Constant-speed geodesics in Wasserstein spaces
March 27, 2011
Let us start from this point: in a metric space (X, d), if a curve ω : [0, 1] → X with ω(0) =
x, ω(1) = y satisfies d(ω(t), ω(s)) ≤ d(x, y)|t − s| then it is a constant-speed geodesic. Moreover,
if X is a length space (i.e. the distance d(x, y) coincides with the infimum of the lengths of the
curves connection x to y), the converse is also true, i.e. any constant speed geodesic satisfies
d(ω(t), ω(s)) = d(x, y)|t − s|.
First, we prove that the space Pp (Ω) endowed with the Wasserstein distance is indeed a length
space, provided Ω is convex.
Theorem 0.1. Suppose that Ω is convex, take µ, ν ∈ Pp (Ω) and γ an optimal transport plan in
Π(µ, ν) for the cost |x − y|p (p ≥ 1). Define πt : Ω × Ω → Ω through πt (x, y) = (1 − t)x + ty. Then
the curve µt := (πt )# γ is a constant speed geodesic in Pp (Ω) connecting µ0 = µ to µ1 := ν.
In the particular case where µ is absolutely continuous, or in general that γ = γT , this very curve
is obtained as ((1 − t)id + tT )# µ.
As a consequence, the space Pp (Ω) endowed with the Wasserstein distance is a geodesic space
(i.e. a length space where the infimum of the length of the curves connecting two given points is a
minimum).
Proof. It is sufficient to prove that Wp (µt , µs ) ≤ Wp (µ, ν)|t − s|. To do that, take γts := (πt , πs )# γ ∈
Π(µt , µs ) and compute
Z
Wp (µt , µs ) ≤
p
|x − y|
1 Z
1
Z
1
p
p
p
p
p
=
|πt (x, y) − πs (x, y)| dγ = |t−s|
|x − y| dγ = |t−s|Wp (µ, ν),
dγts
where we used that |(1 − t)x + ty − (1 − s)x − sy| = |(t − s)(x − y)|.
On can wonder what is the velocity field associated to a geodesic curve µt defined as above, since
it is a Lipschitz curve and hence it must admit the existence of a velocity field vt (at least if p > 1)
satisfying the continuity equation ∂t µt + ∇ · (µt vt ) = 0.
In rough terms this means: take y ∈ spt(µt ) ⊂ Ω, for t ∈]0, 1[, and try to find the speed of the
particle(s) passing at y at time t. This should be the value of vt (y). It would be easier to answer
this question if we had uniqueness of “the particle” passing through y at time t. We will use the
following theorem.
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Theorem 0.2. Let γ be an optimal transport plan for a cost c(x, y) = h(y − x) with h strictly
convex. between two probabilities µ and ν and t ∈]0, 1[. Define µt = (πt )# γ and take y ∈ spt(µt ).
Then there exists a unique pair (x, z) ∈ spt(γ) such that y = (1 − t)x + tz. These values of x and z
will be denoted by Xt (y) and Zt (y). If Ω is compact, the two maps Xt and Zt are also continuous.
In particular, if γ = γT comes from a transport map, then the map Tt := (1 − t)id + tT is
invertible and Tt−1 = Xt .
Proof. Suppose that there exist (x, y) and (x0 , y 0 ) in spt(γ) such that y = (1−t)x+tz = (1−t)x0 +tz 0 .
Set a = z − x and b = z 0 − x0 .
The c−cyclical monotonicity of the support of the optimal γ implies
h(a) + h(b) ≤ h(y 0 − x) + h(y − x0 ) = h(tb + (1 − t)a) + h(ta + (1 − t)b).
Yet, if a 6= b, then strict convexity implies
h(tb + (1 − t)a) + h(ta + (1 − t)b) < th(b) + (1 − t)h(a) + th(a) + (1 − t)h(b) = h(a) + h(b),
which is a contradiction. Hence a = b and from y = x + t(z − x) = x + ta and y = x + t(z 0 − x0 ) =
x0 + tb = x0 + tb one infers x = x0 and finally z = z 0 .
The continuity of Xt and Zt is obtained by compactness. Take yn → y and suppose (up to subsequences) (Xt (yn ), Zt (yn )) → (X, Z). Since spt(γ) is closed then (X, Z) ∈ spt(γ). Yet, the previous
uniqueness proof implies (X, Z) = (Xt (y), Zt (y)). And since any limit of converging subsequences
must coincide with the value at y, in a compact space this gives continuity.
We can now identify the velocity field of the geodesic µt : we know that every particle initially
located at x moves on a straight line with constant speed T (x) − x, which implies vt (y) = (T −
id)(Tt−1 (y). More generally, if γ is not induced by a map, we have vt (y) = Zt (y) − Xt (y).
Proposition 0.3. Let µt = (πt )# γ be the geodesic connecting µ to ν introduced above. Then the
velocity field vt := Zt − Xt is well defined on spt(µt ) for each t ∈]0, 1[ and satisfies
∂t µt + ∇ · (µt vt ) = 0,
||vt ||Lp (µt ) = |µ0 |(t) = Wp (µ, ν).
