MATH 250A: Chapter 4 Lecture Notes 6 December 2017 Aaron Nelson 4.3 Sectional curvature Notation. Here and throughout, let M denote a Riemannian manifold of dimension n ≥ 2, with metric h·; ·i, and R: X (M) × X (M) !L(X (M)) the curvature operator; for a given p 2 M, we will abuse notation and write R: TpM × TpM !L(TpM), which makes sense if we extend vectors v 2 TpM to vector fields V 2 X (M) in the usual way. Next, let V be a real vector space (of dimension at least 2) equipped with an inner-product h·; ·i (differentiating between the two uses of this symbol will be clear from context). Finally, for each x; y 2 V denote the area of the parallelogram determined by x and y by jx ^ yj := pjxj2jyj2 − jhx; yij2: This notation is not surprising if one is familiar with the wedge product; indeed, we may write x ^ y = jx ^ yj(e1 ^ e2) for some orthonormal basis fe1; e2g for spanfx; yg. More generally, if dim(V ) ≥ m ≥ 2, then the wedge product gives x1 ^ · · · ^ xm = jx1 ^ · · · ^ xmj(e1 ^ · · · ^ em) m m for some orthonormal basis fekgk=1 for spanfxkgk=1, where jx1 ^ · · · ^ xmj denotes the volume m of the parallelepiped determined by fxkgk=1. Definition. Let p 2 M and V ⊂ TpM be a two-dimensional subspace. Define K : V × V ! R by ( hR(x;y)x;yi if x; y are linearly independent; K(x; y) := jx^yj2 0 otherwise: Proposition (3.1). Let p 2 M and V ⊂ TpM be a two-dimensional subspace. If (x; y); (u; v) 2 V × V are pairs of linearly independent vectors in V , then K(x; y) = K(u; v). Proof. First, observe that it is possible to transform the basis fx; yg for V into any other basis for V using compositions of the operations (a) fx; yg ! fy; xg; (b) fx; yg ! fλx, yg for some nonzero λ 2 R; (c) fx; yg ! fx + λy; yg for some λ 2 R. Hence, it suffices to prove that K is invariant under these operations. To this end, let x; y 2 V be linearly independent and note the following: (a) Clearly, jy ^xj = jx^yj, and so it suffices to show that hR(y; x)y; xi = hR(x; y)x; yi, which follows by applying part (b) of Proposition (2.5): hR(y; x)y; xi = −hR(x; y)y; xi = hR(x; y)x; yi: 1 (b) Suppose λ 2 Rnf0g. Since jλx^yj = jλjjx^yj, it suffices to note that hR(λx, y)(λx); yi = λ2hR(x; y)x; yi by the bilinearity of R on V × V and linearity of R(·; ·) on V . (c) Suppose λ 2 R. Then we have j(x + λy) ^ yj2 = jx + λyj2jyj2 − jhx + λy; yij2 = jxj2 + 2λhx; yi + λ2jyj2jyj2 − hx; yi2 + λ2jyj4 + 2λhx; yijyj2 = jxj2jyj2 − jhx; yij2 = jx ^ yj2; and so it remains to show that hR(x + λy; y)(x + λy); yi = hR(x; y)x; yi. For this, observe that the bilinearity of R on V × V and linearity of R(·; ·) on V yield hR(x + λy; y)(x + λy); yi = hR(x; y)(x + λy); yi + λhR(y; y)(x + λy); yi = hR(x; y)x; yi + λhR(x; y)y; yi + λhR(y; y)x; yi + λ2hR(y; y)y; yi and, hence, the result follows by applying parts (b) and (c) of Proposition (2.5) to obtain hR(y; y); y; yi = 0 and hR(y; y)x; yi = hR(x; y)y; yi = 0. Remark. An immediate consequence of Proposition (3.1) is that the value of K depends only the linear (in)dependence of its arguments and not on the arguments themselves. In particular, K is constant over the set of pairs of linearly independent vectors in V ; we denote this constant value by K(V ). Definition (3.2). Let p 2 M and V ⊂ TpM be a two-dimensional subspace. The sectional curva- ture of V at p is K(V ). Sectional curvature is important because of its relationship to the curvature operator R. In particu- lar, for any p 2 M, knowing the values fK(V )gV ⊂TpM for all two-dimensional subspaces of TpM completely determines R. We make this precise with the following lemma: Lemma (3.3). Let p 2 M and f : TpM × TpM × TpM ! TpM be a tri-linear mapping satisfying (i) hf(x; y; z); wi + hf(y; z; x); wi + hf(z; x; y); wi = 