• Tangent vectors as derivations
• Connections
  • Christoffel symbols
• Geodesics
• Foonotes

Covariant derivatives are a means of differentiating vectors relative to vectors. First we cover formal definitions of tangent vectors and then proceed to define a means to “covariantly differentiate”.

Tangent vectors as derivations

The most general definition of a vector tangent to a manifold involves derivations. \(\newcommand{\R}{\mathbb{R}} \newcommand{\T}{\mathcal{T}} \newcommand{\vec}{\mathbf} \newcommand{\parens}[1]{\left(#1 \right)} \newcommand{\braces}[1]{\left\{#1 \right\}} \newcommand{\angles}[1]{\langle #1 \rangle} \newcommand{\d}{\text{d}} \newcommand{\dd}[1]{\frac{\d}{\d #1}} \newcommand{\ddd}[2]{\frac{\d #1}{\d #2}} \newcommand{\pd}[1]{\frac{\partial}{\partial #1}} \newcommand{\p}{\partial} \newcommand{\pdd}[2]{\frac{\partial #1}{\partial #2}} \newcommand{\evalat}[1]{\Big|_{#1}}\)

Let $M$ be a smooth manifold; every point $p \in M$ has a neighborhood $U \subset M$ that can be mapped smoothly¹ to $\R^n$ for some $n$ (the dimension of the manifold). Therefore each $p$ can be identified with its coordinates² $\phi(p) = \parens{x^1(p), \dots, x^n(p)}$ where $x^i(p) \in \R$.

Associated with each point is the tangent space $T_p M$. Elements of the tangent space are called tangent vectors and can be intuitively visualized as arrows tangent to the manifold at that point; this can be made rigorous by way of directional derivatives. Let $c(t) = \parens{p^1 +t v^1, \dots, p^n +t v^n}$ be a parameterization of a line through the point $p$ in the direction of $\vec{v} := \parens{v^1, \dots, v^n}$; if $f \in C^{\infty}_p$ in a neighborhood of $p$ then the directional derivative of $f$ in the direction of $\vec{v}$ is defined

\[D_{\vec{v}} f := \lim_{t\rightarrow 0} \frac{f(c(t)) -f(p)}{t} = \dd{t} f(c(t)) \Big|_{t=0}\]

By the chain rule

\[\begin{aligned} D_{\vec{v}} f &= \sum_{i=1}^{n} \pdd{f}{x^i} \Big|_p \ddd{c^i}{t} \Big|_{t=0} \\ &= \sum_{i=1}^{n} \ddd{c^i}{t} \Big|_{t=0} \pdd{f}{x^i} \Big|_p \\ &= \sum_{i=1}^{n} v^i \pdd{f}{x^i} \Big|_p \end{aligned}\]

and hence the directional derivative in the direction of $\vec{v}$ is defined

\[D_{\vec{v}} := \sum_{i=1}^{n} v^i \pdd{}{x^i} \Big|_p\]

This directional derivative gives a map between between vector spaces

\[D_{\vec{v}}: C^{\infty}_p \rightarrow \R\]

that also satisfies the Leibniz rule

\[D_{\vec{v}}(f g) = (D_{\vec{v}}f) g + f (D_{\vec{v}}g)\]

In general any linear map from $C^{\infty}_p \rightarrow \R$ is called a derivation and the set of all derivations is a vector space³. So we can identify tangent vectors⁴ with a legitimate vector space spanned by the basis

\[\braces{\p_1, \dots, \p_n} := \braces{\pd{x^1} \evalat{p}, \dots, \pd{x^n} \evalat{p}}\]

The point being that, while not as geometrically intuitive as arrows, this generalizes to manifolds.

Alternative notation for directional derivative

\[\vec{v}(f) := D_{\vec{v}} f\]

Connections

Invariably (pardon the pun) when doing physics you’ll need derivatives of quantities. This prompts the question: how do you differentiate a vector field⁵? Suppose we have a vector field $\vec{w}(p) = \parens{w^i(p)}$ that varies smoothly with $p$. Notice that each $w_i: M \rightarrow \R$ is a smooth function so it can be differentiated in a straightforward fashion:

\[\vec{v}(w^i) = D_{\vec{v}} w^i\]

Collecting all of these into a column we get

\[\vec{v}(\vec{w}) = \begin{bmatrix} D_{\vec{v}} w^1 \\ \vdots \\ D_{\vec{v}} w^n \end{bmatrix} = \vec{J} \vec{v}\]

where $\vec{J}$ is the Jacobian of $\vec{w}$ as a map from $M$ to $\R^n$. This is called the Euclidean connection. Alternatively in components

\[(\vec{v}(\vec{w}))^i = \vec{v}(w^i) = \sum_j v^j \p_j w^i\]

Unfortuntely this doesn’t generalize so we define abstract differentiation of one vector field with respect to another.

