How can a vector field eat a smooth function?

Question

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?

Answer 1

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".

Answer 2

You can find this in Schuller's lecture on Differential Structure during the aside on derivations and algebras.

Recall that a tangent vector $X_{\gamma,p}\in T_pM$ is nothing but a derivation $D:C^\infty(M)\rightarrow \mathbb{R}$ defined by $X_{\gamma,p}\mapsto X_{\gamma,p}f = (f\circ \gamma)'(0)$.

When you apply $X\in \Gamma(TM)$ to a smooth function on $M$, you obtain a real number for each $p\in M$, i.e. a smooth function on $M$. So a vector field on $M$ is nothing but a derivation $D:C^\infty(M)\rightarrow C^\infty(M)$ defined exactly as in the comments: $X\mapsto Xf$ where $(Xf)(p) = X(p)f=X_{\gamma,p}f$ for all $p\in M$.