Stable hypersurfaces with constant scalar curvature in Euclidean spaces
aa r X i v : . [ m a t h . DG ] S e p Stable hypersurfaces with constant scalar curvature in Euclideanspaces
Hil´ario Alencar ∗ , Walcy Santos ∗ and Detang Zhou ∗ November 21, 2018
Abstract.
We obtain some nonexistence results for complete noncompact stable hyppersurfaces withnonnegative constant scalar curvature in Euclidean spaces. As a special case we prove that there is nocomplete noncompact strongly stable hypersurface M in R with zero scalar curvature S , nonzero Gauss-Kronecker curvature and finite total curvature (i.e. R M | A | < + ∞ ). Key words : scalar curvature, stability, index, hypersurface.
In this paper we study the complete noncompact stable hypersurfaces with constant scalarcurvature in Euclidean spaces. It has been proved by Cheng and Yau [CY] that any com-plete noncompact hypersurfaces in the Euclidean space with constant scalar curvature andnonnegative sectional curvature must be a generalized cylinder. Note that the assumption ofnonnegative sectional curvature is a strong condition for hypersurfaces in the Euclidean spacewith zero scalar curvature since the hypersurface has to be flat in this case. Let M n be acomplete orientable Riemannian manifold and let x : M n → R n +1 be an isometric immersioninto the Euclidean space R n +1 with constant scalar curvature. We can choose a a global unitnormal vector field N and the Riemannian connections ∇ and e ∇ of M and R n +1 , respectively,are related by e ∇ X Y = ∇ X Y + h A ( X ) , Y i N, where A is the second fundamental form of the immersion, defined by A ( X ) = − e ∇ X N. Let λ , ..., λ n be the eigenvalues of A . The r-mean curvature of the immersion in a point p is defined by H r = 1 (cid:0) nr (cid:1) X i <...
0, for all variations with compact support in D and when the scalar curvature is nonzero, we saythat a regular domain D ⊂ M is strongly stable if the critical point is such that ( d A dt ) t =0 ≥ D . It is natural to study the global properties ofhypersurfaces in the Euclidean space with constant scalar curvature. For example we have thefollowing open question (see 4.3 in [AdCE]). Question 1.1
Is there any stable complete hypersurfaces M in R with zero scalar curvatureand nonzero the Gauss-Kronecker curvature? We have a partial answer to the question 1.1.
Theorem A. (see Theorem 3.1)
There is no complete noncompact stable hypersurface M in R n +1 with zero scalar curvature S and -mean curvature S = 0 satisfying lim R → + ∞ R B R S R = 0 , (2) where B R is the geodesic ball in M . When S = 0, S = | A | we have Corollary B.
There is no complete noncompact stable hypersurface M in R with zero scalarcurvature S , nonzero Gauss-Kronecker curvature and finite total curvature (i.e. R M | A | < + ∞ ) .We remark that Shen and Zhu (see [SZ]) proved that a complete stable minimal n -dimensionalhypersurface in R n +1 with finite total curvature must be a hyperplane. The above Corollarycan be seen as a similar result in dimension 3 for hypersurfaces with zero scalar curvature.We also prove the following result for hypersurfaces with positive constant scalar curvaturein Euclidean space. Theorem C. (see Theorem 3.2)
There is no complete immersed strongly stable hypersurface M n → R n +1 , n ≥ , with positive constant scalar curvature and polynomial growth of -volume,that is lim R →∞ R B R S dMR n < ∞ , here B R is a geodesic ball of radius R of M n . As a consequence of the properties of a graph with constant scalar curvature, we have thefollowing corollary:
Corollary D. (see Corollary 4.1)
Any entire graph on R n with nonnegative constant scalarcurvature must have zero scalar curvature. This can be compared with a result of Chern [Ch] which says any entire graph on R n withconstant mean curvature must be minimal. It has been be known by a result of X. Cheng in[Che] (see also [ENR]) that any complete noncompact stable hypersurface in R n +1 with constantmean curvature must be minimal if n <
