Let us say we have a contravariant vector $V^a$,
From here we can write,
$$\nabla_bV^a = \partial_bV^a + \Gamma^a_{bc}V^c$$
I was creating a code to calculate the covariant derivative of a vector.
For $$g_{\mu\nu} = \text{diag}(-1, 1, r^2, r^2sin^2(\theta))$$ and $$V^a = (t^2, tr^2, r^3\theta^2, \phi)$$
my code gives me $$\displaystyle \left[\begin{matrix}2 t & r^{2} & 0 & 0\\0 & 2 r t & 4 r^{2} \theta^{2} & \frac{\phi}{r}\\0 & - r^{4} \theta^{2} & 2 r^{3} \theta + r t & \frac{\phi \cos{\left(\theta \right)}}{\sin{\left(\theta \right)}}\\0 & - \phi r \sin^{2}{\left(\theta \right)} & - \phi \sin{\left(\theta \right)} \cos{\left(\theta \right)} & \frac{r^{3} \theta^{2} \cos{\left(\theta \right)}}{\sin{\left(\theta \right)}} + r t + 1\end{matrix}\right]$$
I wrote the columns and rows as
A)
$$\nabla_bV^a = \displaystyle \left[\begin{matrix} \nabla_0V^0 & \nabla_0V^1 & \nabla_0V^2 & \nabla_0V^3\\ \nabla_1V^0 & \nabla_1V^1 & \nabla_1V^2 & \nabla_1V^3 \\ \nabla_2V^0 & \nabla_2V^1 & \nabla_2V^2 & \nabla_2V^3 \\ \nabla_3V^0 & \nabla_3V^1 & \nabla_3V^2 & \nabla_3V^3 \end{matrix}\right]$$
My question is is this the correct representation ?
Or should it be like this,
B)
$$\nabla_bV^a = \displaystyle \left[\begin{matrix} \nabla_0V^0 & \nabla_1V^0 & \nabla_2V^0 & \nabla_3V^0\\ \nabla_0V^1 & \nabla_1V^1 & \nabla_2V^1 & \nabla_3V^1 \\ \nabla_0V^2 & \nabla_1V^2 & \nabla_2V^2 & \nabla_3V^2\\ \nabla_0V^3 & \nabla_1V^3 & \nabla_2V^3 & \nabla_3V^3 \end{matrix}\right]$$
Hence,
$$\nabla_bV^a = \displaystyle \left[\begin{matrix}2 t & 0 & 0 & 0\\r^{2} & 2 r t & - r^{4} \theta^{2} & - \phi r \sin^{2}{\left(\theta \right)}\\0 & 4 r^{2} \theta^{2} & 2 r^{3} \theta + r t & - \phi \sin{\left(\theta \right)} \cos{\left(\theta \right)}\\0 & \frac{\phi}{r} & \frac{\phi \cos{\left(\theta \right)}}{\sin{\left(\theta \right)}} & \frac{r^{3} \theta^{2} \cos{\left(\theta \right)}}{\sin{\left(\theta \right)}} + r t + 1\end{matrix}\right]$$
To summerize which notations is true $A$ or $B$ ? For me its $A$ but I still wanted to ask