4
$\begingroup$

I'm taking a course in General Relativity, and before getting into Physics itself, the required knowledge of Differential Geometry is being taught.

In that context, a linear connection on a smooth manifold $M$ was introduced as a collection of maps $\nabla^{(r,s)} :\Gamma(TM)\times \Gamma(T^r_s M)\to \Gamma(T^r_s M)$ which takes a vector field $X$ and an $(r,s)$-tensor field $T$ and produce one $(r,s)$-tensor field $\nabla^{(r,s)}_ X T$. We, however, drop the $(r,s)$ and just denote all maps by $\nabla$. This map is required by the definition to satisfy:

  1. It is $C^\infty(M)$-linear on the first entry, that is $X\mapsto \nabla_X T$ for fixed $T$ is $C^\infty(M)$ linear.

  2. It is linear on the second entry, that is $T\mapsto \nabla_X T$ is linear for fixed $X$.

  3. It obeys Liebnitz rule in the second entry, that is, $\nabla_X(T\otimes S)=(\nabla_ X T)\otimes S+ T\otimes (\nabla_X S)$ for fixed $X$.

  4. It reduces to $X$ itself on $(0,0)$-tensors, that is, $\nabla_X f = Xf$ for fixed $X$ and $f\in C^\infty(M)$.

Now given a smooth manifold with linear connection $(M,\nabla)$ one can define the Riemann Curvature Tensor as the tensor field $R : \Gamma(T^\ast M)\times \Gamma(TM)\times \Gamma(TM)\times \Gamma(TM)\to C^\infty(M)$ given by

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

The problem is that this thing was just defined with no motivation whatsoever. It is then said that it gives the change in a vector as paralel transported along a loop formed by $X$ and $Y$. It is also said that this characterizes the curvature of $\nabla$.

My question here is: how can one derive this tensor? I mean, given that we have a connection $\nabla$ and we want to define its curvature, how can we derive this expression, and discover that this tensor field will do? I don't know even why it should be a tensor field, let alone follow some steps to arrive at the correct tensor field.

I just don't like the approach of "define this because it works". I want to be able to find out that this is the correct thing to do.

$\endgroup$
5
  • $\begingroup$ It is a good question and I look forward to there hopefully being some good answers, especially by people who might have taught the subject and therefore the choices they made when first presenting Riemannian curvature. For example even in Do Carmo's Riemannian Geometry where I really like the way the $\nabla$ connection is introduced, and the way geometric intuition is emphasised he still suggests that the reader "gets used to the formal properties of curvature, postponing until Chapter 6 the proof of a more geometric interpretation of curvature" $\endgroup$
    – Nadiels
    Feb 25, 2017 at 15:31
  • $\begingroup$ Maybe its felt that our geometric notion of curvature from surfaces in $\mathbb{R}^3$ is too strong and therefore it necessary to abstract away first? $\endgroup$
    – Nadiels
    Feb 25, 2017 at 15:32
  • $\begingroup$ Unfortunately some professors teach a technique with out explicitly explaining, it is for the student to mimic and eventually develop. Some professors have no conscious knowledge of what they are doing. As if doing physics is like learning to play the drums where some people can do it and some can not. Apparently this has worked and become the norm in both physics and mathematics I feel. $\endgroup$ Feb 25, 2017 at 15:43
  • $\begingroup$ The Riemann tensor "tells you everything there is to say about the metric up to second order". See here. $\endgroup$ Feb 25, 2017 at 16:40
  • $\begingroup$ user1620696, out of your last 30 questions, 22 have at least one answer, but you have only accepted one, and commented on three. I suggest you look back through the questions you've asked and consider accepting any answers that you think have answered your question. If the answers don't answer your question, consider explaining why through comments so that future answerers know what you're looking for. $\endgroup$ Feb 26, 2017 at 23:25

3 Answers 3

4
$\begingroup$

What you are asking essentially comes from Gauss's Theorema Egregium.

The theorem states that the Gaussian curvature is given by

$$K=\frac{R_{1212}}{g}$$ where

$$R_{ijkl}=\frac{\partial \Gamma_{ijk}}{\partial x_l}- \frac{\partial \Gamma_{ijl}}{\partial x_k}+ \Gamma^{r}_{jk}\Gamma_{irl}-\Gamma^{r}_{jl}\Gamma_{irk}$$ is the Riemann curvature tensor.

Gauss's theorem can be expressed by many formulas, this is the expression using the Christoffel symbols. Note that $K$ also need $g$, (but only as a denominator), so its not dependent only on the Christoffel symbols alone.

Geometrically the Christoffel symbols arise from the following equations, \begin{align*} \mathbf{r}_{uu}&=\Gamma_{11}^1 \ \mathbf{r}_{u} + \Gamma_{11}^2 \ \mathbf{r}_{v} +L \ \mathbf{N}\\ \mathbf{r}_{uv}&=\Gamma_{12}^1 \ \mathbf{r}_{u} + \Gamma_{12}^2 \ \mathbf{r}_{v} +M \ \mathbf{N}\\ \mathbf{r}_{vv}&=\Gamma_{22}^1 \ \mathbf{r}_{u} + \Gamma_{22}^2 \ \mathbf{r}_{v} +N \ \mathbf{N}\\ \end{align*}

If you examine these equations you see that these coefficients arise from differentiating the tangent vectors, which results in an vector which is no longer tangent, but then projecting the derivative back onto the surface.

