# Covariant derivative of unit vector/Kronecker delta

Consider: $\nabla_b \delta^a_c=\partial_b\delta^a_c-\Gamma^d_{bc}\delta^a_d+\Gamma^a_{bd}\delta^d_c=0-\Gamma^a_{bc}+\Gamma^a_{bc}=0$ But if I define:

$T^a_{bc}=\nabla_b\delta^a_c$

Then I should have:

$T^a_{bc}=0$

Now I work with a unit vector $v_c$ in a coordinate system such that $v_c\overset{*}{=}\delta^0_c$

Working in this coordinate system if I try to calculate

$\nabla_b v_c\overset{*}{=}\nabla_b\delta^0_c\overset{*}{=}T^0_{bc}=\nabla_b\delta^0_c=\partial_b\delta^0_c-\Gamma^d_{bc}\delta^0_d=-\Gamma^0_{bc}\neq 0\text{ (necessarily) }$

But $T^0_{bc}$ should equal zero! Why the contradiction?

• I gather my problem is somewhere with me setting the upper index in the Kronecker delta to zero and treating that the same as what I'm calling $v_c=\delta^0_c$ since the Kronecker delta tensor has the same representation in every coordinate system and $v_c$ doesn't. I see how the lack of a second term with a connection coefficient means that $\nabla_b\delta^0_c\neq 0$ but it still feels like there's a contradiction somewhere I'm not fully grasping and I must've done something invalid. – Fred Burroughs Oct 21 '17 at 4:31

What you defined as $$\delta^0_c$$ is not a vector because under a coordinate change you have to consider the change of both its indices:

$$\delta^0_c\rightarrow \bar\delta^0_c= J^0_b \delta^b_d(J^{-1})^d\,_c=\delta^0_c$$

Indeed, the Kronecker delta is a tensor of rank $$(1,1)$$ which is isotropic, meaning that the value of its components doesn't change from a coordinate system to another.

The vector you were trying to build would have had this aspect:

$$v_c=(1,0,0,0)$$

in some coordinate system (I think for example to the center of mass system for a time-like vector);

but you should expect it to change according to the usual law of transformation of vectors:

$$\bar v_c=J^{d}\,_cv_d=J^0\,_c$$

where now in principle the components of $$\bar v_c$$ will be different from $$(1,0,0,0)$$ because they depend from the form of the change of coordinates. This tells you that the equality you have imposed cannot last outside of the coordinate system you start with.

To be even clearer, if $$v_c=\delta^0_c$$ was a legitimate vector equality, it should hold in every coordinate frame, while we have proved above that lhs goes in one direction while rhs goes in another (or better, goes nowhere).

Indeed, if you consider the complete derivative of the object $$\delta^0_c$$, you get obviously:

$$\nabla_b\delta_c^0=\partial_b\delta_c^0 + \Gamma_{bc}^\lambda\delta_\lambda^0-\Gamma_{b\lambda}^0\delta^\lambda_c=0+\Gamma_{bc}^0+\Gamma_{bc}^0=0$$

The contradiction is that in the first equation you calculate the covariant derivative of the Kronecker delta, which is zero by definition anyway, and then you apply the covariant derivative to a vector (a covector in reality), which can be any outcome.