# Effective vanishing of the Schwarzian Derivative

Recall for any complex analytic function $f:\mathbb{D}\to \mathbb{C}$ the Schwarzian derivative of $f$ is $$S(f)=\frac{f'''}{f'}-\frac{3}{2} \left( \frac{f''}{f'}\right)^2.$$

It's well known that $S(f)\equiv 0$ if and only if $f$ is a fractional linear transformation, i.e. $$f(z)=\frac{a z+b}{cz+d}.$$

I was wondering if there was an effective version of this result. In other words a result that said that if $S(f)$ was "small" in some sense then $f$ was "close" in some, possible different, sense to a fractional linear transformation.

Of course a major part of the question is deciding what are appropriate notions of "small" and "close". I'd be particularly interested in those that allow $S(f)$ to have poles -- i.e. weaker than $L^\infty$ bounds.

I'm (vaguely) aware of Nehari's work on univalent functions (e.g. if $|S(f)|\leq 2$ then $f$ is univalent) though am not sure how related this is to my question.

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The Schwarzian derivative of $f$ measures the infinitesimal change in cross-ratio caused by $f$ up to second order, so if $S(f)$ is small then $f$ "almost preserves the cross-ratio". Since the fractional linear transformations are precisely the functions which preserve the cross-ratio on the nose, perhaps this is the sort of statement that you're looking for. – Paul Siegel Dec 27 '12 at 22:00
@Paul Siegel I'm more interested in quantitative (rather than qualitative) results. Ideally, something along the lines of there is a constant $C>0$ so that $$||f-\gamma||_X\leq ||S(f)||_Y$$ where $\gamma$ is a fractional linear transformation and $X$ and $Y$ are some natural norms. – Rbega Dec 27 '12 at 22:06
One issue is that for instance $f(z)=z+\epsilon z^2$ has $$S(f)=−6\epsilon^2 \frac{1}{(1+2\epsilon z)^2}$$ Clearly when $\epsilon$ is small then $f$ is "near" the FLT $\gamma(z)=z$. Moreover, $S(f)$ is "small" in some sense but also has double poles (so is not small in, say, an $L^1$ sense let alone a uniform sense). – Rbega Dec 27 '12 at 22:16
@Rbega: I assume that $\mathbb{D}$ means the Poincaré disk. Actually, you will have an even worse problem in that you don't have compactness in any sense. For example, $f(z) = \epsilon z$ will have $S(f)\equiv0$, but, as $\epsilon$ goes to zero, $f$ converges to something that is not a linear fractional transformation (and for which $S(f)$ is not even defined). There is a way to 'break' the $\mathrm{SL}(2,\mathbb{C})$ symmetries that cause noncompactness and get a quantitative statement of the kind that you want, but you may not regard the resulting norms as 'natural'. – Robert Bryant Dec 27 '12 at 23:45
@Robert Bryant You are correct that $\matbb{D}$ is the Poincaré disk and you have a good point. I'm not adverse to breaking the $SL(2,\mathbb{C})$ symmetry -- I'm more curious what is out there. – Rbega Dec 28 '12 at 10:10

Here is a revised and somewhat expanded version of my answer, with a preparatory 'toy version' to help orient the reader.

A simple warmup problem: Before discussing a quantitative variant of the Schwarzian, let me describe the overall idea in a simpler case: Deciding how close two nonconstant meromorphic functions on a Riemann surface are to being constant multiples of each other.

Let $D$ be a connected Riemann surface, and let $f$ and $g$ be nonconstant meromorphic functions on $D$. They will be constant multiples of each other if the meromorphic $1$-form $\omega = f\ dg - g\ df$ vanishes identically. However, this is an exact statement, and it is hard to see how a bound on the 'size' of $\omega$ (assuming that it doesn't vanish outright) could say anything quantitative about how 'close' $f$ and $g$ are to being constant multiples of each other, especially since $f$ and $g$ could have poles (and $\omega$ could have poles, too, for that matter).

As an alternative, consider the singular (i.e., possibly degenerate) conformal metric on $D$ defined by $$ds^2_{f,g} = \frac{4\ (f\ dg - g\ df)\circ\overline{(f\ dg - g\ df)}} {\bigl(|f|^2+|g|^2\bigr)^2}\ .$$ This is the singular metric induced on $D$ by pullback under the holomorphic map $H_{f,g}:D\to\mathbb{CP}^1$ defined by $H_{f,g} = [f,g]$, where the metric on $\mathbb{CP}^1$ is the familiar Fubini-Study metric with constant Gauss curvature 1: $$d\sigma^2 = \frac{4\ (z\ dw - w\ dz)\circ\overline{(z\ dw - w\ dz)}} {\bigl(|z|^2+|w|^2\bigr)^2}\ .$$

Note that $ds^2_{f,g}$ vanishes identically when $\omega$ does but never has poles, so it's easier to define and measure its 'size'. Most importantly, a bound on the size of $ds^2_{f,g}$ implies a bound on the size of the image of $H_{f,g}$. For example, if $R>0$ is the diameter of $D$ under the metric $ds^2_{f,g}$, then the image of $H_{f,g}$ fits into a disk of radius at most $R$ in $\mathbb{CP}^1$. Obviously, the smaller $R$ is, the closer $H_{f,g}$ is to being a constant map, i.e., the closer the ratio of $f$ to $g$ is to being constant. Meanwhile, there are many ways to estimate the diameter of $D$ under the metric $ds^2_{f,g}$; for example, one could have a pointwise bound of the ratio of this metric to another 'reference' metric whose diameter is already known.

Now, the noncanonical thing about this strategy is that there is more than one candidate for the Fubini-Study metric on $\mathbb{CP}^1$. However, these are all equivalent under automorphisms of $\mathbb{CP}^1$, so they form a family parametrized by $\mathrm{SL}(2,\mathbb{C})/\mathrm{SU}(2)$ (which is, of course hyperbolic $3$-space). Moreover, the ratio of any two of them is bounded, so, in a sense, switching to a different member of the family will produce a quantitative measurement that is not essentially different.

A Quantitative version of the Schwarzian: With the above in mind, consider this geometric way to think about the problem of quantifying the Schwarzian: To say that $f$ is a linear fractional function of $z$ is to say that there are constants $(a,b,c,d)$ with $ad-bc\not=0$ such that $$c\ z f + d\ f - a\ z - b = 0.$$ In other words, one is asking whether the mapping $H_{f,z}:\mathbb{D}\to\mathbb{P}^3$ defined by $$H_{f,z} = [1, f, z, fz]$$ has image in a hyperplane $\mathbb{P}^2\subset\mathbb{P}^3$. The condition that $f$ not be constant is what keeps this map from going into a line in $\mathbb{P}^3$.

Meanwhile, note that the image of $H_{f,z}$ always lies in the quadric $\mathbb{Q}\subset\mathbb{P}^3$ defined by $X_0X_3-X_1X_2=0$, and that this quadric is biholomorphic to $\mathbb{P}^1\times\mathbb{P}^1$ in the obvious way: $\bigl([a_0,a_1],[b_0,b_1]\bigr)\mapsto [a_0b_0,a_1b_0,a_0b_1,a_1b_1]$. In particular, we should be thinking of $\mathbb{P}^3$ as the projectivization of the vector space $V\simeq\mathbb{C}^4$ of $2$-by-$2$ matrices endowed with the quadratic form $X_0X_3-X_1X_2$ that is simply the determinant. For use below, I will also fix a volume form on $V$. That way, one can define a triple cross product of vectors $v_1,v_2,v_3\in V$ by letting $v_1\times v_2\times v_3\in V$ be the vector that satisfies $$(v_1\times v_2\times v_3)\cdot w = \det(v_1\wedge v_2\wedge v_3\wedge w)$$ for all $w\in V$.

Now, let $D$ be any connected Riemann surface and let $f$ and $g$ be nonconstant meromorphic functions on $D$. Define a holomorphic curve $H_{f,g}:D\to \mathbb{Q}\subset\mathbb{P}^3$ by $$H_{f,g} = [1, f, g, fg]\ .$$ Then this curve does not have image in a line (since $f$ and $g$ are not constant). The condition that $H_{f,g}$ have image lying in a plane is that the relative Schwarzian $\mathsf{S}(f,g)$ should vanish, where $$\mathsf{S}(f,g) = \left( \frac{f'''}{f'}-\frac32\frac{(f'')^2}{(f')^2} -\frac{g'''}{g'}+\frac32\frac{(g'')^2}{(g')^2}\right)\ dw^2$$ and where the primes denote differentiation with respect to any local coordinate $w$. (It is easy to verify that the meromorphic quadratic differential $\mathsf{S}(f,g)=-\mathsf{S}(g,f)$ is defined independent of the choice of local coordinate $w$.)

