Changfeng Gui,Fengbo Hang,Amir Moradifam and Xiaodong Wang

1Department of Mathematics,University of Texas at San Antonio,San Antonio,TX 78249,USA;

2Courant Institute,New York University,New York,NY 10012,USA;

3Department of Mathematics,University of California at Riverside,Riverside,CA 92521,USA;

4Department of Mathematics,Michigan State University,East Lansing,MI 48824,USA.

Abstract.In this note,we study symmetry of solutions of the elliptic equation

that arises in the consideration of rigidity problem of Hawking mass in general relativity.We provide various conditions under which this equation has only constant solutions,and consequently imply the rigidity of Hawking mass for stable constant mean curvature(CMC)sphere.

Key words:Semilinear elliptic equation,sphere covering inequality,rigidity of Hawking mass.

1 Introduction

The main aim of this note is to study the semilinear elliptic equation

on the standard S2.Here u is a smooth function on S2and α is a positive constant.

When α=1,(1.1)means that the conformal metric e2ugS2has constant curvature 1.Therefore all solutions are given by the pull back of the standard metric by Mobius transformations.This and more general statements also follow from the powerful method of moving plane(see[5,9]).The latter approach can be used to show(1.1)has only constant solution when 0<α<1(see[16]).More recently,the sphere covering inequality was discovered in[12]and applied to show all solutions to equation(1.1)must be constant functions for 1<α≤2.In particular,this confirms a long-standing conjecture of Chang-Yang([3,4])concerning the best constant in Moser-Trudinger type inequalities.Sphere covering inequality and its generalization can also be used to solve many uniqueness and symmetry problems from mathematical physics(see[1,10,12]and many references therein).[10]explains the sphere covering inequality from the point view of comparison geometry and provides some further generalizations.In contrast,for 2<α<3,nontrivial axially symmetric solutions were found in[15].The multiplicity of these nontrivial axially symmetric solutions was carefully discussed in[8].More recently,non-axially symmetric solutions to(1.1)for α>4 but close to 4 were found in[11].In related developments,topological degree of(1.1)for α/∈Z was computed in[6,14,15].We refer the readers to the survey article[20]for more details of mean field equations on a closed surface.

Recently,[18,19]discovered the interesting connection between the equation(1.1)with α=3 and rigidity problems involving Hawking mass in general relativity.Among other results,it was shown in[18]that for 2<α<4,any even solution to(1.1)must be axially symmetric.In particular,when α=3,any even solution,u(x)=u(?x)for all x∈S2,must be a constant function.It is also conjectured in[18,Section 3]that for 2<α≤3,any solution to(1.1)must be axially symmetric.Our note is motivated by this conjecture.Our main result is

Theorem 1.1.Assume 2<α≤3 and u∈C∞?S2?is a solution to

If for some p∈S2,?u(p)=0 and D2u(p)has two equal eigenvalues,then u is axially symmetric with respect to p.In particular,in the case α=3,u must be a constant function.

We may call the point p in the assumption as an umbilical critical point of u.So the theorem reads as:for 2<α≤3,any solution with an umbilical critical point must be axially symmetric with respect to that point.Here we do not know whether the solution is even or not.On the other hand,the approach to Theorem 1.1 can help us relax the even assumption in[18]a little bit.One typical example is

Theorem 1.2.Assume 2<α≤3 and u∈C∞?S2?is a solution to

If every large circle splits S2as two half sphere with equal area under the metric e2ugS2,then u is axially symmetric with respect to some point.In particular,in the case α=3,u must be a constant function.

Note that if u is even,then any large circle clearly splits the area.In Section 3,we will also present several other conditions which is weaker than the even assumption(see Proposition3.1,3.2).It is unfortunate we are not able to remove any of these assumptions.

At last we point out that there is an analogous nonlocal problem on S1,namely

Here ν is the unit outer normal direction and λ is a positive constant.This equation appears in the study of determinant of Laplacian on compact surface with boundary(see[17]).The solutions to the above problem is well understood(see[17,Lemma 2.3]and[21,Theorem 3]).The reason the problem on S1is much simpler than(1.1)is because the Fourier analysis on S1is much easier.

In Section 2,we will describe our main new observation and use it to derive Theorem 1.1.In Section 3,we will apply this new observation to derive several relaxation of the even assumption in[18].In particular,Theorem 1.2 will be proved.

2 Proof of Theorem 1.1

Here

is a smooth function on S2.

