In A First Course in General Relativity, Schutz asks the reader to prove that $\nabla \vec{V}$ is a $(1,1)$-tensor, where $$(\nabla\vec{V})^\alpha_{\ \ \ \ \beta} \equiv V^\alpha_{\ \ \ \ ;\beta} \equiv V^\alpha_{\ \ \ \ ,\beta} + V^\mu\Gamma^\alpha_{\ \ \ \ \mu\beta}.$$

Now, I know that this can be shown if one knows the way $\Gamma^\alpha_{\ \ \ \ \mu\beta}$ transforms with a coordinate transformation. However, Schutz seems to have something else in mind, for he delays asking about how $\Gamma^\alpha_{\ \ \ \ \mu\beta}$ transforms to a later exercise.

So, without taking the traditional route, how does one show $\nabla \vec{V}$ is a tensor?

  • 1
    $\begingroup$ It's a (1,1) tensor if it can eat a vector and produce a 1-form, or if it can eat a 1-form and produce a vector. Is that definition enough to prove what you want? $\endgroup$
    – DanielSank
    Mar 7, 2017 at 0:14
  • $\begingroup$ If you do not know the way $\Gamma$ transforms under coordinate transformations, how did you define it? $\endgroup$
    – ACuriousMind
    Mar 7, 2017 at 12:46
  • 1
    $\begingroup$ @ACuriousMind The Christoffel symbols can be defined as the array of numbers such that $\partial\vec{e}_\alpha/\partial x^\beta = \Gamma^\mu_{\ \ \ \ \alpha\beta}\vec{e}_\mu$, where $\vec{e}_\alpha$ are the coordinate basis vectors. $\endgroup$
    – Doubt
    Mar 27, 2017 at 23:21
  • $\begingroup$ @DanielSank - I think yours is the answer the author has in mind. Would you mind posting this as an answer with more rigor? $\endgroup$
    – Doubt
    Mar 27, 2017 at 23:24
  • $\begingroup$ Isn't this kinda circular? In GR the covariant derivative $\nabla$ is <em>defined</em> such that $\nabla$ applied to a tensor yields a tensor with rank 1 higher, and the form and properties of the Christoffel symbol follow from that definition/requirement, no? $\endgroup$ Aug 7, 2019 at 9:04

1 Answer 1


One can show that the covariant derivative $(\nabla \mathbf{v})^\alpha_{~~\beta}\equiv v^\alpha_{~~;\beta}$ transforms like a $(1,1)$-tensor without using properties of the christoffel symbol but in order to do so one needs to start from a different expression for the covariant derivative: the following holds for the total differential of a vector field $\mathbf{v}=v^\mu(q)\mathbf{e}_\mu$: $$d\mathbf{v}=v^\mu_{~~;\nu}dq^\nu\mathbf{e}_\mu. \tag{1}$$

The total derivative is a physical quantity and therefore is required to be invariant under coordinate transformations. In the following we use the two coordinate systems $\{\alpha^\mu,\beta^\mu\}$ with basis vectors $\{\mathbf{a}_\mu,\mathbf{b}_\mu\}$. The total differential in the different systems reads: $$d\mathbf{v}=\bar v^\mu_{~~;\nu}d\alpha^\nu\mathbf{a}_\mu=v^\mu_{~~;\nu}d\beta^\nu\mathbf{b}_\mu. \tag{2}$$ Plugging standard identities for the transformation of basis vectors and coordinate differentials $$d\beta^\nu=\Lambda^\nu_{~~\mu} d\alpha^\mu=\bar\Lambda_\mu^{~~\nu} d\alpha^\mu,\tag{3a}$$ $$\mathbf{b}_\mu=\Lambda_\mu^{~~\nu}\mathbf{a}_\nu=\bar\Lambda^\nu_{~~\mu}\mathbf{a}_\nu,\tag{3b}$$ into eq. (2) results in $$d\mathbf{v}=v^\mu_{~~;\nu}(\bar\Lambda_\kappa^{~~\nu} d\alpha^\kappa)(\bar\Lambda^\lambda_{~~\mu}\mathbf{a}_\lambda)= \bar\Lambda^\lambda_{~~\mu}\bar\Lambda_\kappa^{~~\nu}v^\mu_{~~;\nu}d\alpha^\kappa\mathbf{a}_\lambda=\bar\Lambda^\mu_{~~\lambda}\bar\Lambda_\nu^{~~\kappa}v^\lambda_{~~;\kappa}d\alpha^\nu\mathbf{a}_\mu\tag{4}$$ and therefore $$\bar v^\mu_{~~;\nu}=\bar\Lambda^\mu_{~~\lambda}\bar\Lambda_\nu^{~~\kappa}v^\lambda_{~~;\kappa} \quad\text{q.e.d.}\tag{5}$$

Eq. (5) is the explicit transformation of a mixed $(1,1)$ tensor of rank 2. The $\Lambda$-tensors perform coordinate transformations between the two systems: $$\Lambda^{\mu}_{~~\nu}=\frac{\partial \alpha^\mu}{\partial \beta^\nu}=\mathbf{a}^\mu \cdot \mathbf{b}_\nu,$$

$$\bar\Lambda^{\mu}_{~~\nu}=\frac{\partial \beta^\mu}{\partial \alpha^\nu}=\mathbf{b}^\mu \cdot \mathbf{a}_\nu,$$



Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.