3.4 More on the Riemann Tensor and its Friends

Recall that the components of the Riemann tensor are given by (3.15),

$$R_{\rho \mu \nu }^{\sigma }={\partial }_{\mu }{\Gamma }_{\nu \rho }^{\sigma }-{\partial }_{\nu }{\Gamma }_{\mu \rho }^{\sigma }+{\Gamma }_{\nu \rho }^{\lambda }{\Gamma }_{\mu \lambda }^{\sigma }-{\Gamma }_{\mu \rho }^{\lambda }{\Gamma }_{\nu \lambda }^{\sigma }\text{\hspace{1em}(3.39)}$$

We can immediately see that the Riemann tensor is anti-symmetric in the final two indices

$$R_{\rho \mu \nu }^{\sigma }=-R_{\rho \nu \mu }^{\sigma }$$

However, there are also a number of more subtle symmetric properties satisfied by the Riemann tensor when we use the Levi-Civita connection. Logically, we could have discussed this back in Section 3.2. However, it turns out that a number of statements are substantially simpler to prove using normal coordinates introduced in Section 3.3.2.

Claim

If we lower an index on the Riemann tensor, and write $R_{\sigma \rho \mu \nu }=g_{\sigma \lambda }{R}_{\rho \mu \nu }^{\lambda }$ then the resulting object also obeys the following identities

$$\begin{array}{rl}& \bullet R_{\sigma \rho \mu \nu }=-R_{\sigma \rho \nu \mu }.\\ & \bullet R_{\sigma \rho \mu \nu }=-R_{\rho \sigma \mu \nu }.\\ & \bullet R_{\sigma \rho \mu \nu }=R_{\mu \nu \sigma \rho }.\\ & \bullet R_{\sigma [\rho \mu \nu ]}=0.\end{array}$$

Proof

We work in normal coordinates, with ${\Gamma }^{\lambda }{}_{\mu \nu }=0$ at a point. The Riemann tensor can then be written as

$$\begin{array}{rl}& R_{\sigma \rho \mu \nu }=g_{\sigma \lambda }\left({\partial }_{\mu }{\Gamma }_{\nu \rho }^{\lambda }-{\partial }_{\nu }{\Gamma }_{\mu \rho }^{\lambda }\right)\\ & =\frac{1}{2}\left({\partial }_{\mu }({\partial }_{\nu }{g}_{\sigma \rho }+{\partial }_{\rho }{g}_{\nu \sigma }-{\partial }_{\sigma }{g}_{\nu \rho })-{\partial }_{\nu }({\partial }_{\mu }{g}_{\sigma \rho }+{\partial }_{\rho }{g}_{\mu \sigma }-{\partial }_{\sigma }{g}_{\mu \rho })\right)\\ & =\frac{1}{2}\left({\partial }_{\mu }{\partial }_{\rho }{g}_{\nu \sigma }-{\partial }_{\mu }{\partial }_{\sigma }{g}_{\nu \rho }-{\partial }_{\nu }{\partial }_{\rho }{g}_{\mu \sigma }+{\partial }_{\nu }{\partial }_{\sigma }{g}_{\mu \rho }\right)\end{array}$$

where, in going to the second line, we used the fact that ${\partial }_{\mu }{g}^{\lambda \sigma }=0$ in normal coordinates. The first three symmetries are manifest; the final one follows from a little playing. (It is perhaps quicker to see the final symmetry if we return to the Christoffel symbols where, in normal coordinates, we have $R^{\sigma }{}_{\rho \mu \nu }={\partial }_{\mu }{\Gamma }^{\sigma }{}_{\rho \nu }-{\partial }_{\nu }{\Gamma }^{\sigma }{}_{\rho \mu }.)$ But since the symmetry equations are tensor equations, they must hold in all coordinate systems.

