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!
[Physics] Covariant Derivative of Kronecker Delta
differential-geometrydifferentiationgeneral-relativityhomework-and-exercisestensor-calculus
Related Solutions
@Prahar is right, the variation of the Christoffel symbol is a tensor, even if the Christoffel itself is not. We have
$\delta \Gamma^\rho_{\mu\nu}=\frac{1}{2}\delta\bigg(g^{\rho\alpha}(2\partial_{(\mu}g_{\nu)\alpha}-\partial_\alpha g_{\mu\nu})\bigg)=\frac{1}{2}\delta g^{\rho\alpha}(2\partial_{(\mu}g_{\nu)\alpha}-\partial_\alpha g_{\mu\nu})+ \frac{1}{2}g^{\rho\alpha}(2\partial_{(\mu}\delta g_{\nu)\alpha}-\partial_\alpha \delta g_{\mu\nu})$
where $A_{(\mu\nu)}=\frac{1}{2}(A_{\mu\nu}+A_{\nu\mu})$. Using $\delta g^{\rho\alpha}=-g^{\rho\gamma}g^{\alpha\delta}\delta g_{\gamma\delta}$ we have:
$\delta \Gamma^\rho_{\mu\nu}=\frac{1}{2}g^{\rho\alpha}(2\partial_{(\mu}\delta g_{\nu)\alpha}-\partial_\alpha \delta g_{\mu\nu}-2\Gamma_{\mu\nu}^\beta\delta g_{\alpha\beta})$
The Christoffel then combines nicely with the standard derivative to give a covariant tensor (the other Christoffel symbols cancel each other)
$\delta \Gamma^\rho_{\mu\nu}=\frac{1}{2}g^{\rho\alpha}(2\nabla_{(\mu}\delta g_{\nu)\alpha}-\nabla_\alpha \delta g_{\mu\nu})$.
So to answer the original question, we finally have:
$\nabla_\mu V_\nu=\nabla_\mu \delta V_\nu-\frac{1}{2}g^{\rho\alpha}(2\nabla_{(\mu}\delta g_{\nu)\alpha}-\nabla_\alpha \delta g_{\mu\nu})A_\rho$
Remember that we did not assume anything on $V_\mu$. Depending on the problem, it is then possible to integrate by parts to isolate $\delta g_{\mu\nu}$ and obtain the energy momentum tensor.
Start with the following form of the Bianchi Identities $$ \nabla^\mu R_{\mu\nu} = \frac{1}{2} \nabla_\nu R $$ Contract both sides with $K^\nu$. We find $$ \frac{1}{2} K^\nu \nabla_\nu R = K^\nu \nabla^\mu R_{\mu\nu} = \nabla^\mu \left( K^\nu R_{\mu\nu} \right) - R_{\mu\nu} \nabla^\mu K^\nu $$ The second term vanishes due to symmetry of $R_{\mu\nu}$. Now, recall that $R_{\mu\nu}K^\nu = \nabla_\nu \nabla_\mu K^\nu$. We now use the following fact $$ \left[ \nabla_\rho, \nabla_\sigma \right] \tau^{\mu\nu} = R^\mu{}_{\lambda\rho\sigma} \tau^{\lambda\nu} + R^\nu{}_{\lambda\rho\sigma}\tau^{\mu\lambda} $$ This implies \begin{equation} \begin{split} \nabla^\mu \left( K^\nu R_{\mu\nu} \right) &= \nabla_\mu\nabla_\nu \nabla^\mu K^\nu = \nabla_{[\mu}\nabla_{\nu ]} \nabla^{[\mu} K^{\nu]} = \frac{1}{2}[\nabla_{\mu}, \nabla_{\nu }] \nabla^{[\mu} K^{\nu]} \\ &=\frac{1}{2}\left(R^\mu_{\;\;\lambda\mu\nu} \nabla^{[\lambda} K^{\nu]}-R^\nu_{\;\;\lambda\nu\mu} \nabla^{[\mu} K^{\lambda]}\right)\\ &=\frac{1}{2}\left(R_{[\lambda\nu ]}\nabla^{[\lambda} K^{\nu]}-R_{[\lambda\mu ]}\nabla^{[\mu} K^{\lambda]}\right) = 0 \end{split} \end{equation} which then implies $$ K^\nu \nabla_\nu R = 0 $$
Best Answer
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 $\nabla_l\delta^i{}_j=0$.