Is there an underlying explanation for the magical powers of the Schwarzian derivative? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T21:28:22Z http://mathoverflow.net/feeds/question/38105 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/38105/is-there-an-underlying-explanation-for-the-magical-powers-of-the-schwarzian-deriv Is there an underlying explanation for the magical powers of the Schwarzian derivative? Paul Siegel 2010-09-08T22:12:45Z 2012-12-09T20:34:02Z <p>Given a function $f(z)$ on the complex plane, define the Schwarzian derivative $S(f)$ to be the function</p> <p>$S(f) = \frac{f'''}{f'} - \frac{3}{2} (\frac{f''}{f'})^2$</p> <p>Here is a somewhat more conceptual definition, which justifies the terminology. Define $[f, z, \epsilon]$ to be the cross ratio $[f(z), f(z + \epsilon); f(z + 2\epsilon), f(z + 3\epsilon)]$, and let $[z, \epsilon]$ denote the cross ratio $[z, z + \epsilon, z + 2\epsilon, z + 3\epsilon]$ (in fact this is just 4, but this notation makes my point clearer). One can ask if $[f, z, \epsilon]$ is well approximated by $[z, \epsilon]$; indeed, it turns out that the error is $o(\epsilon)$. So one pursues the second order error term and finds that $[f, z, \epsilon] = [z, \epsilon] - 2 S(f)(z) \epsilon^2 + o(\epsilon^2)$. So the Schwarzian derivative measures the infinitessimal change in cross ratio caused by $f$. In particular, $S(f)$ is identically zero precisely for Mobius transformations. </p> <p>That's all background. From what I have said so far, the Schwarzian derivative is at best a curiosity. What is not obvious at first glance is the fact that the Schwarzian derivative has magical powers. Here are some examples:</p> <p>First magical power: The Schwarzian derivative is deeply relevant to one dimensional dynamics, stemming from the fact that it behaves in a specific way under compositions. For example, if $f$ is a smooth function from the unit interval to itself with negative Schwarzian derivative and $n$ critical points, then it has at most n+2 attracting periodic orbits.</p> <p>Second magical power: It says something profound about the solutions to the Sturm-Liouville equation, $f''(z) + u(z) f(z) = 0$. If $f_1$ and $f_2$ are two linearly independent solutions, then the ratio $g(z) = f_1(z)/f_2(z)$ satisfies $S(g) = 2u$.</p> <p>Third magical power: The Schwarzian derivative is the unique projectively invariant 1-cocyle for the diffeomorphism group of $\mathbb{R}P^1$. This is probably just a restatement of the conceptual definition I gave above, but I'm not sure; in any event, this gives the Schwarzian derivative a great deal of relevance to conformal field theory (or so I'm told). </p> <p>I'm sure there are more. I'm wondering if all of these powers can be explained by some underlying geometric principle. They all seem vaguely relevant to each other, but the first power in particular seems very hard to relate to the definition in any obvious way. Does anybody have any insights?</p> http://mathoverflow.net/questions/38105/is-there-an-underlying-explanation-for-the-magical-powers-of-the-schwarzian-deriv/38109#38109 Answer by Victor Protsak for Is there an underlying explanation for the magical powers of the Schwarzian derivative? Victor Protsak 2010-09-08T23:03:54Z 2010-09-09T02:07:48Z <p>The following book may contain insights of the type you are looking for, at least for the second and third interpretation:</p> <blockquote> <p>V. Ovsienko, S. Tabachnikov, <em>Projective differential geometry old and new. From the Schwarzian derivative to the cohomology of diffeomorphism groups.</em> Cambridge Tracts in Mathematics, 165. Cambridge University Press, Cambridge, 2005. ISBN: 0-521-83186-5 <a href="http://www.ams.org/mathscinet-getitem?mr=2177471" rel="nofollow">MR</a></p> </blockquote> <p>Although I didn't read the book, earlier articles by the same authors, including a popular exposition in the now defunct "Quant" magazine, were amazingly insightful. Here is how the authors begin their story:</p> <blockquote> <p><em>Every working mathematician has encountered the Schwarzian derivative at some point of his education and, most likely, tried to forget this rather scary expression right away. One of the goals of this book is to convince the reader that the Schwarzian derivative is neither complicated nor exotic, in fact, this is a beautiful and natural geometrical object.</em></p> </blockquote> <hr> <p>I'd like to complement the "third magical power" with the statement that the Schwarzian derivative is a cocycle of the diffeomorphism group of the circle $S^1$ with values in the quadratic differentials on $S^1$, and since the latter space can be identified with the dual space of the vector fields on $S^1,$ it encodes a central extension of $\text{Diff}(S^1);$ the infinitesemial version of this central extension is the Virasoro algebra.