The Schwarzian derivative

Posted on 14 September, 2015 by jesperg

For a diffeomorphism f:\mathbf{E}^1\rightarrow\mathbf{E}^1 between two Euclidean lines, the derivative f'(x)=\lim_{\Delta x\rightarrow 0}\Delta f/\Delta x is an invariant in the sense that to calculate the derivative we fix Euclidean coordinate systems on the domain and on the range and the calculated value will not depend on the choice of coordinate systems.

For a diffeomorphism f:\mathbf{A}^1\rightarrow\mathbf{A}^1 between affine lines, things are not that simple, the calculated value of the derivative f' will be dependent on the affine coordinate systems it is computed in. To find an invariant expression we look at the fundamental invariant of the affine line, the three-point ratio:

\displaystyle [x,x_1,x_\alpha]=\frac{x_\alpha-x}{x_1-x}

We put x_1=x+dx and x_\alpha=x+\alpha dx. Then [x,x_1,x_\alpha]=\alpha and the ratio of the images of the three points is then

\displaystyle \begin{aligned}[] [f(x),&f(x_1),f(x_\alpha)] =\frac{f(x+\alpha dx)-f(x)}{f(x+dx)-f(x)}\\ &=\frac{\alpha f'(x)dx+(\alpha^2/2)f''(x)dx^2+O(dx^3)}{f'(x)dx+(1/2)f''(x)dx^2+O(dx^3)}\\ &= \alpha+\frac{((\alpha^2-\alpha)/2)f''dx^2+O(dx^3)}{f'dx+O(dx^2)}\\ &= \alpha+\frac{\alpha^2-\alpha}{2}\frac{f''}{f'}dx+O(dx^2)\end{aligned}

and so the one-form \omega=(f''/f')dx is an invariant that measures how much the diffeomorphism deviates from being an affine transformation.

Now let y be a coordinate on the range and thus y=f(x). We can express the derivatives of f as

\begin{aligned} f'&=\frac{dy}{dx}\\f''&=\frac{d^2ydx-d^2xdy}{dx^3}\end{aligned}

and the invariant one-form can be reexpressed as \omega=(f''/f')dx=d^2y/dy-d^2x/dx. Each of the two quotients is invariant under affine coordinate changes so in fact any function g(d^2y/dy,d^2x/dx) would define an invariant of the diffeomorphism, but it is only by taking the difference that the second order differentials cancel and we get an invariant that is a one-form on the tangent space.

Let’s carry out this calculation, which is in a sense the last calculation carried out backwards. Since the second differential of f is d^2f=f''dx^2+f'd^2x, we get

\displaystyle \frac{d^2y}{dy}-\frac{d^2x}{dx}=\frac{f''dx^2+f'd^2x}{f'dx}-\frac{d^2x}{dx}=\frac{f''}{f'}dx

as expected.

Moving on to instead study a diffeomorphism f:\mathbf{P}^1\rightarrow\mathbf{P}^1 between projective lines, we now want to see how the cross ratio

\displaystyle [x,x_1,x_a,x_{a^2}]=\left.\frac{x_a-x}{x_1-x}\middle/\frac{x_a-x_{a^2}}{x_1-x_{a^2}}\right.

changes under the diffeomorphism. We put x_1=x+dx, x_a=x+adx, x_{a^2}=x+a^2dx and calculate that [x,x_1,x_a,x_{a^2}]=1+a. On the range space we want to make use of the availability of homogenous coordinates and define a lift of f to a vector \mathbf{v}=(u,v) by

\displaystyle \begin{aligned}u&=\frac{f}{\sqrt{f'}}\\v&=\frac{1}{\sqrt{f'}}\end{aligned}.

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This lift has the property that the curve has constant angular momentum \mathbf{v}'\times\mathbf{v}=u'v-uv'=1 and thus the acceleration vector is parallel to the position vector \mathbf{v}''=\lambda(x)\mathbf{v}. The cross ratio can be calculated in homogenous coordinates as

\displaystyle [\mathbf{v},\mathbf{v}_1,\mathbf{v}_a,\mathbf{v}_{a^2}]=\frac{(\mathbf{v}_a\times\mathbf{v})(\mathbf{v}_1\times\mathbf{v}_{a^2})}{(\mathbf{v}_1\times\mathbf{v})(\mathbf{v}_a\times\mathbf{v}_{a^2})}.

