Does for every vector field there always exist a volume form for which the vector field is a homothety?

Question

Let $v$ be a vector field. Does there exists a volume form $\Omega$ such that its Lie derivative is proportional to itself with a constant coefficient: $$\mathcal{L}_v \Omega= C \cdot \Omega? \ \ \ \ \ (\ast)$$

A simplification of the question: assume that the divergence $\sum_i \frac{\partial v^i}{\partial x_i}$ of the vector field is not zero at all points (in this case $C$ is necessary not zero)?

A related question: assume in addition that the divergence is zero, and require that the constant $C$ in $(\ast)$ is zero?

The questions are local (i.e., we work in an arbitrarily small neighborhood), everything is $C^\infty$-smooth, and is even real-analytic if it makes the life easier. The dimension is arbitrary.

Of course near the points where the vector field does not vanish the existence of such a volume form follows from the existence of a coordinate system such that our vector field is $\partial_{x_1}$. More generally, if a vector field is linearisable near a point, then the existence of such a volume form is also trivial.

Actually, I believe that the answer on the very first question in negative; this belief is because of the divergence of the vector field controls the coefficient $C$ and one can possibly build a counterexample by constructing a vector field such that it vanishes at a convergent sequence of points $a_1,..., a_k, ... \to a$, such that the divegence is zero at the point $a$ and is not zero at all the points $a_k$.

The motivation came from projective differential geometry: it is known (see for example Projectively equivalent connections) that projective structure + a volume form up to a constant coefficient uniquely defines the affine structure. Thus, a positive answer on the very first question would imply that a projective vector field always preserves a affine connection, which would make the investigation of say the singular points of projective vector fields much easier.

Added after the answer and comment of Ben McKay: The comment, and then the answer of Ben McKay does answer two of three questions I pose. The remaining question that I do have hote to get an answer (and, hopefully, a positive one) is whether any vector field with nonzero divergence is homothety vector field for a volume form?

Answer 1

If there is such a volume form $\Omega$, by Moser's theorem we can pick local coordinates in which $\Omega=dx^1 \wedge \dots \wedge dx^n$. If in some coordinates $v$ vanishes to order $k$, for some $k>1$, but not at order $k$, then the same is true in any coordinates. For example, we can suppose that $v=f^2w$ with $f=0$ at some point where $df \ne 0$ and $w \ne 0$ and $w$ not tangent to $f=0$ at that point. This condition is coordinate invariant, and we calculate that $\mathcal{L}_v \Omega=C \Omega$ just when $C=0$ and $2\mathcal{L}_w f + f \, \text{div} w = 0$. But at $f=0$ this forces $\mathcal{L}_w f =0$ tangent, a contradiction. So $v$ does not preserve $\Omega$. But then $v$ does not preserve any volume form.

