Covariant Derivative of Kronecker Delta I am reading Carroll's book on GR right now, and I ran into a little trouble in his chapter 3 on curvature. He is establishing the properties of the covariant derivative, and claims that the fact that the covariant derivative commutes with contraction, i.e $$\nabla_\mu T^\lambda {}_{\lambda \rho} = (\nabla T)_\mu {}^\lambda {}_{\lambda \rho},$$ implies that the covariant derivative of the Kronecker delta is zero: $$\nabla_\mu \delta^\lambda_\sigma = 0.$$ I do not see why this is the case. I would greatly appreciate it if someone could point out what I am missing. Thanks!
 A: Note that $\delta^i{}_j=\delta^i{}_k\delta^k{}_j$. Then 
$$\nabla_l\delta^i{}_j=\nabla_l(\delta^i{}_k\delta^k{}_j)=C(k,m)[\nabla_l(\delta^i{}_k\delta^m{}_j)]=C(k,m)[\delta^i{}_k\nabla_l\delta^m{}_j+\delta^m{}_j\nabla_l\delta^i{}_k]=2\nabla_l\delta^i{}_j$$
where $C(k,m)$[...] is defined as the contraction operator which contracts the $k$ and $m$ indices. Thus $\nabla_l\delta^i{}_j=0$.
A: The partial derivatives $\partial_\lambda \delta^\mu_\nu$ are clearly zero because the components of the Kronecker delta are constant functions of spacetime coordinates (one or zero).
One may always go to a locally Minkowski frame where the Christoffel symbols vanish and there, the covariant derivative is equal to the partial one and vanishes, too. Because the Kronecker delta is a tensor, its covariant derivative must be the same in all other frames, too.
Alternatively, one may explicitly write the Christoffel symbol terms, too:
$$ \nabla_\lambda \delta^\mu_\nu = \partial_\lambda\delta^\mu_\nu - \Gamma_{\lambda\kappa}^\mu \delta^\kappa_\nu - \Gamma_{\lambda\nu}^{\kappa} \delta_{\kappa}^\mu = 0 + \Gamma^\mu_{\lambda\nu}-\Gamma^\mu_{\lambda\nu} = 0.$$
The cancellation between the two connection terms is the same cancellation that also allows to ignore contracted pairs of indices in the tensor. I don't think it's helpful to "derive" the second statement from the first one, however, because the first one doesn't explicitly use the Kronecker delta at all so the claim that the first fact "implies" the other must be taken with a grain of salt.
Well, if I had to say what was meant, it was that
$$ T^\lambda{}_{\lambda\rho} = T^\lambda{}_{\kappa\rho} \delta^\kappa_\lambda $$
and the $\nabla_\mu$ derivative of this sum of products may be written using the Leibniz rule as
$$\nabla_\mu T^\lambda{}_{\lambda\rho} = (\nabla T)_{\mu}{}^\lambda{}_{\kappa\rho} \delta^\kappa_\lambda + T^\lambda{}_{\kappa\rho}  \nabla_\mu \delta^\kappa_\lambda $$
The assumption is the equation without the last $\nabla\delta$ term, so this last term has to vanish for any tensor $T^{\lambda}_{\kappa\rho}$, which implies that the whole $\nabla_\mu \delta^\kappa_\lambda$ object is zero, too.
A: You can write the Kronecker delta tensor as a product of the metric tensor
$$\nabla_a(\delta^a_b) = \nabla_a (g_{bc} g^{ac}) = \nabla_a (g_{bc} g^{ac}) = g_{bc} \nabla_a g^{ac} +  g^{ac}\nabla_a g_{bc} $$
As you may recall, the covariant derivative of the metric tensor is $0$ in general relativity.
Version without using the metricity of the connection : 
$\nabla_a (\delta_b^a T^b) = \nabla_a T^a$, but also $\nabla_a (\delta_b^a T^b) =   \delta_b^a \nabla_a T^b + T^b \nabla_a \delta_b^a$, so contracting the delta with the connection, this gives us 
$$\nabla_a T^a = \nabla_a T^a + T^b \nabla_a \delta_b^a$$
Which will only be true for all $T$ if $\nabla_a \delta_b^a = 0$.
A: Kronecker delta is the (1,1) tensor satisfying
$$\delta^{a}\vphantom{\delta}_{b}v^{b}=v^{a}$$
for an arbitrary vector $v$. By Leibniz rule of affine connection, we get
$$\delta^{a}\vphantom{\delta}_{b}\nabla_{c}v^{b}+v^{b}\nabla_{c}\delta^{a}\vphantom{\delta}_{b}=\nabla_{c}v^{a}$$. Giving arguments vector $u$ and linear form $\omega$ to the above,
$$\omega_{a}u^{c}\left(\delta^{a}\vphantom{\delta}_{b}\nabla_{c}v^{b}+v^{b}\nabla_{c}\delta^{a}\vphantom{\delta}_{b}\right)=\omega_{b}u^{c}\nabla_{c}v^{b}+\omega_{a}u^{c}v^{b}\nabla_{c}\delta^{a}\vphantom{\delta}_{b}=\omega_{a}u^{c}\nabla_{c}v^{a}$$,
we get
$$\omega_{a}u^{c}v^{b}\nabla_{c}\delta^{a}\vphantom{\delta}_{b}=0$$.
Because the above holds for arbitrary $u$, $v$, and $\omega$, we conclude that
$$\nabla_{c}\delta^{a}\vphantom{\delta}_{b}=0$$.