0; (ii) hf(x; y; z); wi = −hf(y; x; z); wi; (iii) hf(x; y; z); wi = −hf(x; y; w); zi; (iv) hf(x; y; z); wi = hf(z; w; x); yi for all x; y; z; w 2 TpM. For each two-dimensional subspace V ⊂ TpM define hf(x; y; x); yi κ(V ) := jx ^ yj2 2 for any pair (x; y) of linearly independent vectors in V . If for each such V we have κ(V ) = K(V ), then f(x; y; z) = R(x; y)z for all x; y; z 2 TpM. Remark. Before we prove Lemma (3.3), it is necessary to remark that κ is well-defined. Indeed, by the assumptions on f, verifying that κ(V ) is constant over the set of pairs of linearly independent vectors in V is identical to the proof of Proposition (3.1). Proof of Lemma (3.3). It suffices to prove hf(x; y; z); wi = hR(x; y)z; wi for any x; y; z; w 2 TpM. To this end, observe that we have hf(x; y; x); yi = hR(x; y)x; yi for all x; y 2 TpM; in- deed, for linearly independent x and y this follows from the assumptions on κ, and for linearly dependent x and y it is easy to verify that hf(x; y; x); yi = 0 = hR(x; y)x; yi. In particular, for any x; y; z 2 TpM, we have hf(x + z; y; x + z); yi = hR(x + z; y)(x + z); yi and so by expanding we obtain hf(x; y; x); yi + 2hf(x; y; z); yi + hf(z; y; z); yi = hR(x; y)x; yi + 2hR(x; y)z; yi + hR(z; y)z; yi; which simplifies to hf(x; y; z); yi = hR(x; y)z; yi. Substituting y + w for y in this expression, expanding, and rearranging then yields hf(x; y; z); wi − hR(x; y)z; wi = hf(y; z; x); wi − hR(y; z)x; wi; from which it follows that hf(x; y; z); wi − hR(x; y)z; wi is invariant under cyclic permutations of (x; y; z). Applying assumption (i) and part (a) of Proposition (2.5), we therefore have 0 = 3hf(x; y; z); wi − hR(x; y)z; wi; the desired result is an immediate consequence of this equality. We now consider the case of constant sectional curvature; that is, for each p 2 M, we require that K(V ) = K(W ) for all two-dimensional subspaces V; W ⊂ TpM. The following proposition (lemma in Do Carmo) provides a characterization of Riemannian manifolds with this property: Proposition (3.4). Let p 2 M and g : TpM × TpM × TpM ! TpM be a tri-linear mapping satisfying hg(x; y; z); wi = hx; zihy; wi − hy; zihx; wi for all x; y; z; w 2 TpM. Then the sectional curvature at p is constant (and equal to K0 2 R) if and only if K0g(x; y; z) = R(x; y)z for all x; y; z 2 TpM. Proof. Before we begin, we observe some simple facts about g: 3 (i) Let x; y; z; w 2 TpM. Then hg(x; y; z); wi + hg(y; z; x); wi + hg(z; x; y); wi = hx; zihy; wi − hy; zihx; wi + hy; xihz; wi − hz; xihy; wi + hz; yihx; wi − hx; yihz; wi = 0: (ii) Let x; y; z; w 2 TpM. Then hg(x; y; z); wi = hx; zihy; wi − hy; zihx; wi = −hy; zihx; wi − hx; zihy; wi = −hg(y; x; z); wi: (iii) Let x; y; z; w 2 TpM. Then hg(x; y; z); wi = hx; zihy; wi − hy; zihx; wi = −hx; wihy; zi − hy; wihx; zi = −hg(x; y; w); zi: (iv) Let x; y; z; w 2 TpM. Then hg(x; y; z); wi = hx; zihy; wi − hy; zihx; wi = hz; xihw; yi − hz; yihw; xi = hg(z; w; x); yi: With these facts in hand, we now prove each implication separately: ()) Assume that the sectional curvature at p is constant (and equal to K0 2 R). Then by definition 2 we have hR(x; y)x; yi = K0jx ^ yj for all x; y 2 TpM. Since we also have hg(x; y; x); yi = jxj2jyj2 − hx; yi2 = jx ^ yj2; it follows that K0hg(x; y; x); yi = hR(x; y)x; yi; that is K hg(x; y; x); yi hR(x; y)x; yi 0 = K = ; jx ^ yj2 0 jx ^ yj2 which is one of the assumptions of Lemma (3.3), provided we take f := K0g. As g (and therefore K0g) satisfies properties (i)-(iv) above, the remaining assumptions of Lemma (3.3) are also satisfied (with f = K0g), and so applying the lemma yields K0g(x; y; z) = R(x; y)z for all x; y; z 2 TpM, as desired. (() Assume that K0g(x; y; z) = R(x; y)z for all x; y; z 2 TpM and some K0 2 R.