5. It is natural to ask that any complete noncompactstable hypersurface in R n +1 with nonnegative constant scalar curvature must have zero scalarcurvature.It should be remarked that Chern[Ch] proved that there is no entire graph on R n with Riccicurvature less than a negative constant. We don’t know whether there exists an entire graphon R n with constant negative scalar curvature.The rest of this paper is organized as follows: we include some results and definitions whichwill be used in the proof of our theorems in Section 2. The proof of main results are given inSection 3 and Section 4 is an appendix in which we prove some stability properties for graphswith constant scalar curvature in the Eucildean space. S = const . We introduce the r ’th Newton transformation, P r : T p M → T p M , which are defined inductivelyby P = I,P r = S r I − A ◦ P r − , r ≥ . The following formulas are useful in the proof (see, [Re], Lemma 2.1).trace( P r ) = ( n − r ) S r , (3)trace( A ◦ P r ) = ( r + 1) S r +1 , (4)trace( A ◦ P r ) = S S r +1 − ( r + 2) S r +2 . (5)From [AdCC] we have the second variation formula for hypersurfaces in a space form ofconstant curvature c , Q n +1 c , with constant 2-mean curvature: d A dt | t =0 = Z D h P ( ∇ f ) , ∇ f i dM − Z D ( S S − S + c ( n − S ) f dM, ∀ f ∈ C ∞ c ( D ) . (6) Definition 2.1
When S = 0 and c = 0 , M is stable if and only if Z M h P ( ∇ f ) , ∇ f i dM ≥ − Z M S f dM, (7)3 or any f ∈ C ∞ c ( M ) . One can see that if P ≡ , then S = 0 and M is stable. When S = const. = 0 , M is stable if and only if Z D h P ( ∇ f ) , ∇ f i dM ≥ Z D ( S S − S + c ( n − S ) f dM, for all f ∈ C ∞ c ( M ) and R M f dM = 0 . We say that M is strongly stable if and only if the aboveinequality holds for all f ∈ C ∞ c ( M ) . Similar to minimal hypersurface we can also define the index I for hypersurfaces with constantscalar curvature. Given a relatively compact domain Ω ⊂ M , we denote by Ind (Ω) the numberof linearly independent normal deformations with support on Ω that decrease A . The index of the immersion are defined asInd ( M ) := sup { Ind (Ω) (cid:12)(cid:12) Ω ⊂ M, Ω relatively compact } . (8) M is strongly stable if Ind ( M ) = 0. The following result has been known in [El]. Lemma 2.1
Let M n → Q n +1 c be a noncompact hypersurface with S = const. > . If M hasfinite index then there exist a compact set K ⊂ M such that M \ K is strongly stable. For hypersurfaces with constant mean curvature, do Carmo and Zhou [dCZ] proved that
Theorem 2.1
Let x : M n → M n +1 be an isometric immersion with constant mean curvature H . Assume M has subexponential volume growth and finite index. Then there exist a constant R such that H ≤ − Ric M \ B R ( N ) , where N is a smooth normal vector field along M and Ric ( N ) is the Ricci curvature of M inthe normal vector N . The technique in [dCZ] was generalized by Elbert [El] to prove the following result:
Theorem 2.2
Let x : M n → Q ( c ) n +1 be an isometric immersion with S = constant > .Assume that Ind M < ∞ and that the 1-volume of M is infinite and has polynomial growth.Then c is negative and S ≤ − c. In particular, it implies that when c = 0 the hypersurfaces in the above theorem must havenonpositive scalar curvature. When S = 0 we know that | S | = | A | . Thus, if S = 0, we have that | A | >
0. Hence S = 0and we can choose an orientation such that P is semi-positive definite. Since | p P A | = trace( A ◦ P )= − S , c = 0, M is stable if Z M h P ( ∇ f ) , ∇ f i dM ≥ Z M | p P A | f dM, (9)for any f ∈ C ∞ c ( M ).By Lemma 4.1 in [AdCC], when S = 0, we know that |∇ A | − |∇ S | ≥
0. In the followinglemma, we characterize the equality case in some special case.
Lemma 3.1
Let M n ( n ≥ be a non-flat connected immersed 1-minimal hypersurface in R n +1 .If |∇ A | = |∇ S | holds on all nonvanishing point of | A | in M , then each component of M with | A | 6 = 0 must be a cylinder over a curve. Proof.
Choose a frame at p so that the second fundamental form is diagonalized. From thecomputations in [SSY], we have | A | = P i h ii , and X i,j,k h ijk − |∇| A || = [( X i,j h ij )( X s,t,k h stk ) − X k ( X i,j h ij h ijk ) ]( X i,j h ij ) − = 12 X i,j,k,s,t ( h ij h stk − h st h ijk ) | A | − = 12 X i,k,s,t ( h ii h stk − h st h iik ) + X s h ss ( X k X i = j h ijk ) | A | − = 12 X i,k,s ( h ii h ssk − h ss h iik ) + X i h ii ( X k X s = t h stk ) | A | − + 12 ( X k X i = j h ijk )= 12 X i,k,s ( h ii h ssk − h ss h iik ) | A | − + ( X k X i = j h ijk )= 12 X i,k,s ( h ii h ssk − h ss h iik ) | A | − +2 X i = j h iij + X i = j,j = k,i = k h ijk ≥ . (10)It is clear that the right hand side is nonnegative and is zero if and only if all terms on theright hand side vanish. X i,j,k h ijk − |∇| A || ≥ . (11)5uppose x : M → R n +1 is the 1-minimal immersion. Since M is not a hyperplane, then | A | is a nonnegative continuous function which does not vanish identically. Let p be such a pointsuch that | A | ( p ) >
0. Then | A | > U containing p . The equality in(11) implies h jji = 0 , for all j = i,h ijk = 0 , for all j = i, j = k, k = ih ii h ssk = h ss h iik , for all i, s, k. So we have h jij = 0 , for all j = i, and from the last equation we claim that at most one i suchthat h iii = 0. Otherwise, without the loss of generality we assume h = 0, and h = 0,we have h h k = h h k for all k . This implies h = h = 0 by choosing k = 1 ,
2. Usingagain the third formula we have h jj h = h h jj for j = 3 , · · · , n . Hence h jj = 0 for all j = 3 , · · · , n , which contradicts to | A | 6 = 0.We now assume h = 0 by continuity we can also assume h = 0. From the last equationof above equation, we have h h ss = h ss h for s = 1. Hence h ss = 0 for all s = 1. Thisimplies that M is a cylinder over a curve. (cid:3) We are now ready to prove
Theorem 3.1
There is no complete noncompact stable hypersurfaces in R n +1 with S = 0 and S = 0 satisfying lim R → + ∞ R B R S R = 0 . Proof.
Assume for the sake of contradiction that there were such a hypersurface M . FromLemma 3.7 in [AdCC], we have L S = |∇ A | − |∇ S | + 3 S S . (12)Since for any φ ∈ C ∞ c ( M ), Z M h P ( ∇ ( φS )) , ∇ ( φS ) i dM = Z M h P (( ∇ φ ) S + φ ∇ S ) , ( ∇ φ ) S ) + φ ∇ S i dM = Z M h P ( ∇ φ ) , ∇ φ i S dM + 2 Z M h P ( ∇ φ ) , ∇ S i φS dM + Z M φ h P ( ∇ S ) , ∇ S i dM, Z M φ S ( |∇ A | − |∇ S | ) dM = Z M ( L S − S S ) φ S dM = − Z M h P ( ∇ S ) , ∇ ( φ S ) i dM − Z M S φ S dM = − Z M φ h P ( ∇ S ) , ∇ S i dM − Z M h P ( ∇ φ ) , ∇ S i φS dM − Z M S φ S dM = − Z M h P ( ∇ ( φS )) , ∇ ( φS ) i dM + Z M h P ( ∇ φ ) , ∇ ( φ ) i S dM − Z M S φ S dM ≤ Z M h P ( ∇ φ ) , ∇ φ i S dM ≤ Z M |∇ φ | S dM, for any φ ∈ C ∞ c ( M ). Here we have used the stability inequality (7) in the fifth line and use thefollowing consequence of (3) in the last inequality: S |∇ φ | ≥ h P ( ∇ φ ) , ∇ φ i . (13)We can choose φ as φ ( x ) = R − r ( x ) R , on B R \ B R ;1 , on B R ;0 , on M \ B R . Thus from the choice of φ we have S ( |∇ A | − |∇ S | ) ≡
0. Therefore the elipticity of L implies L S = 3 S S . From Lemma 3.1, M must be a cylinder over a curve which contradicts S = 0. The proof is complete. (cid:3) The following Lemma is of some independent interest and we include here since its secondpart is useful in the proof of Theorem 3.2.
Lemma 3.2
Let M be a complete immersed hypersurface in Q n +1 c with nonnegative constantscalar curvature S > − n ( n − c and S = 0 .1)If M is strongly stable outside a compact subset, then either M has finite 1-volume, or lim R → + ∞ R Z B R S = + ∞ . lim R → + ∞ R Z B R S = + ∞ . In particular M has infinite 1-volume. roof. We can assume that there exists a geodesic ball B R ⊂ M such that M \ B R is stronglystable. That is, Z M ( S S − S + c ( n − S ) f dM ≤ Z M h P ( ∇ f ) , ∇ f i dM, (14)for all f ∈ C ∞ c ( M \ B R ).Now, since S ≥
0, we have (see [AdCR], p. 392) H H ≥ H , and H ≥ H / . By using that S = nH , S = n ( n − H and S = n ( n − n − H , it follows that( n − n S S ≥ S , that is, − S ≥ − ( n − n S S . (15)We also have that S n ≥ (cid:18) S n ( n − (cid:19) / , which implies S ≥ (cid:18) nn − (cid:19) / S / . (16)By using inequality (15) in (14), it follows that Z M (cid:18) S S − n − n S S + c ( n − S (cid:19) f dM ≤ Z M h P ( ∇ f ) , ∇ f i dM, that is, Z M (cid:18) S + n ( n − c (cid:19) S f dM ≤ n Z M h P ( ∇ f ) , ∇ f i dM. By using (13), we obtain that Z M S |∇ f | dM ≥ Z M h P ( ∇ f ) , ∇ f i dM Therefore, there exists a constant
C > Z M S |∇ f | dM ≥ C Z M S f dM. (17)8) When M is strongly stable outside B R . We can choose f as f ( x ) = r ( x ) − R , on B R +1 \ B R ;1 , on B R + R +1 \ B R +1 ; R + R +1 − r ( x ) R , on B R + R +1 \ B R + R +1 ;0 , on M \ B R + R +1 , where r ( x ) is the distance function to a fixed point. Then1 R Z B R + R \ B R + R S dM + Z B R \ B R S dM ≥ C Z B R + R \ B R S dM. If the 1-volume is infinite, we can choose R large such that C Z B R + R \ B R S dM > Z B R \ B R S dM, hence lim R → + ∞ R Z B R + R \ B R + R S dM = + ∞ .
2) When M is strongly stable we can choose a simpler test function f as f ( x ) = , on B R ; R − r ( x ) R , on B R \ B R ;0 , on M \ B R , which implies that when S = 0, lim R → + ∞ R Z B R S dM = + ∞ . The proof is complete. (cid:3)
Theorem 3.2
There is no complete immersed strongly stable hypersurface M n → R n +1 , n ≥ ,with positive constant scalar curvature and polynomial growth of -volume, that is lim R →∞ R B R S dMR n < ∞ , where B R is a geodesic ball of radius R of M n . Proof.
Suppose that M is a complete immersed strongly stable hypersurface M n → R n +1 , n ≥
3, with positive constant scalar curvature. From Theorem 2.2, it suffices to show that the1-volume R M S dM is infinite which is the part (2) of Lemma 3.2. (cid:3) Graphs with S = const in Euclidean space In this section we include some stability properties and estimates for entire graphs on R n whichmay be known to experts ant not easy to find a reference. Using these facts we give the proofof Corollary 4.1. Let M n a hypersurface of R n +1 given by a graph of a function u : R n → R ofclass C ∞ ( R n ). For such hypersurfaces we have: Proposition 4.1
Let M n a graph of a function u : R n → R of class C ∞ ( R n ) . Then1. If S = 0 and S does not change sign on M , then M n is a stable hypersurface.2. If M has S = C > , then M n is strongly stable. Proof.
Considerer and f : M → R a C ∞ function with compact support and let W = p |∇ u | . In order to calculate h P ( ∇ f ) , ∇ f i , write g = f W . Thus h P ( ∇ f ) , ∇ f i = h P ( ∇ ( gW )) , ∇ ( gW ) i = h P ( g ∇ W + ∇ g W ) , g ∇ ( 1 W ) + 1 W ∇ g i = h gP ( ∇ W ) + 1 W P ( ∇ g ) , g ∇ ( 1 W ) + 1 W ∇ g i = g h P ( ∇ W ) , ∇ W i + gW h P ( ∇ W ) , ∇ g i + gW h P ( ∇ g ) , ∇ W i + 1 W h P ( ∇ g ) , ∇ g i . By using that P is selfadjoint, we have: h P ( ∇ f ) , ∇ f i = g h P ( ∇ W ) , ∇ W i + 2 gW h P ( ∇ W ) , ∇ g i + 1 W h P ( ∇ g ) , ∇ g i . (18)On the other hand, if { e , ..., e n } is a geodesic frame along M ,div( f gP ( ∇ W ) = n X i =1 h∇ e i ( f gP ( ∇ W )) , e i i = n X i =1 h f g i P ( ∇ W ) + f i gP ( ∇ W ) + f g ∇ e i ( P ( ∇ W )) , e i i = n X i =1 { f g i h P ( ∇ W ) , e i i + f i g h P ( ∇ W ) , e i i + f g h∇ e i ( P ( ∇ W )) , e i i} . Since f = gW , we get f i = g i W + g (cid:18) W (cid:19) i , gf i = gg i W + g (cid:18) W (cid:19) i = f g i + g (cid:18) W (cid:19) i Hence,div( f gP ( ∇ W ) = n X i =1 { f g i h P ( ∇ W ) , e i i + ( f g i + g (cid:18) W (cid:19) i ) h P ( ∇ W ) , e i i} + f gL ( 1 W )= n X i =1 { f g i h P ( ∇ W ) , e i i + g (cid:18) W (cid:19) i h P ( ∇ W ) , e i i} + f gL ( 1 W )= 2 f h P ( ∇ W ) , ∇ g i + g h P ( ∇ W ) , ∇ ( 1 W ) i + f gL ( 1 W )= 2 gW h∇ W , P ( ∇ g ) i + g h P ( ∇ W ) , ∇ ( 1 W ) i + f W L ( 1 W ) . Thus, 2 gW h∇ W , P ( ∇ g ) i = div( f gP ( ∇ W )) − g h P ( ∇ W ) , ∇ ( 1 W ) i − f W L ( 1 W ) . (19)Now, by using (19) into equation (18), we get h P ( ∇ f ) , ∇ f i = div( f gP ( ∇ W )) − f W L ( 1 W ) + 1 W h P ( ∇ g ) , ∇ g i . Now, the divergence theorem implies that Z M h P ( ∇ f ) , ∇ f i dM = − Z M f W L ( 1 W ) dM + Z M W h P ( ∇ g ) , ∇ g i dM. Choose the orientation of M in such way that S ≥
