TOPICS IN DIFFERENTIAL GEOMETRY MINIMAL SUBMANIFOLDS MATH 286, SPRING RICHARD SCHOEN

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1 TOPICS IN DIFFERENTIAL GEOETRY INIAL SUBANIFOLDS ATH 286, SPRING RICHARD SCHOEN NOTES BY DAREN CHENG, CHAO LI, CHRISTOS ANTOULIDIS Contents 1. Background on the 2D mapping problem Hopf Differential General existence theorem 4 2. inimal submanifolds and Bernstein theorem First variation of area functional Second variation of area functional, Bernstein theorem 5 3. Bernstein s theorem in higher codimensions Complexifying the stability operator Stable minimal surfaces in R 4 and T Stable minimal genus-0 surfaces in R n, n Positive isotropic curvature High homotopy groups of PIC manifolds Fundamental group of PIC manifolds Positive scalar curvature Positive curvature obstructing stability Bonnet-type theorem Some obstructions to positive scalar curvature Asymptotically flat manifolds and AD mass Positive mass theorem Rigidity case Calibrated geometry Definitions and examples The Special Lagrangian Calibration Varitational Problems for (special) Lagrangian Submanifolds inimizing volume among Lagrangians Lagrangian 2D mapping problem Lagrangian cones onotonicity and regularity of minimizers in 2D 52 References 56 Date: January 27,

2 2 NOTES BY DAREN CHENG, CHAO LI, CHRISTOS ANTOULIDIS These are notes from Rick Schoen s topics in differential geometry course taught at Stanford University in the Spring of We would like to thank Rick Schoen for an excellent class. Please be aware that it is likely that we have introduced numerous typos and mistakes in our compilation process, and would appreciate it if these are brought to our attention. This course will focus on applications of the theory of minimal submanifolds. Topics covered include the two dimensional mapping problem and its relevance to the study of positive isotropic curvature, minimal hypersurfaces and scalar curvature as well as the more general theory of marginally outer trapped surfaces (OTS), and calibrated submanifolds and associated problems. 1. Background on the 2D mapping problem The basic setup in the 2D mapping problem is: Question 1.1. Given a map u 0 : 2 ( n, h) from a closed surface to a compact Riemannian manifold, can we homotope u 0 to a map of least area? That is, does there exist u : such that Area(u) = inf{area(v) : v u 0 }? Recall that if u is sufficiently differentiable then by the area formula we have Area(u) = u x 1 u x 2 dx 1 dx 2 where u x 1 u x 2 = u x 1 2 u x 2 2 u x 1, u x 2 2. One drawback of working with the area functional is its diffeomorphism invariance, i.e. Area(u) = Area(u F ) for all F Diff( 2 ), which makes it behave poorly from an analytic point of view. For example even if we re minimizing area, we cannot expect to get good regularity in the limit unless we take care to choose good parametrizations. In two dimensions one way to overcome this is to introduce the energy functional. Definition 1.2. The energy function of a C 1 map u : (, g) (, h) is defined to be (1.1) E(u) = du 2 dv g. From Cauchy-Schwarz we have u x 1 u x 2 1 ( ux u 2 x 2 2) = 1 2 du 2, (assuming we re working at the center point of an exponential chart) with equality happen if and only if u x 1 u x 2, u x 1 = u x 2. In other words, for every C 1 map u : (, g) (, h) we always have the area bounded by half of the energy, with equality only if u is wealky conformal. Definition 1.3. We call a map u : (, g) (, h) harmonic if u is a critical point of the energy functional. When u is simultaneously harmonic and conformal, then any variation {u t } produces two curves depending on the variation: one is the half of its energy, the other is its area. We know u 0 is critical point for energy, then the first curve has vanishing slope at t = 0, which forces the second curve, always lying below the first curve, to have vanishing slope at t = 0. That means u 0 is also a critical point for area functional. In conclusion, we observed the following Fact 1.4. If u 0 is harmonic and conformal, it s also a critical point for the area functional. This observation allows us to study conformal harmonic maps instead of minimizers for area functional. Now we regard energy E as a functional on both the map u and the metric g.

3 286 - TOPICS IN DIFFERENTIAL GEOETRY - LECTURE NOTES 3 Proposition 1.5. The energy functional E(u, g) = du 2 gdv g has the following properties: (1) Conformal invariance: E(u, e 2 λg) = E(u, g). This is so because E(u, g) = g ij u x i, u x j h det gdx 1 dx 2 and a conformal change of the metric g transforms g ij and det g inversely. (2) Diffeomophism invariance: For any diffeomorphism F :, E(u F, F g) = E(u, g) Hopf Differential. Assume u : (, g) (, h) is harmonic, X is a vector field on and F t is the flow generated by X. By diffeomorphism invariance, we have, for small t, E(u F t, F t g) = E(u, g). Take the differential both sides at t = 0. Since u is critical point of energy functional, the differential in the u component is 0. Therefore 0 = d dt E(u F t, Ft g) = 0 + d t=0 dt E(u, Ft g). t=0 In local coordinates this is 0 = d dt t=0 g ij t u x i, u x j det g t dx. Now ġ = L X g = i X j + j X i, so above gives 0 = { (X i,j + X j,i ) u x i, u x j det g + g ij u x i, u x j (div X) det g}dx. Definition 1.6. We define the (stress-energy) tensor to be Then the computation above implies 0 = 2 Therefore we conclude T ij = u x i, u x j 1 2 du 2 g ij. X i,j, T ij dx, X. j T ij = 0, i = 1, 2. And by definition tr g (T ) = 0. So the (stress-energy) tensor T is a transverse traceless tensor. In local coordinates normal at one point, we can write T as a two-by-two matrix: ( 1 (T ij ) = 2 ( u x 1 2 u x 2 2 ) ) u x 1, u x 2 u x 1, u x ( u x 1 2 u x 2 2. ) This reveals the interplay between T and the so called Hopf differential on a surface. Write g = λ 2 dz 2, where z = x 1 + 1x 2 is a local holomorphic coordinate. We define Hopf differential to be φ = u z, u z h dz 2. It s straightforward to check and T is transverse traceless φ is holomorphic T = 0 φ = 0 Note also that T = 0 means u is weakly conformal. So from above we conclude the following Theorem 1.7. The map u is minimal for the area functional if and only if E(u, g) is a critical point jointly in (u, g), where (u, g) takes values in W 1,2 (, ) T r, where T r is the Teichmüller space of genus r.

4 4 NOTES BY DAREN CHENG, CHAO LI, CHRISTOS ANTOULIDIS 1.2. General existence theorem. Now we state the general existence theorem of a minimal map. This is a major analytic tool we use in this course. We ll omit most of the proof due to analytic complexity. Instead, we ll focus on geometric applications. Theorem 1.8. Given u 0 : (, h), denote A 0 = inf{a(v) : v homotopic to u 0 }. There exist 1,..., k and least area maps u i : i (, h) such that: (1) k i=1 genus( i) genus(). (2) A 0 = k i=1 A(u i). In general, when trying to take a converging sequence of maps whose area tends to A 0, two types of singularities may occur: neck-pinches and bubbles. Each neck-pinch degenerates to a 1- dimensional segment between two parts of surfaces, and bubbles happen when area accumulates at one point. The following picture is an illustration of this phenomenon. Bubble Neck-pinch Figure 1. Illustration of limit (, u) attaining minimal area. Each bubble is blown up into a sphere, each neck-pinch degenerates to a segment. However, under some conditions these two types of singularities will not happen. In fact, we have Corollary 1.9. If u 0 is incompressible on simple closed curves then one of i is and all others are genus 0. If further π 2 () = {1} then there exists u homotopic to u 0 attaining the least area A 0 (i.e., there is no bubbling). Here incompressibility on simple closed curves means: for any nontrivial simple closed curve α in π 1 (), u 0 (α) is nontrivial in π 1 (). In most cases harmonic maps are not necessarily conformal, hence not necessarily critical for area functional. However in the case that is the 2-sphere: Theorem If u : (S 2, g S ) (, h) is harmonic, where g S is the standard metric on S 2, then u is also conformal, hence minimal. Proof. By previous section the Hopf differential φ(z)dz 2 is holomorphic on S 2. That is, φ(z)dz 2 is an entire differential on C and extends to. Take ζ = 1/z, then near the Hopf differential is φ(1/ζ)/ζ 4 dζ 2. Near ζ = 0, φ(1/ζ)/ζ 4 is holomorphic. So φ(z)z 4 is an entire function near. Hence φ is bounded by C/ z 4 for every z. By maximum principle we conclude φ 0. Next theorem will be our primary tool for our use. Theorem 1.11 (Sacks-Uhlenbeck [SU81], icallef-oore [88]). If π k () {1} then there exists nonconstant harmonic map u : S 2 and the orse index of u k 2. Remark The orse index is taken with respect to the second variation of the energy functional. In this specific case, the Jacobi operator is: for V Γ(u T ), 2 LV = V + R (u (e i ), V )u (e i ), e 1, e 2 form an orthonormal basis on. i=1

5 286 - TOPICS IN DIFFERENTIAL GEOETRY - LECTURE NOTES 5 Remark Sacks-Uhlenbeck s approach can be (very briefly) sketched as following. For α 1, define ( E α (u) = 1 + du 2 ) α da. S 2 For α > 1, this is a good variational problem and they are able to extract converging subsequences of critical points of E α. icallef-oore further modify E α to make its critical points non-degenerate, and they proved the modified critical points also converge after passing to a subsequence. Remark We here point out that Colding and inicozzi have a different approach for minimizers on S 2, and X. Zhou generalized the result to higher genus surfaces. 2. inimal submanifolds and Bernstein theorem 2.1. First variation of area functional. Let k n be a submanifold. Denote by D the Levi-Civita connection on and by h the vector valued second fundamental form The vector h(x, Y ) = (D X Y ), H = k h(e i, e i ) i=1 X, Y Γ(T ). is the mean curvature, where e 1,..., e k is an orthonormal basis of tangent vector fields. Now if X is a vector field on compactly supported on and F t is a flow with initial velocity X, consider t = F t (). The variation of area functional can be calculated as following δ(x) = d dt t = div Xdµ. t=0 where div (X) = k i=1 D e i X, e i and dµ is the volume measure on. Decompose X into its tangent and normal components X = X T +X, we may write D ei X, e i = D ei X T, e i + D ei X, e i. And the normal component can be further calculated as D ei X, e i = X, (D ei, e i ). Therefore div (X) = div (X T ) X, H. And the first variation of area functional is δ(x) = X, H dµ. Definition 2.1. Call k n minimal if H Second variation of area functional, Bernstein theorem. In many cases it s necessary to consider the second variation of area functional. We have Proposition 2.2. Assume H 0 and X p T p for every p on, and X is compactly supported on. Then the second variation of area functional is given by δ 2 (X, X) = D X 2 h, X 2 with e 1,..., e k being an orthonormal basis on. k R (e i, X, e i, X), Remark 2.3. We split T = T N. Then the ambient connection D gives rise to connections on T and N. If Y Γ(T ) and X Γ(N) then we have D Y X = (D Y X). Then we i=1

6 6 NOTES BY DAREN CHENG, CHAO LI, CHRISTOS ANTOULIDIS may rewrite D X 2 = k i=1 D e i X 2 and h, X 2 = i,j h i,j, X 2 = D T X 2, and the second variation is given as k δ 2 (X, X) = D X 2 R (e i, X, e i, X) D T X 2. Definition 2.4. Define the Jacobi operator L on Γ(N) by i=1 LX = X + R (e i, X)e i + i,j h ij, X h ij. Then L is a second order self-adjoint operator on Γ(N), and δ 2 (X, X) = X, LX dµ. We call the number of negative eigenvalues of L the orse index of. is called stable if the orse index is 0, strictly stable if there are also no Jacobi fields. A famous and important question is to understand the structure of stable minimal surfaces. The first important theorem is given by S. Berstein. Theorem 2.5 (S. Berstein [Ber27]). Let 2 R 3 be a minimal surface and given by a graph x 3 = u(x 1, x 2 ) defined for all (x 1, x 2 ). Then is a plane; i.e., u must be a linear function. Before proving Bernstein s theorem, we first state some important properties of minimal graphs = graph(u) in R n+1, where u : Ω R is a C 2 function. Fact 2.6. is 2-sided. That is, has a unit normal vector field ν. Fact 2.7. is area minimizing in Ω R. The second fact is an easy consequence of calibration theory, which will reappear in later part of the course. We prove this special case here. Extend ν to a unit vector field in Ω R by setting ν(x, y) = ν(x, u(x)). Since ν is parallel in the x n+1 direction, we conclude from the minimal surface equation that div R n+1 ν = 0. In fact, suppose e 1,..., e n is an orthonormal basis tangent to, e n+1 = ν. Then we have n+1 div R n+1 ν = D ei ν, e i. Now ν is of unit length, so D en+1 ν, e n+1 = 0. Therefore n div R n+1 ν = D ei ν, e i = H, ν = 0. The vector field ν gives a calibration in the region Ω R. i=1 i=1 1 ν R Ω Figure 2. Calibration

7 286 - TOPICS IN DIFFERENTIAL GEOETRY - LECTURE NOTES 7 Suppose 1 Ω R and Ω 1 = Ω. Denote ν 1 the outer unit normal vector field on Ω 1. Let Ω be the signed region in R n+1 with 1 = Ω. Then by the divergence theorem, 0 = div R n+1 ν = ν ν ν ν 1. Ω 1 So we conclude = ν ν = ν ν 1 1, 1 Fact 2.8. If is an entire minimal graph, then B R (0) CR n, R 1 by Cauchy-Schwarz. This is an easy consequence of the fact that minimal graphs are area minimizing. Take any R > 0. Then divides B R (0) = S R (0) into two parts 1, 2. Since is a minimal graph over the domain SR n(0) Rn, we have B R (0) min{ 1, 2 } CR n. Now we prove Bernstein s theorem through the following Theorem 2.9. Assume R 3 is stable, proper, orientable minimal surface with Euclidean area growth. That is, B R (0) CR 2 for all R 1. Then is a plane. Proof. Take a normal vector field ν and let X = ϕν, ϕ Cc (). The stability condition gives 0 δ 2 (X, X) = ϕ 2 h 2 ϕ 2, where h is the scalar second fundamental form. So we know h 2 ϕ 2 dµ ϕ 2 dµ, ϕ Lip c (). We use the logarithmic cut-off trick. Denote ρ(x) = x, then ρ is a proper function on and ρ 2 Dρ 2 = 1. Define 1 for ρ R log R ϕ R (ρ) = 2 /ρ log R for R ρ R 2 0 for ρ R 2. Claim: ϕ R 2 C(log R) 1. In fact, we have ϕ R 2 (B R 2 B R ) ρ 2 R2 ( ) = (log R) 2 r 2 dσ dr. (log R) 2 R ρ=r ρ The last equality is got by coarea formula. Here again we use coarea formula just for the constant function 1 on B R (0) to get dσ ρ = d dr B r(0). So (B R 2 B R ) Here we used the area growth of. ρ=r ϕ R 2 (log R) (r 2 2 B r r=r 2 r=r C 1 (log R) 2 + C 2 (log R) 1. ) R3 + 2 r 3 B r (0) dr R

8 8 NOTES BY DAREN CHENG, CHAO LI, CHRISTOS ANTOULIDIS Now take R to we get h 2 dµ B R (0) So h 0 and is a plane. B R 2 h 2 ϕ 2 Rdµ (B R 2 B R ) ϕ R 2 dµ 0. Question We are curious about possible generalization of Berstein s theorem. The following cases have been of great interest for researchers. (1) For higher dimensional n R n+1 entire minimal graphs, can we conclude that is affine space? This question has been answered by many authors over many years. The conclusion is true for n 7 and false for n 8. (2) Can we get a Berstein type theorem when n n+1 where is a curved manifold? In some special cases this question can be answered. We ll get back to this question later. (3) For 2 R n where n 4, can we get a Bernstein type theorem? We ll focus on this direction. The third question is more complicated than it first appears. The fact is, we can construct a family of area minimizing surfaces in higher dimensional Euclidean spaces. Let n = 2m and J : R n R n being a complex structure, meaning J is orthogonal and J 2 = I. For each fixed J take 2 to be a J-holomorphic curve. Then is area minimizing by a similar calibration argument. In particular, consider = {(z, w) : w = f(z)}, Here f is a J-holomorphic function. Then is an area-minimizing surface in R Bernstein s theorem in higher codimensions As mentioned in the previous section, Bernstein s theorem in its full generality fails in higher codimensions due to the presence of J-holomorphic curves, defined as follows. Definition 3.1. Let n = 2m and let J be an orthogonal complex structure on R n, i.e. an orthogonal matrix J with J 2 = I. A J-holomorphic curve is a 2-dimensional surface 2 R n such that J(T x ) = T x, for all x. Proposition 3.2. J-holomorphic curves are area-minimizing among orientable competitors. Proof. Consider the Kähler form ω, defined by ω(x, Y ) = JX Y. Since J is a constant matrix, we observe that ω is closed. Next we show that ω is a calibrating form that restricts to the area form precisely on J-invariant 2-planes. To see this, take any oriented 2-plane Π in R n and let {e 1, e 2 } be a positive orthonormal basis. By the Schwartz inequality, ω(e 1, e 2 ) = Je 1 e 2 1. oreover, ω(e 1, e 2 ) = 1 if and only if Je 1 = e 2, which is equivalent to the J-invariance of Π. To conclude the proof, let 0 be an oriented surface with 0 =, then we can find a region R with 0 = R. Then we have 0 = dω = ω = ω ω R R 0 = ω 0 0 and the proof is complete. Since the hypotheses of Bernstein s theorem certainly doesn t rule out J-holomorphic curves, Proposition 3.2 shows that Bernstein s theorem is generally false in higher codimensions. The best one could hope for is perhaps the following statement.

9 286 - TOPICS IN DIFFERENTIAL GEOETRY - LECTURE NOTES 9 Conjecture 3.3. Let 2 R n be a complete stable minimal surface, possibly with some controlled area growth, then there exists 2k n and a 2k-plane P R n such that is J-holomorphic in P for some complex structure J. It turns out that even this is false in general. Nonetheless, all hope is not lost as there are some interesting special cases in which Conjecture 3.3 is true. Below we list a few positive results. (1) When n = 4 and is oriented with area growth suitably bounded, the conjecture is true. (2) If genus() = 0 and ( K)da <, then the conjecture is true for all n. (3) If the ambient space is replaced by T n, then the conjecture is true for n = Complexifying the stability operator. We ll treat the case (1). A key ingredient in the proof is a complexified version of the second variation formula. We first set up some notations before writing down the formula. As before, let ( 2, g) be an oriented surface in n. Around each point of we can find local isothermal coordinates (x 1, x 2 ), i.e. g = λ 2 ( (dx 1 ) 2 + (dx 2 ) 2), where λ 2 = 2 x 1 = 2 x 2 Next we write (3.1) = 1 ( 2 x 1 i ) x 2 ; = 1 ( 2 x 1 + i ) x 2 Now recall that if is minimal and X Γ(N), then the second variation is given by (3.2) δ 2 (X, X) = D X 2 2 R (e j, X, e j, X) D T X 2 da, j=1 where {e j } is any orthonormal frame for T. Now we complexify T and N and extend the second variation formula to complex vector fields. For X Γ(N C ), we simply write (3.3) δ 2 (X, X) = D X 2 2 R (e j, X, e j, X) D T X 2 da j=1 Of course now D X 2 = D X, D X and likewise for D T X 2. Below we ll use the operators (3.1) to rewrite (3.3). ore precisely, we have the following formula. Proposition 3.4. Let 2 and n be as above and let X Γ(N C ), then (3.4) δ 2 (X, X) = 4 D X 2 R (, X,, X) DT X 2 dx 1 dx 2, Remark 3.5. Notice that the integrand [ D X 2 R (, X, ], X) DT X 2 dx 1 dx 2 is conformally invariant. Thus, even though it s written in terms of coordinates, it makes sense globally on. Proof.

