Let(X, d)be a metric space. A geodesic path joiningx∈X andy ∈X is a map
δfrom[0, l]toX such thatδ(0) =x, δ(l) =yandd(δ(t1), δ(t2)) =|t1−t2|for all
t1, t2∈[0, l]. In particular, l=d(x, y). (X, d)is called a uniquely geodesic space
In this subsection, (X, d)is a uniquely geodesic space with condition K and
γx,y is a map from[0,1]toX defined by
γx,y(t) =δx,y(td(x, y))
for anyt∈[0,1], where δx,y denotes the geodesic path joiningxandy. It is easy
to show thatγx,y(0) =x,γx,y(1) =y and
d(γx,y(t1), γx,y(t2)) =|t1−t2|d(x, y)
for anyt1, t2∈[0,1].
Note thatγx,y(s)is a unique pointcinX which satisfiesd(x, c) =sd(x, y)and
d(c, y) = (1−s)d(x, y)for any0≤s≤1.
Forx∈X, define the map φxonX by the equation
φx(y) =ϕy(x)
for anyy∈X. Then we have
φx(y) =c ⇐⇒ ϕy(x) =c ⇐⇒ mid(x, c) =y.
It implies thatφxis a bijection on X for anyx∈X.
Lemma 5.5. Let x, y, z, c∈X. Then the following holds. (y1) For anys∈R\ {0},
( d(x, z) =|s|d(x, y) d(z, y) =|1−s|d(x, y) ⇐⇒ ( d(x, y) =|1 s|d(x, z) d(y, z) =|1−1 s|d(x, z).
(y2) For any0≤s≤1,
c=γx,y(s) ⇐⇒ ( d(x, c) =sd(x, y) d(c, y) = (1−s)d(x, y). (y3) c=φx(y) ⇐⇒ ( d(x, c) = 2d(x, y) d(c, y) = (2−1)d(x, y).
(y4) For any natural number n,
c=φnx(y) ⇐⇒
(
d(x, c) = 2nd(x, y) d(c, y) = (2n−1)d(x, y).
(y5) For anys >1, c=φnx(γx,y( s 2n)) ⇐⇒ ( d(x, c) =sd(x, y) d(c, y) = (s−1)d(x, y),
wheren∈Nwhich satisfies 2n−1< s≤2n.
(y6) For anys >0,
( d(x, c) =sd(x, y) d(c, y) =|s−1|d(x, y) ⇐⇒ ( d(x, c0) =sd(x, y) d(c0, y) = (1 +s)d(x, y), wherec0 =ϕ x(c).
In particular, for any real number s, there exists a unique point c in X such thatd(e, c) =|s|d(e, x)andd(c, x) =|1−s|d(e, x).
Proof. (y1) and (y2) are obvious. (y3): We have
c=φx(y) ⇐⇒ mid(x, c) =y ⇐⇒ d(x, y) =d(c, y) =1
2d(x, c)
⇐⇒ d(x, c) = 2d(x, y) and d(c, y) =d(x, y).
(y4): We will prove by induction. When n= 1, the argument is true by (y2). Letkbe a natural number and suppose thatφkx(y)is a unique pointck ∈X which
satisfies d(x, ck) = 2kd(x, y) and d(ck, y) = (2k−1)d(x, y). Note that c=φkx+1(y) ⇐⇒ c=φx(ck) ⇐⇒ mid(x, c) =ck ⇐⇒ d(x, ck) =d(c, ck) = 1 2d(x, c) ⇐⇒ d(x, c) = 2d(x, ck) and d(c, ck) =d(x, ck) ⇐⇒ d(x, c) = 2k+1d(x, y) and d(c, ck) = 2kd(x, y). (⇒): Letc=φk+1 x (y). We have d(x, c) = 2k+1d(x, y) and d(y, c)≤d(y, ck) +d(ck, c) = (2k+1−1)d(x, y),
d(y, c)≥d(x, c)−d(x, y) = (2k+1−1)d(x, y).
Therefore
d(y, c) = (2k+1−1)d(x, y).
(⇐): Letcbe a point ofX which satisfies
d(x, c) = 2k+1d(x, y) and d(c, y) = (2k+1−1)d(x, y). Then d(x, y) = 1 2k+1d(x, c) and d(y, c) = (1− 1 2k+1)d(x, c)
as (y1). It implies thaty=γx,c(2k1+1). Putc0= mid(x, c), then we have d(x, c0) =1 2d(x, c) = 2 kd(x, y) and d(y, c0) = d(γx,c( 1 2k+ 1), γx,c( 1 2)) = (1 2 − 1 2k+1)d(x, c) = (2 k−1)d(x, y).
