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(∂η|∇u| + p(∇u, η))dA Z∂6=0U u¯ 1 |∇¯ 2u|2 > 2 (0.4) 2 + (µ + J(ν)+ h − 2hP +2hν, ∇hi) − K dAdt, Zu ZΣt 2 |∇u|  ∇u where ν = |∇u| , K is Gauss curvature of the level set and dA is the area element. Proof. Recall the Bochner formula 1 ∆|∇u|2 = |∇2u|2 + Ric(∇u, ∇u)+ h∇u, ∇∆ui. 2 1 2 For δ > 0 set φδ = (|∇u| + δ) 2 , and use the Bochner formula to find ∆|∇u|2 |∇|∇u|2|2 ∆φδ = − 3 2φδ 4φδ 1 (0.5) > (|∇2u|2 − |∇|∇u||2 + Ric(∇u, ∇u)+ h∇u, ∇∆ui). φδ ∇u On a regular level set Σ, the unit normal is ν = |∇u| and the second fundamental ∇i∇j u form is given by Aij = |∇u| , where ∂i and ∂j are tangent to Σ. We then have |A|2 = |∇u|−2(|∇2u|2 − 2|∇|∇u||2 + [∇2u(ν,ν)]2), and the mean curvature satisfies 2 (0.6) |∇u|H = ∆u − ∇ν u Furthermore by taking two traces of the Gauss equations 2 2 2Ric(ν,ν)= Rg − RΣ − |A| + H , where Rg is the of U and RΣ is the scalar curvature of the level set Σ. Combining these formulas with (0.5) produces 2 1 2 2 2 |∇u| 2 2 ∆φδ > |∇ u| − |∇|∇u|| + h∇u, ∇∆ui + (Rg − RΣ + H − |A| ) φδ  2 

1 2 2 2 2 2 = (|∇ u| + (Rg − RΣ)|∇u| +2h∇u, ∇∆ui + (∆u) − 2(∆u)∇ν u). 2φδ Let us now replace the Hessian with the spacetime Hessian via the relation (0.2), and utilize (0.3) to find

∆φδ

1 2 2 2 > (|∇¯ u − p|∇u|| + (Rg − RΣ)|∇u| +2h∇u, ∇(−P |∇u| + h|∇u|)i 2φδ 2 2 + (−P |∇u| + h|∇u|) − 2(−P |∇u| + h|∇u|)∇ν u). Noting that 1 h∇u, ∇|∇u|i = hν, ∇|∇u|2i = ui∇ u = |∇u|∇2 u, 2 iν ν A PERTURBATION OF SPACETIME LAPLACIAN EQUATION 3 and expanding |∇¯ 2u − p|∇u||2, we have that

∆φδ 1 ¯ 2 2 2 2 2 2 > (|∇ u| − 2hp, ∇ ui|∇u| − |p|g|∇u| + (Rg − RΣ)|∇u| 2φδ +2|∇u|h∇u, −∇P + ∇hi + (−P |∇u| + h|∇u|)2).

2 2 Regrouping using the energy density 2µ = Rg + P − |p|g,

∆φδ

1 2 2 2 2 > [|∇¯ u| − 2hp, ∇ ui|∇u| + (2µ − RΣ)|∇u| − 2|∇u|h∇u, ∇P i 2φδ (0.7) + |∇u|2(h2 − 2hP +2hν, ∇hi]. Consider an open set A⊂ [u, u¯] containing the critical values of u, and let B⊂ [u, u¯] denote the complementary closed set. Then integrate by parts to obtain

h∇φδ , vi = ∆φδ = ∆φδ + ∆φδ. Z∂U ZU Zu−1(A) Zu−1(B) According to Kato type inequality [HKK20, Lemma 3.1] and (0.5) there is a positive constant C0, depending only on Ricg, P − h and its first derivatives such that

∆φδ > −C0|∇u|. An application of the coarea formula to u : u−1(A) → A then produces

(0.8) − ∆φε 6 C0 |∇u| = C0 |Σt|dt, Zu−1(A) Zu−1(A) Zt∈A where |Σt| is the 2-dimensional Hausdorff measure of the t-level set Σt. Next, apply the coarea formula to u : u−1(B) →B together with (0.7) to obtain

∆φδ Zu−1(B) 1 |∇u| |∇¯ 2u|2 2 > 2 2 +2µ − RΣt − (hp, ∇ ui + h∇u, ∇P i) dAdt 2 Zt∈B ZΣt φδ  |∇u| |∇u|  1 1 + (h2 − 2hP +2hν, ∇hi)dAdt. 2 Zt∈B ZΣt φδ Combining with (0.8) and

