We establish a structure on a smooth manifold \((\mathcal{M},\mathcal{O},\mathcal{A})\) that allows us to assign vectors in each tangent space a length and1 an angle between vectors in the same tangent space. Such a structure in each tangent space is a inner product, and the complete structure over all tangent spaces is what we call a metric (i.e. it is a inner produce field).
From this structure, one can then define a notion of length of a curve. Then we can look at shortest (and longest) curves, which are known as geodesics. We will develop this completely independently of the notion of straight curves, i.e. of the covariant derivative, but then shall insist, for obvious reasons, at the end that the two coincide. In doing this we will define what we mean by so-called metric compatible connections.
A metric \(g\) on a smooth manifold \((\mathcal{M},\mathcal{O},\mathcal{A})\) is a \((0,2)\)-tensor field satisfying:
It’s symmetric; \(g(X,Y) = g(Y,X)\), for all \(X,Y\in\Gamma T\mathcal{M}\), and
Non-degeneracy; the map \(\flat : \Gamma T\mathcal{M}\to \Gamma T^*\mathcal{M}\), given by \[\flat(X) :Y := g(X,Y),\] is a \(C^{\infty}\)-isomorphism, i.e. it is invertible and smooth in both directions.
The inverse metric, \(g^{-1}:\Gamma T^*\mathcal{M}\times \Gamma T^*\mathcal{M}\xrightarrow{\sim} C^{\infty}(\mathcal{M})\), w.r.t. a metric \(g\) is the symmetric, \((2,0)\)-tensor field defined by \[g^{-1}(\omega,\sigma) := \omega : \flat^{-1}(\sigma).\]
Remark 10.1.1. One needs to be careful when referring to \(g^{-1}\) as an inverse. It is not an inverse in the sense of a map, but in the matrix sense. That is the map inverse of \(g:\Gamma T\mathcal{M}\times \Gamma T\mathcal{M}\xrightarrow{\sim} C^{\infty}(\mathcal{M})\) would be a map from \(\mathbb{C}^{\infty}(\mathcal{M})\) to \(\Gamma T\mathcal{M}\times\Gamma T\mathcal{M}\), which \(g^{-1}\) is not. If we denote2 the components of \(g^{-1}\) simply as \(g^{ab}\) (so no \(-1\) in it), then what we mean by inverse is that the following holds: \[g^{ac}g_{cb} = \delta^a_b.\]
Remark 10.1.2. It is very common for people to talk about ‘raising/lowering’ indices using the metric/inverse metric. What they mean is the idea that \(\flat(X)\) is a covector and so has a covariant index. However, we’re lazy and so we don’t want to have to keep writing the \(\flat\) bit and so we just write \[X_a := g_{am}X^m,\] and similarly for a covector made into a vector via \(\flat^{-1}\). The clear problem is that, unless specified, we don’t know whether \(T_a\) are the components of a covector defined completely independently of the metric or whether they are the ‘lowered’ components of a vector, and so is dependent on the metric. In these notes we shall never suppress the metric and so shall not talk about ‘raised/lowered’ indices.
Example 10.1.3. Consider the smooth manifold \((S^2,\mathcal{O},\mathcal{A})\) with the chart \((U,x)\) corresponding to spherical coordinates \((\theta,\varphi)\).3 Define the metric as4 \[g_{ij}\big( x^{-1}(\theta,\varphi)\big) := \begin{pmatrix} R^2 & 0 \\ 0 & R^2\sin^2\theta \end{pmatrix}_{ij}.\] This gives us the round sphere of radius \(R\). Note, just as with the connection, defining a metric allows us to give the manifold shape. Note also, though, that, unlike the round sphere obtained using the connection, we can talk about the size of the round sphere obtained from the metric.
Recall the eigenvalue equation \[Av = \lambda\cdot v,\] where \(v\) is an eigenvector. If we want to express this in terms of components, it is clear that \(A\) must be a \((1,1)\)-tensor, otherwise we break Einstein summation convention. That is \[{A^a}_{m}v^m = \lambda \cdot v^a\] is the only valid index placement. If we represent \(A\) as a matrix, it is a well known result of linear algebra that we can bring it to the form \[A = \text{diag}(\lambda_1,\lambda_2,...,\lambda_d),\] where \(d\) is the dimension of the vector space.
We want to have a similar thing for tensors of different ranks. For \((0,2)\)-tensors, in particular the metric, we have the signature of \(g\) which has only \(+1,-1,\) and \(0\) on the diagonal. One should be careful, though, as these are not simply eigenvalues for \(g\); the first reason being we just argued that eigenvalues only make sense for \((1,1)\)-tensors, and besides that it turns out that we can always transform to a chart such that the signature is only \(+1,-1\) and \(0\)s, which you can not do for eingenvalues in general.
