What does derivation $(df_p(v))(g)=v(g \circ f)(p)$ really mean?

Question

What does derivation $(df_p(v))(g)=v(g \circ f)(p)$ really mean?

It's said that $df_p(v) \in T_{f(p)}(N)$ is treated as a derivation which when applied to $g$ results in the directional derivative of $g$. By the R.H.S. this derivative is defined by forming $f \circ g$ and "computing its directional derivative at $p$ using $v$". Or i.e. the directional derivative of $g$ on $N$ is defined as the directional derivative of $f \circ g$.

I don't understand how $v(g \circ f)p$ is a directional derivative. Is it perhaps shorthand for

$$\lim_{h \rightarrow} \frac{g \circ f(x+vh, y+vh,...)-g \circ f(x,y,...)}{h}$$

and $p=(x,y,...)$.

Answer 1

From your comments, it seems to me that you are confusing standard definition of tangent vector in differential geometry and isomorphism $T_pV\cong V$ for $V$ a vector space.

Def. (tangent vector) Let $M$ be a smooth manifold and $C^\infty(M)$ the space of smooth functions on $M$. A linear functional $v$ on $C^\infty(M)$ is called tangent vector at point $p$ if $$v(fg) = v(f)g(p)+f(p)v(g).$$

From this definition it follows that $v(g\circ f)$ should be interpreted as linear functional $v\colon C^\infty(M) \to \mathbb R$ evaluated at function $g\circ f\in C^\infty(M)$.

In a special case when the manifold is a (finite dimensional) vector space $V$, there is an isomorphism $V\to T_pV$ that sends vector $v\in V$ to linear functional $f\mapsto \frac d{dt}f(p + tv)|_{t=0}$, which is precisely directional derivative $D_vf(p)$. It is easy to check that this satisfies Leibniz rule.

Finally, for a smooth map $f\colon M\to N$, we can define a linear map $df_p\colon T_pM\to T_{f(p)}N$ in the following way: if $v\in T_pM$, then $df_p(v)$ should be a linear functional on $C^\infty(N)$, so for a smooth function $g\colon N\to \mathbb R$, we define $$(df_p(v))(g) = v(g\circ f).$$

Answer 2

First we introduce the objects. Let $F\colon M \to N$ be a smooth map between two manifolds. It maps points on $M$ to points on $N$. Also let $g\colon N \to \mathbb{R}$ be a real function on $N$; it maps points on $N$ to real numbers. And finally, let $v \in T_p M$ be a tangent vector at the point $p$ on $M$.

Interpretation of $\bigl(dF_{p}(v)\bigr)(g)$. The tangent map $dF_p$ maps the vector $v\in T_p M$ to a vector $dF_p(v) \in T_{F(p)}N$. The function $g$ is a function on $N$. Since $dF_p (v)$ is a tangent vector to $N$ at the point $F(p)$ and $g$ is a function on $N$, it makes sense to look at the directional derivative $\bigl(dF_{p}(v)\bigr)(g)$. (This is interpreting the definition of a tangent vector on $N$.)

Interpretation of $v(g\circ F)(p)$. The composition $g\circ F$ is a function from the manifold $M$ to $\mathbb{R}$. Since $v\in T_p M$ is a vector tangent to $M$, and $g\circ F$ is a real function on $M$, the directional derivative $v(g \circ F)(p)$ makes sense. (Note that this is also the definition of a tangent vector, but now we are working on $M$.)

These are the interpretations of both sides of the equations. If these are unclear, it might be helpful to revise the definition of a tangent vector.