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The group $H$ acts transitively and primitively on $\mathbb{C}=\mathbb{R}^2$. ('Primitive' means that $H$ preserves no nontrivial foliation.) It's a consequence of the classification of transitive pr …

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 …

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 …

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 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 …

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 …

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+ …

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, …

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+ …

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{\ …

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- …

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 …

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 …