Bäcklund transformations for Darboux integrable differential systems

Abstract

The purpose of this paper is to present a novel group-theoretic method for constructing Bäcklund transformations between systems of differential equations. Our approach is based upon the definition of Bäcklund transformations as integrable extensions of exterior differential systems. The construction of these transformations is obtained using the general theory of symmetry reduction of differential systems. Our method is then applied to differential systems which are integrable by the method of Darboux, and a detailed understanding of the Bäcklund transformations for these systems is obtained.

Appendices

Appendix 1: On the prolongation of integrable extensions

Let $\mathbf {p}:(\mathcal {E}, N) \rightarrow (\mathcal {I}, M)$ be an integrable extension with admissible subbundle $J\subset T^*N$. Let $\pi _N: G_k(TN) \rightarrow N$ and $\pi _M : G_k(TM) \rightarrow M$ be the Grassmann bundles of $k$-planes and let $G_k(\mathcal {E}) \subset G_k(TN)$ and $G_k(\mathcal {I}) \subset G_k(TM)$ be the spaces of $k$-dimensional integral elements for $\mathcal {E}$ and $\mathcal {I}$ respectively. We assume that $\pi _N:G_k(\mathcal {E})\rightarrow N$ and $\pi _M: G_k(\mathcal {I})\rightarrow M$ are smooth bundles and that the inclusion maps $\iota _N : G_k(\mathcal {E}) \rightarrow G_k(TN)$ and $\iota _M :G_k(\mathcal {I}) \rightarrow G_k(TM)$ are smooth immersions. The prolongation spaces for $\mathcal {E}$ and $\mathcal {I}$ are $N^{[1]} = G_k(\mathcal {E})$ and $M^{[1]} =G_k(\mathcal {I})$, and the prolonged differential systems are $\mathcal {E}^{[1]}= \iota _N^* (\mathcal {C}_N)$ and $\mathcal {I}^{[1]}= \iota _M^*( \mathcal {C}_M)$, where $\mathcal {C}_N$ and $\mathcal {C}_M$ are the canonical Pfaffian systems on $G_k(TN)$ and $G_k(TM)$, respectively. Our goal is to prove that $(\mathcal {E}^{[1]}, N^{[1]})$ is an integrable extension of $(\mathcal {I}^{[1]}, M^{[1]})$. This fact is listed in Sect. 2 as IE [iv]. In order to state this result in a more precise manner, we first need the following preliminary lemma.

Lemma 10.1

Let $\mathbf {p}:(\mathcal {E}, N) \rightarrow (\mathcal {I}, M)$ be an integrable extension with admissible subbundle $J\subset T^*N$. Then the map $\mathbf {p}_*:TN\rightarrow TM$ defines a bijection between $k$-dimensional integral elements of $\mathcal {E}$ at $x\in N$ and $k$-dimensional integral elements of $\mathcal {I}$ at $\mathbf {p}(x)$.

Proof

By conditions [i] and [ii] in (2.3), we have $TN = \ker (\mathbf {p}_*)\oplus \mathrm{ann}(J)$. Therefore, $\mathbf {p}_* : \mathrm{ann}(J) \rightarrow TM$ is a bundle map which is an isomorphism on each fibre, that is, $\mathrm{ann}(J)$ is horizontal. We shall use this simple fact repeatedly.

Now, let $x\in N$ and let $P_x\subset T_xN$ be a $k$-dimensional integral element of $\mathcal {E}$ at $x$. If $\theta \in \mathcal {I}_x$, then $\mathbf {p}^* \theta \in \mathcal {E}$, and therefore, $\theta |_{ \mathbf {p}_*(P_x)} = (\mathbf {p}^*\theta )|_{P_x}=0$. Therefore, $\mathbf {p}_*(P_x)$ is an integral element of $\mathcal {I}$ at $\mathbf {p}(x)$. Since $P_x$ is an integral element of $\mathcal {E}$ and $\mathcal {S}(J) \subset \mathcal {E}$, we also have $P_x \subset \mathrm{ann}(J_x)$. Since the restriction of $\mathbf {p}_*$ to $\mathrm{ann}(J_x)$ is an isomorphism, the dimensions of $P_x$ and $\mathbf {p}_*(P_x)$ are the same.

Let $P_{1,x}$ and $P_{2, x}$ are two $k$-dimensional integral elements of $\mathcal {E}$ at $x$ and suppose $\mathbf {p}_*(P_{1,x}) = \mathbf {p}_*(P_{2, x})$. Then, $P_{1,x} \subset \mathrm{ann}(J_x)$ and $P_{2,x} \subset \mathrm{ann}(J_x)$ and again, because the restriction of $\mathbf {p}_*$ to $\mathrm{ann}(J_x)$ is an isomorphism, $P_{1, x} = P_{2,x}$, and therefore, $\mathbf {p}_*$ is a one-to-one mapping on integral elements.

To show that $\mathbf {p}_*$ is onto, let $P'_{\mathbf {p}(x)}\subset T_{\mathbf {p}(x)} M$ be a $k$-dimensional integral element of $\mathcal {I}$ at $\mathbf {p}(x)$. Let

$$\begin{aligned} P_x = \mathbf {p}_*^{-1}(P'_{\mathbf {p}(x)}) \cap \mathrm{ann}(J)\subset T_xN, \end{aligned}$$

(10.1)

that is, let $P_x$ be the subspace of $\mathrm{ann}(J_x)$ which maps by $\mathbf {p}_*$ to $P'_{\mathbf {p}(x)}$. The foregoing arguments imply that $P_x$ and $P'_x$ have the same dimension so that we need only to check that $P_x$ is an integral element of $\mathcal {E}$ to finish the proof. Since $\mathcal {E}$ is generated algebraically by $\mathcal {S}(J)$ and $\mathbf {p}^*(\mathcal {I})$, we need to only check $\mathbf {p}^*\theta (P_x) =0$ for all $\theta \in \mathcal {I}_{\mathbf {p}(x)}$ and $\rho (P_x) = 0$ for all $\rho \in J_x$. These are both trivially true, and therefore, $\mathbf {p}_*$ defines a bijection on $k$-dimensional integral elements.$\square$

Lemma 10.1 implies the following.

Corollary 10.2

The map $\mathbf {p}^{[1]} :N^{[1]} \rightarrow M^{[1]}$ defined by

$$\begin{aligned} \mathbf {p}^{[1]} (P_x) = \mathbf {p}_* (P_x) \end{aligned}$$

(10.2)

is a smooth submersion, gives rise to the commutative diagram

(10.3)

and restricts to a diffeomorphism between the fibres of $\pi _N$ and $\pi _M$.

With these preliminaries in hand, we can now state our theorem on the prolongation of integrable extensions (see IE [iv]).

Theorem 10.3

Let $\mathbf {p}:(\mathcal {E}, N) \rightarrow (\mathcal {I}, M)$ be an integrable extension. Then the submersion $\mathbf {p}^{[1]}:N^{[1]} \rightarrow M^{[1]}$, given by Eq. (10.2), defines $\mathcal {E}^{[1]}$ as an integrable extension of $\mathcal {I}^{[1]}$. Moreover, if $J\subset T^*N$ is an admissible bundle for the extension $\mathcal {E}$ of $\mathcal {I}$, then $\pi _N^*(J)$ is an admissible bundle for the extension $\mathcal {E}^{[1]}$ of $\mathcal {I}^{[1]}$.

To prove Theorem 10.3, we will check the three conditions of Eq. (2.3) for $(\mathcal {E}^{[1]}, N^{[1]})$ to be an integrable extension of $(\mathcal {I}^{[1]}, M^{[1]})$ by showing that $\pi ^*_N(J)$ is an admissible subbundle. To this end, we shall need the following two lemmas.

Lemma 10.4

The maps $\mathbf {p}^{[1]}_* : \ker (\pi _{N,*}) \rightarrow \ker (\pi _{M,*})$ and $\pi _{N,*} : \ker (\mathbf {p}^{[1]}_*) \rightarrow \ker (\mathbf {p}_*)$ are isomorphisms, and

$$\begin{aligned} \ker (\pi _{N,*}) \cap \ker (\mathbf {p}^{[1]}_*) = 0 . \end{aligned}$$

(10.4)

Proof

We use Corollary 10.2 and the commutative diagram (10.3). The first claim follows because $\mathbf {p}^{[1]}$ is a diffeomorphism on each fibre. To prove (10.4), let $Y \in \ker \pi _{N,*} \cap \ker \mathbf {p}^{[1]}_*$. Then $Y$ is $\pi _N$ vertical and $\mathbf {p}^{[1]}_* Y =0$. But $\mathbf {p}^{[1]}_*$ is an isomorphism on vertical vectors, therefore $Y = 0$.

To prove that $\pi _{N,*} : \ker (\mathbf {p}^{[1]}_*) \rightarrow \ker (\mathbf {p}_*)$ is an isomorphism, we begin with the fact that $\ker \mathbf {p}_* = \pi _{N,*} (\ker ( (\pi _M \circ \mathbf {p}^{[1]})_*))$ and use this fact to show that $\pi _{N,*}$ is onto. If $X \in \ker \mathbf {p}_*$, then there exists $Y \in \ker ((\pi _M \circ \mathbf {p}^{[1]})_*)$ with $\pi _{N,*}(Y) = X$. By definition $\mathbf {p}_*^{[1]} Y$ is $\pi _M$ vertical. Since, by Corollary 10.2, $\mathbf {p}^{[1]}$ is a diffeomorphism on the fibres, there exists $Z \in T_p N^{[1]}$ which is $\pi _N$ vertical and such that $\mathbf {p}_*^{[1]}(Z) = \mathbf {p}_*^{[1]}(Y)$. Then, with $Y' = Y-Z$, we have that $\pi _{N,*}(Y') = \pi _{N,*}(Y)=X$ (because $Z$ is $\pi _{N, *}$ vertical), and $\mathbf {p}^{[1]}_*(Y') =\mathbf {p}^{[1]}_*(Y) - \mathbf {p}^{[1]}_*(Z) = 0$. This proves that $\pi _{N,*} : \ker (\mathbf {p}^{[1]}_*) \rightarrow \ker (\mathbf {p}_*)$ is onto. The fact that $\pi _{N,*}$ is one-to-one follows easily from Eq. (10.4).$\square$

Lemma 10.5

The bundles $E^{[1]}$ and $I^{[1]}$ satisfy

$$\begin{aligned} E^{[1]} = \pi _N^*(J) \oplus \mathbf {p}^{[1],*}(I^{[1]}). \end{aligned}$$

(10.5)

Proof

Let $p =P_x\in N^{[1]}$ be an integral $k$-plane for $\mathcal {E}$ at $x\in N$. To prove this lemma, we first note that $\mathbf {p}^* (\mathrm{ann}(\mathbf {p}_* P_x)) \cap J_x = 0$ in which case the annihilator of (10.1) gives

$$\begin{aligned} \mathrm{ann}(P_x) = \mathrm{ann}(\mathbf {p}_*^{-1}(\mathbf {p}_*(P_x)) + J_x = \mathbf {p}^* (\mathrm{ann}(\mathbf {p}_* P_x)) \oplus J_x . \end{aligned}$$

(10.6)

By the definition of the canonical Pfaffian system,

$$\begin{aligned} E^{[1]}_p = \pi _N^*( \mathrm{ann}(P_x)) \end{aligned}$$

(10.7)

while by Eq. (10.2),

$$\begin{aligned} I^{[1]}_{\mathbf {p}^{[1]}(p)} = \pi _M^*\left( \mathrm{ann}( \mathbf {p}_*(P_x)) \right) . \end{aligned}$$

(10.8)

Applying (10.6) and the commutative diagram (10.3) to Eq. (10.7), and using Eq. (10.8), we arrive at

$$\begin{aligned} E^{[1]}_p&= \pi _N^*(J_x \oplus \mathbf {p}^* (\mathrm{ann}( \mathbf {p}_* P_x)) ) = \pi _N^*(J_x ) \oplus \mathbf {p}^{[1],*} \pi _M^* (\mathrm{ann}( \mathbf {p}_* P_x))\\&= \pi _N^* (J_x) \oplus \mathbf {p}^{[1],*} I^{[1]}_{\mathbf {p}^{[1]}(p)}, \end{aligned}$$

which proves the lemma.$\square$

Proof of Theorem 10.3

To prove that $\mathbf {p}^{[1]}: (\mathcal {E}^{[1]}, N^{[1]}) \rightarrow (\mathcal {I}, M^{[1]})$ is an integrable extension, we must check the three conditions of Eq. (2.3) with $\pi ^*_N(J)$ in place of $J$. By the remark at the beginning of the proof of Lemma 10.1 we have that ${\mathrm{rank}}J = \dim N - \dim M$ and therefore, in view of Corollary 10.2,

$$\begin{aligned} {\mathrm{rank}}\pi _N^*(J) = {\mathrm{rank}}J = \dim N - \dim M = \dim N^{[1]} -\dim M^{[1]}. \end{aligned}$$

(10.9)

This proves condition [i] in (2.3).

To prove the transversality condition [ii] in (2.3), let $Y \in \mathrm{ann}(\pi _N^* J_x) \cap \ker \mathbf {p}^{[1]}_*$. This implies $\pi _{N,*}Y\in \mathrm{ann}(J_x)$, while Lemma 10.4 implies $\pi _{N,*} Y \in \ker \mathbf {p}_*$. By the transversality of the admissible subbundle $J$ for $\mathcal {E}$ (see [ii] in Eq. (2.3)), these two conditions on $\pi _{N,*} Y$ show that $\pi _{N,*} Y = 0$. The fact that $\pi _{N,*}: \ker \mathbf {p}^{[1]}_* \rightarrow \mathbf {p}_*$ is an isomorphism (Lemma 10.4) implies $Y=0$. Thus, $\mathrm{ann}(\pi _N^* J_x) \cap \ker \mathbf {p}^{[1]}_* = 0$ and condition [ii] are verified for the prolongation.

It remains to prove condition [iii] in (2.3), which in this case reads,

$$\begin{aligned} \mathcal {E}^{[1]} = \langle \mathbf {p}^{[1],*} \mathcal {I}^{[1]} + \mathcal {S}( \pi _N^* J) \rangle _{\mathrm{alg}}. \end{aligned}$$

(10.10)

Lemma 10.5 shows that (10.10) holds at the level of 1-forms. Since $\mathcal {E}^{[1]}$ is a Pfaffian system, it remains to be shown that the right-hand side of (10.10) is a Pfaffian system. Since $\mathcal {I}^{[1]}$ is a Pfaffian system, it is then sufficient to verify condition (2.6) for a local basis of $\mathcal {S}( \pi _N^* J)$.

Let $\{\, \xi ^u \,\}$ be a set of 1-forms on $N$ which define a local basis for $\mathcal {S}(J)$. Then, the 1-forms $\sigma ^u= \pi _{N}^* \xi ^u$ define a local basis for $\mathcal {S}(\pi _{N}^*(J))$. Since $d \xi ^u \equiv 0 \mod J + \mathbf {p}^* (I^2)$, we immediately have

$$\begin{aligned} d \sigma ^u&\equiv 0 \mod \pi ^*_N(J) + \pi ^*_N( \mathbf {p}^*( I^2))\\&\equiv 0 \mod \pi ^*_N(J) + \mathbf {p}^{[1],*} ( \pi ^*_M(I^2)). \end{aligned}$$

We remark that it is not generally true that $\pi ^*_M(I^\ell ) \subset I ^{[1], \ell }$. However, for $\ell = 2$, one can readily check that $\pi ^*_M(I^2 ) \subset I ^{[1], 2}$ and therefore

$$\begin{aligned} d \sigma ^u = 0 \mod \pi ^*_N (J) + \mathbf {p}^{[1],*} (I^{[1], 2}). \end{aligned}$$

(10.11)

This shows that the right-hand side of (10.10) is differentially closed. This establishes (10.10), and hence, [iii] in (2.3) is proved.$\square$

Appendix 2: A remark on involutivity

In this Appendix, we establish a simple property (see Corollary 11.3 in particular) of the prolongation of an involutive linear Pfaffian systems which we use in the proof of Theorem 7.5.

Let $\mathcal {I}$ be an involutive linear Pfaffian system on a manifold $M$ with independence condition $\omega \in \Omega ^n(M)$. Then, about each point $x\in M$, there is an open set $U$, a local basis of $1$-forms $\{ \theta ^\alpha \} _{1\le \alpha \le m}$ for $\mathcal {I}$, and a local coframe $\{ \theta ^\alpha , \, \omega ^i, \, \pi ^\epsilon \}_{1\le i \le n, 1 \le \epsilon \le p }$ for $M$ such that $\omega = \omega ^1 \wedge \cdots \wedge \omega ^n$, and (see p. 128 in [7])

$$\begin{aligned} d \theta ^ \alpha = A^\alpha _{\epsilon i} \,\pi ^\epsilon \wedge \omega ^i \quad \mod I . \end{aligned}$$

(11.1)

Let $M^{[1]}$ be the space of integral $n$-planes of $\mathcal {I}$ and let $\mathcal {I}^{[1]}$ be the linear Pfaffian system on $M^{[1]}$ which is the prolongation of $\mathcal {I}$. The prolongation of $\mathcal {I}$ is determined locally as follows. Let $S^\epsilon _{i, v}$be smooth functions on the open set $U$ from above which form a basis for the $t$-dimensional solution space of the linear system

$$\begin{aligned} A^\alpha _{\epsilon i} \Sigma ^\epsilon _{j}\, \omega ^j\wedge \omega ^ i = 0 \end{aligned}$$

(11.2)

in the unknowns $\Sigma ^\epsilon _j$. Then, on the open set $U^{[1]}=U\times \mathbf{R}^t\subset M^{[1]}$ define the $1$-forms

$$\begin{aligned} \tilde{\theta }^ \epsilon = \pi ^\epsilon - s^v S^\epsilon _{i,v} \omega ^i, \end{aligned}$$

(11.3)

where $1\le v \le t$ and $s^v$ are coordinates on $\mathbf{R}^t$. On $U^{[1]}$ the prolongation of $\mathcal {I}$ is then given by

$$\begin{aligned} \mathcal {I}^{[1]}= \langle \theta ^\alpha , \tilde{\theta }^\epsilon \rangle _{\mathrm{diff}}. \end{aligned}$$

(11.4)

Let $\pi :M^{[1]}\rightarrow M$ be the projection map. Then, it is clear from Eq. (11.4) that $\pi ^*(\mathcal {I}^1) \subset \mathcal {I}^{[1]}$ (see also p. 150 in [7]). A form $\eta \in \Lambda ^*(M^{[1]})$ is $\pi$ semi-basic if for all $X \in \ker (\pi _*)$. Equations (11.3) and (11.4) show that the one-forms in $I^{[1],1}$ are $\pi$ semi-basic.

The technical result that we need in Theorem 7.5 is now given by the following.

Theorem 11.1

Let $\mathcal {I}$ be an involutive linear Pfaffian system, and let $S^\epsilon _{i, v}$ be the $t$-dimensional basis for the solution space of (11.2). If $k_\epsilon S^\epsilon _{i, v} = 0$, then $k_\epsilon =0$.

The proof of this theorem is based upon the following lemma.

Lemma 11.2

Let $\mathcal {I}$ be an involutive linear Pfaffian system. If $\eta \in \mathcal {I}^{[1],1}$ is a one-form and $d \eta$ is $\pi$ semi-basic, then $\eta \in \mathcal {S}(\pi ^*( I^1))$.

Proof

First, we note that this condition only needs to be checked point-wise. Therefore, let $x\in M$ and let $U^{[1]}\subset M^{[1]}$ be an open set as chosen above. We now use the condition of involutivity to assume that a coframe on $U$ is chosen so that the tableaux $A^\alpha _{\epsilon i}$ has the form given by equation (90) of Chapter IV in [7]. This means we replace the $1$-forms $\pi ^\epsilon$ by the so-called principal components $\bar{\pi }^a_{i}$. Let $s'_k$ be the last nonzero Cartan character. Then, by involutivity $s'_1+\ldots +s'_k = t$, where $t$ is dimension of the kernel in Eq. (11.2), and we set

$$\begin{aligned} \theta ^a_i = \bar{\pi }^a_i - p^a_{ij} \,\omega ^j,\quad \quad i \le k, \ a \le s_i' , \end{aligned}$$

(11.5)

where $p_{ij}=p_{ji}$. The number of independent functions $p^a_{ij}$ is $t=s'_1+ 2 s'_2+ \ldots + k s'_k$, and these define the fibre coordinates for the projection $\pi :U^{[1]}\rightarrow U$. A local basis for the prolongation of $\mathcal {I}^{[1]}$ is then given by

$$\begin{aligned} \mathcal {I}^{[1]}|_{U^{[1]}} = \langle \theta ^\alpha , \, \theta ^a _ i \rangle _{\mathrm{diff}}, \quad \quad i \le k, \ a \le s_i' , \end{aligned}$$

(11.6)

and a local coframe on $U^{[1]}$ is

$$\begin{aligned} \{\, \theta ^\alpha , \, \theta ^a_i, \omega ^i, \, dp^a_{ij} \, \} , \quad 1\le i \le j \le k, \ a \le s_i' . \end{aligned}$$

(11.7)

Suppose that $\eta$ is a one-form in $\mathcal {I}^{[1]}$ satisfying the conditions in the statement of the lemma. Then, using the local coframe as in Eq. (11.7), there exists smooth functions $T^i_a$ and $R_\alpha$ on $U^{[1]}$ such that

$$\begin{aligned} \eta = T^i_a \, \theta ^a_i + R_\alpha \, \theta ^ \alpha . \end{aligned}$$

(11.8)

In order for $d \eta$ in Eq. (11.8) to be $\pi$ semi-basic at $y\in \pi ^{-1}(x)$, it is necessary and sufficient that

(11.9)

Since $\theta ^a_i$, $\omega ^i$, $\theta ^\alpha$, $d\omega ^i$, $d\theta ^\alpha$ and $d \bar{\pi }^a_i$ are $\pi$ semi-basic, we find using this fact along with Eq. (11.5), that (11.9) simplifies to

(11.10)

However, $\theta ^a_i, \omega ^i$, and $\theta ^\alpha$ are point-wise linearly independent, and so this equation implies $T_b^j(y)=0$, and therefore from the expression for $\eta$ in Eq. (11.8), we have $\eta |_y \in \pi ^*( I_x )$.$\square$

Proof of Theorem 11.1

Suppose that $k_\epsilon S^\epsilon _{i, v}(x) = 0$ at some point $x\in U$ where $k_\epsilon \in \mathbf{R}$. Let $\eta \in \mathcal {I}^{[1],1}$ be the $1$-form given using Eq. (11.4) by

$$\begin{aligned} \eta = k_\epsilon \tilde{\theta }^\epsilon = k_\epsilon ( \pi ^\epsilon - s^v S^\epsilon _{i, v} \omega ^i). \end{aligned}$$

(11.11)

The exterior derivative of $\eta$ evaluated at $y\in \pi ^{-1}(x)$ is then

$$\begin{aligned} d\eta |_y&= k_\epsilon ( d\pi ^\epsilon - d s^v \wedge S^\epsilon _{i, v}\, \omega ^i - s^vdS^\epsilon _{ i, v} \wedge \omega ^i - s^vS^\epsilon _{i, v} \,d \omega ^i )|_y \nonumber \\&= k_\epsilon ( d \pi ^\epsilon - s ^vdS^\epsilon _{i, v} \wedge \omega ^i)|_y. \end{aligned}$$

(11.12)

In this computation, we have, as is customary, identified $S^\epsilon _{i, v}$ with $\pi ^*S^\epsilon _{i, v}$. Therefore, $d\eta |_y$ is $\pi$ semi-basic which by Lemma 11.2 implies $\eta |_y \in \pi ^*(I_x)$. From Eqs. (11.11) and (11.4), this is clearly possible only if $k_\epsilon = 0$. This proves Theorem 11.1.$\square$

The geometric content of Theorem 11.1 is contained in the following corollary.

