I'm studying quasi-presymplectic groupoids and I've come across the following proposition which I'm finding challenging to prove. Any help or guidance would be greatly appreciated.
Given a Lie groupoid $\Gamma\rightrightarrows P$ equipped with a two-form $\omega \in \Omega^2(\Gamma)$ and a three-form $\Omega \in \Omega^3(P)$, we define a quasi-presymplectic groupoid under the conditions:
\begin{align} d\Omega &= 0, \ d\omega &= \partial\Omega, \text{ and } \ \partial\omega &= 0. \end{align}
Therefore, $\omega + \Omega$ forms a $3$-cocycle of the total de Rham complex of the groupoid $\Gamma \Rightarrow P$.
Now, consider the graph of the groupoid multiplication, denoted as $\Lambda = {(x, y, z) \mid z = xy, (x, y) \in \Gamma_2}\subset \Gamma\times\Gamma\times\Gamma$.
Proposition: $\Lambda$ is isotropic with respect to the form $(\omega, \omega, -\omega)$.
I understand that this proposition states that the form $(\omega, \omega, -\omega)$ should vanish when acting on any tangent vector at any point in $\Lambda$. However, I'm not sure how to start proving this formally. How can I demonstrate that for any point $(x, y, z) \in \Lambda$ and for any vector $v \in T_{(x, y, z)}\Lambda$, we have $(\omega, \omega, -\omega)(v) = 0$?
Any suggestions or detailed insights would be greatly appreciated.