, has the property that it is consistent, that is, we have for all points p, q in X that im(σp0 q0 ) ⊂ im(σpq ) whenever p0 = σpq (s) and q 0 = σpq (t) with 0 ≤ s ≤ t ≤ 1. Clearly, every consistent conical geodesic bicombing is convex. In Section 2, we show that the converse does not hold by proving the subsequent theorem. Theorem 1.1. There is a compact metric space that admits a convex geodesic bicombing which is not consistent. Although there is a non-consistent convex geodesic bicombing on the space considered in Section 2, this space also admits a consistent convex geodesic bicombing. We suspect that this is a general phenomenon. Question 1.2. Let (X, d) be a proper metric space with a convex geodesic bicombing. Does X also admit a consistent convex geodesic bicombing? The seemingly more general question if every proper metric space with a conical geodesic bicombing admits a consistent conical geodesic bicombing is in fact equivalent to Question 1.2, as every proper metric space with a conical geodesic bicombing also admits a convex geodesic bicombing, cf. [DL15, Theorem 1.1]. A geodesic bicombing σ : D(X) → X is called reversible if σpq (t) = σqp (1 − t) for all points p, q in X and all t ∈ [0, 1]. Every proper metric space with a conical geodesic bicombing admits a reversible conical geodesic bicombing, cf. [Des15, Proposition 1.2]. However, it remained open if there are conical geodesic bicombings that are non-reversible. In Section 2, we modify our non-consistent convex geodesic bicombing from Theorem 1.1 in order to obtain an example of a non-reversible convex geodesic bicombing, see Proposition 2.5. In [Bas15], a barycentric construction has been employed to obtain fixed point results for metric spaces that admit conical geodesic bicombings. This barycentric construction motivated the following definition: A geodesic bicombing σ : D(X) → X has the midpoint property if σpq ( 21 ) = σqp ( 12 ) for all points p, q in X. It seems natural to ask if every conical geodesic bicombing that has the midpoint property is automatically reversible. We show that this is not the case, as we construct in Section 3 a non-reversible conical geodesic bicombing which has the midpoint property. It is a direct consequence of a result of S. G¨ahler and G. Murphy that the only conical geodesic bicombing on a normed vector space is the one that consists of the linear geodesics, cf. [GM81, Theorem 1]. With a mild geometric assumption on the norm, we show in Section 4 that already a conical geodesic bicombing on an open
CONICAL GEODESIC BICOMBINGS ON SUBSETS OF NORMED VECTOR SPACES
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subset of a normed vector space locally consists of linear geodesics. More generally, we get the following result: Theorem 1.3. Let (V, k·k) be a normed vector space with the property that its closed unit ball is the closed convex hull of its extreme points. Suppose that A ⊂ V is a subset of V that admits a conical geodesic bicombing σ : D(A) → A and let p0 in A be a point. If r ≥ 0 is a real number such that B2r (p0 ) ⊂ A, then we have that σ(p, q, t) = (1 − t)p + tq for all points p, q ∈ Br (p0 ) and all real numbers t ∈ [0, 1]. We do not know if Theorem 1.3 remains true if we drop the assumption of the normed vector space (V, k·k) having the property that its closed unit ball is the closed convex hull of its extreme points. But how common is it that the closed unit ball of a normed vector space is the closed convex hull of its extreme points? By invoking the Banach-Alao˘glu theorem and the Kre˘ın-Mil’man theorem it is possible to show that the closed unit ball of a dual Banach space is the closed convex hull of its extreme points. Consequently, we obtain in particular that Theorem 1.3 is valid in every reflexive Banach space. Moreover, using a classification result, due to L. Nachbin, D. Goodner, and J. Kelley, cf. [Kel52], and a result of D. Goodner, cf. [Goo50, Theorem 6.4], it is readily verified that Theorem 1.3 also holds for every injective Banach space. In [Mie15], the second named author generalized the classical Cartan-Hadamard Theorem to metric space that locally admit a consistent convex geodesic bicombing. With Theorem 1.3 on hand, it is possible to use this generalized Cartan-Hadamard Theorem to obtain the following uniqueness result. Theorem 1.4. Let (E, k·k) be a Banach space with the property that its closed unit ball is the closed convex hull of its extreme points. Suppose that C ⊂ E is a closed convex subset of E with non-empty interior. If σ : D(C) → C is a consistent conical geodesic bicombing, then it follows that σ(p, q, t) = (1 − t)p + tq for all points p, q in C and all real numbers t ∈ [0, 1]. The proof of Theorem 1.4 is given in Section 5. Due to Theorems 1.3 and 1.4 it appears that the geometry of a convex subset C of a normed vector space (V, k·k) is very restricted in the sense that it is difficult to construct a conical geodesic bicombing on C that is not given by the linear geodesics. In this perspective, we deem that an affirmative answer to the following question would result in an interesting geometric construction. Question 1.5. Is there a convex subset of a normed vector space that admits two distinct conical geodesic bicombings? In a similar way, one might ask if consistency always implies uniqueness, or in other words: Question 1.6. Is there a metric space with two different consistent conical geodesic bicombings? A possible candidate to an affirmative answer to Question 1.6 is given in [DL15, Example 3.5]. Unfortunately, it is unclear whether the two convex geodesic bicombings given there are consistent or not. Acknowledgments. We would like to thank Urs Lang for helpful remarks and discussions. The authors gratefully acknowledge support from the Swiss National Science Foundation.
CONICAL GEODESIC BICOMBINGS ON SUBSETS OF NORMED VECTOR SPACES
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4
A non-consistent convex geodesic bicombing
The goal of this section is to construct a convex geodesic bicombing that is not consistent and therefore establish Theorem 1.1. To this end, we consider the following norm on R2 : o n √ k(x, y)k := max |x|, 22 k(x, y)k2 , p where k(x, y)k2 = x2 + y 2 is the Euclidean norm. Observe that k(x, y)k = |x| if and only if |y| ≤ |x|. Now define 1 X := (x, y) ∈ R2 : −3 ≤ x ≤ 3, 0 ≤ y ≤ 32 max{0, 1 − x2 } and equip X with the metric d induced by k · k.
-3
-2
-1
0
1
2
3
Figure 1. The metric space X. The space X naturally splits into three pieces, namely X = X− ∪ X0 ∪ X+ with X− := [−3, −1] × {0}, X0 := (x, y) ∈ R2 : −1 < x < 1, 0 ≤ y ≤
1 32 (1
− x2 ) ,
X+ := [1, 3] × {0}. 1 Definition 2.1. For δ ∈ [0, 64 ] we define a geodesic bicombing σ δ : D(X) → X as δ follows. Generally, we take σpq to be the geodesic from p to q which is linear inside X0 , but if both endpoints lie on the antennas X− , X+ we slightly modify it. In more details σ δ is defined as follows: For p := (px , py ), q := (qx , qy ) ∈ X with px ≤ qx let δ σpq (t) := (xpq (t), ypq (t)) ,
with xpq (t) := px + t(qx − px ), 2 δ max{qx − px − 4, 0} max{0, (1 − xpq (t) )}, q y (xpq (t) + 1)}, max{0, qx +1 py ypq (t) := max{0, px −1 (xpq (t) − 1)}, py + t(qy − py ), 0,
for p ∈ X− , q ∈ X+ , for p ∈ X− , q ∈ X0 , for p ∈ X0 , q ∈ X+ , for p, q ∈ X0 , otherwise.
and δ δ σqp (t) := σpq (1 − t). 1 Proposition 2.2. For δ ∈ (0, 64 ], the map σ δ is a reversible convex geodesic bicombing which is not consistent.
Remark 2.3. Observe that for δ = 0 the geodesic bicombing σ δ coincides with the piece-wise linear bicombing which is the unique consistent conical geodesic bicombing on X by Theorem 1.4. Hence we have a family of non-consistent convex geodesic bicombings σ δ converging to the unique consistent convex geodesic bicombing σ 0 .
CONICAL GEODESIC BICOMBINGS ON SUBSETS OF NORMED VECTOR SPACES
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Alternatively, we can modify the geodesics leading from X− to X+ . Definition 2.4. Define σ ˜ δ : D(X) → X by δ δ σ ˜pq (t) = σpq (t),
except for p ∈ X+ , q ∈ X− let δ σ ˜pq = (xpq (t), 0). 1 Proposition 2.5. For δ ∈ (0, 64 ], the map σ ˜ δ is a convex geodesic bicombing which neither is reversible nor consistent.
Let us first show that we defined geodesic bicombings. 1 Lemma 2.6. For δ ∈ [0, 64 ], the maps σ δ and σ ˜ δ are geodesic bicombings.
Proof. The linear case is clear. For the piecewise linear case observe that if p ∈ X− , δ satisfies q ∈ X0 (and similarly in all other cases) we have that the slope m of σpq m=
qy ≤ qx + 1
1 32 (1
− qx2 ) = 1 + qx
1 32 (1
− qx ) ≤
1 16
≤1
and therefore δ δ d(σpq (s), σpq (t)) = |xpq (s) − xpq (t)| = |s − t||qx − px | = |s − t|d(p, q).
Finally, let p ∈ X− , q ∈ X+ . For x, x0 ∈ [−1, 1] we have |δ(qx − px − 4)(1 − x2 ) − δ(qx − px − 4)(1 − x02 )| ≤ δ|qx − px − 4| · |x + x0 | · |x − x0 | ≤
1 16 |x
− x0 |
δ δ (t)) = |xpq (s) − xpq (t)| as before. (s), σpq and hence d(σpq
To prove Propositions 2.2 and 2.5, we will use the following characterization of convexity; see Lemma 3.5 in [LY10]. Lemma 2.7. A continuous function f : [0, 1] → R is convex if and only if for every t ∈ (0, 1) there is some τ0 > 0 such that for all τ ∈ [0, τ0 ] we have 2f (t) ≤ f (t − τ ) + f (t + τ ). Proof of Proposition 2.2. It is immediate that the geodesic bicombing σ δ is nonconsistent and reversible; hence, it remains to prove convexity. δ Given p, q, p0 , q 0 ∈ X we need to show that f (t) := d(σpq (t), σpδ0 q0 (t)) is convex on [0, 1]. Clearly t 7→ |xpq (t) − xp0 q0 (t)| is convex for t ∈ [0, 1]. Hence in the situation δ (t), σpδ0 q0 (t)) = |xpq (t) − xp0 q0 (t)| we have when d(σpq δ 2d(σpq (t), σpδ0 q0 (t)) = 2|xpq (t) − xp0 q0 (t)|
≤ |xpq (t − τ ) − xp0 q0 (t − τ )| + |xpq (t + τ ) − xp0 q0 (t + τ )| δ δ ≤ d(σpq (t − τ ), σpδ0 q0 (t − τ )) + d(σpq (t + τ ), σpδ0 q0 (t + τ )).
