# Morita equivalent Lie groupoids

Suppose $$[X_1\rightrightarrows X_0]$$ and $$[Y_1\rightrightarrows Y_0]$$ are Morita equivalent Lie groupoids. This means, there exists another Lie groupoid $$[Z_1\rightrightarrows Z_0]$$ and Morita morphisms $$[Z_1\rightrightarrows Z_0]\rightarrow [X_1\rightrightarrows X_0]$$ and $$[Z_1\rightrightarrows Z_0]\rightarrow [Y_1\rightrightarrows Y_0]$$.

Another alternative description is, there exists a $$[X_1\rightrightarrows X_0]-[Y_1\rightrightarrows Y_0]$$ biprincipal bibundle" $$X_0\xleftarrow{\phi} P\xrightarrow{\varphi} Y_0$$.

Does it imply some specific condition on the maps $$\varphi$$ and $$\phi$$?

For example, can the differentials $$\phi_{*,a}:T_aP\rightarrow T_{\phi(a)}X_0$$ and $$\varphi_{*,a}:T_aP\rightarrow T_{\varphi(a)}Y_0$$ have some common properties for $$a\in P?$$ In particular, does $$\ker(\varphi_{*,a})=\ker (\phi_{*,a})$$ for all $$a\in P$$?

• What exactly is “it” in “Does it imply”? And what is the meaning of “common properties” in this context? I must say that the current answer by Nesta appears to be perfectly legitimate since the question is not that clear and allows for many interpretations. – Dmitri Pavlov Feb 5 at 15:39
• @DmitriPavlov I don’t understand why it is considered as answer. It is just the definition :) I have mentioned more details in last two lines. As of now that is what “it” means for me. – Praphulla Koushik Feb 5 at 15:47
• I am not sure what you find so funny about the definition of Morita equivalences, but you must be aware that multiple very different (but logically equivalent) definitions can be found in the literature, so unless you explicitly specify what definition you have in mind, you cannot talk about "the" definition. Your new question seems to be quite different from the old question, but nevertheless I posted a (somewhat trivial) answer. – Dmitri Pavlov Feb 5 at 23:04
• I did not mean that there is something funny about the definition. – Praphulla Koushik Feb 6 at 19:00

"Does it imply some specific condition on the maps $$\phi$$ and $$\varphi$$?"

The quick answer: $$\phi$$ and $$\varphi$$ have to be surjective submersions.

I recommend e.g. the paper Stacky Lie groups (arXiv:0702399) by Blohmann for more details, but let me try to explain the basic idea here. The definition of Morita equivalence for Lie groupoids involves the concept of a biprincipal bibundle. We can break this down into the following notions:

• Groupoid actions. Let $$G\rightrightarrows G_0$$ be a Lie groupoid, and let $$l_X:X\to G_0$$ be a smooth map. A groupoid action of $$G$$ on $$X$$ along $$l_X$$ is a smooth map $$G\times_{G_0}^{\mathrm{src},l_X}X\longrightarrow X;\qquad (g,x)\longmapsto gx$$ satisfying some further properties that straightforwardly generalise the notion of a group action. Sometimes you'll see these written as $$G\curvearrowright^{l_X} X$$. Actions from the right are defined analogously.
• Groupoid bundles. This is a notion that will eventually allow us to generalise Lie group principal bundles. A $$G$$-bundle is a smooth map $$\pi:X\to B$$ that is invariant under some action $$G\curvearrowright^{l_X} X$$. You might write this as $$G\curvearrowright^{l_X} X\xrightarrow{\pi}B$$.
• Principality. A Lie groupoid bundle $$G\curvearrowright^{l_X} X\xrightarrow{\pi}B$$ is called principal if $$\pi$$ is a surjective submersion, and the $$G$$-action is free and transitive on the $$\pi$$-fibres.
• Bibundles. A bibundle between two Lie groupoids $$G\rightrightarrows G_0$$ and $$H\rightrightarrows H_0$$ is a pair of actions $$G\curvearrowright^{l_X}X$$ and $$X{~}^{r_X}\curvearrowleft H$$ that interact well together; in particular we must have two groupoid bundles $$G\curvearrowright^{l_X} X\xrightarrow{r_X}H_0$$ and $$G_0\xleftarrow{l_X}X{~}^{r_X}\curvearrowleft H$$. Here the smooth maps $$l_X$$ and $$r_X$$ are the arrows $$\phi$$ and $$\varphi$$ from your question, respectively.
• Biprincipality. A bibundle is called biprincipal if both the groupoid bundles described in the previous bullet point are themselves principal groupoid bundles. From the definition of principality for groupoid bundles it follows that $$l_X$$ and $$r_X$$ (i.e. $$\phi$$ and $$\varphi$$) have to be surjective submersions.

Also note that there is a relation between the formalism of (biprincipal) bibundles and Morita morphisms (also known as weak equivalences). It is known in the literature that (one-sided) principality of bibundles corresponds to essential surjectivity and full faithfulness of Morita morphisms.

• Oh no... There seem to be some misunderstanding... I do know the definition of bibundle... The question is, can we conclude something about the maps $\varphi, \psi$ from the definition.. – Praphulla Koushik Feb 5 at 11:43
• From which definition are you looking to make conclusions about $\phi$ and $\varphi$? Just from the definition of a (possibly non-biprincipal) bibundle? – Nesta Feb 5 at 11:47
• The bibundle $X_0\leftarrow P\rightarrow Y_0$ is a biprincipipal bibundle. I thought it was clear from my question.. Can we say something from here? – Praphulla Koushik Feb 5 at 11:49
• If the bibundle is biprincipal, then (in fact by definition) the maps $\phi$ and $\varphi$ are surjective submersions. Moreover, since the groupoid actions are free and transitive, we know that the $\phi$-fibres are diffeomorphic to $H$-orbits and the $\varphi$-fibres are diffeomorphic to $G$-orbits. I don't know of any other conclusions that you can draw in general! (OK, if the post is not useful I'll consider deleting it.) – Nesta Feb 5 at 11:58
• Yes, fibers are diffeomorphic to corresponding orbits... I did not post a well framed question, so it is not your mistake :) Thanks for your time :) I just hope some one can say something more specific.. – Praphulla Koushik Feb 5 at 12:21

I will answer the new version of the question:

Does $$\ker(\varphi_{*,a})=\ker (\phi_{*,a})$$ for all $$a\in P$$, where $$\phi_{*,a}:T_aP\to T_{\phi(a)}X_0$$ and $$\varphi_{*,a}:T_aP\to T_{\varphi(a)}Y_0$$ are the differentials?

For Morita equivalences, the left and right moment maps $$\phi$$ and $$\varphi$$ are surjective submersions. Accordingly, $$\phi_{*,a}$$ and $$\varphi_{*,a}$$ are surjective linear maps. Hence, if $$X_0$$ and $$Y_0$$ have different dimensions, we cannot have $$\ker(\varphi_{*,a})=\ker (\phi_{*,a})$$.

• That was a bad question :) I will leave it as it it... I hope to make this question interesting someday – Praphulla Koushik Feb 6 at 19:02