Corollary 11.3

If $\mathcal {I}$ is an involutive linear Pfaffian system, then ${I^{ [1] }}' =\pi ^*(I ^1 )$.

Proof

Again note that this a point-wise condition. Since $\pi ^*(I)\subset {I^{[1]}}'$, we need only show that if $\eta \in {I^{[1]}}'$ and has the form $\eta =k _\epsilon \tilde{\theta }^\epsilon$ (using the forms in (11.4)), then $\eta =0$. We therefore compute $d \eta \ \mod I^{[1]}$ which is

$$\begin{aligned} d\eta \equiv k_\epsilon ( d \pi ^\epsilon - S^\epsilon _{i,v} d s^v \wedge \omega ^i ) \quad \mod I^{ [1] }. \end{aligned}$$

Since the forms $d\pi ^\epsilon$ are $\pi$ semi-basic, in order for the right-hand side to be zero at any point $y\in U^{[1]}$, it is necessary that $k_\epsilon S^\epsilon _{i,v} = 0$ at $y$. Therefore, by Theorem 11.1, $k_\epsilon =0$ and hence $\eta =0$.$\square$

Appendix 3: On the definition of Darboux integrability

In this appendix, we prove Theorem 4.6. The key step to proving Theorem 4.6 is to show that the decomposability of $\mathcal {I}$, together with conditions [i] of Definition 4.3, implies that is an integrable Pfaffian system. Then, since we are assuming that , we deduce that .

To prove that is an integrable Pfaffian system, we shall need a generalization of the $0$-adapted coframe defined in [2] (p. 1917) or by (4.10). This new coframe is defined locally in a neighbourhood of any given point and is constructed as follows. First choose independent one-forms ${\varvec{\tau }}= \{\,\tau ^1, \tau ^2, \dots ,\tau ^\ell \,\}$ such that

Extend these by vector-valued (independent) 1-forms $\varvec{{\hat{\eta }}}$ and $\varvec{{\check{\eta }}}$ in such manner that

(12.1)

Then, just as in [2], these forms may in turn be extended (by conditions (4.8)) to a local coframe of vector-valued 1-forms $\{\ {\varvec{\theta }},\, \varvec{{\hat{\sigma }}}, \, \varvec{{\check{\sigma }}}, \varvec{{\hat{\eta }}}, \, \varvec{{\check{\eta }}}, \, {\varvec{\tau }}\, \}$ on $M$ such that

(12.2)

We will call such a coframe $0$-adapted. The first step in the proof of Theorem 4.6 is given by the next lemma. In what follows we will use the convention that bold face Roman letters such as $\varvec{a}$, ${\varvec{A}}$, ${\varvec{\alpha }}, \dots$ denote array-valued functions and differential forms of the appropriate rank and dimensions.

Lemma 12.1

If $\{\ {\varvec{\theta }},\, \varvec{{\hat{\sigma }}}, \, \varvec{{\check{\sigma }}}, \varvec{{\hat{\eta }}}, \, \varvec{{\check{\eta }}}, \, {\varvec{\tau }}\, \}$ is a 0-adapted coframe, then the forms ${\varvec{\tau }}$ (which span ) satisfy the structure equations

$$\begin{aligned} d{\varvec{\tau }}= {\varvec{\alpha }}\wedge {\varvec{\tau }}+ \varvec{a}_1\, \varvec{{\check{\eta }}}\wedge \varvec{{\hat{\eta }}}+\varvec{a}_2\, \varvec{{\check{\sigma }}}\wedge \varvec{{\hat{\eta }}}+ \varvec{a}_3\, \varvec{{\check{\eta }}}\wedge \varvec{{\hat{\sigma }}}. \end{aligned}$$

(12.3)

Proof

The conditions and imply, by the Frobenius condition for integrability and Eq. (12.2), that there exists 1-forms ${\varvec{\alpha }}$, ${\varvec{\beta }}$, ${\varvec{\gamma }}$, ${\varvec{\mu }}$, ${\varvec{\nu }}$, ${\varvec{\xi }}$ such that

$$\begin{aligned} d{\varvec{\tau }}= {\varvec{\alpha }}\wedge {\varvec{\tau }}+ {\varvec{\beta }}\wedge \varvec{{\hat{\eta }}}+ {\varvec{\gamma }}\wedge \varvec{{\hat{\sigma }}}\quad \text {and}\quad d{\varvec{\tau }}= {\varvec{\mu }}\wedge {\varvec{\tau }}+ {\varvec{\nu }}\wedge \varvec{{\check{\eta }}}+ {\varvec{\xi }}\wedge \varvec{{\check{\sigma }}}. \end{aligned}$$

(12.4)