Therefore, it remains to check δ 2kσpq (t) − σpδ0 q0 (t)k2 δ δ ≤ kσpq (t − τ ) − σpδ0 q0 (t − τ )k2 + kσpq (t + τ ) − σpδ0 q0 (t + τ )k2 δ for τ > 0 small, whenever d(σpq (t), σpδ0 q0 (t)) = |xpq (t) − xp0 q0 (t)| ≤ |ypq (t) − yp0 q0 (t)|.
√
2 δ 2 kσpq (t)
− σpδ0 q0 (t)k2 , that is,
CONICAL GEODESIC BICOMBINGS ON SUBSETS OF NORMED VECTOR SPACES
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Observe that for x ∈ [−3, −1] ∪ [1, 3] and (x0 , y 0 ) ∈ X we have that δ d((x, 0), (x0 , y 0 )) = |x − x0 | and therefore we always have d(σpq (t), σpδ0 q0 (t)) = |xpq (t) − xp0 q0 (t)| if xpq (t) ∈ / (−1, 1). Hence, we only need to consider points that satsify xpq (t), xp0 q0 (t) ∈ (−1, 1). δ , σpδ0 q0 are (piece-wise) linear, then locally they are linear First, if both σpq geodesics inside a normed vector space and hence d(σpq (t), σp0 q0 (t)) = kσpq (t) − σp0 q0 (t)k is locally convex, thus convex. δ Let us now assume that σpq is not linear, i.e. p ∈ X− , q ∈ X+ , l := d(p, q) ≥ 4. We look at the different options for σpδ0 q0 separately. But before doing so, let us first fix some notation. We define p0 := σpq (t), p± := σpq (t ± τ ), p∗ = (x∗ , y∗ ) (∗ ∈ {0, +, −}), D := δ(l − 4), ε := τ l and accordingly for σpδ0 q0 . We then get y0 = D(1 − x20 ), x± = x0 ± ε and y± = D(1 − (x0 ± ε)2 ). In each case, we need to consider the situation where x0 , x00 ∈ (−1, 1) and |x0 − x00 | ≤ |y0 − y00 |. Case 1. p0 ∈ X∓ , q 0 ∈ X± and l0 := d(p0 , q 0 ) ∈ [4, l]. 0 0 0 2 0 0 0 0 As above we have y00 = D0 (1 − x02 0 ), x± = x0 ± ε , y± = D (1 − (x0 ± ε ) ) and 0 with λ := ll we get ε0 = λε. We claim that
2kp0 − p00 k2 ≤ kp− − p0− k2 + kp+ − p0+ k2 for ε > 0 (i.e. τ > 0) small enough. First note that kp− − p0− k22 = kp0 − p00 k22 − 2(x0 − x00 )(1 − λ)ε + (1 − λ)2 ε2 + 2(y0 − y00 )aε + a2 ε2 , kp+ − p0+ k22 = kp0 − p00 k22 + 2(x0 − x00 )(1 − λ)ε + (1 − λ)2 ε2 + 2(y0 − y00 )bε + b2 ε2 , for a := 2(x0 D − λx00 D0 ) − (D − λ2 D0 )ε, b := −2(x0 D − λx00 D0 ) − (D − λ2 D0 )ε, with a+b = −2(D −λ2 D0 )ε, a−b = 4(x0 D −λx00 D0 ) and either ab = (D −λ2 D0 )2 ε2 or ab < 0 for ε small. In the following, we assume ab < 0. The other case is similar. Moreover, we have kp− − p0− k22 · kp+ − p0+ k22 = kp0 − p00 k42 + 4ab(y − y 0 )2 − 4(x0 − x00 )2 (1 − λ)2 + 4(x0 − x00 )(1 − λ)(y0 − y00 )(a − b) + 2(1 − λ)2 + a2 + b2 − 4(y0 − y00 )(D − λ2 D0 ) · kp0 − p00 k22 ε2 + O(ε3 ) √ √ and with u + t = u + 2√t u + O(t2 ) and u = kp0 − p00 k42 it follows q 2 kp− − p0− k22 · kp+ − p0+ k22 0 2 ≥ 2kp0 − p0 k2 + 2(1 − λ)2 + a2 + b2 + 4ab − 4(y0 − y00 )(D − λ2 D0 ) 4(x0 − x00 )(y0 − y00 )(1 − λ)(a − b) − 4(x0 − x00 )2 (1 − λ)2 2 + ε + O(ε3 ). (x0 − x00 )2 + (y0 − y00 )2
CONICAL GEODESIC BICOMBINGS ON SUBSETS OF NORMED VECTOR SPACES
We therefore get kp− − p0− k2 + kp+ − p0+ k2
2
q kp− − p0− k22 + kp+ − p0+ k22 + 2 kp− − p0− k22 · kp+ − p0+ k22 0 2 ≥ 4kp0 − p0 k2 + 4(1 − λ)2 + 2(a + b)2 − 8(y0 − y00 )(D − λ2 D0 ) 4(x0 − x00 )(y0 − y00 )(1 − λ)(a − b) − 4(x0 − x00 )2 (1 − λ)2 2 + ε + O(ε3 ) (x0 − x00 )2 + (y0 − y00 )2 =
=
4kp0 − p00 k22 + Cε2 + O(ε3 )
≥
4kp0 − p00 k22 ,
for ε > 0 small enough, provided that 4(1 − λ)2 − 8(y0 − y00 )(D − λ2 D0 )
C= +
16(x0 − x00 )(y0 − y00 )(1 − λ)(x0 D − λx00 D0 ) − 4(x0 − x00 )2 (1 − λ)2 > 0. (x0 − x00 )2 + (y0 − y00 )2
Observe that a + b = O(ε). First, assuming y0 > y00 , we have 2 y0 − y00 = D(1 − x20 ) − D0 (1 − x20 ) = (D − D0 )(1 − x20 ) + D0 (x02 0 − x0 )
≤ δ(l − l0 ) + δ(l0 − 4)(x00 + x0 )(x00 − x0 ) ≤ δ(l − l0 ) + 4δ(y0 − y00 ) and therefore |y0 − y00 | ≤
δ (l − l0 ). 1 − 4δ
Moreover, |D − λ2 D0 |l2 = δ(l3 − 4l − l03 + 4l0 ) = δ(l − l0 )(l2 + ll0 + l02 − 4(l + l0 )) ≤ 60δ(l − l0 ), |x0 D − λx00 D0 |l ≤ |x0 |(D − λD0 )l + |x0 − x00 |D0 l0 ≤ δ(l − l0 )(l + l0 − 4) + 12δ|y0 − y00 | ≤
8δ +
12δ 2 1 − 4δ
(l − l0 ).
Hence, we finally get Cl2 kp0 − p00 k22 =
4(l − l0 )2 (y0 − y00 )2 − 8(y0 − y00 )(D − λ2 D0 )l2 (x0 − x00 )2 + (y0 − y00 )2
+ 16(x0 − x00 )(y0 − y00 )(l − l0 )(x0 D − λx00 D0 )l 192δ 2 960δ 2 − 128δ − (l − l0 )2 (y0 − y00 )2 ≥ 4− 1 − 4δ 1 − 4δ 4 − 144δ − 640δ 2 = (l − l0 )2 (y0 − y00 )2 > 0 1 − 4δ for δ
0 for δ < 0.026. Case 3. σp0 q0 linear with p0 , q 0 ∈ X0 . Let m again denote the slope of σp0 q0 . We distinguish two subcases. (a) If |m| ≤ 1, we have l ∈ [4, 6], l0 ∈ [0, 2] and |ml0 | =
|qy0 − p0y | 0 l = |qy0 − p0y | ≤ |qx0 − p0x |
1 32 .
0
0 Moreover, for ε0 = τ l0 , λ = ss we get x0± = x00 ± ε0 , y± = y00 ± mε0 and ε0 = λε as before and again we get the constant
4(1 − λ)2 − 8(y0 − y00 )D
C= +
8(x0 − x00 )(y0 − y00 )(1 − λ)(λm + 2Dx0 ) − 4(x0 − x00 )2 (1 − λ)2 . (x0 − x00 )2 + (y0 − y00 )2
CONICAL GEODESIC BICOMBINGS ON SUBSETS OF NORMED VECTOR SPACES
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Now, we estimate Cl2 kp0 − p00 k22 =
4(l − l0 )2 (y0 − y00 )2 − 8(y0 − y00 )Dl2 (x0 − x00 )2 + (y0 − y00 )2
+ 8(x0 − x00 )(y0 − y00 )(l − l0 )(ml0 + 2Dx0 l) ≥ 4 − 576δ 2 − 81 − 96δ (l − l0 )2 (y0 − y00 )2 2 31 = (l − l0 )2 (y0 − y00 )2 > 0 8 − 96δ − 576δ for δ < 0.033. (b) If |m| > 1, we have l ∈ [4, 6] and √ q l0 = 22 (qx − px )2 + (qy − py )2 ≤ |qy − py | ≤
1 32 .
Furthermore, let ε0x = x+ − x0 and ε0y = y+ − y0 . Then we have ε0y = mε0x and 2
2
2(τ l0 )2 = ε0x + (mε0x )2 = (1 + m2 )ε0x . 0
√
2 , ε0y = λy ε for λy := mλx = Thus we get ε0x = λx ε for λx := ll √1+m 2 0 x0± = x00 ± ε0x and y± = y00 ± ε0y . We proceed again as before and get the constant
√ l0 √ 2m l 1+m2 ,
4(1 − λx )2 − 8(y0 − y00 )D
C= +
8(x0 − x00 )(y0 − y00 )(1 − λx )(λy + 2Dx0 ) − 4(x0 − x00 )2 (1 − λx )2 , (x0 − x00 )2 + (y0 − y00 )2
with D = δ(l − 4) ≤ δ(l − l0 ) ≤ 6δ(1 − λx ), y0 − y00 ≤ D(1 − x20 ) ≤ D ≤ 6δ(1 − λx ), √ √ l0 2 2 1 ≤ λy = q ≤ (1 − λx ). l 128 64 1 m2 + 1 Now, we estimate Ckp0 − p00 k22 =
4(1 − λx )2 (y0 − y00 )2 − 8(y0 − y00 )D (x0 − x00 )2 + (y0 − y00 )2
+ 8(x0 − x00 )(y0 − y00 )(1 − λx )(λy + 2Dx0 ) 1 2 − 96δ (1 − λx )2 (y0 − y00 )2 ≥ 4 − 576δ − 64 2 255 = (1 − λx )2 (y0 − y00 )2 > 0 64 − 96δ − 576δ 1 for δ < 0.034. Hence this is again true for δ ≤ 64 . √ 0 Observe that for m → +∞ we get λx = 0 and λy = 2 ll , and the same estimates hold.
Proof of Proposition 2.5. Clearly the geodesic bicombing σ ˜ δ is non-consistent and non-reversible.