Actually what we want to define as the signature is the double \((p,q)\) where \(p\) is the number of \(+1\)s and \(q\) is the number of \(-1\)s. This is the definition we shall use in these notes, but it is important to be aware that others refer to different, but related, things as the signature.5
Claim 10.2.1. The signature is independent of the choice of chart. That is the values of \(p\) and \(q\) do not depend on what basis you use in order to write down the matrix components.
Notation. We shall use the standard notation of \(+\)s and \(-\)s as a \(d\)-tuple to indicate the signature. For example if \(d=3\), \(p=2\) and \(q=1\) we write \((+,+,-)\). The position in the tuple corresponds to the corresponding metric components. So for our example \(g_{11}=1\), \(g_{22}=1\) and \(g_{33}=-1\) in this basis.
Remark 10.2.2. What we are really interested in the the relative sign between the components of the metric, and so we could easily have switched \(p\leftrightarrow q\) in the definition and proceeded from there, i.e. our example in the above notation would become \((-,-,+)\). It does not matter which we pick, as long as we are consistent. Our choice of signature is given by the following two definitions.
A metric is called Riemannian if its signature is \((+,+,...,+)\).
A metric with any other signature (apart from \((-,-,...,-)\) of course) is called pseudo-Riemannian. Of particular importance in general relativity is the following case.
A metric is called Lorentzian is its signature is \((+,-,...,-)\).
Remark 10.2.3. The convention given for the Riemannian metric is almost always the one used, however for Lorentzian metrics its about a 50/50 split between people who use \((+,-,...,-)\) and people who use \((-,+,...,+)\). We will use the one given in the definition.6
Show that the metric non-degeneracy condition for a Riemannian metric is equivalent to the non-degeneracy of an inner product. That is \[g(X,Y) = 0, \quad \forall Y\in\Gamma T\mathcal{M} \quad \iff \quad X=0.\] Hint: Think about what it means for a matrix to be invertible and then decompose \(X\) and \(Y\) in a basis.
Let \(\gamma:\mathbb{R}\to\mathcal{M}\) be a smooth curve, then we know its velocity \(v_{\gamma,\gamma(\lambda)}\) at each \(\gamma(\lambda)\in\mathcal{M}\). On a topological manifold this is as far as we can go, but on a metric manifold we have the following.
On a Riemannian metric manifold \((\mathcal{M},\mathcal{O},\mathcal{A},g)\), the speed of a curve at \(\gamma(\lambda)\) is the number \[s(\lambda) := \sqrt{g(v_{\gamma},v_{\gamma})}\Big|_{\gamma(\lambda)}\]
Remark 10.3.1. Although we might expect the velocity components \(v^a\) to have units \(LT^{-1}\), this is not true; they have units \(T^{-1}\). The apparent ‘loss’ of the distance comes from the fact that the components are chart dependent objects and the distance in a chart is physically meaningless and so we cannot attach physical units to it. The speed, however, does have units \(LT^{-1}\), which tells us that the metric components must have units \(L^2\).
Let \(\gamma: (0,1) \to \mathcal{M}\)7 be a smooth curve. Then the length of \(\gamma\) is the number8 \[L[\gamma] := \int_0^1 d\lambda \, s(\lambda).\]
What we have just seen is that the velocity is actually the fundamental object and from it we derive the speed and from that we get the length of a curve. This is entirely opposite to what we learn lower down in school!
Example 10.3.2. Reconsider the round sphere of radius \(R\). Consider its equator, \[\begin{split} \theta(\lambda) & := (x^1\circ \gamma)(\lambda) = \frac{\pi}{2} \\ \varphi(\lambda) & := (x^2\circ \gamma)(\lambda) = 2\pi\lambda^3. \end{split}\] The length of this curve is \[L[\gamma] = \int_0^1 d\lambda \, \sqrt{ g_{ij}\Big(x^{-1}\big(\theta(\lambda),\varphi(\lambda)\big)\Big) (x^1\circ \gamma)'(\lambda)(x^2\circ \gamma)'(\lambda)}.\] Using \[g_{ij} = \text{diag}(R^2,R^2\sin^2\theta), \qquad \theta'(\lambda) = 0, \qquad\text{and}\qquad \varphi'(\lambda) = 6\pi\lambda^2,\] we have \[\begin{split} L[\gamma] & = \int_0^1d\lambda \, \sqrt{R^2\sin^2\big(\theta(\lambda)\big)36\pi^2\lambda^4} \\ & = 6\pi R \int_0^1 \sin(\pi/2) \lambda^2 \\ & = 6\pi R \cdot \frac{1}{3} \\ & = 2\pi R. \end{split}\]
Note that although we used a seemingly funny parameterisation (i.e. \(\lambda^3\) not just \(\lambda\)) in the above example, the answer still came out as we would like. This is obviously because no where in the definitions above did we talk about how we parameterise the curve. Physically it makes sense that the length of the curve is independent of it: the length of your walk does not depend on how quickly you do it, or if you even do it at a constant speed (provided you don’t turn around and walk backwards on yourself at any point). This can be written nicely as the following theorem.
Let \(\gamma:(0,1) \to \mathcal{M}\) be a smooth curve and let \(\sigma:(0,1)\to(0,1)\) be smooth, bijective and increasing, then \(L[\gamma] = L[\gamma\circ\sigma]\).
A curve \(\gamma:(0,1)\to\mathcal{M}\) is called a geodesic on a Riemannian manifold \((\mathcal{M},\mathcal{O},\mathcal{A},g)\) if it is a stationary9 curve w.r.t. the length functional \(L\).
The curve \(\gamma:(0,1)\to\mathcal{M}\) is a geodesic if and only if it satisfies the Euler Lagrange equations for the Lagrangian \[\begin{split} \mathcal{L} : T\mathcal{M} &\to \mathbb{R} \\ X &\mapsto \sqrt{g(X,X)}. \end{split}\] In a chart, this is \[\mathcal{L}(\gamma^i,\Dot{\gamma}^i) = \sqrt{g_{ij}\big(\gamma(\lambda)\big)\Dot{\gamma}^i(\lambda)\Dot{\gamma}^j(\lambda)}.\]
Finding the Euler Lagrange equations proceeds as follows: \[\begin{split} \frac{\partial\mathcal{L}}{\partial\Dot{\gamma}^m} & = \frac{1}{\sqrt{...}} g_{mj}\big(\gamma(\lambda)\big) \Dot{\gamma}^j(\lambda) \\ \therefore \frac{d}{d\lambda}\bigg(\frac{\partial\mathcal{L}}{\partial\Dot{\gamma}^m}\bigg) & = \frac{d}{d\lambda}\bigg(\frac{1}{\sqrt{...}}\bigg) g_{mj}\big(\gamma(\lambda)\big) \Dot{\gamma}^j(\lambda) + \frac{1}{\sqrt{...}} \Big( g_{mj}\big(\gamma(\lambda)\big) \Ddot{\gamma}^j(\lambda) + \Dot{\gamma}^s(\lambda)\big[\partial_s g_{mj}\big(\gamma(\lambda)\big)\big] \Dot{\gamma}^j(\lambda) \Big). \end{split}\] Now we’re stuck with the ugly task of trying to work out \(\frac{d}{d\lambda}\Big(\frac{1}{\sqrt{...}}\Big)\). However, we have already demonstrated that the length of the curve is independent on how we choose to our parameter. We are free, therefore, to choose it to be something convenient, and we simply take it to be such that \(g(\Dot{\gamma},\Dot{\gamma})=1\), that is the speed it one along the whole curve. We then just have \[\frac{d}{d\lambda}\bigg(\frac{\partial \mathcal{L}}{\partial \Dot{\gamma}^m}\bigg) = g_{mj}\big(\gamma(\lambda)\big) \Ddot{\gamma}^j(\lambda) + \Dot{\gamma}^s(\lambda)\Big(\partial_s g_{mj}\big(\gamma(\lambda)\big)\Big) \Dot{\gamma}^j(\lambda).\] We also need to find \[\frac{\partial \mathcal{L}}{\partial \gamma^m} = \frac{1}{2} \Big(\partial_m g_{ij}\big(\gamma(\lambda)\big)\Big) \Dot{\gamma}^i(\lambda) \Dot{\gamma}^j(\lambda),\] where we have already imposed our parameter choice condition. So our Euler Lagrange equations are (dropping the \((\lambda)\)s for notational brevity) \[g_{mj}\Ddot{\gamma}^j + (\partial_ig_{mj})\dot{\gamma}^i\Dot{\gamma}^j - \frac{1}{2}(\partial_mg_{ij})\Dot{\gamma}^i\Dot{\gamma}^j = 0.\] Multiplying both sides by the inverse metric \(g^{mq}\) and using the condition \(g^{mq}g_{mj}=\delta^q_j\), we have \[\Ddot{\gamma}^q + g^{qm}\bigg( \partial_ig_{mj} - \frac{1}{2}\partial_mg_{ij}\bigg) \dot{\gamma}^{(i}\dot{\gamma}^{j)} = 0,\] where the brackets on the last two indices indicate the symmetry \(\dot{\gamma}^i\dot{\gamma}^j = \dot{\gamma}^j\dot{\gamma}^i\). Using this symmetry we can double the first term by switching \(i\leftrightarrow j\), giving us \[\Ddot{\gamma}^q + \frac{1}{2}g^{qm}\big(\partial_ig_{mj}+\partial_jg_{mi} - \partial_mg_{ij}\big)\dot{\gamma}^{(i}\dot{\gamma}^{j)} = 0.\] This is the geodesic equation for the components of \(\gamma\) in a chart. We can write this in the form of an autoparallel10 equation by introducing the following definition.
