5 Curvature

Chapter 5
Curvature

Curvature in Riemannian geometry can seem a little hidden, but we have already encountered its effects. We saw for instance that parallel transport around a loop on the sphere changes a vector. We also saw for the hyperbolic plane that triangles have angles that sum to less than $π$ . Both of these are due to the intrinsic curvature of the space. Our own universe has curvature: in general relativity it is curvature that causes gravity. We commonly interpret an asteroid being deflected as it passes a planet as it being pulled from its straight line path by a force; in truth it is travelling on a geodesic and it is space itself that is bent.

5.1 Symmetries and Identities

We first motivate curvature by locally comparing a Riemannian manifold $M$ to euclidean space $ℝ^{n}$ . Later we will connect it to the more geometric picture presented in Chapter 1. An important feature of euclidean space is that it has (a basis of) parallel vector fields with respect to the Levi-Civita connection. These are vector fields that are parallel along every curve. Because the Levi-Civita connection is uniquely determined by the metric, the property of having a parallel vector field must be a local isometry invariant, i.e. if $M$ at $p$ has a neighbourhood that is isometric to a neighbourhood of $ℝ^{n}$ then it has a parallel vector field.

The obvious way to construct a parallel vector field is to begin with a vector $Z |_{p} \in T_{p} M$ and parallel transport it around. Choose two coordinate direction $\partial 1, \partial 2$ . We parallel transport $Z |_{p}$ along the $x^{1}$ -axis and then from every point in the $x^{2}$ direction. Is the vector field so constructed parallel in the $x^{1}$ direction? By construction $\nabla_{\partial 1} Z = 0$ on the $x^{1}$ -axis, so it is sufficient to have $\nabla_{\partial 2} \nabla_{\partial 1} Z = 0$ .

An alternative way to ask this question is to construct a second vector field $\tilde{Z}$ with $\tilde{Z} |_{p} = Z_{p}$ by first parallel transporting $\tilde{Z} |_{p}$ along the $x^{2}$ -axis and then from every point in the $x^{1}$ direction. By definition $\tilde{Z}$ and $Z$ agree on the $x^{1}$ - and $x^{2}$ -axes. But do they agree at other points? Consider a point $q = p + (h_{1}, h_{2})$ close to $p$ . Then using an approximation

\begin{array}{l} Z |_{p + (h_{1}, 0)} & \approx Z |_{p} + h_{1} (\nabla_{\partial 1} Z) |_{p} \\ Z |_{q} & \approx Z |_{p + (h_{1}, 0)} + h_{2} (\nabla_{\partial 2} Z) |_{p + (h_{1}, 0)} \\ = [Z + h_{1} (\nabla_{\partial 1} Z)] |_{p} + h_{2} [\nabla_{\partial 2} Z + h_{1} (\nabla_{\partial 2} \nabla_{\partial 1} Z)] |_{p + (h_{1}, 0)} \\ = Z |_{p} + h_{1} 0 + h_{2} 0 + h_{1} h_{2} (\nabla_{\partial 2} \nabla_{\partial 1} Z) |_{p + (h_{1}, 0)} . \end{array}

Likewise

\tilde{Z} |_{q} = \tilde{Z} |_{p} + h_{1} 0 + h_{2} 0 + h_{1} h_{2} (\nabla_{\partial 1} \nabla_{\partial 2} \tilde{Z}) |_{p + (0, h_{2})} .

For $h_{1}, h_{2}$ small and since $\tilde{Z} |_{p} = Z |_{p}$ we see that $\tilde{Z} |_{q} = Z |_{q}$ are equal if and only if $\nabla_{\partial 2} \nabla_{\partial 1} Z = \nabla_{\partial 1} \nabla_{\partial 2} Z$ . Thus the lack of a parallel vector field can be measured by the difference $\nabla_{\partial 1} \nabla_{\partial 2} Z - \nabla_{\partial 2} \nabla_{\partial 1} Z$ . We can move away from coordinates by using the properties of the connection to give an equivalent statement for arbitrary vector fields. For $X = X^{i} ∂i, Y = Y^{j} ∂j$

\begin{array}{l} \nabla_{X} \nabla_{Y} Z - \nabla_{Y} \nabla_{X} Z & = X^{i} \nabla_{∂i} (Y^{j} \nabla_{∂j} Z) - Y^{j} \nabla_{∂j} (X^{i} \nabla_{∂i} Z) \\ = X^{i} \frac{\partial Y^{j}}{\partial x^{i}} \nabla_{∂j} Z + X^{i} Y^{j} \nabla_{∂i} \nabla_{∂j} Z - Y^{j} \frac{\partial X^{i}}{\partial x^{j}} \nabla_{∂i} Z - X^{i} Y^{j} \nabla_{∂j} \nabla_{∂i} Z \\ = X^{i} Y^{j} (\nabla_{∂i} \nabla_{∂j} Z - \nabla_{∂j} \nabla_{∂i} Z) + \nabla_{X^{i} (∂i Y^{j}) ∂j} Z - \nabla_{Y^{j} (∂j X^{i}) ∂i} Z \\ = X^{i} Y^{j} (\nabla_{∂i} \nabla_{∂j} Z - \nabla_{∂j} \nabla_{∂i} Z) + \nabla_{X^{i} (∂i Y^{j}) ∂j - Y^{j} (∂j X^{i}) ∂i} Z \\ = X^{i} Y^{j} (\nabla_{∂i} \nabla_{∂j} Z - \nabla_{∂j} \nabla_{∂i} Z) + \nabla_{[X, Y]} Z . \end{array}

Just as for torsion, we see that this ‘commutator’ of vector fields has a part that is due the commutator of the vector fields themselves but also a part that is ‘built in’ to all such commutators.

Definition 5.1. The Riemannian curvature tensor $R$ is a vector valued function

R (X, Y) Z = \nabla_{X} \nabla_{Y} Z - \nabla_{Y} \nabla_{X} Z - \nabla_{[X, Y]} Z .

