# Taking covariant derivative of an expression involving scalars

I want to evaluate an expression of the form $$\nabla_\lambda (Y^\eta \partial_\eta Z^\rho).\tag{1}$$

I'm not sure about how to write down the connection term. My guess is:

$$\nabla_\lambda (Y^\eta \partial_\eta Z^\rho) = \partial_\lambda (Y^\eta \partial_\eta Z^\rho) + Y^\eta \partial_\eta (\Gamma^\rho_{\lambda \alpha} Z^\alpha ) \tag{2}$$

Is this correct?

• Which are the scalars? Also, assuming $Y$ and $Z$ are vectors, why not $\nabla_{\lambda} \left( Y^{\eta} \nabla_{\eta} Z^{\rho}\right)$ instead? Jan 29, 2018 at 17:35
• @secavara The $\eta$ index is contracted so I suppose $Y^\eta \partial_\eta$ can be called a scalar? $\nabla_{\lambda} ( Y^{\eta} \nabla_{\eta} Z^{\rho})$ is indeed the expression I started with, but I expanded the inner covariant derivative term.
– cord
Jan 29, 2018 at 18:11
• Unfortunately you can't ignore the indices of contracted tensors if you want to expand the covariant derivative. This means that you get $\nabla_{\lambda} \left( Y^{\eta} \nabla_{\eta} Z^{\rho}\right) = \partial_\lambda \left(Y^{\eta} \nabla_{\eta} Z^{\rho} \right) + \Gamma^{\eta}_{\lambda \alpha} Y^{\alpha} \nabla_{\eta} Z^{\rho} - \Gamma^{\alpha}_{\lambda \eta} Y^{\eta} \nabla_{\alpha} Z^{\rho} + \Gamma^{\rho}_{\lambda \alpha} Y^{\eta} \nabla_{\eta} Z^{\alpha} .$ Jan 29, 2018 at 18:31
• Okay, the part that confuses me is that in Carroll's book he explicitly writes an equation like $\nabla_{\mu} (\omega_{\lambda} V^\lambda) = \partial_\mu( \omega_{\lambda} V^\lambda)$. That's where I got the idea of not including the connection terms for the contracted indices.
– cord
Jan 29, 2018 at 18:52
• Right, notice that the second and third term in the right hand side of the expression I wrote cancel each other, but notice that this is still a statement about $\nabla_{\lambda} \left( Y^{\eta} \nabla_{\eta} Z^{\rho}\right)$ and not $\nabla_{\lambda} \left( Y^{\eta} \partial_{\eta} Z^{\rho}\right)$ and notice where the connection enters in the last term. Jan 29, 2018 at 19:23

The problem is that $X^{\rho}:= Y^{\eta} \partial_{\eta} Z^{\rho}$ is not components of a vector field (if we assume that $Y^{\eta}$ and $Z^{\rho}$ are components of vector fields). So the covariant derivative $\nabla_{\lambda}X^{\rho}$ in OP's first expression (1) does not make geometric sense.
• In general you are correct but it might be from a commutator $X:=[Y,Z]$, written in local coordinates, that is just confusing. Jan 29, 2018 at 19:23