By the inductive assumption, we havec0=φk
x(y)and hencec=φkx+1(y).
By the principle of induction, the proof of (y4) is complete.
(y5): Lets >1. Then there existn∈Nsuch that2n−1< s≤2n. Puts0= s 2n,
then 1
2 < s 0≤1.
(⇒): Letc0 be a point inX that satisfies
d(x, c0) =s0d(x, y) and d(c0, y) = (1−s0)d(x, y).
Following (y1) we havec0=γx,y(s0). Letcbe a point in X that satisfies
d(x, c) = 2nd(x, c0) and d(c, c0) = (2n−1)d(x, c0).
Since (y3), we havec=φnx(c0). Putb= mid(x, y),then
d(c0, b) =d(γx,y(s0), γx,y( 1 2)) = (s 0−1 2)d(x, y) = (1− 1 2s0)d(x, c0) and d(x, b) =1 2d(x, y) = 1 2s0d(x, c0) = 1 2n+1s0d(x, c) = 1 2sd(x, c). Thus, d(b, c)≤d(b, c0) +d(c0, c) = (2n− 1 2s0)d(x, c0) = (1− 1 2n+1s0)d(x, c) = (1− 1 2s)d(x, c),
d(b, c)≥d(x, c)−d(x, b) = (1− 1 2n+1s0)d(x, c) = (1− 1 2s)d(x, c) and hence d(b, c) = (1− 1 2s)d(x, c).
Therefore, we haveb=γx,c(21s). Sinceb= mid(x, y), we havey=γx,c(1s), that is,
d(x, y) = 1 sd(x, c) and d(y, c) = (1− 1 s)d(x, c). By (y1) we have d(x, c) =sd(x, y) and d(c, y) = (s−1)d(x, y).
(⇐): Letcbe a point in X which satisfies
d(x, c) =sd(x, y) and d(c, y) = (s−1)d(x, y).
By (y1) and (y2) we havey =γx,c(1s). Letc1= mid(x, c)and ck+1= mid(x, ck)
for anyk∈N. Thenck =γx,c(21k)andφ
k
x(ck) =c for anyk∈N. Thus we have d(y, cn) = ( 1 s− 1 2n)d(x, c) = (1− s 2n)d(x, y) and d(x, cn) = 1 2nd(x, c) = s 2nd(x, y).
It implies thatcn=γx,y(2sn)and hencec=φ
n
x(γx,y(2sn)).
(y6): Lets >0.
(⇒): Letc∈X be a point inX such that
d(x, c) =sd(x, y) and d(c, y) =|1−s|d(x, y).
Sinceϕxis an isometry andϕx(x) =x, we have
d(x, ϕx(c)) =d(x, c) =sd(x, y)
and
d(ϕx(c), ϕx(y)) =d(c, y) =|1−s|d(x, y).
By the definition ofϕx, we haved(a, ϕx(a)) = 2d(x, a)for anya∈X and hence
d(ϕx(y), y) = 2d(x, y), d(ϕx(c), c) = 2d(x, c) = 2sd(x, y).
We first assume that0< s≤1. Then
and
d(ϕx(c), y)≥d(ϕx(y), y)−d(ϕx(y), ϕx(c)) = (1 +s)d(x, y).
It follows that
d(ϕx(c), y) = (1 +s)d(x, y).
Next, we assume that1< s. Then
d(ϕx(c), y)≤d(ϕx(c), x) +d(x, y) = (1 +s)d(x, y)
and
d(ϕx(c), y)≥d(ϕx(c), c)−d(c, y) = (1 +s)d(x, y).
It follows that
d(ϕx(c), y) = (1 +s)d(x, y).
(⇐): Since (y2) to (y5), there exists a pointcin X such that
d(x, c) =sd(x, y) and d(c, y) =|s−1|d(x, y).
We have
d(x, ϕx(c)) =sd(x, y) and d(ϕx(c), y) = (1 +s)d(x, y)
as opposite direction. Letc0∈X be a point such that
d(x, c0) =sd(x, y) and d(c0, y) = (1 +s)d(x, y).
Putt= 1 +sthent >1 and
d(y, c0) =td(y, x) and d(c0, x) = (t−1)d(y, x).
Since (y4), such a pointc0 is unique inX. Thusc0 =ϕx(c).
