Last time we have seen that in any Riemannian manifold $(M,\g)$ there is a unique connection $\nabla$, called the Levi-Civita connection, that is symmetric and compatible with the metric, and which gives rise to the Riemannian geodesics and the exponential map with nice naturality properties, i.e., preserved under isometries.

We now discuss the question of whether a Riemannian manifold is locally isometric to the Euclidean space. This will give rise to the notion of curvature, which measures the deviation from being a flat space.

As before, this is following Lee’s Riemannian Manifold, $\S7$.


Local invariants

Recall that Riemannian geometry is concerned with properties that are preserved under isometries, which are the isomorphisms in the category of Riemannian manifolds. We are thus interested in the question of whether there are nontrivial local invariants that are preserved under isometries, i.e., whether we can distinguish some Riemannian manifolds, or whether they are all locally equivalent.

As Lee notes, many structures in differential geometry do not have local invariants. For example, every nonvanishing vector field can be written locally as $v = \partial / \partial x^1$ in suitable coordinates, so they are all locally equivalent. As another example, every $1$-dimensional Riemannian manifold is locally isometric to $\R$ by the inverse of the unit speed parameterization. Finally, all symplectic forms are locally equivalent since by Darboux’s theorem they can be written in canonical form as $\sum dx^i \wedge dy^i$.

In contrast, the situation with Riemannian manifold is rather different. We know that at each point $p \in M$ we can choose a local coordinate such that the metric at $p$ is the Euclidean metric $g_{ij} = \delta_{ij}$ (we can do this via the normal coordinates, which we did not discuss last time, but can be constructed from the exponential map, see Lee, Proposition 5.11). However, this does not mean that we can make the metric on a neighborhood of $p$ the Euclidean metric, and indeed we cannot always guarantee that.

For example, in Lee, Problem 5-4 it is shown that the Euclidean plane $\R^2$ is not locally isometric to the $2$-sphere $S^2 \subset \R^3$. The issue, as Lee explains, is that every tangent vector in the plane (and indeed in $\R^n$) can be extended to a parallel vector field, so any Riemannian manifold that is locally isometric to $\R^2$ must have the same property locally, but this is not true for the sphere.


Flatness criterion

Concretely, given a Riemannian $2$-manifold $M$ and a tangent vector $w_p \in T_pM$ at a point $p \in M$, we can attempt to construct a parallel extension of $w_p$ as follows. Choose a local coordinate $(x^1,x^2)$ centered at $p$. We first parallel translate $w_p$ along the $x^1$-axis, then parallel translate the result along the coordinate lines parallel to the $x^2$-axis. This yields a vector field $w$ that by construction is parallel along every $x^2$-coordinate line, i.e., $\nabla_{\partial_2} w = 0$. We also know $w$ is parallel along the $x^1$-axis, i.e., $\nabla_{\partial_1} w = 0$ when $x^2 = 0$, but it is not clear whether $w$ is parallel along every $x^1$-coordinate line, i.e., whether $\nabla_{\partial_1} w = 0$ always.

To show $\nabla_{\partial_1} w = 0$, by the uniqueness of parallel translates, it suffices to show that $\nabla_{\partial_2} \nabla_{\partial_1} w = 0$. When $M = \R^2$ with the Euclidean metric, this is true because we in fact have

and we already know that $\nabla_{\partial_2} w = 0$. Indeed, by direct computation, the above is equal to $\partial_2 \partial_1 w^k \partial_k = \partial_1 \partial_2 w^k \partial_k$, since ordinary second partial derivatives commute. This says that the construction above is the same as if we are parallel translating along $x^2$ first, and then along $x^1$.

However, the relation above fails to hold for an arbitrary Riemannian metric, in particular on the sphere. As Lee explains (p. 117): “It is precisely the noncommutativity of such second covariant derivatives that forces this construction to fail on the sphere. Lurking behind this noncommutativity is the fact that the sphere is ‘curved’.” See also Susskind’s General Relativity, Lecture 3 (starting at 1:21:45) for another explanation of this phenomenon on the rounded cone.

More generally, we can examine the quantity $\nabla_u \nabla_v w - \nabla_v \nabla_u w$ for arbitrary vector fields $u,v,w$. On $M = \R^n$ with the Euclidean metric, we have $\nabla_u \nabla_v w = \nabla_u (vw^k \partial_k) = uvw^k \partial_k$, and similarly for $\nabla_v \nabla_u w = vuw^k \partial_k$, so their difference is $(uvw^k - vuw^k) \partial_k = [u,v] (w^k \partial_k) = \nabla_{[u,v]} w$. Our preceding discussion above is when $u = \partial_1$ and $v = \partial_2$, in which case their commutator vanishes, $[\partial_1,\partial_2] = 0$.

