# Intrinsic derivative of a vector field at a zero

I am having trouble interpreting Exercise 3.1.16 In McDuff Salamon "Introduction to Symplectic Topology". In particular, what is the intrinsic derivative of a vector field at a zero?

Let $(M, ω)$ be a symplectic manifold and let $\phi \in Ham(M, ω)$ be generated by a time-independent Hamiltonian function $H : M \rightarrow R$.

(ii) Let p be a critical point of $H$. Prove that $d\phi(p) = exp(A)$, $A := D_XH(p)\in sp(T_pM, ω_p)$. Here $D_X(p) : T_pM → T_pM$ denotes the intrinsic derivative of a vector field $X \in \chi (M)$ at a point $p \in M$ where the vector field vanishes and $sp$ denotes the symplectic Lie algebra as in (2.2.2)

I consider those sort of 'intrinsically/(coordinates-)invariantly/geometrically' defined objects to be difficult to digest, especially because there is some implicit meaning to those adverbs that prevent the new-comer to really grasp what is at stake. My answer will thus try to progressively move towards what could be meant by 'intrinsic derivative of $X$', so I apologize for its length.

One would colloquially say that there is in general no coordinates independent way to differentiate a vector field $X$ defined on a manifold $M$ at some point $p \in M$. The meaning of this is that if we read $X$ into two different coordinates systems around $p$, differentiate $X$ component-wise in each system respectively and then look at how the two notions of 'differential of $X$' are related to one another via the jacobian of the change of coordinates, we discover that in general the relation is not 'simple' because it would involve higher order derivatives of the change of coordinates.

To give a coordinates independent meaning to the derivative of $X$ at $p$, one either needs some particular properties of $X$ at $p$ or to introduce additional structure on $M$ (such as a linear connection).

When a smooth vector field $X$ vanishes at $p$, then it turns out that there is a coordinates independent way to differentiate $X$ at $p$, the result of which we denote $DX(p)$. In order to compute this thing, we don't need to argue abstractly without reference to a choice of coordinates; on the contrary, we are quite welcome to work in some choice of coordinates, as the result in any two different coordinates systems are related to one another by a 'simple' expression.

Concretely, consider two coordinates systems $(x_i)$ and $(x'_j)$ on $M$ around $p$, and assume for simplicity that $p$ is $x_i = 0 = x'_j$ for all $i, j$. We read $X$ in the first coordinates system as $X_i \frac{\partial}{\partial x_i}$ and we read it in the second system as $X'_j \frac{\partial}{\partial x'_j}$. (The convention of summation over repeated indices is implied.) We would naively define the derivative of $X$ in the first coordinates system in the $k$-th direction as $\frac{\partial X_i}{\partial x_k} \frac{\partial}{\partial x_i}$, and similarly define the derivative of $X$ in the second coordinates system in the $l$-th direction as $\frac{\partial X'_j}{\partial x'_l} \frac{\partial}{\partial x'_j}$. Now, the two readings of $X$ are related via the chain derivation formula as follows:

$$X'_j \frac{\partial}{\partial x'_j} = \underset{X_i}{\underbrace{X'_j \frac{\partial x_i}{\partial x'_j}}} \frac{\partial}{\partial x_i},$$

so that

\begin{align} \frac{\partial X_i}{\partial x_k} \frac{\partial}{\partial x_i} &= \frac{\partial x'_j}{\partial x_k} \frac{\partial}{\partial x'_j}\left( X'_l \frac{\partial x_i}{\partial x'_l} \right) \frac{\partial}{\partial x_i} \\ &= \frac{\partial x'_j}{\partial x_k} \left( \frac{\partial X'_l}{\partial x'_j} \frac{\partial x_i}{\partial x'_l} + X'_l \frac{\partial^2 x_i}{\partial x'_j\partial x'_l} \right) \frac{\partial}{\partial x_i} \\ &= \frac{\partial x'_j}{\partial x_k} \left\lbrack \frac{\partial X'_l}{\partial x'_j} \frac{\partial }{\partial x'_l} \right\rbrack + X'_l \frac{\partial x'_j}{\partial x_k} \frac{\partial^2 x_i}{\partial x'_j\partial x'_l} \frac{\partial}{\partial x_i} \, . \end{align}

We recognize inside the square brackets the naive derivative of $X$ in the second coordinates system, transformed via terms involving only the jacobian of the change of coordinates. There is this second term which involves the second derivatives of the change of coordinates, suggesting that the above naive notions of derivative of $X$ are not coordinates independent. However, if every $X'_l$ vanishes at a given point (i.e. if $X$ vanishes at a point), then the second term disappears and so the transformation law corresponds to the way a linear map (namely, the proposed intrinsic derivative) would transform under (linear) change of basis of the tangent plane $T_pM$.

A somewhat more invariant way to describe the above discussion goes as follows. A vector field $X$ is invariantly defined as a derivation on smooth functions i.e. $X : C^{\infty}(M) \to C^{\infty}(M) : f \mapsto Xf$ is $\mathbb{R}$-linear and satisfies Leibniz's law. We could want to invariantly define the derivative of $X$ along a vector $Y$ at $p$; if such a thing existed, we might expect that the result of considering $Y(Xf)$ would be independent of $f$. However we compute in coordinates that $$Y(Xf) = Y_i \frac{\partial}{\partial x_i} \left( X_j \frac{\partial f}{\partial x_j} \right) = Y_i \frac{\partial X_j}{\partial x_i}\frac{\partial f}{\partial x_j} + Y_i X_j \frac{\partial^2 f}{\partial x_i \partial x_j} .$$

We thus see that the result at some point $p$ depends on the choice of $f$ except if $X$ vanishes at that point.

Now the right way to define $DX(p)$ at a point $p$ where $X(p) = 0$ is as follows. The vector field $X$ can be seen as a section of the tangent bundle, hence as a smooth map $X : M \to TM$. Ideally, we would be interested in the 'vertical variation' of $X$, that is the variation 'along' the fibers of $\pi : TM \to M$. We can consider the differential $dX : TM \to T(TM)$ and all we would need to know is how to project $T(TM)$ to the vertical bundle $V(TM) := Ker(d\pi)$. Such a projection corresponds to a splitting $T(TM) = V(TM) \oplus H$ where $H$ is a choice of transverse subspace at each $\xi \in TM$. This asks for a choice of $H$, except along the zero section of $\pi$ where $H$ can be readily chosen as the tangent space to the graph of the zero section. This abstract way to define $DX$ at a point $p$ where $X(p)=0$ turns out, in coordinates, to be precisely what we obtained above: naive component-wise differentiation.

This hopefully answer your first concern above about the interpretation of the intrinsic derivative. The symplectic exercise might be difficult for other reasons; if you have trouble solving it, I would recommend to post another question on MSE more targeted on this peculiar exercise.