By Lemma 5.5, for any x, y∈ X and any s∈ R, there exists a unique point
c ∈X such thatd(x, c) =|s|d(x, y) and d(c, y) = |1−s|d(x, y). We will denote the such pointc byγx,y(s).
Lemma 5.6. For anyx, y∈X ands, t∈R, the equation d(γx,y(s), γx,y(t)) =|s−t|d(x, y)
holds.
Proof. Putar=γx,y(r)for anyr∈R. We can assume thats≤t.
(the case: 0≤s≤t≤1): trivial.
(the case: 0≤s≤1≤t): By the definition of ar, we have
d(x, as) = |s|d(x, y) =sd(x, y),
d(as, y) = |1−s|d(x, y) = (1−s)d(x, y) d(x, at) = |t|d(x, y) =td(x, y),
Then
d(as, at) ≤ d(as, y) +d(y, at) = (t−s)d(x, y), d(as, at) ≥ d(x, at)−d(x, as) = (t−s)d(x, y)
and hence
d(as, at) = (t−s)d(x, y).
(the case: 1≤s≤t): Since (y1), we havey=γx,at( 1 t). Letc=γx,at( s t), then, since0≤ 1 t, s t ≤1, we have d(x, c) = s td(x, at) =sd(x, y), d(y, c) = |1 t − s t|d(x, at) =|1−s|d(x, y).
It implies thatc=as. Thus
d(as, at) =|s
t −1|d(x, at) =|s−t|d(x, y).
As the above part of the proof, we have
0≤s, t⇒d(γx,y(s), γx,y(t)) =|s−t|d(x, y).
(the case: s≤0≤1≤t): By the definition of at we have
d(x, at) = td(x, y), d(at, y) = (t−1)d(x, y). It follows that d(at, x) = t t−1d(at, y), d(x, y) = (1− t t−1)d(at, y). It implies thatx=γat,y( t t−1). Letc=γat,y( t−s t−1). Since0≤ t t−1, t−s t−1, we have d(y, c) = |1−t−s t−1|d(y, at) = (1−s)d(x, y), d(c, x) = |t−s t−1− t t−1|d(y, at) =|s|d(x, y). Hencec=as. Thus d(at, as) =d(at, c) = t−s t−1d(at, y) = (t−s)d(x, y).
(the case: s≤0 ≤t ≤1): By the definition ofar and triangle inequality, we have d(as, at) ≤ d(as, x) +d(x, at) = −sd(x, y) +td(x, y) = (t−s)d(x, y), d(as, at) ≥ d(y, as) +d(y, at) = (1−s)d(x, y) + (1−t)d(x, y) = (t−s)d(x, y) and hence d(as, at) = (t−s)d(x, y).
(the case: s≤t≤0): For anyr∈R, we have
γx,y(r) =γy,x(1−r)
by the definition. Since0≤1−s,1−t, we have
d(γx,y(s), γx,y(t)) =d(γy,x(1−s), γy,x(1−t)) = (t−s)d(y, x).
Lemma 5.7. Let x, y∈X. The equation
ϕγx,y(s)(γx,y(t)) =γx,y(2s−t) holds for anys, t∈R.
Proof. Since Lemma 5.6, we have
d(γx,y(2s−t), γx,y(s)) = |s−t|d(x, y), d(γx,y(s), γx,y(t)) = |s−t|d(x, y)
and
d(γx,y(2s−t), γx,y(t)) =|2s−2t|d(x, y).
Thus
d(γx,y(2s−t), γx,y(s)) =d(γx,y(s), γx,y(t)) = 1
2d(γx,y(2s−t), γx,y(t))
and hence
γx,y(s) = mid(γx,y(t), γx,y(2s−t)).
Therefore,
5.3
A Normed Gyrolinear Space Induced by a Metric Space
In this subsection, let (X, d) be a uniquely geodesic space which satisifies the conditionK. Forx, y∈X ands∈R,γe,x(s)denote the unique point c∈X that
satisfiesd(x, c) =|s|d(x, y)andd(c, y) =|1−s|d(x, y).
Definition 5.8. Let(X, d)be a uniquely geodesic space that satisfies the condition
K. For fixede∈X, we define the scalar multiplication⊗eonX by s⊗ex=γe,x(s)
for any x∈ X and s ∈R. We call ⊗e the scalar multiplication induced by the
metricdonX at e.
In the following part of this subsection e is a point in X. ⊕e is the binary
operation induced by the metricdonX at e, and⊗e is the scalar multiplication
induced by the metricdonX at e.
