Cinf-structures to integrate involutive distributions

My collaborators and I have recently published two papers (this one and this other) in which  we develop a method to obtain the integral manifolds of involutive distributions. Exploring the integral manifolds of involutive distributions contributes to a broader understanding within differential geometry, an area with ties to many other mathematical branches. Furthermore, the study of these distributions and manifolds holds relevance in physics, especially within classical mechanics and field theory. A deeper grasp of these mathematical constructs can be beneficial for ongoing research in both mathematics and physics.

 Given a distribution, for example \(\mathcal{Z}=\{Z_1,Z_2\}\) in \(\mathbb R^n\), the idea of our work is to complete it with a sequence of \(n-2\) vector fields \(Y_1, Y_2,Y_3,\ldots\) in such a way that

- \(Y_1\) is a \(\mathcal{C}^{\infty}\)-symmetry of \(\mathcal{Z}\).
- \(Y_2\) is a \(\mathcal{C}^{\infty}\)-symmetry of \(\mathcal{Z}\oplus \{X_1\}\).
- \(Y_3\) is a \(\mathcal{C}^{\infty}\)-symmetry of \(\mathcal{Z}\oplus \{X_1,X_2\}\),
- and so on. 

The sequence is called a \(\mathcal{C}^{\infty}\)-structure for \(\mathcal{Z}\).

But, what is a \(\mathcal{C}^{\infty}\)-symmetry? Certainly, it is not a symmetry of distribution, but retains some of its characteristics. It is known that the flow of a symmetry of distributions, let's call it \(X_1\), carry integral manifolds of \(\mathcal{Z}\) into integral manifolds. 

 

(Here the integral manifolds of \(\mathcal{Z}\) are represented as curves, but represent higher dimensional manifolds.)

Well, a \(\mathcal{C}^{\infty}\)-symmetry doesn't do this, but at least its flow strings together an entire family of integral manifolds.


The following \(\mathcal{C}^{\infty}\)-symmetry in the sequence of a \(\mathcal{C}^{\infty}\)-structure would look like this:

where \(X_2\) would be a symmetry.

Here we proved, as a theoretical curiosity, that every \(\mathcal{C}^{\infty}\)-symmetry can be "corrected" to an autentic symmetry, but the process is by no means straightforward, and is not necessary for our method to work.

More specifically, a \(\mathcal{C}^{\infty}\)-symmetry of a distribution \(\mathcal{Z}\) is a vector field \(Y\) such that \(\mathcal{Z}\oplus \{Y\}\) is involutive. I.e., instead of the condition \([Y,\mathcal{Z}] \in \mathcal{Z}\), which characterizes symmetries, we now have 

\[[Y,\mathcal{Z}] \in \mathcal{Z}\oplus\{Y\}.\]

If the sequence of vector fields \(Y_1, Y_2,\ldots\) were made of proper symmetries instead of \(\mathcal{C}^{\infty}\)-symmetries, then we would have the important notion of solvable structure instead of a \(\mathcal{C}^{\infty}\)-structure. The key advantage of \(\mathcal{C}^{\infty}\)-structures is that \(\mathcal{C}^{\infty}\)-symmetries are easier to find than symmetries, and even if the process to obtain the integral manifolds of \(\mathcal{Z}\) have more obstacles than with solvable structures, in many cases it reaches a successful conclusion.

Let's show the method with a trivial example. Consider the distribution in \(\mathbb R^4\) given by the vector fields

\[
\begin{array}{rl}
    Z_1=&x_4 \partial_{x_1}+(x_2-x_3 x_4) \partial_{x_2}-x_3 \partial_{x_3}+x_4 \partial_{x_4},\\
    Z_2=&\partial_{x_1}+x_3 \partial_{x_2}+\partial_{x_4},
\end{array}
\]
which is involutive since \( [Z_1,Z_2]=-Z_2 \).

