# Absolutely continuous invariant measures for a minimal flow

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?

• 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. – Alejandro Feb 27 at 12:41
• 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. – Rohil Prasad Feb 27 at 21:20
• 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. – Alejandro Feb 27 at 22:07
• Ah, you're right. Thanks! – Rohil Prasad Feb 28 at 2:49

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.