10 10 NOTES BY DAREN CHENG, CHAO LI, CHRISTOS ANTOULIDIS 1. We start from (3.3). Introducing isothermal coordinates as above, the area element da becomes λ 2 dx 1 dx 2. Plugging this into (3.2) and using the orthonormal frame e j = λ 1, j = 1, 2, we find that x j 2 2 (3.5) δ 2 (X, X) = D X 2 R ( x j x j, X, 2 x j, X) D T X 2 dx 1 dx 2 x j (3.6) j=1 2. Next we notice that j=1 D X 2 + D X 2 = 1 4 D X id x 1 X, D x 2 X + id x 1 X x 2 Likewise, we also have and D T X 2 + D T X 2 = 1 2 j= D X + id x 1 X, D x 2 X id x 1 = 1 ( ) D X 2 + D 2 x 1 X 2 x 2 ( ) D T X 2 + D T x 1 X 2 x 2 R (, X,, X) + R (, X,, X) = 1 ( R ( 2 x 1, X, x 1, X) + R ( ) x 2, X, x 2, X) Plugging these into (3.5), we obtain δ 2 (X, X) = 2 D X 2 + D X 2 R (, X,, X) x 2 X + R (, X,, X) DT X 2 D T X 2 dx 1 dx 2 3. Take the term D X 2 dx 1 dx 2. We want to integrate by parts to write it in terms of D X 2 dx 1 dx 2, a curvature term and some other stuff. To do so, we observe D X 2 = D X 2 D T X 2 = D X, D X D T X 2 (3.7) = D X, X D D X, X D T X 2 = D X, X D D X, X R (,, X, X) DT X 2 = D X, X D X, X + D X, D X R (,, X, X) DT X 2 = D X, X D X, X + D X + D T X 2 R (,, X, X) DT X 2 Integrating over, using the fact that X has compact support and plugging into (3.6), we get δ 2 (X, X) = 2 2 D X 2 R (,, X, X) R (, X,, X)

11 286 - TOPICS IN DIFFERENTIAL GEOETRY - LECTURE NOTES 11 (3.8) R (, X,, X) 2 DT X 2 dx 1 dx 2 By the first Bianchi identity, R (,, X, X) + R (, X,, X) = R (X,,, X) = R (, X,, X). Therefore from (3.8) we get δ 2 (X, X) = 2 2 D X 2 2R (, X,, X) 2 DT X 2 dx 1 dx 2 = 4 D X 2 R (, X,, X) DT X 2 dx 1 dx 2 as stated. The proof is now complete. In the case where the ambient manifold is R n, (3.4) simplifies and we have the following beautiful stability criterion. Corollary 3.6. Suppose 2 R n is a stable oriented minimal surface, then (3.9) D T X 2 dx 1 dx 2 D X 2 dx 1 dx 2, for all X Γ(N C ) Proof. Each section X Γ(N C ) can be written as X = X 1 + ix 2, where X 1, X 2 are sections of the real normal bundle N. Then we have δ 2 (X, X) = δ 2 (X 1, X 1 ) + δ 2 (X 2, X 2 ) 0, where the last inequality is true by stability. The corollary now follows from Proposition Stable minimal surfaces in R 4 and T 4. Let s come back to complete oriented stable minimal surfaces in R 4. Recall that our goal is to construct an orthogonal complex structure J on R 4 with respect to which is holomorphic. We introduce some notations before describing the construction. We will roughly be following [ic84]. For clarity, below we suppose is the image of an isometric stable minimal immersion F : 2 R 4, where 2 is a complete oriented surface. Let E R 4 denote the pullback of T R 4 and its metric structure via F. Then we can view T as a sub-bundle of E and use the metric to define the orthogonal complement bundle, which we denote by N. Since is oriented, the pullback metric induces a complex structure J T on. Also, still by orientability, we can define a complex structure J on N by rotation by 90 in the clockwise or counterclockwise direction (notice that we have a choice here). We then define J : Hom(E) as follows: for each p, given a vector v E p, we define, J p (v) = J T p (v T ) + J p (v ), where v T and v denote the orthogonal projections of v onto T p and N p, respectively. The triviality of E allows us to view J as a map from to Hom(R 4 ). What we want to demonstrate now is that J is constant, so that J : Hom(R 4 ) extends as a complex structure on all of R 4. To see this, we first complexify E, T and N and extend J T and J to be complex linear maps. Then J T gives rise to a splitting T C = T 1,0 T 0,1. Likewise, N C splits as N 1,0 N 0,1. We denote N 1,0 by V ; then N 0,1 = V. With these notations, we form the following sub-bundle of E C C 4 : W = T 1,0 V.

12 12 NOTES BY DAREN CHENG, CHAO LI, CHRISTOS ANTOULIDIS For each p, the fiber W p is a subspace of C 4. The constancy of J is then translated into the constancy of W. Proposition 3.7. If Γ(W ) is closed under the usual directional derivatives in C 4, then W p is independent of p. Proof. Take w W p C 4 and extend it as a constant vector field on. Note that since C 4 = W p W p, for each q we can decompose w q = wq 1,0 + wq 0,1, with wq 1,0 W q and wq 0,1 W q. We will show that w 0,1 is constantly zero. To see this, observe that since w is a constant vector field, letting denote a directional derivative, we have (3.10) 0 = w = w 1,0 + w 0,1 = ( w) 1,0 + ( w) 0,1 where the last equality follows from the assumption that Γ(W ) is closed under differentiation. Since W p W p is a direct sum, (3.10) immediately implies that both w 1,0 and w 0,1 are constant. In particular, since w 0,1 p = 0, we see that w 0,1 is constantly zero. To check that Γ(W ) is closed under differentiation, we will use the following proposition. Proposition 3.8. Let Fzz denote the projection of F zz onto N C and let Fzz 1,0, Fzz 0,1 be the projection of Fzz onto V, V, respectively. If Fzz 0,1 = 0 then Γ(W ) is closed under differentiation. Proof. Recall that F is minimal. Introducing isothermal coordinates, F is also conformal. Thus F is harmonic and we have (3.11) F z F z = 0 (Conformality) (3.12) F zz = 0 (Harmonicity) Next take a local positive orthonormal frame {e 3, e 4 } of N such that J (e 3 ) = e 4 ; J (e 4 ) = e 3, and let ε = 1 2 (e 3 ie 4 ). Then V = span C (ε) and W = span C (ε, F z ). Now let s Γ(W ) and write s = a(z)f z + b(z)ε. To save notations, below we simply write X Y if X Y mod W. Now we compute (3.13) s af zz + b ε, and expand the two terms on the right using the basis {F z, F z, ε, ε}. The first term becomes where we used the fact that F 1,0 zz vanishes. Using the assumption F 0,1 zz F zz = F zz F z F z 2 F z + F zz F z F z 2 F z + (F zz ε)ε + (F zz ε)ε F zz F z F z 2 F z + (F zz ε)ε = F zz F z F z 2 F z + (Fzz 0,1 ε)ε, F zz 0; that is, F zz V. Next we look at the second term in (3.13). Then we have ε = 0 in the last equality. Now by (3.11), the first term above = 0, we see that the second term vanishes as well. Thus ε ε F z F z 2 F z + ( ε ε)ε

