Dunajski M. Solitons, Instantons, and Twistors

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336 C: Overdetermined PDEs

Cauchy form is

∂u

∂t

= F

, t, u

∂u

∂x

), (C34)

t=t

= g

Theorem C.6.1 (Cauchy–Kowalewska) If the equation (C34) and the initial data

are real-analytic then there exists a unique solution in the form of a power series

(t, x)=g

)+g

)(t −t

)

+ ···

which converges on some domain containing t = t

This theorem can be reﬁned if the ﬁrst derivatives of u with respect to t are speciﬁed

only for the ﬁrst r components of u, that is, if α =1,..., r < N in (C34). In this

case the system is underdetermined as there are fewer equations than unknowns.

The general analytic solution to (C34) depends on (N −r) arbitrary functions. This

is quite obvious, a choice of (N − r) functions is needed to put the equation in the

‘determined form’ (C34) with α =1,...,N.

Deﬁnition C.6.2 A k-dimensional subspace E ⊂ T

M is an integral element of I if

θ(e

,...,e

)=0

for all θ ∈ I

and e

∈ E, i =1,...,k.

The set of all k-dimensional integral elements is denoted V

(I). It is clear that

tangent space to any k-dimensional integral manifold is an integral element. We

aim to answer the following:

Question. When is an integral element tangent to an integral manifold?

Certainly not always, as obstructions can arise from the Frobenius theorem.

Example. The EDS

I = <dx ∧ dz, dy ∧ (dz − ydx)>

diff

has a two-dimensional integral element

{∂

+ y∂

,∂

}

at each point, but no two-dimensional integral manifolds as the vectors spanning

E do not satisfy the Frobenius condition (C9).

If E ⊂ V

(I) and G ⊂ E is a p-dimensional subspace of E then G ⊂ V

(I). Thus

restrictions of integral elements are integral elements. But the converse is not true,

and not every extension of integral element may be an integral element.

C.6 Cartan–Kähler theorem 337

Deﬁnition C.6.3 Let E ⊂ V

(I) be spanned by {e

,...,e

}. The polar space of E is

H(E)={v ∈ T

M,θ(v, e

,...,e

)=0, ∀θ ∈ I

k+1

}⊂T

The polar space is a vector space containing E, but it does not have to be an

integral element. However if v ∈ H(E) and v is not an element of E then the direct

sum E ⊕ span{v} isa(k + 1)-dimensional integral element. Thus H(E) is the space

of possible one-dimensional extensions of a given integral element. Constructing H

from a given E comes down to solving a set of linear homogeneous equations for

components of v. In practice to compute the polar space of a k-dimensional integral

element E, contract all (k + 1)-forms in the ideal with all vectors in E. The resulting

one-forms should be annihilated by all vectors in H(E). An integral element E is

called regular if the dimension of the polar space is constant in a neighbourhood

of E in V

(I). Moreover E is called ordinary if the intersection of V

(I) with an

open neighbourhood of E is a smooth submanifold of the Grassmanian Gr

(TM)

of all k planes in TM.

For a given E ∈ V

(I) deﬁne

r(E)=dim

[

H(E)

]

− k −1

to be the dimension of the set of (k + 1) integral elements that contain E with

r(E)=−1 if there are no such elements.

Example [25]. Let

I = <dx ∧ dz, dy ∧ (dz − ydx)>

diff

be an EDS on

. One-dimensional integral element E is spanned by e

= a∂

b∂

+ c∂

. The vector v = f ∂

+ g∂

+ h∂

is in H(E) if two linear equations

cf − ah = 0 and −ybf − (c − ya)g + bh =0

for ( f, g, h) hold. If c − ya = 0 we get H(E)=E and thus r (E)=−1. If c − ya =0

then dim

[

H(E)

]

= 2 and r(E) = 0. In particular E is not a regular integral

element.

An integral manifold S ⊂ M is called ordinary/regular iff all of its tangent spaces

are ordinary/regular elements. For regular integral manifolds we deﬁne r(S)=

r(T

S).

Theorem C.6.4 (Cartan–Kähler) Let (M, I) be a real analytic EDS and let  ⊂ M

be an n-dimensional analytic submanifold whose tangent spaces are regular inte-

gral elements such that dim

[

H(T

)

]

= n +1. Then there exists an open neighbour-

hood of x ∈  and a unique analytic (n +1)-dimensional integral manifold S ⊂ U

containing  ∩ U.

