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

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?

• On a compact oriented manifold the integral of $\mathcal{L}_X \Omega$ is zero, since it is a difference (infinitesimally) of homologous forms. So one obstruction. But locally, I don't know. – Ben McKay Jan 4 '14 at 17:34
• Try $v(x)=x^2\partial_x$. If $\Omega=f(x) \, dx$ is $C^1$ then you get $\mathcal{L}_v \Omega=C\Omega$ just when $(C-2x)/x^2=f'/f$ so must have $f=0$ at $x=0$, i.e. not a nonvanishing volume form. – Ben McKay Jan 4 '14 at 17:41
• Just a word about the motivation from projective geometry: Even for projective vector fields this won't work. The vector field $$X = x^2\ \frac{\partial\ \ }{\partial x} + xy\ \frac{\partial\ \ }{\partial y}$$ is projective for the standard projective structure on $\mathbb{R}^2$, but it doesn't preserve a non-vanishing area form up to scale on any neighborhood of $(x,y)=(0,0)$. – Robert Bryant Jan 4 '14 at 19:56
• Thanks, Robert. Actually I new that flat projective structures admit vector fields that are not affine (for any connection). In your example, the vector field vanish with its derivatives at the point (0,0). If (in dim >2) the projective curvature tensor does not vanish at a point, the values of a projective vector field and the first derivatives at the point determine the projective vector field on the whole manifold. All examples I know (most of them come from metric projective structure) indicate that there always exists an affine connection such that the vector field preserves it. – Vladimir S Matveev Jan 4 '14 at 22:23
• Sorry for not commenting on your comment earlier, Ali Taghavi. I do not know the answer and actually do not consider the question as a natural one: taking the star hodge requires that we have an additional structure on the manifold (say, metric in the canonical situation, or actually if you do the hodge star with the volume form your need only the volume form). In the case this additional structure is invariant with respect to our vector field, we are done, since we have an invariant volume form. If it is not invariant, it does not really help (or I do not see how it could help). – Vladimir S Matveev Jan 14 '14 at 19:28

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.

• Thank you, Ben. You answer answers two of three questions; I leave the question unanswered for a while with a hope that somebody (may be you or Robert) answers the remaining question. – Vladimir S Matveev Jan 4 '14 at 22:26
• I think if you look at the Poincare-Dulac normal form for a strictly contracting vector field, you see that it preserves the standard volume form up to a constant multiple. So this suggests that every vector field with nonzero divergence at its zeroes preserves a volume form up to a constant multiple. – Ben McKay Jan 5 '14 at 10:26

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?)

• Robert, I do not know whether answer on your question is true. I thought that I found an example when it is wrong (which of course will imply nonlinearisability of projective vector fields) but found a possible flaw there. I will continue to think about it tomorow – Vladimir S Matveev Jan 8 '14 at 22:47
• I am giving up for a while: I can not prove that any projective vector field is locally linearizable near points where the projective curvature does not vanish. All examples indicate that this is the case though: at the present point I can show this statement for projective structures coming from any riemannian metric, and also for the projective structure with submaximal dimension of the space of projective vector fields (the projective structure was found by Egorov and the description of their projective vector fields can be found in the recent paper of Boris Kruglikov and Dennis The. – Vladimir S Matveev Jan 14 '14 at 19:23

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.

• Thanks, Marcos. The geometry of (or the idea behind) your examples is essentially the same as in the example of @Ben McKay – Vladimir S Matveev Oct 14 '14 at 10:41

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 $$\leq -2$$ 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 counterexamples for vector fields without equilibria.)