Parallel transport on non-collapsed 𝖱𝖢𝖣(𝐾, 𝑁) spaces

Lemma A.1 (Lions)

Let 𝐸 and 𝐻 be a normed and a Hilbert space respectively. Assume that 𝐸 is continuously embedded in 𝐻, with ∥ v ∥ H ≤ ∥ v ∥ E for every v ∈ E . Let B : H × E → R be a bilinear form such that B ⁢ ( ⋅ , v ) is continuous for every v ∈ E . If 𝐵 is coercive, namely there exists c > 0 such that B ⁢ ( v , v ) ≥ c ⁢ ∥ v ∥ E 2 for every v ∈ E , then for all l ∈ E ′ , there exists h ∈ H such that B ⁢ ( h , v ) = l ⁢ ( v ) for every v ∈ E and

Definition A.2 (Parallel transport in W 1 , 2 ⁢ ( π ) )

Let ( X , d , m ) be an RCD ⁢ ( K , ∞ ) space and 𝝅 a Lipschitz test plan on X . Given V ̄ ∈ e 0 ∗ ⁢ L 2 ⁢ ( T ⁢ X ) , we say that V ∈ W 1 , 2 ⁢ ( π ) is a parallel transport in W 1 , 2 ⁢ ( π ) of V ̄ along 𝝅 if D π ⁢ V = 0 and, for every Z ∈ TestVF ⁡ ( π ) ,

where we denote by t ↦ R ⁢ ( Z ) t the absolutely continuous representative of t ↦ ∫ ⟨ V t , Z t ⟩ ⁢ d π .

Theorem A.3 (Existence of PT in W 1 , 2 ⁢ ( π ) )

Let ( X , d , m ) be an RCD ⁢ ( K , ∞ ) space and 𝝅 a Lipschitz test plan on X . Let V ̄ ∈ e 0 ∗ ⁢ L 2 ⁢ ( T ⁢ X ) be given. Then there exists V ∈ W 1 , 2 ⁢ ( π ) that is a parallel transport in W 1 , 2 ⁢ ( π ) of V ̄ along 𝝅.

Proof

Fix ε ∈ ( 0 , 1 / 2 ) . Consider the Hilbert space H : = ( H 1 , 2 ( π ) , ∥ ⋅ ∥ H 1 , 2 ⁢ ( π ) ) . Define also

Clearly, ∥ Z ∥ H ≤ ∥ Z ∥ E for every Z ∈ E . Now let us define B : H × E → R and ℓ : E → R as

respectively. The map 𝐵 is bilinear by construction. Moreover, for some constant C > 0 , we have that | B ⁢ ( V , Z ) | ≤ C ⁢ ∥ V ∥ H ⁢ ∥ Z ∥ H for every V ∈ H and Z ∈ E ; thus, in particular, B ⁢ ( ⋅ , Z ) is continuous for any Z ∈ E . The Leibniz rule grants that

holds for every Z ∈ E , whence coercivity of the map 𝐵 follows: given any Z ∈ E , we have

Furthermore, it holds that ℓ ∈ E ′ and ∥ ℓ ∥ E ′ ≤ ∥ V ̄ ∥ e 0 ∗ ⁢ L 2 ⁢ ( T ⁢ X ) . Therefore, Lemma A.1 yields the existence of an element V ε ∈ H such that ∥ V ε ∥ H ≤ ∥ V ̄ ∥ e 0 ∗ ⁢ L 2 ⁢ ( T ⁢ X ) / ε and B ⁢ ( V ε , Z ) = ℓ ⁢ ( Z ) for every Z ∈ E , which explicitly reads as

for every Z ∈ E . Given any Z ∈ TestVF ⁡ ( π ) and φ ∈ LIP ⁡ ( [ 0 , 1 ] ) with spt ⁡ ( φ ) ⊆ [ 0 , 1 ) , it holds that t ↦ φ ⁢ ( t ) ⁢ Z t belongs to 𝐸 and D π ⁢ ( φ ⁢ Z ) t = φ ′ ⁢ ( t ) ⁢ Z t + φ ⁢ ( t ) ⁢ D π ⁢ Z t for a.e. t ∈ [ 0 , 1 ] . Then