If w is not identically zero,then classical results(see[2,7,13])imply that the nodal set of w consists of finitely many smooth curves which only intersects at critical points of w.Moreover w behaves like a harmonic polynomial near each critical point,i.e.nodal set locally looks like straight lines with equal angles at critical points.

If ??Hyis a simply connected nodal domain, then it follows from the sphere covering inequality([12,Theorem 1.1]),or more precisely,the formulation with standard S2as background metric([10,Proposition 3.1]),that

This inequality and(2.5)implies Hycan not contain 3 or more simply connected nodal domains.

The crucial step to prove symmetry of solutions in[12,18]is counting the number of simply connected nodal domains.As observed in[12,Section 4.2],if we have a critical point of u on Cy,namely q∈Cy,and w is not identically zero,then the order of w at q(i.e.the order of the first nonvanishing term in Taylor expansion of w at q)is at least 2.Hence in Hy,at least one nodal line emanates from q with equal angle.This implies Hycontains at least two simply connected nodal domains.

Let z be the unit tangent vector of Cyat q.Our new observation is:if z is an eigenvector of D2u(q),and w is not identically zero,then the order of w at q is at least 3.If the order is larger than or equal to 4,then Hycontains at least 3 simply connected nodal domains,and it contradicts with(2.5).When the order of w at q is 3,the nodal set of w emanates two lines from q with anglein between,and w takes alternating signs in each angle.Since we can not have 3 or more simply connected nodal domains,the only possibility is we have only two nodal domains(i.e.the two emanating nodal line from q form a loop in Hy).It follows from Hopf principle thatis nonzero and of a fixed sign on Cy{q},here ν is the unit outer normal vector of Hy(in fact,ν=?y).In particular,there is no critical point on Cy{q}.We state this conclusion as a lemma.

3 Some relaxation of even assumption

Here we want to show the discussion in Section 2 can help us relax the even assumption in[18].

Proof of Theorem 1.2.Note that the equal area assumption can be written as:for any y∈S2,

Assume q is a critical point of u and z∈TqS2is an eigenvector of D2u(q).Denote y=q×z,v=u?Ryand w=u?v.Then w must be identically zero i.e.u is symmetric with respect to Cy.Indeed if w is not identically zero,it follows from Lemma 2.1 thatis nonzero and of a fixed sign on Cy{q},here ν=?y is the unit outer normal vector of Hy.Using

This contradicts with the equal area assumption.

By rotation we can assume e3is a critical point of u,and D2u(e3)has e1,e2as eigenvectors.It follows from previous discussion that u is symmetric with respect to Ce1and Ce2.

We will show u must be symmetric with respect to Ce3.One this is known it follows from[18,Lemma 8]that u must be axially symmetric.

To continue we let v=u?Re3and w=u?v,then using the equal area assumption,same argument as above shows

In all the cases,we know u must be axially symmetric,and hence it must be constant when α=3([18,Proposition 1]).

Along the same line we have the following Proof.By rotation we can assume p=e3and D2u(e3)has e1,e2as eigenvector.It follows from Lemma 2.1 and the fact?u(?e3)=0 that u is symmetric with respect to Ce1and Ce2.Now using

the argument in the proof of Theorem 1.2 tells us u is also symmetric with respect to Ce3.It follows from[18,Lemma 8]that u must be axially symmetric.

Proposition 3.2.Assume 2<α≤3 and u∈C∞?S2?is a solution to

If there exists p∈S2such that?u(p)=0,?u(?p)=0 and D2u(p)=D2u(?p)(here we identify TpS2with T?pS2naturally),then u must be axially symmetric.If α =3,then u is a constant function.

Proof.By rotation we can assume p=e3and D2u(e3)has e1,e2as eigenvector.It follows from Lemma 2.1 and the fact?u(?e3)=0 that u is symmetric with respect to Ce1and Ce2.It follows from the equation that

Hence u(e3)=u(?e3).Let w=u?u?Re3,then because w is symmetric with respect to Ce1and Ce2,we see w vanishes at least to order 4(does not include 4)at e3.If w is not identically zero,then it will have at least 3 simply connected nodal domains.This contradicts with the sphere covering inequality by the discussion in Section 2.It follows from[18,Lemma 8]that u must be axially symmetric.

Acknowledgments

C.Gui is partially supported by National Science Foundation Grant DMS-1601885.A.Moradifam is supported by National Science Foundation grant DMS-1715850.X.Wang is partially supported by Simons Foundation Collaboration Grant for Mathematicians no.312820.