Why is the covariant derivative of a one-form $\nabla_{i}v_j=\frac{\partial v_j}{\partial u^{i}}-\Gamma^k_{~ij}v_k$?

I understood that the covariant derivative of a vector field is $$\nabla_{i}v^j=\frac{\partial v^j}{\partial u^{i}}+\Gamma^j_{~ik}v^k$$ Then why is the covariant derivative of a covector field $$\nabla_{i}v_j=\frac{\partial v_j}{\partial u^{i}}-\Gamma^k_{~ij}v_k$$ I tried from the first formula by lowering indices, but I just do not get the minus sign before the $$\Gamma$$. Probably a trivial question of some of you, but not for a beginner of tensor calculus like myself!

• Hint: The covariant derivative of a scalar is a partial derivative. Try doing the derivative of $v_j v^j$ and using the known expression for the covariant derivative of a vector field to arrive at the desired result. Feb 8, 2021 at 20:40
• If you want to try the brute force approach, you can break the two terms in $\nabla_i v_m = g_{m j} \nabla_i v^j = g_{m j} \partial_i v^j + g_{m j} \Gamma^j_{i k} v^k$, and find $g_{m j} \partial_i v^j = \partial_i v_m - v^j \partial_i g_{mj}$ and $g_{m j} \Gamma^j_{i k} v^k = - \Gamma^n_{im} v_n + v^j \partial_i g_{mj}$. Sep 26, 2021 at 10:10

I follow Carroll's method of deriving this. So, we postulate that the covariant derivative is the partial derivative plus some correction (see Wald for a proof). We then have the following $$\nabla_\mu V^\nu=\partial_\mu V^\nu+\Gamma^\nu_{\mu\lambda}V^\lambda$$ Using similar reasoning to how we got the above, we postulate the following: $$\nabla_\mu V_\nu=\partial_\mu V_\nu+\gamma^\lambda_{\mu\nu}V_\lambda$$ However, we as of yet have no justification on equating $$\gamma$$ and $$\Gamma$$ just yet. They transform the same (which I will not prove here), but that's all. To prove a relation between the two, we assume two more things about the covariant derivative in addition to linearity and the Leibniz product rule: that the covariant derivative of the Kronecker delta vanishes $$\nabla_\mu \delta^\nu_\lambda=0$$ and that it reduces to the partial derivative on scalars $$\nabla_\mu\phi=\partial_\mu\phi$$. Using these, let's calculate the following: $$\nabla_\mu(V_\sigma W^\sigma)$$
\begin{align} \nabla_\mu(V_\sigma W^\sigma)&=\nabla_\mu(\delta_\lambda^\sigma V_\sigma W^\lambda) \\ &=V_\sigma W^\lambda\nabla_\mu\delta^\sigma_\lambda+V_\sigma\nabla_\mu W^\sigma+W^\sigma\nabla_\mu V_\sigma \\ &=V_\sigma\partial_\mu W^\sigma+W^\sigma\partial_\mu V_\sigma+V_\sigma\Gamma^\sigma_{\mu\lambda}W^\lambda+W^\sigma\gamma^\lambda_{\mu\sigma}V_\lambda \end{align} But, $$V_\sigma W^\sigma$$ is a scalar! Thus, by our second new property, it should reduce to the partial derivative. So, we also have \begin{align} \nabla_\mu(V_\sigma W^\sigma)&=\partial_\mu(V_\sigma W^\sigma)\\ &=V_\sigma\partial_\mu W^\sigma+W^\sigma\partial_\mu V_\sigma \end{align} Equating terms, we immediately see that $$0=\Gamma^\sigma_{\mu\lambda}V_\sigma W^\lambda+\gamma^\lambda_{\mu\sigma}W^\sigma V_\lambda$$ Relabeling indices, we see that $$\gamma^\lambda_{\mu\sigma}W^\sigma V_\lambda=-\Gamma^\lambda_{\mu\sigma}W^\sigma V_\lambda$$ Since $$V_\sigma$$ and $$W^\sigma$$ are completely arbitrary, we have in general that $$\bbox[5px,border:1.5px solid black] { \gamma^\lambda_{\mu\sigma}=-\Gamma^\lambda_{\mu\sigma} }$$ So, after much ado, we finally have, for the covariant derivative of a one-form, $$\nabla_\mu V_\nu=\partial_\mu V_\nu-\Gamma^\lambda_{\mu\nu}V_\lambda$$

• In fact we can rearrange $(\nabla_\mu-\partial_\mu)(V_\sigma W^\sigma)=0$ by the product rule to an expression for $(\nabla_\mu-\partial_\mu)V_\lambda$ without starting from an Ansatz.
– J.G.
Sep 26, 2021 at 9:10

The covariant derivative is at first a map $$\nabla:\Gamma(TM)\times\Gamma(TM)\to\Gamma(TM)$$ taking two vector fields $$(X,Y)$$ to a third $$\nabla_X Y$$ obeying some properties. In particular it has to be a tensor in the first entry $$\nabla_{fX}Y=f\nabla_XY$$ and must be a derivation in the second entry $$\nabla_{X}(fY)=X(f)Y+f\nabla_X Y$$.

