Skip to main content
5 events
when toggle format what by license comment
Jan 3, 2011 at 10:59 comment added Jeffrey Giansiracusa As I would use the terminology, an $n$-type is an equivalence class of spaces under the relation of being $n$-equivalent, so yes, 'any 4-manifold' would be a perfectly good answer as far as I am concerned.
Jan 2, 2011 at 19:09 comment added HJRW In that case, I don't think I understand the question. Why isn't 'any 4-manifold' a good answer?
Jan 2, 2011 at 9:19 comment added Jeffrey Giansiracusa you're right, it is a wedge of spheres, but then the quotient by the mapping class group is the space with an interesting 2-type, since it is an approximation to the classifying space of the mapping class group. So $\pi_2$ would be trivial, but having an interesting fundamental group makes a 2-type eligible for being considered interesting in my book.
Jan 2, 2011 at 5:17 comment added HJRW Doesn't the curve complex have the homotopy type of a wedge of (typically higher dimensional) spheres? I think Harer showed this and computed the dimension. In particular, isn't $\pi_2$ usually trivial? As you say, Andy will know.
Jan 1, 2011 at 21:24 history answered Jeffrey Giansiracusa CC BY-SA 2.5