- #1

Bacle

- 662

- 1

Hi, Everyone:

I am curious as to the relationship between general knots and homology in the

following respect:

Say an orientable surface S is knotted in X^4, an orientable 4-manifold. Then there are

two non-isotopic embeddings e, e' of S in X^4 . Does it follow that H_2(X^4)=/0?

If e(S), e'(S) are both orientable in X^4, then they define respective homology

classes a and b. How can we tell if a~b? I imagine we would use the respective

induced maps e* and e'* , but I don't see where to go from there. I think there

may be an issue of bordism here ( AFAIK, for dimensions n<=4 , homology and

bordism coincide, i.e., if a~b , for a,b in H_k ; k<=4, then a-b bounds a (k+1)-manifold) -- and not

just some subspace. ) Conversely: if I knew that H_2(X^4)=0 . Does it follow that there aren't any S-knots, i.e., that there

is only one embedding of S in X^4 , up to isotopy? . Again, it seems to come down to determining if

we can have a 3-manifold whose boundaries are e(S) and e'(S). I don't see why this could not happen. Any Ideas?

Thanks.

I am curious as to the relationship between general knots and homology in the

following respect:

Say an orientable surface S is knotted in X^4, an orientable 4-manifold. Then there are

two non-isotopic embeddings e, e' of S in X^4 . Does it follow that H_2(X^4)=/0?

If e(S), e'(S) are both orientable in X^4, then they define respective homology

classes a and b. How can we tell if a~b? I imagine we would use the respective

induced maps e* and e'* , but I don't see where to go from there. I think there

may be an issue of bordism here ( AFAIK, for dimensions n<=4 , homology and

bordism coincide, i.e., if a~b , for a,b in H_k ; k<=4, then a-b bounds a (k+1)-manifold) -- and not

just some subspace. ) Conversely: if I knew that H_2(X^4)=0 . Does it follow that there aren't any S-knots, i.e., that there

is only one embedding of S in X^4 , up to isotopy? . Again, it seems to come down to determining if

we can have a 3-manifold whose boundaries are e(S) and e'(S). I don't see why this could not happen. Any Ideas?

Thanks.

Last edited: