Asymptotic behavior of flows by powers of the Gaussian curvature

Transcription

1 Acta Math., 29 (207), 6 DOI: 0.430/ACTA.207.v29.n.a c 207 by Institut Mittag-Leffler. All rights reserved Asymptotic behavior of flows by powers of the Gaussian curvature by Simon Brendle Columbia University New Yor, NY, U.S.A. Kyeongsu Choi Columbia University New Yor, NY, U.S.A. Panagiota Dasalopoulos Columbia University New Yor, NY, U.S.A.. Introduction Parabolic flows for hypersurfaces play an important role in differential geometry. One fundamental example is the flow by mean curvature (see [8]). In this paper, we consider flows where the speed is given by some power of the Gaussian curvature. More precisely, given an integer n 2 and a real number α>0, a -parameter family of immersions F : M n [0, T )!R n+ is a solution of the α-gauss curvature flow if, for each t [0, T ), F (M n, t)=σ t is a complete convex hypersurface in R n+ and F (, t) satisfies t F (p, t) = Kα (p, t)ν(p, t). Here, K(p, t) and ν(p, t) are the Gauss curvature and the outward-pointing unit normal vector of Σ t at the point F (p, t), respectively. Theorem. Let Σ t be a family of closed, strictly convex hypersurfaces in R n+ moving with speed K α ν, where α /(n+2). Then, either the hypersurfaces Σ t converge to a round sphere after rescaling, or we have α=/(n+2) and the hypersurfaces Σ t converge to an ellipsoid after rescaling. Flows by powers of the Gaussian curvature have been studied by many authors, starting with the seminal paper of Firey [4] in 974. Tso [2] showed that the flow The first-named author was supported in part by the National Science Foundation under grant DMS The third-named author was supported in part by the National Science Foundation under grant DMS

2 2 s. brendle,. choi and p. dasalopoulos exists up to some maximal time, when the enclosed volume converges to zero. In the special case α=/n, Chow [9] proved convergence to a round sphere. Moreover, Chow [0] obtained interesting Harnac inequalitites for flows, by powers of the Gaussian curvature (see also [7]). In the affine invariant case α=/(n+2), Andrews [] showed that the flow converges to an ellipsoid. This result can alternatively be derived from a theorem of Calabi [7], which asserts that the only self-similar solutions for α=/(n+2) are ellipsoids. The arguments in [7] and [] rely crucially on the affine invariance of the equation, and do not generalize to other exponents. In the special case of surfaces in R 3 (n=2), Andrews [2] proved that flow converges to a sphere when α=; this was later extended in [4] to the case n=2 and α [ 2, 2]. The results in [2] and [4] rely on an application of the maximum principle to a suitably chosen function of the curvature eigenvalues; these techniques do not appear to wor in higher dimensions. However, it is nown that the flow converges to a self-similar solution for every n 2 and every α /(n+2). This was proved by Andrews [3] for α [/(n+2), /n]; by Guan and Ni [6] for α=; and by Andrews, Guan, and Ni [5] for all α (/(n+2), ). One of the ey ingredients in these results is a monotonicity formula for an entropy functional. This monotonicity was discovered by Firey [4] in the special case α=. Thus, the problem can be reduced to the classification of self-similar solutions. In the affine invariant case α=/(n+2), the self-similar solutions were already classified by Calabi [7]. In the special case when α and the hypersurfaces are invariant under antipodal reflection, it was shown in [5] that the only self-similar solutions are round spheres. Very recently, the case /n α<+/n was solved in [8] as part of K. Choi s Ph. D. thesis. In particular, this includes the case α= conjectured by Firey [4]. Finally, we note that there is a substantial literature on other fully non-linear parabolic flows for hypersurfaces (see e.g. [3], [6], [20]) and for Riemannian metrics (cf. [2]). In particular, Gerhardt [5] studied convex hypersurfaces moving outward with speed K α ν, where α<0. Note that, for α<0, one can show using the method of moving planes that any convex hypersurface satisfying K α = x, ν, where α<0, is a round sphere. Using an a-priori estimate in [], Gerhardt [5] proved that the flow converges to a round sphere after rescaling. We now give an outline of the proof of Theorem. In view of the discussion above, it suffices to classify all closed self-similar solutions to the flow. The self-similar solutions Σ=F (M n ) satisfy the equation K α = F, ν. ( α ) To classify the solutions of ( α ), we distinguish two cases.

3 flows by powers of the gaussian curvature 3 First, suppose that α [ /(n+2), 2]. In this case, we consider the quantity Z = K α tr(b) nα 2α F 2, where b denotes the inverse of the second fundamental form. The motivation for the quantity Z is that Z is constant when α=/(n+2) and Σ is an ellipsoid. Indeed, if α=/(n+2) and Σ={x R n+ : Sx, x =} for some positive definite matrix S with determinant, then K /(n+2) = F, ν and Z = K /(n+2) tr(b)+ F 2 = tr(s ). Hence, in this case Z is constant, and equals the sum of the squares of the semi-axes of the ellipsoid. Suppose now that Σ=F (M n ) is a solution of ( α ) for some α [ /(n+2), 2]. We show that Z satisfies an inequality of the form αk α b ij i j Z +(2α )b ij i K α j Z 0. The strong maximum principle then implies that Z is constant. By examining the case of equality, we are able to show that either h=0, or α=/(n+2) and the cubic form vanishes. This shows that either Σ is a round sphere, or α=/(n+2) and Σ is an ellipsoid. Finally, we consider the case α ( 2, ). As in [8], we consider the quantity W = K α nα 2nα F 2. By applying the maximum principle, we can show that any point where W attains its maximum is umbilical. From this, we deduce that any maximum point of W is also a maximum point of Z. Applying the strong maximum principle to Z, we are able to show that Z and W are both constant. This implies that Σ is a round sphere. We first recall the notation: 2. Preliminaries The metric is given by g ij = F i, F j, where F i := i F. Moreover, g ij denotes the inverse of g ij, so that g ij g j =δ i. Also, we use the notation F i =g ij F j. We denote by H and h ij the mean curvature and second fundamental form, respectively. For a strictly convex hypersurface, we denote by b ij the inverse of the second fundamental form h ij, so that b ij h j =δ i. Moreover, tr(b) will denote the trace of b, i.e. the reciprocal of the harmonic mean curvature.

4 4 s. brendle,. choi and p. dasalopoulos We denote by L the operator L=αK α b ij i j. We denote by C ij the cubic form C ij = 2 K (n+2) h ij + 2 h j i K /(n+2) + 2 h i j K /(n+2) + 2 h ij K /(n+2). We next derive some basic equations. Proposition 2. Given a strictly convex smooth solution F : M n!r n+ of ( α ), the following equations hold: i b j = b jl b m i h lm, () L F 2 = 2αK α b ij (g ij h ij K α ) = 2αK α tr(b) 2nαK 2α, (2) i K α = h ij F, F j, (3) LK α = F, F i i K α +nαk α αk 2α H, (4) Lh ij = K α i K α j K α +αk α b pr b qs i h rs j h pq (5) + F, F h ij +h ij +(nα )h i h j K α αk α Hh ij, Lb pq = K α b pr b qs r K α s K α +αk α b pr b qs b ij b m r h i s h jm (6) + F, F i i b pq b pq (nα )g pq K α +αk α Hb pq. Proof. The relation i (b j h l )= i δ j l =0 gives h l i b j = b j i h l. This directly implies (). We next compute i j F 2 = 2 i F, j F +2 F, i j F = 2g ij 2h ij F, ν = 2g ij 2K α h ij, and hence L F 2 = 2αK α tr(b) 2nαK 2α. This proves (2). To derive equation (3), we differentiate ( α ): i K α = h i F, F. If we differentiate this equation again, we obtain i j K α = F, F i h j +h ij h i h j F, ν = F, F h ij +h ij K α h i h j, and hence LK α = F, F K α +nαk α αk 2α H.

5 flows by powers of the gaussian curvature 5 On the other hand, using () we compute i j K α = i (αk α b pq j h pq ) = αk α b pq i j h pq +α 2 K α b rs b pq i h rs j h pq αk α b pr b qs i h rs j h pq. Using the commutator identity i j h pq = i p h jq = p i h jq +R ipjm h m q +R ipqm h m j = p q h ij +(h ij h pm h im h jp )h m q +(h iq h pm h im h pq )h m j, we deduce that αk α b pq i j h pq = αk α b pq p q h ij +αk α Hh ij nαk α h im h m j = Lh ij +αk α Hh ij nαk α h im h m j. Combining the equations above yields Lh ij = α 2 K α b rs b pq i h rs j h pq +αk α b pr b qs i h rs j h pq + F, F h ij +h ij +(nα )h i h j K α αk α Hh ij. This completes the proof of (5). Finally, using (), we obtain Lb pq = αk α b ij i ( b pr b qs j h rs ) = 2αK α b ij b p b rm b qs i h m j h rs b pr b qs Lh rs. Applying (5), we conclude that Lb pq = α 2 K α b pr b qs b ij b m r h ij s h m +αk α b pr b qs b ij b m r h i s h jm + F, F b pq b pq (nα )g pq K α +αk α Hb pq. Since K α =αk α b ij h ij, the identity (6) follows. 3. Classification of self-similar solutions: the case α [ /(n+2), 2 In this section, we consider the case α [ /(n+2), 2]. We begin with an algebraic lemma. ]

6 6 s. brendle,. choi and p. dasalopoulos Lemma 3. Assume α [ /(n+2), 2], λ,..., λ n are positive real numbers (not necessarily arranged in increasing order), and σ,..., σ n are arbitrary real numbers. Then n Q := σi 2 +2 λ n i σi 2 ( n ( n ) n )( n ) 4αλ n i σ i + (nα ) n α i σ i σ i ( n )( n ) 2 ( n ) 2 2α 2 λ n i σ i +(2nα 2 +()α ) σ i 0. Moreover, if equality holds, then we either have σ =...=σ n =0, or we have α=/(n+2) and σ =...=σ = 3 σ n. Proof. If n σ i=0, the assertion is trivial. Hence, it suffices to consider the case n σ i 0. By scaling, we may assume n σ. Let us define { σi α, for i =,...,, τ i = σ n +()α, for i = n. Then and n i τ i = n n τ i = σ i = 0 n Therefore, the quantity Q satisfies i Q = (τ i +α) 2 +(τ n + ()α) 2 +2 = = 4α n n λ n i τ i 2α 2 n n σ i +(nα ) α n λ n i (τ i +α) 2 λ n i +2nα 2 +()α i. τi 2 +2α τ i +2( ()α)τ n +()α 2 +( ()α) α 2 n λ n i τi 2 +2α (τi 2 +2ατ i +α 2 ) 4α λ n i 2α 2 +2nα 2 +()α τ i +2( (n+)α)τ n +2 λ n i τ i 4ατ n λ n i τi 2 +()α((n+2)α ).

7 flows by powers of the gaussian curvature 7 Using the identity n τ i=0, we obtain Q = n τi 2 +2( (n+2)α)τ n +2 λ n i τi 2 +()α((n+2)α ). Moreover, the identity n τ i=0 gives n τi 2 = Thus, ( τ i + ) 2 τ n 2 τ n Q = τ i + n 2 τ 2 n = n ( τ n + n ( (n+2)α) +2 λ n i ) 2 + ( τ i + τ n ( τ i + τ n τi 2 + n ( 2α)((n+2)α ). ) 2 ) 2 + n τ n. 2 The right-hand side is clearly non-negative. Also, if equality holds, then τ =...=τ n =0 and α=/(n+2). This proves the lemma. Theorem 4. Assume α [ /(n+2), 2] and Σ is a strictly convex closed smooth solution of ( α ). Then, either Σ is a round sphere, or α=/(n+2) and Σ is an ellipsoid. Proof. Taing the trace in equation (6) gives L tr(b) = K α b pr b s p r K α s K α +αk α b pr b s pb ij b m r h i s h jm + F, F i i tr(b) tr(b) n(nα )K α +αk α H tr(b). Using equation (4), we obtain L(K α tr(b)) = K α L tr(b)+lk α tr(b)+2αk α b ij i K α j tr(b) = b pr b s p r K α s K α +αk 2α b pr b s pb ij b m r h i s h jm + F, F i i (K α tr(b))+(nα )K α tr(b) n(nα )K 2α +2αK α b ij i K α j tr(b). Using (2), it follows that the function satisfies Z = K α tr(b) nα 2α F 2 (7) LZ = b pr b s p r K α s K α +αk 2α b pr b s pb ij b m r h i s h jm + F, F i i (K α tr(b))+2αk α b ij i K α j tr(b). (8)