Now, the 1-forms ${\varvec{\tau }}\in \mathcal {I}^1 = {\mathrm{span}}\{\, \tilde{\theta }^e\}$, and so, by the decomposability condition (4.3), there are no $\varvec{{\hat{\sigma }}}\wedge \varvec{{\check{\sigma }}}$ terms in either of these structure equations. Hence, after absorbing the ${\varvec{\tau }}$, $\varvec{{\hat{\eta }}}$ terms in ${\varvec{\gamma }}$ and the ${\varvec{\tau }}$, $\varvec{{\check{\eta }}}$ terms in ${\varvec{\xi }}$ into the other coefficients, we may re-write Eq. (12.4) in expanded form as

$$\begin{aligned} \begin{aligned} d{\varvec{\tau }}&= {\varvec{\alpha }}\wedge {\varvec{\tau }}+ {\varvec{\beta }}\wedge \varvec{{\hat{\eta }}}+ (\varvec{c}_1 {\varvec{\theta }}+ \varvec{c}_2 \varvec{{\check{\eta }}}+\varvec{c}_3 \varvec{{\hat{\sigma }}}) \wedge \varvec{{\hat{\sigma }}}\quad \text {and}\\ d{\varvec{\tau }}&= {\varvec{\mu }}\wedge {\varvec{\tau }}+ {\varvec{\nu }}\wedge \varvec{{\check{\eta }}}+ (\varvec{C}_1 {\varvec{\theta }}+\varvec{C}_2 \varvec{{\hat{\eta }}}+\varvec{C}_3 \varvec{{\check{\sigma }}}) \wedge \varvec{{\check{\sigma }}}, \end{aligned} \end{aligned}$$

(12.5)

where the $\varvec{c}_i$ and $\varvec{C}_i$ are $3$-dimensional arrays of locally defined smooth functions. For the right-hand sides of these equations to be equal, we find immediately that $\varvec{c}_1=\varvec{c}_3=\varvec{C}_1=\varvec{C}_3=0$. Upon absorbing the ${\varvec{\tau }}$ terms in ${\varvec{\beta }}$ and ${\varvec{\nu }}$ into ${\varvec{\alpha }}$ and ${\varvec{\mu }}$, we can re-write Eq. (12.5) in expanded form as

$$\begin{aligned} \begin{aligned} d{\varvec{\tau }}&= {\varvec{\alpha }}\wedge {\varvec{\tau }}+ (\varvec{b}_1 {\varvec{\theta }}+\varvec{b}_2 \varvec{{\hat{\eta }}}+\varvec{b}_3\varvec{{\check{\eta }}}+\varvec{b}_4 \varvec{{\check{\sigma }}}) \wedge \varvec{{\hat{\eta }}}+ \varvec{c}_2 \varvec{{\check{\eta }}}\wedge \varvec{{\hat{\sigma }}}, \quad \text {and}\\ d{\varvec{\tau }}&= {\varvec{\mu }}\wedge {\varvec{\tau }}+ (\varvec{B}_1 {\varvec{\theta }}+\varvec{B}_2 \varvec{{\hat{\eta }}}+\varvec{B}_3\varvec{{\check{\eta }}}+\varvec{B}_4 \varvec{{\hat{\sigma }}}) \wedge \varvec{{\check{\eta }}}+ \varvec{C}_2\varvec{{\hat{\eta }}}\wedge \varvec{{\check{\sigma }}}. \end{aligned} \end{aligned}$$

(12.6)

For the right-hand sides of these equations to be equal, we further find that $\varvec{b}_1= \varvec{b}_2=\varvec{B}_1=\varvec{B}_3=0$ in which case Eq. (12.6) reduces to (12.3), as required.$\square$

To complete the proof of Theorem 4.6, we must show that the coefficients $\varvec{a}_1$, $\varvec{a}_2$ and $\varvec{a}_3$ in (12.3) vanish, and to this end, we shall need a refined coframe, generalizing the 1-adapted coframe of [2].

Lemma 12.2

Let $\{\ {\varvec{\theta }},\, \varvec{{\hat{\sigma }}}, \, \varvec{{\check{\sigma }}},\, \varvec{{\hat{\eta }}}, \, \varvec{{\check{\eta }}}, \, {\varvec{\tau }}\, \}$ be a 0-adapted coframe. There exists vector-valued functions $\varvec{I}_1,\varvec{I}_2,\varvec{I}_3$, and $\varvec{J}_1,\varvec{J}_2,\varvec{J}_3$ and matrix-valued functions $\varvec{E}$ and $\varvec{F}$ such that the forms

$$\begin{aligned} \varvec{{\hat{\sigma }}}_0 = d \varvec{I}_3, \quad \varvec{{\hat{\eta }}}_0 = d\varvec{I}_2 + \varvec{E}\varvec{{\hat{\sigma }}}_0 , \quad \varvec{{\check{\sigma }}}_0 = d \varvec{J}_3 \quad \varvec{{\check{\eta }}}_0 = d \varvec{J}_2+\varvec{F}\varvec{{\check{\sigma }}}_0 \end{aligned}$$

(12.7)

may be used to define a 0-adapted coframe $\{\, {\varvec{\theta }}, \, \varvec{{\hat{\eta }}}_0, \, \varvec{{\check{\eta }}}_0, \, \varvec{{\hat{\sigma }}}_0,\, \varvec{{\check{\sigma }}}_0, \, {\varvec{\tau }}\, \}$. Moreover, the 1-forms ${\varvec{\tau }}$ can be expressed as

$$\begin{aligned} {\varvec{\tau }}= \varvec{R}\,( d\varvec{I}_1 + \varvec{S}\varvec{{\hat{\eta }}}_0+ \varvec{T}\varvec{{\hat{\sigma }}}_0) = {\varvec{A}}\, ( d\varvec{J}_1 + \varvec{B}\varvec{{\check{\eta }}}_0+ \varvec{C}\varvec{{\check{\sigma }}}_0), \end{aligned}$$

(12.8)

where $\varvec{R}, {\varvec{A}}, \varvec{S}, \dots$ are smooth matrix-valued functions. The matrices $\varvec{R}, {\varvec{A}}$ are invertible matrices of dimension .

Proof

We shall use the following simple observation to choose the functions $\varvec{I}$ and $\varvec{J}$—if $K = \{\ \omega ^1,\, \omega ^2,\,\dots , \,\omega ^k\ \}$ is any completely integrable Pfaffian (where the $\omega ^i$ are independent 1-forms), then there are (locally defined) functions whose differentials will complete any given subset of the generators $\{\omega ^{i_1},\dots , \omega ^{i_p}\}$ to a basis of $K$.