Claim

The Riemann tensor also obeys the Bianchi identity

$${\nabla }_{[\lambda }{R}_{\sigma \rho ]\mu \nu }=0\text{\hspace{1em}(3.40)}$$

Alternatively, we can anti-symmetrise on the final two indices, in which case this can be written as $R^{\sigma }{}_{\rho [\mu \nu ;\lambda ]}=0$

Proof

We again use normal coordinates, where ${\nabla }_{\lambda }{R}_{\sigma \rho \mu \nu }={\partial }_{\lambda }{R}_{\sigma \rho \mu \nu }$ at the point $p.$ Schematically, we have $R=\partial \Gamma +\Gamma \Gamma$ , so $\partial R={\partial }^2\Gamma +\Gamma \partial \Gamma$ and the final Γ∂Γ term is absent in normal coordinates. This means that we just have $\partial R={\partial }^2\Gamma$ which, in its full coordinated glory, is

$${\partial }_{\lambda }{R}_{\sigma \rho \mu \nu }=\frac{1}{2}{\partial }_{\lambda }\left({\partial }_{\mu }{\partial }_{\rho }{g}_{\nu \sigma }-{\partial }_{\mu }{\partial }_{\sigma }{g}_{\nu \rho }-{\partial }_{\nu }{\partial }_{\rho }{g}_{\mu \sigma }+{\partial }_{\nu }{\partial }_{\sigma }{g}_{\mu \rho }\right)$$

Now anti-symmetrise on the three appropriate indices to get the result.

For completeness, we should mention that the identities $R_{\sigma [\rho \mu \nu ]}=0$ and ${\nabla }_{[\lambda }{R}_{\sigma \rho ]\mu \nu }=$ 0 (sometimes called the first and second Bianchi identities respectively) are more general, in the sense that they hold for an arbitrary torsion free connection. In contrast, the other two identities, $R_{\sigma \rho \mu \nu }=-R_{\rho \sigma \mu \nu }$ and $R_{\sigma \rho \mu \nu }=R_{\mu \nu \sigma \rho }$ hold only for the Levi-Civita connection.

3.4.1 The Ricci and Einstein Tensors

There are a number of further tensors that we can build from the Riemann tensor.

First, given a rank (1, 3) tensor, we can always construct a rank (0, 2) tensor by contraction. If we start with the Riemann tensor, the resulting object is called the Ricci tensor. It is defined by

$$R_{\mu \nu }=R_{\mu \rho \nu }^{\rho }$$

The Ricci tensor inherits its symmetry from the Riemann tensor. We write $R_{\mu \nu }=$ $g^{\sigma \rho }{R}_{\sigma \mu \rho \nu }=g^{\rho \sigma }{R}_{\rho \nu \sigma \mu }$ , giving us

$$R_{\mu \nu }=R_{\nu \mu }$$

We can go one step further and create a function R over the manifold. This is the Ricci scalar,

$$R=g^{\mu \nu }{R}_{\mu \nu }$$

The Bianchi identity (3.40) has a nice implication for the Ricci tensor. If we write the Bianchi identity out in full, we have

$$\begin{array}{rlr}& & {\nabla }_{\lambda }{R}_{\sigma \rho \mu \nu }+{\nabla }_{\sigma }{R}_{\rho \lambda \mu \nu }+{\nabla }_{\rho }{R}_{\lambda \sigma \mu \nu }=0\\ & \times g^{\mu \lambda }{g}^{\rho \nu }& \Rightarrow {\nabla }^{\mu }{R}_{\mu \sigma }-{\nabla }_{\sigma }R+{\nabla }^{\nu }{R}_{\nu \sigma }=0\end{array}$$

which means that

$${\nabla }^{\mu }{R}_{\mu \nu }=\frac{1}{2}{\nabla }_{\nu }R$$

This motivates us to introduce the Einstein tensor,

$$G_{\mu \nu }=R_{\mu \nu }-\frac{1}{2}R{g}_{\mu \nu }$$

which has the property that it is covariantly constant, meaning

$${\nabla }^{\mu }{G}_{\mu \nu }=0\text{\hspace{1em}(3.41)}$$

We’ll be seeing much more of the Ricci and Einstein tensors in the next section.

3.4.2 Connection 1-forms and Curvature 2-forms

Calculating the components of the Riemann tensor is straightforward but extremely tedious. It turns out that there is a slightly different way of repackaging the connection and the torsion and curvature tensors using the language of forms. This not only provides a simple way to actually compute the Riemann tensor, but also offers some useful conceptual insight.

Vielbeins

Until now, we have typically worked with a coordinate basis $\{e_{\mu }\}=\{{\partial }_{\mu }\}$ . However, we could always pick a basis of vector fields that has no such interpretation. For example, a linear combination of a coordinate basis, say

$${\hat{e}}_a=e_a^{\mu }{\partial }_{\mu }$$

will not, in general, be a coordinate basis itself.