CONICAL GEODESIC BICOMBINGS ON SUBSETS OF NORMED VECTOR SPACES
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For convexity, the same arguments as in the proof of Proposition 2.2 apply. The only new case is p0 ∈ X+ and q 0 ∈ X− . With the notions from above with 0 x0± = x00 ∓ ε0 for ε0 = τ l0 and λ = ll we obtain the constant 4(1 + λ)2 − 8y0 D
C= +
16(x0 − x00 )y0 (1 + λ)x0 D − 4(x0 − x00 )2 (1 + λ)2 . (x0 − x00 )2 + y02
With the inequalities D = δ(l − 4) ≤ 2δ and |y0 | ≤ Ckp0 − p00 k22 =
1 32
we get
4(1 + λ)2 y02 − 8y0 D (x0 − x00 )2 + y02
+ 16(x0 − x00 )y0 (1 + λ)x0 D
for δ
0
≥
(4 − 33δ) (1 + λ)2 y02 > 0
1 64 .
A non-reversible conical geodesic bicombing that has the midpoint property
In the first part of this section we construct a non-reversible conical geodesic bicombing. Afterwards, we modify this non-reversible conical geodesic bicombing to satisfy the midpoint property. To begin, we define some auxiliary maps. Consider R2 equipped with the maximum norm k·k∞ and let s : R2 → R2 denote the map given by (x, y) 7→ (x, −y). We define X1 := (x, y) ∈ R2 : x ∈ [−2, 1] and |x| − 1 ≤ y ≤ ||x| − 1| , A1 := (x, y) ∈ R2 : |x + 1| ≤ y ≤ 1 . and X2 := s(X1 ), A2 := s(A1 ). The set readily verified that the map f : X2 → X1 ( (x, y), (x, y) 7→ s(x, y),
X1 ∪ X2 is depicted in Figure 2. It is given by if x ∈ [−1, 1], if x ∈ [−2, −1]
is an isometry. Let f¯: X1 ∪ X2 → X1 be the map that is equal to IdX1 on X1 and equal to f on X2 . Observe that the map f¯ is 1-Lipschitz. We set Y1 := X1 ∪ A1 and Y2 := X2 ∪ A2 . Further, we define the map π : Y1 ∪ Y2 → X1 ∪ X2 through the assignment (x, y) 7→ x, sgn(y) min |y| , ||x| − 1| . Observe that π is a 1-Lipschitz retraction that maps Yk to Xk for each k ∈ {1, 2}. Let λ : D(R2 ) → R2 be the conical geodesic bicombing on R2 that is given by the linear geodesics. Lemma 3.1. The map σ : D(X1 ) → X1 given by ( π ◦ λ(p, q, t), (p, q, t) 7→ f ◦ π ◦ λ f −1 (p), f −1 (q), t ,
if px ≤ qx , if qx ≤ px .
is a non-reversible conical geodesic bicombing on (X1 , k·k∞ ).
CONICAL GEODESIC BICOMBINGS ON SUBSETS OF NORMED VECTOR SPACES
p
11
1.
0.5 q
−2.5
−2.
−1.5
−1.
−0.5
0.5
1.
0 −0.5 p0
−1.
−1.5
Figure 2. The blue line corresponds to σpq and the red line corresponds to the image of σqp under the isometry f −1 .
Proof. Observe that both maps σ (1) := π ◦ λ and σ (2) := f ◦ π ◦ λ ◦ f −1 × f −1 × Id[0,1]
define conical geodesic bicombings on X1 . Thus, it follows that σ : D(X1 ) → X1 is a geodesic bicombing. In the following we show that σ is conical. Let p, q, p0 , q 0 ∈ X1 be points. As (2) (1) both maps σ (1) and σ (2) are conical geodesic bicombings on X1 with σpq = σpq if px , qx ≤ −1 or px , qx ≥ −1, it remains to check inequality (1.1) in the two cases if px , qx0 ≤ −1 and qx , p0x ≥ −1 or p0x , qx ≤ −1 and qx0 , px ≥ −1. Now, suppose that px , qx0 ≤ −1 and qx , p0x ≥ −1. The other case is treated analogously. Let t ∈ [0, 1] be a real number. Since the map f¯ ◦ π is 1-Lipschitz, we compute
kσpq (t) − σp0 q0 (t)k∞ = f¯ ◦ π ◦ λ(p, q, t) − f¯ ◦ π ◦ λ(f −1 (p0 ), f −1 (q 0 ), t) ∞
≤ (1 − t) p − f −1 (p0 ) ∞ + t q − f −1 (q 0 ) ∞ . By our assumptions on the points p, q, p0 , q 0 , it follows that
p − f −1 (p0 ) = kp − p0 k , ∞
∞
q − f −1 (q 0 ) = f −1 (q) − f −1 (q 0 ) = kq − q 0 k . ∞ ∞ ∞ Hence, by putting everything together, we obtain that σ is a conical geodesic bicombing on X1 . By construction, it follows that σ is non-reversible; see Figure 2. Now, we may use σ to construct a non-reversible conical geodesic bicombing that has the midpoint property.
CONICAL GEODESIC BICOMBINGS ON SUBSETS OF NORMED VECTOR SPACES
12
Lemma 3.2. The map τ : D(X1 ) → X1 given by the assignment ( σ p, 21 σ(p, q, 12 ) + σ(q, p, 12 ) , 2t , if t ∈ [0, 12 ], (p, q, t) 7→ 1 1 1 σ 2 σ(p, q, 2 ) + σ(q, p, 2 ) , q, 2t − 1 , if t ∈ [ 12 , 1], is a conical geodesic bicombing on (X1 , k·k∞ ) that has the midpoint property but is not reversible. Proof. It is straightforward that τ is a conical geodesic bicombing with the midpoint property. To see that τ is non-reversible, take for instance p := (− 32 , 12 ), q := (0, 12 ) and observe that 1 7 5 ) = (− 78 , 18 ) 6= (− 87 , 48 ) = τ (q, p, 12 ); τ (p, q, 12
compare Figure 3.
1.
0.5
p
q
m
−2.5
−2.
−1.5 p
−1.
−0.5
0.5
1.
0
0
−0.5
−1.
−1.5
Figure 3. The blue line corresponds to τpq |[0, 12 ] and the red line corresponds to the image of τqp |[ 12 ,1] under the isometry f −1 . The point m is equal to 21 σpq ( 21 ) + σqp ( 12 ) .
4
Conical geodesic bicombings on subsets of normed vector spaces
Let (V, k·k) be a normed vector space, let p0 ∈ V be a point and let r ≥ 0 be a real number. We set Ur (p0 ) := {z ∈ V : kp0 − zk < r}, Br (p0 ) := {z ∈ V : kp0 − zk ≤ r}, Sr (p0 ) := {z ∈ V : kp0 − zk = r}. To ease notation, we abbreviate Br := Br (0) and Sr := Sr (0). The goal of this section is to establish the following rigidity result.
CONICAL GEODESIC BICOMBINGS ON SUBSETS OF NORMED VECTOR SPACES
13
Theorem 4.1. Let (V, k·k) be a normed vector space. Suppose that A ⊂ V is a subset of V that admits a conical geodesic bicombing σ : D(A) → A and let p, q be points of A. If there are points eP 1 , . . . , en ∈ B1 that are extreme points of B1 and a n tuple (λ1 , . . . , λn ) ∈ [0, 1]n with k=1 λk = 1 such that n
p−q kp − qk X = λk ek and 2 2 k=1 ( n ) p + q kp − qk X + (−1)εk λk ek : (ε1 , . . . , εn ) ∈ {0, 1}n ⊂ A, 2 2
(4.1)
(4.2)
k=1
then it follows that σ(p, q, t) = (1 − t)p + tq for all t ∈ [0, 1]. Theorem 1.3 then is a direct consequence. Proof of Theorem 1.3. Let p, q ∈ Br (p0 ) be two points. As p+q ∈ Br (p0 ) and 2 kp−qk p+q ≤ r, the ball B kp−qk ( 2 ) is contained in A. Hence, since the unit ball of V 2 2 is the closed convex hull of its extreme points, it follows that σ(p, q, t) = (1−t)p+tq for all t ∈ [0, 1] by Theorem 4.1 and a simple limiting argument. In order to derive Theorem 4.1 we need some preparatory lemmas and definitions. Let t ∈ [0, 1] be a real number and let p, q be points in V . We define M (t) (p, q) := {z ∈ V : kz − pk = t kp − qk , kz − qk = (1 − t) kp − qk} By construction, we have M (t) (p, −p) = S2tkpk + p ∩ S(1−t)2kpk − p ; hence, 1 1−t 1 (t) p− Skpk , (4.3) p − M (p, −p) = Skpk ∩ 2t t t provided that t ∈ (0, 1]. For each t ∈ (0, 1] we define the map E (t) : V → P(V ) via the assignment 1 1−t p 7→ Skpk ∩ p− Skpk . t t The first lemma of this section is basically well-known, although it is hardly ever stated in this fashion. Lemma 4.2. Let (V, k·k) be a normed vector space and let p ∈ V be a point. If p is an extreme point of Bkpk , then it holds E (t) (p) = {p} for all real numbers t ∈ (0, 1) and therefore M (t) (p, −p) = {(1 − 2t)p} for all t ∈ [0, 1]. Proof. Note that by the use of the identity (4.3) the second part of the conclusion is a direct consequence of the first part of the conclusion. Thus, we are left to show that if p is an extreme point of Bkpk , then E (t) (p) = {p} for all t in (0, 1). We argue by contraposition. Suppose that there is a real number t in (0, 1) and a point p0 ∈ E (t) (p) with p0 6= p. As p0 ∈ E (t) (p), it follows that p0 ∈ Skpk and that there is a point q ∈ Skpk such that p0 = 1t p − 1−t t q. Observe that q 6= p and 1 1−t 0 (1 − t)q + tp = (1 − t)q + t p− q = p. t t Hence the point p is not extreme in Bkpk , as desired. By putting everything together, the lemma follows.
CONICAL GEODESIC BICOMBINGS ON SUBSETS OF NORMED VECTOR SPACES
14
Next, we define the map λ : D(V ) → V via the assignment (p, q, t) 7→ (1 − t)p + tq. The subsequent lemma is the key ingredient in the proof of Theorem 4.1. Lemma 4.3. Let (V, k·k) be a normed vector space and let A ⊂ V be a subset that admits a conical geodesic bicombing σ : D(A) → A. Let p be a point in A such that −p ∈ A. If there is a point z in V such that the points 2z − p and p − 2z are contained in A and such that σ(p, p − 2z, ·) = λ(p, p − 2z, ·) and σ(2z − p, −p, ·) = λ(2z − p, −p, ·), then we have that σ(p, −p, t) ∈ (1 − 2t)z + M (t) (p − z, z − p) . for all real numbers t ∈ [0, 1]. Proof. Let t ∈ [0, 1] be a real number. We compute kσ(p, −p, t) − λ(p, p − 2z, t)k ≤ 2t kp − zk kσ(p, −p, t) − λ(2z − p, −p, t)k ≤ 2(1 − t) kp − zk . Note that kλ(p, p − 2z, t) − λ(2z − p, −p, t)k = 2 kp − zk. Therefore, it follows that σ(p, −p, t) ∈ M (t) (λ(p, p − 2z, t), λ(2z − p, −p, t)) . It is readily verified that M (t) (u + h, v + h) = h + M (t) (u, v) for all t in [0, 1] and u, v, h ∈ V . Consequently, we obtain that M (t) (λ(p, p − 2z, t), λ(2z − p, −p, t)) = (1 − 2t)z + M (t) (p − z, z − p) . Thus, the lemma follows.