Answer 2

There may be obstructions, even in the projective case and even when the vector field does not vanish to second order. Suppose given a projective structure on an $n$-manifold $M$ and a projective vector field $X$ that vanishes at a point $p\in M$. Then, in projective normal coordinates (either the Thomas-Veblen version or the Cartan version) $x = (x^i)$ centered on $p$, the vector field $X$ will have the form $$ X = a^i_j\ x^j\ \frac{\partial\ \ }{\partial x^i} + (b_j x^j)\ \left(x^i\ \frac{\partial\ \ }{\partial x^i}\right) $$ for some constants $a^i_j$ and $b_j$. If $\Omega = f\,dx^1\wedge\cdots\wedge dx^n$ is an $n$-form on a neighborhood of $p$, then the condition that ${\mathcal{L}}_X\Omega = C\ \Omega$ is that $f$ satisfy the partial differential equation $$ a^i_j\ x^j\ \frac{\partial f}{\partial x^i} + (b_j x^j)\ \left(x^i\ \frac{\partial f}{\partial x^i}\right) +\bigl( (n{+}1) (b_j x^j) + a^i_i - C\bigr) f = 0. $$ Evaluating this at $x=0$, gives $(a^i_i-C)f(0) = 0$, so if $\Omega$ is to be nonvanishing at $0$, one must have $C = a^i_i$. Supposing this, the equation simplifies to $$ a^i_j\ x^j\ \frac{\partial f}{\partial x^i} + (b_j x^j)\ \left(x^i\ \frac{\partial f}{\partial x^i}\right) +(n{+}1) (b_j x^j) f = 0. $$ Writing $f = f_0 + f_1 + \cdots$, where $f_k$ is the $k$-th homogeneous term in the Taylor series, the above equation now implies the recursive relations $$ a^i_j\ x^j\ \frac{\partial f_k}{\partial x^i} + (n{+}k) (b_j x^j) f_{k-1} = 0 \qquad (k\ge 1). $$ Obviously, this cannot be satisfied for $k=1$ with $f_0\not=0$ unless $b_j = a^i_jc_i$ for some constants $c_i$. (Note, in particular, that a vector field $X$ of the above form that does not satisfy this condition provides a negative answer to Vladimir's remaining question; the divergence at $p$ is not really relevant.) Conversely, if this condition holds, then $$ f = (1 + c_ix^i)^{-(n+1)} $$ satisfies the equation, so the desired $n$-form $\Omega$ does exist, with $C = a^i_i$, at least in an open neighborhood of $p$.

Thus, following on Vladimir's comments above, the answer to the motivating question from projective geometry comes down to whether, when the projective curvature does not vanish at $p$, one always has $b_j=a^i_jc_i$ for some constants $c_i$ when $X$ is a projective vector field vanishing at $p$. (Perhaps Vladimir knows whether this is true. Vladimir?)

Answer 3

For a vectorfield $v$ in a region of $\mathbb R^n$ and a volume element $\Omega=f\,\Lambda$ (where $\Lambda$ is the Lebesgue volume element and $f$ is a strictly positive smooth function), the equation $$\mathcal L_v\Omega=C\,\Omega$$ (with $C\in\mathbb R$ constant) is equivalent to the fact that for each measurable set $A$ we have $$|v^t(A)|_\Omega=\mathrm e^{C\,t}|A|_\Omega,$$ where $v^t$ is the diffeomorphism obtained by following the vectorfield $v$ during a time $t$, and $|A|_\Omega:=\int_A\Omega$ is the measure of $A$ according to $\Omega$. Observe that in this formulation we may allow more general density functions $f$.

Consider near $0\in\mathbb R^2$ the vectorfield $$v(x,y)=y\,(x\,\partial_y-y\,\partial x).$$ This vectorfield flows along the semicircles $$\sqrt{x^2+y^2}=const,\quad y­­>0$$ towards the left, from a geometrically repelling fixedpoint on the positive $x$-axis to a similar attracting fixedpoint on the negative $x$-axis. Each semiannulus $$\epsilon\leq\sqrt{x^2+y^2}\leq 2\epsilon,\quad y\geq 0$$ is preserved, but its content is pressed towards the left (at an exponential rate). We see immediately that the volume element $\Omega$ cannot exist for this $v$, because the right half of the semiannulus is being expanded and the left half is being contracted. But this vectorfield still has zero divergence at the origin.

To get an example with nonzero divergence we go to $\mathbb R^3$ and define $$v(x,y,z)=y\,(x\,\partial_y-y\,\partial x)-z\,\partial_z,$$ which has nonzero divergence in a neighbourhood of the origin. The flow of this vectorfield sends each semi-solidtorus $$S_k=\left\{(x,y,z):\epsilon\leq\sqrt{x^2+y^2}\leq 2\epsilon,\ y\geq 0,\ z\in\left[\frac \epsilon{\mathrm e^{k+1}},\frac \epsilon{\mathrm e^k}\right]\right\}$$ to the following semi-solidtorus $S_{k+1}$ after one unit of time. But the content is again compressed towards the left. So the density function $f$ has singularities on the plane $z=0$.

