The vector field of a given flow - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T05:58:31Z http://mathoverflow.net/feeds/question/39907 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/39907/the-vector-field-of-a-given-flow The vector field of a given flow Denis Serre 2010-09-24T21:30:37Z 2010-09-26T08:02:24Z <p>Let $f:(0,1)\rightarrow(0,1)$ be a map with some regularity (${\mathcal C}^1$, ${\mathcal C}^2$, ${\mathcal C}^\infty$, analytic ?). We assume that $f(t)> t$ for every $t$, and that $f'> 0$.</p> <p>Does there exist a vector field $X$ over $(0,1)$, whose flow at time $t=1$ is $f$ ?</p> <p>If the answer is yes (as I bet), it will complete the existence proof in MO question <a href="http://mathoverflow.net/questions/34187/nth-root-of-a-b-mapsto-gm-am/39906#39906" rel="nofollow">link text</a></p> http://mathoverflow.net/questions/39907/the-vector-field-of-a-given-flow/39915#39915 Answer by Pietro Majer for The vector field of a given flow Pietro Majer 2010-09-24T23:31:40Z 2010-09-26T08:02:24Z <p>I assume your map is surjective, thus an increasing $C^k$ diffeo $f:(0,1)\to(0,1)$ (say $1\le k\le\infty).$ The latter, as a discrete dynamical system, turns out to be $C^k$ conjugated with the shift by translation $t\mapsto t+1$ on $\mathbb{R}$. Now if $h:(0,1)\to\mathbb{R}$ is such an (increasing) conjugation, define a field $X$ by taking for all $t\in (0,1)$ $$X(t)=\frac{1}{h^'(t)}.$$ Then $f$ is the flow of $X$ at time $1$, just because $u:=h^{-1}:\mathbb{R}\to(0,1)$ solves the autonomous ODE $u'=X(u).$ So the solution at time $1$ corresponding to the initial value $t$ at time $0$ is actually $u\left(h(t)+1\right)=f(t).$</p> <p><em>Rmk</em>. The general fact behind is: for a diffeo on a manifold, being a time-one map of a flow, is a property invariant by smooth conjugation -the flow transforms by conjugation, and its generator is the pull back of $X$. And the shift $x\mapsto x+1$ is, of course, the map at time 1 of a flow.</p> <p><em>Construction of the conjugation</em>. Note that here a conjugation is an increasing diffeo $h:(0,1)\to\mathbb{R}$ solving the linear functional equation $h\left(f(x)\right) = h(x)+1.$ A solution of class $C^k$ can be constructed fixing any smooth diffeo $h_0:\left[\frac{1}{2},f(\frac{1}{2})\right]\to[0,1]$ with the convenient $k$-jets at the end-points of its domain, and extending it (uniquely) to a diffeo $h:(0,1)\to\mathbb{R}$ by means of the functional equation -the condition is that the $k$-jet of $h_0\left(f(x)\right)$ at $x=\frac{1}{2}$ and the $k$-jet of $h_0(x)+1$ at $x=f(\frac{1}{2})$ should coincide, in order that the glueing be $C^k$. </p> <p>A maybe more clear way of achieving that, is: first fix any $C^k$-germ $H$ of local diffeo with $H(1/2)=0$ and $H'(0)>0$. Then take $h_0$ as a smooth diffeo from a nbd of $J:=\left[\frac{1}{2},f(\frac{1}{2})\right]$ to a nbd of $[0,1]$, whose germs at $\frac{1}{2}$, and at $f(\frac{1}{2})$, are respectively $H$ and $H\circ f^{-1} + 1$. The extension of $h_0$ from the two germs is of course immediate by using some cut-off function. Then we may proceed as said before, from ${h_0}_ { |J}$.</p> <p>(Clearly, these naive constructions immediately fall, in the case of $C^\omega$ conjugation of $C^\omega$ diffeo's on open intervals with no fixed points.) </p> http://mathoverflow.net/questions/39907/the-vector-field-of-a-given-flow/39919#39919 Answer by Bill Thurston for The vector field of a given flow Bill Thurston 2010-09-25T00:22:12Z 2010-09-25T04:51:38Z <p>I'll rephrase Pietro Majer's answer, and also give some more information.</p> <p>For any $f: (0,1) \to (0,1)$ with $f(t) > t$, the quotient space $(0,1) / \left &lt; f \right >$ is diffeomorphic to the circle. Choose any diffeomorphism to $S^1 =\mathbb R /\mathbb Z$, and use the vector field $d/dt$ lifted back to $(0,1)$.