Vector field on manifold

Question

I've only seen a vector field $V$ on a manifold $M$ as a mapping $V:M\to TM$. Is it true that they can also be seen as a mapping $V:C^{\infty}\left(M\right)\to C^{\infty}\left(M\right)$? How would $V$ work in the second case?

Answer 1

There are a couple of things to point out here. Let $M$ be a smooth manifold.

• Every tangent vector at $p$ may be thought of as a derivation at $p$: indeed, if $v_p$ is a tangent vector at $p$ and $f: U \to \mathbf{R}$ is a smooth function in a neighbourhood $U \subseteq M$ of $p$, then, $v_p(f)$ can be thought of as the derivative of $f$ at $p$ in the direction of $v$. More precisely, for a curve $\gamma$ defined around $0$ (so that $\gamma(0) = p$) which represents the tangent vector $v_p$, one sets:

$$v_p(f) = \left.\frac{\mathrm{d}(f \circ \gamma)}{\mathrm{d}t}\right|_{t=0}.$$

Firstly, this is well-defined: that is, the definition of $v_p$ is independent of the curve $\gamma$ that "represents" $v_p$. (This is "per definitionem", like everything is.)

The name comes from the fact that $v_p: C^\infty(U) \to \mathbf{R}$ satisfies the identity $$v_p(fg) = f(p) v_p(g) + g(p) v_p(f).$$

• Equipped with the above interpretation as derivation at a point $p$, we may view a smooth vector field $X$ as a derivation $\mathcal{L}_X: C^\infty(M) \to C^\infty(M)$ on the algebra $C^\infty(M)$ by setting $\mathcal{L}_X(f)(p) = X_p(f)$. (Check that this makes sense!)

See Lee's Introduction to Smooth Manifolds, a wonderful book on differential topology.