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Apr 30 at 7:38 vote accept Troubled Shallows
Apr 30 at 6:51 vote accept Troubled Shallows
Apr 30 at 7:38
Apr 29 at 20:37 comment added Kevin Walker @WillSawin There are regular MO users who are much more knowledgeable on these matters than I am, so I hesitate to give a detailed answer to your question. The reason I hedged with "in some sense" is that while in higher dimensions it is (relatively) easy to reduce topological questions to homotopical or algebraic questions, the resulting homotopical/algebraic questions might themselves be difficult. Ryan Budney's answer alludes to this, I think.
Apr 29 at 19:19 comment added Will Sawin Would it be possible to give more detail on "in some sense". In other words, is it fair to say that higher-dimensional manifolds are easier to study and classify specifically in the setting of smooth manifolds that are simply-connected (or have sufficiently small fundamental group)? Here I mean difficulty relevant for present research - i.e. Freedman's theorem and Pereleman's theorem were hard to prove but low-dimensional topology is mostly not concerned with extending the techniques of that proof.
Apr 29 at 13:53 history edited Kevin Walker CC BY-SA 4.0
typo
Apr 29 at 13:52 comment added Kevin Walker @Municipal-Chinook-7 Yes, your original question had it backwards (in my and many others' opinion). Higher dimensional manifolds are in some sense easier to study and classify.
Apr 29 at 5:46 comment added Troubled Shallows @RyanBudney yes, good point.
Apr 29 at 5:21 comment added Ryan Budney @Municipal-Chinook-7: Don't you mean to say it's more difficult to study low-dimensional topology? Because you appear to be re-stating your previously-stated assumption.
Apr 29 at 2:22 comment added David Roberts "Convincing sub-speculations", for vague philosophising below the level of actual concrete ideas, but that everything takes as a good motivating idea
Apr 29 at 1:06 comment added Troubled Shallows So do I have it backwards? Is it that, contra my assumption, it is often easier to study 'low-dimensional topology' due to the number of dimensions required to perform the Whitney trick and perform surgery?
Apr 28 at 23:26 comment added Sam Hopkins Now I want to write a paper with a section titled "Sub-convincing speculations"
Apr 28 at 20:39 comment added Daniel Asimov Making smooth submanifolds of dimensions K and L disjoint by a small perturbation in a smooth N manifold is an easy consequence of transversality when K + L < N.
Apr 28 at 17:48 comment added Neal Thank you for the response and reference!
Apr 28 at 15:40 comment added Misha @Neal: No, that’s not what the Whitney trick is. It’s a geometric move for cancelling intersections; see here: celebratio.org/Whitney_H/article/220
Apr 28 at 13:44 comment added Neal Making embedded surfaces disjoint via perturbation is the "Whitney trick," right?
Apr 28 at 13:10 history answered Kevin Walker CC BY-SA 4.0