Given a metric, there is a non-coordinate basis that will prove particularly useful for computing the curvature tensor. This is the basis such that, on a Riemannian manifold,

$$g({\hat{e}}_a,{\hat{e}}_b)=g_{\mu \nu }{e}_a^{\mu }{e}_b^{\nu }={\delta }_{ab}$$

Alternatively, on a Lorentzian manifold we take

$$g({\hat{e}}_a,{\hat{e}}_b)=g_{\mu \nu }{e}_a^{\mu }{e}_b^{\nu }={\eta }_{ab}\text{\hspace{1em}(3.42)}$$

The components ${e_a}^{\mu }$ are called vielbeins or tetrads. (On an n-dimensional manifold, these objects are usually called “German word for $n^{\prime }$ -beins. For example, onedimensional manifolds have einbeins; four-dimensional manifolds have vierbeins.)

The is reminiscent of our discussion in Section 3.1.2 where we mentioned that we can always find coordinates so that any metric will look flat at a point. In (3.42), we’ve succeeded in making the manifold look flat everywhere (at least in a patch covered by a chart). There are no coordinates that do this, but there’s nothing to stop us picking a basis of vector fields that does the job. In what follows, $\mu \nu$ indices are raised/lowered with the metric $g_{\mu \nu }$ while $a,b$ indices are raised/lowered with the flat metric ${\delta }_{ab}$ or ${\eta }_{ab}$ We will phrase our discussion in the context of Lorentzian manifolds, with an eye to later applications to general relativity.

The vielbeins aren’t unique. Given a set of vielbeins, we can always find another set related by

$${\tilde{e}}_a^{\mu }=e_b^{\mu }({\Lambda }^{-1}{)}_a^b\mathrm{with}{\Lambda }_a^c{\Lambda }_b^d{\eta }_{cd}={\eta }_{ab}\text{\hspace{1em}(3.43)}$$

These are Lorentz transformations. However now they are local Lorentz transformation, because Λ can vary over the manifold. These local Lorentz transformations are a redundancy in the definition of the vielbeins in (3.42).

The dual basis of one-forms $\{{\hat{\theta }}^a\}$ is defined by ${\hat{\theta }}^a({\hat{e}}_b)={\delta }_b^a$ . They are related to the coordinate basis by

$${\hat{\theta }}^a=e_{\mu }^{a}d{x}^{\mu }$$

Note the different placement of indices: $e^a{}_{\mu }$ is the inverse of ${e_a}^{\mu }$ , meaning it satisfies $e^a{}_{\mu }{e}_b{}^{\mu }={\delta }_b^a$ and $e^a{}_{\mu }{e}_a{}^{\nu }={\delta }_{\mu }^{\nu }$ . In the non-coordinate basis, the metric on a Lorentzian manifold takes the form

$$g=g_{\mu \nu }d{x}^{\mu }\otimes d{x}^{\nu }={\eta }_{ab}{\hat{\theta }}^a\otimes {\hat{\theta }}^b\Rightarrow g_{\mu \nu }=e_{\mu }^a{e}_{\nu }^b{\eta }_{ab}$$

For Riemannian manifolds, we replace ${\eta }_{ab}$ with ${\delta }_{ab}$ .

The Connection One-Form

Given a non-coordinate basis $\{{\hat{e}}_a\}$ , we can define the components of a connection in the usual way (3.8)

$${\nabla }_{{\hat{e}}_c}{\hat{e}}_b={\Gamma }_{cb}^a{\hat{e}}_a$$

Note that, annoyingly, these are not the same functions as ${\Gamma }_{\rho \nu }^{\mu }$ , which are the components of the connection computed in the coordinate basis! You need to pay attention to whether the components are Greek $\mu ,$ ν etc which tells you that we’re in the coordinate basis, or Roman a, b etc which tells you we’re in the vielbein basis.

We then define the matrix-valued connection one-form as

$${\omega }_b^a={\Gamma }_{cb}^a{\hat{\theta }}^c\text{\hspace{1em}(3.44)}$$

This is sometimes referred to as the spin connection because of the role it plays in defining spinors on curved spacetime. We’ll describe this in Section 4.5.6.