To get a quantitative sense of how close $H_{f,g}$ is to lying in a plane, one needs a quantitative sense of how close the 'normal' map $$K_{f,g} = [{\hat H}_{f,g}\times {\hat H}'_{f,g}\times {\hat H}''_{f,g}]: D\to\mathbb{P}^3$$ is to being constant. (Here, I am taking a (local if necessary) meromorphic lifting ${\hat H}_{f,g}:D\to V$. This lifting is unique up to a multiple, and this multiple goes away under projectivization. Also, the choice of local parameter $w$ used to compute the derivatives does not matter, so that $K_{f,g}$ is well-defined.)

To get a sense of how close $K_{f,g}$ is to being constant, fix some metric on $\mathbb{P}^3$; for the purposes of this argument, say we fix a Fubini-Study metric with Kähler form $\Omega$ on $\mathbb{P}^3$. (This is where we are introducing symmetry breaking, since there is no canonical metric on $\mathbb{P}^3$, but rather a family of Fubini-Study metrics, all differing by projective transformations, so the moduli space of such is $\mathrm{SL}(4,\mathbb{C})/\mathrm{SU}(4)$. One can do a little better than this because one can choose one that is invariant under the maximal compact in the automorphism group of the inner product, and that smaller moduli space is $\mathrm{SO}(4,\mathbb{C})/\mathrm{SO}(4)$.)

At any rate, consider the nonnegative real $(1,1)$-form on $D$ defined by $$\Omega_{f,g} = (K_{f,g})^*(\Omega).$$ This $(1,1)$-form vanishes identically if and only if $H_{f,g}$ has image lying in a plane, otherwise, it vanishes only at isolated points. Indeed, when it doesn't vanish identically, relative to a local coordinate $w$ on $D$, it takes the form $$\Omega_{f,g} = \frac{i}{2}|w|^{2k}p(w,\bar w)\ dw\wedge d\bar w.$$ where $p$ is nonvanishing and $k\ge0$. In particular, the (possibly degenerate) metric given in local coordinates by $$ds^2_{f,g} = |w|^{2k}p(w,\bar w)\ dw\circ d\bar w$$ is well-defined on $D$. Calculation shows that $ds^2_{f,g}$ is essentially the square norm of $\mathsf{S}(f,g)$ times a factor that 'clears the denominators', so that $ds^2_{f,g}$ does not have 'poles'. Thus, in a natural (but not canonical) sense, a bound on $ds^2_{f,g}$ is equivalent to a bound on $\mathsf{S}(f,g)$.

In particular, if you were to take the 'norm' on $\mathsf{S}(f,g)$ to be the diameter of $D$ with respect to the singular metric $ds^2_{f,g}$ (which is well-defined), then that norm would give you an upper bound on the diameter of the image of $K_{f,g}$, i.e., it would say that, for example, the distance from $K_{f,g}(p)$ to $K_{f,g}(p_0)$ is bounded, so that the image of $H_{f,g}$ would lie at most a bounded distance from the plane that $K_{f,g}(p_0)$ defines.

In geometric terms, this is exactly the kind of bound you were asking for. The only difference is that you aren't using a norm directly on the Schwarzian, but on a closely related quantity $ds^2_{f,g}$, one that depends on an arbitrary choice (so it's not quite canonical) but that has better compactness properties (so that you can actually get estimates).

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Awesome! I definitely need to read this more careful (and wonder if there is a more fleshed out reference), but this is very much along the lines of what I was interested in. – Rbega Dec 28 '12 at 16:54
@Rbega: Now that I know that this is the kind of thing you want, I can probably add a few words of introduction to clarify the idea and clean up the exposition, which is currently a little off-the-cuff and informal. I don't know where to send you for further reading, but maybe people who are expert in complex analysis (which doesn't include me) can make some suggestions. I'll add an analysis of a simpler case that one can understand more easily at the beginning so that you can more easily see the overall picture by analogy. – Robert Bryant Dec 29 '12 at 16:47

Specific estimate depends on the way you want to measure the distance from a function to fractional-linear transformations.

But the general idea is to reduce the Schwarz equation to a linear one. Namely, $S(f)=g$ is equivalent to the linear equation $y^{\prime\prime}+(1/2)gy=0$ in the sense that $f=y_1/y_2$ where $y_i$ are two linearly independent solutions of the linear equation. Now suppose that $|g|$ is uniformly small in a disc. Choose your favorite normalization of $y_1,y_2$ at the center of this disc, and use a standard estimate for solutions of linear the equation. (Either the majorant method, or reduction to the integral equation). You obtain that $y_1,y_2$ are uniformly close to linear (affine) functions. And their ratio is close to a fractional-linear function.

But depending on context, you have to choose the measure of closeness to avoid difficulties when the denominator is zero or close to it.

I recommend Is there an underlying explanation for the magical powers of the Schwarzian derivative? for explanation about Schwarzian derivative

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