8 Picard–Vessiot theoryOne of Lie’s motivations for studying Lie groups was to extend Galois theory todifferential equations, by studying the symmetry groups of differential equations.We will give a very sketchy account of this, missing out most proofs (and forthat matter most definitions).The theorem in Galois theory that a polynomial in characteristic 0 is solv-able by radicals if and only if its Galois group is solvable has an analogue fordifferential equaitons: roughly speaking, a differential equation is solvable byradicals, integration, and exponentiation if and only if its group of symmetriesis a solvable algebraic group. This theory was initiated by Picard and Vessiotbut it is sometimes hard to tell exactly what they proved as their definitions aresomewhat vague. Kolchin gave a rigorous reformulation of their results using thetheory of algebraic groups (which he created for this purpose). In particular oneneeds to distinguish between nilpotent and semisimple abelian groups (whichlook the same as Lie groups, but are quite different as algebraic groups). Thecorrect analogue of nilpotent Lie algebras is not nilpotent groups but unipotentgroups (those such that all eigenvalues of all elements are 1): for example, thegroup of diagonal matrices is nilpotent but not unipotent.In this extension of Galois theory, one replaces fields by differential fields:fields with a derivation D. Just as adjoining a root of a polynomial equation toa field gives an extension of fields, adjoining a root of a differential equation toa field gives an extension of differential fields. As in Galois theory, one can formthe differential Galois group of an extension k ⊂ Kof differential fields as thegroup of automorphisms of the differential field K fixing all elements of k . Muchof the theory of differential Galois groups is quite similar to usual Galois theory:for example, one gets a Galois correspondence between algebraic subgroups ofthe differential Galois group of an extension and sub differential fields.Example 90 Suppose we adjoin a root of the equation df/dx = p(x) to thefield k = Q(x) of rational functions over Q. This extension has a group ofautomorphisms given by the additive group of Q, because we can change f tof + c for some constant of integration c to get an automorphism.Example 91 Suppose we adjoin a root of the equation df/dx = p(x)f (withsolution exp(Rp)) to the field k = Q(x) of rational functions over Q. This ex-tension has a group of automorphisms given by the multiplicative group of Q,because we can change f to cf for some nonzero constant c to get an automor-phism.The theory applies to homogeneous linear differential equations, so that theset of solutions is a finite-dimensional vector space acted on by the differentialGalois group. Equations such as df/dx = 1/x with solutions log x are not ho-mogeneous so the theory does not apply directly to them, but we can easily turnthem into homogeneous equations such as (d/dx)x(d/df)f = 0, at the expenseof making the space of solutions 2-dimensional rather than 1-dimensional.Exercise 92 Find the Lie group of automorphisms of the solutions of(d/dx)x(d/df)f = 0and describe its action on the space of solutions.40We will sketch the proof of one of the results of Picard-Vessiot theory, whichsays roughly that a linear homogeneous differential equation can be solved byradicals, exponentials, and integration if and only if its differential Galois groupis solvable.In one direction this follows by calculating the differential Galois group: eachtime we take radicals we get a finite cyclic group, each time we take an integralwe get a differential Galois group isomorphic to the additive group, and eachtime we take an exponential we get a differential Galois group isomorphic to themultipicative group. So by repeating such extensions we get a group built outof additive groups, which is solvable.Conversely, suppose the differential Galois group is solvable. The quotientby the connected component is a finite solvable group, which corresponds torepeatedly taking radicals just as in ordinary Galois theory, so we can assumethat the differential Galois group is connected and solvable. Now we applyLie’s theorem on solvable Lie algebras (or more precisely Kolchin’s version ofit for solvable algebraic groups) so see that the differential Galois group has aneigenvector f in the space of solutions of the differential equation. Also Dfhas the same eigenvalue as D commutes with the differential Galois group, soDf/f is fixed by the differential Gaois group and so is in the base field. So fsatisfies the differential equation Df = af for some a, which can be solved byexponentials and integration.Example 93 A typical application of differential Galois theory is that Bessel’sequation x2d2y/dx2+xdy/dx+(x2−ν2)y = 0 cannot be solved using integrationand elementary functions unless ν − 1/2 is integral. Except for these specialvalues, the differential Galois group is SL2which is not solvable. Finding thedifferential Galois group is rather too much of a digression, but we can at leastget non-trivial upper and lower bounds for it as follows. First, we can show thatit lies in SL2(rather than just GL2) by using the Wronskian of the equation,given by W =f gf′ g′for two independent solutions f and g. The Wronskianget multiplied by the determinant of a matrix of the differential Galois group,so the elements of the differential Galois group have determinant 1 if and onlyif the Wronskian is in the base field.Exercise 94 Show that the Wronskian of d2y/dx2+ p(x)dy/dx + q(x)y = 0satisfies the differential equation dW/dx + p(x)W = 0. Use this to find theWronskian of Bessel’s equation, and deduce that the differential Galois grouplies in the special linear group.To find a lower bound for the differential Galois group, we observe that mon-odromy gives elements of this group. (Monodromy means go around a branchpoint.) Bessel’s equation has a branch point at 0, and the two solutions Jν=xν× (something holomorphic) and J−ν= x−ν× · · · for ν not an integer are mut-liplied by e±2π iνby the monodromy, so the differential Galois group containsthe diagonal matrix with these entries. When ν is an integer the monodromyis unipotent instead of semisimple: in this case the solutions are Jνwith trivialmonodromy, and Yν= (something holomorphic)+Jν×(something holomorphic)×log that has a logarithmic singularity and is changed by a multiple of Jνby