The connection one-forms don’t transform covariantly under local Lorentz transformations (3.43). Instead, in the new basis, the components of the connection one-form are defined as ${\nabla }_{{\hat{\tilde{e}}}_b}{\hat{\tilde{e}}}_c={\tilde{\Gamma }}_{bc}^a{\hat{\tilde{e}}}_a$ . You can check that the connection one-form transforms as

$${\tilde{\omega }}_b^a={\Lambda }_c^a{\omega }_d^c({\Lambda }^{-1}{)}_b^d+{\Lambda }_c^a(d{\Lambda }^{-1}{)}_b^c\text{\hspace{1em}(3.45)}$$

The second term reflects the fact that the original connection components ${\Gamma }_{\nu \rho }^{\mu }$ do not transform as a tensor, but with an extra term involving the derivative of the coordinate transformation (3.13). This now shows up as an extra term involving the derivative of the local Lorentz transformation.

There is a rather simple way to compute the connection one-forms, at least for a torsion free connection. This follows from the first of two Cartan structure relations:

Claim

For a torsion free connection,

$$d{\hat{\theta }}^a+{\omega }_b^a\wedge {\hat{\theta }}^b=0\text{\hspace{1em}(3.46)}$$

Proof

We first look at the second term,

$${\omega }_b^a\wedge {\hat{\theta }}^b={\Gamma }_{cb}^a(e_{\mu }^{c}d{x}^{\mu })\wedge (e_{\nu }^{b}d{x}^{\nu })$$

The components ${\Gamma }^a{}_{cb}$ are related to the coordinate basis components by

$${\Gamma }_{cb}^a=e_{\rho }^a{e}_c^{\mu }\left({\partial }_{\mu }{e}_b^{\rho }+e_b^{\nu }{\Gamma }_{\mu \nu }^{\rho }\right)=e_{\rho }^a{e}_c^{\mu }{\nabla }_{\mu }{e}_b^{\rho }\text{\hspace{1em}(3.47)}$$

So

$$\begin{array}{rl}& {\omega }_b^a\wedge {\hat{\theta }}^b=e_{\rho }^a{e}_c^{\lambda }{e}_{\mu }^c{e}_{\nu }^b({\partial }_{\lambda }{e}_b^{\rho }+e_b^{\sigma }{\Gamma }_{\lambda \sigma }^{\rho })d{x}^{\mu }\wedge d{x}^{\nu }\\ & =e_{\rho }^a{e}_{\nu }^b{\partial }_{\mu }{e}_b^{\rho }d{x}^{\mu }\wedge d{x}^{\nu }\end{array}$$

where, in the second line we’ve used $e_c{}^{\lambda }{e}^c{}_{\mu }={\delta }_{\mu }^{\lambda }$ and the fact that the connection is torsion free so ${\Gamma }_{[\mu \nu ]}^{\rho }=0$ . Now we use the fact that $e_{\nu }^b{e}_b{}^{\rho }={\delta }_{\nu }^{\rho },\mathrm{so}{e}_{\nu }^b{\partial }_{\mu }{e}_b{}^{\rho }=-e_b{}^{\rho }{\partial }_{\mu }{e}_{\nu }^b$ We have

$$\begin{array}{rl}& {\omega }_b^a\wedge {\hat{\theta }}^b=-e_{\rho }^a{e}_b^{\rho }{\partial }_{\mu }{e}_{\nu }^{b}d{x}^{\mu }\wedge d{x}^{\nu }\\ & =-{\partial }_{\mu }{e}_{\nu }^{a}d{x}^{\mu }\wedge d{x}^{\nu }=-d{\hat{\theta }}^a\end{array}$$

which completes the proof.

The discussion above was for a general connection. For the Levi-Civita connection, we have a stronger result

Claim

For the Levi-Civita connection, the connection one-form is anti-symmetric

$${\omega }_{ab}=-{\omega }_{ba}\text{\hspace{1em}(3.48)}$$

Proof

This follows from the explicit expression (3.47) for the components ${\Gamma }_{bc}^a$ . Lowering an index, we have

$${\Gamma }_{abc}={\eta }_{ad}{e}_{\rho }^d{e}_b^{\mu }{\nabla }_{\mu }{e}_c^{\rho }=-{\eta }_{ad}{e}_c^{\rho }{e}_b^{\mu }{\nabla }_{\mu }{e}_{\rho }^d=-{\eta }_{cf}{e}_{\sigma }^f{e}_b^{\mu }{\nabla }_{\mu }({\eta }_{ad}{g}^{\rho \sigma }{e}_{\rho }^d)$$

where, in the final equality, we’ve used the fact that the connection is compatible with the metric to raise the indices of $e_{\rho }^d$ inside the covariant derivative. Finishing of the derivation, we then have

$${\Gamma }_{abc}=-{\eta }_{cf}{e}_{\rho }^f{e}_b^{\mu }{\nabla }_{\mu }{e}_a^{\rho }=-{\Gamma }_{cba}$$

