　Covariant derivative

Since direction of basis vector might differ between two distinct points on curvature, we have to compare the difference of two vectors by setting their starting points on a single point.

Let $A^i+dA^i(A^i+\delta A^i)$ be a vector which we moved vector $A^i$ (parallel to curvature) from $P_1$ to $P_2$ .

The difference we are looking for is

$DA^i=A^i+dA^i-(A^i+\delta A^i)=dA^i-\delta A^i$

　　　　　Let us define　　　 $\delta A^i=-\Gamma_{kl}^iu^kdx^l$ 　　　and　　　 $A_{;l}^i=\frac{DA^i}{dx^l}$ .

(We will see later that this $\Gamma_{kl}^i$ corresponds to the Cristoffel symbol we saw in geodesic equation.)

then　　　　　　　　　　　　　　　　　　 $A_{;l}^i=A_{,l}^i+\Gamma_{ik}^iA^k$

since scalar is invariant,　　　 $\delta(u_iu^i)=0,　　A_{i;l}=A_{i,l}-\Gamma_{il}^kA_k$ .

　Riemann tensor

Let us calculate difference the vector make while we move it parallel to infinitely small closed curve on curvature. It is,

$\Delta A_k=\oint\Gamma_{kl}^iA_idx^l,$ 　　then　　 $\frac{\partial A_i}{\partial x^l}=\Gamma_{il}^nA_n$ .

Thus　　　　　 $\Delta A_k=\frac{1}2(\frac{\partial \Gamma_{km}^lA_i}{\partial x^l}-\frac{\partial \Gamma_{kl}^iA_i}{\partial x^m})\Delta f^{lm}　　　(\because Stokes　theorem)$ 　

　　　　　　　　　 $=\frac{1}2(\frac{\partial \Gamma_{km}^i}{\partial x^l}-\frac{\partial \Gamma_{kl}^i}{\partial x^m}+\Gamma_{nl}^i\Gamma_{km}^n-\Gamma_{nm}^i\Gamma_{kl}^n)A_i\Delta f^{lm}$

and here we define

Riemann tensor:　 $R_{klm}^i=\frac{1}2(\frac{\partial \Gamma_{km}^i}{\partial x^l}-\frac{\partial \Gamma_{kl}^i}{\partial x^m}+\Gamma_{nl}^i\Gamma_{km}^n-\Gamma_{nm}^i\Gamma_{kl}^n)$