If $M$ is a smooth manifold, how do I interpret an expression such as $f X$ where $f \in C^{\infty}(M)$ and $X$ is a vector field, i.e. a map $M \rightarrow TM$?
To my understanding, a vector field is a derivation, a map with certain properties. I imagine it as the set of "directions" in each point of the manifold.
Now I came accross the definition of a covariant derivative of (smooth) vector fields $X, Y$, which confuses me. The properties satisfied by the covariant derivative are the following:
- $\nabla_{fX + gY} Z = f \nabla_X Z + g \nabla_Y Z$
- $\nabla_X (Y + Z) = \nabla_X Y + \nabla_X Z$
- $\nabla_X (fY) = X(f)Y + f \nabla_X Y$
My question: What does the $f X$ says? If $f$ just takes points from $M$ and assigns them a value (I guess in $\mathbb{R}^n$?), then how can it be multiplied or composed with a vector field that goes from $M$ to the tangent space? And what is the difference between $X(f)$ and $fX$ in the conditions?
Thank you very much.
$f \in C^\infty(M)$ means that $f$ is a smooth function $f:M \to \Bbb R$.
Now if we have a vector field $X:M \to TM$, then we can define $fX:M \to TM$ as follows: for each point $p \in M$, the space $T_pM$ is a vector space and $X(p)$ lives in that vector space. So we may scale $X(p) \in T_pM$ by the scalar $f(p) \in \Bbb R$ and define $(fX)(p):=f(p)X(p)$ to obtain a smooth map $fX:M \to TM$ such that $(fX)(p) \in T_pM$ for all $p \in M$, i.e. a vector field.
Intuitively, imagine you have at each point of the manifold a direction, i.e. a tangent vector (your $p \mapsto X(p)$) and also a scalar ($p \mapsto f(p)$), then you can scale at each point the tangent vector you have by the scalar to obtain a new tangent vector.
The notation $X(f)$ denotes the directional derivative of $f$ with respect to $X$, see this question for an equivalent definition of vector fields that makes this construction clear.