The result then follows from the definition ${\omega }_{ab}={\Gamma }_{acb}{\hat{\theta }}^c$

The Cartan structure equation (3.46), together with the anti-symmetry condition (3.48), gives a quick way to compute the spin connection. It’s instructive to do some counting to see how these two equations uniquely define ${\omega }^a{}_b$ . In particular, since ${\omega }_{ab}$ is anti-symmetric, one might think that it has $\frac{1}{2}n(n-1)$ independent components, and these can’t possibly be fixed by the n Cartan structure equations (3.46). But this is missing the fact that ${\omega }_{ab}$ are not numbers, but are one-forms. So the true number of components in ${\omega }_{ab}$ is $n\times \frac{1}{2}n(n-1)$ . Furthermore, the Cartan structure equation is an equation relating 2-forms, each of which has $\frac{1}{2}n(n-1)$ components. This means that it’s really $n\times \frac{1}{2}n(n-1)$ equations. We see that the counting does work, and the two fix the spin connection uniquely.

The Curvature Two-Form

We can compute the components of the Riemann tensor in our non-coordinate basis,

$$R_{bcd}^a=R({\hat{\theta }}^a;{\hat{e}}_c,{\hat{e}}_d,{\hat{e}}_b)$$

The anti-symmetry of the last two indices, $R^a{}_{bcd}=-R^a{}_{bdc}$ , makes this ripe for turning into a matrix of two-forms,

$${\mathcal{R}}_b^a=\frac{1}{2}{R}_{bcd}^a{\hat{\theta }}^c\wedge {\hat{\theta }}^d\text{\hspace{1em}(3.49)}$$

The second of the two Cartan structure relations states that this can be written in terms of the curvature one-form as

$${\mathcal{R}}_b^a=d{\omega }_b^a+{\omega }_c^a\wedge {\omega }_b^c\text{\hspace{1em}(3.50)}$$

The proof of this is mechanical and somewhat tedious. It’s helpful to define the quantities $[{\hat{e}}_a,{\hat{e}}_b]:=:f_{ab}^{:c}{\hat{e}}_c$ along the way, since they appear on both left and right-hand sides.

3.4.3 An Example: the Schwarzschild Metric

The connection one-form and curvature two-form provide a slick way to compute the curvature tensor associated to a metric. The reason for this is that computing exterior derivatives takes significantly less effort than computing covariant derivatives. We will illustrate this for metrics of the form,

$$d{s}^2=-f(r{)}^{2}d{t}^2+f(r{)}^{-2}d{r}^2+r^2(d{\theta }^2+{\sin }^2\theta d{\varphi }^2)\text{\hspace{1em}(3.51)}$$

For later applications, it will prove useful to compute the Riemann tensor for this metric with general $f(r)$ . However, if we want to restrict to the Schwarzschild metric we can take

$$f(r)=\sqrt{1-\frac{2GM}{r}}\text{\hspace{1em}(3.52)}$$

The basis of non-coordinate one-forms is

$${\hat{\theta }}^0=fdt,{\hat{\theta }}^1=f^{-1}dr,{\hat{\theta }}^2=rd\theta ,{\hat{\theta }}^3=r\sin \theta d\varphi \text{\hspace{1em}(3.53)}$$

Note that the one-forms $\hat{\theta }$ should not be confused with the angular coordinate $\theta !$ In this basis, the metric takes the simple form

$$d{s}^2={\eta }_{ab}{\hat{\theta }}^a\otimes {\hat{\theta }}^b$$

We now compute $d{\hat{\theta }}^a$ . Calculationally, this is straightforward. In particular, it’s substantially easier than computing the covariant derivative because there’s no messy connection to worry about. The exterior derivatives are simply

$$d{\hat{\theta }}^0=f^{\prime }dr\wedge dt,d{\hat{\theta }}^1=0,d{\hat{\theta }}^2=dr\wedge d\theta ,d{\hat{\theta }}^3=\sin \theta dr\wedge d\varphi +r\cos \theta d\theta \wedge d\varphi$$

