I am currently enrolled in a General Relativity course, and was taught about the connection but I can't really wrap my head around it qualitatively. All I can think of is that it must have something to do with the coordinate system that one uses to describe a space(or space-time) but I can't give it a geometric interpretation**(Check the 2nd EDIT)**.
Thank you.

EDIT 1: I am searching for an explanation in terms of the curvature of the space and the coordinates.

EDIT 2: Upon searching for an answer, I found that the relation of the connection with the covariant derivative offers some insight: The connection term in the covariant derivative is an extra term to the normal derivative that is there in order to account for the changes in the coordinate basis vectors. If anybody could use this type of logic to give a complete geometric interpretation of the connection, it would be great!

  • $\begingroup$ I don't know if you're familiar with concepts like vector bundles so I'm not sure if the mathematical definition will help you. The geometric intuition should be that a connection allows you to transport vectors along smooth curves. In particular this allows you to differentiate vector fields on a manifold, or in general sections of vector bundles. $\endgroup$
    – Qidi
    Nov 19, 2016 at 1:28
  • $\begingroup$ @Qidi unfortunately I am not familiar with vector bundles. So, why does the connection allows us to transport vectors along smooth curves? What does it do, geometrically? $\endgroup$ Nov 19, 2016 at 1:33
  • $\begingroup$ I'm not sure how you exactly define a connection in your course, so I'm not sure how to explicitly explain it. Usually by a connection(or covariant derivative) one means something that takes 2 vector fields and produce a vector field, this can be interpreted as differentiating a vector field along another. For instance in $\mathbb{R}^n$ the derivative defines a connection. Assume you have a vector field along some curve, then requiring such a field to be parallel exactly means the derivative along the curve is zero. You can try to do the same for e.g. curved surfaces. $\endgroup$
    – Qidi
    Nov 19, 2016 at 1:50
  • $\begingroup$ @Qidi Well, we initially defined the connection via the geodesic equation. Afterwards, we defined the covariant derivative and connected the two. But, still I don't understand how to interpret it graphically using this knowledge. For instance, through the definition of the covariant derivative, the connection gives an extra term besides the normal derivative but I do not understand why it is needed except that it makes the covariant derivative a tensor(transforms properly under general transformations). $\endgroup$ Nov 19, 2016 at 1:56
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    $\begingroup$ You may find the following discussion to be helpful: math.stackexchange.com/q/47095/272316c $\endgroup$ Nov 19, 2016 at 3:56

3 Answers 3


Maybe it helps to look at some simple example.

In this example, I just take the normal Euclidean plane with normal Cartesian coordinates. However I attach a non-standard basis to each point. Namely, to describe vectors at point $(x,y)$ I use the basis $$e_1 = \begin{pmatrix} \cos x \\\sin x \end{pmatrix}, e_2 = \begin{pmatrix} -\sin x \\\cos x \end{pmatrix}$$ So you see, the basis in general is different at different points (and not the slightest related to the metric!). It is, however, differentiable.

Now let's further assume that on my Euclidean plane, I have a constant vector field $$v(x) = \begin{pmatrix}a\\b\end{pmatrix}$$ So nothing very interesting. Except that I'm now going to express it in my local basis. Then I get $$v(x) = v^1 e_1 + v^2 e_2 = (a\cos x + b\sin x) e_1 + (-a\sin x + b \cos x) e_2$$ That is, although the vector field is constant, the components of the vector field in my contrieved basis are not; instead they depend on $x$ (the reason why they don't also depend on $y$ is only because I chose the basis that way).

In particular, we get for the partial derivatives of the components: \begin{align} v^1{}_{,x} &= -a\sin x + b\cos x & v^1{}_{,y} &= 0\\ v^2{}_{,x} &= -a\cos x - b\sin x & v^2{}_{,y} &= 0 \end{align} This apparent variability of the vector field is only due to the change of the basis; we have actually defined the vector field to be constant. But of course, not all vector fields are constant, so the question arises: How can we distinguish between actually changing vector fields and component changes that are just artefacts of the vectors?

Well, we are still in an Euclidean space, so we can just calculate the true change. So let's say we start at point $(x,y)$ and move along the path $(x+t,y)$, so we recover the derivative in $x$ direction. Then we have \begin{align} \frac{\mathrm dv}{\mathrm dt} &= \frac{\mathrm d}{\mathrm dt}(v^1 e_1 + v^2 e_2)\\ &= v^1{}_{,x} e_1 + v^1 e_{1,x} + v^2{}_{,x} e_2 + v^2 e_{2,x} \end{align} If we write this "true derivative" with semicolon instead of comma, we + therefore get for the components of $\frac{\mathrm dv}{\mathrm dt}$: $$v^i{}_{;x} = v^i{}_{,x} + v^1 \omega^i(e_{1,x}) + v^2 \omega^i(e_{2,x}) = v^i{}_{,x} + \omega^i(e_{j,x})v^j$$ where I used the notation $\omega^i$ for the dual basis of $e_i$, that is $\omega^i(e_j)=\delta^i_j$, and in the last step the Einstein summation convention.

As you see, the "true" derivative decomposes into two parts: The component derivative, and a correction term. The correction is the contraction of the vector with an object with three indices: An upper and a lower index for the basis, and an additional lower index for the coordinate direction in which we are differentiating. These three-index objects are the connection coefficients, $\Gamma^i_{jk}=\omega^i(e_{j,k})$.

Now of course nobody would choose such a strange basis. Instead, the basis is chosen to point in the direction of the coordinates. This means that the basis indices coincide with the coordinate indices; that is, all indices correspond to the same set of directions. Therefore you don't have to distinguish between those two types of indices (unlike above, where the basis indices were $1$ and $2$, while the coordinate indices were $x$ and $y$).


The key to understanding this cycle of ideas is parallel transport. In classical physics we are use to moving vectors around. Take a vector starting at one point and transport the vector parallel to itself so that it now starts at another point. This also comes into play when we differentiate a vector field. For that we must compare two vectors with different (but close by) initial points. Here again we move one of the vectors parallel to itself so that the two vectors have the same initial point and then we take the difference etc.

However if we traverse a path along a curved surface how do we move a vector parallel with itself ? What is the CONNECTION between vectors at different points ? It turns out that there is a well defined notion of parallel transport along any curve on a surface, although it is somewhat counterintuitive. Take the case of traversing a line of latitude different from the equator. Start with a vector pointing toward the north pole, as you move taking the vector parallel to itself it will no longer point to the pole. And if you make a circuit of the line of latitude and return to your starting point the vector will no longer coincide with the vector you started with. And the closer that line of latitude is to the pole the greater the discrepancy.

Covariant differentiation is the differentiation of a vector field along a path where in order to compare nearby vectors we move them parallel according to the parallel transport on the surface. If for example we start with one vector and create a vector field along a curve by transporting that vector parallel, the the covariant derivative of this field will be zero. In fact that is one way to define the concept of parallel transport. Covariant differentiation and parallel transport are expression of the same thing. May books take covariant differentiation as the basic concept.

Finally, how does a connection arise ? It is defined by the Christoffel symbols, as you know, and these come from differentiating the vectors on the surface in three dimensional space, thus no longer getting a vector tangent to the surface, and then projecting onto the tangent plane of the surface. This is seen in the defining 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*}

Where $\mathbf{N}$ is the normal to the surface.

A few final comments, a connection is weaker than a Riemann metric. A metric allows one to define both a connection and a curvature. But the connection alone does not define the curvature. It is a mistake to think it has to do with the coordinate system, it is a geometric idea. And I have been speaking of surfaces as if they were imbedded in three dimensional space, this is a source of intuition but the ideas generalise to higher dimensional manifolds, not imbedded in space.

Well, there is more to be said but I hope this helps and gives you an intuition with which to move forward.

  • $\begingroup$ Thanks for the great answer. I have two question though if you are willing to answer them: What information does the connection lack that the metric has that makes it weaker than the metric? Also, "It is a mistake to think it has to do with the coordinate system, it is a geometric idea": I did not understand it; doesn't the connection has to do with the coordinate system? $\endgroup$ Nov 27, 2016 at 21:08
  • $\begingroup$ Both good questions. Not sure I have good answers. But I can say that curvature is defined by $K=\frac{R_{1212}}{|g|}$, and $R_{1212}$ involves only the christoffel symbols, and $|g|$ is an area measure, so I would suggest that this is the missing ingredient. As for the second remark, I just meant that the ideas of differential geometry are independent of the coordinate systems used, on the other hand the christoffel symbols are not a tensor, so maybe that remark should be taken with a grain of salt. $\endgroup$ Nov 27, 2016 at 21:18
  • $\begingroup$ Oh ok, good. Thanks again. You said that you have a lot more to say on this topic, so would it be too much to ask to edit your answer to include some more of your great insights! I am missing intuition at the moment and it would really help! $\endgroup$ Nov 27, 2016 at 21:21
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    $\begingroup$ At the moment I cannot think of anything I can reasonably add to my answer. I hope that reading the first paragraphs will give you some intuition, check out Differential Geometry of Curves and Surfaces by DoCarmo. $\endgroup$ Nov 27, 2016 at 21:25


If we combine GR and QT by adding their "co-variant" derivatives then we get a double curvature tensor.

f;n = f,n + [Hn , f]– +[Gn , f]+ = f,n + Hn f – f Hn + Gn f+ f Gn (to be called the QTGR-derivative of f in 4-D spacetime)

for short f;n = f,n+ Xn f + f Yn and similarly f;p = f,p+ Xp f + f Yp (where X (=G+H) and Y (=G–H) )

This gives, after some algebraic effort using the classical curvature tensor derivation :

f;np – f;pn = Rnp f + f Řnp

Rnp = Gn,p – Gp,n +[Gp,Gn]– + Hn,p – Hp,n +[Hp,Hn]– +[Hp,Gn]– +[Gp,Hn]–

Řnp = Gn,p – Gp,n +[Gn,Gp]– + Hp,n – Hn,p +[Hn,Hp]– +[Hp,Gn]– +[Gp,Hn]–

(The n and p above are subscripts, +/- signs after brackets are for anti-commutators and commutators)

G represents the part played by Christoffel connection Gamma through an anti-commutator, and H represents the part played by the 4-D Hamiltonian through a commutator.

f is any 4*4 matrix representing a space-time dynamics of the system.

We have thus two curvature tensor-operators - classical GR has missed out the second Řnp. Hence classical GR is an incomplete theory.

Mass should get a complex phase as m0.A.exp(i.theta)

The energy momentum tensor is then : Tjk= m0.A.exp(i.theta) ujuk/v0

(j, k and 0 are subscripts, u is 4-velocity and the ratio m0/v0 the rest-mass/rest-volume of a "particle", A is "wave"-amplitude, with j, k, n and p = 0,1,2,3 )

The phase theta includes all non-gravity forces, fields and gauges.

A theory formulated thus will combine GR and QT in 4-D.


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