Proof. We already saw that Xt and Zt are well-defined, so we only need to check the continuity
equation and the Lp estimate. For the continuity equation, take φ ∈ C 1 and compute
Z
Z
Z
d
d
φ dµt =
φ((1 − t)x + tz) dγ(x, z) = ∇φ((1 − t)x + tz) · (z − x) dγ(x, z)
dt
dt
Z
Z
=
∇φ(πt (x, z)·(Zt (πt (x, z))−Xt (πt (x, z))dγ(x, z) = ∇φ(y) · (Zt (y) − Xt (y)) dµt (y).
To compute the Lp norm we have
Z
Z
Z
|vt |p dµt = |Zt (y) − Xt (y)|p dµt (y) = |z − x|p dγ(x, z) = Wp (µ, ν)p ,
and we used in both the computations Zt (πt (x, z))−Xt (πt (x, z) = z − x for every (x, z) ∈ spt(γ).
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We now want to prove that, at least for p > 1, all the geodesics for the distance Wp have this
form, i.e. they are given by µt = (πt )# γ for an optimal γ. Non-uniqueness would stay true in case
γ is not unique.
We first prove the following proposition.
Proposition 0.4. Let µt be a constant speed geodesic between µ and ν. Suppose p > 1, consider
the transport cost c(x, y) = |x − y|p and take t ∈]0, 1[. Then
• In the transport between µt and µ there is a unique optimal transport plan γ − , and the same
is true between µt and ν, with a unique optimal transport γ + .
• Take σ ∈ P(Ω × Ω × Ω) with (πx,y )# σ = γ − and (πy,z )# σ = γ + and define γ := (πx,z )# σ.
Then γ is optimal between µ and ν.
• The measure µt satisfies µt = (πt )# γ (but this optimal transport plan γ may a priori depend
on t, since it has been obtained by gluing γ − and γ + .
• both γ − and γ + are induced by transport maps.
Proof. Let us first pick two optimal plans γ − and γ + in Π(µ, µt ) and Π(µt , ν), which are not
necessarily unique for the moment, and “compose” them by building σP(Ω × Ω × Ω) as in the gluing
lemma that we used for the triangle inequality. Notice that
Wp (µ, ν) ≤ ||x − z||Lp (γ) = ||x − z||Lp (σ) ≤ ||x − y||Lp (σ) + ||y − z||Lp (σ)
= ||x − y||Lp (γ − ) + ||y − z||Lp (γ + ) = Wp (µ, µt ) + Wp (µt , ν) = Wp (µ, ν).
This implies that all the inequality are equality, and in particular
• that γ is optimal, since Wp (µ, ν) = ||x − z||Lp (γ) ,
• that we have equality in the triangle inequality for the Lp norm and, since p > 1 and the Lp
norm is strictly convex for p > 1, we get x − y = λ(x − z) and y − z = (1 − λ)(x − z), for
λ ∈ [0, 1]. By computing the Lp (σ) norms we find λ = t.
This proves that σ−a.e. we have x − y = t(x − z), which may be read as y = (1 − t)x + tz. This
also gives µt = (πy )# σ = (πt (x, z))# σ = (πt )# γ.
We now need to prove that γ − and γ + are unique and induced by transport maps. First, let
us prove that γ − is induced by a map (which goes in this case in the direction µt 7→ µ. Take
(x, y) ∈ spt(γ − ). There exist at least a point z such that (x, y, z) ∈ spt(σ) (otherwise (x, y) ∈
/
spt(πx, y)#σ). Analogously, if we take (x0 , y) ∈ spt(γ − ) there exists z 0 such that (x0 , y, z 0 ) ∈ spt(σ).
Hence y = (1 − t)x + tz = (1 − t)x0 + tz 0 and (x, z), (x0 , z 0 ) ∈ spt(γ). Since γ is optimal, Theorem
0.2 implies x0 = x. This proves that γ − comes from a transport map. Moreover, as always, proving
that any optimal transport plan must be induced by a map also proves uniqueness. The same proof
may be performed on γ + .
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Our main problem, if we want to characterize constant speed geodesic via µt = (πt )# γ, is the
fact that we obtained something of this kind in the previous proposition, but the optimal γ could
depend on t. Yet, we can fix this problem :
Proposition 0.5. Let µt be a constant speed geodesic between µ and ν, and suppose p > 1. Then
there exists an optimal plan γ ∈ Π(µ, ν) such that, for every t, we have µt = (πt )# γ.
Proof. Take t = 1/2 and pick two optimal plans γ − and γ + in Π(µ, µ1/2 ) and Π(µ1/2 , ν). The
previous theorem guarantees that this plans are actually uniquely defined. Consider also γ =
(πx,z )# σ as before. This plan γ is optimal and it will be optimal plan we will use (it is indeed
uniquely defined, but we will not use it).
Consider the curve νs = µs/2 which is a constant speed geodesic defined on the interval [0, 1] and
connecting µ to µ1/2 ). For every s ∈]0, 1[ we have νs = (πs )# γ̃, where γ̃ is an optimal plan from µ
to µ1/2 ) which could a priori depend on s. Yet, this optimal plan is unique and hence it coincides
with γ − . This means that, for every s, we have
s
s
µs/2 = νs = (πs )# γ̃ = ((1 − s)x + sy)# σ = ((1 − )x + z)# σ = (πs/2 )# γ,
2
2
where we used the fact that σ−a.e. y = 21 x + 12 z.
This proves that µt coincides with (πt )# γ for t ∈]0, 21 [. For t = 0, 12 and 1 this is true as well,
and for t ∈] 12 , 1[ this may be proven in the very same way (using γ + instead of γ.
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