The first Cartan structure relation, $d{\hat{\theta }}^a=-{\omega }^a{}_b\wedge {\hat{\theta }}^b$ , can then be used to read of the connection one-form. The first equation tells us that ${\omega }^0{}_1=f^{\prime }fdt=f^{\prime }{\hat{\theta }}^0$ . We then use the anti-symmetry (3.48), together with raising and lowering by the Minkowski metric $\eta =\mathrm{diag}(-1,+1,+1,+1)$ to get ${\omega }^1{}_0={\omega }_{10}=-{\omega }_{01}={\omega }^0{}_1$ . The Cartan structure equation then gives $d{\hat{\theta }}^1=-{{\omega }^1}_0\wedge {\hat{\theta }}^0+...$ . and the ${\omega }^1\circ \wedge {\hat{\theta }}^0$ contribution happily vanishes because it is proportional to ${\hat{\theta }}^0\wedge {\hat{\theta }}^0=0$

Next, we take ${\omega }^2{}_1=fd\theta =(f/r){\hat{\theta }}^2$ to solve the $d{\hat{\theta }}^2$ structure equation. The antisymmetry (3.48) gives ${\omega }^1{}_2=-{\omega }^2{}_1=-(f/r){\hat{\theta }}^2$ and this again gives a vanishing contribution to the $d{\hat{\theta }}^1$ structure equation.

Finally, the $d{\hat{\theta }}^3$ equation suggests that we take ${\omega }^3{}_1=f$ sin $\theta d\varphi =(f/r){\hat{\theta }}^3$ and ${\omega }^3{}_2=$ cos $\theta d\varphi =(1/r)$ cot $\theta {\hat{\theta }}^3$ . These anti-symmetric partners ${\omega }^1{}_3=-{\omega }^3{}_1$ and ${\omega }^2{}_3=-{\omega }^3{}_2$ do nothing to spoil the $d{\hat{\theta }}^1$ and $d{\hat{\theta }}^2$ structure equations, so we’re home dry. The final result is

$$\begin{array}{rl}{\omega }_1^0={\omega }_0^1=f^{\prime }{\hat{\theta }}^0,& {\omega }_1^2=-{\omega }_2^1=\frac{f}{r}{\hat{\theta }}^2\\ {\omega }_1^3=-{\omega }_3^1=\frac{f}{r}{\hat{\theta }}^3,& {\omega }_2^3=-{\omega }_3^2=\frac{\cot \theta }{r}{\hat{\theta }}^3\end{array}$$

Now we can use this to compute the curvature two-form. We will focus on

$${\mathcal{R}}_1^0=d{\omega }_1^0+{\omega }_c^0\wedge {\omega }_1^c$$

We have

$$d{\omega }_1^0=f^{\prime }d{\hat{\theta }}^0+f^{\prime \prime }dr\wedge {\hat{\theta }}^0=((f^{\prime }{)}^2+f^{\prime \prime }f)dr\wedge dt$$

The second term in the curvature 2-form is ${\omega }^0{}_c\wedge {\omega }^c{}_1={\omega }^0{}_1\wedge {\omega }^1{}_1=0$ . So we’re left with

$${\mathcal{R}}_1^0=\left((f^{\prime }{)}^2+f^{\prime \prime }f\right)dr\wedge dt=\left((f^{\prime }{)}^2+f^{\prime \prime }f\right){\hat{\theta }}^1\wedge {\hat{\theta }}^0$$

The other curvature 2-forms can be computed in a similar fashion. We can now read of the components of the Riemann tensor in the non-coordinate basis using (3.49).

(We should remember that we get a contribution from both $R_{101}^0$ and $R_{110}^0=-R_{101}^0$ which cancels the factor of $1/2$ in (3.49).) After lowering an index, we find that the non-vanishing components of the Riemann tensor are

$$\begin{array}{rl}& R_{0101}=f{f}^{\prime \prime }+(f^{\prime }{)}^2\\ & R_{0202}=\frac{f{f}^{\prime }}{r}\\ & R_{0303}=\frac{f{f}^{\prime }}{r}\\ & R_{1212}=-\frac{f{f}^{\prime }}{r}\\ & R_{1313}=-\frac{f{f}^{\prime }}{r}\\ & R_{2323}=\frac{1-f^2}{r^2}\end{array}$$

We can also convert this back to the coordinates $x^{\mu }=(t,r,\theta ,\varphi )$ using

$$R_{\mu \nu \rho \sigma }=e_{\mu }^a{e}_{\nu }^b{e}_{\rho }^c{e}_{\sigma }^d{R}_{abcd}$$