In more detail, by comparing the Lebesgue measures $|S_k|_\Lambda=c\,e^{-k}$ with the $\Omega$-measure $|S_k|_\Omega=c'\,C^k$ we see that we need to have $C=\frac 1{\mathrm e}$ to hold a hope that the density factor $f$ is bounded above and below by strictly positive constants. But this value is actually not important because we needn't care know how much each $S_k$ measures according to $\Omega$. It's enough to know that the density on the left is much (exponentially in $k$) greater than the density on the right. We will have (exponential) singularities of $f$ along the positive $x$-axis or the negative $x$-axis.

Answer 4

Here's an easy obstruction. Let $v$ be a smooth vector field having two equilibria $p_1,p_2$ with $\text{trace}(D_{p_1}v) \neq \text{trace}(D_{p_2}v)$. Then there is no volume form $\Omega$ making $v$ a homothety, because $\mathcal{L}_{v}\Omega(p_i) = \text{trace}(D_{p_i}v)\Omega_{p_i}$.

So, e.g., the vector field $v(x,y)= \sin(x)\partial_x - 2y \partial_y$ on $\mathbb{R}^2$ has divergence $\leq -1$ everywhere with respect to the Euclidean volume form, but there is no volume form making $L_v$ a homothety since $\text{trace}D v$ is equal to $-1$ at some equilibria and $-3$ at others.

Consideration of more general Lyapunov exponents, e.g. Floquet multipliers of periodic orbits, yields similar obstructions for vector fields without equilibria.

Answer 5

If the vector field V is non-singular (i.e., has no stationary points), one obvious obstruction is that some nonempty open set U of the manifold M (here assumed compact) is carried by one diffeomorphism ø_t (of the flow of the vector field) to a proper subset ø_t(U) of itself.

This implies that the set difference U - ø_t(U) would contain a nonempty open set, so this difference would have positive volume, which implies that Vol(U) = Vol(ø_t(U)) + Vol(U - ø_t(U)), implying that Vol(U) were the sum of itself and a positive number. But if V preserved volume, then Vol(ø_t(U)) would equal Vol(U). So this obstruction excludes the possibility that V preserves a positive volume form.

This may be the only obstruction; I'm not sure.

Answer 6

If a vector field $V$ has a semi stable limit cycle then there is no any volum form $\Omega$ such that $Div_{\Omega} X=C$, a constant.

Explanation:

A semi stable limit cycle is an isolated periodic orbit which is attractor from the interior and repellor from the exterior(Or attractor from exterior and repellor from interior).

Because if divergence is a non zero constant, say positive(negative, resp.) so every closed orbit $\gamma$ is strongly hyperbolic hence is a repellor (attractor, resp.) limit cycle so it attracts a FULL neighborhood of $\gamma$ in negative or positive time. Any such closed orbit can not be a semi stable limit cycle. Recall that a closed orbit $\gamma$ is called a strongly atractive hyperbolic if the divergence of the field is negative at all point of $\gamma$ and is strongly repellor hyperbolic if the divergence is postive at all points of $\gamma$.

On the other hand if $C=0$ then the flow is volume preserving so this contradicts to the fact that a half neighborhood of a closed orbit $\gamma$ tends to $\gamma$ as times goes to $\pm \infty$

Note: Semi stable limit cycles exists via concrte examples:

For example the following system has a semi stable limit cycle $\gamma: x^2+y^2=1$:

$$\begin{cases} x'=y+x(x^2+y^2-1)^2\\y'=-x+y(x^2+y^2-1)^2 \end{cases}$$

but they are very sensitive in the sense that they disappear or they would be twice by very small perturbation. In the foliation language, a semi stable limit cycle of a vector field on a surface is a closed leaf whose holonomy map $h$ satisfies either $h(x)-x>0 \quad\forall x\in (-\epsilon, \epsilon)\setminus \{0\}$ or $h(x)-x<0 \quad \forall x\in (-\epsilon, \epsilon)\setminus \{0\}$ for $\epsilon$ sufficiently small.

For a related post on semi stable limit cycle see this question on quadratic systems