If xp≡a∂x+b∂y+c∂zx_p \equiv a \partial_x + b \partial_y + c \partial_z is a vector at p∈R3p \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:

xp∣−Y≡(xp⋅Y[1],xp⋅Y[2],xp⋅Y[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,y2−x,x+z3)Y \equiv (xy^2 + 4z, y^2 - x, x + z^3), then the computation yields:

xp∣−Y≡(xp⋅Y[1],xp⋅Y2,xp⋅Y3)((a∂x+b∂y+c∂z)⋅(xy2+4z),(a∂x+b∂y+c∂z)⋅(y2−x),(a∂x+b∂y+c∂z)⋅(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)=xp∣−Y+xp∣−Zx_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=(xp∣−Y)+(xp′∣−Y)(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)(xp∣−y)(f(p) x_p) |- Y = f(p) (x_p |- y)

§ Property 3: Scaling of RHS

xp∣−(fY)=(xpf(p))Y+f(xp∣−Y)x_p |- (fY) = (x_p f(p)) Y + f (x_p |- Y)

xp∣−(fY)≡(xp⋅fY[1],xp⋅fY2,xp⋅fY3)=((a∂x∣px+b∂y∣py+c∂z∣pz)⋅(fY[1]),…,… )=((aY[1]∂x∣pxf+af∂x∣pxY[1]+bY[1]∂y∣pyf+bf∂y∣pyY[1]cY[2]∂z∣pzf+cf∂y∣pzY[2],…,… )=((fa∂xY[1]+fb∂yY[2]+fc∂zY[3])Y+(Y[1]a∂x+Y[2]b∂y+Y[3]c∂z)⋅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 xp∣−Yx_p |- Y

We can compute xp∣−Yx_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 xp∣−Yx_p |- Y as:

d(Y∘σ(t))dt∣t=0Y′(σ(t))∣t=0⋅σ′(t)∣t=0Y′(σ(0))⋅σ′(0)Y′(p)⋅xpxp[1]∂xY[1]+xp[2]∂yY[2]+xp[3]∂yY[3]xp∣−Y \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 tp∣−Y=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 C∞C^\infty curve.