## September 7, 2020

### 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.

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.
Posted at September 7, 2020 5:09 PM UTC

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### Re: Ricatti Equations and the Projective Line

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

Posted by: Blake Stacey on September 7, 2020 8:06 PM | Permalink | Reply to this

Slightly funny in Italian, since “ricatti” is the plural of “ricatto”, that is, blackmail.

Posted by: Daniele A. Gewurz on September 7, 2020 9:42 PM | Permalink | Reply to this

### Re: Riccati Equations and the Projective Line

Whoops! Fixed! It reminds me of “ricotta” so I tend to get mixed up.

Posted by: John Baez on September 7, 2020 10:43 PM | Permalink | Reply to this

### Re: Ricatti Equations and the Projective Line

The first physics topic where I’d seen Riccati equations invoked was supersymmetric quantum mechanics. In this subject, one studies a Hamiltonian like $H = \frac{p^2}{2m} + V(x) = -\frac{\hbar^2}{2m} \partial_x^2 + V(x)$ by believing really hard that it can be treated like the simple harmonic oscillator. Because if we could factor $H$ into the product of an operator $A$ and its adjoint, then we could find the ground state of $H$ by figuring out what $A$ annihilates, thereby turning a second-order differential equation into a first-order one. Moreover, if $H_1 = A^\dagger A$, then its partner Hamiltonian $H_2 = A A^\dagger$ will be isospectral with $H_1$, except for a zero-energy ground state. In cases of interest, it turns out that $H_2$ has the same functional form as $H_1$ up to some parameter adjustments, so we can iterate this trick, building up the spectrum and the eigenstates of a physically meaningful Hamiltonian. This even works for the hydrogen atom, once you factor apart the radial and angular dependences.

If we want to split a Hamiltonian into an operator and its adjoint, we should probably start with an operator which is linear in the derivative of $x$, thus: $A = \frac{\hbar}{\sqrt{2m}} \partial_x + W(x),$ where $W(x)$ is called the superpotential. Taking the adjoint of $A$ flips the sign on the derivative (recall that $p$ must be self-adjoint): $A^\dagger = -\frac{\hbar}{\sqrt{2m}} \partial_x + W(x).$ The partner potentials $V_1(x)$ and $V_2(x)$ are, in terms of the superpotential, $V_1(x) = W^2(x) - \frac{\hbar}{\sqrt{2m}} \partial_x W(x)$ and $V_2(x) = W^2(x) + \frac{\hbar}{\sqrt{2m}} \partial_x W(x).$ We have the first derivative of $W$ depending on $W^2$ and another function of $x$, which means that these are Riccati equations without the linear term.

Posted by: Blake Stacey on September 7, 2020 8:16 PM | Permalink | Reply to this

### Re: Riccati Equations and the Projective Line

Neat! I suppose you can solve them in nice examples, like the radial equation for a hydrogen atom?

You’ve mentioned this story before to me, more than once in face, but since I was blind to the charms of Riccati equations I didn’t really absorb it very well.

Posted by: John Baez on September 9, 2020 6:21 AM | Permalink | Reply to this

### Re: Ricatti Equations and the Projective Line

The treatments I remember (like those I linked above) solve for the superpotential in a very guess-and-check way. “Oh, we need something with a Coulomb part, and with an angular-momentum barrier, so let’s try this.” It is apparently possible to bring “shape invariance”, the property of potentials that makes them amenable to this approach, together with continued fractions, but I’m not at all familiar with the details.

Posted by: Blake Stacey on September 10, 2020 4:13 AM | Permalink | Reply to this

### Re: Riccati Equations and the Projective Line

I think the answer to Question 1 is that this theory must exist, at least in Platonic heaven — so I just want to know if it’s made its way down here to Earth.

Here’s how the theory starts:

If $M = G/H$ then we get a map from the Lie algebra of $G$ to vector fields on $M$:

$V \colon \mathfrak{g} \to Vect(M)$

in fact a Lie algebra homomorphism. Then, given any smooth function

$f \colon \mathbb{R} \to \mathfrak{g}$

we get a differential equation describing the motion of a point $y(t)$ on $M$:

$\frac{d}{d t} y(t) = V(f(t))_{y(t)}$

If $M$ is compact this will have solutions for all smooth $f$ and all initial data $y(0) \in M$, and the solutions will be of the form

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

where

$T \colon \mathbb{R} \to G$

is some smooth function obeying

$T(0) = 1 \in G,$

$\frac{d}{d t} T(t) = f(t) T(t)$

The space $C^\infty(\mathbb{R},\mathfrak{g})$ of smooth functions $f \colon \mathbb{R} \to \mathfrak{g}$ serves as a Lie algebra for the infinite-dimensional Lie group $C^\infty(\mathbb{R},G)$, so this group has an action — the adjoint action — on $C^\infty(\mathbb{R},\mathfrak{g})$. Thus, this group acts on the space of ‘generalized Riccati equations’

$\frac{d}{d t} v(t) = V(f(t))_{y(t)}$

where $f \in C^\infty(\mathbb{R},\mathfrak{g})$.

