How can vector field simultaneously be a function and also an operator that acts on a function?

Question

In elementary calculus we have definition:

A vector field is a function that assigns a vector to each point in $\mathbb{R}^2$ or $\mathbb{R}^3$

i.e. F(x,y) = P(x,y) $\hat i$ + Q(x,y,) $\hat j$

In differential geometry we have notation:

$vf$ where $v$ = $v_k \partial^k$ so $vf$ = $v_k \partial^k f$

more precisely $v : C^\infty(M) \to C^\infty(M) $

In the former case, a vector field is defined as $F(x,y)$, in the latter case, vector field is an operator $v = v_k \partial^k$

I am very new to differential geometry. Is there any reconciliation between the two concepts?

Answer 1

A (smooth) vector field $X$ on some space can give you either

1. For every point $p$ in the space a vector $\vec v_p$
2. For every (smooth) scalar function $f$ on the space a new (smooth) scalar function $f_X$

The reconciliation is that $f_X(p)$ is the directional derivative of $f$ at $p$ along $\vec v_p$.

Answer 2

First I want to point out that your $F(x,y)$ can be identified with the operator,

$$ \vec{F}(x,y) \cdot \vec{\nabla} f,$$

which is consistent with the second definition.

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In differential geometry, a vector field is a smooth assignement of tangent vectors on a manifold. Tangent vectors can defined as linear derivations of functions in $C^\infty(M)$.

Locally we can use coordinates to identify tangent vectors with linear combinations of partial derivatives. Which means that your second definition does say that a vector field assigns tangent vectors to each point (The $v_k$ will vary from point to point).

Answer 3

You will see it when you consider the definition of a vector field via the tangent bundle of a manifold. This is often not well explained in physics textbooks. Have a look at Arnold: Mathematical Methods of Classical Mechanics for an excellent explanation.