We see immediately from the calculation above that $R$ only depends on the pointwise values of $X$ and $Y$ . We can continue to get an formula for the curvature tensor in terms of the Christoffel coefficients.

\begin{array}{l} R (∂i, ∂j) Z & = \nabla_{∂i} \nabla_{∂j} (Z^{k} \partial k) - \nabla_{∂j} \nabla_{∂i} (Z^{k} \partial k) \\ = \nabla_{∂i} (∂j Z^{k} \partial k + Z^{k} \nabla_{∂j} ∂k) - \nabla_{∂j} (∂i Z^{k} \partial k + Z^{k} \nabla_{∂i} ∂k) \\ = \nabla_{∂i} (∂j Z^{k} \partial k + Z^{k} Γ_{jk}^{l} \partial l) - \nabla_{∂j} (∂i Z^{k} \partial k + Z^{k} Γ_{ik}^{l} \partial l) \\ = (∂i∂j Z^{k} \partial k + ∂j Z^{k} Γ_{ik}^{l} \partial l + ∂i Z^{k} Γ_{jk}^{l} \partial l + Z^{k} \partial i Γ_{jk}^{l} \partial l + Z^{k} Γ_{jk}^{m} Γ_{im}^{l} \partial l) \\ - (∂j∂i Z^{k} \partial k + ∂i Z^{k} Γ_{jk}^{l} \partial l + ∂j Z^{k} Γ_{ik}^{l} \partial l + Z^{k} \partial j Γ_{ik}^{l} \partial l + Z^{k} Γ_{ik}^{m} Γ_{jm}^{l} \partial l) \\ = Z^{k} (\partial i Γ_{jk}^{l} + Γ_{jk}^{m} Γ_{im}^{l} - \partial j Γ_{ik}^{l} - Γ_{ik}^{m} Γ_{jm}^{l}) \partial l . \end{array}

This proves¹ that the curvature tensor only depends pointwise on the value of $Z$ , even though it is constructed out of derivatives. The expression in the bracket is called $R_{ijk}^{l}$ . Because it is often a pain to work with a vector valued function, it is common to define a curvature quadlinear form

Rm (X, Y, Z, W) = g (R (X, Y) Z, W) .

Clearly the metric depends on $W$ only pointwise, so it makes sense to express this in a chart as

Rm (X, Y, Z, W) = X^{i} Y^{j} Z^{k} W^{l} g_{lm} R_{ijk}^{m} = : X^{i} Y^{j} Z^{k} W^{l} {Rm}_{ijkl} .

Note, different authors use different conventions about the order of the indices. We follow Lee, whereas Jost and Wikipedia use the order $lkij$ . Petersen refuses to choose a side by only ever using $Rm (∂i, ∂j, ∂k, ∂l)$ . It is also common to use $R$ for both objects and let the position of the indices distinguish them. On paedogogical grounds we avoid this.

Example 5.2 (Euclidean Space). The plane (or any euclidean space) in its standard chart has Christoffel coefficients identically equal to zero. Therefore its curvature vanishes at all points.

Example 5.3. On any one dimensional manifold, there is only one coefficient of the Riemannian curvature tensor:

R_{111}^{1} = \partial 1 Γ_{11}^{1} + Γ_{11}^{1} Γ_{11}^{1} - \partial 1 Γ_{11}^{1} - Γ_{11}^{1} Γ_{11}^{1} = 0 .

Therefore one dimensional manifolds have no intrinsic curvature. In fact we have seen this in Chapter 1: curves can be reparameterised by arc-length, which makes the induced metric the euclidean metric.

Example 5.4 (Stereographic Projection). In Exercise 3.32 we calculated the Christoffel coefficients for $𝕊^{2}$ in stereographic coordinates.

\begin{array}{l} \frac{2 x^{1}}{∥ x ∥^{2} + 1} & = - Γ_{11}^{1} = - Γ_{21}^{2} = - Γ_{12}^{2} = Γ_{22}^{1}, \\ \frac{2 x^{2}}{∥ x ∥^{2} + 1} & = Γ_{11}^{2} = - Γ_{21}^{1} = - Γ_{12}^{1} = - Γ_{22}^{2} . \end{array}

We can use that with the above formula to calculate $R_{ijk}^{l}$ . We prepare ourselves by calculating the partial derivatives

\begin{array}{l} \frac{\partial}{\partial x^{1}} Γ_{22}^{1} & = \frac{- 2 {(x^{1})}^{2} + 2 {(x^{2})}^{2} + 2}{{(∥ x ∥^{2} + 1)}^{2}} & \frac{\partial}{\partial x^{2}} Γ_{22}^{1} & = \frac{- 4 x^{1} x^{2}}{{(∥ x ∥^{2} + 1)}^{2}} \\ \frac{\partial}{\partial x^{1}} Γ_{11}^{2} & = \frac{- 4 x^{1} x^{2}}{{(∥ x ∥^{2} + 1)}^{2}} & \frac{\partial}{\partial x^{2}} Γ_{11}^{2} & = \frac{2 {(x^{1})}^{2} - 2 {(x^{2})}^{2} + 2}{{(∥ x ∥^{2} + 1)}^{2}} \end{array}

Many of the curvature coefficients are zero just by definition.

R_{iik}^{l} = \partial i Γ_{ik}^{l} - \partial i Γ_{ik}^{l} + Γ_{ik}^{m} Γ_{im}^{l} - Γ_{ik}^{m} Γ_{im}^{l} = 0,

so of the sixteen coefficients there are at most eight non-zero entries.

\begin{array}{l} R_{ijk}^{l} & = \partial i Γ_{jk}^{l} - \partial j Γ_{ik}^{l} + Γ_{jk}^{m} Γ_{im}^{l} - Γ_{ik}^{m} Γ_{jm}^{l} \\ R_{121}^{1} = - R_{211}^{1} & = \partial 1 Γ_{21}^{1} - \partial 2 Γ_{11}^{1} + Γ_{21}^{1} Γ_{11}^{1} + Γ_{21}^{2} Γ_{12}^{1} - Γ_{11}^{1} Γ_{21}^{1} - Γ_{11}^{2} Γ_{22}^{1} \\ = - \partial 1 Γ_{11}^{2} + \partial 2 Γ_{22}^{1} + Γ_{11}^{2} Γ_{22}^{1} + Γ_{22}^{1} Γ_{11}^{2} - Γ_{22}^{1} Γ_{11}^{2} - Γ_{11}^{2} Γ_{22}^{1} \\ = 0 . \end{array}