Note also that all the solutions of a generalized Riccati equation are encapsulated in a function $T \in C^\infty(\mathbb{R}, G)$.

Someone must have investigated this theory.

Posted by: John Baez on September 7, 2020 10:56 PM | Permalink | Reply to this

I believe this is the theory of “equations of Lie type” and the general theory of such equations dates back to the work of Lie himself. I guess this is where the terminology “solvable Lie group” comes from. If the group G is solvable, then any equation of Lie type for G is solvable by quadratures.

Posted by: Gavin Ball on September 8, 2020 6:54 PM | Permalink | Reply to this

### Re:

Yes! Equations of Lie type are very interesting, and the theory is simply very satisfying. Robert Bryant has notes on the topic in the first third of the book “Geometry and Quantum Field Theory” which one can find here: https://bookstore.ams.org/pcms-1

There are other topics related to symplectic geometry, mechanics, and Lie groups and is a very enjoyable read.

-TK

Posted by: Taylor J Klotz on September 14, 2020 11:35 PM | Permalink | Reply to this

### Re: Riccati Equations and the Projective Line

On Twitter, Francis has a nice remark about the connection between Riccati equations and 2nd-order equations:

The way I understand this is that we are projectivizing a second order linear differential equation. Namely, if $u$ satisfies a 2nd order ODE, then $y=u/u'$ satisfies Riccati. We could therefore view $y$ as the point $[u : u']$ in projective space.

He means a 2nd-order linear ODE. All this makes tons of sense to me now — it’s really nice!

If you have a 2nd-order linear ODE with smooth coefficients it has a 2d vector space of solutions $u$. For each $t \in \mathbb{R}$, the time evolution map

$T(t) : (u(0), u'(0)) \mapsto (u(t),u'(t))$

is a linear map from $\mathbb{C}^2$ to itself. Projectivizing we get a smooth one-parameter family of projective linear maps

$P T(t) :\mathbb{C}\mathrm{P}^1 \to \mathbb{C}\mathrm{P}^1$

But this precisely a Ricatti equation!

But remember, given a point $[a : b] \in \mathbb{C}\mathrm{P}^1$ we can treat it as a point in $\mathbb{C}$, namely $a/b$, as long as $b \ne 0$. So, in low-brow terms, the ratio $u(t)/u'(t)$ obeys a Ricatti equation.

The converse works just as easily.

Posted by: John Baez on September 9, 2020 1:05 AM | Permalink | Reply to this

### Thomas

As a first step, the relations among a Riccati equation with constants multiplying the sl(2) diff op generators, the Schwarzian derivative, and the KdV equation are informative and rife with special number arrays in combinatorics. Look at g(h(x)) in my answer to the MO-Q “Is there an underlying explanation for the magical powers of the Schwarzian derivative?

A natural extension of the Riccati equation is the general Lie infinitesimal generator/vector related to the inviscid Burgers-Hopf equation and solutions of autonomous ODEs for evolution/flow equations. See the MO-Q “Why is there a connection between enumerative geometry and nonlinear waves?

Posted by: Tom Copeland on September 17, 2020 5:53 PM | Permalink | Reply to this

### Re: Thomas

Posted by: John Baez on September 17, 2020 6:34 PM | Permalink | Reply to this

### Re: Riccati Equations and the Projective Line

On the Bernoullis and an early (maybe first) Riccati equation, see the MO-Q (and linked MSE-Q) “Link of a power series by the Bernoullis for a Riccati equation to zonotopes?”

Posted by: Tom Copeland on September 17, 2020 6:48 PM | Permalink | Reply to this

### Re: Riccati Equations and the Projective Line

Almost forgot. Look at the 2014 formulas in OEIS A008292 on the Eulerian polynomials. Their bivariate e.g.f. is the solution of a Riccati equation.

Posted by: Tom Copeland on September 17, 2020 7:21 PM | Permalink | Reply to this

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