We have that

\displaystyle \begin{aligned} \mathbf{v}_1&=\mathbf{v}(x+dx)=\mathbf{v}+\mathbf{v}'dx+\frac{1}{2}\mathbf{v}''dx^2+\frac{1}{6}\mathbf{v}'''dx^3+O(dx^4)\\ &=\left(1+\frac{\lambda}{2}dx^2+\frac{\lambda'}{6}dx^3\right)\mathbf{v}+\left(dx+\frac{\lambda}{6}dx^3\right)\mathbf{v}'+O(dx^4)\\ \mathbf{v}_a&=\mathbf{v}(x+adx)\\ &=\left(1+\frac{\lambda}{2}a^2dx^2+\frac{\lambda'}{6}a^3dx^3\right)\mathbf{v}+\left(adx+\frac{\lambda}{6}a^3dx^3\right)\mathbf{v}'+O(dx^4)\\ \mathbf{v}_{a^2}&=\mathbf{v}(x+a^2dx)\\ &=\left(1+\frac{\lambda}{2}a^4dx^2+\frac{\lambda'}{6}a^6dx^3\right)\mathbf{v}+\left(a^2dx+\frac{\lambda}{6}a^6dx^3\right)\mathbf{v}'+O(dx^4)\end{aligned}

The cross products become

\displaystyle \begin{aligned}\mathbf{v}_1\times\mathbf{v}&=dx+\frac{1}{6}\lambda dx^3+O(dx^4)\\ \mathbf{v}_a\times\mathbf{v}&=adx+\frac{a^3}{6}\lambda dx^3+O(dx^4)\\ \mathbf{v}_1\times\mathbf{v}_{a^2}&=(1-a^2)dx+\left(\frac{1}{6}-\frac{a^2}{2}+\frac{a^4}{2}-\frac{a^6}{6}\right)\lambda dx^3+O(dx^4)\\ \mathbf{v}_a\times\mathbf{v}_{a^2}&=(a-a^2)dx+\left(\frac{a^3}{6}-\frac{a^4}{2}+\frac{a^5}{2}-\frac{a^6}{6}\right)\lambda dx^3+O(dx^4)\end{aligned}

and the cross ratio of the image points is

\displaystyle \begin{aligned}[][\mathbf{v},\mathbf{v}_1,\mathbf{v}_a,\mathbf{v}_{a^2}]&=\frac{1-a^2+\left(\frac{1}{6}-\frac{a^2}{3}+\frac{a^4}{3}-\frac{a^5}{6} \right) \lambda dx^2+O(dx^3)}{1-a+\left(\frac{1}{6}-\frac{a}{6}+\frac{a^2}{6}-\frac{a^3}{2}+\frac{a^4}{2}-\frac{a^5}{6}\right) \lambda dx^2+O(dx^3)}\\&=1+a+\left( -\frac{a^2}{3}+\frac{3a^4-2a^5}{6-6a} \right) \lambda dx^2+O(dx^3)\end{aligned}.

So we get a quadratic form, the Schwarzian derivative, S(f)=-2\lambda dx^2 that tells us how much the diffeomorphism deviates from being a projective transformation. The coefficient \lambda is calculated as

\displaystyle\lambda = \frac{u''}{u}=\frac{v''}{v}=-\frac{1}{2}\frac{f'''f'-\frac{3}{2}(f'')^2}{(f')^2}

and the Schwarzian derivative when directly expressed in derivatives of f becomes

\displaystyle S(f)=\left(\frac{f'''}{f'}-\frac{3}{2}\left(\frac{f''}{f'}\right)^2\right)dx^2.

We have already given above the first and second derivative of f expressed as differentials of x and y. The third derivative then can also be reexpressed as

\displaystyle f'''=\frac{d^3ydx^2-d^3xdydx-3d^2yd^2xdx+3d^2x^2dy}{dx^5}

and the Schwarzian derivative then becomes

\displaystyle S(f)=\frac{d^3y}{dy}-\frac{3}{2}\frac{d^2y^2}{dy^2}-\frac{d^3x}{dx}+\frac{3}{2}\frac{d^2x^2}{dx^2}.

Thus we see, just as in the affine case, that the invariant form for the diffeomorphism is a difference between two separately invariant higher order forms, one on the range and one on the domain.

We now define s(x)=d^3x/dx-(3/2)d^2x^2/dx^2 and s(y|x)=s(y)-s(x). For a composition of diffeomorphisms f=g\circ h, we get

S(f)=s(y|x)=s(y)-s(x)=s(y)-s(h)+s(h)-s(x)=s(y|h)+s(h|x)=S(g)+S(h)

and we have shown the cocycle condition for the Schwarzian derivative.

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