Question

Recall the definition of a symplectic groupoid. Roughly this is a Lie groupoid such that the object manifold is Poisson, and the arrow manifold is symplectic such that the symplectic form is compatible with the groupoid structure. There is naturally a forgetful 2-functor $SymplGpd \to LieGpd$.

Clearly not every Lie groupoid is in the image of this functor. Pick, for instance a manifold which is not symplectic and consider the trivial Lie groupoid with this as objects. But given that a Lie groupoid is in the image of the forgetful 2-functor, what does the fibre over it look like? To ask a question off the top of my head, would it be too much to ask that symplectic groupoids with identical underlying Lie groupoids are Morita equivalent (in the sense of Xu see e.g. this or this, or this for the quasisymplectic version, in case that helps.)

I'm cheating a bit here, because I am not specifying the 1- and 2-arrows of the 2-category of symplectic groupoids. I do this so that answers can clarify what these might be and how this relates to my question.

Answer 1

Let me state a simple observation regarding the fibre of the 2-functor that arose in a conversation with Rui Loja Fernandes.

A symplectic groupoid $(\Sigma, \omega) \Rightarrow P$ induces a Poisson bivector $\pi$ on $P$, which is completely determined by the fact that the source and target maps are Poisson and anti-Poisson respectively (beware of the fact that not all Poisson structures arise in this fashion). Moreover, the foliation induced by the groupoid on $P$ corresponds to the symplectic foliation $\mathcal{F}$ of $\pi$. The symplectic groupoid contains all information regarding the Poisson geometry of $(P,\pi)$. A way to reformulate the question is to ask the following: given a Poisson manifold $(P,\pi)$ with symplectic groupoid $(\Sigma,\omega) \Rightarrow (P,\pi)$, what other Poisson structures on $P$ are there whose symplectic groupoid (viewed as a Lie groupoid) is $\Sigma \Rightarrow P$? Note that the requirement that the underlying Lie groupoid does not change fixes the symplectic foliation $\mathcal{F}$.

Fix a symplectic groupoid $(\Sigma,\omega) \Rightarrow (P,\pi)$. It is helpful to view $\pi$ as a Dirac structure on $P$ as follows. Set $\mathbb{T}P = \mathrm{T}P \oplus \mathrm{T}^*P$; consider

• the non-degenerate, symmetric bilinear product on $\mathbb{T}P$ defined by $$ \langle (X,\alpha),(Y,\beta) \rangle = X(\beta) + Y(\alpha);$$
• the Courant bracket $[.,.]_C:\Gamma(\mathbb{T}P) \times \Gamma(\mathbb{T}P) \to \Gamma(\mathbb{T}P)$ defined by $$ [(X,\alpha ) , (Y,\beta)]_C = \left([X,Y], L_X \beta-L_{Y} \alpha + \frac{1}{2}\mathrm{d}(\alpha(Y) - \beta(X)\right), $$ where $[.,.]$ denotes the standard Lie bracket and $L_X, L_Y$ denotes Lie derivatives.

A Dirac structure on $P$ is a subbundle $L \subset \mathbb{T}P$ such that

1. $L$ is maximally isotropic with respect to the above bilinear product;
2. $L$ is involutive under $[.,.]_C$.

Like Poisson bivectors, Dirac structures come equipped with a (possibly singular) foliation given by the projection onto the first factor of $\mathbb{T}P$.

The Poisson bivector $\pi$ defines a Dirac structure by

$$ L_{\pi} :=\{(\pi^{\sharp}(\alpha),\alpha) \,:\,\alpha \in \Omega^1(P) \}, $$

where $\pi^{\sharp} :\mathrm{T}^*P \to \mathrm{T}P$ is the bundle morphism defined by

$$ (\pi^{\sharp}(\alpha))(\beta) :=\pi(\alpha,\beta) $$

for $\alpha,\beta \in \Omega^1(P)$. The symplectic foliation $\mathcal{F}$ of $\pi$ coincides with the foliation associated to the Dirac structure.

Given a 2-form $B$ on $P$, define an endomorphism $\mathbb{T}P \to \mathbb{T}P$ by

$$ (X,\alpha) \mapsto (X,\alpha +B^{\flat}(X)), $$

where $(B^{\flat}(X))(Y) = B(X,Y)$. Denote the image of a Dirac structure $L$ under this endomorphism by $e^{B}L$; if $\mathrm{d}B=0$, then $e^B L$ is also a Dirac structure (this is known as a gauge transformation ). Note that gauge transformations do not change the underlying foliation associated to the initial Dirac structure $L$.

Let $L_{\pi}$ denote the Dirac structure associated to $\pi$ and, for a closed 2-form $B$ on $P$, consider $L' = e^B L_{\pi}$. For $L'$ to come from a bivector (which is necessarily Poisson), it is necessary that the endomorphism

$$ \mathrm{I} + B^{\flat} \circ \pi^{\sharp} : \mathrm{T}P \to \mathrm{T}P$$

be invertible. Under these conditions, obtain a new Poisson bivector $\pi'$ on $P$. It turns out that $(P,\pi')$ comes from a symplectic groupoid which, as a Lie groupoid, is isomorphic to the symplectic groupoid $(\Sigma,\omega) \Rightarrow (P,\pi)$. If $\omega'$ denotes the symplectic form on $\Sigma$ inducing the Poisson bivector $\pi'$, then $\omega' = \omega + \Omega$, where

$$\Omega = s^\ast B - t^\ast B $$

and $s,t :\Sigma \to P$ are the source and target maps respectively.

The above shows that the fibre of the 2-functor over $\Sigma \Rightarrow P$ that is considered in the question above contains the space of closed 2-forms $B$ on $P$ such that $\mathrm{I} + B^{\flat} \circ \pi^{\sharp}$ is invertible, where $\pi$ is a Poisson structure induced on $P$ by some multiplicative symplectic form on $\Sigma$. It remains to understand `how much more' there is to this fibre, but I have no good insights in this direction.

Answer 2

This is more of an extended comment, rather than an answer. Consider the Lie trivial groupoid of the form $M\Rightarrow M$, where $M$ is a closed oriented surface. Then any two symplectic forms on $M$ with the same total integral over $M$ are symplectomorphic (Moser deformation argument). Hence the volume of your area form is a symplectic invariant. So the fiber of your map contains the category is equivalent to the set $\mathbb{R} \smallsetminus {0}$. But there is more: take a covering space $\tilde{M}$ of $M$. The fundamental group $\pi_1 (M)$ acts on $\tilde{M}$ and the action groupoid $\pi_1 (M)\times \tilde{M}\Rightarrow \tilde{M}$ is Morita equivalent to $M$. It is also a symplectic groupoid in your fiber for any choice of a $\pi_1(M)$-invariant symplectic form on $\tilde{M}$...

(edit) Unfortunately the form on $\pi_1 (M)\times \tilde{M}\Rightarrow \tilde{M}$ does not look multiplicative, as Daniele Sepe points out. Oops.