$h_V^{r\,s} \: {\mathrm{\Ω}}^{\left(r\,s\right)}\left(J^{\mathrm{\∞}}\left(E\right)\right) \to {\mathrm{\Omega }}^{\left(r\,s-1\right)}\left(J^{\mathrm{\∞}}\left(E\right)\right)$. 

These operators satisfy$h_v^{r\, s\+1} \left(d_{V }\mathrm{\ω}\right) \+ d_V\left(h_V^{r\,s} \left(\mathrm{\ω}\right)\right) \= \mathrm{\ω}$so that if $d_{V }\mathrm{\ω}\=0\,$then $\mathrm{\ω} \= d_{V }\mathrm{\η}$where $\mathrm{\η} \= h_V^{r\,s} \left(\mathrm{\omega }\right)$.

Here are the explicit formulas for the vertical homotopy operators. Let $\(x^i\, u^{\mathrm{\α}}\, u_{i_{}}^{\mathrm{\α}}\, u_{i_{}j}^{\mathrm{\α}}$, ..., $u_{\mathrm{ij} \cdot \cdot \cdot k}^{\mathrm{\α}}\, ....\)$be a local system of jet coordinates and let ${\mathrm{\Θ}}_{i_1{i}_2\cdot \cdot \cdot i_k}^{\mathrm{\α}} \= {\mathrm{du}}^{\mathrm{\α}}-u_{i_1{i}_2\cdot \cdot \cdot i_k\mathrm{\ℓ}}^{\mathrm{\α}} {\mathrm{dx}}^{\mathrm{\ℓ}}$be the contact forms. The vertical radial vector field on $E$is $R \= u^{\mathrm{\α}} \frac{\partial }{\partial u_{}^{\mathrm{\α}}}$ and its prolongation to jet space is

pr $R \= u^{\mathrm{\alpha }} \frac{\partial }{\partial u_{}^{\mathrm{\alpha }}} \+ u_i^{\mathrm{\alpha }} \frac{\partial }{\partial u_i^{\mathrm{\α}}} \+ u_{\mathrm{ij}}^{\mathrm{\alpha }} \frac{\partial }{\partial u_{\mathrm{ij} }^{\mathrm{\alpha }}} \+ \cdot \cdot \cdot \cdot$

The flow of the vector field pr $R$ is the transformation ${\mathrm{\Φ}}_t\:J^{\infty }\left(E\right)\to J^{\infty }\left(E\right)$ given by ${\mathrm{\Phi }}_t\left(x^i\, u^{\mathrm{\α}}\, u_{i_{}}^{\mathrm{\α}}\, u_{i_{}j}^{\mathrm{\α}}\, ..\.\right)$= $\left(x^i\, {e^{t}u}^{\mathrm{\alpha }}\, {e^{t}u}_{i_{}}^{\mathrm{\alpha }}\, {e^{t}u}_{i_{}j}^{\mathrm{\alpha }}\, ..\.\right)$. The vertical homotopy operators are then defined in terms of pr $R$and ${\mathrm{\Phi }}_t$  and the interior product operator $\mathrm{\ι}$(see Hook) by

$h_V^{r\,s}\left(\mathrm{\ω}\right) \= {\int }_0^1\frac{1}{t} {\mathrm{\Φ}}_{\log \left(t\right)}^{\*}\left( {\mathrm{\ι}}_{\mathrm{pr} R}\left(\mathrm{\ω}\right)\right) \mathrm{dt}$.

As a concrete example, if $\mathrm{\ω} \in {\mathrm{\Ω}}^{\left(1\,2\right)}\left(J^{\infty }\left(E\right)\right)$is given by $\mathrm{\ω} \= A_{\mathrm{\ℓ} \mathrm{\α} \mathrm{\β}}^{ I J}$($x^i\, u^{\mathrm{\α}}\, u_{i_{}}^{\mathrm{\α}}\, u_{i_{}j}^{\mathrm{\α}}$, ...)${\mathrm{dx}}^{\mathrm{\ℓ}}\wedge {\mathrm{\Θ}}_I^{\mathrm{\α}}\wedge {\mathrm{\Theta }}_J^{\mathrm{\β}}$, then

   $h_V^{r\,s}\left(\mathrm{\omega }\right) \= {\int }_0^1 t A_{\ell \mathrm{\alpha } \mathrm{\beta }}^{ I J}\left(x^i\, {\mathrm{tu}}^{\mathrm{\alpha }}\, {\mathrm{tu}}_{i_{}}^{\mathrm{\alpha }}\, {\mathrm{tu}}_{i_{}j}^{\mathrm{\α}}\, ..\.\right) {\mathrm{dx}}^{\ell }\wedge {\mathrm{\Theta }}_I^{\mathrm{\alpha }}\wedge {\mathrm{\Theta }}_J^{\mathrm{\beta }} \ⅆt$.

Thus the formulas for the vertical homotopy operators are essentially the same as that for the standard de Rham homotopy operators.

Example 1.

Create the jet space $J^3\left(E\right)$ for the bundle $E \=$${\mathrm{\ℝ}}^2\times \mathrm{\ℝ}\mathit{\to }\mathrm{\ℝ}$ with coordinates $\left(x\,y\,u\right)\to \left(x\,y\right)$.

Show that the form ${\mathrm{\ω}}_1$is $d_V$ closed.

Apply the vertical homotopy operator to ${\mathrm{\ω}}_1\.$

Check that the vertical exterior derivative of ${\mathrm{\η}}_{1 }$gives ${\mathrm{\ω}}_1\.$

Alternatives to ${\mathrm{\eta }}_{1 }$can be obtained using the path = "zigzag" option for the VerticalHomotopy command. See DeRhamHomotopy for more details.