# How to prove $\nabla\vec{V}$ is a tensor without transformation properties?

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?

• 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? Mar 7, 2017 at 0:14
• If you do not know the way $\Gamma$ transforms under coordinate transformations, how did you define it? Mar 7, 2017 at 12:46
• @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. Mar 27, 2017 at 23:21
• @DanielSank - I think yours is the answer the author has in mind. Would you mind posting this as an answer with more rigor? Mar 27, 2017 at 23:24
• 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? Aug 7, 2019 at 9:04

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