(FWIW, I hadn’t heard of that theorem before but don’t feel that surprised by the statement. But I’m quite curious if the proofs provide an intuitive understanding for why we need 4 dimensions.
Maybe this is hindsight bias, but I feel like if you had asked me “Can we get any member of [broad but appropriately restricted class of groups] as fundamental group of a [sufficiently general class of manifolds]?” my immediate reply would have been “uh, I’d need to think about this, but it’s at least plausible that the answer is yes”, whereas there is no way I’d have intuitively said “yes, but we need at least four dimensions”.)
I think that often the topology of things in low dimensions ends up interestingly different to in high dimensions—roughly when your dimensionality gets big enough (often 3, 4, or 5 is “big enough”) there’s enough space to do the things you want without things getting in the way.
One of the proofs I know takes advantage of the fact that f D3×S1 (which is not simply connected) has boundary S2×S1 , which is also the boundary of D2×S2 (which is simply connected); there isn’t room for the analogous trick a dimension down.
(FWIW, I hadn’t heard of that theorem before but don’t feel that surprised by the statement. But I’m quite curious if the proofs provide an intuitive understanding for why we need 4 dimensions.
Maybe this is hindsight bias, but I feel like if you had asked me “Can we get any member of [broad but appropriately restricted class of groups] as fundamental group of a [sufficiently general class of manifolds]?” my immediate reply would have been “uh, I’d need to think about this, but it’s at least plausible that the answer is yes”, whereas there is no way I’d have intuitively said “yes, but we need at least four dimensions”.)
I think that often the topology of things in low dimensions ends up interestingly different to in high dimensions—roughly when your dimensionality gets big enough (often 3, 4, or 5 is “big enough”) there’s enough space to do the things you want without things getting in the way.
One of the proofs I know takes advantage of the fact that f D3×S1 (which is not simply connected) has boundary S2×S1 , which is also the boundary of D2×S2 (which is simply connected); there isn’t room for the analogous trick a dimension down.