</p> http://mathoverflow.net/questions/38105/is-there-an-underlying-explanation-for-the-magical-powers-of-the-schwarzian-deriv/38125#38125 Answer by Bill Thurston for Is there an underlying explanation for the magical powers of the Schwarzian derivative? Bill Thurston 2010-09-09T01:26:35Z 2010-09-12T01:53:42Z <p>Like may people (but not all people), I have trouble thinking in terms of formulas such as that for the Schwarzian. For me, a geometric image works much better. I'll describe a geometric picture, similar to what I discussed in my paper "Zippers and univalent functions" (and can be found elsewhere as well, but I don't have a good sense of references).</p> <p>The group of Moebius transformations is 3 dimensional, and for any locally-defined diffeomorphism f of R or any locally-defined holomorphic map of C, you can fit the value, the first derivative and the 2nd derivative at any point by a unique Moebius transformation, the osculating Moebius transformation at the point. From this, you can make a recipe to extend the diffeomorphism into the upper half plane or upper half space models for hyperbolic geometry: map each vertical line according to the osculating Moebius transformation at its base. When you do this, vertical lines are mapped isometrically, but the metric is necessarily distorted in the horizontal directions unless f is a Moebius transformation. The Schwarzian derivative gives the asymptotic behavior of this distortion. For real maps, if the Schwarzian is negative, the hyperbolic rays are bent away from each other. A hyperbolic line perpendicular to the vertical lines is mapped to a curve that (in terms of the hyperbolic metric) bends downward. If you consider any interval on R, it has a natural projectively-invariant metric, called the Hilbert metric, and identified with the 1-dimensional hyperbolic metric in this case. The bending implies that the metric is expanded by f, relative to the Hilbert metric of its image. For example, $\log(t)$ is an arc-length parametrization for $(0,\infty)$. The map $x \rightarrow x^k$ expands the parameter by a factor of $k$. The expanding property is highly significant for analyzing dynamics.</p> <p>In the complex analytic case, the Schwarzian is a holomorphic quadratic differential, and can be geometrically indicated by two perpendicular foliations: one set of streamlines where the quadratic form takes positive real values, and a perpendicular set of lines where it takes negative real values. Whenever the quadratic form has a simple zero, the foliations have singularities where the lines make a Y pattern, branching in 3 ways. At any critical point of f, there is a double pole of the Schwarzian, where the positive real foliation circles around and the negative real foliation is radial. These lines show how the extension of f bends surfaces asymptotically near the complex plane; if you start with an umbilic surface such as a plane, horosphere or equidistant surface, they show the asymptotic pattern for lines of curvature for the image of the surfaces via the extension of f. For example, you can visualize z -> z^2 as extending by mapping the hyperbolic cylinders around the z axis (they appear as cones in upper half space), wrapping them twice around the vertical axis (thus stretching the circumference by a factor of 2) , and stretching vertically by a factor of 2. The curvature along the meridians is increased, and the curvature in directions parallel to the axis is decreased. Of course, the Poincare metric of a small disk is preserved, but you can still see the metric effect in terms of the nonconformality of the behavior on surfaces in hyperbolic space. You can also see it by the shape of the image of a small circle on C. To 2nd order, it remains round, but there is a 3rd order effect that makes it elliptical, where the short axis is the direction in which the Schwarzian is negative. </p> <p>When you look at computer plots of the quadratic differentials for holomorphic maps, they pop into 3 dimensions, strongly suggesting the geometry of some families of surfaces that can be associated to a holomorphic map of C.</p> <p><img src="http://dl.dropbox.com/u/5390048/3prongSchwarzian.jpg" alt="alt text"></p> <p>The Schwarzian for the rational function $f(z)(= (z^3-3 z -1)/(z^3+1)$. It's challenging to make a revealing plot for a complex rational function like $f$, since it maps 3 times over the Riemann sphere, but the Schwarzian is easy to draw, and shows how an extension of f bends hyperbolic space. The critical points of f(z) are surrounded by circular positive-real circles where the Schwarzian has a double pole, indicating how they are wrapped twice around a core singularity. A typical zero is visible in the center, where the bending branches three ways. The zeros and poles of the function f itself are not visible, since $0$ and $\infty$ have no special significance in the geometry of $S^2 = CP^1$.</p> <p>Enough ... there are endless mysteries to the Schwarzian, but these geometric images are helpful to me. I'll second Victor's suggestion: I haven't read the Osvienko- Tabachnikov book either, but I'm confident from prior experience that it's interesting material.