This is particularly easy in this case because the matrices ${e_a}^{\mu }$ defining the one-forms (3.53) are diagonal. We then have

$$\begin{array}{rl}& R_{trtr}=f{f}^{\prime \prime }+(f^{\prime }{)}^2\\ & R_{t\theta t\theta }=f^3{f}^{\prime }r\\ & R_{t\varphi t\varphi }=f^3{f}^{\prime }r{\sin }^2\theta \\ & R_{r\theta r\theta }=-\frac{f^{\prime }r}{f}\\ & R_{r\varphi r\varphi }=-\frac{f^{\prime }r}{f}{\sin }^2\theta \\ & R_{\theta \varphi \theta \varphi }=(1-f^2)r^2{\sin }^2\theta \end{array}\text{\hspace{1em}(3.54)}$$

Finally, if we want to specialise to the Schwarzschild metric with $f(r)$ given by (3.52), we have

$$\begin{array}{rl}& R_{trtr}=-\frac{2GM}{r^3}\\ & R_{t\theta t\theta }=\frac{GM(r-2GM)}{r^2}\\ & R_{t\varphi t\varphi }=\frac{GM(r-2GM)}{r^2}{\sin }^2\theta \\ & R_{r\theta r\theta }=-\frac{GM}{r-2GM}\\ & R_{r\varphi r\varphi }=-\frac{GM{\sin }^2\theta }{r-2GM}\\ & R_{\theta \varphi \theta \varphi }=2GMr{\sin }^2\theta \end{array}$$

Although the calculation is a little lengthy, it turns out to be considerably quicker than first computing the Levi-Civita connection and subsequently motoring through to get the Riemann tensor components.

3.4.4 The Relation to Yang-Mills Theory

It is no secret that the force of gravity is geometrical. However, the other forces are equally as geometrical. The underlying geometry is something called a fibre bundle, rather than the geometry of spacetime.

We won’t describe fibre bundles in this course, but we can exhibit a clear similarity between the structures that arise in general relativity and the structures that arise in the other forces, which are described by Maxwell theory and its generalisation to Yang-Mills theory.

Yang-Mills theory is based on a Lie group G which, for this discussion, we will take to be $SU(N)$ or $U(N)$ . If we take $G=U(1)$ , then Yang-Mills theory reduces to Maxwell theory. The theory is described in terms of an object that physicists call a gauge potential. This is a spacetime “vector” $A_{\mu }$ which lives in the Lie algebra of G. In more down to earth terms, each component is an anti-Hermitian $N\times N$ matrix, $(A_{\mu }{)}_b^a,$ with $a,b=1,\dots ,N$ . In fact, as we saw above, this “vector” is really a one-form. The novelty is that it’s a Lie algebra-valued one-form.

Mathematicians don’t refer to $A_{\mu }$ as a gauge potential. Instead, they call it a connection (on a fibre bundle). This relationship becomes clearer if we look at how $A_{\mu }$ changes under a gauge transformation

$${\tilde{A}}_{\mu }=\Omega A_{\mu }{\Omega }^{-1}+\Omega {\partial }_{\mu }{\Omega }^{-1}$$

where $\Omega (x)\in G$ . This is identical to the transformation property (3.45) of the one-form connection under local Lorentz transformations.

In Yang-Mills, as in Maxwell theory, we construct a field strength. In components, this is given by

$$(F_{\mu \nu }{)}_b^a={\partial }_{\mu }(A_{\nu }{)}_b^a-{\partial }_{\nu }(A_{\mu }{)}_b^a+[A_{\mu },A_{\nu }{]}_b^a$$

Alternatively, in the language of forms, the field strength becomes

$$F_b^a=d{A}_b^a+A_c^a\wedge A_b^c$$

Again, there is an obvious similarity with the curvature 2-form introduced in (3.50). Mathematicians refer to the Yang-Mills field strength the “curvature”.

A particularly quick way to construct the Yang-Mills field strength is to take the commutator of two covariant derivatives. It is simple to check that

$$[{\mathcal{D}}_{\mu },{\mathcal{D}}_{\nu }]=F_{\mu \nu }$$

where I’ve suppressed the $a,b$ indices on both sides. This is the gauge theory version of the Ricci identity (3.16): for a torsion free connection,

$$[{\nabla }_{\mu },{\nabla }_{\nu }]Z^{\sigma }=R_{\rho \mu \nu }^{\sigma }{Z}^{\rho }$$

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