Fix a Lebesgue point s ∈ ( 0 , 1 ) of t ↦ ∫ − ⟨ V t ε , Z t ⟩ + ε ⁢ ⟨ D π ⁢ V t ε , Z t ⟩ ⁢ d ⁢ π . Define φ n as

for all n ∈ N , n > 1 / ( 1 − s ) . Note that ( φ n ) n ⊆ LIP ⁡ ( [ 0 , 1 ] ) is a bounded sequence in L ∞ ⁢ ( 0 , 1 ) , spt ⁡ ( φ n ) ⊆ [ 0 , 1 ) for all 𝑛, and φ n → χ [ 0 , s ] pointwise as n → ∞ . Moreover, it holds that

Therefore, by plugging φ = φ n into (A.3) and letting n → ∞ , we deduce that

Given that V ε ∈ H 1 , 2 ⁢ ( π ) , we can find a sequence ( Z n ) n ⊆ TestVF ⁡ ( π ) that W 1 , 2 ⁢ ( π ) -converges to V ε . We start noting that, from [24, Theorem 3.23], there exists a continuous injection i : H 1 , 2 ⁢ ( π ) → C ⁢ ( π ) such that ∥ V ε − Z n ∥ C ⁢ ( π ) ≤ 2 ⁢ ∥ V ε − Z n ∥ W 1 , 2 ⁢ ( π ) , which grants that lim n → ∞ ∥ V 0 ε − Z 0 n ∥ e 0 ∗ ⁢ L 2 ⁢ ( T ⁢ X ) = 0 . Therefore, it follows that

By plugging Z = Z n into (A.4), letting n → ∞ , and using the Leibniz rule in H 1 , 2 ⁢ ( π ) , we get

Since 1 2 ⁢ ∫ | V 0 ε | 2 ⁢ d π − ∫ ⟨ V ̄ , V 0 ε ⟩ ⁢ d π = 1 2 ⁢ ∫ | V ̄ − V 0 ε | 2 ⁢ d π − 1 2 ⁢ ∫ | V ̄ | 2 ⁢ d π , we can rewrite the former expression as

Therefore, we obtain that

By integrating the above inequality over the interval [ 0 , 1 ] , multiplying by 2, and then applying Young’s inequality a ⁢ b ≤ ε ⁢ a 2 + b 2 / ( 4 ⁢ ε ) , we infer that

whence accordingly

Observe also that { ε ⁢ V ε } ε ∈ ( 0 , 1 / 2 ) is bounded in 𝐻. Therefore, there exist V ∈ L 2 ⁢ ( π ) , W ∈ H , and a sequence ε n ↘ 0 such that V ε n ⇀ V weakly in L 2 ⁢ ( π ) and ε n ⁢ V ε n ⇀ W weakly in 𝐻. In particular, it must hold that W = 0 . Hence, by letting n → ∞ in the identity

which holds for every n ∈ N and Z ∈ TestVF c ⁡ ( π ) by (A.2), we can finally conclude that

is satisfied for every Z ∈ TestVF c ⁡ ( π ) . This grants that V ∈ W 1 , 2 ⁢ ( π ) and D π ⁢ V = 0 . Finally, let us prove (A.1). Fix any Z ∈ TestVF ⁡ ( π ) and denote by t ↦ R ⁢ ( Z ) t the absolutely continuous representative of t ↦ ∫ ⟨ V t , Z t ⟩ ⁢ d π (that belongs to W 1 , 1 ⁢ ( 0 , 1 ) ). Write (A.4) with ε = ε n , integrate over s ∈ [ 0 , 1 ] , and then let n → ∞ : by exploiting the weak convergence V ε n ⇀ V in L 2 ⁢ ( π ) (and by using the dominated convergence theorem), we obtain that

where we applied the Leibniz rule with one vector field in H 1 , 2 ⁢ ( π ) and the other in W 1 , 2 ⁢ ( π ) . Hence the statement is achieved. ∎