Accordingly, choose independent functions $\varvec{I}_3$ such that and let $\varvec{{\hat{\sigma }}}_0 = d\varvec{I}_3$. The first equation in (12.7) therefore holds. Then choose independent functions $\varvec{I}_2$ such that . The 1-forms $\varvec{{\hat{\eta }}}$ can therefore be written as

$$\begin{aligned} \varvec{{\hat{\eta }}}= \varvec{P}d\varvec{I}_2 + \varvec{G}{\varvec{\tau }}+ \varvec{H}\varvec{{\hat{\sigma }}}_0, \end{aligned}$$

where $\varvec{P}$ is an invertible matrix of functions. Let

$$\begin{aligned} \varvec{{\hat{\eta }}}_0 = \varvec{P}^{-1}(\varvec{{\hat{\eta }}}- \varvec{G}{\varvec{\tau }}) = d\varvec{I}_2 + \varvec{E}\varvec{{\hat{\sigma }}}\end{aligned}$$

so that the second equation in (12.7) holds. Note, by (12.1), that and . Finally, if we choose $\varvec{I}_1$ such that , then the first equation in (12.8) holds. Similar arguments allow us to choose $\varvec{J}_1,\varvec{J}_2,\varvec{J}_3$ so that the remaining Eqs. (12.7) and (12.8) hold.$\square$

Remark 12.3

The number of invariants $\varvec{I}_1$ and $\varvec{J}_1$ are the same and equals the rank of the bundle .

Lemma 12.4

For the coframe $\{\,{\varvec{\theta }}, \varvec{{\hat{\sigma }}}_0,\varvec{{\check{\sigma }}}_0, \varvec{{\hat{\eta }}}_0,\varvec{{\check{\eta }}}_0, {\varvec{\tau }}\, \}$ we have structure equations

$$\begin{aligned} d \varvec{{\hat{\eta }}}_0 = \varvec{a}\, {\varvec{\tau }}\wedge \varvec{{\hat{\sigma }}}_0 + \varvec{b}\, \varvec{{\hat{\eta }}}_0 \wedge \varvec{{\hat{\sigma }}}_0 + \varvec{c}\, \varvec{{\hat{\sigma }}}_0 \wedge \varvec{{\hat{\sigma }}}_0 . \end{aligned}$$

(12.9)

Proof

Equation (12.7) in Lemma 12.2 gives

$$\begin{aligned} d \varvec{{\hat{\eta }}}_0 = d\varvec{E}\wedge \varvec{{\hat{\sigma }}}_0 . \end{aligned}$$

But, by definition, the 1-forms $\varvec{{\hat{\eta }}}_0$ belong to $\mathcal {I}^1$, and therefore, by the decomposability condition (4.3), the 1-forms $d \varvec{E}$ contain no $\varvec{{\check{\sigma }}}_0$ terms and hence . Therefore, (see Lemma 4.5), and since , we have $d \varvec{E}= \varvec{a}{\varvec{\tau }}+ \varvec{b}\varvec{{\hat{\eta }}}_0 + \varvec{c}\varvec{{\hat{\sigma }}}_0$ and (12.9) holds.$\square$

Proof of Theorem 4.6

Equation (12.3) holds for any 0-adapted coframe; in particular, it holds for the 0-adapted coframe $\{\, {\varvec{\theta }}, \varvec{{\hat{\sigma }}}_0, \varvec{{\check{\sigma }}}_0, \varvec{{\hat{\eta }}}_0,\varvec{{\check{\eta }}}_0, {\varvec{\tau }}\, \}$ constructed in Lemma 12.2. We now compare Eq. (12.3) with the result of taking the exterior derivative of Eqs. (12.8) and utilizing (12.7) and (12.9),

$$\begin{aligned} d {\varvec{\tau }}&= (d\varvec{R})\varvec{R}^{-1} \wedge {\varvec{\tau }}+ \varvec{R}(\ d\varvec{S}\wedge \varvec{{\hat{\eta }}}_0+\varvec{S}d\varvec{{\hat{\eta }}}_0 + d \varvec{T}\wedge \varvec{{\hat{\sigma }}}_0\ )\nonumber \\&= (d\varvec{R})\varvec{R}^{-1} \,{\wedge }\, {\varvec{\tau }}\!+\! \varvec{R}(\ d\varvec{S}\,{\wedge }\, \varvec{{\hat{\eta }}}_0\!+\!\varvec{S}(\varvec{a}{\varvec{\tau }}\,{\wedge }\, \varvec{{\hat{\sigma }}}_0 \!+\! \varvec{b}\varvec{{\hat{\eta }}}_0 \,{\wedge }\, \varvec{{\hat{\sigma }}}_0 \!+\! \varvec{c}\varvec{{\hat{\sigma }}}_0 \,{\wedge }\, \varvec{{\hat{\sigma }}}_0) \!+\! d \varvec{T}\,{\wedge }\, \varvec{{\hat{\sigma }}}_0\ ).\nonumber \\ \end{aligned}$$

(12.10)

Now, decomposability implies $d \varvec{T}$ is independent of the forms $\varvec{{\check{\sigma }}}_0$. Therefore, . Hence, from (12.10), we deduce that there are no terms of the form $\varvec{{\check{\eta }}}_0 \wedge \varvec{{\hat{\sigma }}}_0$ in $d {\varvec{\tau }}$. Therefore, $\varvec{a}_3=0$ in Eq. (12.3) (in the current 0-adapted coframe). A similar argument using the expression for ${\varvec{\tau }}$ in terms of the invariants $\varvec{J}$ gives $\varvec{a}_2=0$ in Eq. (12.3), leaving

$$\begin{aligned} d{\varvec{\tau }}= {\varvec{\alpha }}\wedge {\varvec{\tau }}+\varvec{a}_1 \varvec{{\check{\eta }}}_0 \wedge \varvec{{\hat{\eta }}}_0 . \end{aligned}$$

(12.11)

Returning to Eq. (12.10), we now conclude, from the absence of the terms $\varvec{{\hat{\eta }}}_0 \wedge \varvec{{\check{\sigma }}}_0$ in (12.11), that $d\varvec{S}$ is free of $\varvec{{\check{\sigma }}}_0$ terms and so, again by Lemma 4.5, . Consequently, Eq. (12.10) implies that $d {\varvec{\tau }}$ contains no terms $\varvec{{\hat{\eta }}}_0\wedge \varvec{{\check{\eta }}}_0$ and therefore $\varvec{a}_1=0$ in Eq. (12.11). This proves that is an integrable Pfaffian system which, by the remarks made at the beginning of this appendix, shows that .$\square$

Appendix 4: Infinitesimal equivariant mappings

In this appendix, we prove the following theorem which is required for the proof of Theorem 8.2.