\begin{array}{l} R_{122}^{1} = - R_{212}^{1} & = \partial 1 Γ_{22}^{1} - \partial 2 Γ_{12}^{1} + Γ_{22}^{1} Γ_{11}^{1} + Γ_{22}^{2} Γ_{12}^{1} - Γ_{12}^{1} Γ_{21}^{1} - Γ_{12}^{2} Γ_{22}^{1} \\ = \partial 1 Γ_{22}^{1} + \partial 2 Γ_{11}^{2} - Γ_{22}^{1} Γ_{22}^{1} + Γ_{11}^{2} Γ_{11}^{2} - Γ_{11}^{2} Γ_{11}^{2} + Γ_{22}^{1} Γ_{22}^{1} \\ = \partial 1 Γ_{22}^{1} + \partial 2 Γ_{11}^{2} \\ = \frac{4}{{(∥ x ∥^{2} + 1)}^{2}} . \end{array}

\begin{array}{l} R_{121}^{2} = - R_{211}^{2} & = \partial 1 Γ_{21}^{2} - \partial 2 Γ_{11}^{2} + Γ_{21}^{1} Γ_{11}^{2} + Γ_{21}^{2} Γ_{12}^{2} - Γ_{11}^{1} Γ_{21}^{2} - Γ_{11}^{2} Γ_{22}^{2} \\ = - \partial 1 Γ_{22}^{1} - \partial 2 Γ_{11}^{2} - Γ_{11}^{2} Γ_{11}^{2} + Γ_{22}^{1} Γ_{22}^{1} - Γ_{22}^{1} Γ_{22}^{1} + Γ_{11}^{2} Γ_{11}^{2} \\ = - \partial 1 Γ_{22}^{1} - \partial 2 Γ_{11}^{2} \\ = - \frac{4}{{(∥ x ∥^{2} + 1)}^{2}} . \end{array}

\begin{array}{l} R_{122}^{2} = - R_{212}^{2} & = \partial 1 Γ_{22}^{2} - \partial 2 Γ_{12}^{2} + Γ_{22}^{1} Γ_{11}^{2} + Γ_{22}^{2} Γ_{12}^{2} - Γ_{12}^{1} Γ_{21}^{2} - Γ_{12}^{2} Γ_{22}^{2} \\ = - \partial 1 Γ_{11}^{2} + \partial 2 Γ_{22}^{1} + Γ_{22}^{1} Γ_{11}^{2} + Γ_{11}^{2} Γ_{22}^{1} - Γ_{11}^{2} Γ_{22}^{1} - Γ_{22}^{1} Γ_{11}^{2} \\ = 0 . \end{array}

We see that many of the coefficients are zero, and the non-zero ones are equal up to a sign.

From Exercise 3.8 we also have the coefficients of the metric in this chart. In particular, they form a diagonal matrix.

g_{i j} = \frac{4}{{(∥ x ∥^{2} + 1)}^{2}} δ_{i j} .

Then the other form of the curvature is

{Rm}_{ijkl} = g_{lm} R_{ijk}^{m} = \frac{4}{{(∥ x ∥^{2} + 1)}^{2}} R_{ijk}^{l} .

The non-zero coefficients are

{Rm}_{1221} = - {Rm}_{2121} = - {Rm}_{1212} = {Rm}_{2112} = \frac{16}{{(∥ x ∥^{2} + 1)}^{4}} .

These local expressions show us that the curvature tensor determines $n^{4}$ smooth functions ${Rm}_{ijkl}$ . However some symmetries are apparent already from the definition, such as $R (X, Y) Z = - R (Y, X) Z$ . Here are the others

Theorem 5.5 (Symmetries).

(i)

$Rm$ is antisymmetric in the first pair and last pair of entries:

Rm (X, Y, Z, W) = - Rm (Y, X, Z, W) = - Rm (X, Y, W, Z) .

(ii)

$Rm$ is symmetric under the exchange of the first and last pair:

Rm (X, Y, Z, W) = Rm (Z, W, X, Y) .

(iii)

$R$ has the following cyclic symmetry, called the first or algebraic Bianchi identity:

R (X, Y) Z + R (Z, X) Y + R (Y, Z) X = 0 .

Proof. (i) seems the logical place to start. We have already noted that antisymmetry in the first pair comes from the definition of $R$ .

R (X, Y) Z = \nabla_{X} \nabla_{Y} Z - \nabla_{Y} \nabla_{X} Z - \nabla_{[X, Y]} Z .

For antisymmetry in the last pair we first compute for $Z = W$ and use metric-compatibility

\begin{array}{l} g (R (X, Y) Z, Z) & = g (\nabla_{X} \nabla_{Y} Z, Z) - g (\nabla_{Y} \nabla_{X} Z, Z) - g (\nabla_{[X, Y]} Z, Z) \\ = X (g (\nabla_{Y} Z, Z)) - g (\nabla_{Y} Z, \nabla_{X} Z) - Y (g (\nabla_{X} Z, Z)) + g (\nabla_{X} Z, \nabla_{Y} Z) \\ - \frac{1}{2} [X, Y] (g (Z, Z)) \\ = X (g (\nabla_{Y} Z, Z)) - Y (g (\nabla_{X} Z, Z)) - \frac{1}{2} [X, Y] (g (Z, Z)) \\ = X (\frac{1}{2} Y (g (Z, Z))) - Y (\frac{1}{2} X (g (Z, Z))) - \frac{1}{2} [X, Y] (g (Z, Z)) = 0 . \end{array}

Any bilinear function that is zero on $(Z, Z)$ is antisymmetric:

\begin{array}{l} 0 & = g (R (X, Y) Z + W, Z + W) - g (R (X, Y) Z - W, Z - W) \\ = 2 g (R (X, Y) Z, W) + 2 g (R (X, Y) W, Z) . \end{array}

Next we prove (iii). The follows from a long calculation, but one that can be shortened using the following piece of notation from Petersen:

S T (X, Y, Z) = T (X, Y, Z) + T (Z, X, Y) + T (Y, Z, X) .

\begin{array}{l} S R (X, Y) Z & = S \nabla_{X} \nabla_{Y} Z - S \nabla_{Y} \nabla_{X} Z - S \nabla_{[X, Y]} Z \\ = S \nabla_{Z} \nabla_{X} Y - S \nabla_{Z} \nabla_{Y} X - S \nabla_{[X, Y]} Z \\ = S \nabla_{Z} (\nabla_{X} Y - \nabla_{Y} X) - S \nabla_{[X, Y]} Z \\ = S (\nabla_{Z} [X, Y] - \nabla_{[X, Y]} Z) \\ = S [Z, [X, Y]], \end{array}

using twice that

\nabla

is torsion-free. This expression is always zero, a fact known as the Jacobi identity, which is easily proved using a chart:

S [Z, [X, Y]] = [Z, [X, Y]] + [X, [Y, Z]] + [Y, [Z, X]] = 0 .