Proposition 5.9. Let (X, d)be a uniquely geodesic space which satisfies the con- ditionK. Let e∈X and put kxke=d(e, x)for any x∈X. Assume that (X, d)
satisfies the following condition:
(K4) x→y impliesϕx(a)→ϕy(a)for any x, y, a∈X.
Then(X,⊕e,⊗e,k · ke, id)is a normed gyrolinear space with gyrometricd. Proof. Recall that(X,⊕e)is a gyrocommutative gyrogroup by Lemma 5.3.
(GL1): It is an immediate consequence of γe,x(1) =x. (GL2): By Lemma 5.7, we have
(r⊗ex)⊕e(s⊗ex) = ϕmid(e,r⊗ex)ϕe(s⊗ex)
= ϕmid(e,γe,x(r))((−s)⊗ex)
= ϕγe,x(r2)(γe,x(−s)) = γe,x( 2r 2 −(−s)) = γe,x(r+s) = (r+s)⊗ex
(GL3): Putz=r⊗e(s⊗ex), thenz=γe,s⊗ex(r). Since (y1),x=γe,s⊗ex( 1 s). We havez= (rs)⊗exas d(e, z) =d(e, r⊗e(s⊗ex)) =|r|d(e, s⊗ex) =|rs|d(e, x) and d(x, z) = d(γe,s⊗ex( 1 s), γe,s⊗ex(r)) = |r−1 s|d(e, s⊗ex) =|rs−1|d(e, x).
(GL4): Sincegyr[x, y]is an automorphism,gyr[x, y]e=e. By equation (10) of Proposition 5.4,gyr[x, y]is a isometry. Hence we have
c=r⊗ea ⇐⇒ ( d(e, c) =|r|d(e, a) d(c, a) =|1−r|d(e, a) ⇐⇒ (
d(gyr[x, y]e,gyr[x, y]c) =|r|d(gyr[x, y]e,gyr[x, y]a) d(gyr[x, y]c,gyr[x, y]a) =|1−r|d(gyr[x, y]e,gyr[x, y]a) ⇐⇒
(
d(e,gyr[x, y]c) =|r|d(e,gyr[x, y]a)
d(gyr[x, y]c,gyr[x, y]a) =|1−r|d(e,gyr[x, y]a) ⇐⇒ gyr[x, y]c=γe,gyr[x,y]a(r)
⇐⇒ gyr[x, y]c=r⊗eγ[x, y]a
(GL5): For anyx∈X, since(X,⊕e)satisfies the condition (G5), we have gyr[n⊗ex, x] = gyr[((n−1)⊗ex)⊕ex, x]
= gyr[(n−1)⊗ex, x]
for any integern. It follows that
gyr[n⊗ex, x] = gyr[0⊗ex, x] = gyr[e, x] =idX
for any integern. Also, we have
gyr[x, m⊗ex] = gyr−1[m⊗ex, x] =idX
for any integerm. Since gyrocommutative gyrogroup satisfies the equation
gyr[a, b] gyr[b, c]gyr[c, a] =id
for anya, b, c([6]Theorem 3.31), we have
gyr[n⊗ex, m⊗ex] = gyr[n⊗ex, x] gyr[x, m⊗ex] gyr[m⊗ex, n⊗ex] =idX
for any integersn, m. It follows that
gyr[n m ⊗ey, y] = gyr[n⊗e( 1 m⊗ey), m⊗e( 1 m ⊗ey)] =idX (11)
for anyy∈X and rational number n
m.
Let{kn}be a sequence of rational numbers such thatkn→α. Then we have
kn⊗ex=γe,x(kn)→γe,x(α) =α⊗exby Lemma 5.6. By the condition (K4) we
have
for anyx, a∈X. Thus we have gyr[kn⊗ex, x](a) = λ (kn⊗ex⊕ex)λkn⊗exλx(a) = λ(−kn−1)⊗exλkn⊗exλx(a) → λ(−α−1)⊗exλα⊗exλx(a) = λ (α⊗ex⊕ex)λα⊗exλx(a) = gyr[α⊗ex, x](a)
for any real numberαanda, x∈X, where{kn}is a sequence of rational numbers such thatkn→α. Following (11) we havegyr[α⊗ex, x] =idX for anyx∈X and α∈R. Thus we have
gyr[r⊗ex, s⊗ex] = gyr[ r
s⊗e(s⊗ex), s⊗ex] =idX
for anyx∈X andr, s∈R.