8 8 s. brendle,. choi and p. dasalopoulos Using (3), we obtain and hence This gives and 2 i F 2 = F, F i = b ij j K α, i Z = K α i tr(b)+tr(b) i K α nα α bj i jk α. F, F i i (K α tr(b)) = b ij i K α j Z + nα α bi b j ik α j K α 2αK α b ij i K α j tr(b) = 2αb ij i K α j Z (2αb ij tr(b) 2(nα )b i b j ) ik α j K α. Substituting these identities into (8) gives LZ +(2α )b ij i K α j Z = 4αb ij i K α j Z +αk 2α b pr b s pb ij b m r h i s h jm +( 2αb ij tr(b)+(2nα+ α )b i b j ) ik α j K α. (9) Let us fix an arbitrary point p. We can choose an orthonormal frame so that h ij (p)=λ i δ ij. With this understood, we have i K α = αk α n j= j i h jj and i tr(b) = n j= j i h jj. (0) Let D denote the set of all triplets (i, j, ) such that i, j and are pairwise distinct. Then, by using (9) and (0), we have α K 2α (LZ +(2α )b ij i K α j Z) = D i + + j i ( ih j ) 2 +4α λ 2 i ( h ii ) 2 +2 ( log K)(K α Z) i λ 3 i ( h ii ) 2 ( 2α2 tr(b)+(2nα 2 +()α ) )( log K) 2. () We claim that, for each, the following holds: λ 2 i ( h ii ) 2 +2 λ 3 i ( h ii ) 2 +4α( log K)(K α Z) i i +( 2α 2 tr(b)+(2nα 2 +()α ) )( log K) 2 0. (2)

9 flows by powers of the gaussian curvature 9 Notice that K α Z = i i h ii +(α tr(b) (nα ) ) log K. After relabeling indices, we may assume =n. If we put σ i := i n h ii, then the assertion follows from Lemma 3. This proves (2). Combining () and (2), we conclude that LZ +(2α )b ij i K α j Z 0 at each point p. Therefore, by the strong maximum principle, Z is a constant. Hence, the left-hand side of () is zero. Therefore, i h j =0 if i, j and are all distinct. Moreover, since we have equality in the lemma, we either have i h ii =0 for all i and, or we have α=/(n+2) and i h ii = 3 λ h for i. In the first case, we conclude that i h j =0 for all i, j and, and thus Σ is a round sphere. In the second case, we obtain This gives i h ii = n+2 log K for i and h = 3 n+2 log K. C ij = 2 K /(n+2) h ij + 2 h j i K /(n+2) + 2 h i j K /(n+2) + 2 h ij K /(n+2) = 0 for all i, j and. Since the cubic form C ij vanishes everywhere, the surface is an ellipsoid by the Berwald Pic theorem (see e.g. [9, Theorem 4.5]). This proves the theorem. 4. Classification of self-similar solutions: the case α ( 2, ) We now turn to the case α ( 2, ). In the following, we denote by λ... λ n the eigenvalues of the second fundamental form, arranged in increasing order. Each eigenvalue defines a Lipschitz continuous function on M. Lemma 5. Suppose that ϕ is a smooth function such that λ ϕ everywhere and λ =ϕ at p. Let µ denote the multiplicity of the smallest curvature eigenvalue at p, so that λ =...=λ µ <λ µ+... λ n. Then, at p, i h l = i ϕδ l for, l µ. Moreover, i i ϕ i i h 2 l>µ(λ l λ ) ( i h l ) 2. at p.

10 0 s. brendle,. choi and p. dasalopoulos Proof. Fix an index i, and let γ(s) be the geodesic satisfying γ(0)= p and γ (0)=e i. Moreover, let v(s) be a vector field along γ such that v(0) span{e,..., e µ } and v (0) span{e µ+,..., e n }. Then, the function s!h(v(s), v(s)) ϕ(γ(s)) v(s) 2 has a local minimum at s=0. This gives 0 = d ds (h(v(s), v(s)) ϕ(γ(s)) v(s) 2 ) s=0 = i h(v(0), v(0))+2h(v(0), v (0)) i ϕ v(0) 2 2 v(0), v (0) = i h(v(0), v(0)) i ϕ v(0) 2. Since v(0) span{e,..., e µ } is arbitrary, it follows that i h l = i ϕδ l for, l µ at the point p. We next consider the second derivative. To that end, we choose v(0)=e, v (0) = l λ ) l>µ( i h l e l, and v (0)=0. Since the function s!h(v(s), v(s)) ϕ(γ(s)) v(s) 2 has a local minimum at s=0, we obtain 0 d2 ds 2 (h(v(s), v(s)) ϕ(γ(s)) v(s) 2 ) s=0 = i i h(v(0), v(0))+4 i h(v(0), v (0))+2h(v (0), v (0)) i i ϕ v(0) 2 4 i ϕ v(0), v (0) 2ϕ v (0) 2 = i i h 4 (λ l λ ) ( i h l ) 2 +2 λ l (λ l λ ) 2 ( i h l ) 2 l>µ l>µ i i ϕ 2 λ (λ l λ ) 2 ( i h l ) 2 l>µ = i i h 2 l>µ(λ l λ ) 2 ( i h l ) 2 i i ϕ. This proves the assertion. Theorem 6. Assume α ( 2, ) and Σ is a strictly convex closed smooth solution of ( α ). Then Σ is a round sphere. Proof. Let us consider the function W = K α nα 2nα F 2. (3)

11 flows by powers of the gaussian curvature Let us consider an arbitrary point p where W attains its maximum. As above, we denote by µ the multiplicity of the smallest eigenvalue of the second fundamental form. Let us define a smooth function ϕ such that W ( p) = K α ϕ nα 2nα F 2. Since W attains its maximum at p, we have λ ϕ everywhere and λ =ϕ at p. Therefore, we may apply the previous lemma. Hence, at the point p, we have i h l = i ϕδ l for, l µ. Moreover, ϕ h 2 l>µ (λ l λ ) ( h l ) 2 at p. We multiply both sides by αk α By (5), we have Lϕ Lh 2αK α and sum over. This gives l>µ (λ l λ ) ( h l ) 2. Lh ij = K α i K α j K α +αk α b pr b qs i h rs j h pq + F, F h ij +h ij +(nα )h i h j K α αk α Hh ij. Thus, Lϕ 2αK α + l>µ (λ l λ ) ( h l ) 2 K α ( K α ) 2 +αk α,l F, F h +λ +(nα )λ 2 K α αk α Hλ. λ l ( h l ) 2 Using the estimate 2αK α l>µ = αk α αk α l>µ = 2αK α >µ (λ l λ ) ( h l ) 2 +αk α,l l>µ λ l ( h l ) 2 (2(λ l λ ) l )( h l ) 2 +αk α (2(λ l λ ) l )( h l ) 2 +αk α λ ( h ) 2 λ ( h ) 2 ((λ λ ) )( h ) 2 +αk α ( h ) 2,

12 2 s. brendle,. choi and p. dasalopoulos we obtain Lϕ 2αK α + ((λ λ ) )( h ) 2 +αk α ( h ) 2 K α ( K α ) 2 >µ F, F h +λ +(nα )λ 2 K α αk α Hλ. Since ϕ= h, it follows that L(ϕ ) 2αK α λ 3 +2αK α λ 3 ( h ) 2 ((λ λ ) )( h ) 2 >µ αk α λ 4 ( h ) 2 +K α ( K α ) 2 + = 2αK α λ 3 F, F ( ) λ (nα )Kα +αk α H (λ λ ) ( h ) 2 >µ +αk α λ 4 ( h ) 2 +K α ( K α ) 2 + F, F ( ) λ (nα )Kα +αk α H. This gives L(K α ϕ ) = K α L(ϕ )+ϕ L(K α )+2αK α 2α K α (K α ϕ ) 2α +2αK 2α λ 3 (λ λ ) ( h ) 2 >µ +αk 2α λ 4 ( h ) 2 + ( K α ) 2 + K α (ϕ ) ( K α ) 2 F, F (K α ϕ )+(nα )K α (nα )K2α. By assumption, the function is constant. Consequently, K α ϕ nα F 2 2nα (K α ϕ ) = nα 2nα F 2,

