As a corollary of Alexander duality, if M is a connected orientable, compact manifold of dimension n, and if H_1(M,Z) = (0) (the first homology group with coeffs in Z vanishes), then no nonorientable compact manifold N of dimension n -1 can be embebded in M. (for example, see Bredon, Topology and Geometry, Corollary 8.9, Page 353). This implies that K can’t be embedded in S^3. But this is not NOT Dan’s question! Unfortunately, by the Kunneth formula H_1(S^1xS^2, Z) = Z. (H_0(S^1,Z) = H_1(S^1,Z) = Z, H_0(S^2,Z) = Z, H_1(S^2,Z) = (0), H_2(S^2,Z) = Z). It appears that S^1 x S2 is a bit “bigger” than S^3, so I don’t know whether K can be embedded in S^1 x S^2. I would lean for no. Best, — Jean
On Nov 25, 2020, at 10:35 PM, Dan Asimov <asimov@msri.org> wrote:
I'll just say that, so far, nothing correct has been mentioned about the answer to either puzzle.
—Dan _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun