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Your conjecture is true, and it can be proved by making a few observations. First, for each $\alpha\in\mathbb{R}$, let $V_\alpha$ be the vector space of germs of smooth functions $f$ at the origin th …

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 …

First, it's not finite dimensional, even in the case of a torus. Just let $x,y$ be the $2\pi$-periodic functions on the torus and take $\alpha = \mathrm{d} x$, and you'll see that $H^0$ is all the fun …

I don't know a reference, but, as you've stated it, this is a trivial result: If $p$ is a point on a smooth surface $S\subset\mathbb{R}^3$ at which the Gauss curvature is negative, then $p$ is non- …

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 …

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, …

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 …

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+ …

The only thing one really needs to define the operation $\mathrm{Div}:{\frak{X}}(S^3)\to C^\infty(S^3)$, i.e., mapping vector fields on $S^3$ to functions on $S^3$, is a volume form on $S^3$. (One ca …

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 …

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 …

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+ …

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 …

I'll just add a few more references to what Ben McKay mentioned. Phillip Griffiths and I wrote a paper on invariants of parabolic equations in 1 space variable (Characteristic cohomology of differenti …

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{\ …