Suppose that A is a subset of a normed vector space (V, k·k) and assume that A admits a conical geodesic bicombing σ : D(A) → A. The translation Tz : A → Tz (A) about the vector z ∈ V given by the assignment x 7→ x + z is an isometry and the map (Tz )∗σ : D(Tz (A)) → Tz (A) given by (x, y, t) 7→ Tz (σ(T−z (x), T−z (y), t)) is a conical geodesic bicombing on Tz (A). Now, we have everything on hand to prove Theorem 4.1. Proof of Theorem 4.1. We proceed by induction on n ≥ 1. If n = 1, then Lemma 4.2 tells us that p − q p − q p−q ,− , t = (1 − 2t) T− p+q σ 2 2 2 2 ∗ for all t ∈ [0, 1]. Thus, we obtain that σ(p, q, t) = (1 − t)p + tq for all t ∈ [0, 1]. Suppose now that n > 1 and that the statement holds for n − 1. We may 1 assume that λ1 ∈ (0, 1). We define (λ01 , . . . , λ0n−1 ) := 1−λ (λ2 , . . . , λn ) and 1 0 0 (e1 , . . . , en−1 ) := (e2 , . . . , en ). Observe that n X k=1
λk ek = λ1 e1 + (1 − λ1 )
n−1 X
λ0k e0k .
Further, note that
n−1
n
X
X
λk ek < 1, λ0k e0k = 1, as otherwise (4.4) implies
k=1
(4.4)
k=1
k=1
(4.5)
CONICAL GEODESIC BICOMBINGS ON SUBSETS OF NORMED VECTOR SPACES
which is not possible due to (4.1). We abbreviate r := z := r(1 − λ1 )
n−1 X
λ0k e0k ,
p0 :=
k=1
kp−qk 2
p−q , 2
15
and we set q 0 := p0 − 2z.
Note that n−1
X p0 − q 0 = r(1 − λ1 ) λ0k e0k . 2 k=1
Hence, by the use of (4.5) it follows that kp0 − q 0 k = r(1 − λ1 ). 2
(4.6)
We have that n
n−1
k=1
k=1
X p−q p0 + q 0 (4.1) X (4.4) = −z = r λk ek − r(1 − λ1 ) λ0k e0k = rλ1 e1 2 2 and therefore (n−1 ) p0 + q 0 kp0 − q 0 k X + (−1)εk λ0k e0k : (ε1 , . . . , εn−1 ) ∈ {0, 1}n−1 2 2 k=1 ( ) n X (4.2) (4.6) εk n−1 = r λ1 e1 + (−1) λk ek : (ε2 . . . , εn ) ∈ {0, 1} ⊂ T− p+q (A). 2
k=2
Thus, we can apply the induction hypothesis to p0 , q 0 ∈ T− p+q (A) and obtain that 2
T− p+q
σ (p0 , p0 − 2z, ·) = λ (p0 , p0 − 2z, ·) .
∗
2
Similarly, we obtain T− p+q σ (2z − p0 , −p0 , ·) = λ (2z − p0 , −p0 , ·) . 2
∗
Now, by the use of Lemma 4.3 it follows that T− p+q σ (p0 , −p0 , t) ∈ (1 − 2t)z + M (t) (p0 − z, z − p0 ) 2
∗
for all real numbers t ∈ [0, 1]; consequently, we get T− p+q σ (p0 , −p0 , t) = (1 − 2t)p0 , 2
∗
since p0 − z = rλ1 e1 is an extreme point in Brλ1 and thus we can use Lemma 4.2 to deduce that M (t) (p0 − z, z − p0 ) = {(1 − 2t)(p0 − z)}. Hence, we have p+q σ(p, q, t) = T− p+q σ (p0 , −p0 , t) + = (1 − t)p + tq, 2 2 ∗ as desired.
CONICAL GEODESIC BICOMBINGS ON SUBSETS OF NORMED VECTOR SPACES
5
16
Proof of Theorem 1.4
Before we start with the proof of Theorem 1.4, we recall some notions from [Mie15] in a slightly modified way. Let (X, d) be a metric space, let p ∈ X be a point and let r > 0 be a real number. We set Ur (p) := {q ∈ X : d(p, q) < r}. Let U ⊂ D(X) be a subset. A map σ : U → X is a convex local geodesic bicombing if for every point p ∈ X there is a real number rp > 0 such that [ U= D(Urp (p)). p∈X
and if the restriction σ|D(Urp (p)) : D(Urp (p)) → Urp (p) is a consistent conical geodesic bicombing for each point p ∈ X. Note that, contrary to [Mie15], we do not assume that σ|D(Urp (p)) is reversible, since this property is not used to prove [Mie15, Theorem 1.1]. We can then restate [Mie15, Theorem 1.1] as follows: Theorem 5.1. Let X be a complete, simply-connected metric space with a convex local geodesic bicombing σ. Then the induced length metric on X admits a unique convex geodesic bicombing σ ˜ which is consistent with σ. The uniqueness of σ ˜ follows immediately from the proof of [Mie15, Theorem 1.1]. With Theorem 5.1 on hand it is possible to derive Theorem 1.4 by the use of Theorem 1.3. Proof of Theorem 1.4. Let int(C) denote the interior of C and let p, q be two points in int(C). We abbreviate [p, q] := (1 − t)p + tq : t ∈ [0, 1] . As int(C) is convex, we have that [p, q] ⊂ int (C). For each point z ∈ C we set ( min{kz − wk : w ∈ [p, q]} if z ∈ C \ int(C) rz := 1 if z ∈ int(C). 2 inf {kz − wk : w ∈ C \ int(C)} Note that rz > 0 for all points z ∈ C and we have that Urz (z) ∩ [p, q] = ∅ if z ∈ C \int(C). Further, for every point z ∈ int(C) it follows that B2rz (z) ⊂ C; thus, we may invoke Theorem 1.3 to deduce that if z ∈ int(C), then σz1 z2 (t) = (1 − t)z1 + tz2 for all points z1 , z2 ∈ Brz (z) and all real numbers t ∈ [0, 1]. We define [ U := D(Urz (z)). z∈C loc
Note that the map σ := σ|U defines a convex local bicombing on C. The geodesic σpq (·) and the linear geodesic from p to q are both consistent with the local bicombing σ loc . Hence, by Theorem 5.1, we conclude that σpq (·) must be equal to the linear geodesic from p to q, that is, we have σpq (t) = (1 − t)p + tq for all real numbers t ∈ [0, 1]. Now, suppose that p, q ∈ C. As C is convex, it is well-known that C = int (C), cf. [AB06, Lemma 5.28]. Let (pk )k≥1 , (qk )k≥1 ⊂ int (C) be two sequences such that pk → p and qk → q with k → +∞. It is readily verified that σpk qk (·) → σpq (·) with k → +∞, since σ is a conical geodesic bicombing. As a result, we obtain that the geodesic σpq (·) is equal to the linear geodesic from p to q, as desired.
CONICAL GEODESIC BICOMBINGS ON SUBSETS OF NORMED VECTOR SPACES
17
References [AB95] J. Alonso and M. Bridson. Semihyperbolic groups. Proceedings of the London Mathematical Society, 3(1):56–114, 1995. [AB06] C. Aliprantis and K. Border. Infinite dimensional analysis: a hitchhiker’s guide. Springer, 2006. [Bas15] G. Basso. Fixed point theorems for metric spaces with a conical geodesic bicombing. arXiv preprint arXiv:1509.01691, 2015. [Des15] D. Descombes. Asymptotic rank of spaces with bicombings. arXiv preprint arXiv:1510.05393, 2015. [DL15] D. Descombes and U. Lang. Convex geodesic bicombings and hyperbolicity. Geometriae Dedicata, 177(1):367–384, 2015. [DL16] D. Descombes and U. Lang. Flats in spaces with convex geodesic bicombings. Analysis and Geometry in Metric Spaces, 4(1):68–84, 2016. [Dre84] A. W. M. Dress. Trees, tight extensions of metric spaces, and the cohomological dimension of certain groups: a note on combinatorial properties of metric spaces. Advances in Mathematics, 53(3):321–402, 1984. [GM81] S. G¨ ahler and G. Murphy. A metric characterization of normed linear spaces. Mathematische Nachrichten, 102(1):297–309, 1981. [Goo50] D. B. Goodner. Projections in normed linear spaces. Transactions of the American Mathematical Society, 69(1):89–108, 1950. [Kel52] J. L. Kelley. Banach spaces with the extension property. Transactions of the American Mathematical Society, 72(2):323–326, 1952. [Kel16] M. Kell. Sectional curvature-type conditions on finsler-like metric spaces. arXiv preprint arXiv:1601.03363, 2016. [Lan13] U. Lang. Injective hulls of certain discrete metric spaces and groups. Journal of Topology and Analysis, 5(03):297–331, 2013. [LY10] Y.-C. Li and C.-C. Yeh. Some characterizations of convex functions. Computers & mathematics with applications, 59(1):327–337, 2010. [Mie15] B. Miesch. The Cartan-Hadamard Theorem for metric spaces with local bicombings. arXiv preprint arXiv:1509.07001, 2015. [RS90] S. Reich and I. Shafrir. Nonexpansive iterations in hyperbolic spaces. Nonlinear Analysis: Theory, Methods and Applications, 15(6):537–558, 1990. [Tak70] W. Takahashi. A convexity in metric space and nonexpansive mappings. I. Kodai Mathematical Seminar Reports, 22(2):142–149, 1970.
¨ rich, Ra ¨ mistrasse 101, 8092 Zu ¨ rich, Schweiz Mathematik Departement, ETH Zu
E-mail adress: [email protected] E-mail adress: [email protected]
CONICAL GEODESIC BICOMBINGS ON SUBSETS OF NORMED VECTOR SPACES GIULIANO BASSO AND BENJAMIN MIESCH
Abstract. In this paper we establish existence and uniqueness results for conical geodesic bicombings on subsets of normed vector spaces. Concerning existence, we give a first example of a convex geodesic bicombing that is not consistent. Furthermore, we show that under a mild geometric assumption on the norm a conical geodesic bicombing on an open subset of a normed vector space locally consists of linear geodesics. As an application, we obtain by the use of a Cartan-Hadamard type result that if a closed convex subset of a Banach space has non-empty interior, then it admits a unique consistent conical geodesic bicombing.
