4.3 Sectional Curvature Notation

Home , Ricci curvature, Sectional curvature

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 ∈ M, we will abuse notation and write R: TpM × TpM → L(TpM), which makes sense if we extend vectors v ∈ TpM to vector ﬁelds V ∈ 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 ∈ V denote the area of the parallelogram determined by x and y by

|x ∧ y| := p|x|2|y|2 − |hx, yi|2.

This notation is not surprising if one is familiar with the wedge product; indeed, we may write x ∧ y = |x ∧ y|(e1 ∧ e2) for some orthonormal basis {e1, e2} for span{x, y}. More generally, if dim(V ) ≥ m ≥ 2, then the wedge product gives x1 ∧ · · · ∧ xm = |x1 ∧ · · · ∧ xm|(e1 ∧ · · · ∧ em) m m for some orthonormal basis {ek}k=1 for span{xk}k=1, where |x1 ∧ · · · ∧ xm| denotes the volume m of the parallelepiped determined by {xk}k=1.

Deﬁnition. Let p ∈ 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) := |x∧y|2 0 otherwise.

Proposition (3.1). Let p ∈ M and V ⊂ TpM be a two-dimensional subspace. If (x, y), (u, v) ∈ 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 {x, y} for V into any other basis for V using compositions of the operations

(a) {x, y} → {y, x};

(b) {x, y} → {λx, y} for some nonzero λ ∈ R;

(c) {x, y} → {x + λy, y} for some λ ∈ R. Hence, it suﬃces to prove that K is invariant under these operations. To this end, let x, y ∈ V be linearly independent and note the following:

(a) Clearly, |y ∧x| = |x∧y|, 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 ∈ R\{0}. Since |λx∧y| = |λ||x∧y|, 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 ∈ R. Then we have |(x + λy) ∧ y|2 = |x + λy|2|y|2 − |hx + λy, yi|2 = |x|2 + 2λhx, yi + λ2|y|2|y|2 − hx, yi2 + λ2|y|4 + 2λhx, yi|y|2 = |x|2|y|2 − |hx, yi|2 = |x ∧ y|2,

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 ∈ M and V ⊂ TpM be a two-dimensional subspace. The sectional curvature of V at p is K(V ).

Sectional curvature is important because of its relationship to the curvature operator R. In particular, for any p ∈ M, knowing the values {K(V )}V ⊂TpM for all two-dimensional subspaces of TpM completely determines R. We make this precise with the following lemma:

Lemma (3.3). Let p ∈ 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 ∈ TpM. For each two-dimensional subspace V ⊂ TpM deﬁne hf(x, y, x), yi κ(V ) := |x ∧ y|2

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 ∈ 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 ∈ TpM. To this end, observe that we have hf(x, y, x), yi = hR(x, y)x, yi for all x, y ∈ TpM; indeed, 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 ∈ 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 ∈ 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 ∈ 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 ∈ TpM. Then the sectional curvature at p is constant (and equal to K0 ∈ R) if and only if K0g(x, y, z) = R(x, y)z for all x, y, z ∈ TpM. Proof. Before we begin, we observe some simple facts about g:

3 (i) Let x, y, z, w ∈ 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 ∈ 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 ∈ 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 ∈ 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 ∈ R). Then by definition 2 we have hR(x, y)x, yi = K0|x ∧ y| for all x, y ∈ TpM. Since we also have hg(x, y, x), yi = |x|2|y|2 − hx, yi2 = |x ∧ y|2, 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 = , |x ∧ y|2 0 |x ∧ y|2 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 ∈ TpM, as desired. (⇐) Assume that K0g(x, y, z) = R(x, y)z for all x, y, z ∈ TpM and some K0 ∈ R. Then, as we 2 have already seen that hg(x, y, x), yi = |x ∧ y| , for any two-dimensional subspace V ⊂ TpM and any pair (x, y) of linearly independent vectors in TpM we have hR(x, y)x, yi K(V ) = = K |x ∧ y|2 0 and, hence, the sectional curvature at p is constant (and equal to K0).

4 n Corollary (3.5). Let p ∈ M and {ek}k=1 be an orthonormal basis for TpM. For each i, j, k, ` ∈ {1, . . . , n}, deﬁne Rijk` := hR(ei, ej)ek, e`i. Then the sectional curvature at p is constant (and equal to K0 ∈ R) if and only if Rijk` = K0(δikδj` − δjkδi`) for all i, j, k, ` ∈ {1, . . . , n}, where δab denotes the Kronecker delta.

Proof. By Proposition (3.4), the sectional curvature at p is constant (and equal to K0) if and only if K0g(x, y, z) = R(x, y)z for all x, y, z, w ∈ TpM, where g : TpM × TpM × TpM → TpM is a tri-linear mapping satisfying hg(x, y, z), wi = hx, zihy, wi − hy, zihx, wi for all x, y, z, w ∈ TpM. Hence, the sectional curvature is constant if and only if

Rijk` = hR(ei, ej)ek, eì = K0hg(ei, ej, ek), eì = K0 hei, ekihej, eì − hej, ekihei, eì = K0(δikδj` − δjkδi`), which is the desired result.

Remark. The condition Rijk` = K0(δikδj` − δi`δjk) for all i, j, k, ` ∈ {1, . . . , n} is equivalent to having, for each pair {i, j} ⊂ {1, . . . , n}, Rijij = −Rijji = K0(1 − δij) and Rijk` = 0 for all k, ` ∈ {1, . . . , n}\{i, j}.

Examples. Suppose the sectional curvature at p ∈ M is constant, with value K0 ∈ R. Then, modulo scaling the metric on M, there are only three unique cases to consider:

n (1) If K0 = 0, then M is locally isometric to R , n-dimensional Euclidean space;

n (2) If K0 = 1, then M is locally isometric to S , the unit n-sphere;

n (3) If K0 = −1, then M is locally isometric to H , n-dimensional hyperbolic space. (See Figure 1 below for an illustration when n = 2.) Riemannian manifolds of constant sectional curvature are sometimes referred to as space forms.

Figure 1. A geodesic triangle. Three points around p ∈ M are connected via geodesic curves and 2 ∼ 2 ∼ 2 ∼ 2 projected onto R . In black, M = R (K0 = 0); in blue, M = S (K0 = 1); in red, M = H (K0 = −1).

5 4.4 Ricci and scalar curvature The following is a brief introduction to Ricci and scalar curvature. Only basic deﬁnitions and re- sults are provided, and all proofs are omitted. The notation of section 4.3 is assumed throughout.

n−1 Notation. Let p ∈ M and x0 ∈ TpM such that |x0| = 1. Denote by {xk}k=1 ⊂ TpM an orthonormal basis for the hyperplane in TpM which is orthogonal to span{x0}.

Deﬁnition. The Ricci curvature at p in the direction x0 is deﬁned by

n−1 1 X Ric (x ) := hR(x , x )x , x i. p 0 n − 1 0 k 0 k k=1 Moreover, the scalar curvature at p is deﬁned by

n−1 n−1 n−1 1 X 1 X X S(p) := Ric (x ) = hR(x , x )x , x i n p j n(n − 1) j k j k j=0 j=0 k=0

Proposition. The Ricci curvature at p in the direction x0 is independent of the choice of orthonormal n−1 n−1 basis for the hyperplane in TpM which is orthogonal to span{x0}. That is, if {xk}k=1 and {zk}k=1 are two such orthonormal bases, then

n−1 n−1 X X hR(x0, xk)x0, xki = hR(x0, zk)x0, zki. k=1 k=1

Corollary. The scalar curvature at p is independent of the choice of orthonormal basis for TpM.