13 286 - TOPICS IN DIFFERENTIAL GEOETRY - LECTURE NOTES 13 = ε F z F z 2 F z (ε had unit length) = ε F zz F z 2 F z (integrate by parts in the first term) = ε F zz 0,1 F z 2 F z (Fzz 1,0 ε = 0) = 0 (by assumption). Thus for each section s of W, we ve shown that s 0. Similarly we can show that s 0. Thus Γ(W ) is preserved by differentiation. To verify the assumption of Proposition 3.8, we suppose in addition that is parabolic. Definition 3.9. Given a Riemannnian surface, we say that is parabolic if every positive superharmonic function on is constant. Below we give some examples of parabolic manifolds. Example (3.14) (1) The complex plane C is parabolic. On the other hand, the unit disk D C is not parabolic. (2) Any compact Riemann surface with finitely many punctures is parabolic. (3) If is a complete surface with B R CR 2 for R large, then is parabolic. Proof. Suppose u > 0 is a positive superharmonic function on. Letting w = log u, we have w = u u u 2 u 2 u u w 2 w 2 (since u 0). Next we test the inequality (3.14) against ϕ 2, where ϕ is any test function ϕ CC 1 (), getting ϕ 2 w 2 d vol ϕ 2 wd vol = 2 φ ϕ, w d vol 1 ϕ 2 w 2 d vol +2 ϕ 2 d vol 2 Hence we get ϕ 2 w 2 d vol 4 ϕ 2 d vol. Applying the logarithmic cut-off trick as in the proof of the Bernstein theorem in the last section, we conclude that w, and thus u, is constant. (4) If 2 R n is an entire minimal graph, then with the induced metric is parabolic. Proof. We will prove that is conformally equivalent to C. By the uniformization theorem, we know that is conformally equivalent either to C or to D. Assume by contradiction that the latter holds and let F : D be a biholomorphic map. Since is isometrically and minimally embedded, F is harmonic as a map of D into R n. odifying F by an automorphism of D is necessary, we may assume that F (0) = (0, 0, u(0, 0)). Next denote F (x 1, x 2 ) = (F 1 (x 1, x 2 ), F 2 (x 1, x 2 )). By the previous paragraph, F is a harmonic diffeomorphism from D to (R 2, h), where h is obtained by pulling back the induced metric on via (x 1, x 2 ) (x 1, x 2, u(x 1, x 2 )).

14 14 NOTES BY DAREN CHENG, CHAO LI, CHRISTOS ANTOULIDIS Recall that in complex coordinates, the Jacobian of F can be written as (3.15) J( F ) = F z 2 F z 2, which is everywhere strictly positive since F is a diffeomorphism. This implies that F z is everywhere non-zero, so we can define a metric g on D by g = F z 2 dz 2. Now since F is harmonic, we have F zz = 0 and hence F z 2 = 0, which means that (D, g) is flat (Gauss curvature zero). Using (3.15) again, we see that d F is dominated by F z and hence F (h) c g, where c is a dimensional constant. Now for an arbitrary R > 0, we can choose r such that dist F (h) (0, D r) = dist h (0, (F (D r ))) R. Combining this with the previous inequality, we get (3.16) c dist g (0, D r ) R Next we take the coordinate function x on D. Then { g x = 0 x 1 so by the harmonic function estimates in [?] and the definition of g, we have 1 F z 2 = gx 2 (0) c dist g (0, D r ) 2 This in turn gives us d F 2 (0) F z 2 (0) cr 2. Since R is arbitrary, we obtain a contradiction, so is conformally equivalent to C and hence parabolic. After this little digression into parabolic manifolds we return to our problem and give the precise statement of the main result of this section. Theorem Assume F : 2 R 4 is an oriented, stable, parabolic, complete minimal surface. Then F is J-holomorphic for some orthogonal complex structure J on R 4. Proof. By Proposition 3.8, the proof reduces to showing that Fzz 0,1 vanishes. We will demonstrate this by plugging special test functions into the stability inequality (3.4) and using parabolicity. To that end we consider a test function of the form fs, where f is a real-valued smooth function with compact support on, and s Γ(N C ). Then we have (3.17) δ 2 (fs, fs) = f z 2 s 2 f 2 (Re s, D zz s ) f 2 ( T s) 2 dx 1 dx 2, where we re using D to denote D. To derive this formula, we recall that by (3.4) we have (3.18) δ 2 (fs, fs) = D z (fs) 2 T (fs) 2 dx 1 dx 2 To handle the first term we compute D z (fs) 2 = D z (fs) D z (fs) = (f z s + fd z s) (f z s + fd z s) = f z 2 s 2 + f 2 D z s 2 + ff z D z s s + (complex conjugate of the previous term) c R 2.

15 (3.19) = TOPICS IN DIFFERENTIAL GEOETRY - LECTURE NOTES 15 f z 2 s 2 + f 2 D z s 2 + 2Re(ff z D z s s) Now notice that ff z D z s s = 1 (f 2 ) z D z s s 2 = 1 f 2 D z D z s s f 2 D z s 2 Plugging this back into (3.19), we get (3.20) D z (fs) 2 = f z 2 s 2 f 2 Re(D zz s s) For the second term in (3.18), we notice that z (fs) = f z s + f z s. Since s is a normal section, projection onto T kills the first term and we re left with (3.21) T z (fs) = f( T z s) (3.17) now follows by plugging (3.20) and (3.21) back into (3.18). To proceed, we take a vector a C 4 and denote by a 1,0 (p) its projection onto V p. Applying (3.17) with a 1,0 in place of s and using the stability of in R 4, the result we get is the following (3.22) f 2 q(a)da f 2 a 1,0 2 da a f 2 da, where q is the following expression: (3.23) q(a) = 2 F z 4 Re { (F 1,0 zz a)(f 1,0 zz a) }. Take an orthonormal basks {a 1,..., a 4 } of C 4, denote q(a j ) by q j and sum over j, we obtain 4 (3.24) q j = 2 { } F z 4 Re Fzz 1,0 F 1,0 zz = 0 j=1 Now by [FCS80], the inequality (3.22) with q j in place of q(a) implies the existence of a positive function u j on solving (3.25) u j + q j u j = 0 Letting w j = log u j, an easy calculation shows that Thus we get w j = q j w j 2. (q j + w j 2 )f 2 = ( w j )f 2 = 2 1 f 2 w j and therefore (q j w j 2 )f 2 2 f 2 Summing over j and using (3.24), we deduce that 1 2 j=1 4 w j 2 f 2 8 f 2. f f w j f 2,

