Covariant derivative of four-position

May someone confirm or deny that covariant derivative of four-position is just metric tensor?

I mean: $$\nabla_{\gamma}X_{\alpha} = g_{\gamma \alpha}$$

When I try to rewrite it with base vectors it is $$\nabla_{\gamma}X_{\alpha} = \frac{\partial x_{\alpha} \vec{g}^{\alpha}}{\partial q_{\gamma}} = \frac{\partial x_{\alpha}}{\partial q_{\gamma}} \vec{g}^{\alpha} + x_{\alpha} \frac{\partial \vec{g}^{\alpha}}{\partial q_{\gamma}}$$

Does above two parts of equation sum to metric tensor?

• $X_\mu$ is not a tensor or a tensor density. Its covariant derivative is not defined at all. – Prahar Apr 9 at 20:29

In the general case, 4-position is not a vector, so it does not have a covariant derivative. Describing it as a vector seems to assume Minkowski metric, in which case the covariant derivative is simply the partial derivative

We do have $$x^i_{,j} = {\partial x^i \over \partial x^j} = g^i_j$$

$$x_{i,j} = g_{ij}$$

In general relativity, we don't equip spacetime itself with a vector space structure, which means that the "four-position" is not a four-vector. As a result, the object $$x_\mu$$ whose components are given by $$x_{\mu}=g_{\mu \nu} x^\nu$$ is not a covector. One can show this by its transformation properties.

If we change coordinate systems from $$x$$ to $$y$$, then $$g_{\mu\nu}^{(x)} \rightarrow g^{(y)}_{\mu \nu} = \frac{\partial x^\alpha}{\partial y^\mu} \frac{\partial x^\beta}{\partial y^\nu}g^{(x)}_{\alpha\beta}$$ $$x^{\nu}\rightarrow y^\nu$$

and so

$$x_\mu = g^{(x)}_{\mu \nu}x^\nu \rightarrow g^{(y)}_{\mu \nu}y^\nu = \frac{\partial x^\alpha}{\partial y^\mu} \frac{\partial x^\beta}{\partial y^\nu}g^{(x)}_{\alpha\beta} y^\nu = \frac{\partial x^\alpha}{\partial y^\mu}g^{(x)}_{\alpha \beta} \left(\frac{\partial x^\beta}{\partial y^\nu} y^\nu \right)$$

If this object were a covector, it would transform as $$x_\mu \rightarrow y_\mu = \frac{\partial x^\alpha}{\partial y^\mu} x_\alpha = \frac{\partial x^\alpha}{\partial y^\mu} g^{(x)}_{\alpha\beta} x^\beta$$

Comparing this to what we found above, this is only true if $$\frac{\partial x^\beta}{\partial y^\nu}y^\nu = x^\beta$$ i.e. if the coordinate transformation is linear. Since coordinate transformations are, in general, not linear, then this equality does not hold, and $$x_\nu$$ are not the components of a covector.

Since covariant derivatives are defined by their actions on tensor fields (scalar fields, vector fields, etc), then we're dead in the water - the collection of symbols

$$\nabla_\gamma x_\alpha$$

does not correspond to a well-defined object.

• If the transformation is linear, then it's tensor by definition. A tensor transforms as a multilinear object. The covariant derivative of a $(r,s)$ tensor generates a $(r,s+1)$ tensor. The $X_{\alpha}$ is a $(0,1)$ tensor and becomes a $(0,2)$ tensor - it's just not a metric tensor - and the OP's calculation of the covariant derivation of a $(0,1)$ tensor is wrong. – Cinaed Simson Apr 9 at 20:56
• @CinaedSimson I believe you are missing the point that I made - $x_\mu$ are not the components of a tensor at all. The transformation whose linearity (or lack thereof) I refer to in my answer is a general coordinate transformation, not the action of a tensor. – J. Murray Apr 9 at 20:58