Is the covariant derivative $\nabla$ in a Riemannian a one-form?

Question

Is the covariant derivative $\nabla$ in a Riemannian abstract manifold $\mathcal{M}$ a one-form operator?

Answer 1

Given a function $f$ on a Riemannian manifold and a point $p$ on the manifold, you can define its directional derivative of $f$ at $p \in \mathcal{M}$ as a function of the tangent vector $v \in T_p\mathcal{M}$. This is often denoted $D_vf(p)$ or $d_vf(p)$. With $f$ and $p$ fixed, this is a linear function on $T_p\mathcal{M}$ and therefore an element of $T^*_p\mathcal{M}$, which we denote $df_p$ and call the differential of $f$ at $p$.

A connection is a generalization of this to vector fields. Given a vector field $F$, a connection $\nabla$, and a point $p \in \mathcal{M}$, one can define the directional derivative of $F$ as a map from a tangent vector $v \in T_p\mathcal{M}$ to a tangent vector, which we usually denote by $\nabla_vX(p) \in T_p\mathcal{M}$. If $F$, $\nabla$, and $p$ are fixed, then this map is a linear map that is usually written as \begin{align*} \nabla F(p): T_p\mathcal{M} &\rightarrow T_p\mathcal{M}\\ v &\mapsto \nabla_vF(p). \end{align*} If we now fix a frame $(e_1, \dots, e_n)$ of vector fields and a dual frame $(\omega^1, \dots, \omega^n)$ of $1$-forms, then, given any $v$, we can write $\nabla_vF(p) \in T_p\mathcal{M}$ with respect to the basis $(e_1(p), \dots, e_n(p))$ of $T_p\mathcal{M}$. In other words, if $F$, $p$, and the frame are fixed, then there are scalar functions $\phi^1, \dots, \phi^n$ of $v$ such that \begin{equation} (\nabla_vF)(p) = \phi^1(v)e_1(p) + \cdots + \phi^n(v)e_n(p). \tag{*} \end{equation} Since $\nabla_vF(p)$ is a linear function of $v$, so are the functions $\phi^1, \dots, \phi^n$. Therefore, they are elements of $T^*_p\mathcal{M}$. Since they also depend on $p$, they are in fact $1$-forms.

On the other hand, since $(e_1(p), \dots, e_n(p))$ is a basis of $T_p\mathcal{M}$, the vector $F(p) \in T_p\mathcal{M}$ can be written as $$ F(p) = a^1(p)e_1(p) + \cdots + a^ne_n(p). $$ Since $\nabla$ is a derivation, it follows that $$ \nabla_vF(p) = (d_va^1)(p)e_1 + \cdots + (d_va^n)(p)e_n + a^1\nabla_ve_1(p) + \cdots a^n\nabla_ve_n(p). $$ If we apply equation (*) to each vector field $e_1$, we get $1$-forms $\phi^i_j$ such that for each $1 \le j \le n$, $$ \nabla_ve_j(p) = \phi^1_j(v)e_1(p) + \cdots + \phi^n_j(v)e_n(p). $$ Putting this all together, we see that given any vector field $$ F = a^1e_1 + \cdots + a^ne_n, $$ its covariant derivative in the direction of a tangent vector $v \in T_p\mathcal{M}$ at $p \in \mathcal{M}$ is given by $$ \nabla_vF = \langle v, da^i + a^i\phi^i_j\rangle e_j. $$ In particular, a connection uniquely determines a matrix of $1$-forms $\phi^i_j$ and vice versa.

In this sense, a connection is a $1$-form operator.