16 16 NOTES BY DAREN CHENG, CHAO LI, CHRISTOS ANTOULIDIS Again by [FCS80], we get a positive function v such that 8 v ( w j 2 )v = 0 j=1 v 0; that is v is a positive superharmonic function. By the parabolicity of, v must be a (nonzero) 4 constant. Looking back at the PDE satisfied by v, we immediately deduce that w j 2 = 0, so each w j, and hence each u j, is constant. By (3.25), we see that each q j is zero. Since the a j s form a basis for C 4, we conclude that (3.26) q(a) = 2 { } F z 4 Re (Fzz 1,0 a)(f 1,0 zz a) = 0, for all a C 4. Now at a point p where F zz (p) 0, we can let a = F zz F zz and plug it into (3.26). Then we get (3.27) F 1,0 zz (p) F 0,1 zz (p) = 0. Thus at each p, one of Fzz 1,0 (p) and Fzz 0,1 (p) must vanish. The fact that F is conformal and harmonic implies that Fzz 1,0 dz 2 and Fzz 0,1 dz 2 are holomorphic quadratic differentials with values in V and V, respectively. Hence we conclude, by unique continuation, that either Fzz 1,0 or Fzz 0,1 vanishes identically. In the latter case, the proof is complete by invoking Proposition 3.8. In the former case we simply change the complex structure J on N. (Recall that we had a choice when constructing J. See the remarks before Proposition 3.7.) ore or less the same argument establishes the same theorem in the compact setting of ambient flat 4-tori instead of R 4. Theorem Assume F : 2 T 4 is an oriented, stable, compact minimal surface and that T 4 is a flat torus. Then F is J-holomorphic for some orthogonal complex structure J on T 4. Proof sketch. By arguing as in 3.11 we get λ 0 ( + q j ) 0 on for all j {1, 2, 3, 4}. u j = e w j > 0 be the lowest eigenfunction so that, as before, (q j w j 2 )f 2 2 f 2 for j {1, 2, 3, 4}. Summing over j and recalling the definition of the q j we conclude 1 w j 2 f 2 8 f 2. 2 j Picking f = 1 (since compact) we see that each w j is constant, so each u j is constant, so q j = λ 0 ( + q j ) is constant. Since the q j sum to zero they must then all be zero and the result follows like before. In the proof of Theorem 3.11 we made use of identity (3.23) in (3.22). Let s justify that now: Claim We can rewrite 2 F z 2 where a C 4, a = 1. [ a 1,0 F zz 2 F z 2 + Re j=1 ( a 1,0 D z D z a 1,0)] = 2 F z 4 Re ( (F 1,0 zz a)(f 1,0 zz a) ) Let

17 286 - TOPICS IN DIFFERENTIAL GEOETRY - LECTURE NOTES 17 Proof of claim. Recall that ε = 1 2 (e 3 ie 4 ) is such that {ε, ε} forms an orthonormal frame for N C = V V. Note that ε ε = ε ε = 0 and ε ε = 1. By differentiating a 1,0 = (a ε)ε once and using the product rule, (3.28) D z a 1,0 = z (a ε)ε + (a ε)d z ε = (a ( z ε) T )ε + (a D z ε)ε + (a ε) [ (D z ε ε)ε + (D z ε ε)ε ] = (a ( z ε) T )ε + [ a ((D z ε ε)ε + (D z ε ε)ε )] ε + (a ε) [ (D z ε ε)ε + (D z ε ε)ε ] = (a ( z ε) T )ε + (a ε)(d z ε ε)ε + (a ε)(d z ε ε)ε, = (a ( z ε) T )ε, because D z ε ε = D z ε ε = 0, as ε ε = ε ε = 0 because ε ε = 1 = [ a (( z ε (F z / F z 2 ))F z + ( z ε (F z / F z 2 )] ))F z ε ( ) F z = a F z 2 (ε F zz ) ε. where the last equality follows from the product rule and minimality, F zz = 0. We will differentiate again in z, but before doing so, first we observe that ( ) Fz z F z 2 F z = 0 by conformality, F z F z = 0, and ( ) Fz z F z 2 F z = 0 by minimality, F zz = 0. Consequently, z (F z / F z 2 ) is purely normal and thus ( ) Fz z F z 2 = F zz F z 2. Plugging this into (3.28) and exploiting similar cancelations among the derivatives of ε, ε, we get ( D z D z a 1,0 F ) ( zz F ) zz = a F z 2 (ε F zz ) ε = a F z 2 F 1,0 zz and ( a 1,0 D z D z a 1,0 = a Fzz ) F z 2 (a 1,0 F 1,0 zz (a ) = Fzz ) F z 2 (a F 1,0 zz ) by replacing a 1,0 with a in the dot product with F 1,0 zz. By replacing F zz = Fzz 1,0 + Fzz 0,1 using Fzz 0,1 = F 1,0 zz we get a 1,0 D z D z a 1,0 = a F zz 1,0 1,0 (a F F z 2 zz ) a F zz 0,1 1,0 (a F F z 2 = a F zz 1,0 0,1 1,0 a Fzz 2 (a F F z 2 zz ) F z 2 = a F zz 1,0 1,0 (a F F z 2 zz ) a1,0 F zz 2 F z 2. zz ) and then

18 18 NOTES BY DAREN CHENG, CHAO LI, CHRISTOS ANTOULIDIS From this we conclude Re (a 1,0 D z D z a 1,0 + a1,0 F zz 2 ) F z 2 which gives the required result. = 1 ( ) F z 2 Re (a Fzz 1,0 )(a F 1,0 zz ) 3.3. Stable minimal genus-0 surfaces in R n, n 5. We now try to see what we can prove when R 4 (or T 4 ) is replaced by R n, n 5. We will show that: Theorem Let F : 2 R n, n 5, with complete, oriented, stable, genus 0, and finite total curvature. Then there exists an affine subspace A 2k R n such that F is J-holomorphic for some J. Remark The requirement of finite total curvature might appear to be too strong but in fact it isn t. One can check using Gauss-Bonnet that, provided F is proper, quadratic area growth, χ() < finite total curvature. We will appeal to a theorem by Chern and Osserman [CO67]: Theorem 3.16 ([CO67]). Suppose 2 R n is a complete orientable minimal surface with finite total curvature, i.e., ( K) da <. Then is conformally equivalent to a punctured compact surface and the Gauss map extends through the punctures, meaning p T p, N p extend smoothly to. We will also need the following consequence of the stability inequality (3.29) 2 [f 2 ( zs) T 2 F z 2 + f 2 ] F z 2 Re (s D zd z s) da f 2 s 2 da, for all f C c (). Lemma Suppose is complete, oriented, stable, parabolic, and that s is a bounded section of N C with D z s = 0. Then ( z s) T = 0. Proof. From (3.29) with D z s = 0 and s bounded we conclude that f 2 ( zs) T 2 F z 2 da c f 2 da for all f, so by elliptic theory there exists u > 0 with u + ( zs) T 2 c F z 2 u = 0. By parabolicity u needs to be constant, and therefore ( z s) T = 0. Proof of Theorem By invoking the Chern-Osserman theorem we can construct a complex (n 2)-plane bundle E extending N C, where S 2 in view of our genus 0 assumption, and we can also extend the connection D from before to a connection on E. By [K58] and the fact that dim = 2 it follows that E is a holomorphic vector bundle; i.e., for all p S 2 there exists a local basis s 1,..., s n 2 of E which is holomorphic (D z s j = 0). By [Gro57], the holomorphic vector bundle E necessarily decomposes as a direct sum E = (L 1 L p ) (L p+1 L r ) (L r+1 L n 2 ) of complex line bundles order so that: (1) L 1,..., L p have c 1 (L) > 0,

