Fix $(X, g)$ to be some compact Riemannian manifold and $V \in \Gamma(TX)$ a smooth, non-vanishing vector field. Suppose the flow is minimal, i.e. every orbit is dense in $X$, and volume-preserving.

Suppose there exists a measure $\sigma$ that is both invariant under the flow of $V$ and absolutely continuous with respect to the Riemannian volume measure. In other words, it is given by integration against $f \text{dvol}_g$ for $f \in L^1(X)$.

If $f$ is not merely $L^1$ but continuous, then our topological assumptions imply immediately that $f$ is constant, so actually any invariant measures are a constant multiple of the volume measure (the minimality precludes the existence of any singular invariant measures).

Can we construct an example such that $f$ is not constant almost everywhere?

  • $\begingroup$ I don't think the minimality precludes the existence of singular invariant measures. You have the celebrated Furstenber's example, in Furstenberg's paper "Strict erogidicity and transformations of the torus", which is a minimal smooth area-preserving diffeomorphism in the 2-torus and is not uniquely ergodic. In that example every ergodic measure is singular respect to Lebesgue. $\endgroup$ – Alejandro Feb 27 at 12:41
  • $\begingroup$ Ah yes, you're right. The singular invariant measures could be supported on a Lebesgue measure zero, yet dense subset. I did read a bit about the Furstenberg example, but it's explicitly noted that the transformation here is not isotopic to the identity - in particular it can't arise from a flow like in my question. $\endgroup$ – Rohil Prasad Feb 27 at 21:20
  • $\begingroup$ Furstenberg's example is indeed isotopic to the identity. However, that is not the point. The suspension flow of any minimal diffeomorphism produces a minimal flow and if the diffeomorphism preserves a smooth volume form (in this case an area form), the suspended flow preserves one as well. So, one can produce such a flow from Furstenberg's example. $\endgroup$ – Alejandro Feb 27 at 22:07
  • $\begingroup$ Ah, you're right. Thanks! $\endgroup$ – Rohil Prasad Feb 28 at 2:49

It seems to me that the answer to your question is "yes", and in fact, Furstenberg's example can be used to construct such an example.

The idea of Furstenberg's construction in the paper Strict Ergodicity and Transformation of the Torus consists in finding an irrational number $\alpha$ and a $\phi\in C^\infty(\mathbb{T}^1,\mathbb{R})$ with $\int\phi(x) d\mathrm{Leb}(x) = 0$ such that there is a measurable $L^1$ function $u\colon \mathbb{T}\to\mathbb{R}$ satisfying $\phi(x) = u(x+\alpha) - u(x)$ for Lebesgue a.e. $x\in\mathbb{T}$, but there is no continuous function $u$ satisfying this property.

Then, the diffeomorphism $g : \mathbb{T}^2 \ni (x,y) \mapsto (x+\alpha,y+\phi(x))$ is minimal, preserves the Lebesgue measure of $\mathbb{T}^2$, but this is not an ergodic measure. In fact, the map $h : (x,y)\mapsto (x,y+u(x))$ leaves invariant the Lebesgue measure of $\mathbb{T}^2$ and $g\circ h = h \circ k$, where $k : (x,y)\mapsto (x+\alpha, y)$; and Lebesgue is clearly not ergodic for $k$.

However, there are infinitely many $k$-invariant measures which are absolutely continuous with respect to Lebesgue. To see this, one can consider for instance a non trivial interval $I\subset\mathbb{T}$ (i.e. $I$ and its complement contain more than one point), and define the measure $\mu:= \mathrm{Leb} \otimes \frac{1}{\mathrm{Leb}(I)}\mathrm{Leb}\big|_I$. Then, since $h$ preserves Lebesgue, we conclude that $h_*\mu$ is $g$ invariant, is absolutely continuous with respect to Lebesgue and is different from Lebesgue.

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