Definition of differential on manifolds

Question

I'm studying some differential Geometry at the moment and I'm getting a bit stuck with the definition of the differential. It's defined as follows

\begin{array}{cl} \phi_{\star,m} : T_{m}M \rightarrow T_{\phi(m)}N\\ v \mapsto \phi_{\star,m}(v)f \mapsto v(f \circ \phi) \end{array}

Where $f$ is some element of the germ at $\phi(m)$. What I'm confused about is to me it looks like $v(f\circ \phi) \in T_{m}M$ which obviously can't be correct. So to summarize I'm asking is how can $v(f \circ \phi)$ be in $T_{\phi(m)}N$? All heuristic arguments welcome.

Answer 1

Remember that tangent vectors eat smooth functions on their respective manifold. We have a tangent vector $v\in T_mM$, it only knows how to eat functions in $C^\infty (M)$. We can't just stick in a function $f \in C^\infty (N)$ . So first we precompose $f$ with $\phi$ which is a function in $C^\infty(M)$ (Think of the domain of $\phi$ and the range of $f$). $v$ knows how to differentiate $C^\infty(M)$ functions so we can know safely feed $f\circ \phi$ into $v$.

Now $\phi_{\star,m}(v)(\cdot )$ inherits all the properties necessary to be a tangent vector from $v$.

Answer 2

You might be a bit confused. You have said $v(f\circ \phi)\in T_m M$. This is not true. Tangent vectors map functions to real numbers. So this quantity would be a real number. Not a tangent vector.

Let $F:M\rightarrow N$ be a smooth map between smooth manifolds. The differential of $F$ at a point $p$, denoted $dF_p$ (your notation is common as well), is a map:

$ \begin{align*} dF_p:T_p M&\rightarrow T_{F(p)} N\\ v&\mapsto dF_p(v). \end{align*} $

As written above, tangent vectors eat smooth functions on their respective manifold. Since $dF_p(v)$ lies in $T_{F(p)}N$, it can only take in smooth functions defined on $N$. The natural question to then ask is what does it do to said functions. Well, if $f\in C^{\infty}(N)$, then we define its action on $f$ as

$ \begin{align*} (dF_p(v))(f)=v(f\circ F) \end{align*} $

Note that $v\in T_p M$. So it only makes sense to eat functions defined on $M$. This is okay, though, since $f\circ F$ has domain $M$ and since it is a composition of smooth functions, it is also smooth. So $v(f\circ F)$ makes sense.