Multiplicative forms
Let $\mathcal{G}\rightrightarrows M$ be a Lie groupoid. Let $\mathrm{m}$ be the groupoid multiplication map $$ \mathrm{m}:\mathcal{G}^{(2)}\to \mathcal{G}: (g,h)\mapsto g\circ h. $$ Consider a $0$-form, namely, a function $f\in C^\infty(\mathcal{G}$. In order for the groupoid structure to be compatible with the $0$-form, $f\colon\mathcal{G} \to \mathbb{R}$ should be a homomorphism of groupoids: $$ f(g\cdot h) = f(g) + f(h), \forall (g,h)\in \mathcal{G}^{(2)}, $$ or equivalently $$ \mathrm{m}^*(f) = \operatorname{pr}^\ast_1 f+ \operatorname{pr}^\ast_2 f. $$
Multiplicativity condition
Two properties that might illustrate the character of multiplicativity condition.
- Multiplicative forms are right/left-invariant on $\mathsf{s}$-/$\mathsf{t}$-fibers. Note that here right actions are precomposition of morphisms (we use the right-to-left convention for morphisms).
- Multiplicativity is a cohomological condition.
The first property: In exactly the same way as the above bi-invariance of multiplicative $1$-forms on Lie groups is proved, one can see that
The second property. Define maps from base to morphisms and then to the fiber product: $$ \delta:\Omega^k(\mathcal{G})\to \Omega^k(\mathcal{G}^{[2]}): \omega \mapsto (\operatorname{pr}^*_1 -\mathrm{m}^\ast + \operatorname{pr}^\ast_2)\omega $$ $$ \delta:\Omega^k(M)\to \Omega^k(\mathcal{G}): \alpha \mapsto (\mathsf{s}^\ast-\mathsf{t}^\ast)\alpha $$ then $\delta\circ\delta =0$. This can be checked by evaluating $\delta\circ\delta \alpha$ at $(g_2,g_1)\in \mathcal{G}^{[2]}$: $$ p_2 \xleftarrow{g_2} p_1 \xleftarrow{g_1} p_0. $$
The condition that $\omega\in\Omega^k(\mathcal{G})$ is multiplicative means that $\omega$ is a cocycle, $$ \delta \omega = 0. $$ The boundaries, or multiplicatively exact forms, are those $\omega \in \Omega^{k-1}(\mathcal{G})$ of the form $$ \omega = \delta(\alpha) = (\mathsf{s}^\ast -\mathsf{t}^\ast)\alpha,\quad (\alpha \in \Omega^{k-1}(M)). $$
Symplectic Groupoid
A good example will be the action groupoid associated with the coadjoint action. Let $G$ be a Lie group and let $\Sigma = G\rtimes \mathfrak{g}^\ast \rightrightarrows \mathfrak{g}^\ast$ be the action groupoid. Spelled out, this is the groupoid with $\mathsf{s}(g,\xi)=\xi,\mathsf{t}(g,\xi)=g\cdot\xi = \operatorname{Ad}_g^\ast \xi$.
Symplectic groupoid actions
The space $(S,\omega)$ is called a Hamiltonian $(\Sigma,\Omega)$-space with the moment map $\mu:S\to M$.