Conley index for isolated invariant sets with no exit points - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T00:01:57Z http://mathoverflow.net/feeds/question/118574 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/118574/conley-index-for-isolated-invariant-sets-with-no-exit-points Conley index for isolated invariant sets with no exit points Pablo 2013-01-10T21:17:27Z 2013-01-11T12:49:57Z <p>Conlay described in $\textit{Isolated Invariant Sets and the Morse Index (1976)}$ the bases of what would be known as Conley Index Theory. </p> <p>For the sake of simplicity let's think of vector fields defined on manifolds (a more general situation is enough by just considering locally compact spaces). Im going to pose a few basic notions needed for the question so even someone who is not familiar with the topic may give an answer.</p> <blockquote> <p>A set (in the phase space) is called invariant if it is the union of solution curves. It is isolated if it is the maximal invariant set in some neighborhood of itself. A compact such neighborhood is called an isolating neighborhood for the invariant set.</p> <p>An isolating neighborhood is an isolating block if the integral curves through each boundary point of the neighborhood goes immediately out of it in one or the other time direction.</p> </blockquote> <p>And finally:</p> <blockquote> <p>The $\textit{(Conley)}$ index is the homotopy type of the pointed space obtained from a block on collapsing the set of exit points (points in the boundary where integral curves go out the neighborhood) to one point.</p> </blockquote> <p>Well, I've read that when we have an isolating block, say $N$, that has no exit points, we would have to collapse the empty set $\emptyset$ to one point (?), so there's a convention that the resulting space is $$(N \bigcup \lbrace \star \rbrace, \star)$$ that is, the disjoint union of the space $N$ and an external point, all of it pointed at that external point).</p> <p>So the questions are</p> <ul> <li>Why this convention makes sense (apart from that $\textit{it just works}$)? Is that the natural way of defining the operation "collapse to a point'' when all you have to collapse is $\textit{nothing}$ (i.e. $\emptyset$)?</li> <li>Is there a bigger frame where we don't need this convention because a more general rule contains it as a particular case?</li> </ul> http://mathoverflow.net/questions/118574/conley-index-for-isolated-invariant-sets-with-no-exit-points/118591#118591 Answer by alvarezpaiva for Conley index for isolated invariant sets with no exit points alvarezpaiva 2013-01-10T23:40:43Z 2013-01-10T23:40:43Z <p>The Conley index is a generalization of the Morse index and it must give basically "the same" answer when the Morse index is defined: when the invariant set is an isolated non-degenerate critical point and the isolating neighborhood is a small neighborhood where the Morse lemma gives you a normal form for the function. If you work out the dictionary between Morse index and Conley index, you will see why in the case of an isolated minimim the convention comes up.</p> http://mathoverflow.net/questions/118574/conley-index-for-isolated-invariant-sets-with-no-exit-points/118610#118610 Answer by Nicola Ciccoli for Conley index for isolated invariant sets with no exit points Nicola Ciccoli 2013-01-11T09:28:33Z 2013-01-11T09:28:33Z <p>I maybe wrong but a way of seeing is the following: from the universal property of quotients you would like that any continuous map $f:N\to Y$ to a topological space $Y$ defines a continuous map of pointed topological spaces between $N$ collapsed with nothing and $(Y,y_0)$ no matter what $y_0$ is. Since you "collapse with nothing" you do not want to lose some continuous functions, do you?</p> <p>And I would say that this definition does exactly this, since it just requires that you declare $f(\star)=y_0$.</p> http://mathoverflow.net/questions/118574/conley-index-for-isolated-invariant-sets-with-no-exit-points/118623#118623 Answer by Tirasan Khandhawit for Conley index for isolated invariant sets with no exit points Tirasan Khandhawit 2013-01-11T12:49:57Z 2013-01-11T12:49:57Z <p>The fundamental result of Conley index theory states that a homotopy type of a pointed space $N_1 / N_2$, with $[N_2]$ as a basepoint, is independent of the choice of an index pair $(N_1 , N_2)$. I feel like you are mainly asking why $N / \emptyset = N \coprod \lbrace * \rbrace$. </p> <p>This might not be entirely correct but one way I see the quotient $N_1 / N_2$ is the same as $N_1$ attached with a cone of $N_1$ on $N_2$. So when $N_2$ is empty you don't glue anything and have an extra basepoint. </p> <p>Alternatively, there is an extension of Conley index by Mrozek, Reineck, Srzednickiy, and Matematyki in their paper "The Conley index over a base." Let $X$ be your space with a flow and let $Z$ be a space with a map $\omega : X \rightarrow Z$. This kind of makes $X$ a parametrized space (or ex-space over $Z$). The Conley index over $Z$ is given by $N_1 \cup_{\omega_{|_{N_2}}} Z$. This recovers your situation when $Z$ is a point.</p>