</p> <p>This construction is unlikely to be smooth at either endpoint, unless special care is taken. In many cases, there are established techniques to embed the germ of a diffeomorphism in a flow near a fixed point. The most widely used case in dynamics is that if a diffeomorphisms of $\mathbb R^n$ (say, $C^\infty$ or $C^\omega$) has a fixed point $p$ at which the first derivative has no zeros on the unit circle, then it is $C^\infty$ or $C^\omega$ conjugate to a linear diffeomorphism near that point. The conjugacy is unique up to a linear map. There are many other cases where the structure is known. I don't have a reference handy, but I think any text on dynamical systems will have this, and the proof is reasonably elementary. In the analytic case, you can solve recursively for the Taylor series coefficients, and check local convergence. </p> <p>In the 1-dimensional case at hand, if $f'(0) > 1$, then a microscopic picture of $f$ near $0$ looks very close to the linear map $x \to f'(0) x$. We can conjugate a well-fitting microscopic coordinate system by a high power of $f$ to make a fixed-size coordinate in which $f$ is nearly linear: the formula for the coordinate is $u_n(x) = (f'(0))^n f^{-n}(x)$. These converge in the $C^\infty$ topology to a coordinate that $f$ maps linearly. Another way to think of it is that to calculate the ratios of intervals in the new measure of length, send them both near $0$ by a high power of $f^{-1}$, and measure ratios there. It is not hard to show that the limit $u(x))$ is smooth --- it's the limit of long compositions, but most of the composing takes place near 0 where $f$ is nearly linear.</p> <p>Conjugating an orientation preserving map to a linear map gives an embedding in a flow near the fixed point, since orientation preserving linear maps are embedded in linear flows ( i.e, in the image of the the exponential map in $GL(n,\mathbb R)$).</p> <p>For a diffeomorphism of the closed interval <code>$[0,1]$</code>, if the derivative is not 1 at either endpoint, then their is a unique smooth conjugacy to a linear map near each endpoint (since the exponential map for $GL(1,\mathbb R)$ is univalent.) An embedding in a flow near one endpoint extends uniquely over the entire open interval: for any point, apply $f^{-1}$ enough times to get it to inside the last fundamental domain where the flow is defined.</p> <p>The definitions from the two endpoints typically conflict. Generic diffeomorphisms of a <strong>closed</strong> interval are <strong>not</strong> embeddable in a smooth flow. A diffeomorphism of an interval has a smooth invariant that measures the discrepancy between these: a pair of smooth probability measures on $S^1$ that coincide if and only if the overall diffeomorphism is embeddable in a flow. In other words: the exponential map is not locally surjective in the group of diffeomorphisms of an interval.</p> <p>Even when a diffeomorphism of a closed interval is not embeddable in a flow, it might still have a functional square root, a cube root, or whatever. A similar analysis applies --- near a hyperbolic fixed point, it is uniquely conjugate to a linear map, and it has a unique square root. This determines a square root in the adjoining open intervals. You can compare definitions from different fixed points and see if they match.</p> <p>There is a lot more known about algebraic relationships that can and cannot occur in groups of diffeomorphisms of the interval, and in higher dimensional analogues as well. (The question concerns the simplest case, the groups $\mathbb Z$ and $\mathbb R$). This theory has had considerable attention motivated by trying to understand the structure of smooth foliations. For example, any finitely generated group of diffeomorphisms of a closed interval must be residually torsion-free nilpotent (the generalized Reeb stability theorem). From this it is possible to deduce that there are many groups of homeomorphisms of the interval that are not topologically equivalent to any smooth action.</p>