The Riemann curvature tensor can be expressed using the Christoffel symbols like this:

$R^m{}_{jkl} = \partial_k\Gamma^m{}_{lj} - \partial_l\Gamma^m{}_{kj} + \Gamma^m{}_{ki}\Gamma^i{}_{lj} - \Gamma^m{}_{li}\Gamma^i{}_{kj}$

How did they come up with this? What was the idea?

I searched the web but the descriptions I found were too formal, and I was unable to decipher what the author tries to describe.

So I'm looking for some thoughts or an easy paper I can start from and derive this formula myself.

  • 1
    $\begingroup$ It can be explained using paralell transport along a closed path or calculating the commutator of the covariant derivate. $\endgroup$
    – jinawee
    Oct 17, 2013 at 18:06
  • 1
    $\begingroup$ Probably not going to be an easy thing to do, but why not start here "they" started and try solving $f^\alpha_{;\beta\gamma}-f^\alpha_{;\gamma\beta}$ as best you. $\endgroup$
    – Kyle Kanos
    Oct 17, 2013 at 18:16
  • $\begingroup$ The idea is to express the curvature of a manifold totally intrinsically, i.e. without reference to a normal vector. The fact that this is possible is sort of amazing. $\endgroup$ Oct 17, 2013 at 18:22
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    $\begingroup$ It seems more suitable for math.SE. $\endgroup$
    – MBN
    Oct 17, 2013 at 19:18
  • $\begingroup$ @MBN The reason why I want to understand this is that I'm building my pathway towards the general relativity. I thought there are more people here who can help with this than there. $\endgroup$
    – Calmarius
    Oct 17, 2013 at 19:52

5 Answers 5


The idea is that we want to define some notion of curvature for a manifold that intuitively agrees with the intuition we have about curvature.

The genius insight that leads to the desired definition is the notion of parallel transport. Speaking non-rigorously here, the basic idea is that if you transport a tangent vector on a manifold parallel to itself all the way around a closed curve, then the vector will come back to itself in flat spaces, but it will become a different vector in a curved space.

To see why the notion of parallel transport has anything to do with curvature, think for example, of the Euclidean plane $\mathbb R^2$ versus the two-dimensional sphere $S^2$.

Consider the curve consisting of an equilateral triangle with one vertex at the origin. Now imagine placing a vector emanating from the origin, and imagine moving that vector along the triangle, keeping its "tail" on the triangle, and making sure to keep the vector parallel to itself the whole time. If you transport the vector once around the triangle back to the origin in this way, then you get the same vector back.

Something drastically different happens if you do the same thing on the sphere as the following diagram from the wiki page on parallel transport indicates

enter image description here

If you move a vector from point A back to itself along the curve indicated in the diagram, the vector does not return to itself. This happens because the sphere is curved.

In fact, the notion of parallel transport can be used to completely characterize what we mean by curvature. The logic you'll find in many books on GR and differential geometry is roughly as follows:

  1. Define the notion of a connection (basically this defines what you mean by taking derivatives on the manifold).

  2. Use the connection to define the notion of parallel transport which agrees with our intuition of parallel transport in, for example, the sphere example above.

  3. Show that there is a tensor that measures precisely how much the components of a vector change when it is parallel transported along a small closed curve on the manifold.

  4. Call this tensor the Riemann tensor, and use it as the object that captures the notion of curvature.

There is a great discussion of this in a lot of books. I personally like the discussion on pages 36-38 of Wald's General Relativity.

Addendum. Wald actually shows that if you consider a curve bounding a small two-dimensional patch parameterized by coordinates $s$ and $t$ on the given manifold, then the change $\delta v^a$ in the components of a vector transported along the boundary of this patch satisfies \begin{align} \delta v^a = \delta t\,\delta s\, v^dT^cS^bR_{cbd}^{\phantom{cbd}a} \end{align} where $\delta t\,\delta s$ is the area of the patch, and $T^c$ and $S^b$ are the tangents to the curves of constant $s$ and $t$ respectively.

  • $\begingroup$ I read somewhere that the idea of the curvature is that you transport the vector around a small patch, then divide the vector's deviation by the area of the patch. Both the devation and area will be tiny, by using the L'Hôpital's rule you can get a curvature vector. Is this correct? Do I need to transport the vector along geodesics to get the correct result? How do you apply the idea for higher dimensions? Eg how do you transport a vector around a small cube for example? $\endgroup$
    – Calmarius
    Oct 17, 2013 at 20:10
  • $\begingroup$ @Calmarius For your first question, see the addendum to my answer. You do not need to parallel transport along geodesics. The idea applies to all dimensions in the same way; the curve along which you parallel transport is one-dimensional in all cases. $\endgroup$ Oct 17, 2013 at 22:06
  • $\begingroup$ So basically let's have 2 vectors to describe the sides of a parallelogram at a point, and have a third to be transported along the parallelogram. So If I do the math and subtract the resulting vector from the original I should get the Riemann curvature tensor, am I right? $\endgroup$
    – Calmarius
    Oct 22, 2013 at 10:28
  • $\begingroup$ @Calmarius That's basically it with the caveat that you should consider a small parallelogram. $\endgroup$ Oct 23, 2013 at 1:32

See Joshphysic's answer for the details; I'd like to add some "overview" comments. The essential, fundamental ideas here are:

  1. Deviation from Euclid's parallel postulate (see Wiki page "Parallel postulate");
  2. How "badly" vector fields in the manifold in question fail to be integrable (see the foreword to the Wiki page on "Riemann Curvature Tensor" ) to an isometry with a truly Euclidean manifold (i.e. $\mathbb{R}^N$)

The deviation in (1) is nought if and only if the curvature tensor and torsion tensor vanish. The deviation in (2) is measured by the curvature tensor as in Joshphysics's answer by its "non holonomy" i.e. how much a parallel transport of a test vector around a small loop varies divided by that loop's "area". The fundamental theorem of Riemannian geometry shows that we can always define a unique connexion (the Levi-Civita connexion) that absorbs the torsion into the curvature, so that both concepts above are addressed fully by $\mathbf{R}(X,Y)$. Most GR is done with this choice, so torsion isn't discussed much. But it's still worth reading up on this as you learn about curvature. Torsion takes on a fundamental role in Einstein-Cartan theory, but I'm just dropping names here as, like Sergeant Schultz, I know nothing about this - this is a future intellectual project for me.

Joshphysic's Wald reference is good, I also like Schutz's treatment of the ideas as given in Chapter 6 of his "Geometrical Methods of Mathematical Physics". His latest version of "A First Course in General Relativity" is a bit light on on these concepts as he has had to shift some material out to his "Geometrical Methods" book to make way for discussions of experimental GR, which is an exciting field at the moment.

In passing, check out some lovely diagrams see this answer that Bakhoda wrote for me on Maths SE.

If you're willing to do a bit of work, you can turn to chapter 14 of Roger Penrose's "Road to Reality" (called "Calculus on Manifolds""). Simply reading this will give you a good top-level understanding. If you go back and do all the exercises, your understanding will be pretty thorough - although this is quite a project.

Another basic description of these ideas is given in Chapter 3 of Wulf Rossmann's "Lectures on Differential Geometry". You can download from there: Rossmann is a bit of a mathematical Feynman - working tirelessly to seek the clearest and most elementary descriptions of things.

For another read with a Penrose-esque flavor with the most magnificent and lovingly drawn diagrams you ever saw, the relevant parts of Misner, Thorne and Wheeler is good, but this is monstrous volume and I don't have it before me so I can't tell you wherein to find it. But it ought to be pretty obvious if you get your hands on a copy.

  • $\begingroup$ @dj_mummy IMO the co-ordinate free notation is essential for understanding and learning - indices are just too much information but you do need them - a rule of thumb that works for me is that you'll need indices for detailed calculations (it's like programming, you can write them and understand them yourself, but much less so something done by someone else!), index-free to talk about ideas and proofs about properties ("qualitative proofs"). I've never practised Penrose's funky diagram notation - they look rather promising. $\endgroup$ Oct 18, 2013 at 7:53

Thought I'd take a slightly different approach couching the above in terms of a first course in vector calculus.

Suppose you have a curve $\vec{r}(t)$ in $R^3$. It could be a straight line, a circle, a helix, cycloid, etc.

You can associate to any point on a well behaved curve a unit tangent vector $\hat{T}=\frac{\frac{d\vec{r}}{dt}}{|\frac{d\vec{r}}{dt}|}$

In turn you can have a unit normal such that , $\kappa(t) \hat{N}=\frac{d\hat{T}}{dt}$ where $\hat{N}$ is the unit normal and $\kappa$ is the curvature,

and Unit Binormal vector $\hat{B}=\hat{N}\times\hat{T}/|\hat{N}\times\hat{T}|$

It can be shown that $d\hat{B}/dt=-\tau\hat{N}$ where $\tau$ is Torsion.

