A quasi-symplectic groupoid, introduced by P. Xu in [1], consists of a Lie groupoid $\Gamma \rightrightarrows P$ together with a two-form $\omega$ on $\Gamma$ and a three-form $\Omega$ on $P$ satisfying some compatibility and non-degenerateness assumptions. Quasi-symplectic groupoids include, in particular, Weinstein's notion of symplectic groupoids [2] when $\omega$ is non-degenerate and $\Omega = 0$.
I've heard people make the following claim:
Claim. A symplectic groupoid is equivalent to a quasi-symplectic groupoid $(\Gamma \rightrightarrows P, \omega, \Omega)$ with $\Omega = 0$.
The implication $(\Rightarrow)$ is immediate. But I don't see why $(\Leftarrow)$ should be true, and this claim doesn't seem to appear in [1]. Specifically, I don't see why $\omega$ should be non-degenerate, i.e. a symplectic form.
Question. Is the above claim true? If not, what are examples of quasi-symplectic groupoids with $\Omega = 0$ but $\omega$ not symplectic?
References
[1] Xu, P.: Momentum maps and Morita equivalence. J. Differential Geom. 67 (2004), no. 2, 289–333.
[2] Weinstein, A.: Symplectic groupoids and Poisson manifolds. Bull. Amer. Math. Soc. (N.S.) 16 (1987), no. 1, 101–104.