The Cartan–Kähler theorem states when an n-dimensional integral manifold can

be thickened to an (n + 1)-dimensional integral manifold. This theorem needs to

be modiﬁed by introducing a so-called restraining manifold if the dimension of

the polar space of T

 is greater than (n + 1). This is needed for uniqueness. The

restraining manifold R is an analytic submanifold of M of co-dimension r()

338 C: Overdetermined PDEs

such that  ⊂ R and T

R ∩ H(T

) has dimension (n + 1) for all x ∈ . Then

there exists a unique connected (n + 1) analytic integral manifold S which satisﬁes

 ⊂ S ⊂ R. The reader is referred to [23] where this is discussed.

The proof of the Cartan–Kähler theorem is obtained by adopting local coor-

dinates and reducing the problem to a solution of system of PDEs of the form

(C34). This uses the Cauchy–Kowalewska theorem and so one needs to require

real-analyticity. Again, consult [23] or [90] for details.

The integral manifolds can in principle be constructed successively using the

Cartan–Kähler theorem. At each step the integral manifold is determined by a

choice of restraining manifolds and the arbitrary functions in the maximal integral

manifold parameterize these choices. We shall now discuss the Cartan test which

gives a handle on how to calculate this freedom in the ‘general solution’. Applying

the Cartan–Kähler theorem successively, starting from one-dimensional integral

manifolds gives a sufﬁcient condition for the existence of an integral manifold

tangent to a given integral element: If E ⊂ V

(I) contains a ﬂag of subspaces

{0} = E

⊂ E

⊂···⊂E

= E ⊂ T

where the integral elements E

⊂ V

(I) are regular, then there exists a real-analytic

n-dimensional integral manifold S ⊂ M passing through x and satisfying T

P = E.

This corollary from Theorem C.6.4 is not of great practical signiﬁcance, as the

regularity assumption needs to be checked at each step. Also, it gives a sufﬁcient

condition which is not necessary as not all integral manifolds have tangent spaces

which are ﬁnal objects in a ﬂag of regular integral elements.

To get around this, consider the integral ﬂag F =(E

,...,E

), not necessarily

regular, and set

c(E

):=dim(T

M) − dim H(E

), k =1, 2,...,n

and let c(E

−1

) = 0. The Cartan characters of the ﬂag F are the non-negative

numbers deﬁned by

(F):=c(E

) − c(E

k−1

Theorem C.6.5 (Cartan test) Let (M, I) be an EDS and let F =(E

,...,E

) be

an integral ﬂag of I. Then V

(I) has co-dimension at least

c(F ):=c(E

)+c(E

)+···+ c(E

n−1

)

in the Grassmannian

(TM) at E

. Moreover V

(I) is a smooth submanifold

of Gr

(TM) of co-dimension c(F) iff the ﬂag F is regular.

Recall that the Grassmannian Gr

(E)isthesetofk-dimensional subspaces of a vector

space E. It is a smooth manifold of dimension k

dim(E −k)

.Thesetofallk-dimensional

subspaces in T

M as x varies over M is denoted Gr

(TM). It is a manifold of dimension

dim (M)+k

dim(M − k)

. Given a k-plane E in T

M on which dx

∧···∧dx

= 0 we can choose

coordinates (x

,...,x

, u

,...u

)onM such that E is spanned by vectors

∂

∂x



α=1

(E)

∂

∂u

, i =1,...,k,

where p

= p

(E) are coordinates on the ﬁbres of Gr

(TM) → M.

C.6 Cartan–Kähler theorem 339

Performing Cartan test on a given ﬂag is just a matter of linear algebra. If the

ﬂag passes the test and therefore is regular, the Cartan–Kähler theorem implies the

existence of at least one real-analytic n-dimensional integral manifold S ⊂ M such

that T

S = E

(there will be exactly one such manifold if the r(E

n−1

) = 0. Otherwise

the restraining manifold has to be chosen). Of course for a given integral element

there may be more than one ﬂag which terminates at E

. In practice it makes

sense to choose the ﬁrst element in a ﬂag such that the ﬁrst Cartan character s

as large as possible, then choose the second element such that the next character

is as large as possible, etc. The sum s

+ s

+ ···+ s

is ﬁxed regardless of these

choices. In what follows we shall drop the reference to the ﬂag and write s

instead

of s

(F). The highest k such that s

= 0 is called the Cartan character. Moreover let

c(E

)=s. Using the deﬁnitions of Cartan characters and c

= dim(M − n)wecan

write

+ s

+ ···+ s

= c

and rewrite the inequality in Cartan test as

dim

[

(I)

]

− dim (M) ≤ s

+2s

+ ···+ ns

, (C35)

where the LHS is the ﬁbre dimension of V

(I).