It can be easily checked that the vector fields
\[
\begin{array}{rl}
    Y_1=&x_4 \partial_{x_2}+\partial_{x_3}\\
    Y_2=&\partial_{x_4}
\end{array}
\]
constitute a \(\mathcal{C}^{\infty}\)-structure for \( \mathcal{Z} \), since
\[
\begin{array}{rl}
    [Y_1,Z_1]=&-Y_1,\\
    [Y_1,Z_2]=&0.\\
\end{array}
\]

Observe that we don't worry about the last vector field in a \(\mathcal{C}^{\infty}\)-structure, since any vector field linearly independent (pointwise) satisfies the condition.


We construct the following 1-forms:
\[
\begin{array}{rl}
    \omega_1=& \det(Z_1,Z_2,-,Y_2)=\\
                   =&-x^3 dx_1+x_3 dx_2+(x_2-2x_3 x_4)dx_3,\\
    \omega_2=& \det(Z_1,Z_2,Y_1,-)=\\
                  =&(x_2-x_3 x_4)dx_1-(x_2-x_3 x_4) dx_4.
\end{array}
\]
Then, since the distribution determined by \(\{Z_1,Z_2,Y_1\}\) is involutive, the 1-form \(\omega_2\) is Frobenius integrable, so we solve the completely integrable Pfaffian equation
\[
\omega_2\equiv 0,
\]
\[
(x_2-x_3 x_4)(-dx_1+ dx_4)\equiv 0,
\]
obtaining \( I_2(x_1,x_2,x_3,x_4)=-x_1+x_4 \). 

The submanifolds
\[
\Sigma_{(C_2)}=\{x\in \mathbb{R}^4:-x_1+x_4=C_2\}
\]
contains the integral manifolds of \(\mathcal{Z}\) so we restrict to them to continue our search. They are parametrized by
\[
\begin{array}{rccc}
\iota_2:& N_1 & \to & \mathbb{R}^4\\
        &(x_1,x_2,x_3) & \mapsto &(x_1,x_2,x_3,C_2+x_1)
\end{array}
\]
being \( N_1=\mathbb{R}^3 \). By pulling back with \( \iota_2 \) we obtain the 1-forms \( \langle \omega_1|_{\Sigma_{(C_2)}},\omega_2|_{\Sigma_{(C_2)}} \rangle \) defined on \( \Sigma_{(C_2)} \). We have that
\[
\omega_2|_{\Sigma_{(C_2)}}=0
\]
and
\[
\omega_1|_{\Sigma_{(C_2)}}=\iota_2^*(\omega_1)=-x_3^2 dx_1+x_3dx_2-(2 x_3C_2+2x_3x_1-x_2)dx_3.
\]

The Pfaffian equation \( \omega_1|_{\Sigma_{(C_2)}} \equiv 0 \) is also completely integrable, and a solution is given by
\[
I_1(x_1,x_2,x_3;C_2)=C2x_3^2+x_1 x_3^2-x_2 x_3.
\]

The submanifolds
\[
\Sigma_{(C_1,C_2)}=\{x\in N_1:I_1(x_1,x_2,x_3;C_2)=C_1\}
\]
are indeed the integral submanifold we are looking for. They are parametrized by
\[
\begin{array}{rccc}
\iota_1:& N_0 & \to & N_1\\
   & (x_1,x_2)& \mapsto & \left(x_1,x_2,\frac{x_2+\sqrt{4C_1C_2+4C_1 x_1+x_2^2}}{2 (C_2+x_1)} \right)
\end{array}
\]
being now the domain \( N_0 \) an open set of \( \mathbb{R}^2 \) which depends on \( C_1 \) and \( C_2 \) in such a way that \( \iota_1 \) is well defined.

This way, an explicit expression for the integral manifolds of \( \mathcal{Z} \) is
\[
\iota_2\circ \iota_1 : (x_1,x_2) \mapsto \left( x_1,x_2,\frac{x_2+\sqrt{4C_1C_2+4C_1 x_1+x_2^2}}{2 (C_2+x_1)}, C_2+x_1\right).
\]
The constants \( C_1, C_2 \) let us prescribe the point \( p\in \mathbb{R}^4 \) to be crossed by the integral manifold.
 

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