The Lorentzian Lichnerowicz conjecture for real-analytic, three-dimensional manifolds

Proposition A.2 (see [18, Proposition 6.3])

Let ( M , g ) be a smooth Lorentzian manifold. Let { f k } < Conf ⁡ ( M , [ g ] ) with holonomy sequence { p k } along x ̂ k → x ̂ in M ̂ . Assume that there exists a conformal geodesic segment β : [ 0 , 1 ] → Ein 1 , n − 1 emanating from 𝑜 such that lim k → ∞ p k . [ β ] = o . Then any pointwise holonomy sequence of { p k } at β ⁢ ( 1 ) admits a subsequence which is a holonomy sequence for { f k } with respect to some converging sequence y ̂ k → y ̂ in M ̂ .

Let ( M , g ) be a smooth, 3-dimensional, Lorentzian manifold. Let { p k } be a holonomy sequence for { f k } along x ̂ k → x ̂ in M ̂ . Suppose there exists a conformal geodesic segment β : [ 0 , 1 ] → Ein 1 , 2 emanating from 𝑜 such that lim k → ∞ p k . [ β ] = o . If { p k } admits a pointwise holonomy sequence at β ⁢ ( 1 ) which is stable, then a nonempty open subset U ⊂ M is conformally flat.

Let { p k } be a sequence of P + . After passing to a subsequence, there exists a conformal geodesic segment β : [ 0 , 1 ] → Ein 1 , 2 emanating from 𝑜 such that

and such that { p k } admits a pointwise holonomy sequence at β ⁢ ( 1 ) which is stable.

Proof

Denote the Euclidean norm on R 3 by ∥ ⋅ ∥ . Write

Here v k is a sequence of R 1 , 2 satisfying ∥ v k ∥ = 1 , and t k ≥ 0 . The expression v k * stands for v k t ⁢ I , where 𝕀 is as in Section 4.1. Observe that { p k } , hence { t k } , is unbounded because { f k } is unbounded. After taking a subsequence, we assume that t k → ∞ and that there is a vector v = lim v k .

Recall the conformal immersion j o : R 1 , 2 → Ein 1 , 2 from Section A.1 above, and let x k → x ∈ R 1 , 2 . For u ∈ [ 0 , 1 ] and k ∈ N , define

Observe that each β k and 𝛽 are conformal geodesic segments emanating from 𝑜. Define

From the matrix expression of p k and the formula for j o ,

After possibly taking a further subsequence of { p k } , we may assume that t k ⁢ | q ⁢ ( v k ) | converges in [ 0 , ∞ ] . There are two subcases.

First case: t k ⁢ | q ⁢ ( v k ) | → ∞ . In this case, q ⁢ ( v k ) is nonzero for 𝑘 large enough so that, after perhaps taking a subsequence of { p k } , we may assume that the sign of q ⁢ ( v k ) is constant. Choose ϵ = ± 1 so that ϵ ⁢ q ⁢ ( v k ) ≥ 0 for all sufficiently large 𝑘. Choose 𝑥 such that

1. ϵ ⁢ q ⁢ ( x ) is positive,

2. ⟨ x , v ⟩ is positive.

Let u ∈ ( 0 , 1 ] and write

Under the assumption t k ⁢ | q ⁢ ( v k ) | → ∞ ,

so that, for 𝑘 big enough,

If t k ⁢ u ≥ 1 , we can infer from (A.1) and the previous inequalities that

Observe that (A.2) also holds trivially if u = 0 .

Suppose t k ⁢ u ≤ 1 , and note that conditions (a) and (b) on 𝑥 imply that, for 𝑘 big enough,

Then, for 𝑘 large enough,

From (A.3), we infer

Taking x k ≡ x gives lim k → ∞ p k ⁢ ( [ β ] ) = o . Moreover, (A.4) shows that

for any x k → x . In particular, { p k } is stable at j o ⁢ ( x ) = β ⁢ ( 1 ) in the sense of [17, Definition 4.1]. It was proved in [17, Lemma 4.3] that if { p k } has this stability property at some point 𝑧, it admits a pointwise stable holonomy sequence at 𝑧. Lemma A.4 is thus proved in this case.

Second case: lim k → ∞ t k ⁢ q ⁢ ( v k ) = a ∈ R . This time, take x ∈ R 1 , 2 such that

1. q ⁢ ( x ) = 0 ,

2. ⟨ x , v ⟩ > 0 .

Because ⟨ x k , v k ⟩ + u 4 ⁢ t k ⁢ q ⁢ ( v k ) ⁢ q ⁢ ( x k ) tends to ⟨ x , v ⟩ as k → ∞ ,

We infer from (A.1) that

This inequality holds trivially for u = 0 , giving

Taking x k ≡ x gives lim k → ∞ p k ⁢ ( [ β ] ) = o . Moreover, if { x k } is any sequence converging to 𝑥, inequality (A.5) shows that lim k → ∞ p k ⁢ ( j o ⁢ ( x k ) ) = o . As in the first case, this implies that { p k } admits a pointwise, stable holonomy sequence at j o ⁢ ( x ) = β ⁢ ( 1 ) , and Lemma A.4 is proved. ∎