I wanted to confirm that my explicit (symbolic) computations of the covariant derivative of a constant inner product is (carefully) done right and correct.

For the physicists (includes me), this is useful when thinking about 4-velocities.

The defining property of a 4-velocity $u$ is this:

$ u\cdot u \equiv u^{\flat}(u) \equiv g(u,u) = u_{\mu} u^{\mu} = g_{\mu \nu}u^{\mu} u^{\nu} = -1 $

For $f \in C^{\infty}(M)$ (smooth function on manifold $M$), $\nabla_u f = u(f) = u_i\frac{ \partial f}{ \partial x^i} \, \text{(locally)}$. So clearly $\nabla_u (-1) =0$.

Then is the chain rule or Leibnitz rule still valid for the so-called inner product $u\cdot u = u_{\mu} u^{\mu}$ so that

$ \nabla_v (u_{\gamma} u^{\gamma} ) = u_{\gamma} \nabla_v u^{\gamma} + u^{\gamma} \nabla_v u_{\gamma} = 0 $ ?

For the special case of $v=u$ (taking the covariant derivative with respect to $u$, or however it's called), and for a Levi Civita connection $\nabla$, so that one (of 2) of its defining properties being

$\nabla g = 0$, so that $\nabla_{\gamma} g_{\mu \nu}=0$

Then $ u^{\gamma} \nabla_u u_{\gamma} = u^{\gamma}\nabla_u g_{\gamma \nu} u^{\nu} = u^{\gamma} g_{\gamma \nu} u^{\sigma} \nabla_{\sigma}u^{\nu} = u_{\nu} \nabla_u u^{\nu}$

So then

$\nabla_u (u_{\gamma} u^{\gamma}) = 2 u_{\gamma} \nabla_u u^{\gamma} = 0 $.

Then I can conclude that, assuming a Levi-Civita connection $\nabla$, $u\cdot u =-1$ also means that

$u_{\nu} \nabla_u u^{\nu} =0$

Is this all correct?

Update: EY 20150530

Yvonne Choquet-Bruhat, General Relativity and the Einstein Equations : Oxford Mathematical Monographs, Oxford University Press, 2009. ISBN-13: 978-0199230723 Chapter 13 Singularities, the 3rd section, Equation (3.6), confirms that $u^{\beta} \nabla_{\alpha} u_{\beta}=0$ is true, since $u^{\beta}u_{\beta}=-1$

I just want to know if I did the steps correctly.


1 Answer 1


Yes, that's all correct. The "inner product Leibniz rule" $\partial_X (g(Y,Z)) = g(\nabla_X Y, Z) + g(Y, \nabla_X Z)$ is often used to define the Levi-Civita connection.

  • $\begingroup$ Thank you Anthony Carapetis! $\endgroup$ Jun 23, 2015 at 18:57

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