1
Introduction
In [DL15], D. Descombes and U. Lang initiated the systematic study of metric spaces that obey a weak form of non-positive curvature which is encapsulated in the existence of a conical geodesic bicombing. Essentially, a conical geodesic bicombing on a metric space is a selection of geodesics such that the distance function between each two selected geodesics is bounded from above by the linear interpolation of its values at the endpoints. Conical geodesic bicombings arise naturally in the context of injective metric spaces, cf. [Lan13], and the class of metric spaces that admit a conical geodesic bicombing is closed under 1-Lipschitz retractions and ultralimits. The term bicombing originates from W. Thurston and is prominently used by J. Alonso and M. Bridson to define semihyperbolic metric spaces, cf. [AB95]. Notions related to conical geodesic bicombings also occur in metric fixed point theory, most notably W-convexity mappings, cf. [Tak70], and hyperbolic spaces in the sense of S. Reich and I. Shafrir, cf. [RS90]. Recently, classical results from the theory of CAT(0) spaces have been transferred to metric spaces that admit conical geodesic bicombings, cf. [Bas15, Des15, DL16, Kel16] and [Mie15]. In this article we proceed with the systematic study of metric spaces that admit conical geodesic bicombings and we enlarge the collection of examples from [DL15]. Let (X, d) denote a metric space. We abbreviate D(X) := X × X × [0, 1]. A map σ : D(X) → X is said to be a geodesic bicombing if the path σpq (·) := σ(p, q, ·) is a constant speed geodesic from p to q for all points p, q in X, that is, we have σpq (0) = p, σpq (1) = q and d(σpq (t), σpq (s)) = |t − s| d(p, q) for all real numbers s, t ∈ [0, 1]. Furthermore, a geodesic bicombing σ : D(X) → X is called conical if it satisfies d(σpq (t), σp0 q0 (t)) ≤ (1 − t)d(p, p0 ) + td(q, q 0 ) Date: April 15, 2016. 1
(1.1)
CONICAL GEODESIC BICOMBINGS ON SUBSETS OF NORMED VECTOR SPACES
2
for all points p, q, p0 , q 0 in X and all real numbers t ∈ [0, 1]. Examples of metric spaces that admit conical geodesic bicombings include convex subsets of normed vector spaces, Busemann spaces and injective metric spaces. Our first result deals with convex geodesic bicombings. A geodesic bicombing σ : D(X) → X is convex if the map t 7→ d(σpq (t), σp0 q0 (t)) is convex on [0, 1] for all points p, q, p0 , q 0 in X. Note that if the underlying metric space is not uniquely geodesic, then a conical geodesic bicombing is not necessarily convex. Examples of conical geodesic bicombings that are not convex are ubiquitous; for instance, nonconvex conical geodesics bicombings may be obtained via 1-Lipschitz retractions of linear geodesics, see [DL15, Example 2.2] or Lemma 3.1. In [DL15], it is shown that metric spaces of finite combinatorial dimension in the sense of Dress, cf. [Dre84], possess at most one convex geodesic bicombing. If it exists, this unique convex geodesic bicombing, say σ : D(X) → X, has the property that it is consistent, that is, we have for all points p, q in X that im(σp0 q0 ) ⊂ im(σpq ) whenever p0 = σpq (s) and q 0 = σpq (t) with 0 ≤ s ≤ t ≤ 1. Clearly, every consistent conical geodesic bicombing is convex. In Section 2, we show that the converse does not hold by proving the subsequent theorem. Theorem 1.1. There is a compact metric space that admits a convex geodesic bicombing which is not consistent. Although there is a non-consistent convex geodesic bicombing on the space considered in Section 2, this space also admits a consistent convex geodesic bicombing. We suspect that this is a general phenomenon. Question 1.2. Let (X, d) be a proper metric space with a convex geodesic bicombing. Does X also admit a consistent convex geodesic bicombing? The seemingly more general question if every proper metric space with a conical geodesic bicombing admits a consistent conical geodesic bicombing is in fact equivalent to Question 1.2, as every proper metric space with a conical geodesic bicombing also admits a convex geodesic bicombing, cf. [DL15, Theorem 1.1]. A geodesic bicombing σ : D(X) → X is called reversible if σpq (t) = σqp (1 − t) for all points p, q in X and all t ∈ [0, 1]. Every proper metric space with a conical geodesic bicombing admits a reversible conical geodesic bicombing, cf. [Des15, Proposition 1.2]. However, it remained open if there are conical geodesic bicombings that are non-reversible. In Section 2, we modify our non-consistent convex geodesic bicombing from Theorem 1.1 in order to obtain an example of a non-reversible convex geodesic bicombing, see Proposition 2.5. In [Bas15], a barycentric construction has been employed to obtain fixed point results for metric spaces that admit conical geodesic bicombings. This barycentric construction motivated the following definition: A geodesic bicombing σ : D(X) → X has the midpoint property if σpq ( 21 ) = σqp ( 12 ) for all points p, q in X. It seems natural to ask if every conical geodesic bicombing that has the midpoint property is automatically reversible. We show that this is not the case, as we construct in Section 3 a non-reversible conical geodesic bicombing which has the midpoint property. It is a direct consequence of a result of S. G¨ahler and G. Murphy that the only conical geodesic bicombing on a normed vector space is the one that consists of the linear geodesics, cf. [GM81, Theorem 1]. With a mild geometric assumption on the norm, we show in Section 4 that already a conical geodesic bicombing on an open
CONICAL GEODESIC BICOMBINGS ON SUBSETS OF NORMED VECTOR SPACES
3
subset of a normed vector space locally consists of linear geodesics. More generally, we get the following result: Theorem 1.3. Let (V, k·k) be a normed vector space with the property that its closed unit ball is the closed convex hull of its extreme points. Suppose that A ⊂ V is a subset of V that admits a conical geodesic bicombing σ : D(A) → A and let p0 in A be a point. If r ≥ 0 is a real number such that B2r (p0 ) ⊂ A, then we have that σ(p, q, t) = (1 − t)p + tq for all points p, q ∈ Br (p0 ) and all real numbers t ∈ [0, 1]. We do not know if Theorem 1.3 remains true if we drop the assumption of the normed vector space (V, k·k) having the property that its closed unit ball is the closed convex hull of its extreme points. But how common is it that the closed unit ball of a normed vector space is the closed convex hull of its extreme points? By invoking the Banach-Alao˘glu theorem and the Kre˘ın-Mil’man theorem it is possible to show that the closed unit ball of a dual Banach space is the closed convex hull of its extreme points. Consequently, we obtain in particular that Theorem 1.3 is valid in every reflexive Banach space. Moreover, using a classification result, due to L. Nachbin, D. Goodner, and J. Kelley, cf. [Kel52], and a result of D. Goodner, cf. [Goo50, Theorem 6.4], it is readily verified that Theorem 1.3 also holds for every injective Banach space. In [Mie15], the second named author generalized the classical Cartan-Hadamard Theorem to metric space that locally admit a consistent convex geodesic bicombing. With Theorem 1.3 on hand, it is possible to use this generalized Cartan-Hadamard Theorem to obtain the following uniqueness result. Theorem 1.4. Let (E, k·k) be a Banach space with the property that its closed unit ball is the closed convex hull of its extreme points. Suppose that C ⊂ E is a closed convex subset of E with non-empty interior. If σ : D(C) → C is a consistent conical geodesic bicombing, then it follows that σ(p, q, t) = (1 − t)p + tq for all points p, q in C and all real numbers t ∈ [0, 1]. The proof of Theorem 1.4 is given in Section 5. Due to Theorems 1.3 and 1.4 it appears that the geometry of a convex subset C of a normed vector space (V, k·k) is very restricted in the sense that it is difficult to construct a conical geodesic bicombing on C that is not given by the linear geodesics. In this perspective, we deem that an affirmative answer to the following question would result in an interesting geometric construction. Question 1.5. Is there a convex subset of a normed vector space that admits two distinct conical geodesic bicombings? In a similar way, one might ask if consistency always implies uniqueness, or in other words: Question 1.6. Is there a metric space with two different consistent conical geodesic bicombings? A possible candidate to an affirmative answer to Question 1.6 is given in [DL15, Example 3.5]. Unfortunately, it is unclear whether the two convex geodesic bicombings given there are consistent or not. Acknowledgments. We would like to thank Urs Lang for helpful remarks and discussions. The authors gratefully acknowledge support from the Swiss National Science Foundation.
CONICAL GEODESIC BICOMBINGS ON SUBSETS OF NORMED VECTOR SPACES
2
4
A non-consistent convex geodesic bicombing
The goal of this section is to construct a convex geodesic bicombing that is not consistent and therefore establish Theorem 1.1. To this end, we consider the following norm on R2 : o n √ k(x, y)k := max |x|, 22 k(x, y)k2 , p where k(x, y)k2 = x2 + y 2 is the Euclidean norm. Observe that k(x, y)k = |x| if and only if |y| ≤ |x|. Now define 1 X := (x, y) ∈ R2 : −3 ≤ x ≤ 3, 0 ≤ y ≤ 32 max{0, 1 − x2 } and equip X with the metric d induced by k · k.
-3
-2
-1
0
1
2
3
Figure 1. The metric space X. The space X naturally splits into three pieces, namely X = X− ∪ X0 ∪ X+ with X− := [−3, −1] × {0}, X0 := (x, y) ∈ R2 : −1 < x < 1, 0 ≤ y ≤
1 32 (1
− x2 ) ,
X+ := [1, 3] × {0}. 1 Definition 2.1. For δ ∈ [0, 64 ] we define a geodesic bicombing σ δ : D(X) → X as δ follows. Generally, we take σpq to be the geodesic from p to q which is linear inside X0 , but if both endpoints lie on the antennas X− , X+ we slightly modify it. In more details σ δ is defined as follows: For p := (px , py ), q := (qx , qy ) ∈ X with px ≤ qx let δ σpq (t) := (xpq (t), ypq (t)) ,
with xpq (t) := px + t(qx − px ), 2 δ max{qx − px − 4, 0} max{0, (1 − xpq (t) )}, q y (xpq (t) + 1)}, max{0, qx +1 py ypq (t) := max{0, px −1 (xpq (t) − 1)}, py + t(qy − py ), 0,
for p ∈ X− , q ∈ X+ , for p ∈ X− , q ∈ X0 , for p ∈ X0 , q ∈ X+ , for p, q ∈ X0 , otherwise.
and δ δ σqp (t) := σpq (1 − t). 1 Proposition 2.2. For δ ∈ (0, 64 ], the map σ δ is a reversible convex geodesic bicombing which is not consistent.
Remark 2.3. Observe that for δ = 0 the geodesic bicombing σ δ coincides with the piece-wise linear bicombing which is the unique consistent conical geodesic bicombing on X by Theorem 1.4. Hence we have a family of non-consistent convex geodesic bicombings σ δ converging to the unique consistent convex geodesic bicombing σ 0 .
