John Carlos Baez on Nostr: npub12n4k2…c5dmn - interesting puzzle! To me the obvious approach is to start with ...

npub12n4k27w8jyj29f0jrpmupmzx8rpvttg6c4z0jws7xfu9kky0cp6q8c5dmn (npub12n4…5dmn) - interesting puzzle! To me the obvious approach is to start with some low 𝑛.

Let's compare SO(3) to S² × SO(2). Since π₁(SO(3)) ≅ ℤ/2 while π₁(S² × SO(2)) ≅ ℤ, they're not homeomorphic. By the way, SO(3) ≅ ℝP² since it's SU(2)/±1 and SU(2) is a 3-sphere.

Next let's compare SO(4) to S³ × SO(3). Since π₁(SO(n)) ≅ ℤ/2 for all n ≥ 3 now both these spaces have π₁ ≅ ℤ/2 so the above fundamental group argument no longer works, and we see it will never work to distinguish SO(n+1) from Sⁿ × SO(n) for n ≥ 3: they both have π₁ ≅ ℤ/2.

So we need a deeper idea.

(1/n)