The Christoffel symbols are symmetric with respect to the lower indices. This property is based on the geometric condition of path independence of the incremental vector. (Geometry of the flat space supports this property.)

The Christoffel symbol formula is derived based on the condition of symmetry of basis vectors. The Christoffel symbol therefore becomes symmetric with respect to the lower indices, for a curved space-time defined by the Schwarzschild metric, even though we do not have a geometrical picture for the four-dimensional curved space-time.

See the attached image for the verification of symmetry of basis vector in a curved space. But this leads to contradictory results for the unit vector derivatives.

It therefore becomes difficult to write the expressions for the unit vectors and define a coordinate system for the curved space. Hence, it is difficult to write a force vector or any other vector. To overcome this problem, it is assumed that the space is normally curved and approximately flat in a local inertial frame. This assumption can help us to write the vectors locally, but still not sufficient to provide the geometrical justification for the Christoffel symbol symmetry in a curved space.

Question: If it is incorrect to presume the symmetry condition for the conventional Christoffel symbol in a curved space-time, then do we need to first draw a geometrical picture of the four-dimensional curved space-time to write an appropriate Christoffel symbol formula?


Should we attempt to get desired results (similar to the Schwarzschild metric results) for planetary orbits from the three-dimensional flat-space Lagrangian analysis, by making appropriate modifications in the potential function?

enter image description here

  • $\begingroup$ What source are you using which incorporates the imaginary unit into the definition of your basis vectors? $\endgroup$
    – J. Murray
    Commented Apr 27, 2021 at 12:06
  • $\begingroup$ Hi NSRG. Do you know about torsion? $\endgroup$
    – Qmechanic
    Commented Apr 27, 2021 at 12:25
  • $\begingroup$ Can you please avoid including images of text, type in the relevant parts and equations, to keep questions searchable. $\endgroup$
    – ohneVal
    Commented Apr 27, 2021 at 12:59
  • $\begingroup$ You can incorporate imaginary units in the incremental vector to get -ve terms in the metric. Results of formula (D) from the image attached are same as the curvature tensor components of tensor analysis. The imaginary units disappear while calculating Riemann curvature tensor coefficients or Ricci tensor coefficients. $\endgroup$
    – NSRG
    Commented Apr 27, 2021 at 13:00
  • $\begingroup$ I have already raised a question about torsion. But not answered so far. $\endgroup$
    – NSRG
    Commented Apr 27, 2021 at 13:01

1 Answer 1


I don't actually understand what it means to "draw a geometrical picture of the four-dimensional curved space-time to write an appropriate Christoffel symbol formula." The answer to the title question,

Is it incorrect to assume the Christoffel symbol symmetry (with respect to the lower indices) for a curved space-time?

is no. The Christoffel symbols are symmetric in their lower indices by definition.

The problem you're running into is that the Christoffel symbols are the connection coefficients in a coordinate (or holonomic) basis. That is, given some coordinate system $x^\mu$, the coordinate-induced basis vectors are $\frac{\partial}{\partial x^\mu}$, and it is with respect to this basis that the Christoffel symbols are defined. The symmetry of these basis vectors - by which I assume you mean the fact that they commute with one another - is an inherent property of partial derivatives, which leads to the symmetry of the lower indices of the Christoffel symbols.

Now, this basis is very convenient for a lot of things, but it is not orthonormal; the metric components $g_{\mu\nu} \equiv \mathbf g(\frac{\partial}{\partial x^\mu},\frac{\partial}{\partial x^\nu})\neq \eta_{\mu\nu}$ in general. For that reason, you may wish to define orthonormal basis vectors$^\dagger$ $\mathbf e_\mu \equiv a_\mu^{\ \ \nu} \frac{\partial}{\partial x^\nu}$ such that $\mathbf g(\mathbf e_\mu,\mathbf e_\nu) = \eta_{\mu\nu}$. These vector fields are often called a tetrad, vierbein, or orthonormal frame. The orthonormal spherical coordinates you quote are an example of such a frame field.

You're always free to do this. However, in general the basis $\mathbf e_\mu$ is not a coordinate-induced basis, meaning that there is no possible coordinate system $y^\mu$ such that $\mathbf e_\mu = \frac{\partial}{\partial y^\mu}$. As a result, the basis vectors may not commute with one another, which implies that the connection coefficients as expressed in this basis will not have symmetry in its lower indices. However, note that using the definition I gave above, these connection coefficients should not be called Christoffel symbols, because the latter specifically refer to a coordinate basis; instead, the name spin connection is typically used. For more information about this, see this nice answer by ACuriousMind.

$^\dagger$Whatever resource you're using apparently incorporates the imaginary unit such that the metric components are $\delta_{\mu\nu}$ rather than $\eta_{\mu\nu}$. This seems misguided to me, as it obscures the geometrical nature of the theory.

  • $\begingroup$ Thanks for your reply. In my earlier question about torsion, I have stated the standard Christoffel symbols formula. The derivation assumes basis vector symmetry. This symmetry requires support of geometry of space and can not be merely a mathematical assumption. A geometrical framework which supports this assumption can be described as a geometrical picture. However, I shall go through your reply in detail. Thanks $\endgroup$
    – NSRG
    Commented Apr 27, 2021 at 13:06
  • $\begingroup$ @NSRG whether basis vectors commute is a property of that basis unrelated to the curvature of space. In flat 2D space, the Cartesian basis vectors commute but the orthonormal polar basis vectors do not. Coordinate bases always commute as a result of the symmetry of mixed partial derivatives, and the curvature of the space has no bearing on this fact. $\endgroup$
    – J. Murray
    Commented Apr 27, 2021 at 13:14
  • $\begingroup$ I have written the 4 basis vectors for the 4 coordinates. You can obtain the derivatives of unit vectors from Christoffel symbol values. All the basis vectors follow the symmetry of basis vectors principle amongst each other. This leads to contradictions as discussed in image and it is not possible to write unit vectors. Therefore how can we even define a coordinate system for curved space? $\endgroup$
    – NSRG
    Commented Apr 27, 2021 at 13:32
  • $\begingroup$ The same unit vector has two contradictory results in two derivatives. Please see image. It is not possible to write any expression for the unit vector which can satisfy both the contradictory statements. If we can't write unit vectors then it is not possible to draw a geometrical picture of the space $\endgroup$
    – NSRG
    Commented Apr 27, 2021 at 13:52
  • 1
    $\begingroup$ @NSRG The unit vectors you describe are not a coordinate basis as defined in my answer. Their derivatives are not symmetric for precisely this reason, and so the connection coefficients you obtain by computing their derivatives are not Christoffel symbols. They are the coefficients of a spin connection, again mentioned in my answer. $\endgroup$
    – J. Murray
    Commented Apr 27, 2021 at 13:59

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