Then we want to actually have maps $$\nabla : \Gamma(TM)\times\Gamma(T^r_sM)\to\Gamma(T^r_s M)$$ which take now a vector field and a tensor field $$(X,T)$$ and give you a tensor field $$\nabla_X T$$. These maps must clearly be built from the initial one you had for vector fields only.

Now all tensor fields with $$r$$ indices up and no index down, i.e., elements of $$\Gamma(T^r_0M)$$ can be generated from vector fields by the tensor product. In that case if you demand that $$\nabla_X(T\otimes S)=\nabla_X T\otimes S+T\otimes \nabla_X S,\tag{1}$$

then the initial map, which acts only on vector fields, already fixes all the maps $$\nabla:\Gamma(TM)\times \Gamma(T^r_0M)\to \Gamma(T^r_0M).\tag{2}$$

Moreover property (1) is just again saying that $$\nabla$$ acts as a derivative.

Now we also have all the tensor fields with $$s$$ indices down and no index up, i.e., elements of $$\Gamma(T^0_sM)$$. All of these can be generated from the one-forms $$\Gamma(T^0_1M)$$ and tensor products. So if we fix $$\nabla$$ acting on forms, property (1) again fixes it once and for all in all $$\Gamma(T^0_sM)$$.

The way we go here is by demanding that if $$\omega\in \Gamma(T^0_1M)$$ is a one-form and $$Y\in \Gamma(TM)$$ is a vector field, we have $$\nabla_{X}[\omega(Y)]=(\nabla_X \omega)(Y)+\omega(\nabla_X Y)\tag{3}.$$

This is again a Liebnitz rule of sorts, where we view the contraction $$\omega(Y)$$ as a kind of product. Observe from (3) that since $$\nabla_X Y$$ is already defined, and since $$\omega(Y)$$ is just a smooth function, if we say how $$\nabla$$ acts on smooth functions we are done, this formula fully defines $$\nabla_X \omega$$.

Now the most natural definition is to take $$\nabla_X f = X(f)$$ for any smooth function. In particular, comparing the initial definition of $$\nabla$$ on vector fields which demanded $$\nabla_X(fY)=X(f)Y+f\nabla_X Y$$ and property (1), we see that these are fully compatible once we identify that $$f\otimes Y = fY$$.

Once this is done we have that $$\nabla_X\omega(Y) = X(\omega(Y))-\omega(\nabla_X Y),\tag{4}$$

and as I said from (1) this extends to all $$T^0_sM$$. Clearly now everything also extends to all $$T^r_sM$$.

Now to compare. Introduce a coordinate system $$x^\mu$$ with coordinate basis $$\partial_\mu$$. Denote $$\nabla_\mu := \nabla_{\partial_\mu}$$. Then if $$X$$ is a vector field $$X = X^\nu\partial_\nu$$ we have $$\nabla_\mu X = \nabla_\mu (X^\nu \partial_\nu)=\partial_\mu X^\nu + X^\nu \nabla_\mu \partial_\nu=\partial_\mu X^\nu \partial_\nu + X^\nu\Gamma_{\mu\nu}^\lambda \partial_\lambda=(\partial_\mu X^\lambda + \Gamma_{\mu\nu}^\lambda X^\nu)\partial_\lambda\tag{5}.$$

Where I just have used the rules for $$\nabla$$ and defined $$\Gamma_{\mu\nu}^\lambda$$ by means of $$\nabla_\mu \partial_\nu = \Gamma_{\mu\nu}^\lambda \partial_\lambda$$.

Finally using (4) we can evaluate the components of the one-form $$\nabla_\mu \omega$$ in the coordinate basis $$dx^\nu$$ by applying it to $$\partial_\nu$$. Recalling that $$\omega(\partial_\nu)=\omega_\nu$$ we get $$\nabla_\mu\omega(\partial_\nu)=\partial_\mu \omega_\nu-\omega(\nabla_\mu \partial_\nu)=\partial_\mu\omega_\nu-\omega(\Gamma_{\mu\nu}^\lambda\partial_\lambda)=\partial_\mu\omega_\nu-\Gamma_{\mu\nu}^{\lambda}\omega_\lambda\tag{6}$$

as you suggested.

Finally all this would work for any covariant derivative $$\nabla$$. It is only when you impose vanishing torsion and metric compatibility that you fix uniquely the Levi-Civita one and you are able to determine, in local coordinates, $$\Gamma_{\mu\nu}^\lambda$$ in terms of the metric.

So in summary it all boils down in how to define a covariant derivative. In particular in how we first define it on vectors and extends to tensors in the simples and most natural way.