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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:

  1. $\nabla_{fX + gY} Z = f \nabla_X Z + g \nabla_Y Z$
  2. $\nabla_X (Y + Z) = \nabla_X Y + \nabla_X Z$
  3. $\nabla_X (fY) = X(f)Y + f \nabla_X Y$

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.

2 Answers2

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$f \in C^\infty(M)$ 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.

Lukas Heger
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A more dynamical interpretation for $fX$ is that, considering the local flow $\phi$ of $X$, $f$ gives a velocity (or time) change of $\phi$. The flows of $X$ and $fX$ will have the same trajectories, but the trajectories will be parameterized differently (this is an instance of "orbit equivalence"). Thus for instance if for some reason one ought to slow down the flow $\phi$ on an open subset $U$, one can multiply $X$ with a sufficiently smooth positive function $f$ such that $f(p)\approx 0$ for $p\in U$. E.g. see Importance of finitude in the Poincaré-Bendixson theorem for an instantiation of this framework.

Alp Uzman
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  • Could you explain why the flows of $X$ and $fX$ will have the same trajectories? – Sunny. Y Oct 12 '22 at 19:56
  • @Sunny.Y Heuristically $X$ gives the velocities (so both direction and magnitude). Multiplying by $f$ does not change the direction since $f$ is scalar valued, however it might change the magnitude. More rigorously one can also write down the orbit equivalence in terms of $f$ in this case, by taking antiderivatives. – Alp Uzman Oct 18 '22 at 09:07
  • Thank you for your explanation! – Sunny. Y Oct 23 '22 at 04:31