Geometry of Curved Space, Part 1: Prerequisites for General Relativity

I'm TAing a course on general relativity this semester, and I'm covering some of the geometry background in tutorials. Since I need to prepare some material for these, I thought there was no harm in putting it up on this blog. So here we go.

Throughout, we'll let $M$ be a smooth manifold equipped with a metric $g$. Whenever it makes life easier, I'll assume that $g$ is positive definite (rather than Lorentzian). For a point $p \in M$, denote its tangent space by $T_p M$.

Theorem 1 For any $p \in M$, there exist a neighborhood $U$ of 0 in $T_p M$ and a neighborhood $V$ of $p$ in $M$, and a diffeomorphism $\exp_p: U \to V$ called exponential map. This map takes lines through the origin in $T_p M$ to geodesics in $M$ passing through $p$.

Theorem 2 In exponential coordinates, the components of the metric are
$$g_{ij} = \delta_{ij} + O(|x|^2)$$

Corollary 3 In exponential coordinates, the Christoffel symbols vanish at $p$.

Corollary 4 The Christoffel symbols are not the components of a tensor.

Corollary 5 Not all metrics are equivalent: there is a local invariant, called the curvature.

Construction of exponential map. The metric on $M$ induces a (constant) metric $g_{ij}$ on $T_p M$. By a linear change of coordinates on $T_p M$, we can assume that this induced metric is just $g_{ij} = \delta_{ij}$. Now the geodesic equation on $M$ is a 2nd order ODE which has a unique solution once we specify an initial condition. An initial condition is just a pair $(p,v)$ consisting of a point $p \in M$ and tangent vector $v \in T_p M$. Since we have fixed $p$, each $v \in T_p M$ gives a unique geodesic through $p$. Call it $\gamma_v(t)$. Then define the exponential map as follows:
$$\exp_p(v) := \gamma_v(1)$$
The fact that this map is well-defined, smooth, and 1-1 (at least locally) follows from the standard existence and uniqueness theorem for ODEs. So to see that it is a diffeomorphism near $0$, we can just compute its differential and apply the inverse function theorem.

The easy way out. By construction, every geodesic through 0 is of the form $\gamma_v(t) = tv$. Plugging this into the geodesic equation,
$$\dot{v} + \Gamma(x)^i_{jk} v^j v^k = 0$$
we see that at 0, $\Gamma^i_{jk}$ vanishes, and in particular, the first partial derivatives of the metric vanish.

The hard way: the differential of $exp$ at 0. First, taylor expand the velocity of a geodesic, and evaluate at time $t = 1$:
$$v(t) = v + \dot{v} + \frac{1}{2} \ddot{v} + \cdots$$
Now, by the geodesic equation, $\dot{v}$ is $O(v^2)$. Similarly, by differentiating the geodesic equation, we find that all of the higher time derivatives are also $O(v^2)$. So we find that the exponential map is just the identity + $O(v^2)$, and hence its differential at 0 is just the identity.

Now, we have argued that in exponential coordinates, the Christoffel symbols vanish at 0. Recall that for any tensor $T$, if the components of $T$ vanish at some point $p$ in one coordinate system, then $T$ is identically 0 at that point (i.e. its components vanish at that point in all coordinate systems). If the Christoffel symbols were a tensor, then, the above shows that they must be identically zero at all points in $M$, in all coordinate systems. But this is absolutely not the case--even in flat $\mathbb{R}^n$, we can pick coordinates so that the Christoffel symbols do not vanish. Hence they are not a tensor.

Aside Though the Christoffel symbols are not a tensor, they are the components of something which does have a coordinate indepdendent definition: a connection 1-form. A connection 1-form is not a tensor but rather a section of a certain associated bundle. More on this in future posts.

Claim Suppose $\gamma$ is a curve in $M$ with tangent vector $T$, and suppose $V$ is a vector field defined on $\gamma$. Then $\nabla_T V$ is well-defined, independent of the smooth extension of $V$.

Proof Suppose $V$ and $W$ are two smooth vector field that agree on $\gamma$. We would like to show that
$$\nabla_T V = \nabla_T W$$
i.e., this directional derivative depends only on their restriction to $\gamma$. It suffices to prove this pointwise. In coordinates, we have
$$\nabla_T V = T^k \frac{\partial V^i}{\partial x^k} + \Gamma^i_{jk} V^j T^k$$
and similarly for $W$. The terms involving Christoffel symbols do not depend on derivatives of $V$ or $W$, so they agree by assumption. If $T$ is zero at a point, there is nothing to show. So assume that $T$ is nonzero at some point. Then near this point, we can choose coordinates $x^i$ such that
$$\gamma(t) = (t, 0)$$
so that
$$T = (1, 0)$$
Then $V$ and $W$ agree when $x^i = 0$ for $i \geq 2$, and hence their partials agree. We have
$$T^k \frac{\partial V^i}{\partial x^k} = \frac{V^i}{\partial x^1} = T^k \frac{\partial W^i}{\partial x^k}$$

Explicit Formulas for Christoffel Symbols. Using properties of covariant derivatives, we find
$$0 = \nabla_k g_{ij} = \frac{\partial g_{ij}}{\partial x^k} - \Gamma^l_{ki} g_{lj} - \Gamma^l_{kj} g_{il}$$
So
$$\Gamma_{kij} + \Gamma_{kji} = g_{ij,k}$$
$$\Gamma_{ijk} + \Gamma_{ikj} = g_{jk,i}$$
$$\Gamma_{jki} + \Gamma_{jik} = g_{ki,j}$$
Taking (2) + (3) - (1) gives
$$2\Gamma_{ijk} = g_{jk,i} + g_{ki,j} - g_{ij,k}$$
Hence
$$\Gamma_{ij}^k = \frac{g^{kl}}{2}\left(g_{jl,i} + g_{li,j} - g_{ij,l} \right)$$