What is the Derivative of a Vector Field in a Manifold?

Question

I'm studying the book "Geometric Theory of Dynamical Systems An Introduction" - Jacob Palis, Jr. Welington de Melo.

On page 10, the author defines:

Let $M^m\subset \mathbb{R}^k$ be a differentiable manifold. A vector field of class $C^r$ on M is a $C^r$ map $X: M \rightarrow \mathbb{R}^k$ which associates a vector $X(p) \in T_pM$ to each point $p \in M$. This corresponds to a $C^r$ map $X: M \rightarrow TM$ such that $\pi X$ is the identity on $M$ where $\pi$ is the natural projection from $TM$ to $M$. We denote by $\mathfrak{X}^r (M)$ the set of $C^r$ vector fields on $M$.

And on page 58, comes the definition that I'm having problem

Let $X \in \mathfrak{X}^r(M)$ and let $p\in M$ be a singularity of $X$. We say that $p$ is a hyperbolic singularity if $DX_p: T_p M \rightarrow T_p M $ is a hyperbolic linear vector field, that is, $DX_p$ has no eigenvalue on the imaginary axis.

I don't understand why $DX_p: T_pM \rightarrow T_pM$. My knowledge of Differential Topology just allows me to differentiate functions like $f: M \rightarrow N$, implying that $Df_p: T_pM \rightarrow T_{f(p)} N$, so (in my mind) the correct would be $DX_p:T_pM \rightarrow T_{X(p)}\mathbb{R}^k$, so talk about eigenvalue on $DX_p$ doesn't make sense, because $DX_p$ is not a endomorphism.

The most strange part is that $T_pM$ isn't even isomorphic to $T_{X(p)}\mathbb{R}^k$. Can anyone tell me what am I confusing?

Answer 1

View $X$ as a smooth map $X \colon M \to TM$ with the property that $\pi X = 1_M$ where $\pi \colon TM \to M$ is the projection map. As $X$ is a smooth map between manifolds it has a derivative $DX_p \colon T_p M \to T_{X(p)} TM$ defined just like the derivative of any smooth map. This does indeed look like a problem: the vector space on the left has dimension $\dim(M)$ while the one on the right has dimension $2 \dim(M)$, so it doesn't make sense to speak of eigenvalues!

But $X$ isn't just any old smooth map: we haven't used the equation $\pi X = 1_M$ yet. Let us differentiate both sides of this equation using the chain rule:

$$D\pi_{X(p)} \circ DX_p = 1_{T_p M}$$

Trivializing everything over a coordinate neighborhood of $p$, we have $T_{X(p)} TM \cong \mathbb{R}^n \times \mathbb{R}^n$ and $T_p M \cong \mathbb{R}^n$; in this trivialization $D\pi_{X(p)}$ is just the projection map onto the first factor. Thus the equation above says that $DX_p$ has to have the form

$$DX_p(v) = (v, Lv)$$

for some linear transformation $L \colon \mathbb{R}^n \to \mathbb{R}^n$. It is a fairly standard abuse of notation to think of this object $L$ as "the derivative" of $X$, even though the derivative is really $(I, L)$ where $I$ is the identity.

Answer 2

Since $M$ is a manifold, there is a neighborhood $U$ of $p$ and a chart $U\to\Bbb R^m$. In particular, it determines an isomorphism $T_pM\cong\Bbb R^m$, and the tangent manifold over $U$ is isomorphic to the trivial fibration $U\times T_pM$. (We can flatten it locally.)

So, locally (i.e., over $U$) we can regard the vector field $X$ as $U\to T_pM$.
(We could as well write $U\to\Bbb R^m$ if that's clearer.)
Finally, as $T_pM$ is a linear space, its tangent space (on any point) is identified with itself.

Answer 3

It is possible to minorly adapt the answer by Paul Siegel in order to avoid trivializations. We just use the natural split over a point $(p,0)$ of $T_{(p,0)}TM$.

Again, a vector field is a map $X: M \to TM$ such that $\pi \circ X=\mathrm{Id}$.

If $X(p)=0$, we have that $dX_p:T_pM \to T_{(p,0)}TM \simeq T^h_{(p,0)}TM\oplus T^v_{(p,0)}TM,$ where $$T^h_{(p,0)}TM:=\{\dot{\gamma}(0) \mid \gamma \text{ is a curve through $(p,0)$ such that } \gamma \subset M\},$$ $$T^v_{(p,0)}TM:=\{\dot{\gamma}(0) \mid \gamma \text{ is a curve through $(p,0)$ such that } \gamma \subset T_{p}M\}.$$

Note that the existence of such a natural decomposition uses heavily the fact that $X(p)=0$, otherwise we would need to rely on a metric (more directly, a connection) on $M$ to furnish it (for more information, see here).

There is a natural isomorphism $i: T^v_{(p,0)}TM \to T_pM $ (It is similar to the isomorphism that exists from $T_pV \to V$, where $V$ is a vector space). The "derivative" which the text is alluding to is then $$DX_p=\iota \circ \pi_2 \circ dX_p.$$