19 286 - TOPICS IN DIFFERENTIAL GEOETRY - LECTURE NOTES 19 (2) L p+1,..., L r have c 1 (L) = 0, and (3) L r+1,..., L n 2 have c 1 (L) < 0. Roughly speaking, we will show that if there are no flat bundles then F is going to be J- holomorphic; conversely, flat bundles will correspond to direction of vanishing of the second fundamental form and will help determine the affine space A 2k from the statement of the theorem. Seeing as to how E was initially constructed as a complexification of a real bundle, the real pairing E x E x R, (s 1, s 2 ) s 1 s 2, gives rise to a holomorphic isomorphism E = E. According to this isomorphism the signs of the first Chern classes flip and therefore our original decomposition has to have as many positive line bundles as it does negative ones; namely, p = n 2 r. By definition of c 1 ( ), the bundles L 1,... L r (whose first Chern class is non-negative) all admit nontrivial global holomorphic sections s 1,..., s n 2. Since S 2 is compact, these sections are additionally bounded. By Lemma 3.17 above, ( z s j ) T = 0 for all j {1,..., r}. There are two cases to consider. First, suppose that all L i have c 1 = 0. Then s 1,..., s n 2 is a global basis of holomorphic sections which we have showed satisfy ( z s j ) T = 0 and therefore the second fundamental form of vanishes: s = j a j s j (s F zz)f z F z 2 = zs F z F z 2 = ( z s) T = 0. Therefore is totally geodesic and we re done. Now suppose that p > 0, n 2 r = p > 0. For convenience we set up the following table of index notation: 1 µ, ν p, r + 1 a, b n 2, 1 i, j r, p + 1 A, B n 2. In other words, indices µ, ν run over positive line bundles, i, j run over non-negative line bundles, and so on. We list some properties of s 1,..., s n 2 that we will need. (1) s µ s j = 0, since z (s µ s j ) = D z s µ s j + s µ D z s j = 0 because we know that our sections are holomorphic. Therefore s µ s j is a holomorphic function on S 2, thus constant. However, the section s µ belongs to a positive line bundle and necessarily vanishes somewhere. The claim follows. As a consequence, we get: (3.30) span{l 1,..., L r } = span{l 1,..., L p }. (2) z s j s k = 0, since z ( z s j s k ) = z z s j s k + z s j z s k = z ( z s j s k ) z s j z s k + z s j z s k. Since s j is holomorphic, z s j is purely tangential so the first term drops out by orthogonality. Next, s k is a bounded holomorphic section so by Lemma 3.17 (which relies on stability), z s k is purely normal, the second term drops out by orthogonality. The same goes for the third term. Therefore the expression above vanishes, so z s j s k dz is a holomorphic 1-form. The claim follows since Riemann-Roch forces such a differential to vanish identically. From this it follows that z s j span{l 1,..., L p }, and since ( z s j s k ) T = 0 by stability (Lemma 3.17), we get (3.31) z s j, D z s j span{l 1,..., L p }. Now we check the following Claim The bundle ξ = L 1 L r (T C ) 1,0 is parallel.

20 20 NOTES BY DAREN CHENG, CHAO LI, CHRISTOS ANTOULIDIS Proof of claim. By Proposition 3.7 we need to check that z, z map Γ(ξ) into itself. By linearity this amounts to showing z s j, z s j, z F z, z F z Γ(ξ). By minimality z F z Γ(ξ) is clear, while z s j Γ(ξ) is just (3.31). For the other two cases we compute z s j = ( z s j ) T because D z s j = 0 by holomorphicity ( ) ( ) F z F z = z s j F z 2 F z + z s j F z 2 F z ( ) F z = z s j F z 2 F z, the last equality following from minimality, F zz = 0, and therefore z s j Γ(ξ). Likewise we find ( ) ( ) F z F z z F z = z F z F z 2 F z + z F z F z 2 F z + ( z F z ) ( ) F z = z F z F z 2 F z + ( z F z ), seeing as to how the first term drops out in view of conformality, F z F z = 0. Now we observe z F z s k = F z z s k = 0 by stability, and we conclude z F z Γ(ξ). This completes the proof of the claim. Next we check the following Claim We have dim(ξ ξ) = r p. Proof of claim. Recall that ξ = L 1 L r (T C ) 1,0. For brevity write V = L 1 L r, so that ξ = V (T C ) 1,0. From (3.30) we see that V V, so span{ξ, ξ} = C n. Observe that n = dim C n = dim span{ξ, ξ} = 2r + 2 dim(ξ ξ) which gives dim(ξ ξ) = 2r+2 n. From the decomposition of E into line bundles by Grothendieck s theorem we further have p + r = n 2 r = n 2 p. Combining these two relations we conclude which is the required result. dim(ξ ξ) = 2(n 2 p) + 2 n = n 2 2p = r p The proof of the theorem is now completed via the following sequence of steps: (1) Since ξ is parallel, let s write ξ = Λ for a complex (r + 1)-dimensional vector space Λ. Notice that the complex (r p)-dimensional vector space T = Λ Λ is (by definition) preserved by complex conjugation and therefore the complexification W R C of a real (r p)-dimensional vector space W. (2) Seeing as to how (T C ) 1,0 is manifestly not preserved by conjugation we get that W is a parallel subbundle of the real normal bundle N, or in other words, that = F () is a subset of an affine subspace P R n perpendicular to W, the dimension of which is evidently dim R n dim W = n (r p) = n r + p = 2p + 2. That is, we have constructed an affine subspace P 2p+2 R n that contains the surface. (3) From the decomposition N C = ( T ) ( T ), the being taken within N C of course, we characterize ( T ) as the complexified normal bundle of viewed as a surface within P 2p+2.

21 286 - TOPICS IN DIFFERENTIAL GEOETRY - LECTURE NOTES 21 (4) From (3.30) we find that ( T ) L 1 L p L 1 L p and, by dimension counting, this inclusion is actually an exact equality. Namely, ( T ) = L 1 L p L 1 L p. (5) By restricting to the context 2 P 2p+2 and the parallel nature of ξ we find that there exists a constant almost complex structure on P 2p+2 with respect to which 2 is J-holomorphic as in the proof of Theorem Positive isotropic curvature We ll see that a number of the techniques developed in the previous section will extend to nonflat ambient spaces and thereby give important geometric consequences. Instead of studying the second variation operator for area, however, we will study the second variation operator for energy: E(F ) = df 2 h da h where F : 2 ( n, g). For the purposes of computing the energy integral, the Riemann surface 2 is thought of as being a Riemannian manifold ( 2, h), though the Dirichlet energy is conformally invariant as we have seen before. By a computation similar to that for second variation of area, we find: Proposition 4.1. If X Γ(F (T )) and F is a critical point for the energy functional, then 1 2 δ2 E(X, X) = X 2 2 R(e i, X, e i, X) da h. Remark 4.2. This is reminiscent of the formula for the second variation of energy on geodesics γ, 1 2 δ2 E(X, X) = γ X 2 R(γ, X, γ, X) ds. γ We will complexify the (ambient) tangent bundle and the stability operator like we did before. For X Γ(F (T C )) of the form X = X 1 + ix 2, we define i=1 1 2 δ2 E(X, X) = X, X and by arguing as in Proposition 3.4 we get: 2 R(e i, X, e i, X) da h, Proposition 4.3. If X Γ(F (T C )) and F is a critical point for the energy functional, then in complex coordinates z = x + iy we have 1 8 δ2 E(X, X) = z X 2 R( z, X, z, X) dx dy. Remark 4.4. In general variations of energy and area behave differently. Critical points of the prior are harmonic maps, and critical points of the latter are minimal surfaces. (Recall that we ve seen that these coincide on a round S 2.) The second variation of energy and the second variation of area behave differently, too. The stability operator for energy is easier to work with since it has one less term in it but is also coarser for example, every harmonic map into flat space is clearly stable. i=1