Define

\[TM := \bigsqcup_{p \in M} T_p M\]

as the tangent bundle of $M$; an element of the tangent bundle is of the form $(p, \vec{v})$ with $p \in M$ and $\vec{v} \in T_p M$. In fact $TM$ has a smooth manifold structure⁶; let $\pi: TM \rightarrow M$ be the bundle projection. A section of $TM$ is a map $F:M \rightarrow TM$ is an assignment of a vector to every $p \in M$, i.e. vector field on $M$. Thus, a vector field is smooth if the section $F$ is a smooth map between manifolds.

Let $\mathcal{T}(M)$ denote the space of smooth sections of $TM$, i.e. the space of smooth vector fields on $M$, and define a connection as the map $\nabla: \mathcal{T}(M) \times \mathcal{T}(M) \rightarrow \mathcal{T}(M)$. Written $(\vec{v},\vec{w}) \mapsto \nabla_{\vec{v}} \vec{w}$, the connection satisfies

1. $\nabla_{\vec{v}} \vec{w}$ is linear in $\vec{v}$ over $f,g \in C^{\infty}(M)$:

   \[\nabla_{f \vec{v}_1 + g \vec{v}_2} \vec{w} = f \nabla_{\vec{v}_1} \vec{w} + g \nabla_{\vec{v}_2} \vec{w}\]

2. $\nabla_{\vec{v}} \vec{w}$ is linear in $\vec{w}$ over $a,b \in \R$:

   \[\nabla_{\vec{v}} \parens{a \vec{w}_1 + b \vec{w}_2} = a \nabla_{\vec{v}}\vec{w}_1 + b \nabla_{\vec{v}}\vec{w}_2\]

3. For all $f \in C^{\infty}(M)$ the connection satisfies a product rule

   \[\nabla_{\vec{v}} \parens{f \vec{w}} = f \nabla_{\vec{v}} \vec{w} + (vf)\vec{w}\]

$\nabla_{\vec{v}} \vec{w}$ is also called the covariant derivative of $\vec{w}$ in the direction $\vec{v}$.

Christoffel symbols

Suppose we have a local frame $\braces{\vec{e}_i}$ on a manifold $M$⁷. Since $\braces{\vec{e}_i}$ is a basis and $\nabla$ maps pairs of vector fields to a vector field we can, for each pair $i,j$, expand $\nabla _{\vec{e}_i} \vec{e}_j$ in terms of the same basis/frame

\[\begin{aligned} \nabla _{\vec{e}_i} \vec{e}_j &= \Gamma_{ij}^1 \vec{e}_1 + \cdots + \Gamma_{ij}^n \vec{e}_n \\ &= \Gamma_{ij}^k \vec{e}_k \end{aligned}\]

where the $\Gamma_{ij}^k$ expansion coefficients, are called Christoffel symbols. Then for arbitrary vector fields $\vec{v} := v^i \vec{e}_i$ and $\vec{w} := w^j \vec{e}_j$ and by the product rule above

\[\nabla_{\vec{v}} \vec{w} = \nabla_{\vec{v}} \parens{w^j \vec{e}_j} = w^j \nabla_{\vec{v}} \vec{e}_j + {\vec{v} (w^j)} \vec{e}_j\]

By linearity in $\vec{v}$ the first term can be rewritten

\[w^j \nabla_{\vec{v}} \vec{e}_j = w^j \nabla_{v^i \vec{e}_i} \vec{e}_j = v^i w^j \nabla_{\vec{e}_i} \vec{e}_j = v^i w^j \Gamma_{ij}^k \vec{e}_k\]

Renaming dummy indices we get

\[\nabla_{\vec{v}} \vec{w} = \parens{ v^i w^j \Gamma_{ij}^k + \vec{v}(w^k) } \vec{e}_k\]

Notice that we recover the Euclidean connection by setting $\Gamma_{ij}^k=0$.

Geodesics

Let $\gamma: I \rightarrow M$, where $I = [0,1]$, be a smooth curve; the velocity vector $\dot{\pmb{\gamma}}$ is defined as the push-forward⁸ $\gamma_{*}$ of

\[\pd{t} \equiv \dd{t}\]

for $T_t I$. Writing $\gamma(t) := \parens{\gamma^1(t), \dots, \gamma^n(t)}$ in coordinates we get that

\[\dot{\pmb{\gamma}}(f) = \pdd{f}{\gamma^i} \pdd{\gamma^i}{t} = (\dot{\gamma}^i \partial_i)(f)\]

Let $\mathcal{T}(\gamma)$ denote the space of vector fields along $\gamma$ (i.e. $\vec{v} \in T_{\gamma(t)}M$ for all $\vec{v}$). Then it can be shown⁹ that any connection $\nabla$ on $M$ determines a covariant derivative $D_t \colon \T(\gamma) \to \T(\gamma)$ along $\gamma$ satisfying the properties

1. Linearity for $a,b \in \R$:

   \[D_t(a\vec{v} + b\vec{w}) = aD_t\vec{v} + bD_t\vec{w}\]

2. Product rule for $f \in C^\infty (I)$:

   \[D_t(f\vec{v}) = \dot{f} \vec{v} + f D_t\vec{v}\]

3. If $\vec{v}$ can be extended to $\tilde{\vec{v}}$ in a neighborhood of $\gamma(t)$ then this covariant derivative equals the connection (in the neighborhood) with respect to the velocity of $\gamma$ (on $\gamma(t)$)

   \[D_t(\vec{v}) \equiv \nabla_{\dot{\pmb{\gamma}}} \tilde{\vec{v}}\]

Using uniqueness we can compute the coordinate representation of $D_t \vec{v}$. Let $\vec{v} = v^j \partial_j$; then by linearity and the product rule

\[D_t \vec{v} = D_t \parens{v^j \partial_j} = \dot{v}^j \partial_j + v^j D_t \partial_j\]

where $\partial_j$ is the basis vector of/induced by $\gamma$. Since $\partial_j$ extends to the coordinate basis for the manifold in a neigbhorhood of $\gamma(t)$, the second term becomes

\[\begin{aligned} v^j D_t \partial_j &= v^j \nabla_{\dot{\pmb{\gamma}}} \partial_j \\ &= v^j \nabla_{\dot{\gamma}^i \partial_i} \partial_j \\ &= v^j \dot \gamma^i \nabla_{ \partial_i} \partial_j \\ &= v^j \dot \gamma^i \Gamma_{ij}^k \partial_k \end{aligned}\]

where the Christoffel symbols are with respect to $\partial_i$ (the basis along $\gamma$). Thus

\[D_t \vec{v} = \parens{\dot{v}^k + v^j \dot \gamma^i \Gamma_{ij}^k} \partial_k\]

Now define the acceleration of $\gamma$ to be

\[D_t \dot{\pmb{\gamma}} = \nabla_{\dot{\pmb{\gamma}}} \dot{\pmb{\gamma}}\]

This might seem strange but it is owing to the fact that $\dot{\pmb{\gamma}} = \gamma_{*} (d/dt)$ and so it plays the role of the derivative with respect to time on $\gamma$. Using the above, in local coordinates

\[D_t \dot{\pmb{\gamma}} = \parens{\ddot{\gamma}^k + \gamma^j \dot \gamma^i \Gamma_{ij}^k} \partial_k\]

A geodesic with respect to the connection $\nabla$ is a curve with zero acceleration, i.e. $D_t \dot{\pmb{\gamma}} \equiv 0$. This is the case iff (in coordinates)

\[\ddot{\gamma}^k + \gamma^j \dot \gamma^i \Gamma_{ij}^k = 0\]

which is a set of coupled second-order differential equations called the geodesic equation(s). For example, in the Euclidean connection (where $\Gamma_{ij}^k \equiv 0$) this reduces to the familiar

\[\ddot{\gamma}^k = 0\]

with equally as familiar solution $\gamma(t) = \gamma(0) + \dot \gamma(0) t$, i.e. straight lines. In general (as with most differential equations) there is no solution to the geodesic equation.

Foonotes