Theorem 13.1

Let $H$ be a connected Lie group acting freely and regularly on $N$. Let $G$ be a Lie group acting freely and regularly on $M$ and let $\Gamma _H$ and $\Gamma _G$ be the corresponding Lie algebras of infinitesimal generators. Suppose $\Phi :N \rightarrow M$ is a smooth map whose differential satisfies

$$\begin{aligned} \Phi _* : \Gamma _H \rightarrow \Gamma _G \end{aligned}$$

(13.1)

and is a monomorphism of the Lie algebras. Then there exists a unique homomorphism $\phi :H \rightarrow G$ such that $\Phi$ is $H$ equivariant, that is, $\Phi (h \cdot p) = \phi (h) \cdot \Phi (p)$ for all $p\in N$ and $h \in G$. Moreover $\phi _*:{\mathfrak h}\rightarrow \mathfrak g$ is a Lie algebra monomorphism.

Proof

Write $\mu : H \times N \rightarrow N$ and $\tau : G \times M \rightarrow M$ for the actions of $H$ on $N$ and $G$ on $M$ and fix a point $p\in N$. Then, by the freeness and regularity hypothesis for the actions of $H$ and $G$, the maps $\mu _p: H \rightarrow N$ and $\tau _{\Phi (p)}:G \rightarrow M$ given by (see (2.2))

$$\begin{aligned} \mu _p(h) = h\cdot p \quad \text {and} \quad \tau _{\Phi (p)}(g) = g \cdot \Phi (p) \end{aligned}$$

(13.2)

are imbeddings. Equation (13.1) implies $\Phi \circ \mu _p :H \rightarrow M$ satisfies

$$\begin{aligned} (\Phi \circ \mu _p)_* TH \subset \varvec{\Gamma }_G. \end{aligned}$$

(13.3)

Since $H$ is connected, Eq. (13.3) shows that $\Phi (\mu _p(H))$ is contained in the maximal integral manifold of the integrable distribution $\varvec{\Gamma }_G$ through $\Phi (p)$. But the maximal integral manifold of $\Gamma _G$ through $\Phi (p)$ is just the connected component of the orbit of $G$ through $\Phi (p)$ and therefore

$$\begin{aligned} \Phi (\mu (H)) \subset G\cdot \Phi (p). \end{aligned}$$

(13.4)

Equation (13.4), together with the fact that the action of $G$ is free on $M$, implies there exists a unique function $\phi : H \rightarrow G$ such that the following diagram commutes

(13.5)

Since the maps $\mu _p:H\rightarrow N$ and $\tau _{\Phi (p)}:G \rightarrow M$ are imbeddings, the function $\Phi$ restricts to a smooth map from the orbit of $H$ through $p$ to the orbit of $G$ though $\Phi (p)$. This implies that $\phi :H \rightarrow G$ is also smooth.

To prove the theorem, we will prove that the map $\phi :H \rightarrow G$ defined in diagram (13.5) is a Lie group homomorphism with injective differential. We also show that the definition of $\phi :H \rightarrow G$ is independent of the choice of the point $p$.

From the commutative diagram (13.5), we have $\Phi (h \cdot p) = \phi (h) \cdot \Phi (p)$ which, on account of the freeness of the action of $G$, implies that $\phi (e) = \epsilon$, where $e$ is the identity element of $H$ and $\epsilon$ is the identity element of $G$. By taking differentials of the maps in (13.5) and applying the chain rule, we therefore obtain the following commutative diagram

(13.6)

Recall that ${\mathfrak h}=T_{e}H$ and $\mathfrak g=T_{\epsilon }G$ are the Lie algebras of $H$ and $G$ defined by brackets of right-invariant vector fields and that the maps $\mu$ and $\tau$ induce Lie algebra isomorphisms $\rho :{\mathfrak h}\rightarrow \Gamma _H$ and $\lambda : \mathfrak g\rightarrow \Gamma _G$ by $\rho (Z_{e}) = \mu _{x*}( Z_{e})$ and $\lambda (W_{\epsilon }) = \mu _{y*} (W_{\epsilon })$ (see (2.12)). Hence, there is a uniquely defined Lie algebra monomorphism $\psi : {\mathfrak h}\rightarrow \mathfrak g$ such that the following diagram

(13.7)

commutes. The combination of the diagrams (13.6) and (13.7) yields

$$\begin{aligned} \tau _{\Phi (p), *} (\phi _{*}(Z_e))&= \Phi _{*}(\mu _{p, *}(Z_e)) = \Phi _{*}(\rho ((Z_e)_p) = (\Phi _*(\rho (Z_e))(p)\\&= \lambda (\psi (Z_e))(\Phi (p)) = \tau _{\Phi (p), *}(\psi (Z_e)) \end{aligned}$$

and therefore, by the injectivity of $\tau _{\Phi (p), *}$, we have $\phi _{*}(Z_e) = \psi (Z_e)$ and therefore $\phi _{*} = \psi$. This proves that $\phi _{*} : T_eH \rightarrow T_\epsilon G$ is a Lie algebra monomorphism. It then follows from Exercise 9, p. 134 of [27] that $\phi :H \rightarrow G$ is a homomorphism (with injective differential). Moreover, since the map $\psi$ does not depend upon the point $p$, then $\phi _{*}(Z_e) = \psi (Z_e)$ does not depend upon $p$. By the uniqueness theorem for homomorphisms (Theorem 3.16, p. 92 [27]), the homomorphism $\phi$ does not depend on $p$.$\square$