Is volume--preserving an intrinsic property? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T07:48:17Z http://mathoverflow.net/feeds/question/54594 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/54594/is-volume-preserving-an-intrinsic-property Is volume--preserving an intrinsic property? Pengfei 2011-02-07T02:10:52Z 2011-02-07T18:09:50Z <p>Let $M$ be a compact smooth manifold without boundary. A Riemannian metric $g$ on $M$ induces a volume measure (or Lebesgue measure) $m_g$ on $M$.</p> <p>A diffeomorphism $f:M\to M$ is said to be {volume--preserving} if $f_*(m_g)=m_g$, that is, for each Borel subset $A\subset M$, $f_*(m_g)(A):=m_g(f^{-1}A)=m_g(A)$. Or equivalently the Jacobian (determinant of the tangent map $Df_x:T_xM\to T_{fx}M$) satisfies $J(f,m_g)(x)=1$ for every point $x\in M$.</p> <p>If we change the Riemannian metric to $g'$ and the induced measure $m_{g'}$, the volume--preserving property with respect to $g$ is slightly distorted: there exists a uniform constant $C\ge1$ such that</p> <p>$C^{-1}\le J(f^n,m_{g'})(x)\le C$, for every point $x\in M$ and every time $n\in\mathbb{Z}$.---- $(*)$</p> <p>My question is: is $(*)$ a characterization of volume--preserving? </p> <p>That is, for a given $f\in\mathrm{Diff}(M)$, if $(*)$ holds for some arbitrarily chosen Riemannian metric $g$, does there exist a Riemannian metric $g'$ such that $J(f,m_g)=1$?</p> <p>Thanks!</p> http://mathoverflow.net/questions/54594/is-volume-preserving-an-intrinsic-property/54605#54605 Answer by Bill Thurston for Is volume--preserving an intrinsic property? Bill Thurston 2011-02-07T04:47:19Z 2011-02-07T15:21:45Z <p>If the volume form for $M^n$ is $\Omega$, note that the average $\Omega_k$ of $f^{k*}(\Omega)$ for $0 \le k \le n$ is a volume form that has much tighter bounds for invariance by $f$ than $\Omega$, since all terms except the first and the last are equal. You can pass to the limit of the $\Omega_k$ as a measure, and obtain an invariant measure that has the form $g * \Omega$, where $|\log(g)|$ is bounded. </p> <p>Thus, the only issue is smoothness, but this is a real issue. Here's how to make a counterexample. I'll describe it as a flow on $S^2$; to get a diffeomorphism, take the time one map of the flow. Begin with the flow lines of a rotation of the sphere about the north pole-south pole axis. Now distort the flow lines near a point on the equator by a non-smooth homeomorphism that squeezes the northern hemisphere toward the point with equator, compressing say by a factor of two near that point, but is the identity on the southern hemisphere. The rotation flow conjugated by the homeomorphism is not smooth, but multiply the image of the rotation vector field by a function that depends only on the flow line and has $C^\infty$ contact to 0 at the equator. The resulting vector field $X$ is smooth. Invariant measures for the flow are determined by the spacing between flow lines, so they are equivalent to measures on $S^2$ modulo flow lines: for a smooth invariant 2-form $\Omega'$, the identity is that the 1-form $i_X(\Omega') = 0$ would be closed. This 1-form is 0 in the direction of $X$, that is, it would have the form $d(h)$, for some function $h$ that is constant on flow lines. But for any such measure, the ratio to the standard area form will have a discontinuity somewhere along the equator.</p> <p>This particular example is topologically conjugate to a volume preserving flow for a smooth Riemannian metric, but I think a little more work would give examples that are not even topologically conjugate.</p> http://mathoverflow.net/questions/54594/is-volume-preserving-an-intrinsic-property/54674#54674 Answer by rpotrie for Is volume--preserving an intrinsic property? rpotrie 2011-02-07T18:09:50Z 2011-02-07T18:09:50Z <p>Just to add a positive result. For circle diffeomorphisms ($C^1$) one has that if the rotation number is irrational, a necessary and sufficient condition to be linearizable (by a $C^1$ map, which in particular implies you will preserve a $C^1$-length) is to have bounded derivatives (that $log |(f^n)'|$ is uniformly bounded). So, in this context you have your answer. This is a consequence of Gottshalk-Hedlund theorem, see for example the book by deMelo-VanStrien. </p>