Now (ii) follows from (i) and (iii)

\begin{array}{l} Rm (X, Y, Z, W) & = - Rm (Z, X, Y, W) - Rm (Y, Z, X, W) \\ = Rm (Z, X, W, Y) + Rm (Y, Z, W, X) \\ = Rm (W, Z, X, Y) - Rm (X, W, Z, Y) - Rm (W, Y, Z, X) - Rm (Z, W, Y, X) \\ = 2 Rm (Z, W, X, Y) + Rm (X, W, Y, Z) + Rm (W, Y, X, Z) \\ = 2 Rm (Z, W, X, Y) - Rm (X, Y, Z, W) . & □ \end{array}

Exercise 5.6. Show that these symmetries restrict the number of independent coefficients of $Rm$ to $n^{2} (n^{2} - 1) ∕ 12$ . In particular, for $n = 2$ there is essentially only one coefficient.

The algebraic Bianchi identity was first written down by Ricci. However it is so named because it looks similar to a cyclic identity discovered by Bianchi. It’s general form requires additional definitions of a kind we have avoided, so we give a special form that is suitable to applications.

Theorem 5.7. For any $p \in M$ take the normal chart centered at $p$ such that $g_{ij} (p) = δ_{ij}$ . The second or differential Bianchi identity states that at $p$ :

∂m {Rm}_{ijkl} + ∂k {Rm}_{ijlm} + ∂l {Rm}_{ijmk} = 0 .

Proof. In the normal chart at $p$ this point is the origin. Due to Lemma 4.37 we know in the normal chart at $p$ that $g_{ij} (p) = δ_{ij}$ and $∂k g_{ij} (p) = 0$ . In the proof of that lemma it was shown, and it follows easily from $∂k g_{ij} (0) = 0$ , that $Γ_{ij}^{k} (0) = 0$ . Hence

\begin{array}{l} (∂m {Rm}_{ijkl}) (0) & = (∂m (g_{\ln} R_{ijk}^{n})) (0) = 0 + δ_{\ln} (∂m R_{ijk}^{n}) (0) = \partial m R_{ijk}^{l} (0) \\ = ∂m (∂i Γ_{jk}^{l} - \partial j Γ_{ik}^{l} + Γ_{jk}^{m} Γ_{im}^{l} - Γ_{ik}^{m} Γ_{jm}^{l}) (0) \\ = ∂m∂i Γ_{jk}^{l} (0) - \partial m∂j Γ_{ik}^{l} (0) + (0 + 0) - (0 + 0) . & (5.8) \end{array}

Taking the cyclic permutations of $ijm$ proves

(∂m {Rm}_{ijkl}) (0) + (∂i {Rm}_{jmkl}) (0) + (∂j {Rm}_{mikl}) (0) = 0 .

And swapping the first and last pair gives the identity as stated in the theorem. □

We motivated the introduction of the curvature tensor by asking whether a space was locally isometric to euclidean space, specifically whether there existed a parallel vector field in a neighbourhood of a point. Historically this question was approached through the lens of coordinate transformation: does there exist a coordinate transformation that makes the metric coefficients constant and equal to $δ_{ij}$ ? The relevance of normal coordinates, where $g_{ij} (p) = δ_{ij} + 0 + O (∥ p ∥^{2})$ , to the question now seems obvious. What Riemann found was that there was an obstacle in the second order of the Taylor expansion that could not be removed. To see the relation between our definition and Riemann’s observation, start with the formula for $R_{ijk}^{l}$ in terms of the Christoffel coefficients and substitute in the expression for them in terms of the metric coefficients. Suppose you have a chart where $g_{ij} (p) = δ_{ij} + 0 + O (∥ p ∥^{2})$ . The normal chart has this property, but there may be others.

\begin{array}{l} 2 ∂i Γ_{jk}^{l} & = \partial i g^{n l} (\partial k g_{j n} + ∂j g_{k n} - ∂n g_{k j}) + g^{n l} \partial i (∂k g_{j n} + ∂j g_{k n} - ∂n g_{k j}) \\ 2 ∂i Γ_{jk}^{l} (0) & = 0 + \partial i∂k g_{j l} + ∂i∂j g_{k l} - ∂i∂l g_{k j} \\ 2 ∂j Γ_{ik}^{l} (0) & = \partial j∂k g_{i l} + ∂j∂i g_{k l} - ∂j∂l g_{k i} \\ R_{ijk}^{l} & = \partial i Γ_{jk}^{l} - \partial j Γ_{ik}^{l} + Γ_{jk}^{m} Γ_{im}^{l} - Γ_{ik}^{m} Γ_{jm}^{l} \\ R_{ijk}^{l} (0) & = \partial i Γ_{jk}^{l} (0) - \partial j Γ_{ik}^{l} (0) + 0 - 0 \\ = \frac{1}{2} (∂i∂k g_{j l} (0) - ∂i∂l g_{k j} (0) - ∂j∂k g_{i l} (0) + ∂j∂l g_{k i} (0)) . \end{array}

If in this coordinate chart all the second derivatives of $g$ also vanished at some point, then the right hand side would be zero. This implies $R (X, Y) Z = 0$ for all vectors $X, Y, Z \in T_{p} M$ . But curvature is defined independent of coordinates charts. If the curvature tensor is non-zero at some point $p$ in some directions, logically it is impossible in any chart for the Taylor series of $g_{ij}$ at $p$ to additionally vanish in the second order. In particular curvature is an obstruction to being locally euclidean.

As we saw in Theorem 5.5, the numerous symmetries of $R$ mean that there is a lot of redundancy in its coefficients. It makes sense therefore to ask if there is a way to distil the information of the curvature tensor into a simpler object. We provide two such simplifications now, and will look at a third in the last section.