CONICAL GEODESIC BICOMBINGS ON SUBSETS OF NORMED VECTOR SPACES
5
Alternatively, we can modify the geodesics leading from X− to X+ . Definition 2.4. Define σ ˜ δ : D(X) → X by δ δ σ ˜pq (t) = σpq (t),
except for p ∈ X+ , q ∈ X− let δ σ ˜pq = (xpq (t), 0). 1 Proposition 2.5. For δ ∈ (0, 64 ], the map σ ˜ δ is a convex geodesic bicombing which neither is reversible nor consistent.
Let us first show that we defined geodesic bicombings. 1 Lemma 2.6. For δ ∈ [0, 64 ], the maps σ δ and σ ˜ δ are geodesic bicombings.
Proof. The linear case is clear. For the piecewise linear case observe that if p ∈ X− , δ satisfies q ∈ X0 (and similarly in all other cases) we have that the slope m of σpq m=
qy ≤ qx + 1
1 32 (1
− qx2 ) = 1 + qx
1 32 (1
− qx ) ≤
1 16
≤1
and therefore δ δ d(σpq (s), σpq (t)) = |xpq (s) − xpq (t)| = |s − t||qx − px | = |s − t|d(p, q).
Finally, let p ∈ X− , q ∈ X+ . For x, x0 ∈ [−1, 1] we have |δ(qx − px − 4)(1 − x2 ) − δ(qx − px − 4)(1 − x02 )| ≤ δ|qx − px − 4| · |x + x0 | · |x − x0 | ≤
1 16 |x
− x0 |
δ δ (t)) = |xpq (s) − xpq (t)| as before. (s), σpq and hence d(σpq
To prove Propositions 2.2 and 2.5, we will use the following characterization of convexity; see Lemma 3.5 in [LY10]. Lemma 2.7. A continuous function f : [0, 1] → R is convex if and only if for every t ∈ (0, 1) there is some τ0 > 0 such that for all τ ∈ [0, τ0 ] we have 2f (t) ≤ f (t − τ ) + f (t + τ ). Proof of Proposition 2.2. It is immediate that the geodesic bicombing σ δ is nonconsistent and reversible; hence, it remains to prove convexity. δ Given p, q, p0 , q 0 ∈ X we need to show that f (t) := d(σpq (t), σpδ0 q0 (t)) is convex on [0, 1]. Clearly t 7→ |xpq (t) − xp0 q0 (t)| is convex for t ∈ [0, 1]. Hence in the situation δ (t), σpδ0 q0 (t)) = |xpq (t) − xp0 q0 (t)| we have when d(σpq δ 2d(σpq (t), σpδ0 q0 (t)) = 2|xpq (t) − xp0 q0 (t)|
≤ |xpq (t − τ ) − xp0 q0 (t − τ )| + |xpq (t + τ ) − xp0 q0 (t + τ )| δ δ ≤ d(σpq (t − τ ), σpδ0 q0 (t − τ )) + d(σpq (t + τ ), σpδ0 q0 (t + τ )).
Therefore, it remains to check δ 2kσpq (t) − σpδ0 q0 (t)k2 δ δ ≤ kσpq (t − τ ) − σpδ0 q0 (t − τ )k2 + kσpq (t + τ ) − σpδ0 q0 (t + τ )k2 δ for τ > 0 small, whenever d(σpq (t), σpδ0 q0 (t)) = |xpq (t) − xp0 q0 (t)| ≤ |ypq (t) − yp0 q0 (t)|.
√
2 δ 2 kσpq (t)
− σpδ0 q0 (t)k2 , that is,
CONICAL GEODESIC BICOMBINGS ON SUBSETS OF NORMED VECTOR SPACES
6
Observe that for x ∈ [−3, −1] ∪ [1, 3] and (x0 , y 0 ) ∈ X we have that δ d((x, 0), (x0 , y 0 )) = |x − x0 | and therefore we always have d(σpq (t), σpδ0 q0 (t)) = |xpq (t) − xp0 q0 (t)| if xpq (t) ∈ / (−1, 1). Hence, we only need to consider points that satsify xpq (t), xp0 q0 (t) ∈ (−1, 1). δ , σpδ0 q0 are (piece-wise) linear, then locally they are linear First, if both σpq geodesics inside a normed vector space and hence d(σpq (t), σp0 q0 (t)) = kσpq (t) − σp0 q0 (t)k is locally convex, thus convex. δ Let us now assume that σpq is not linear, i.e. p ∈ X− , q ∈ X+ , l := d(p, q) ≥ 4. We look at the different options for σpδ0 q0 separately. But before doing so, let us first fix some notation. We define p0 := σpq (t), p± := σpq (t ± τ ), p∗ = (x∗ , y∗ ) (∗ ∈ {0, +, −}), D := δ(l − 4), ε := τ l and accordingly for σpδ0 q0 . We then get y0 = D(1 − x20 ), x± = x0 ± ε and y± = D(1 − (x0 ± ε)2 ). In each case, we need to consider the situation where x0 , x00 ∈ (−1, 1) and |x0 − x00 | ≤ |y0 − y00 |. Case 1. p0 ∈ X∓ , q 0 ∈ X± and l0 := d(p0 , q 0 ) ∈ [4, l]. 0 0 0 2 0 0 0 0 As above we have y00 = D0 (1 − x02 0 ), x± = x0 ± ε , y± = D (1 − (x0 ± ε ) ) and 0 with λ := ll we get ε0 = λε. We claim that
2kp0 − p00 k2 ≤ kp− − p0− k2 + kp+ − p0+ k2 for ε > 0 (i.e. τ > 0) small enough. First note that kp− − p0− k22 = kp0 − p00 k22 − 2(x0 − x00 )(1 − λ)ε + (1 − λ)2 ε2 + 2(y0 − y00 )aε + a2 ε2 , kp+ − p0+ k22 = kp0 − p00 k22 + 2(x0 − x00 )(1 − λ)ε + (1 − λ)2 ε2 + 2(y0 − y00 )bε + b2 ε2 , for a := 2(x0 D − λx00 D0 ) − (D − λ2 D0 )ε, b := −2(x0 D − λx00 D0 ) − (D − λ2 D0 )ε, with a+b = −2(D −λ2 D0 )ε, a−b = 4(x0 D −λx00 D0 ) and either ab = (D −λ2 D0 )2 ε2 or ab < 0 for ε small. In the following, we assume ab < 0. The other case is similar. Moreover, we have kp− − p0− k22 · kp+ − p0+ k22 = kp0 − p00 k42 + 4ab(y − y 0 )2 − 4(x0 − x00 )2 (1 − λ)2 + 4(x0 − x00 )(1 − λ)(y0 − y00 )(a − b) + 2(1 − λ)2 + a2 + b2 − 4(y0 − y00 )(D − λ2 D0 ) · kp0 − p00 k22 ε2 + O(ε3 ) √ √ and with u + t = u + 2√t u + O(t2 ) and u = kp0 − p00 k42 it follows q 2 kp− − p0− k22 · kp+ − p0+ k22 0 2 ≥ 2kp0 − p0 k2 + 2(1 − λ)2 + a2 + b2 + 4ab − 4(y0 − y00 )(D − λ2 D0 ) 4(x0 − x00 )(y0 − y00 )(1 − λ)(a − b) − 4(x0 − x00 )2 (1 − λ)2 2 + ε + O(ε3 ). (x0 − x00 )2 + (y0 − y00 )2
CONICAL GEODESIC BICOMBINGS ON SUBSETS OF NORMED VECTOR SPACES
We therefore get kp− − p0− k2 + kp+ − p0+ k2
2
q kp− − p0− k22 + kp+ − p0+ k22 + 2 kp− − p0− k22 · kp+ − p0+ k22 0 2 ≥ 4kp0 − p0 k2 + 4(1 − λ)2 + 2(a + b)2 − 8(y0 − y00 )(D − λ2 D0 ) 4(x0 − x00 )(y0 − y00 )(1 − λ)(a − b) − 4(x0 − x00 )2 (1 − λ)2 2 + ε + O(ε3 ) (x0 − x00 )2 + (y0 − y00 )2 =
=
4kp0 − p00 k22 + Cε2 + O(ε3 )
≥
4kp0 − p00 k22 ,
for ε > 0 small enough, provided that 4(1 − λ)2 − 8(y0 − y00 )(D − λ2 D0 )
C= +
16(x0 − x00 )(y0 − y00 )(1 − λ)(x0 D − λx00 D0 ) − 4(x0 − x00 )2 (1 − λ)2 > 0. (x0 − x00 )2 + (y0 − y00 )2
Observe that a + b = O(ε). First, assuming y0 > y00 , we have 2 y0 − y00 = D(1 − x20 ) − D0 (1 − x20 ) = (D − D0 )(1 − x20 ) + D0 (x02 0 − x0 )
≤ δ(l − l0 ) + δ(l0 − 4)(x00 + x0 )(x00 − x0 ) ≤ δ(l − l0 ) + 4δ(y0 − y00 ) and therefore |y0 − y00 | ≤
δ (l − l0 ). 1 − 4δ
Moreover, |D − λ2 D0 |l2 = δ(l3 − 4l − l03 + 4l0 ) = δ(l − l0 )(l2 + ll0 + l02 − 4(l + l0 )) ≤ 60δ(l − l0 ), |x0 D − λx00 D0 |l ≤ |x0 |(D − λD0 )l + |x0 − x00 |D0 l0 ≤ δ(l − l0 )(l + l0 − 4) + 12δ|y0 − y00 | ≤
8δ +
12δ 2 1 − 4δ
(l − l0 ).
Hence, we finally get Cl2 kp0 − p00 k22 =
4(l − l0 )2 (y0 − y00 )2 − 8(y0 − y00 )(D − λ2 D0 )l2 (x0 − x00 )2 + (y0 − y00 )2
+ 16(x0 − x00 )(y0 − y00 )(l − l0 )(x0 D − λx00 D0 )l 192δ 2 960δ 2 − 128δ − (l − l0 )2 (y0 − y00 )2 ≥ 4− 1 − 4δ 1 − 4δ 4 − 144δ − 640δ 2 = (l − l0 )2 (y0 − y00 )2 > 0 1 − 4δ for δ
0 for δ < 0.026. Case 3. σp0 q0 linear with p0 , q 0 ∈ X0 . Let m again denote the slope of σp0 q0 . We distinguish two subcases. (a) If |m| ≤ 1, we have l ∈ [4, 6], l0 ∈ [0, 2] and |ml0 | =
|qy0 − p0y | 0 l = |qy0 − p0y | ≤ |qx0 − p0x |
1 32 .