Given a ﬂag F which passes the test it is possible to chose a coordinate system

,...,x

, u

,...,u

)

centreed at x ∈ U ⊂

n+s

such that E

is spanned by

{

∂

∂x

,...,

∂

∂x

}, 0 ≤ k < n

and elements H(E

) are annihilated by the one-forms

{du

,...,du

Let S be the collection of real-analytic integral manifolds near S. This means that

S ∈ S if it can be represented by

= F

,...,x

where the analytic functions F

are deﬁned in the neighbourhood of x = 0. Then

the collection S depends on s

constants, s

functions of one variable, ..., s

functions of n variables. Thus the integers (s

, s

,...,s

) measure the arbitrariness

of the general integral manifold.

Example. A Lagrangian submanifold of R

is an integral manifold of the ideal

generated by a symplectic structure

θ = dx

∧ du

+ dx

∧ du

+ ···+ dx

∧ du

Choose a ﬂag

{0}⊂{

∂

∂x

}⊂{

∂

∂x

∂

∂x

}⊂···⊂{

∂

∂x

∂

∂x

,...,

∂

∂x

340 C: Overdetermined PDEs

For this ﬂag H(E

) is the whole tangent space, so c

=0.ThenH(E

) consists of

all vectors which annihilate du

and more generally

H(E

)={du

,...,du

}

⊥

Thus

=0, c

=1, c

=2,...,c

= n

which implies

=0, s

= s

= ···= s

=1.

To calculate the ﬁbre dimension of V

(I) note that the vectors spanning E

annihilate the one-forms du

, k =1,...,n. The nearby integral planes are given



j=1

, k =1,...,n,

and the total number of symmetric coefﬁcients p

= p

is the ﬁbre dimension of

(I) which appears on the LHS of the inequality (C35). This number is



n +1



which is also equal to the RHS of (C35). Thus we have equality and the ﬂag is

regular. The general integral manifold depends on one function of n variables

and functions of lower number of variables. Explicitly

∂ f

∂x

, k =1,...,n,

where f = f (x

, x

,...,x

Example. Consider the EDS (C5) in R

<θ

= du − pdx − qdy,θ

= dp ∧ dq − dx ∧ dy>

diff

for the Monge–Ampere equation. The four-dimensional space V

(I)={θ

}

⊥

one-dimensional integral elements is spanned by

{∂

+ p∂

,∂

+ q∂

,∂

Pick E

= {∂

} to be the ﬁrst element in the ﬂag. The two-forms in the ideal are

, dθ

= dx ∧ dp + dy ∧ dq, and θ

∧ γ,

where γ is any one-form. Therefore the polar space H(E

) will consist of all

vectors annihilating ∂

,∂

dθ

, and θ

. Thus H(E

)={dq, dx,θ

}

⊥

. This

space is two-dimensional and there is a unique extension of E

to an integral

element

= {∂

,∂

+ q∂

}

C.6 Cartan–Kähler theorem 341

(we would have got the same integral element E

if we had picked E

= {∂

q∂

}). Contracting the two vectors in E

with all three-forms in the ideal shows

that this cannot be further extended, that is, H(E

)=E

. Therefore we pick a

ﬂag

{0}⊂{∂

}⊂{∂

,∂

+ q∂

} = E ⊂ T

This ﬂag has c

=5− 4=1, c

=5− 2=3, c

=5− 2 = 3 and so s

=1, s

=2.

To perform the Cartan test we need to compute the co-dimension of V

(I)inthe

Grassmannian of two-planes. The two-planes close to E

are spanned by

= ∂

+ α(∂

+ p∂

)+β∂

+ γ∂

and v

= ∂

+ q∂

+ δ(∂

+ p∂

)+∂

+ φ∂

for some (α,β,...,φ). The conditions

)=0,θ

)=0, dθ

)=0, and θ

)=0

give four linear equations

γ =0,φ=0,− α =0, and β + δ =0.

Thus the ﬁbre co-dimension of V

(I) is four which is equal to c

+ c

.The

Cartan test is satisﬁed and the general solution to the Monge–Ampere equation

depends on two functions of one variable.