What I have just described is exactly covariant differentation. Thus Christoffel symbols and connections are basically the same thing, in fact

$$\nabla_{X_i}(X_i)=\sum_k \Gamma^k_{ji}X_k$$ they are just the coefficients of the connection. Historically the curvature formula came first and latter connections were introduced. In fact it can be proven that the only invariants of a Riemann metric $g_{ij}$ are the Riemann Curvature tensor $R_{ijkl}$ and its covariant derivatives. Thus one way or another the Curvature tensor is forced upon us.

It might be helpful to you to prove the formula in the question in local coordinates using the equations in this answer.

Hope this helps.

$\endgroup$
4
$\begingroup$

As you seem to understand, the idea is that the choice of covariant derivative is equivalent to a choice of parallel transport, which in turn defines a concept of intrinsic curvature.

Consider a curve $\gamma : \mathbb{R} \to M$. Its velocity defines a vector field along its image $U=\gamma(\mathbb{R})$, which I'll denote by $\dot\gamma$. Now, consider a map we will call the parallel transport map of $\gamma$, $\tau_\gamma(\lambda,t,X) : \mathbb{R}\times \mathbb{R}\times TU \to TU$. It takes a tangent vector at $\gamma(t)$ and maps it to a (possibly) different tangent vector at the same point. Intuitively, the map is supposed to push the vector field along in such a way that all the vectors are parallely transported along the curve as $\lambda$ is increased. At the level of a smooth manifold structure, we have quite a bit of freedom in how we choose to define the parallel transport. In terms of the connection, the parallel transport map is chosen to be the solution to the following differential equation:

$$ \frac{\text{d}}{\text{d}\lambda} \tau_\gamma(\lambda, t, X) = \nabla_{\dot\gamma_t} X$$

That's quite messy, so let me break it down. The equation is supposed to hold for all $\lambda$, for any fixed real number $t$ and vector field $X$ defined on the curve $\gamma$. On the LHS, since $X$ and $t$ are fixed, we essentially have a function $\tau_\gamma(\lambda) : \mathbb{R} \to T_{\gamma(t)}M$. Notice that the codomain of the function is a single tangent space, and hence it's fine to subtract the values of $\tau_\gamma$ for different $\lambda$, and take the derivative. In other words, the covariant derivative of a vector field with respect to the tangent field of a curve can be thought of as the rate at which a vector changes as it is parallely transported along that curve.

So why does the Riemann tensor take the form that it does? The way you've written it makes it a bit hard to see the intuition behind it. Remember that a linear map between finite dimensional vector spaces $L : V\to V$ can be associated with a $(1,1)-$tensor $\tilde L : V^*\times V \to K$. In the same way, the Riemann tensor as you've written it can be associated with an alternative map $\tilde R : \Gamma(TM)\times \Gamma(TM)\times \Gamma(TM)\to\Gamma(TM)$. In other words, a map that eats three vector fields and produces a new vector field. Its geometric intuition is that it provides the difference between the initial and final value of the third vector after it is parallely transported around an infinitesimal parallelogram with sides given by flows of the first two vector fields. More precisely, it measures the rate at which the initial and final vectors become more different as the parallelogram increases in size.

The reason that the Riemann tensor is a tensor comes directly from what it's supposed to do. It measures a type of rate of change of a vector field, and as such, shares all the linear properties that rates of change have.

$\endgroup$
3
$\begingroup$

Since you're taking a general relativity course, let me answer this from the point of view of a physicist.

Have you heard of tidal forces in general relativity? The intuition is that, in a spacetime with gravity, a family of objects that are initially moving at the same velocity but are positioned slightly displaced from one another will deviate from one another as they move over time. Said another way, gravity is something that pulls objects apart. Think of astronauts being stretched like spaghetti when they fall into black holes.

To describe this physical intuition mathematically, let us parametrise the positions of our of family of object as $x^\mu(s,\tau)$, where $\tau$ is proper time and $s$ is a spacetime parameter that parameterises the various objects in our family (i.e. $s$ parametrises the different body parts of our poor astronaut). Assuming that these objects travel along geodesics, which is true of freely falling objects in general relativity, you can prove that the deviation of the paths of these various objects in our family is given by, $$\frac{D^2 x^\mu} {d \tau^2} = R^\mu_{\nu\rho\sigma} \frac{dx^\mu}{d\tau} \frac{dx^\rho}{d\tau} \frac {dx^\sigma}{ds}.$$ In other words, the Riemann curvature tensor $R^\mu_{\nu\rho\sigma}$ , as defined in your course, is precisely the thing that measures the strength of gravitational tidal forces! The bigger the Riemann curvature tensor, the more strongly neighbouring objects are pulled apart by gravity.

For a proof of this tidal force formula, look up "geodesic deviation" in wikipedia.

Now, rather than starting with your definition of the Riemann curvature tensor and showing that it obeys this tidal force property, why not turn the argument the other way round and define the Riemann curvature tensor to be the quantity that appears in this tidal force equation, which, after some calculation, turns out to agree with definition from your course?

Finally, you asked why it is that the Riemann tensor must be a tensor. The tensorial property is crucial if you're going to use this quantity in general relativity. The whole point of GR is to write equations of motion that look the same in any coordinate system, and tensors are the mathematical quantities that do this for you.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .