§ Covariant derivative

If xpax+by+czx_p \equiv a \partial_x + b \partial_y + c \partial_z is a vector at pR3p \in \mathbb R^3 and YY is a vector field, then the covariant derivative of YY in the direction XX is given by taking the directional derivative of each component of YY along XX:
xpY(xpY[1],xpY[2],xpY[3]) x_p |- Y \equiv (x_p \cdot Y[1], x_p \cdot Y[2], x_p \cdot Y[3])
The notation |- is meant to suggest that XpX_p is acting on YY. For a concrete example, if Xp(a,b,c)X_p \equiv (a, b, c) and Y(xy2+4z,y2x,x+z3)Y \equiv (xy^2 + 4z, y^2 - x, x + z^3), then the computation yields:
xpY(xpY[1],xpY2,xpY3)((ax+by+cz)(xy2+4z),(ax+by+cz)(y2x),(ax+by+cz)(x+z3))=(ay2+2bxy+4,a+2by,a+3cz2) \begin{aligned} &x_p |- Y \equiv (x_p \cdot Y[1], x_p \cdot Y_2, x_p \cdot Y_3) \\ & ((a \partial_x + b \partial_y+ c \partial_z ) \cdot (xy^2 + 4z), (a \partial_x+ b \partial_y+ c \partial_z) \cdot (y^2 - x), (a \partial_x+ b \partial_y+ c \partial_z) \cdot (x + z^3)) &= (ay^2 + 2bxy + 4, -a + 2by, a + 3cz^2) \end{aligned}

§ Property 1: Linearity in RHS

We have that xp(Y+Z)=xpY+xpZx_p |- (Y + Z) = x_p |- Y + x_p |- Z. This is proven by the linearity of the partial derivative.

§ Property 2: Linearity in LHS: (xp+xp)Y=(xpY)+(xpY)(x_p + x'_p)|- Y = (x_p |- Y) + (x'_p |- Y).

This follows as vector addition is linear, and the action of the directional derivative is linear.

§ Property 3: Scaling of LHS

(f(p)xp)Y=f(p)(xpy)(f(p) x_p) |- Y = f(p) (x_p |- y)

§ Property 3: Scaling of RHS

xp(fY)=(xpf(p))Y+f(xpY)x_p |- (fY) = (x_p f(p)) Y + f (x_p |- Y)
xp(fY)(xpfY[1],xpfY2,xpfY3)=((axpx+bypy+czpz)(fY[1]),,)=((aY[1]xpxf+afxpxY[1]+bY[1]ypyf+bfypyY[1]cY[2]zpzf+cfypzY[2],,)=((faxY[1]+fbyY[2]+fczY[3])Y+(Y[1]ax+Y[2]by+Y[3]cz)f,,)=(fxp)Y+(Y(p) \begin{aligned} &x_p |- (fY) \equiv (x_p \cdot fY[1], x_p \cdot fY_2, x_p \cdot fY_3) \\ &= ((a \partial_x|_{p_x} + b \partial_y|_{p_y} + c \partial_z|_{p_z} ) \cdot (fY[1]), \dots, \dots) &= ((a Y[1] \partial_x|_{p_x} f + af \partial_x|_{p_x} Y[1] + b Y[1] \partial_y|_{p_y} f + b f \partial_y|_{p_y} Y[1] c Y[2] \partial_z|_{p_z} f + c f \partial_y|_{p_z} Y[2], \dots, \dots) \\ &= ((fa \partial_x Y[1] + fb\partial_y Y[2] + fc \partial_z Y[3]) Y + (Y[1] a \partial_x + Y[2] b \partial_y + Y[3] c \partial_z) \cdot f, \dots, \dots) \\ &= (fx_p) |- Y + (Y(p) \end{aligned}

§ Computing xpYx_p |- Y

We can compute xpYx_p |- Y once we have a curve σ\sigma that is compatible with xpx_p. So if we have a curve σ(0)=p\sigma(0) = p, and σ(0)=xp\sigma'(0) = x_p, and we know YY, we can then compute xpYx_p |- Y as:
d(Yσ(t))dtt=0Y(σ(t))t=0σ(t)t=0Y(σ(0))σ(0)Y(p)xpxp[1]xY[1]+xp[2]yY[2]+xp[3]yY[3]xpY \begin{aligned} \frac{d (Y \circ \sigma(t))}{dt}|_{t = 0} \\ &Y'(\sigma(t))|_{t=0} \cdot \sigma'(t)|_{t = 0} \\ &Y'(\sigma(0)) \cdot \sigma'(0) \\ &Y'(p) \cdot x_p \\ &x_p[1] \partial_x Y[1] + x_p[2] \partial_y Y[2] + x_p[3] \partial_y Y[3] \\ &x_p |- Y \end{aligned}
Realy, we only need to know YY along σ\sigma to compute the derivative, no more. So it's enough to have (1) a curve σ\sigma that is compatible with xpx_p, and (2) knowledge of the vector field YY along σ\sigma.

§ Parallel vector fields

Say that a curve σ\sigma is tangent to a vector tpt_p if σ(0)=p\sigma(0) = p and σ(0)=t\sigma'(0) = t. Then a vector field YY defined a along σ\sigma is parallel along tt iff tpY=0t_p |- Y = 0. Intuitively, this means that the vector field does not change in the direction of tpt_p, so it keeps its value constant along tpt_p. It is as if the values of Y(0)Y(0) have been transported "parallely"/ "with no distortion" along the tangent tpt_p.

§ Parallel vector fields

Let σ\sigma be a CC^\infty curve.