How is the action of a vector field on a function defined?

Question

Questions of definition:

1) How do we make sense of $X(f)$ with $X$ a vector field on $M$ and $f$ a smooth function on $M$?

One idea I have is that it is not actually $X(f)$ where we apply the vector field on $f$ but actually the canonically associated derivation. That is $X(f)$ is the following function $$ X(f):M\rightarrow \mathbb{R}:x\rightarrow X_x(f)$$

2) How do we make sense of $Xg(Y,Z)$ where $g$ is a riemanian metric and $X,Y,Z$ vector fields.

One idea is that $Xg(Y,Z)$ is the fonction defined as follows $$ Xg(Y,Z):M\rightarrow \mathbb{R}:x\rightarrow X_x\cdot g_x(X_x,Y_x)$$

Answer 1

To get this answered, the short answer to both questions is yes. A vector field $X$ is a section of the tangent bundle $TM$, in particular $X:M\to TM$. So $X$ maps points to vectors, and vectors map functions to real numbers.

So when people say $Xf$, they actually mean $\tilde X(f)$ where $\tilde X$ is implicitly defined as follows: $\tilde X: C^\infty(M)\to C^\infty(M)$, $\, \tilde X(f)(p):=X(p)(f)$.

The second question follows easily from this.

Answer 2

You might be used to thinking of a vector field as a map from an open neighborhood $U$ of $M$ to $\mathbb{R}^n$. For instance, if $M = \mathbb{R}^3$ (the US "Calc III" setting), we write $$ \mathbf{F}(x,y,z) = P(x,y,z) \mathbf{i} + Q(x,y,z) \mathbf{j} + R(x,y,z) \mathbf{k} $$ In the modern setting, we associate $\mathbf{i}$ to $\frac{\partial}{\partial x}$, $\mathbf{j}$ to $\frac{\partial}{\partial y}$, and $\mathbf{k}$ to $\frac{\partial}{\partial z}$. So the vector field above becomes $$ X = P\frac{\partial}{\partial x} + Q\frac{\partial}{\partial y} + R\frac{\partial}{\partial z} $$ $X$ maps functions to functions by $$ X(f) = P\frac{\partial f}{\partial x} + Q\frac{\partial f}{\partial y} + R\frac{\partial f}{\partial z} $$ and is a derivation by the product rule. On the level of points, it is, as you guessed, $$ X(f)(x,y,z) = P(x,y,z) \frac{\partial f}{\partial x}(x,y,z) + Q(x,y,z) \frac{\partial f}{\partial y}(x,y,z) + R(x,y,z) \frac{\partial f}{\partial z}(x,y,z) $$

For your second question, let's pass to $M = \mathbb{R}^n$, or, $U$ is a coordinate patch of $M$ with coordinates $x^1, \dots, x^n$. Then given three vector fields $X$, $Y$, and $Z$ on $U$, we can write \begin{align*} X &= P^i \frac{\partial}{\partial x^i} \\ Y &= Q^i \frac{\partial}{\partial x^i} \\ Z &= R^i \frac{\partial}{\partial x^i} \\ \end{align*} where the $P$'s, $Q$'s and $R$'s are functions, and we use the Einstein summation convention. Also, $g$ can be decomposed as $g_{ij} dx^i \,dx^j$. So $$ g(Y,Z) = g_{ij} Q^i R^j $$ This is a sum of products of functions. If we hit this with $X$, then, we have \begin{align*} Xg(Y,Z) &= P^k\frac{\partial}{\partial x^k}\left(g_{ij} Q^i R^j\right)\\ &= P^k\left(\frac{\partial g_{ij}}{\partial x^k}Q^i R^j + g_{ij} \frac{\partial Q^i}{\partial x^k} R^j + g_{ij} Q^i \frac{\partial R^j}{\partial x^k}\right) \end{align*}

I don't think it's valid to think of $g$ evaluated at $x$, then paired with $Y_x$ and $Z_x$, then acted upon by $X_x$. Because, as you can see from the above, the components of $g$ itself get differentiated. The proper order is that the tensor $g$ gets paired with the vector fields $Y$ and $Z$, resulting in a function, which is acted upon by $X$.