0
0 Moreover, for ε0 = τ l0 , λ = ss we get x0± = x00 ± ε0 , y± = y00 ± mε0 and ε0 = λε as before and again we get the constant
4(1 − λ)2 − 8(y0 − y00 )D
C= +
8(x0 − x00 )(y0 − y00 )(1 − λ)(λm + 2Dx0 ) − 4(x0 − x00 )2 (1 − λ)2 . (x0 − x00 )2 + (y0 − y00 )2
CONICAL GEODESIC BICOMBINGS ON SUBSETS OF NORMED VECTOR SPACES
9
Now, we estimate Cl2 kp0 − p00 k22 =
4(l − l0 )2 (y0 − y00 )2 − 8(y0 − y00 )Dl2 (x0 − x00 )2 + (y0 − y00 )2
+ 8(x0 − x00 )(y0 − y00 )(l − l0 )(ml0 + 2Dx0 l) ≥ 4 − 576δ 2 − 81 − 96δ (l − l0 )2 (y0 − y00 )2 2 31 = (l − l0 )2 (y0 − y00 )2 > 0 8 − 96δ − 576δ for δ < 0.033. (b) If |m| > 1, we have l ∈ [4, 6] and √ q l0 = 22 (qx − px )2 + (qy − py )2 ≤ |qy − py | ≤
1 32 .
Furthermore, let ε0x = x+ − x0 and ε0y = y+ − y0 . Then we have ε0y = mε0x and 2
2
2(τ l0 )2 = ε0x + (mε0x )2 = (1 + m2 )ε0x . 0
√
2 , ε0y = λy ε for λy := mλx = Thus we get ε0x = λx ε for λx := ll √1+m 2 0 x0± = x00 ± ε0x and y± = y00 ± ε0y . We proceed again as before and get the constant
√ l0 √ 2m l 1+m2 ,
4(1 − λx )2 − 8(y0 − y00 )D
C= +
8(x0 − x00 )(y0 − y00 )(1 − λx )(λy + 2Dx0 ) − 4(x0 − x00 )2 (1 − λx )2 , (x0 − x00 )2 + (y0 − y00 )2
with D = δ(l − 4) ≤ δ(l − l0 ) ≤ 6δ(1 − λx ), y0 − y00 ≤ D(1 − x20 ) ≤ D ≤ 6δ(1 − λx ), √ √ l0 2 2 1 ≤ λy = q ≤ (1 − λx ). l 128 64 1 m2 + 1 Now, we estimate Ckp0 − p00 k22 =
4(1 − λx )2 (y0 − y00 )2 − 8(y0 − y00 )D (x0 − x00 )2 + (y0 − y00 )2
+ 8(x0 − x00 )(y0 − y00 )(1 − λx )(λy + 2Dx0 ) 1 2 − 96δ (1 − λx )2 (y0 − y00 )2 ≥ 4 − 576δ − 64 2 255 = (1 − λx )2 (y0 − y00 )2 > 0 64 − 96δ − 576δ 1 for δ < 0.034. Hence this is again true for δ ≤ 64 . √ 0 Observe that for m → +∞ we get λx = 0 and λy = 2 ll , and the same estimates hold.
Proof of Proposition 2.5. Clearly the geodesic bicombing σ ˜ δ is non-consistent and non-reversible.
CONICAL GEODESIC BICOMBINGS ON SUBSETS OF NORMED VECTOR SPACES
10
For convexity, the same arguments as in the proof of Proposition 2.2 apply. The only new case is p0 ∈ X+ and q 0 ∈ X− . With the notions from above with 0 x0± = x00 ∓ ε0 for ε0 = τ l0 and λ = ll we obtain the constant 4(1 + λ)2 − 8y0 D
C= +
16(x0 − x00 )y0 (1 + λ)x0 D − 4(x0 − x00 )2 (1 + λ)2 . (x0 − x00 )2 + y02
With the inequalities D = δ(l − 4) ≤ 2δ and |y0 | ≤ Ckp0 − p00 k22 =
1 32
we get
4(1 + λ)2 y02 − 8y0 D (x0 − x00 )2 + y02
+ 16(x0 − x00 )y0 (1 + λ)x0 D
for δ
0
≥
(4 − 33δ) (1 + λ)2 y02 > 0
1 64 .
A non-reversible conical geodesic bicombing that has the midpoint property
In the first part of this section we construct a non-reversible conical geodesic bicombing. Afterwards, we modify this non-reversible conical geodesic bicombing to satisfy the midpoint property. To begin, we define some auxiliary maps. Consider R2 equipped with the maximum norm k·k∞ and let s : R2 → R2 denote the map given by (x, y) 7→ (x, −y). We define X1 := (x, y) ∈ R2 : x ∈ [−2, 1] and |x| − 1 ≤ y ≤ ||x| − 1| , A1 := (x, y) ∈ R2 : |x + 1| ≤ y ≤ 1 . and X2 := s(X1 ), A2 := s(A1 ). The set readily verified that the map f : X2 → X1 ( (x, y), (x, y) 7→ s(x, y),
X1 ∪ X2 is depicted in Figure 2. It is given by if x ∈ [−1, 1], if x ∈ [−2, −1]
is an isometry. Let f¯: X1 ∪ X2 → X1 be the map that is equal to IdX1 on X1 and equal to f on X2 . Observe that the map f¯ is 1-Lipschitz. We set Y1 := X1 ∪ A1 and Y2 := X2 ∪ A2 . Further, we define the map π : Y1 ∪ Y2 → X1 ∪ X2 through the assignment (x, y) 7→ x, sgn(y) min |y| , ||x| − 1| . Observe that π is a 1-Lipschitz retraction that maps Yk to Xk for each k ∈ {1, 2}. Let λ : D(R2 ) → R2 be the conical geodesic bicombing on R2 that is given by the linear geodesics. Lemma 3.1. The map σ : D(X1 ) → X1 given by ( π ◦ λ(p, q, t), (p, q, t) 7→ f ◦ π ◦ λ f −1 (p), f −1 (q), t ,
if px ≤ qx , if qx ≤ px .
is a non-reversible conical geodesic bicombing on (X1 , k·k∞ ).
CONICAL GEODESIC BICOMBINGS ON SUBSETS OF NORMED VECTOR SPACES
p
11
1.
0.5 q
−2.5
−2.
−1.5
−1.
−0.5
0.5
1.
0 −0.5 p0
−1.
−1.5
Figure 2. The blue line corresponds to σpq and the red line corresponds to the image of σqp under the isometry f −1 .
Proof. Observe that both maps σ (1) := π ◦ λ and σ (2) := f ◦ π ◦ λ ◦ f −1 × f −1 × Id[0,1]
define conical geodesic bicombings on X1 . Thus, it follows that σ : D(X1 ) → X1 is a geodesic bicombing. In the following we show that σ is conical. Let p, q, p0 , q 0 ∈ X1 be points. As (2) (1) both maps σ (1) and σ (2) are conical geodesic bicombings on X1 with σpq = σpq if px , qx ≤ −1 or px , qx ≥ −1, it remains to check inequality (1.1) in the two cases if px , qx0 ≤ −1 and qx , p0x ≥ −1 or p0x , qx ≤ −1 and qx0 , px ≥ −1. Now, suppose that px , qx0 ≤ −1 and qx , p0x ≥ −1. The other case is treated analogously. Let t ∈ [0, 1] be a real number. Since the map f¯ ◦ π is 1-Lipschitz, we compute
kσpq (t) − σp0 q0 (t)k∞ = f¯ ◦ π ◦ λ(p, q, t) − f¯ ◦ π ◦ λ(f −1 (p0 ), f −1 (q 0 ), t) ∞
≤ (1 − t) p − f −1 (p0 ) ∞ + t q − f −1 (q 0 ) ∞ . By our assumptions on the points p, q, p0 , q 0 , it follows that
p − f −1 (p0 ) = kp − p0 k , ∞
∞
q − f −1 (q 0 ) = f −1 (q) − f −1 (q 0 ) = kq − q 0 k . ∞ ∞ ∞ Hence, by putting everything together, we obtain that σ is a conical geodesic bicombing on X1 . By construction, it follows that σ is non-reversible; see Figure 2. Now, we may use σ to construct a non-reversible conical geodesic bicombing that has the midpoint property.
CONICAL GEODESIC BICOMBINGS ON SUBSETS OF NORMED VECTOR SPACES
12
Lemma 3.2. The map τ : D(X1 ) → X1 given by the assignment ( σ p, 21 σ(p, q, 12 ) + σ(q, p, 12 ) , 2t , if t ∈ [0, 12 ], (p, q, t) 7→ 1 1 1 σ 2 σ(p, q, 2 ) + σ(q, p, 2 ) , q, 2t − 1 , if t ∈ [ 12 , 1], is a conical geodesic bicombing on (X1 , k·k∞ ) that has the midpoint property but is not reversible. Proof. It is straightforward that τ is a conical geodesic bicombing with the midpoint property. To see that τ is non-reversible, take for instance p := (− 32 , 12 ), q := (0, 12 ) and observe that 1 7 5 ) = (− 78 , 18 ) 6= (− 87 , 48 ) = τ (q, p, 12 ); τ (p, q, 12
compare Figure 3.
1.
0.5
p
q
m
−2.5
−2.
−1.5 p
−1.
−0.5
0.5
1.
0
0
−0.5
−1.
−1.5
Figure 3. The blue line corresponds to τpq |[0, 12 ] and the red line corresponds to the image of τqp |[ 12 ,1] under the isometry f −1 . The point m is equal to 21 σpq ( 21 ) + σqp ( 12 ) .
4
Conical geodesic bicombings on subsets of normed vector spaces
Let (V, k·k) be a normed vector space, let p0 ∈ V be a point and let r ≥ 0 be a real number. We set Ur (p0 ) := {z ∈ V : kp0 − zk < r}, Br (p0 ) := {z ∈ V : kp0 − zk ≤ r}, Sr (p0 ) := {z ∈ V : kp0 − zk = r}. To ease notation, we abbreviate Br := Br (0) and Sr := Sr (0). The goal of this section is to establish the following rigidity result.
