Editing Jet bundle (section)

===Example===
Consider the case ''(E, π, M)'', where ''E'' ≅ '''R'''<sup>2</sup> and ''M'' ≃ '''R'''. Then, ''(J<sup>1</sup>(π), π, E)'' defines the first jet bundle, and may be coordinated by ''(x, u, u<sub>1</sub>)'', where

:<math>\begin{align}
    x(j^1_{p}\sigma) &= x(p) = x  \\
    u(j^1_{p}\sigma) &= u(\sigma(p)) = u(\sigma(x)) = \sigma(x) \\
  u_1(j^1_{p}\sigma) &=  \left.\frac{\partial \sigma}{\partial x}\right|_{p} = \dot{\sigma}(x)
\end{align}</math>

for all ''p'' ∈ ''M'' and ''σ'' in Γ<sub>''p''</sub>(''π''). A contact form on ''J<sup>1</sup>(π)'' has the form

:<math>\theta = du - u_1 dx</math>

Consider a vector ''V'' on ''E'', having the form

:<math>V = x \frac{\partial}{\partial u} - u \frac{\partial}{\partial x}</math>

Then, the first prolongation of this vector field to ''J<sup>1</sup>(π)'' is

:<math>\begin{align}
  V^1 &= V + Z \\
      &= x \frac{\partial}{\partial u} - u \frac{\partial}{\partial x} + Z \\
      &= x \frac{\partial}{\partial u} - u \frac{\partial}{\partial x} + \rho(x, u, u_1) \frac{\partial}{\partial u_1}
\end{align}</math>

If we now take the Lie derivative of the contact form with respect to this prolonged vector field, <math>\mathcal{L}_{V^1}(\theta),</math> we obtain

:<math>\begin{align}
  \mathcal{L}_{V^1}(\theta)
    &= \mathcal{L}_{V^1}(du - u_1dx) \\
    &= \mathcal{L}_{V^1}du - \left(\mathcal{L}_{V^1}u_1\right)dx - u_1\left(\mathcal{L}_{V^1}dx\right) \\
    &= d\left(V^1u\right) - V^1 u_1 dx - u_1 d\left(V^1x\right) \\
    &= dx - \rho(x, u, u_1)dx + u_1 du \\
    &= (1 - \rho(x, u, u_1))dx + u_1 du \\
    &= [1 - \rho(x, u, u_1)]dx + u_1(\theta + u_1 dx) && du = \theta + u_1 dx \\
    &= [1 + u_1u_1 - \rho(x, u, u_1)]dx + u_1\theta  
\end{align}</math>

Hence, for preservation of the contact ideal, we require

:<math>1 + u_1 u_1 - \rho(x, u, u_1) = 0 \quad \Leftrightarrow \quad \rho(x, u, u_1) = 1 + u_1 u_1.</math>

And so the first prolongation of ''V'' to a vector field on ''J<sup>1</sup>(π)'' is

:<math>V^1 = x \frac{\partial}{\partial u} - u \frac{\partial}{\partial x} + (1 + u_1u_1)\frac{\partial}{\partial u_1}.</math>

Let us also calculate the second prolongation of ''V'' to a vector field on ''J<sup>2</sup>(π)''. We have <math>\{x, u, u_1, u_2\}</math> as coordinates on ''J<sup>2</sup>(π)''. Hence, the prolonged vector has the form

:<math> V^2 = x \frac{\partial}{\partial u} - u \frac{\partial}{\partial x} + \rho(x, u, u_1, u_2)\frac{\partial}{\partial u_1} + \phi(x, u, u_1, u_2)\frac{\partial}{\partial u_2}.</math>

The contact forms are

:<math>\begin{align}
    \theta &= du - u_1dx \\
  \theta_1 &= du_1 - u_2dx
\end{align}</math>

To preserve the contact ideal, we require

:<math>\begin{align}
    \mathcal{L}_{V^2}(\theta) &= 0 \\
  \mathcal{L}_{V^2}(\theta_1) &= 0
\end{align}</math>

Now, ''θ'' has no ''u''<sub>2</sub> dependency. Hence, from this equation we will pick up the formula for ''ρ'', which will necessarily be the same result as we found for ''V<sup>1</sup>''. Therefore, the problem is analogous to prolonging the vector field ''V<sup>1</sup>'' to ''J''<sup>2</sup>(π). That is to say, we may generate the ''r''-th prolongation of a vector field by recursively applying the Lie derivative of the contact forms with respect to the prolonged vector fields, ''r'' times. So, we have

:<math>\rho(x, u, u_1) = 1 + u_1 u_1</math>

and so

:<math>\begin{align}
  V^2 &= V^1 + \phi(x, u, u_1, u_2)\frac{\partial}{\partial u_2} \\
      &= x \frac{\partial}{\partial u} - u \frac{\partial}{\partial x} + (1 + u_1 u_1)\frac{\partial}{\partial u_1} + \phi(x, u, u_1, u_2)\frac{\partial}{\partial u_2}
\end{align}</math>

Therefore, the Lie derivative of the second contact form with respect to ''V<sup>2</sup>'' is

:<math>\begin{align}
  \mathcal{L}_{V^2}(\theta_1)
    &= \mathcal{L}_{V^2}(du_1 - u_2dx) \\
    &=  \mathcal{L}_{V^2}du_1 - \left(\mathcal{L}_{V^2}u_2\right)dx - u_2\left(\mathcal{L}_{V^2}dx\right) \\
    &= d(V^2 u_1) - V^2u_2dx - u_2d(V^2x) \\
    &= d(1 + u_1 u_1) - \phi(x, u, u_1, u_2)dx + u_2du \\
    &= 2u_1du_1 - \phi(x, u, u_1, u_2)dx + u_2du \\
    &= 2u_1du_1 - \phi(x, u, u_1, u_2)dx + u_2 (\theta + u_1dx)              &   du &= \theta + u_1 dx \\
    &= 2u_1(\theta_1 + u_2dx) - \phi(x, u, u_1, u_2)dx + u_2(\theta + u_1dx) & du_1 &= \theta_1 + u_2 dx \\
    &= [3u_1u_2 - \phi(x, u, u_1, u_2)]dx + u_2\theta + 2u_1\theta_1 
\end{align}</math>

Hence, for <math>\mathcal{L}_{V^2}(\theta_1)</math> to preserve the contact ideal, we require

:<math>3u_1 u_2 - \phi(x, u, u_1, u_2) = 0 \quad \Leftrightarrow \quad \phi(x, u, u_1, u_2) = 3u_1 u_2.</math>

And so the second prolongation of ''V'' to a vector field on ''J''<sup>2</sup>(π) is

:<math> V^2 = x \frac{\partial}{\partial u} - u \frac{\partial}{\partial x} + (1 + u_1 u_1)\frac{\partial}{\partial u_1} + 3u_1 u_2\frac{\partial}{\partial u_2}.</math>

Note that the first prolongation of ''V'' can be recovered by omitting the second derivative terms in ''V<sup>2</sup>'', or by projecting back to ''J<sup>1</sup>(π)''.