0. Since S −| A | = 2 S ≥
0, we obtain that S ≥ | A | . Thus, h P ( ∇ g ) , ∇ g i = S |∇ g | − h A ∇ g, ∇ g i ≥ ( S − | A | ) |∇ g | ≥
0, which impliesthat Z M h P ( ∇ f ) , ∇ f i dM ≥ − Z M f W L ( 1 W ) dM. (20)When S is constant, we will use the following formula proved by Reilly (see [Re], PropositionC). L ( 1 W ) = L ( h N, e n +1 i ) + ( S S − S ) h N, e n +1 i = 0 , where N is the normal vector of M and e n +1 = (0 , ..., , ± , according to our choice of theorientation of M .Thus, Z M h P ( ∇ f ) , ∇ f i dM ≥ − Z M f W L ( 1 W ) dM = 0for all function f with compact support. Hence M is stable if S = 0 and strongly stable in thecase S = 0. (cid:3) emark 4.1 We would like to remark that the operator L need not to be elliptic in the aboveproof. Proposition 4.2
Let M n a graph of a function u : R n → R of class C ∞ ( R n ) , with S ≥ . Let B R be a geodesic ball of radius R in M . Then Z B θR S dM ≤ C ( n )1 − θ R n , where C ( n ) and θ are constants, with < θ < . In particular, Z M S dM has polynomialgrowth. Proof.
Let f : M → R be a be a function in C ∞ ( M ), that is a smooth function with compactsupport. Observe that div (cid:18) f ∇ uW (cid:19) = f div (cid:18) ∇ uW (cid:19) + (cid:28) ∇ f, ∇ uW (cid:29) , where W = p |∇ u | . By using the fact that S is given by S = div (cid:18) ∇ uW (cid:19) , we have that Z M f S dM = Z M f div (cid:18) ∇ uW (cid:19) dM = − Z M (cid:28) ∇ f, ∇ uW (cid:29) dM. (21)Now, choose a family of geodesic balls B R that exhausts M . Fix θ , with 0 < θ < f : M → R be a continuous function that is one on B θR , zero outside B R and linear on B R \ B θR . Therefore, from equation (21) we obtain Z B θR S dM ≤ Z B R f S dM ≤ Z B R (cid:28) ∇ uW , ∇ f (cid:29) dM. By using Cauchy-Schwarz inequality and the fact that |∇ u | W ≤
1, it follows that Z B θR S dM ≤ Z B R |∇ f | dM ≤ Z B r \ B θR − θ ) R dM ≤ − θ ) R vol( B R ) . We observe that since M is a graph, if Ω R = { ( x , .., x n +1 ) ∈ R n +1 | − R ≤ x n +1 ≤ R ; p x + ... + x n ≤ R } , thenvol( B R ) ≤ Z Ω R dx ...dx n +1 = C ( n ) R n +1 . Hence, Z B θR S dM ≤ − θ ) R vol( B R ) = C ( n )1 − θ R n . (cid:3)
12e have the following Corollary of Theorem 3.2
Corollary 4.1
Any entire graph on R n with nonnegative constant scalar curvature must havezero scalar curvature. Proof.
Suppose by sake of contradiction that there exist a entire graph with S = const > S >
0, we get that S = | A | + 2 S >
0, we obtain that S does not change sign and we can choose the orientation in such way that S >
0. Thus thegraph has polynomial growth of the 1-volume. Thus we have a contradiction with Theorem3.2. Thus it follows that S = 0. (cid:3) References [AdCC]
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