${\mathrm{\Θ}}^{\mathrm{\α}} \= {\mathrm{du}}^{\mathrm{\α}}-u_{\mathrm{\ℓ}}^{\mathrm{\α}}{\mathrm{dx}}^{\mathrm{\ℓ}}\,$ ${\mathrm{\Θ}}_i^{\mathrm{\α}} \= {\mathrm{du}}_i^{\mathrm{\α}}-u_{i\mathrm{\ℓ}}^{\mathrm{\α}}{\mathrm{dx}}^{\mathrm{\ℓ}\mathit{ }}\, .... \,$ ${\mathrm{\Θ}}_{\mathrm{ij}\cdot \cdot \cdot k}^{\mathrm{\α}} \= {\mathrm{du}}_{\mathrm{ij}\cdot \cdot \cdot k}^{\mathrm{\α}}-u_{\mathrm{ij}\cdot \cdot \cdot \mathrm{k\ℓ}}^{\mathrm{\α}} {\mathrm{dx}}^{\mathrm{\ℓ}} \, ....$ .

Note that exterior derivatives of the contact 1-forms are

${\mathrm{d\Θ}}^{\mathrm{\α}} \= {\mathrm{dx}}^{\mathrm{\ℓ}} \wedge {\mathrm{\Θ}}_{\mathrm{\ℓ}\mathit{ }}^{\mathrm{\alpha }}\,$ d${\mathrm{\Θ}}_i^{\mathrm{\α}} \= {\mathrm{dx}}^{\ell } \wedge {\mathrm{\Theta }}_{\mathrm{i\ℓ}\mathit{ }}^{\mathrm{\α}}\, .... \,$ ${\mathrm{d\Θ}}_{\mathrm{ij}\cdot \cdot \cdot k}^{\mathrm{\α}} \= {\mathrm{dx}}^{\ell } \wedge {\mathrm{\Theta }}_{\mathrm{ij}\cdot \cdot \cdot \mathrm{k\ℓ}\mathit{ }}^{\mathrm{\α}}\.$

A differential $p-$form $\mathrm{\ω} \in {\mathrm{\Ω}}^p\left(J^{\mathrm{\∞}}\right)$ is called a bi-form of degree $\left(r\,s\right)$ if  it is a sum of wedge products of  $r 1$-forms on $M$ and $s$contact 1-forms, that is,

$\mathrm{\ω} \= A_{i_1{i}_2\cdot \cdot \cdot i_r a_1 \cdot \cdot \cdot a_s}^{ }{\mathrm{dx}}^{i_1}\wedge {\mathrm{dx}}^{i_2 }\wedge \cdot \cdot \cdot \wedge {\mathrm{dx}}^{i_r} \wedge C^{a_1}\wedge C^{a_2} \cdot \cdot \cdot \wedge C^{a_s}\,$where each $C^{a_k}$ is a contact 1-form.

The space of all $p$-forms then decomposes as a direct sum of bi-forms

${\mathrm{\Ω}}^p\left(J^{\mathrm{\∞}} \right) \= \underset{r\+s \=p}{\overset{}{⨁}} {\mathrm{\Ω}}^{\left(r\,s\right)}\left(J^{\mathrm{\∞}}\left(E\right)\right)$

The above formulas for the exterior derivative of the contact forms shows that $d\:{\mathrm{\Omega }}^{\left(r\,s\right)}\left(J^{\infty }\left(E\right)\right)\to {\mathrm{\Omega }}^{\left(r\+1\,s\right)}\left(J^{\infty }\left(E\right)\right) \oplus {\mathrm{\Omega }}^{\left(r\,s\+1\right)}\left(J^{\infty }\left(E\right)\right)$and therefore $d \= d_{H } \+ d_V\,$  where

 $d_{H }\:{\mathrm{\Omega }}^{\left(r\,s\right)}\left(J^{\infty }\left(E\right)\right)\to {\mathrm{\Omega }}^{\left(r\+1\,s\right)}\left(J^{\infty }\left(E\right)\right)$ and   $d_{V }\:{\mathrm{\Omega }}^{\left(r\,s\right)}\left(J^{\infty }\left(E\right)\right)\to {\mathrm{\Omega }}^{\left(r\,s\+1\right)}\left(J^{\infty }\left(E\right)\right)\.$

The differential operator $d_{H }$is called the horizontal exterior derivative and the differential operator $d_V$is called the vertical exterior derivative. One has that

$d_H\circ d_H \=0\, d_H\circ d_V \+ d_V\circ d_H \=0\,$and  $d_V\circ d_V \=0$.

The coordinate formulas for the horizontal exterior derivative are

$d_H\left(x^i\right) \= {\mathrm{dx}}^{i }\, d_H\left(u_{\mathrm{ij} \cdot \cdot \cdot k}^{\mathrm{\alpha }}\right) \= u_{\mathrm{ij} \cdot \cdot \cdot k\mathrm{\ℓ}}^{\mathrm{\α}} {\mathrm{dx}}^{\mathrm{\ℓ}}\, d_H\left({\mathrm{dx}}^i\right) \= 0\, d_H\left({\mathrm{\Θ}}_{\mathrm{ij}\cdot \cdot \cdot k}^{\mathrm{\alpha }}\right) \= {\mathrm{dx}}^{\mathrm{\ℓ}} \wedge {\mathrm{\Θ}}_{\mathrm{ij}\cdot \cdot \cdot k\mathrm{\ℓ}}^{\mathrm{\alpha }}$.

The coordinate formulas for the vertical exterior derivative are

$d_V\left(x^i\right) \=0\, d_V\left(u_{\mathrm{ij} \cdot \cdot \cdot k}^{\mathrm{\alpha }}\right) \= {\mathrm{\Θ}}_{\mathrm{ij} \cdot \cdot \cdot k}^{\mathrm{\α}} \, d_V\left({\mathrm{dx}}^i\right) \= 0\, d_V\left({\mathrm{\Θ}}_{\mathrm{ij}\cdot \cdot \cdot k}^{\mathrm{\alpha }}\right) \= 0.$

Example 1.

Create the jet space $J^2\left(E\right)$for the bundle $E$$$with coordinates $\left(x\, y\, u\, v\right) \to \left(x\, y\right)$.

Calculate the horizontal exterior derivative of a function.

Calculate the horizontal exterior derivative of a type (1, 0) bi-form.

Calculate the horizontal exterior derivative of a type (0, 2) bi-form.