Cfrgtkky

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 $M^msubset mathbb{R}^k$ 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 $pin M$ 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?

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:

$$Dpi_{X(p)} circ DX_p = 1_{T_p M}$$

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 $Dpi_{X(p)}$ 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.

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 $dX_p:T_pM to T_{(p,0)}TM simeq T^h_{(p,0)}TMoplus T^v_{(p,0)}TM,$
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
$$DX_p=iota circ pi_2 circ dX_p.$$

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

So, locally (i.e., over $U$) we can regard the vector field $X$ as $Uto T_pM$.

(We could as well write $UtoBbb R^m$ if that's clearer.)

Finally, as $T_pM$ is a linear space, its tangent space (on any point) is identified with itself.