# How can a vector field eat a smooth function?

I was watching this Frederic Schuller's lecture on Grassmann algebra and de Rham cohomology. I don't understand. At 43:40 it looks like a vector field is eating a smooth function.

The $X_i$ should be a vector field, that is a section of a smooth tangent bundle. So it should map a point on the manifold to one of the point's tangent vectors. But omega is a $0$-$n$ tensor field, that is, a function that takes n vector fields and outputs a smooth function (manifold $\to \mathbb R$). So how can you give the $X_i$ a smooth function when it should eat a point of the manifold? What am I missing?

• $X(f)$ at $p$ is the result of tangent vector $X_p$ acting on function $f$ at $p$, I suspect he covers this in an earlier lecture – Calvin Khor Jul 22 '18 at 17:22
• @CalvinKhor Is this just notation, like, lets call X(f), where X is a vector field and f is a smooth function, the function defined by p in M is mapped to X(p)(f)? Or is it true that the functions X that takes a smooth function and outputs a smooth function taking p to X(f)(p) are isomorphic to the functions X that takes a point on the manifold and map it to a tangent vector? – p.h.f. Jul 22 '18 at 18:13
• en.wikipedia.org/wiki/Lie_derivative – Qiaochu Yuan Jul 22 '18 at 18:20
• @p.h.f. it's the first case. I don't know if there is a sense in which they are isomorphic. – Calvin Khor Jul 22 '18 at 18:25
• The X(f)(p) in my comment should be X(p)(f). – p.h.f. Jul 22 '18 at 18:26

Given $\vec{v} = (v^1,...,v^n) \in \mathbb{R}^n$ and $f: \mathbb{E}^n \to \mathbb{R}$. Then we define $\vec{v}[f] = \sum_j \frac{\partial f}{\partial x^j} v^j$. Given a vector field $X: U \to \mathbb{R}^n$ we have $X_p \in \mathbb{R}^n_p$. We can choose a basis $e^1_p,...,e^n_p$ at $\mathbb{R}^n_p$ such that,
$$X_p = \sum_i a(i,p) \ e^i_p$$
• We then identify $X_p$ with its coordinate form i.e $(a(1,p),...,a(n,p)) \in \mathbb{R}^n$. Then we let $X[f]$ be the vector field such that,
$$X[f] (p) := X_p[f] = \sum_j \frac{\partial f}{\partial x^j} \cdot a(j,p)$$
• Here we write $[f]$ to be the germ class of $f$. The germ class of a smooth function $f$ at $p$ are all the other smooth functions which agree with $f$ on a neighborhood of $p$. Recall that such functions have the same directional derivative.
• The whole reason behind all this is that it can be shown that derivations at $p$ are isomorphic to the tangent space at a point, which are just vectors. To see this full carried out, see Loring Tu's "Introduction to Manifolds".