13 flows by powers of the gaussian curvature 3 and ( 0 = L K α ϕ nα 2nα nα n +2αK 2α λ 3 F 2 ) K α F 2 2α (λ λ ) ( h ) 2 >µ +αk 2α λ 4 ( h ) 2 + ( K α ) 2 + nα ( F, F F 2 +(nα )K α 2nα ( K α ) 2 n tr(b) ). Recall that 2 F 2 = F, F = K α by (3). Moreover, using the identity ϕ= h, we obtain 0 = (K α ϕ ) nα ( 2nα F 2 = nα nα λ ) K α K α h at p. Note that, if 2 l µ, then h l =0 for all. Putting = gives l h =0 and l K =0 for 2 l µ. Putting these facts together, we obtain Using the identities 2(nα ) n 0 2(nα ) n +2αλ >µ ( K α ) 2 2α (λ λ ) ( nα nα + n 2 α λ 2 ( K α ) 2 + ( K α ) 2 + nα nα 2αλ λ ( K α ) 2 λ ) 2 ( K α ) 2 ( K α ) 2 +(nα )K α ( n tr(b) ). +2αλ (λ λ ) ( = nα nα λ ( nα nα + 2 n + 2 n 2 α λ (λ λ ) ) 2 + nα ) nα λ 2 and 2(nα ) n 2αλ 2 + n 2 α λ 2 +λ 2 + nα nα λ 2 = n 2 α (2nα )λ 2,

14 4 s. brendle,. choi and p. dasalopoulos the previous inequality can be rewritten as follows: 0 ( nα nα + 2 n + 2 ) n 2 α λ (λ λ ) ( K α ) 2 >µ + ( n 2 α (2nα )λ 2 ( K α ) 2 +(nα )K α ). n tr(b) Since α>/n, it follows that p is an umbilical point. As p is an umbilical point and α> 2, there exists a neighborhood U of p with the property that α K 2α (LZ (2α+)b ij i K α j Z) = D i + + j i ( ih j ) 2 λ 2 i ( h ii ) 2 +2 i λ 3 i ( h ii ) 2 ( 2α2 tr(b)+(2nα 2 +()α ) )( log K) 2 0 at each point in U. (Indeed, if n 3, the last inequality follows immediately from the fact that ()α 0. For n=2 the last inequality follows from a straightforward calculation.) Now, since p is an umbilical point, we have Z(p) nw (p) nw ( p)=z( p) for each point p U. Thus, Z attains a local maximum at p. By the strong maximum principle, Z(p)=Z( p) for all points p U. This implies that W (p)=w ( p) for all points p U. Thus, the set of all points where W attains its maximum is open. Consequently, W is constant. This implies that Σ is umbilical, and hence a round sphere. 5. Proof of Theorem Suppose that we have any strictly convex solution to the flow with speed K α ν, where α [/(n+2), ). It is nown that the flow converges to a soliton after rescaling; for α>/(n+2), this follows from [5, Theorem 6.2], while for α=/(n+2) this follows from results in [3, 9]. By Theorems 4 and 6, either the limit is a round sphere, or α=/(n+2) and the limit is an ellipsoid.

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16 6 s. brendle,. choi and p. dasalopoulos Simon Brendle Department of Mathematics Columbia University 2990 Broadway New Yor, NY 0027 U.S.A. Panagiota Dasalopoulos Department of Mathematics Columbia University 2990 Broadway New Yor, NY 0027 U.S.A. Kyeongsu Choi Department of Mathematics Columbia University 2990 Broadway New Yor, NY 0027 U.S.A. Received December 6, 206