22 22 NOTES BY DAREN CHENG, CHAO LI, CHRISTOS ANTOULIDIS It is important to be able to understand the effect of curvature on stabiliity. In the context of the area functional, we know that positive curvature gives rise to instability. Likewise, we can force instability in Proposition 4.3 provided we can construct global holomorphic sections X and that the complex sectional curvatures R( z, X, z, X) are positive. This section aims to pursue these ideas further. Let s set up our notation. Recall that n is a real Riemannian manifold with real metric,. We complexify T C = T R C and extend, to T C, mimicking the extension of the dot product X Y on R n to a dot product on C n. Namely, for X = X 1 + ix 2 T C we set X, X = X 1, X 1 X 2, X 2 + 2i X 1, X 2. Notice that this is not a Hermitian metric, just symmetric and bilinear over C. Definition 4.5. A vector X T C is called isotropic if X, X = 0; i.e., if X 1 = X 2 and X 1, X 2 = 0. A plane Π 2 T C is isotropic if every X Π is isotropic. Example 4.6. If F is conformal, then F z = df ( z ) is isotropic. We made extended use of this fact in the previous section. Lemma 4.7. If Π 2 T C is isotropic then there exist real vectors e 1, e 2, e 3, e 4 T, orthonormal with respect to the real metric, such that Π 2 = span{e 1 + ie 2, e 3 + ie 4 }. Proof. The pairing (X, Y ) = X, Y is Hermitian, and by Gram-Schmidt over C we may arrange for a basis X, Y of Π 2 to be such that (X, X) = (Y, Y ) = 1 and (X, Y ) = 0. Write X = 1 2 (e 1 + ie 2 ), Y = 1 2 (e 3 + ie 4 ). We make the following observations: (1) The isotropy of X and Y and the fact that X, X = Y, Y = 1 together give e 1, e 1 = e 2, e 2 = e 3, e 3 = e 4, e 4 = 1, and e 1, e 2 = e 3, e 4 = 0. (2) The isotropy of X + Y = 1 2 (e 1 + e 3 + i(e 2 + e 4 )) gives e 1 + e 3, e 1 + e 3 = e 2 + e 4, e 2 + e 4 e 1, e 3 = e 2, e 4, and 0 = e 1 + e 3, e 2 + e 4 = e 1, e 4 + e 2, e 3. (3) The complex orthogonality X, Y = 0 gives 0 = e 1 + ie 2, e 3 ie 4 = e 1, e 3 + e 2, e 4 + i[ e 2, e 3 e 1, e 4 ]. These facts put together show that e 1, e 2, e 3, e 4 are real orthonormal vectors. Definition 4.8. A (real) Riemannian manifold ( n, g) is called PIC (short of positive isotropic curvature, or originally positive curvature on totally isotropic 2-planes) if every isotropic 2-plane Π 2 T C and every complex orthonormal basis X, Y for Π satisfy R(X, Y, X, Y ) > 0. Remark 4.9. Just for the sake of comparison, we recall that a Riemannian manifold is said to have positive (sectional) curvature if R(X, Y, X, Y ) > 0 for every real orthonormal basis X, Y of every real 2-plane Π 2 T. We make the following observations regarding the definition of PIC: (1) PIC manifolds are not necessarily Ricci positive. In particular, round products S 1 S n 1 are always PIC but not Ricci positive. (2) We can perturb the spherical metrics above in such a way that S 1 S n 1 is still PIC and yet has negative Ricci curvature somewhere. (3) PIC manifolds are always scalar positive.

23 286 - TOPICS IN DIFFERENTIAL GEOETRY - LECTURE NOTES 23 (4) All 2- and 3-manifolds are vacuously PIC, because they have no isotropic complex 2-planes, since isotropic subspaces can be checked to take up no more than half the total dimension of their ambient vector space. There are a number of interesting PIC manifolds: Theorem 4.10 ([88]). The following manifolds are PIC: (1) ( n, g) with positive curvature operator R : Λ 2 T Λ 2 T ; i.e., R(ξ), ξ > 0 for all ξ Λ 2 T p \{0}. 1 In fact, it s enough for R to be (2,2)-positive, i.e. the positivity condition be met for 2-vectors ξ with tensor-rank at most 2. (2) ( n, g) with positive pointwise strictly 1 4-pinched curvature; i.e., there exists a continuous κ : (0, ) such that 1 4 κ(p) < K Π κ(p) for all Π 2 T p. The following theorem by icallef and Wang shows that the class of PIC manifolds is rich enough to support connected sums. Theorem 4.11 ([W93]). If (1 n, g 1), (2 n, g 2) have isotropic curvatures bounded from below by a positive constant (e.g., if they are compact and PIC), then 1 n# 2 n supports a PIC metric. We proceed by proving the fact that a manifold is PIC if the curvature operator is positive definite or it s 1/4-pinched. Proof. By previous lemma any isotropic plane Π is spanned by vectors X, Y with So Then the complexified curvature is given by X = 1 2 e 1 + ie 2, Y = 1 2 (e 3 + ie 4 ). R(X, Y, X, Ȳ ) = R(X Y, X Ȳ ) = 1 4 R((e 1 + ie 2 ) (e 3 + ie 4 ), (e 1 ie 2 ) (e 3 ie 4 )). K(Π) = 1 4 [R(e 1 e 3 e 2 e 4, e 1 e 3 e 2 e 4 ) + R(e 1 e 4 + e 2 e 3, e 1 e 4 + e 2 e 3 )] > 0, if the curvature operator R is positive definite. We also see that it suffices to require the curvature operator is positive operator on the sum of 2 wedges, which is called (2, 2) positive by ichallef- oore. We next prove pointwise strict 1/4-pinching condition implies PIC. We fix a point on the manifold and let e 1,..., e 4 be 4 orthonormal vectors in the tangent space. Further expanding the above equation, we have K(Π) = 1 4 [K 13 + K 24 2R(e 2, e 4, e 1, e 3 ) + K 23 + K R(e 1, e 4, e 2, e 3 )] = 1 4 (K 13 + K 24 + K 14 + K 23 2R 1234 ). Here we ve used the first Bianchi identity. We conclude the proof by the following property of 1/4-pinched manifold. Proposition If p and k(p) > 0 such that 1 4k(p) < K(Π) k(p) for all two-plane Π T p, then R 1234 < 1 2 k(p). The proof is straightforward consequence of the following two identities. Let u, v, w, x be 4 orthonormal vectors in T p. (i) 4R(u, v, w, v) = R(u + w, v, u + w, v) R(u w, v, u w, v) 1 Recall that if ξ = u v is a simple 2-vector, with u, v orthonormal, then R(u v), u v = R(u, v, u, v) = K span{u,v}.