How difficult is Morse theory on stacks? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T07:24:26Z http://mathoverflow.net/feeds/question/15942 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/15942/how-difficult-is-morse-theory-on-stacks How difficult is Morse theory on stacks? Theo Johnson-Freyd 2010-02-21T00:11:31Z 2010-05-08T18:36:39Z <p>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.</p> <p>Suppose I have a <strong>Lie groupoid</strong> $G \rightrightarrows G_0$, by which I mean the following data:</p> <ul> <li>two finite-dimensional (everything is smooth) manifolds $G,G_0$, </li> <li>two surjective submersions $l,r: G \to G_0$,</li> <li>an embedding $e: G_0 \hookrightarrow G$ that is a section of both the maps $l,r$,</li> <li>a <strong>composition law</strong> $m: G \times_{G_0} G \to G$, 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 $G_0$.</li> <li>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,</li> <li>$m(e(l(g)),g) = g = m(g,e(r(g)))$ for all $g\in G$, </li> <li>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 $m(g,i(g)) = e(l(g))$.</li> </ul> <p>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 &mdash; (certain) bibundles rather than functors, for example. But I digress.</p> <p>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.</p> <p>Oh, let me also recall the notion of <strong>tangent Lie algebroid $A \to G_0$ to a Lie groupoid</strong>. 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 <strong>anchor</strong> 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 $G_0$.</p> <p>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 <strong>Hessian</strong>, 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 <strong>nondegenerate</strong> 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.</p> <p>Nondegeneracy rules out some singular behavior of $[y]$, like the irrational line in the torus.</p> <p>Anyway, my question is as follows:</p> <blockquote> <p>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 <em>point</em> $\bar y \in V_0$?</p> </blockquote> <p>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. $\bar f^{(2)}_{\bar y} = f^{(2)}_y$.</p> http://mathoverflow.net/questions/15942/how-difficult-is-morse-theory-on-stacks/23945#23945 Answer by AndrĂ© Henriques for How difficult is Morse theory on stacks? AndrĂ© Henriques 2010-05-08T18:36:39Z 2010-05-08T18:36:39Z <p>Yes, it's possible to find such a neighbourhood $U_0$ of $[y]$.</p> <p>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 <i>only</i> critical point of $f$ in $V_0$. </p> <p>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. <hr> Note: <i>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 $V_0\times V_0 \to G_0\times G_0\leftarrow G$.</i></p>