(NG1): kxke= 0 ⇐⇒ d(e, x) = 0 ⇐⇒ x=e
(NG2): Following Proposition 5.4, we have
kx⊕eyke = d(e, x⊕ey)
≤ d(e, x) +d(x, x⊕ey)
= d(e, x) +d(e, y) =kxke+kyke
for anyx, y∈X.
(NG3): For anyx∈X andr∈R, we have kr⊗exke = d(e, r⊗ex)
= d(γe,x(0), γe,x(r)) = |0−r|d(e, x) =|r|kxke
by Lemma 5.6.
(NG4): Since any gyroautomorphism preserves the identityeand Proposition 5.4, we have
kgyr[x, y]ake = d(e,gyr[x, y]a)
= d(gyr[x, y]e,gyr[x, y]a) = d(e, a) =kake
for anya, x, y∈X.
Finally, since Proposition 5.4, we have
The following Corollary 5.10 is a immediately consequence of Proposition 5.9 and Proposition 3.4.
Corollary 5.10. Let X be a set and%be a function%:X×X →X that satisfies
%(x, y) = 0 if and only ifx=y. Let e∈X and putkxk0
e=%(e, x)for anyx∈X.
Let f be a monotone increasing bijectionf :kXk0
e→R≥0, wherekXk0e={kxk0e: x∈X}. Putd=f %. Suppose that(X, d)is a uniquely geodesic space that satisfies the conditionK and the condition (K4). Then (X,⊕e,⊗e,k · k0
e, f) is a normed
gyrolinear space with gyrometric%.
Proof. Putkxke=d(e, x). By Proposition 5.9, (X,⊕e,⊗e,k · ke, id)is a normed
gyrolinear space. Sincef is a monotone increasing bijectionf :kXk0
e→R≥0, we
have
kXke={kxke:x∈X}={fkxk0e:x∈X}=R≥0
and hence f−1 is a monotone increasing bijection f−1 : kXk → kXk0
e. Note
that 0 ∈ kXk0 as kek0 = 0. Since f−1 is strictly monotone increasing, we have
f−1(0) = 0. By Proposition 3.4, we have (X,⊕
e,⊗e,k · k0, f). Finally, we have %(x, y) =f−1d(x, y) =f−1kx eyk=kx eyk0.
5.4
Examples
Example 5.11. Let k · k be the Euclidean norm and dbe the Euclidean metric onRn. Then the Euclidean space(Rn, d)is a uniquely geodesic metric space that
satisfies the conditionK with
mid(x, y) = x+y 2
and
ϕx(y) = 2x−y
for anyx, y∈R. In this case,(Rn,⊕0) = (Rn,+) as x⊕0y = ϕmid(0,x)ϕ0(y)
= ϕx
2(−y)
= x+y
for any x, y∈X. Moreover,⊗0 coincides with the usual scalar multiplication on
Rn as
d(0, rx) =krxk=|r|kxk=rd(0, x)
and
d(x, rx) =kx−rxk=|1−r|kxk=|1−r|d(0, x)
for anyx∈Rn andr∈ R.
Example 5.12. LetD={z∈C:|z|<1}. The Möbius addition⊕inDis given
by the equation
a⊕b= a+b 1 +ab
for any a, b ∈ D. (D,⊕) is a gyrocommutative gyrogroup (see [6]) and is called
the Möbius gyrogroup. The identity of(D,⊕)is the origin ofCand a=−afor everya∈D. Moreover, the Möebius multiplication is given by
r⊗a= tanh(rtanh−1|a|) a |a|
for anya∈Dandr∈R. The Möbius gyrometric%is given by the equation %(a, b) =|( a)⊕b| = −a+b 1−ab
for every a, b∈D. (D, %)is a metiric space in itself, and (D,tanh−1%) is a met-
ric space again. (D, %) doesn’t satisfy the condition K. However, (D,tanh−1%)
satisfies the conditionK with
mid(a, b) =1
2 ⊗(ab)
and
ϕa(b) = (2⊗a)⊕(−b)
for anya, b∈D. Let⊕0 be the binary operation onDinduced bytanh−1%at 0
then ⊕0 =⊕. Moreover, (D,tanh−1%)is a uniquely geodesic metric space. Let ⊗0 be the scalar multiplication onDinduced bytanh−1%at0then⊗0=⊗.
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[10] A. A. Ungar, Analytic Hyperbolic Geometry in n Dimensions: An Introduc- tion, CRC Press, Boca Raton, FL, 2015.
Toshikazu Abe
Information Engeineering, Niigata University, Niigata, Japan