Thus, we have the following relation for arbitrary vector fields $u,v,w$ on $\R^n$:

By the naturality of the Levi-Civita connection, this must also hold for any Riemannian manifold that is locally isometric to $\R^n$. This is the flatness criterion.


Riemann curvature

Let $(M,\g)$ be a Riemannian manifold. Motivated by the flatness criterion above, we define the (Riemann) curvature endomorphism to be the map

given by

We can show that $R$ is a $\binom{3}{1}$-tensor field, which means it is multilinear over $C^\infty(M)$. Then we can write $R$ in terms of local coordinates $(x^i)$ using one upper (contravariant) and three lower (covariant) indices:

where the coefficients ${R_{ijk}}^l$ are defined by

We can also define the (Riemann) curvature tensor as the covariant $4$-tensor field

obtained by “lowering” the last index using the metric, so it acts on vector fields by

and in coordinates we have

where the coefficients are

(Here $\phantom{.}^\flat$ is the flat operator in the musical isomorphism that lowers an index, just like how it lowers the pitch in music.)

What is nice about the Riemann curvature endomorphism and curvature tensor is that they are local isometry invariants (Lee, Lemma 7.2). This means if $\varphi \colon (M, \g) \to (\tilde M, \tilde \g)$ is a local isometry, then

and


Flat manifolds

We say a Riemannian manifold $(M,\g)$ is flat if it is locally isometric to Euclidean space, that is, if every point $p \in M$ has a neighborhood that is isometric to an open set in $\R^n$ with the Euclidean metric. A qualitative geometric meaning of the curvature tensor is that it is precisely the obstruction to being flat. More precisely, we have the following result (Lee, Theorem 7.3):

Theorem: A Riemannian manifold is flat if and only if its curvature tensor vanishes identically.

We have seen above that the Euclidean metric satisfies the flatness criterion, so its curvature tensor vanishes. So if a Riemannian manifold is flat, then its curvature tensor, which is the pullback of the curvature tensor of the Euclidean metric, also vanishes.

Conversely, suppose a Riemannian manifold $(M,\g)$ has a vanishing curvature tensor, so its curvature endomorphism vanishes as well, which means the flatness criterion above holds. We begin by showing that $\g$ admits a parallel orthonormal frame in a neighborhood of any point.

Let $p \in M$, and choose any orthonormal basis $(E_1|_p, \dots, E_n|_p)$ for $T_pM$. Let $(x^i)$ be any coordinates (for example, the normal coordinates) centered at $p$ such that $E_i|_p = \partial_i$. We parallel translate each $E_i|_p$ successively along the $x^1$-axis, then the $x^2$-axis, and so on until the $x^n$-axis. This gives $n$ vector fields $(E_1,\dots,E_n)$ on a neighborhood of $p$, which form an orthonormal frame since parallel translation preserves inner products. Then we can use the flatness criterion, similar to our preceding discussion above, to show inductively that they are parallel (see Lee, p. 120).

Finally, because the Levi-Civita connection is symmetric, we have

which means the vector fields $(E_1,\dots,E_n)$ form a commuting orthonormal frame. Then by the “normal form for commuting vector fields”, we can find coordinates $(y^i)$ on a (possibly smaller) neighborhood of $p \in M$ such that $E_i = \partial / \partial y^i$. Thus we are done, since in any such coordinates the metric is given by $\g_{ij} = \g(\partial_i,\partial_j) = \g(E_i,E_j) = \delta_{ij}$, since $(E_1,\dots,E_n)$ are orthonormal, so the map $y = (y^1,\dots,y^n)$ is the desired local isometry between a neighborhood of $p$ and an open subset of the Euclidean space.


Symmetries of the curvature tensor

The curvature tensor has many symmetries, including the following (Lee, Proposition 7.4). First, from the definition, it is clear that the curvature tensor is skew-symmetric in the first two arguments:

or in components, $R_{ijkl} = -R_{jikl}$.

Second, by the compatibility with the metric, which recall means the Levi-Civita connection satisfies the product rule with respect to the inner product, we can show that $Rm(u,v,w,w) = 0$ always. Then by expanding $Rm(u,v,w+z,w+z) = 0$, we find that the curvature tensor is also skew-symmetric on the last two arguments:

or in components, $R_{ijkl} = -R_{ijlk}$.

Third, by the symmetry of the Levi-Civita connection, which recall means the torsion vanishes, we can show that:

where the last equality follows from Jacobi identity. This implies that when we cyclically permute the first three arguments the sum is zero:

or in components, $R_{ijkl} + R_{jikl} + R_{kijl} = 0$. This is the algebraic (or first) Bianchi identity.

Fourth, by writing the algebraic Bianchi identity four times, and using the first two properties above, we find that the curvature tensor stays the same if we swap the first two and the last two arguments:

or in components, $R_{ijkl} = R_{klij}$. Using the four properties above, we can also show that the three-term sum obtained by cyclically permuting any three indices of $Rm$ is also zero.

Fifth, the total covariant derivative of $Rm$ satisfies the following differential (or second) Bianchi identity:

or in components, $R_{ijkl;m} + R_{ijlm;k} + R_{ijmk;l} = 0$. Recall that the total covariant derivative is defined by $\nabla Rm(u,v,x,y,z) = \nabla_z Rm(u,v,x,y)$. We prove the identity above by working in normal coordinates, and choosing the vectors $u,v,x,y,z$ to be the coordinate basis vectors, which simplifies the computation since their commutators and covariant derivatives vanish identically at the point. The rest follows from direct calculation and using the compatibility of the Levi-Civita connection with the metric.


Ricci and scalar curvatures

We can “summarize” the curvature endomorphism by taking its trace along the first and last indices. The result is a covariant $2$-tensor called the Ricci curvature (or Ricci tensor), denoted by $Rc$ (or $Ric$):

where the coefficients $R_{ij} = R_{ji}$ are symmetric and are given by:

We can summarize this further by taking the trace of $Rc$, where we first need to raise the last index using the other musical isomorphism, the sharp operator, so $Rc^{\sharp}$ has components ${R_i}^k = g^{jk}R_{ij}$. Upon taking trace, we obtain the scalar curvature:

Then we can show (Lee, Lemma 7.7) that the covariant derivatives of the Ricci and scalar curvatures satisfy the following contracted Bianchi identity:

where $\textrm{div}$ is the divergence operator. In components, this is ${R_{ij;}}^j = \frac{1}{2} S_{;i}$.

A Riemannian metric is called an Einstein metric if its Ricci tensor is a scalar multiple of the metric:

at each point, for some function $\lambda \in C^\infty(M)$. By taking traces on both sides and noting that the trace of $\g$ is equal to $\g_{ij} \g^{ji} = \delta_i^i = n$, the dimension of $M$, we find that $\lambda = \frac{1}{n} S$, so the Einstein metric condition can also be written as:

This implies that if $\g$ is an Einstein metric on a connected manifold of dimension $n \ge 3$, then its scalar curvature is constant (Lee, Proposition 7.8). This follows by taking the covariant derivative of each side above, noting that the covariant derivative of the metric is zero, then taking the trace of the result and using the contracted Bianchi identity to obtain $\frac{1}{2} S_{;i} = \frac{1}{n} S_{;i}$. So when $n > 2$, this implies $S_{;i} = 0$. Since $S_{;i}$ is the component of $\nabla S = dS$ and since $M$ is connected, this implies that $S$ is constant.

Finally, it is interesting to note that Hilbert showed that Einstein metrics are the critical points for the total scalar curvature functional

on the space of all metrics on $M$ with fixed volume, so Einstein metrics are optimal in a certain sense. As Lee explains (p. 126), “they form an appealing higher-dimensional analogue of the metrics of constant Gaussian curvature on $2$-manifolds.”

This also has to do with the Einstein’s field equation $Rc - \frac{1}{2}S\g = T$ from general relativity, which also involves the stress-energy tensor $T$. When $T = 0$, this reduces to the vacuum Einstein field equation $Rc = \frac{1}{2}S\g$, which implies $S = 0$ since this holds in dimension $n = 4$ (spacetime with the Lorentz metric). Thus, the vacuum Einstein equation is equivalent to $Rc = 0$, which means $\g$ is a (pseudo-Riemannian) Einstein metric $Rc = \lambda \g$ with $\lambda = 0$. But interestingly, as Lee notes at the end of $\S7$: “There is no direct connection between the physicists’ version of the Einstein equation and the mathematicians’ version. Mathematically, Einstein metrics are interesting not because of their relation to physics, but because of their potential applications to uniformization in higher dimension.” But I think the physicists’ version is also very interesting because of how it describes our physical universe, which we will hopefully discuss further in a future post.