The title is a little tongue-in-cheek, since I have a very particular question, but I don't know how to condense it into a pithy title. If you have suggestions, let me know.

Suppose I have a Lie groupoid $G \rightrightarrows G_0$, by which I mean the following data:

  • two finite-dimensional (everything is smooth) manifolds $G,G_0$,
  • two surjective submersions $l,r: G \to G_0$,
  • an embedding $e: G_0 \hookrightarrow G$ that is a section of both the maps $l,r$,
  • a composition law $m: G \times_{G_0} G \to G$, where the fiber product is the pull back of $G \overset{r}\to G_0 \overset{l}\leftarrow G$, intertwining the projections $l,r$ to $G_0$.
  • Such that $m$ is associative, by which I mean the two obvious maps $G \times_{G_0} G \times_{G_0} G \to G$ agree,
  • $m(e(l(g)),g) = g = m(g,e(r(g)))$ for all $g\in G$,
  • and there is a map $i: G \to G$, with $i\circ l = r$ and $i\circ i = \text{id}$ and $m(i(g),g) = e(r(g))$ and $m(g,i(g)) = e(l(g))$.

Then it makes sense to talk about smooth functors of Lie groupoids, smooth natural transformations of functors, etc. In particular, we can talk about whether two Lie groupoids are "equivalent", and I believe that a warm-up notion for "smooth stack" is "Lie groupoid up to equivalence". Actually, I believe that the experts prefer some generalizations of this — (certain) bibundles rather than functors, for example. But I digress.

Other than that we know what equivalences of Lie groupoids are, I'd like to point out that we can work also in small neighborhoods. Indeed, if $U_0$ is an open neighborhood in $G_0$, then I think I can let $U = l^{-1}(U_0) \cap r^{-1}(U_0)$, and then $U \rightrightarrows U_0$ is another Lie groupoid.

Oh, let me also recall the notion of tangent Lie algebroid $A \to G\_0$ to a Lie groupoid. The definition I'll write down doesn't look very symmetric in $l\leftrightarrow r$, but the final object is. The fibers of the vector bundle $A \to G\_0$ are $A\_y = {\rm T}\_{e(y)}(r^{-1}(y))$, the tangent space along $e(G\_0)$ to the $r$-fibers, and $l: r^{-1}(y) \to G\_0$ determines a God-given anchor map $\alpha = dl: A \to {\rm T}G\_0$, and because $e$ is a section of both $l,r$, this map intertwines the projections, and so is a vector bundle map. In fact, the composition $m$ determines a Lie bracket on sections of $A$, and $\alpha$ is a Lie algebra homomorphism to vector fields on $G_0$.

Suppose that I have a smooth function $f: G_0 \to \mathbb R$ that is constant on $G$-orbits of $G_0$, i.e. $f(l(g)) = f(r(g))$ for all $g\in G$. I'd like to think of $f$ as a Morse function on "the stack $G_0 // G$". So, suppose $[y] \subseteq G_0$ is a critical orbit, by which I mean: it is an orbit of the $G$ action on $G_0$, and each $y \in [y]$ is a critical point of $f$. (Since $f$ is $G$-invariant, critical points necessarily come in orbits.) If $y$ is a critical point of $f$, then it makes sense to talk about the Hessian, which is a symmetric pairing $({\rm T}\_yG\_0)^{\otimes 2} \to \mathbb R$, but I'll think of it as a map $f^{(2)}\_y : {\rm T}\_yG\_0 \to ({\rm T}\_yG\_0)^*$. In general, this map will not be injective, but rather the kernel will include $\alpha\_y(A\_y) \subseteq {\rm T}\_yG\_0$. Let's say that the critical orbit $[y]$ is nondegenerate if $\ker f^{(2)}_y = \alpha\_y(A\_y)$, i.e. if the Hessian is nondegenerate as a pairing on ${\rm T}\_yG\_0 / \alpha\_y(A\_y)$. I'm pretty sure that this is a condition of the orbit, not of the individual point.

Nondegeneracy rules out some singular behavior of $[y]$, like the irrational line in the torus.

Anyway, my question is as follows:

Suppose I have a Lie groupoid $G \rightrightarrows G_0$ and a $G$-invariant smooth function $f: G_0 \to \mathbb R$ and a nondegenerate critical orbit $[y]$ of $f$. Can I find a $G$-invariant neighborhood $U_0 \supseteq [y]$ so that the corresponding Lie groupoid $U \rightrightarrows U_0$ is equivalent to a groupoid $V \rightrightarrows V_0$ in which $[y]$ corresponds to a single point $\bar y \in V_0$? I.e. push/pull the function $f$ over to $V_0$ along the equivalence; then can I make $[y]$ into an honestly-nondegenerate critical point $\bar y \in V_0$?

I'm assuming, in the second phrasing of the question, that $f$ push/pulls along the equivalence to a $V$-invariant function $\bar f$ on $V_0$. I'm also assuming, so if I'm wrong I hope I'm set right, that ${\rm T}\_{\bar y}V\_0 \cong {\rm T}\_yG\_0 / \alpha\_y(A\_y)$ canonically, so that e.g. $\bar f^{(2)}_{\bar y} = f^{(2)}_y$.

  • $\begingroup$ There is a literature on "equivariant Morse theory" (of which my ignorance is nearly total). Have you looked into this? In particular, your question seems to specialize to a pretty natural-looking question about a G-invariant morse function on a manifold equipped with a G-action. $\endgroup$ Feb 21, 2010 at 2:13
  • $\begingroup$ @Charles: That's a good suggestion. This isn't my area, and more generally I don't really care about the actual "Morse Theory" of stacks. All I really care about is the local question of whether I can make my critical point nondegenerate. But, yes, the literature very well might answer it. $\endgroup$ Feb 21, 2010 at 2:37
  • 1
    $\begingroup$ Similar motivation is in Hepworth $\endgroup$ Apr 6, 2010 at 21:27
  • 1
    $\begingroup$ You might also want to look at Hepworth's other paper, arXiv:0712.2432 'Morse Inequalities for Orbifold Cohomology' though I think it doesn't quite answer the question that you've posed. $\endgroup$ Apr 12, 2010 at 9:15

1 Answer 1


Yes, it's possible to find such a neighbourhood $U_0$ of $[y]$.

Here's how you do it. Pick a submanifold $M\subset G_0$, $y\in M$, transverse to $[y]$. By your Morse-ness assumption, the restriction $f|_M$ is Morse, with critical point $y$. Pick a neighborhood $V_0\subset M$ of $y$, such that $y$ is the only critical point of $f$ in $V_0$.

Let $U_0$ be the orbit of $V_0$ under $G$. Clearly, $U_0$ is a neighborhood of $[y]$. Now consider the restriction of $G$ to $U_0$. This is defined to be the groupoid with object space $U_0$, and morphism space $U_0\times_{G_0}G\times_{G_0}U_0$. That's your groupoid $U\rightrightarrows U_0$. Similarly, you can consider the restriction $V\rightrightarrows V_0$ of $G$ to $V_0$. The inclusion $(V\rightrightarrows V_0) \hookrightarrow (U\rightrightarrows U_0)$ is a Morita equivalence because it's essentially surjective and fully faithfull.

Note: The notions "essentially surjective" and "fully faithfull" for Lie groupoids are somewhat stronger than what you might initially guess. The first one also requires the existence of locally defined smooth maps $U_0\to V_0$, while "fully faithfull" means that $V$ is the pullback of $V_0\times V_0 \to G_0\times G_0\leftarrow G$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.