Timeline for Second Homotopy Group of Graph Manifolds
Current License: CC BY-SA 3.0
12 events
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| Jul 14, 2012 at 15:21 | comment | added | Vitali Kapovitch | @Lee Mosher, Thanks again. you are absolutely right. when I was thinking about it yesterday it wasn't immediately clear to me how to slide the sphere off the pieces in the graph of contractible spaces (it's easy to do geometrically in the npc case which is what Agol was using in both of his arguments) but you are right - this is a purely topological statement once you know that both the edge and the vertex spaces are contractible. | |
| Jul 14, 2012 at 9:01 | vote | accept | Malte | ||
| Jul 14, 2012 at 7:03 | comment | added | Lee Mosher | I guess all I am really trying to say is that Agol's second argument can be recast without any 3-manifold topology, as a general statement about graphs of $K(\pi,1)$'s (well, no 3-manifold topology except for the statement that Seifert fibered manifolds are $K(\pi,1)$'s). From a homotopy theory point of view, it might be simpler to think of the proof that way. From a 3-manifold topology point of view, there's no improvement on Agol's wording. | |
| Jul 14, 2012 at 6:07 | comment | added | Lee Mosher | @Vitali: Upstairs in the universal cover of the graph of $K(\pi,1)$ spaces, one sees a tree of contractible spaces. Such a space is contractible, for if you map a sphere in, the image of its projection hits only a finite number of edges, the union of which is a finite subtree. One proceeds by induction on the number of edges in that subtree, using contractibility of edge and vertex spaces to homotope the sphere so that the number of edges it hits is decreased. Once the image of the sphere hits only 1 edge, it is contained in an edge space, and then is easily homotoped to a point. | |
| Jul 14, 2012 at 1:30 | history | edited | Ian Agol | CC BY-SA 3.0 |
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| Jul 13, 2012 at 23:32 | comment | added | Vitali Kapovitch | @Lee Mosher thanks. but my question really was about that last easy to deduce step :=) | |
| Jul 13, 2012 at 23:08 | comment | added | Lee Mosher | @Vitali: A different simple proof uses graphs of spaces (as in Scott-Wall). Start from the fact that a Seifert fibered manifold is a $K(\pi,1)$. Any graph manifold with $\pi_1$-injective tori is a graph of spaces whose vertex spaces (Seifert fibered manifolds) are $K(\pi,1)$'s and whose edge spaces (tori) are $K(\pi,1)$'s. It is simple to deduce that the total space must then be a $K(\pi,1)$. | |
| Jul 13, 2012 at 22:24 | comment | added | Ian Agol | @ Vitali: See p. 17 of the slides: homepages.math.uic.edu/~agol/cover/cover18.html You can't necessarily match up the NPC metrics downstairs (although the cases that you can are classified by BKN equations), but upstairs you can, using $GL(R^2)$ is homotopic to $O(2)$. | |
| Jul 13, 2012 at 21:36 | comment | added | Vitali Kapovitch | Thanks! I understand the second argument (it's very nice and simple) but not the first. how do you put a npc metric on the universal cover? not all graph manifolds admit npc metrics. | |
| Jul 13, 2012 at 21:24 | comment | added | Ian Agol | You can see the notes to my talk linked above for a geometric proof. The point is that you can put a non-positively curved metric on each Seifert piece with geodesic boundary. Then in the universal cover, you can put these metrics together to get a global npc metric. Alternatively, you can map $S^2$ into the space, and make it transverse to the canonical tori. Then an innermost disk in the preimage must be homotopically trivial, so you can homotope it off, leaving fewer components of the preimage. If the sphere misses the tori, then it is contractible. | |
| Jul 13, 2012 at 21:04 | comment | added | Vitali Kapovitch | what is the simplest proof that a graph manifold with $\pi_1$-injective tori/Klein bottle is aspherical? I know this is classical (I think originally due to Whitehead?) but is there an elementary proof of this? | |
| Jul 13, 2012 at 20:50 | history | answered | Ian Agol | CC BY-SA 3.0 |