## § Covariant derivative

If $x_p \equiv a \partial_x + b \partial_y + c \partial_z$ is a vector at $p \in \mathbb R^3$ and $Y$ is a vector field, then the covariant derivative of $Y$ in the direction $X$ is given by taking the directional derivative of each component of $Y$ along $X$:
$x_p |- Y \equiv (x_p \cdot Y, x_p \cdot Y, x_p \cdot Y)$
The notation |- is meant to suggest that $X_p$ is acting on $Y$. For a concrete example, if $X_p \equiv (a, b, c)$ and $Y \equiv (xy^2 + 4z, y^2 - x, x + z^3)$, then the computation yields:
\begin{aligned} &x_p |- Y \equiv (x_p \cdot Y, 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 $x_p |- (Y + Z) = x_p |- Y + x_p |- Z$. This is proven by the linearity of the partial derivative.

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

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

#### § Property 3: Scaling of LHS

$(f(p) x_p) |- Y = f(p) (x_p |- y)$

#### § Property 3: Scaling of RHS

$x_p |- (fY) = (x_p f(p)) Y + f (x_p |- Y)$
\begin{aligned} &x_p |- (fY) \equiv (x_p \cdot fY, 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), \dots, \dots) &= ((a Y \partial_x|_{p_x} f + af \partial_x|_{p_x} Y + b Y \partial_y|_{p_y} f + b f \partial_y|_{p_y} Y c Y \partial_z|_{p_z} f + c f \partial_y|_{p_z} Y, \dots, \dots) \\ &= ((fa \partial_x Y + fb\partial_y Y + fc \partial_z Y) Y + (Y a \partial_x + Y b \partial_y + Y c \partial_z) \cdot f, \dots, \dots) \\ &= (fx_p) |- Y + (Y(p) \end{aligned}

#### § Computing $x_p |- Y$

We can compute $x_p |- Y$ once we have a curve $\sigma$ that is compatible with $x_p$. So if we have a curve $\sigma(0) = p$, and $\sigma'(0) = x_p$, and we know $Y$, we can then compute $x_p |- Y$ as:
\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 \partial_x Y + x_p \partial_y Y + x_p \partial_y Y \\ &x_p |- Y \end{aligned}
Realy, we only need to know $Y$ along $\sigma$ to compute the derivative, no more. So it's enough to have (1) a curve $\sigma$ that is compatible with $x_p$, and (2) knowledge of the vector field $Y$ along $\sigma$.

#### § Parallel vector fields

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

#### § Parallel vector fields

Let $\sigma$ be a $C^\infty$ curve.