Do Taylor series of analytic vector field depend on metric tensor?

Question

Let $\mathcal{M}$ be $n$-dimensional Riemannian manifold. And let $F$ be an analytic vector field on it. By definition this means that $$f^i(x(q)) = f^i \vert_{x(p)} + \frac{\partial f^{i}}{\partial x_{j_1}} \vert_{x(p)} x^{j_1} + \frac{1}{2}\frac{\partial^2 f^i}{\partial x_{j_1} \partial x_{j_2}} \vert_{x(p)} x^{j_1} x^{j_2} + O(||x(q) - x(p)||^3)$$ where $f^i$ is a representation of $F$ in map $x$; $p \in \mathcal{M}$ and $q$ lies in some small enough neighbourhood of $p$. The question is: am I correct that since $f^i$ in chart can be treated just like simple function from $\mathbb{R}^n \to \mathbb{R}^n$, the partial derivatives and norm in "O" are just standard partial derivatives and Euclidean norm from $\mathbb{R}^n$? The question arise in discussion with my colleague who is sure that norm in "O" is dependent on metric tensor and so are partial derivatives. For me it is quite strange since analytic vector fields can be built on manifolds where metric tensor even does not exist(they can be built because definition of tangent vector does not depend on metric tensor). Maybe I am missing something?

Answer 1

Yes, the expression itself is independent of the choice of coordinate (and metric used to define the norm), even though the ingredients are not.

Given a vector field $F$ on a manifold, its local coordinate representation $(f^i) = (f^1, \ldots, f^n)$ depends on the choice of coordinates, as do its partial derivatives $\partial_j f^i, \partial_k \partial_j f^i, \ldots$. That does not preclude, however, certain algebraic combinations of the $f^i$ and their derivatives from being independent of the choice of components, perhaps, as in our case, modulo a suitable equivalence relation. This is the case, for example, for coordinate expressions for tensor invariants for curvature tensors, which of course requires the additional data of a connection, which we neither have nor need in our case.

If we have two local charts $(x^i), (y^i)$ in which $f$ has respective representations, say, $(\bar{f}^i), (\hat{f}^i)$, we can form the second-order Taylor expansions of the components $\bar{f}^i$ and $\hat{f}^i$ in their respective coordinate charts around a common point $p$ in the manifold; for sanity we can translate our charts so that $x(p) = y(p) = \mathbf 0$. Explicitly, these expansions are given by

$$\bar{j}_2^i := \bar{f}^i(\mathbf{0}) + (\partial_{x_j} \bar{f}^i)(\mathbf{0}) x^j + (\partial_{x^k} \partial_{x^j} \bar{f}^i)(\mathbf{0}) x^j x^k$$

and the obvious analogue $\hat{j}_2^i$. Now, we can form the vector fields $\bar{j}_2^i \partial_{x^i}$ and $\hat{j}_2^i \partial_{y^i}$, which in their respective coordinate charts approximate $F$ to second order. In fact, using coordinate transformation rules shows that in any coordinate chart the coordinate representation of the vector field

$$\Delta := \bar{j}_2^i \partial_{x^i} - \hat{j}_2^i \partial_{y^i}$$

vanishes to the desired order, that is, the Taylor expansion of $\Delta$ in any coordinate system has zero constant, linear, and quadratic terms---in other words, the property of vanishing to this (analogously, to any fixed order) is independent of the choice of coordinate systems! This lets us conclude that the vector field $\bar{j}_2^i \partial_{x^i}$ is independent of the choice of coordinate system modulo third-order terms; in other words, the expression in the question is well-defined provided we interpret the $O(\cdots)$ as indicating third- and higher-order terms in an arbitrary coordinate chart.

This leads to a natural definition: A $k$-jet of a vector field is a germ of a vector field modulo agreement to $k$th other, that is, the germs $X, Y$ represent the same $k$-jet if all of the components up through the $k$th degree of the Taylor expansion are zero. We denote the $k$-jet of a vector field (or a germ of a vector field $X$) at $p$ by $j^k_p(X)$, and a representative is given in any chart $(x^i)$ by $(\bar{j}^i_k)$, whose components are defined in the obvious way generalizing the definition of $\bar{j}_2^i \partial_{x^i}$ above. This all works just as well, by the way, for smooth functions or other vector bundle sections on a manifold.

Finally, since any two metrics are locally comparable, we may as well write the error term $O(\cdots)$ in the way indicated. This isn't great notation, in that it suggests there is a norm when we have none available, but probably this sort of notation is used commonly because $O(\cdots)$ notation doesn't mesh especially well with multiple variables in the first place. If you wanted to be more precise and avoid the norm notation, you could define $\mathcal{I}$ to be the polynomial ideal in $\mathbb{R}[x^1, \ldots, x^n]$ generated by the monic monomials of total order $3$ and write $O(\mathcal{I})$; in fact, this hints nicely at the jet formulation above.

Remark The question asked about analytic vector fields, but everything I've said so far applies to smooth fields in general. Because jets and analytic objects can both be framed in terms of coordinate Taylor series, though, there's something important that we can say relating them: We know that a vector field is analytic iff at every point on the manifold its Taylor series (in any, equivalently every, local chat) converges locally to the vector field itself---in the language of jets, we say that an analytic vector field is determined by its $\infty$-jet. An $\infty$-jet of a vector field (germ) is what you might guess it is, namely an equivalence class of vector field germs, where two germs are equivalent if their $k$-jets agree for all $k \in \mathbb{Z}_{\geq 0}$.