CONICAL GEODESIC BICOMBINGS ON SUBSETS OF NORMED VECTOR SPACES
13
Theorem 4.1. Let (V, k·k) be a normed vector space. Suppose that A ⊂ V is a subset of V that admits a conical geodesic bicombing σ : D(A) → A and let p, q be points of A. If there are points eP 1 , . . . , en ∈ B1 that are extreme points of B1 and a n tuple (λ1 , . . . , λn ) ∈ [0, 1]n with k=1 λk = 1 such that n
p−q kp − qk X = λk ek and 2 2 k=1 ( n ) p + q kp − qk X + (−1)εk λk ek : (ε1 , . . . , εn ) ∈ {0, 1}n ⊂ A, 2 2
(4.1)
(4.2)
k=1
then it follows that σ(p, q, t) = (1 − t)p + tq for all t ∈ [0, 1]. Theorem 1.3 then is a direct consequence. Proof of Theorem 1.3. Let p, q ∈ Br (p0 ) be two points. As p+q ∈ Br (p0 ) and 2 kp−qk p+q ≤ r, the ball B kp−qk ( 2 ) is contained in A. Hence, since the unit ball of V 2 2 is the closed convex hull of its extreme points, it follows that σ(p, q, t) = (1−t)p+tq for all t ∈ [0, 1] by Theorem 4.1 and a simple limiting argument. In order to derive Theorem 4.1 we need some preparatory lemmas and definitions. Let t ∈ [0, 1] be a real number and let p, q be points in V . We define M (t) (p, q) := {z ∈ V : kz − pk = t kp − qk , kz − qk = (1 − t) kp − qk} By construction, we have M (t) (p, −p) = S2tkpk + p ∩ S(1−t)2kpk − p ; hence, 1 1−t 1 (t) p− Skpk , (4.3) p − M (p, −p) = Skpk ∩ 2t t t provided that t ∈ (0, 1]. For each t ∈ (0, 1] we define the map E (t) : V → P(V ) via the assignment 1 1−t p 7→ Skpk ∩ p− Skpk . t t The first lemma of this section is basically well-known, although it is hardly ever stated in this fashion. Lemma 4.2. Let (V, k·k) be a normed vector space and let p ∈ V be a point. If p is an extreme point of Bkpk , then it holds E (t) (p) = {p} for all real numbers t ∈ (0, 1) and therefore M (t) (p, −p) = {(1 − 2t)p} for all t ∈ [0, 1]. Proof. Note that by the use of the identity (4.3) the second part of the conclusion is a direct consequence of the first part of the conclusion. Thus, we are left to show that if p is an extreme point of Bkpk , then E (t) (p) = {p} for all t in (0, 1). We argue by contraposition. Suppose that there is a real number t in (0, 1) and a point p0 ∈ E (t) (p) with p0 6= p. As p0 ∈ E (t) (p), it follows that p0 ∈ Skpk and that there is a point q ∈ Skpk such that p0 = 1t p − 1−t t q. Observe that q 6= p and 1 1−t 0 (1 − t)q + tp = (1 − t)q + t p− q = p. t t Hence the point p is not extreme in Bkpk , as desired. By putting everything together, the lemma follows.
CONICAL GEODESIC BICOMBINGS ON SUBSETS OF NORMED VECTOR SPACES
14
Next, we define the map λ : D(V ) → V via the assignment (p, q, t) 7→ (1 − t)p + tq. The subsequent lemma is the key ingredient in the proof of Theorem 4.1. Lemma 4.3. Let (V, k·k) be a normed vector space and let A ⊂ V be a subset that admits a conical geodesic bicombing σ : D(A) → A. Let p be a point in A such that −p ∈ A. If there is a point z in V such that the points 2z − p and p − 2z are contained in A and such that σ(p, p − 2z, ·) = λ(p, p − 2z, ·) and σ(2z − p, −p, ·) = λ(2z − p, −p, ·), then we have that σ(p, −p, t) ∈ (1 − 2t)z + M (t) (p − z, z − p) . for all real numbers t ∈ [0, 1]. Proof. Let t ∈ [0, 1] be a real number. We compute kσ(p, −p, t) − λ(p, p − 2z, t)k ≤ 2t kp − zk kσ(p, −p, t) − λ(2z − p, −p, t)k ≤ 2(1 − t) kp − zk . Note that kλ(p, p − 2z, t) − λ(2z − p, −p, t)k = 2 kp − zk. Therefore, it follows that σ(p, −p, t) ∈ M (t) (λ(p, p − 2z, t), λ(2z − p, −p, t)) . It is readily verified that M (t) (u + h, v + h) = h + M (t) (u, v) for all t in [0, 1] and u, v, h ∈ V . Consequently, we obtain that M (t) (λ(p, p − 2z, t), λ(2z − p, −p, t)) = (1 − 2t)z + M (t) (p − z, z − p) . Thus, the lemma follows.
Suppose that A is a subset of a normed vector space (V, k·k) and assume that A admits a conical geodesic bicombing σ : D(A) → A. The translation Tz : A → Tz (A) about the vector z ∈ V given by the assignment x 7→ x + z is an isometry and the map (Tz )∗σ : D(Tz (A)) → Tz (A) given by (x, y, t) 7→ Tz (σ(T−z (x), T−z (y), t)) is a conical geodesic bicombing on Tz (A). Now, we have everything on hand to prove Theorem 4.1. Proof of Theorem 4.1. We proceed by induction on n ≥ 1. If n = 1, then Lemma 4.2 tells us that p − q p − q p−q ,− , t = (1 − 2t) T− p+q σ 2 2 2 2 ∗ for all t ∈ [0, 1]. Thus, we obtain that σ(p, q, t) = (1 − t)p + tq for all t ∈ [0, 1]. Suppose now that n > 1 and that the statement holds for n − 1. We may 1 assume that λ1 ∈ (0, 1). We define (λ01 , . . . , λ0n−1 ) := 1−λ (λ2 , . . . , λn ) and 1 0 0 (e1 , . . . , en−1 ) := (e2 , . . . , en ). Observe that n X k=1
λk ek = λ1 e1 + (1 − λ1 )
n−1 X
λ0k e0k .
Further, note that
n−1
n
X
X
λk ek < 1, λ0k e0k = 1, as otherwise (4.4) implies
k=1
(4.4)
k=1
k=1
(4.5)
CONICAL GEODESIC BICOMBINGS ON SUBSETS OF NORMED VECTOR SPACES
which is not possible due to (4.1). We abbreviate r := z := r(1 − λ1 )
n−1 X
λ0k e0k ,
p0 :=
k=1
kp−qk 2
p−q , 2
15
and we set q 0 := p0 − 2z.
Note that n−1
X p0 − q 0 = r(1 − λ1 ) λ0k e0k . 2 k=1
Hence, by the use of (4.5) it follows that kp0 − q 0 k = r(1 − λ1 ). 2
(4.6)
We have that n
n−1
k=1
k=1
X p−q p0 + q 0 (4.1) X (4.4) = −z = r λk ek − r(1 − λ1 ) λ0k e0k = rλ1 e1 2 2 and therefore (n−1 ) p0 + q 0 kp0 − q 0 k X + (−1)εk λ0k e0k : (ε1 , . . . , εn−1 ) ∈ {0, 1}n−1 2 2 k=1 ( ) n X (4.2) (4.6) εk n−1 = r λ1 e1 + (−1) λk ek : (ε2 . . . , εn ) ∈ {0, 1} ⊂ T− p+q (A). 2
k=2
Thus, we can apply the induction hypothesis to p0 , q 0 ∈ T− p+q (A) and obtain that 2
T− p+q
σ (p0 , p0 − 2z, ·) = λ (p0 , p0 − 2z, ·) .
∗
2
Similarly, we obtain T− p+q σ (2z − p0 , −p0 , ·) = λ (2z − p0 , −p0 , ·) . 2
∗
Now, by the use of Lemma 4.3 it follows that T− p+q σ (p0 , −p0 , t) ∈ (1 − 2t)z + M (t) (p0 − z, z − p0 ) 2
∗
for all real numbers t ∈ [0, 1]; consequently, we get T− p+q σ (p0 , −p0 , t) = (1 − 2t)p0 , 2
∗
since p0 − z = rλ1 e1 is an extreme point in Brλ1 and thus we can use Lemma 4.2 to deduce that M (t) (p0 − z, z − p0 ) = {(1 − 2t)(p0 − z)}. Hence, we have p+q σ(p, q, t) = T− p+q σ (p0 , −p0 , t) + = (1 − t)p + tq, 2 2 ∗ as desired.
CONICAL GEODESIC BICOMBINGS ON SUBSETS OF NORMED VECTOR SPACES
5
16
Proof of Theorem 1.4
Before we start with the proof of Theorem 1.4, we recall some notions from [Mie15] in a slightly modified way. Let (X, d) be a metric space, let p ∈ X be a point and let r > 0 be a real number. We set Ur (p) := {q ∈ X : d(p, q) < r}. Let U ⊂ D(X) be a subset. A map σ : U → X is a convex local geodesic bicombing if for every point p ∈ X there is a real number rp > 0 such that [ U= D(Urp (p)). p∈X
and if the restriction σ|D(Urp (p)) : D(Urp (p)) → Urp (p) is a consistent conical geodesic bicombing for each point p ∈ X. Note that, contrary to [Mie15], we do not assume that σ|D(Urp (p)) is reversible, since this property is not used to prove [Mie15, Theorem 1.1]. We can then restate [Mie15, Theorem 1.1] as follows: Theorem 5.1. Let X be a complete, simply-connected metric space with a convex local geodesic bicombing σ. Then the induced length metric on X admits a unique convex geodesic bicombing σ ˜ which is consistent with σ. The uniqueness of σ ˜ follows immediately from the proof of [Mie15, Theorem 1.1]. With Theorem 5.1 on hand it is possible to derive Theorem 1.4 by the use of Theorem 1.3. Proof of Theorem 1.4. Let int(C) denote the interior of C and let p, q be two points in int(C). We abbreviate [p, q] := (1 − t)p + tq : t ∈ [0, 1] . As int(C) is convex, we have that [p, q] ⊂ int (C). For each point z ∈ C we set ( min{kz − wk : w ∈ [p, q]} if z ∈ C \ int(C) rz := 1 if z ∈ int(C). 2 inf {kz − wk : w ∈ C \ int(C)} Note that rz > 0 for all points z ∈ C and we have that Urz (z) ∩ [p, q] = ∅ if z ∈ C \int(C). Further, for every point z ∈ int(C) it follows that B2rz (z) ⊂ C; thus, we may invoke Theorem 1.3 to deduce that if z ∈ int(C), then σz1 z2 (t) = (1 − t)z1 + tz2 for all points z1 , z2 ∈ Brz (z) and all real numbers t ∈ [0, 1]. We define [ U := D(Urz (z)). z∈C loc
Note that the map σ := σ|U defines a convex local bicombing on C. The geodesic σpq (·) and the linear geodesic from p to q are both consistent with the local bicombing σ loc . Hence, by Theorem 5.1, we conclude that σpq (·) must be equal to the linear geodesic from p to q, that is, we have σpq (t) = (1 − t)p + tq for all real numbers t ∈ [0, 1]. Now, suppose that p, q ∈ C. As C is convex, it is well-known that C = int (C), cf. [AB06, Lemma 5.28]. Let (pk )k≥1 , (qk )k≥1 ⊂ int (C) be two sequences such that pk → p and qk → q with k → +∞. It is readily verified that σpk qk (·) → σpq (·) with k → +∞, since σ is a conical geodesic bicombing. As a result, we obtain that the geodesic σpq (·) is equal to the linear geodesic from p to q, as desired.
CONICAL GEODESIC BICOMBINGS ON SUBSETS OF NORMED VECTOR SPACES
17
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¨ rich, Ra ¨ mistrasse 101, 8092 Zu ¨ rich, Schweiz Mathematik Departement, ETH Zu
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