Using the chain rule, $\frac{d\vec{r}}{dt}=\frac{d\vec{r}}{ds}\frac{ds}{dt}$ where $ds$ is infinitesimal arc length. It's typically assumed $ds/dt=1$ to keep the math easier. It also has some interesting physical implications regarding pseudo forces which can help give an intuitive understanding of gravitational effects of a curved space.

Curvature is defined as $\kappa(t)=\frac{d\theta}{ds}$ where $d\theta$ is a measure if infinitesimal change in the direction of the unit tangent vector, and again $ds$ is the infinitesimal arc length.

$\hat{N}, \hat{B}$, and $\hat{T}$ form an orthonormal, curve centric, basis. Certain relationships between them hold no matter how your coordinates change, if the curve is rotated about the z axis, reflected across some plane, moved somewhere else in $R^3$. These include curvature, vertices, and other geometric features.

A $d\theta$ is implied by the change in any unit vector. Vectors representing both magnitude and direction, holding the magnitude constant only allows for change in direction which can be represented as an angle change. Here there are 3 unit vectors to choose from, so there are 3 possible $d\theta$s. The derivatives of these unit vectors are vectors themselves expressible as those vectors: Frenet-Seret Curvature relations

Notice in these equations that the derivative of vectors on the left is a linear combination of vectors on the right. If we had a column vector made of the basis vectors $T_{ij}=<\vec{N},\vec{B},\vec{T}>$, the derivative of this column vector with respect to $ds$ would be some "matrix" $M$ multipled by $T_{ij}$. The double index is needed because $i$ selects which of the 3 vectors we care about, and $j$ represents which component of that vector we are interested in.

So $\frac{dT_{ij}}{ds}=M\cdot T_{ij}$ is a very compact form of the Frenet-Serret equations. It represents curvature by giving information on the derivatives of the unit vectors.

Roughly speaking this 2 indexed entity (called a rank 2 tensor) is a vector of vectors, or a nested vector. So, in a similar sense, is a matrix. They appear all over for example, there's the Maxwell Stress Tensor in Electromagnetism or the stress tensor in materials science.

A vector field associates with points in $R^n$ another element in $R^n$.

In non-cartesian coordinate systems, unit vectors can change from point to point. This means they have non-zero derivatives implying some concept of curvature in play. This in turn means their components change.

$\Gamma^a_{bc}=$ the $ath$ component of coordinate $c$ derivative of unit vector $b$. For example in spherical coordinates, $\Gamma^r_{\theta \theta}=\frac{-1}{r}$ These are Christofel Symbols of the Second Kind.

The Christofel Symbols have their own derivatives which also have implications regarding curvature.

So curvature can be categorized by Christofel symbols and their derivatives. Whereas matrix elements are referred to be a row/column pair, for Christofel symbols, we need which component of which derivative of which vector to be specified, implying 3 indices. (Despite requiring 3 indices, it is not itself a tensor, but that can be deferred). Taking the derivative of a tensor creates a tensor having an additional lower index. Rank is the number of indices of a tensor. So the derivative of the Christofel symbol has Rank 4.

Notice the Riemann Curvature Tensor is of rank 4. Also notice the form it takes and compare to the expression of curvature for an implicit curve: Curvature of implicit Curve. You'll find the more formal treatments use curves to illustrate the principles in play, basically generalizing from the primitive concepts of curvature of a curve to curvature of a Manifold.

This hasn't been especially rigorous, but hopefully helps develop an intuition for the concepts in play.


Curvature is such an important concept in mathematics and physics that there are a quite a number of ways of thinking about it. The oldest is Gaussian curvature and the most modern uses concepts in vector and principal bundles.

Gaussian Curvature

Let's start with the simplest curve. This is the straight line and obviously the curvature should be zero, that is it has no curvature. The next simplest curve is the circle. It's the simplest here because it's 1d and whatever curvature we assign it we can see that this curvature is the same at all points of the circle.

The question is what curvature should we assign it? A circle has only one number associated with it and that is the radius. Of course there are other numbers associated with the circle like its area or it's circumference but these are all wholly dependent on the radius. The obvious measure of the curvature we can take is just the reciprocal of the curvature, that is 1/r. We see that when the radius increases to infinity then the curvature goes to zero. And this makes sense because the arc of the circle at that point approaches a straight line.

Now we can see how to evaluate the curvature of an arbitrary curve. We choose a point and we fit a circle to that point that is tangent to it, in some sense, and then we say the curvature of the curve is the curvature of the circle that we fitted to it.

The next question to tackle is how to evaluate the curvature of a surface at a point p. Well, we take a plane through the surface there. This will give us a curve and we can take the curvature of this curve. However, this plane was arbitrary, so let's take every plane through this point and evaluate the curvature of every curve that this gives. Obviously, there is a minimum and maximum of these curvatures and they are called the principal curvatures, $\kappa_1, kappa_2$. The Gaussian curvature is their product, $\kappa_1, kappa_2$. Now, what Gauss discovered, is that this curvature is intrinsic to the surface and doesn't depend on how the surface is embedded in space. This impressed him so much he called it the Theorem Egregium (the remarkable theorem). He was right to be impressed as it's this property of the Gaussian curvature, generalised in many ways that informs the modern day concept, and in particular Einsteins GR and further afield, in particle physics: all the forces in the standard model - the electromagnetic, the weak and strong force can all be understood in terms of curvature (of principal bundles).

Obviously for GR we require evaluating the curvature for 4-dimensional manifolds and not just surfaces. We use a trick similar to what we've just done. Take a point p in a manifold, of any dimension. Then at this point it has a tangent space. Now we choose a plane through this point and this determines a surface in the manifold that goes through this point and which has this plane as it's tangent plane and we take the Gaussian curvature of this surface at this point. This gives us a set of curvatures associated with all the planes we can place through this point at this point. This curvature is called the sectional curvature. And it turns out the sectional curvature determines and is determined by the Riemannian curvature.

Riemannian Curvature

This has a simple but abstract definition. It's defined as

$R(u,v):=[\nabla_u,\nabla_v] - \nabla_{[u,v]}$

It's measuring the non-commutativity of tangent vectors on the manifold; or alternatively, the loss of parallel transport as we move around a small parallelogram.

It turns out that at a point p if we choose two tangent vectors u,v there then

$(R(u,v)w,z) =$ $(1/6)(d^2/ds.dt) [K(u+sz, v+tw) - K(u+sw, v+tz)]$ at the point $(s,t) = 0$

Where K is the sectional curvature and we can see that the left hand side is measuring non-commutativity in some sense too.

However, the Riemann curvature tensor is much more conveniant - it's a tensor and it satisfies three simple symmetry properties and it's abstract definition - though abstract - is straight-forward and easy to generalise (in the appropriate sense!) to other contexts unlike the the concrete and geometric definition of Gaussian curvature which is quite involved. This sort of thing occurs pretty often in mathematics and physics.

Another way to compare the two is to evaluate it on a simple situation and the simplest is that of a surface where:

$R_abcd = K(g_ac.g_db - g_ad.g_cb)$

Where K=R/2 is the Gaussian curvature, and in fact also, the sectional curvature. It equals half the scalar curvature R, which is just the complete contraction of the Riemann tensor. (Note, that $g_ab$ is the metric.)


To define the Riemannian curvature we require a metric on the manifold. This is a way of measuring lengths and angles. However, it turns out we can get away with less. All we need is a connection which tells us how to parallel transport from one point to another, that is it connects tangent spaces. An equivalent notion is holonomy which tells us how tangent vectors change if we transport them around small loops.


As per @R.Romero's answer to this, in 3D space Christoffel symbols specify the step-wise small change in the direction and length of each of the 3 unit basis vectors Tangent, Normal, and Binormal as measured from the object itself, with origin [0,0,0] also at the object itself:$$\vec T, \vec N, \vec B$$

cleverly expressed in terms of the lengths measured using these same T, N, B unit vectors themselves at this exact iteration moment, thereby giving us a total 3x3 matrice which will give us the next set of T, N, B axis direction and length that we can use at the next time-step iteration, is this correct? Then this process is repeated, so we have constantly changing numbers on a 3x3 matrice (which is defined as the Riemann tensor) as we move along in space.

So, in 4D spacetime, this will generalize to a 4x4 matrice since there are 4 basis vectors specified at each point in spacetime.

But since the basis vectors $$\vec T, \vec N, \vec B$$ keep changing every instant following the orientation of the object itself (ie. at any moment the object will always be facing forward along the Tangent (velocity) unit vector, with the Normal and Bi-normal unit vectors completing a orthonormal grid around it), how is it possible to relate this object-centered constantly changing coordinate system to a fixed coordinate system of a "camera" that is looking at the object moving in space?


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