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If I understand correctly, there is already a counterexample on the torus: On the $xy$-plane $\mathbb{R}^2$, let $X$ be the vector field $$ X = \sin x\,\frac{\partial\ }{\partial x} + \cos x\,\frac{\ …

Any surface of revolution in $3$-space with poles will have this property. The reason is that, in this case, any geodesic either goes through a pole (i.e., a point where the axis of revolution meets …

This is really more of a response to the OP's request for an 'insightful way to view the computation' than it is an answer to the specific problem; so keep this in mind. On the other hand, as I'll po …

Actually, no matter what $A$ is, there will be nonzero solutions that will converge to zero, so you can't prove unboundedness, even though it may be true that the 'generic' solution is unbounded. Her …

As Ben's argument suggests, the proof that, if the flow of $X$ preserves $J$ then the flow of $JX$ preserves $J$ does depend on the integrability of $J$. As a concrete example of an almost-complex ma …

No, this is not possible. In fact a more general result holds: If a vector field $X$ has an isolated singularity at $x\in M$ for which the linearization $X'(x):T_xM\to T_xM$ has no real eigenvalues, …

It is not hard to concoct such an example in sufficiently high degree. For an example of degree $5$, take $$ \begin{align} x' &= x\,(1-x^2-y^2)(x^2+y^2-3) - y\,(2-x^2-y^2)\\ y' &= y\,(1-x^2-y^2)(x^2+ …

Here is how one can construct an example: Consider the smooth, nonvanishing $1$-form $$ \omega = y^3(1{-}y)^2\,\mathrm{d}x + \big(y^3-2(1{-}y)^2\bigr)\,\mathrm{d}y. $$ Note: This $\omega$ came from …

Since the metric doesn't have to extend to the origin, take the flat metric $$ g = \frac{\bigl(\mathrm{d}\left(x\sqrt{1+x^2/2}\right)\bigr)^2 + \mathrm{d}y^2}{x^2+x^4/2+y^2}. $$ The level curves $x^2+ …

The vector fields $X$ that satisfy this condition are highly constrained, as there exist only a finite-dimensional family of $C^5$ solutions in a neighborhood of any given point in the plane. Here is …

Actually, it's easier than the general curvature case: Let $X^\flat$ be the $1$-form dual to $X$ via the metric $g$. Then the orthogonal plane field to the integral curves of $X$ is integrable if an …

This isn't an answer, other than a general set of comments to explain why there is no answer in the form that the OP wants. First of all, when $V$ is a vector space of dimension $n$ and $k$ is an int …

This ODE has some very interesting properties. If one clears fractions and writes it out as $$ x(x+2y)(x-2y+1)\,y'' = (4x^2-8y^2+3x+4y)\,y' + x(4y-1)\,(y')^2, \tag1 $$ one recognizes this as the equa …

This is an amplification of my comment on Vladimir's answer. It's actually not at all hard to see the generality of the surface metrics that admit a $k$-th degree polynomial first integral of their g …

This is a comment, not an answer, but it's too long to fit into a comment window. I don't know of a 'generic' condition on metrics on the torus that would guarantee that there are no null-homotopic c …