1
$\begingroup$

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 $$ \nabla_{i}v_j=\frac{\partial v_j}{\partial u^{i}}-\Gamma^j_{~ik}v_j $$ 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!

$\endgroup$
4
  • $\begingroup$ I did not add the tag special relativity - I had added "relativity". No idea how that wrong tag got attachen!!! $\endgroup$ – Fuzzy Feb 8 at 20:29
  • $\begingroup$ Don't worry Fuzzy, that's because the relativity tag was made a synonym of the special-relativity tag, and as such, it gets redirected to the special-relativity tag every time someone uses it. $\endgroup$ – Urb Feb 8 at 20:39
  • 2
    $\begingroup$ 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. $\endgroup$ – scaphys Feb 8 at 20:40
  • 1
    $\begingroup$ scphys, gimme a minute, I try, thanks for the hint! $\endgroup$ – Fuzzy Feb 8 at 20:45
2
$\begingroup$

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, $$\nabla_\mu V_\nu=\partial_\mu V_\nu-\Gamma^\lambda_{\mu\nu}V_\lambda$$

$\endgroup$
0
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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