Given a metric \(g\) and a chart \((U,x)\), we define the Christoffel symbols (or Levi-Civita connection coefficients) as \[^{LC}{\Gamma^q}_{ij}\big(\gamma(\lambda)\big) := \frac{1}{2}g^{qm}\big(\partial_ig_{mj}+\partial_jg_{mi} - \partial_mg_{ij}\big),\] where the components of the metric (and its inverse, obviously) are taken in the chart given.
Notation. We can lighten the notation slightly by defining \[g_{ij,m} := \partial_mg_{ij},\] and similarly for any other tensor rank. That is, we simply denote a partial derivative in a chart by a comma and the index follows it is the derivative entry. This notation is very useful, as it can be used along side the semi-colon notation for the covariant derivative \[T_{jk;i} := (\nabla_i T)_{jk}\] We shall adopt this notation in these notes, however the reader is warned that Dr. Schuller does not use this notation, and so to make sure they can transition between the two when comparing these notes to the lectures.
This process, specifically the point at which we say that the \(^{LC}\Gamma\)s come from a connection, \(^{LC}\nabla\), identifies the shortest11 curves (geodesics) with straight curves (autoparallels). This is clearly a physically very reasonable, and correct, thing to do. It is important to note, though, that up until this point, geodesics and autoparallels are completely separate entities.
Note by making this identification, we obtain the connection from the metric. That is, we do not need to provide both a metric and a connection, but by simply providing a metric we can obtain a unique connection such that the shortest curves and the straight curves coincide. This sounds like a chart dependent thing, and therefore not a good thing to do. However the following theorem puts our minds to rest on this point, letting us know everything is OK.
Let \((\mathcal{M},\mathcal{O},\mathcal{A},g,\nabla)\) be a topological manifold equipped with both a metric and a connection. If
\(\nabla\) is torsion free, and
\(\nabla g = 0\), known as metric compatibility,
then we can conclude \(\nabla = ^{LC}\nabla\).
Proof. See tutorials. ◻
Show that the metric compatibility condition allows us to ‘move the metric in and out of the covariant derivative’. That is, \[g\cdot \nabla T = \nabla g\cdot T.\]
Finally for this lecture, let’s introduce some definitions. As we see, all of them are directly related to the metric.
Let \((\mathcal{M},\mathcal{O},\mathcal{A},g)\) be a metric manifold. The components of the Riemann Christoffel curvature are defined by \[\text{Riem}_{abcd} := g_{am}{\text{Riem}^m}_{bcd},\] where \({\text{Riem}^m}_{bcd}\) are the Riemann tensor components obtained from the Levi-Civita connection \(^{LC}\nabla\).
Let \((\mathcal{M},\mathcal{O},\mathcal{A},g)\) be a metric manifold and let \(\text{Riem}\) be the Riemann tensor obtained from the Levi-Civita connection. We then define the Ricci scalar as \[R = g^{ab}\text{Ric}_{ab},\] where \(\text{Ric}_{ab}:={\text{Riem}^c}_{acb}\) are the components of the Ricci curvature tensor.
Let \((\mathcal{M},\mathcal{O},\mathcal{A},g)\) be a metric manifold and let \(\text{Riem}\) be the Riemann tensor obtained from the Levi-Civita connection. We define the components of the Einstein curvature as \[G_{ab} := \text{Ric}_{ab} - \frac{1}{2}g_{ab}R,\] where \(\text{Ric}\) and \(R\) are the Ricci curvature and Ricci scalar, respectively.
It is important to note that these quantities are not only related to the metric through its direct appearance in the expressions, but also through the fact that, in order to define the Riemann curvature tensor we need a connection and for all of them we used the Levi-Civita connection, a metric dependent object. For this latter reason, the Ricci curvature tensor (defined previously) is also a metric dependent object.
As all of the names above suggest, we have just established a link between the curvature of the spacetime and the metric structure.12 This is the first major step into understanding the main principles of general relativity: that matter generates curvature on the spacetime.
Remark 10.4.3. The above definitions can, of course, all be expressed in a chart free manner as they are tensors, however the notation can be a bit confusing and it’s much easier to see in component form, hence why we have defined it this way.
Show that \(\text{Riem}_{abcd}\) have the correct transformation behaviour the components of a \((0,4)\)-tensor field. From this is follows analogously that the remaining definitions are indeed tensors.