### Riccati Equations and the Projective Line

#### Posted by John Baez

Riccati equations are a natural next step after you’ve studied *linear* differential equations. Linear first-order ordinary differential equations look like this:

$\frac{d y}{d t} = f_1(t) y + f_0(t)$

Riccati equations look like this:

$\frac{d y}{d t} = f_2(t) y^2 + f_1(t) y + f_0(t)$

I think I finally get why they’re interesting. Riccati equations are to the *projective* geometry as linear first-order ordinary differential equations are to *affine* geometry!

Since I’m really not talking about analysis here I’ll assume all my functions are smooth. A homogeneous linear first-order ordinary differential equation

$\frac{d y}{d t} = f_1(t) y$

has solutions $y(t)$ depending in a *linear* manner on the initial data:

$y(t) = a(t) y(0)$

where $a$ is a smooth function independent of $y(0)$. More generally, a linear first-order ordinary differential equation

$\frac{d y}{d t} = f_1(t) y + f_0(t)$

has solutions $y(t)$ depending in an *affine* manner on the initial data:

$y(t) = a(t) y(0) + b(t)$

where $a$ and $b$ are smooth functions independent of $y(0)$.

Riccati equations are the next chapter in this story! A Riccati equation

$\frac{d y}{d t} = f_2(t) y^2 + f_1(t) y + f_0(t)$

has solutions $y(t)$ depending in a *projective linear* manner on the initial data:

$y(t) = \frac{a(t) y(0) + b(t)}{c(t) y(0) + d(t)}$

for functions $a,b,c,d$ independent of $y(0)$.

Unlike solutions of first-order linear equations, solutions of Riccati equations can blow up in finite time. However, these solutions become defined for all time if we let $y$ take values in the projective line rather than the affine line: that is, $\mathbb{R}\mathrm{P}^1$ or $\mathbb{C}\mathrm{P}^1$ rather than $\mathbb{R}$ or $\mathbb{C}$.

Indeed, the group of fractional linear transformations, the projective general linear group $\mathrm{PGL}(2,\mathbb{R})$ or $\mathrm{PGL}(2,\mathbb{C})$, acts naturally on the projective line, with the group of affine transformations being the subgroup that preserves the affine line. So, Riccati equations naturally extend the theory of linear first-order differential equations in a manner dictated by geometry.

Even better, in both the real and complex case, the group of smooth $\mathrm{PGL}(2)$-valued functions on the real line acts on the space of all Riccati equations! First, note that this group acts on $\mathbb{P}^1$-valued functions on the real line as follows:

$\left( \begin{array}{cc} a(t) & b(t) \\ c(t) & d(t) \end{array} \right) \colon y(t) \mapsto \tilde{y}(t) = \frac{a(t) y(t) + b(t)}{c(t) y(t) + d(t)}$

Then, given any Riccati equation

$\frac{d y}{d t} = f_2(t) y^2 + f_1(t) y + f_0(t) \qquad (\star)$

there is another Riccati equation such that $\tilde{y}$ is a solution of this new Riccati equation iff $y$ is a solution of $(\star)$. You can figure out this new Riccati equation explicitly with a calculation.

This is one of the main tricks for solving Riccati equations. Iterating this trick, it leads to some methods for solving Riccati equations using continued fractions! Indeed, this seems to be why Euler got interested in continued fractions in the first place.

We might hope to generalize this whole story to equations with cubic nonlinearities:

$\frac{d y}{d t} = f_3(t) y^3 + f_2(t) y^2 + f_1(t) y + f_0(t)$

but I’m pretty sure it doesn’t work — at least, not in an easy way. The point is that linear vector fields

$a z \frac{d}{d z}$

on the line form a Lie algebra: they’re closed under commutators. They form the Lie algebra of the group of linear transformations of the line. So do affine vector fields:

$(a z + b) \frac{d}{d z}$

They form the Lie algebra of the group of affine transformations of the affine line. So do quadratic vector fields:

$(a z^2 + b z + c) \frac{d}{d z}$

They form the Lie algebra of the group of projective transformations of the projective line, $PGL(2)$. But cubic vector fields don’t form a Lie algebra! They are not closed under commutators. (Do the computation.)

I have a couple of questions:

**Question 1.** Is there a known generalization of these ideas to other homogeneous spaces? That is, given a Lie group $G$ and a Lie group $H$ is there a theory of first-order differential equations for functions $y \colon \mathbb{R} \to G/H$ whose solutions are given by

$y(t) = T(t) y(0)$

where $T \colon \mathbb{R} \to G$? And does the group $C^\infty(\mathbb{R},G)$ act on the space of such differential equations?

**Question 2.** Is there a finite-dimensional Lie group bigger than $PGL(2,\mathbb{R})$ that acts on $\mathbb{R}\mathrm{P}^1$ in a nontrivial way?

**Question 3.** Is there a finite-dimensional Lie group bigger than $PGL(2,\mathbb{C})$ that acts on $\mathbb{C}\mathrm{P}^1$ in a nontrivial way? Here I mean $\mathbb{C}\mathrm{P}^1$ as a *real* smooth manifold: I’m obviously not expecting any *complex-analytic* maps from the Riemann sphere to itself other than fractional linear transformations.

If the answers to Questions 2 or 3 were yes, we might get a generalization of Riccati equations to some larger class of first-order ODE for $\mathbb{P}^1$-valued functions.

For more information on this stuff, try these:

Vaughn Climenhaga, Riccati equations and fractional linear transformations, January 23, 2013.

Vaughn Climenhaga, More on Riccati equations and fractional linear transformations, January 24, 2013.

The business about $C^\infty(\R, PGL(2))$ acting on the space of Riccati equations is my own interpretation of material here:

- Lisa Lorentzen and Haakon Waadeland,
*Continued Fractions with Applications*, North Holland, 1992.

## Re: Ricatti Equations and the Projective Line

Typo: “Ricatti” instead of “Riccati” in the first three instances.