Definition 5.9. For every point $p \in M$ and vectors $Y, Z \in T_{p} M$ , we consider the linear map $X \mapsto R (X, Y) Z$ from $T_{p} M$ to itself. The Ricci curvature $Ric (Y, Z)$ is the trace of this map. It is bilinear in $Y, Z$ so can be expressed nicely using coefficients

Ric (Y, Z) : = Y^{j} Z^{k} {Ric}_{jk} = Y^{j} Z^{k} δ_{l}^{i} R_{ijk}^{l} = Y^{j} Z^{k} g^{im} {Rm}_{ijkm} .

Likewise the scalar curvature $S$ is the trace (in the sense of bilinear forms) of the Ricci curvature with respect to the metric:

S = {tr}_{g} Ric = g^{jk} {Ric}_{jk} = g^{jk} g^{im} {Rm}_{ijkm} .

It may seem more natural to take the trace of $Z \mapsto R (X, Y) Z$ . However this is zero because $Rm$ is antisymmetric in the last pair. Likewise antisymmetry of $R$ in $X, Y$ means taking the trace of $X \mapsto R (Y, X) Z$ just gives a negative sign. The Ricci curvature is in fact a symmetric bilinear form:

{Ric}_{kj} = g^{im} {Rm}_{ikjm} = g^{im} {Rm}_{jmik} = g^{mi} {Rm}_{mjki} = g^{im} {Rm}_{ijkm} = {Ric}_{jk} .

Example 5.10 (Stereographic Projection). In Example 5.4 we computed the coefficients $R_{ijk}^{l}$ in the chart $U_{N}$ . Most were zero. Therefore the coefficients of the Ricci tensor in this chart are

\begin{array}{l} {Ric}_{jk} & = δ_{l}^{i} R_{ijk}^{l} = R_{1 jk}^{1} + R_{2 jk}^{2} \\ {Ric}_{11} & = R_{111}^{1} + R_{211}^{2} = R_{211}^{2} = \frac{4}{{(∥ x ∥^{2} + 1)}^{2}} \\ {Ric}_{12} & = R_{112}^{1} + R_{212}^{2} = 0 \\ {Ric}_{21} & = R_{121}^{1} + R_{221}^{2} = 0 \\ {Ric}_{22} & = R_{122}^{1} + R_{222}^{2} = R_{122}^{1} = \frac{4}{{(∥ x ∥^{2} + 1)}^{2}} . \end{array}

As expected, this is a symmetric matrix.

For the scalar curvature we need the the inverse of the matrix of the metric

g^{i j} = \frac{{(∥ x ∥^{2} + 1)}^{2}}{4} δ^{i j} .

Hence

\begin{array}{l} S & = g^{jk} {Ric}_{jk} = \frac{{(∥ x ∥^{2} + 1)}^{2}}{4} ({Ric}_{11} + {Ric}_{22}) = 2 . \end{array}

Not unreasonably, the scalar curvature for the sphere is at every point $2$ .

We will not go deeply into the theory of Ricci and scalar curvature, but we will mention some special cases of interest. Spaces with $Ric \equiv 0$ are called Ricci-flat. A slightly more general class of Riemannian manifolds are Einstein manifolds. These have the property that $Ric = λg$ for a function $λ : M \to ℝ$ . Since the Ricci curvature must be a symmetric bilinear form, this is more-or-less the simplest form it could take. Taking trace of both sides shows that for Einstein manifolds,

S = g^{jk} {Ric}_{jk} = λ g^{jk} g_{jk} = λ \sum_{k} δ_{k}^{k} = λ dim M .

Example 5.11 (Stereographic Projection). Observe that the sphere is an Einstein manifold with $λ \equiv 1$ as

{Ric}_{jk} = \frac{4}{{(∥ x ∥^{2} + 1)}^{2}} δ_{ij} = g_{ij} .

We also see that its scalar curvature is $S = λ dim 𝕊^{2} = 1 \times 2 = 2$ .

These are named for Einstein because the equation for the curvature of space-time in the theory of general relativity is

{Ric}_{ij} - \frac{1}{2} S g_{ij} = T_{ij},

where the right hand side is a function representing matter-energy. If you allow on the left hand side an additional ‘cosmological constant’

{Ric}_{ij} - \frac{1}{2} S g_{ij} + Λ g_{ij} = T_{ij},

then Einstein manifolds are models of a vacuum universe (no matter-energy). Einstein originally published his theory without a cosmological constant. At the time it was thought that the universe was static and eternal, and in a subsequent publication he argued for $Λ$ to permit this. A decade later, the observations of distant galaxies by Hubble showed that the universe was expanding. He would call this his “biggest blunder”, as trusting the simplicity of his original derivation would have meant another successful prediction of the theory. It seems a little mean naming this class of manifolds after a man’s biggest blunder.

Theorem 5.12 (Schur). On a connected Einstein manifold with dimension three or greater, the scalar curvature is constant.

Proof. The key to this proof is the differential Bianchi identity 5.7. In normal coordinates at a point $p$

(∂m {Ric}_{jk}) (p) = (∂m (g^{il} {Rm}_{ijkl})) (p) = 0 + δ^{il} (\partial m ({Rm}_{ijkl})) (p) .

On the other hand

(∂m (λ g_{jk})) (p) = (∂mλ) (p) δ_{jk} + 0 .

Putting this into the Bianchi identity gives (leaving evaluation at $p$ implicit)

\begin{array}{l} 0 & = ∂m {Rm}_{ijkl} - ∂k {Rm}_{ijml} + ∂l {Rm}_{ijmk} \\ 0 & = δ^{il} \partial m {Rm}_{ijkl} - δ^{il} \partial k {Rm}_{ijml} + δ^{il} \partial l {Rm}_{ijmk} \\ = ∂mλ δ_{jk} - ∂kλ δ_{jm} + δ^{il} \partial l {Rm}_{ijmk} \\ 0 & = δ^{jk} \partial mλ δ_{jk} - δ^{jk} \partial kλ δ_{jm} + δ^{jk} δ^{il} \partial l {Rm}_{ijmk} \\ = ∂mλ dim M - ∂mλ - δ^{il} δ^{jk} \partial l {Rm}_{jimk} \\ = ∂mλ (dim M - 1) - δ^{il} \partial lλ δ_{im} \\ = ∂mλ (dim M - 2) . \end{array}

If $dim M > 2$ then this forces $λ$ to have zero derivative in every direction. Morever, we can do this for every point. Therefore $λ = {(dim M)}^{- 1} S$ is constant. □

5.2 Hypersurfaces

A hypersurface is the embedding of an $n$ -dimensional manifold $M$ in an $(n + 1)$ -dimensional manifold $N$ (codimension one). Additionally, assume that both a Riemannian manifolds and that the embedding is Riemannian. Alternatively, you may begin with a Riemannian manifold $N$ and any manifold $M$ and then put the pullback metric on $M$ , which will make the embedding Riemannian. Recall Definition 3.59 and Theorem 3.60. They explain how the Levi-Civita connection $\nabla^{⊤}$ of $M$ can be calculated using the Levi-Civita connection $\nabla^{N}$ ; essentially $\nabla^{⊤}$ is the projection of $\nabla^{N}$ .

