Is the Schwarzian Derivative a connection?

Question

The Schwarzian derivative is the quadratic differential

$$ S(f) = \Bigg( \frac{f'''}{f'} - \frac{3}{2} \Big( \frac{f''}{f'} \Big)^2 \Bigg) (dz)^2 .$$

Bill Thurston, in his Math overflow answer here and his paper "Zippers and Univalent Functions" presents one interpretation of the Schwarzian derivative as being the "rate of change of the best approximating Mobius transformation of $f$."

More rigorously, define $$ M(f, w)(z)$$ to be the Mobius transformation $$ M(f, w)(z) = \frac{az + b}{cz + d} $$ that has the same value, first derivative, and second derivative of $f(z)$ at $z = w$. There are four unknowns, $a,b,c,d$ and the four equations $$ ad - bc = 1 $$ $$ M(f,w)(w) = f(w) $$ $$ (\partial_z M(f,w) )(w) = f'(w) $$ $$ (\partial_z^2 M(f,w) )(w) = f''(w) $$ which allows us to solve for $a,b,c,d$ explicitly (but the expressions are not enlightening so I will not write them here).

Thurston's "rate of change" of the best approximating Mobius transformation is defined to be

$$ \frac{d}{dt} M(f, w)^{-1} \circ M(f, w + t) \Big\rvert_{t = 0}. $$ Note that because, for any Mobius transformation $A$, $$ M(A \circ f, w) = A \circ M(f, w). $$ This implies that if we send $f \mapsto A \circ f$, our rate of change becomes $$ \frac{d}{dt} M(A \circ f, w)^{-1} \circ M(A \circ f, w + t) \Big\rvert_{t = 0} $$ $$= \frac{d}{dt} M(f, w)^{-1} \circ A^{-1} \circ A\circ M(f, w + t) \Big\rvert_{t = 0}. $$ and the overall expression doesn't change.

In fact, it can be explicitly shown that this is actually just the Schwarzian derivative:

$$ \frac{d}{dt} M(f, w)^{-1} \circ M(f, w + t) \Big\rvert_{t = 0} = \frac{1}{2} (z - w)^2 \Bigg( \frac{f'''}{f'} - \frac{3}{2} \Big( \frac{f''}{f'} \Big)^2 \Bigg). $$

Here is my question: the schematic expression $\partial_t M^{-1} M(t)$ looks an awful lot to me like the expression for a connection. Can the Schwarzian derivative be thought of, in any way, as a connection? What is the base space and the Principal bundle? Also, connections are usually thought of as "Lie algebra valued one forms," but the Schwarzian has a $(dz)^2$ instead of a $dz$. What is the relevance of that?

Answer 1

The Schwarzian derivative shows up in the setting of projective structures on $\mathbf{R}$. Let $f:\mathbf{R}\rightarrow\mathbf{R}$ and think of the value of $f$ as an affine coordinate on $\mathbf{RP}^1$ so that $f$ is a curve in $\mathbf{RP}^1$. This curve can be lifted to a areavelocity("angular momentum")-preserving curve $\phi(t)$ in $\mathbf{R}^2$ (homogenous coordinates on $\mathbf{RP}^1$) which satisfies the Sturm-Liouville-equation $$ \phi''+u(t)\phi=0 $$ where $$ u(t)=\frac{1}{2}S(f) $$ with Schwarzian derivative $$ S(f)=\frac{f'''}{f'}-\frac{3}{2}\left(\frac{f''}{f'}\right)^2 $$ A projective structure is equivalent to a projective connection so this can also be stated as bundle over the base $\mathbf{R}$ with fiber $\mathbf{RP}^1$ and an $\mathrm{sl}(2)$-valued connection one-form. A projective connection has an associated developing map onto $\mathbf{RP}^1$ and this map is the given $f:\mathbf{R}\rightarrow\mathbf{RP}^1$.