</p> <p><strong>Addendum.</strong> Since you expressed interest in the real case, here's one way to illustrate it. The idea is simple: take a standard family of horocycles in the domain, in this case circle of constant height tangent to the real line, and push them forward by the osculating Moebius transformation. The image is actually determined by just $f$ and $f'$, but to calculate (rather than just see) the envelope would require $f''$. This picture is for $x^3 - 3 x$ (which folds $[-2,2]$ 3 times over its image), in the interval $[-2.1,2.1]$ The fat shape with downard hyperbolic curvature of the envelope in contrast to upward curvature of the envelope in the domain, suggests a caterpillar outgrowing its skin and demonstrates the negative Schwarzian. <img src="http://dl.dropbox.com/u/5390048/HoroMetricCircles.jpg" alt="alt text"></p> http://mathoverflow.net/questions/38105/is-there-an-underlying-explanation-for-the-magical-powers-of-the-schwarzian-deriv/38133#38133 Answer by unknown (google) for Is there an underlying explanation for the magical powers of the Schwarzian derivative? unknown (google) 2010-09-09T02:15:02Z 2010-09-09T03:23:15Z <p>It's a cocycle in group cohomology.</p> <p>$S[f\circ g]= S[f]^g + S[g]$</p> <p>which is the cocycle condition! The group is Aut(meromorphic functions) and the other group is meromorphic functions.</p> http://mathoverflow.net/questions/38105/is-there-an-underlying-explanation-for-the-magical-powers-of-the-schwarzian-deriv/38249#38249 Answer by VA for Is there an underlying explanation for the magical powers of the Schwarzian derivative? VA 2010-09-09T23:53:02Z 2010-09-09T23:53:02Z <p>Nobody pointed out <a href="http://www.ams.org/notices/200901/tx090100034p.pdf" rel="nofollow">What is ... Schwarzian Derivative</a>? (Notices of AMS Jan 2009), which succinctly explains quite a lot.</p> http://mathoverflow.net/questions/38105/is-there-an-underlying-explanation-for-the-magical-powers-of-the-schwarzian-deriv/49504#49504 Answer by Engineers are better at maths for Is there an underlying explanation for the magical powers of the Schwarzian derivative? Engineers are better at maths 2010-12-15T09:53:37Z 2010-12-15T09:53:37Z <p>Let $f:\mathbf{R}\mapsto\mathbf{P}^1$ be a function realised as $y=f(x)$, where $y$ is an affine coordinate on $\mathbf{P}^1$. Since the target space is the projective line it is desirable to lift $f$ to homogeneous coordinates, i.e. to choose $\mathbf{v}=(u(x), v(x))$ such that $u/v=f$. (Maybe it is not quite right to call this homogenous coordinates, rather this is a lifting of $f$ to the line bundle over $\mathbf{P}^1$, i.e. converting $f$ to a parameterised curve in $\mathbf{R}^2$.) Differentiating the constraint gives $$\frac{u'v-uv'}{v^2} = f',$$ which suggests taking $u'v-uv'=1$ as an additional constraint (assuming $f'>0$). This fixes $u$ and $v$ to be $$u=\frac{f}{\sqrt{f'}}$$ $$v=\frac{1}{\sqrt{f'}}$$</p> <p>The added constraint says that the angular momentum of the lifted curve is constant, $\mathbf{v}'\times\mathbf{v}=1$ and this means that the acceleration vector is parallel to the position vector, $\mathbf{v''}=\lambda\mathbf{v}$. The proportionality factor $\lambda = \lambda(x)$ is $$\lambda = \frac{u''}{u}=\frac{v''}{v}=-\frac{1}{2}\frac{f'''f'-\frac{3}{2}(f'')^2}{(f')^2},$$ which is the Schwarzian with an extra factor of $-1/2$.</p> <p>The proportionality factor $\lambda$ is invariant under linear transformations of $\mathbf{R}^2$. Since a projective coordinate change on the target space lifts to such a linear transformation, the Schwarzian is independent of the choice of the affine coordinate $y$ and is a projective invariant.</p> http://mathoverflow.net/questions/38105/is-there-an-underlying-explanation-for-the-magical-powers-of-the-schwarzian-deriv/115919#115919 Answer by Peter Michor for Is there an underlying explanation for the magical powers of the Schwarzian derivative? Peter Michor 2012-12-09T20:34:02Z 2012-12-09T20:34:02Z <p>The Schwarzian derivative encodes the adjoint action of the Bott-Virasoro group. One version of it is ${Diff}_{\mathcal S}(\mathbb R)\times \mathbb R$ (here $\mathcal S$ stands for "rapidly falling towards the identity") with multiplication $$\binom{\phi}{\alpha}.\binom{\psi}{\beta} = \binom{\phi\circ\psi}{\alpha+\beta+c(\phi,\psi)}$$ where the Bott cocycle is: $$c(\phi,\psi) = \frac12\int_{\mathbb R} \log(\phi'\circ \psi)\,d\log(\psi').$$ This is alluded to on page 22ff of the book of Ovsienko and Tabashnikov mentioned in Viktor Protsaks answer. A short exposition along the lines of this answer is on page 55 of <a href="http://www.mat.univie.ac.at/~michor/geom-evolution.pdf" rel="nofollow">(here).</a> I think that this encodes much of the magic of the Schwartzian derivative. Dualize it to the coadjoint action and note that many coadjoint orbits are of the form $Diff_{\mathcal S}(\mathbb R)/PSL(2,\mathbb R)$ to see the projective properties. Etc.</p>