We can also ask what information is lost by this projection.

Definition 5.13. The second fundamental form of $M$ in $N$ is the function

I I (X, Y) = \nabla_{X}^{N} Y - \nabla_{X}^{⊤} Y .

This is a function from tangent vector fields on $M$ to a normal vector field on $M$ .

It is tensorial despite being defined in terms of derivatives.

Theorem 5.14 (Codazzi-Mainardi). The second fundamental form $I I$ is a symmetric $C^{\infty}$ -bilinear function.

Proof. Consider the antisymmetric part of $I I$ . For tangent vector fields $X, Y$ to $M$ , we compute

\begin{array}{l} I I (X, Y) - I I (Y, X) = \nabla_{X}^{N} Y - \nabla_{X}^{⊤} Y - \nabla_{Y}^{N} X + \nabla_{Y}^{⊤} X = T^{N} (X, Y) - T^{⊤} (X, Y) = 0, \end{array}

where $T^{N}$ and $T^{⊤}$ are torsion. But Levi-Civita connections are torsion-free. This proves the symmetry.

By definition covariant derivatives are $C^{\infty}$ -linear in the direction. Hence $I I (X, Y)$ is $C^{\infty}$ -linear in $X$ . But by symmetry it is also $C^{\infty}$ -linear in $Y$ . □

We should explain the relation between this new definition of the second fundamental form and the definition of Section 1.5. In the case that $M$ is a hypersurface in $N$ , there is an up-to-sign unique unit normal vector field $ν$ of $M$ .² In this case the signed length of the second fundamental form can be computed as

h (X, Y) = g (I I (X, Y), ν) = g (\nabla_{X}^{N} Y, ν) .

The other ingredient that we need is Meusnier’s theorem 1.29. The theorem tells us for a arc-length parameterised curve that $h (α^{'}, α^{'})$ is equal to the normal curvature

κ_{n} = α^{″} \cdot ν = g (α^{″}, ν) = g (\nabla_{α^{'}} α^{'}, ν),

since for $ℝ^{3}$ the metric is the dot product and the covariant derivative is just the usual directional derivative. Symmetric bilinear forms are determined on their diagonal, so the new definition is a generalisation of the old to submanifolds that are not hypersurfaces.

Notice that the Gauss formula only applies to tangent vector fields of $M$ . We can also ask about the derivative of a vector field $ν$ orthogonal to $M$ . For any vector field $Y$ tangent to $M$ , we have that $g (Y, ν) = 0$ and thus

\begin{matrix} 0 = X (g (Y, ν)) = g (\nabla_{X}^{N} Y, ν) + g (Y, \nabla_{X}^{N} ν) = g (\nabla_{X}^{⊤} Y + I I (X, Y), ν) + g (Y, \nabla_{X}^{N} ν) \\ \Rightarrow g (Y, \nabla_{X}^{N} ν) = - g (I I (X, Y), ν) . \end{matrix}

The above is called the Weingarten formula. To understand the component of $\nabla_{X}^{N} ν$ perpendicular to $M$ , we can write $ν = ∥ ν ∥ \hat{ν}$ and apply the product rule:

\nabla_{X}^{N} ν = X (∥ ν ∥) \hat{ν} + ∥ ν ∥ \nabla_{X}^{N} \hat{ν} .

Further

0 = X (1) = X (g (\hat{ν}, \hat{ν})) = 2 g (\hat{ν}, \nabla_{X}^{N} \hat{ν})

shows that $\nabla_{X}^{N} \hat{ν}$ is tangent to $M$ . By considering all $Y \in T_{p} M$ in the Weingarten formula, $\nabla_{X}^{N} \hat{ν} \in T_{p} M$ is completely determined. In sum this tells us that the covariant derivative of any vector field in $N$ in a tangent direction of $M$ can be calculated with $\nabla^{⊤}$ and $I I$ alone. In the special case of a hypersurface, the right hand side is $- h (X, Y)$ and this is the analogue of the working following Exercise 1.30.

From these formulae follows a particularly nice formula relating the Riemann curvature tensors of $M$ and $N$ . Originally the Levi-Civita connection was simply defined to the be tangent connection, so this formula was called the Gauss formula. We first use the fact that $W$ is tangent to $M$ to kill off as many normal components as possible, then apply the Weingarten formula

\begin{array}{l} {Rm}^{N} (X, Y, Z, W) & = g (R^{N} (X, Y) Z, W) = g (\nabla_{X}^{N} \nabla_{Y}^{N} Z - \nabla_{Y}^{N} \nabla_{X}^{N} Z - \nabla_{[X, Y]}^{N} Z, W) \\ = g (\nabla_{X}^{T} (\nabla_{Y}^{⊤} Z + I I (Y, Z)) - \nabla_{Y}^{T} (\nabla_{X}^{⊤} Z + I I (X, Z)) - \nabla_{[X, Y]}^{⊤} Z, W) \\ = {Rm}^{M} (X, Y, Z, W) + g (W, \nabla_{X}^{T} I I (Y, Z)) - g (W, \nabla_{Y}^{T} I I (X, Z)) \\ = {Rm}^{M} (X, Y, Z, W) - g (I I (X, W), I I (Y, Z)) + g (I I (Y, W), I I (X, Z)) . \end{array}

For surfaces in euclidean space, the Riemann curvature form ${Rm}^{ℝ^{3}}$ vanishes and

g (I I (X, W), I I (Y, Z)) = g (h (X, W) ν, h (Y, Z) ν) = h (X, W) h (Y, Z),

as well as $g (I I (Y, W), I I (X, Z)) = h (Y, W) h (X, Z)$ , giving us a relation between the curvature tensor of $M$ and the second fundamental form.

Theorem 5.15 (Theorema Egregium). Let $M$ be a Riemannian embedded surface in $ℝ^{3}$ . Let $X, Y \in T_{p} M$ be orthonormal vectors. The Gauss curvature $K$ is related to the curvature tensor by

K = Rm (X, Y, Y, X) .

Therefore the Gauss curvature of a surface is an isometry invariant.

There exists a modification of this formula for arbitrary tangent vectors, but it amounts to applying Gram-Schmidt orthogonalisation to the vectors $X, Y$ . Therefore we work exclusively with this more elegant form.

Proof. If $X, Y \in T_{p} M$ are orthonormal, then we can use them as a basis of $T_{p} M$ . With respect to this basis

h = (\begin{matrix} h_{11} & h_{12} \\ h_{21} & h_{22} \end{matrix}) = (\begin{matrix} h (X, X) & h (X, Y) \\ h (Y, X) & h (Y, Y) \end{matrix}) .

Using Lemma 1.38 and the Gauss formula for curvature gives us therefore that

K = h_{11} h_{22} - h_{12}^{2} = h (Y, Y) h (X, X) - h (X, Y) h (Y, X) = {Rm}^{M} (Y, X, X, Y) . □

Example 5.16 (Stereographic Projection). We know the Riemann curvature of $𝕊^{2}$ in the $U_{N}$ chart already, so we should be able to use that to calculate the Gauss curvature. First we need orthonormal vectors at every point of the chart $U_{N}$ . The coordinate vectors are orthogonal to one another, but not unit length. Therefore take

X = \frac{∥ x ∥^{2} + 1}{2} \partial 1, Y = \frac{∥ x ∥^{2} + 1}{2} \partial 2 .

Then

K = Rm (X, Y, Y, X) = {(\frac{∥ x ∥^{2} + 1}{2})}^{4} {Rm}_{1221} = 1 .

As we observed in Example 1.35, the normal curvature of a sphere at every point and in every direction is the inverse of its radius, here $1$ . Hence the principal curvatures are $1$ and directly from Definition 1.37 we see that the Gauss curvature is $1$ , in agreement with the above calculation.

Definition 5.17. If $M$ is a 2-dimensional Riemannian manifold, we define the Gauss curvature $K (p) = Rm (X, Y, Y, X)$ for any orthonormal basis $X, Y$ of $T_{p} M$ .

Exercise 5.18. Implicit in this definition is the claim that this quantity is independent of the choice of orthonormal basis of $T_{p} M$ . Prove this claim.

Example 5.19 (Hyperbolic Plane). Although we have not exhibited the hyperbolic plane Riemannian immersed in a euclidean space (and such an immersion does not exist), we can use this definition to find its Gaussian curvature. The first thing that we need is an orthonormal basis at each point. We know that $\partial_{x}$ and $\partial_{y}$ are orthogonal to one another, and by adjusting their lengths we have $X = y \partial_{x}$ , $Y = y \partial_{y}$ . Then

\begin{array}{l} K & = Rm (y \partial_{x}, y \partial_{y}, y \partial_{y}, y \partial_{x}) = y^{4} g_{1 m} R_{122}^{m} = y^{2} R_{122}^{1} \\ = y^{2} (\partial_{1} Γ_{22}^{1} - \partial_{2} Γ_{12}^{1} + Γ_{22}^{m} Γ_{1 m}^{1} - Γ_{12}^{m} Γ_{2 m}^{1}) \\ = y^{2} (0 - \partial_{y} (- y^{- 1}) + 0 + {(- y^{- 1})}^{2} - {(- y^{- 1})}^{2} - 0) = - 1 . \end{array}

Thus the hyperbolic plane is a surface that has constant negative Gaussian curvature, in comparison to the sphere that has constant positive Gaussian curvature.

5.3 Sectional Curvature

The Ricci and scalar curvatures are natural simplifications of the curvature tensor and have nice properties, but we are yet to see any geometric intuition for these so-called curvatures. In this section we finally address this question.

In Chapter 1 we defined the curvature of a surfaces by reference to the curvature of curves in that surface. In particular the normal curvature was important, which we now understand is the curvature of a geodesic. Likewise we can define a type of curvature using surfaces within our manifold. These surfaces should be special in some way so that their curvature reflects the curvature of the manifold.

Definition 5.20. For any point $p \in M$ let us use the normal chart at $p$ . Given a pair of orthonormal vectors $X, Y \in T_{p} M$ , they span a plane in the chart, called the plane section $P$ . The sectional curvature $K (P) = K (X, Y)$ of $M$ at $p$ is defined to be the Gauss curvature at $p$ of the plane section.

Example 5.21. Consider a two dimensional manifold. Then in the chart there is only one possible plane, namely the chart itself. In this case the definitions are somewhat trivial and the sectional curvature is just the Gauss curvature of the manifold from Definition 5.17.

In general it is important to use normal coordinates to define the plane section. In some sense, this plane is constructed out of geodesics. We first show that the second fundamental form of the plane section in $M$ is zero at $p$ . Choose any vector $v \in T_{p} P$ . Let $γ_{v}$ be the geodesic through $p$ in the direction of $v$ . The definition of the normal chart is that this is a ray, hence it lies in $P$ . By the Gauss formula,

0 = \nabla_{γ^{'}}^{M} γ^{'} = \nabla_{γ^{'}}^{P} γ^{'} + I I (γ^{'}, γ^{'}) .

The tangent and normal directions are linearly independent, so both terms on the right must vanish. In particular $0 = I I (γ^{'} (0), γ^{'} (0)) = I I (v, v)$ for all $v \in T_{p} M$ . Since $I I$ is symmetric it must vanish at $p$ . Together with Theorema Egregium 5.15 and the Gauss formula for curvature, we have proved

Theorem 5.22. If $X, Y \in T_{p} M$ are orthonormal then the sectional curvature of $M$ at $p$ is

K (X, Y) = {Rm}^{P} (X, Y, Y, X) = {Rm}^{M} (X, Y, Y, X) .

It may seem as though the sectional curvature is just another simplification of the full curvature tensor. But this is not the case.

Theorem 5.23. The sectional curvatures uniquely determine the curvature tensor. Moreover if at some point $p \in M$ the sectional curvature is constant $K (X, Y) = κ$ for all orthonormal $X, Y \in T_{p} M$ then the curvature tensor is given by

Rm (X, Y, Z, W) = κ (g (X, W) g (Y, Z) - g (X, Z) g (Y, W)) .

Proof. Suppose two curvature tensors have the same sectional curvatures at a point $p$ . Then their difference $R$ is a quadlinear map on $T_{p} M$ with the same symmetries as a curvature tensor. It is sufficient to prove $R (X, Y, Z, W) = 0$ for $X, Y, Z, W$ elements of an orthonormal basis for $T_{p} M$ . We now use the symmetries:

\begin{array}{l} 0 & = R (X + W, Y, Y, X + W) \\ = R (X, Y, Y, X) + R (X, Y, Y, W) + R (W, Y, Y, X) + R (W, Y, Y, X) \\ = 0 + 2 R (X, Y, Y, W) + 0, \end{array}

and

\begin{array}{l} 0 & = R (X, Y + Z, Y + Z, W) \\ = R (X, Y, Y, W) + R (X, Y, Z, W) + R (X, Z, Y, W) + R (X, Z, Z, W) \\ = 0 + R (X, Y, Z, W) + R (X, Z, Y, W) + 0 . \end{array}

In other words, in addition to being antisymmetric in the first and last pair, $R$ is also antisymmetric in the middle pair. Finally we apply the algebraic Bianchi identity:

\begin{array}{l} 0 & = R (X, Y, Z, W) + R (Z, X, Y, W) + R (Y, Z, X, W) \\ = R (X, Y, Z, W) - R (X, Z, Y, W) - R (Y, X, Z, W) \\ = R (X, Y, Z, W) + R (X, Y, Z, W) + R (X, Y, Z, W) . \end{array}

Hence $R (X, Y, Z, W) = 0$ as required.

For the second claim, we observe that if a tensor has the symmetries of a curvature tensor and has constant sectional curvature then from the preceding argument it must be the unique such curvature tensor. One can readily check that the given formula has the necessary symmetries. For orthonormal $X, Y \in T_{p} M$ we have

K (X, Y) = Rm (X, Y, Y, X) = κ (g (X, X) g (Y, Y) - g (X, Y) g (Y, X)) = κ . □

Example 5.24 (Stereographic Projection). We see now that the curvature of the sphere $𝕊^{2}$ has exactly this form. The Gauss curvature at every point is $1$ . Then

{Rm}_{ijkl} = Rm (∂i, ∂j, ∂k, ∂l) = 1 (g_{il} g_{jk} - g_{ik} g_{jl}) = {(\frac{4}{{(∥ x ∥^{2} + 1)}^{2}})}^{2} (δ_{il} δ_{jk} - δ_{ik} δ_{jl}) .

We had in Theorem 5.12 that Einstein manifolds, which have a special form of the Ricci curvature, have constant scalar curvature. There is a similar result in the case of sectional curvature.

Theorem 5.25 (Schur). Suppose that a connected manifold of dimension three or more has pointwise constant sectional curvature. This means there is a function $κ : M \to ℝ$ such that for all orthonormal $X, Y \in T_{p} M$ we have $K (X, Y) = κ (p)$ . Then $κ$ is constant.

Proof. At any point $p \in M$ we can use a normal chart. In this chart

∂m Rm (∂i, ∂j, ∂k, ∂l) |_{p} = ∂m (κ (g_{il} g_{jk} - g_{ik} g_{jl})) |_{p} = ∂mκ (p) (δ_{il} δ_{jk} - δ_{ik} δ_{jl})

The differential Bianchi identity at $p$ says

\begin{array}{l} 0 & = ∂m {Rm}_{ijkl} |_{p} + ∂k {Rm}_{ijlm} |_{p} + ∂l {Rm}_{ijmk} |_{p} \\ = ∂mκ (p) (δ_{il} δ_{jk} - δ_{ik} δ_{jl}) + ∂kκ (p) (δ_{im} δ_{jl} - δ_{il} δ_{jm}) + ∂lκ (p) (δ_{ik} δ_{jm} - δ_{im} δ_{jk}) . \end{array}

In particular, choose $l = i$ , $k = j$ , and $m, i, j$ distinct. It is possible for three indices to be distinct because the dimension is at least three.

\begin{array}{l} 0 = ∂mκ (p) (1 - 0) + ∂kκ (p) (0 - 0) + ∂lκ (p) (0 - 0) = ∂mκ (p) . \end{array}

Thus $κ$ has zero derivative at $p$ , and hence at every point. It must be constant. □

Spaces with constant sectional curvature are called space forms. In every dimension the space forms are the euclidean space, the sphere (with scalings), and hyperbolic plane (with scalings), this classification is due to Killing and Hopf. Thus these three spaces, which have been the main focus of our examples, are in terms of Riemannian geometry the nicest spaces. We have in a previous section mentioned the study of manifolds with special curvature.

Another direction of research is to impose a bound on the curvature. For example, a theorem of Myers states that if a Riemannian manifold is complete as a metric space and the infimum of its sectional curvatures is positive, then it is compact. Or a result of Synge says that a compact orientable even-dimensional Riemannian manifold with positive sectional curvatures must be simple connected.

Finally, both theorems of Schur rely on the dimension being three or greater. This is not a limitation of the proof: Riemann surfaces really are special. There is only one intrinsic curvature for them, the Gauss curvature, and integral of the Gauss curvature over the whole manifold is closely connected with its topology (the Gauss-Bonnet theorem). Moreover every Riemann surface is conformally equivalent to a Riemann surface with constant curvature, which then are the three space forms. More can be learnt about the theory of Riemann surfaces in Complex Analysis II (Funktiontheorie II).

¹Can you imagine calculating this without the summation convention?

²If $M$ is orientable this field is global. If $M$ is non-orientable then this can only be chosen locally.

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Chapter 5Curvature

5.1 Symmetries and Identities

5.2 Hypersurfaces

5.3 Sectional Curvature

Chapter 5
Curvature