∆φδ = ∆φδ − ∆φδ = ∂ηφδ − ∆φδ, Zu−1(B) ZU Zu−1(A) Z∂U Zu−1(A) we obtain

∂ηφδ + C0 |Σt|dt Z∂U Zt∈A 1 |∇u| |∇¯ 2u|2 2 > 2 2 +2µ − RΣt − (hp, ∇ ui + h∇u, ∇P i) dAdt 2 Zt∈B ZΣt φδ  |∇u| |∇u|  1 1 (0.9) + (h2 − 2hP +2hν, ∇hi)dAdt. 2 Zt∈B ZΣt φδ 4 XIAOXIANG CHAI

On the set u−1(B), we have that |∇u| is uniformly bounded from below. In addition, on ∂6=0U it holds that |∇u| ∂ηφδ = ∂η|∇u| → ∂η|∇u| as δ → 0. φδ Therefore, the limit δ → 0 may be taken in (0.9), resulting in the same bulk expression except that φε is replaced by |∇u|, and with the boundary integral taken over the restricted set. Furthermore, by Sard’s theorem (see [HKK20, Remark 3.3]) the measure |A| of A may be taken to be arbitrarily small. Since the map t 7→ |Σt| is integrable over [u, u¯] in light of the coarea formula, by then taking |A| → 0 we obtain Lastly integration by parts gives

2 ij i j hp, ∇ ui = p ∇ij u = p(∇u, η) − u ∇ pij , ZU ZU Z∂U ZU and recalling that J = divg(p − Pg) and RΣt =2KΣt , yields the desired result.  Now we discuss two special boundary conditions of the equation (0.3) namely Dirichlet and Neumann boundary conditions and its geometric implications on the boundary contribution

(∂η|∇u| + p(∇u,υ))dA Z∂6=0U in (0.4). We assume that u ∈ C2(U ∪ S) where S is a relatively open portion of ∂U. This is a valid assumption due to an existence theorem of [HKK20, Section 4]. We have the following: Lemma 0.2. Assume the solution u to (0.3) takes constant values on S, then

∂η|∇u| + p(∇u, η) = (−HS − trS p + h)|∇u|. Proof. Since u is constant on S, then by using the decomposition of the Laplacian ∆, 2 −P |∇u| + h|∇u| = ∆u = HSh∇u, ηi + ∇ u(η, η). The decomposition is already used in (0.6). Since u is constant on S, ∇u either point outward or inward of U. We calculate only the case when ∇u points outward, that is η = ∇u/|∇u|,

∂η|∇u| + p(∇u, η) 1 2 = |∇u| (∇ u)(∇u, η)+ p(∇u, η) =(∇2u)(η, η)+ p(η, η)|∇u|

= − HS |∇u|− P |∇u| + p(η, η)|∇u| + h|∇u|

=(−H − trS p + h)|∇u|. 

∂u Lemma 0.3. Suppose that ∂η =0 on S, then 1 ∂η|∇u| = |∇u| B(∇u, ∇u), where B is the second fundamental form of S in U. A PERTURBATION OF SPACETIME LAPLACIAN EQUATION 5

Proof. First,

∂η|∇u| 1 i j = |∇u| ηj ∇iu∇ ∇ u 1 i j i j i = |∇u| [∇iu∇ (ηj ∇ u) − ∇ u∇ u∇ ηj ] 1 = − |∇u| B(∇u, ∇u), we have used the boundary condition ∂u/∂η = 0. 

Note that we have not used u is a solution to (0.3). On a regular level set Σt, ∇u let e1 be a unit tangent vector of ∂Σt, then {η,ν = |∇u| ,e1} forms an orthonormal basis. We see that the geodesic curvature of ∂Σt in Σt is given by

κ∂Σt = h∇e1 η,e1i

∂u since ∂η = 0 implies that η is orthogonal to ∂Σt in Σt and points outward of Σt. So

B(ν,ν)= h∇ν η,νi

=(h∇ν η,νi + h∇e1 η,e1i) − h∇e1 η,e1i

=HS − κ∂Σt .

Therefore, we conclude the following.

Lemma 0.4. Assume the solution u to (0.3) takes constant values on S, at a point in S ∩ ∂Σt with Σt being a regular level set, then

∇u ∂η|∇u| + p(∇u, η) = (−HS + p( |∇u| ,η)+ κ∂Σt )|∇u|. So to study rigidity questions on initial data sets with boundary, it is natural to assume

µ + J(ν)+ h2 − 2hP +2|∇h| > 0, and the convexity condition on the boundary

HS > | trS p − h|, or

HS > p(η,e0) where e0 is any unit vector on S. With h = 0, the conditions on the boundary are termed boundary dominant energy conditions by [AdLM19] in their study of initial data sets with a noncompact boundary.

Acknowledgment Research of Xiaoxiang Chai is supported by KIAS Grants under the research code MG074402. I would also like to thank Tin Yau Tsang (UCI) and Sven Hirsch (Duke) for discussions on the level set technique of harmonic functions. 6 XIAOXIANG CHAI

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Korea Institute for Advanced Study, Seoul 02455, South Korea Email address: [email protected]