In theory one could always reduce a problem to the analysis of a Pfafﬁan system

(i.e. one where I is generated by one-forms) as any EDS can be prolonged to such

system. If a Pfafﬁan system is generated by

<θ

,...,θ

then the vectors {e

, e

, ···, e

} spanning E

in a ﬂag

{0}⊂E

⊂ E

⊂···⊂E

⊂ T

are found by solving the system

= 0 and e

dθ

)=0, i, j =1,...,k,α=1,...,N.

This however comes at the price of introducing more variables and working in

spaces of high dimension: For a system of r PDEs of order k for N functions of n

unknowns



, u

∂u

∂x

,...,

∂

∂x

...∂x



=0,ρ=1,...,r,

α =1,...N, i =1,...,n

the Pfafﬁan system is generated by one-forms

− p

, dp

− p

, ..., dp

···i

k−1

− p

···i

on the manifold M given by the zero locus

, u

, p

, ···, p

···i

342 C: Overdetermined PDEs

in the kth jet space J

, R

). There are however more economical tricks to reduce

a problem to a Pfafﬁan system. The following example, modiﬁed from [17], shows

one such trick.

Example. Consider the Ricci-ﬂat Kähler equation (C7) in four dimensions. We

are interested in the real-analytic solutions, so we can complexify the dependent

and independent variables and regard (w, z, ¯w,

z) as independent holomorphic

coordinates on an open ball in

. The equation (C7) is modelled by the ideal

generated by one one–form and one four–form on

<d − pdw −

pd ¯w − qdz −

z, dp ∧ dq ∧ dw ∧ dz − d ¯w ∧ d

z ∧ dw ∧ dz>

diff

together with the independence condition dw ∧ dz ∧d ¯w ∧ d

z = 0. To reformu-

late the problem as a Pfafﬁan system rewrite the vanishing of the four-form as

d( pdq − ¯wd

z) ∧ dw ∧ dz =0.

The independence condition implies dw ∧ dz = 0. Thus locally there exist func-

tions a, b, such that

pdq − ¯wd

z = d − adz − bdw

integral manifolds. Conversely equation (C7) can be modelled as a Pfafﬁan

EDS

I = <θ

= d − pdw −

pd ¯w −qdz −

z,θ

= d −adz−bdw − pdq +¯wd

diff

with coordinates (w, z, ¯w,

z, p, q,

q, a, b,,).

The space of one-dimensional integral elements {θ

,θ

}

⊥

is 10-dimensional, thus

= 2 and s

= 2. Let E

= {e

}. The polar space of E

is the eight-dimensional

vector space

H(E

)={θ

,θ

dθ

, e

dθ

}

⊥

Thus c

= 4 and s

= c

− c

=2.LetE

= {e

, e

}. Then

H(E

)={θ

,θ

, e

dθ

, e

dθ

, e

dθ

, e

dθ

}

⊥

so c

= 6 and s

= 2. We continue looking for polar spaces and extending the

integral elements. Let E

= {e

, e

}. This gives

=8, s

= 2. Pick some e

∈

H(E

) and set E

= {e

, e

}. Now

H(E

)={θ

,θ

, e

dθ

, e

dθ

}

⊥

, i =1,...,4

and dim

[

H(E

)

]

≤ 4. However E

⊂ H(E

) and we must have H(E

)=E

and

the integral element E

is not extendable. We have c

=12− 4=8ands

=0.

Thus the maximal integral manifolds may be at most four-dimensional if we

can pick a regular ﬂag. We can verify the computations of Cartan characters by

The ﬂag must be chosen carefully for this to be true. The choice (C36) will do.

C.6 Cartan–Kähler theorem 343

choosing the ﬂag with

= ∂

+ ∂

¯w

+(p +

p)∂



+ b∂



, e

= ∂

− ∂

, e

= ∂

+ ∂

, and e

= ∂

(C36)

Then

H(E

)={θ

,θ

, dp + d

p, db + d

⊥

H(E

)={θ

,θ

, dp + d

p, db + d

z, dw − d ¯w, dq}

⊥

H(E

)={θ

,θ

, dp + d

p, db + d

z, dw − d ¯w, dq, d

z, dz}

⊥

H(E

)=E

The co-dimension of V

(I) around E = E

can be now computed as in the

last example. The Cartan test holds and thus the general real-analytic Ricci-ﬂat

Kähler metric in four dimensions depends on two arbitrary functions of three

variables.

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