How Do Knots in 4-Manifolds Influence Homology Classes?

[Bacle]

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.

 

[lavinia]

I am confused. Aren't there knotted circles in R^3?

 

[Bacle]

I am confused. Aren't there knotted circles in R^3?

Yes, you are right. I was wondering if the results generalized to codimension-2
manifolds; R^n is in general an extreme example, since it is too nice in too many ways.

 

[lavinia]

Yes, you are right. I was wondering if the results generalized to codimension-2
manifolds; R^n is in general an extreme example, since it is too nice in too many ways.

I don't know anything about knots but let's see if we can construct a knotted torus in R^4.

 

[Tinyboss]

Are you only interested in the second homology group, or is that just the example you chose?

I think it fails for the first homology group, which is the abelianization of the fundamental (first homotopy) group, and therefore is trivial if the fundamental group is perfect. So for example a space with fundamental group A_5 by definition of the fundamental group has non-isotopic embeddings of S^1, but its first homology group is trivial.

 

[CharmedQuark]

At Tinyboss- His example is for 4-manifolds, so the second homology group is the only one that matters to any significant degree.

At OP- As to the second part of your question, If the second homology group is trivial then certainly any two embeddings of S are the boundary of some 3-manifold. However I'm not sure that we can say that these two embeddings are isotopic. In fact I would probably say no. If limit ourselves to a case more familiar, consider S^3. The first homology group of S^3 is trivial, however if we embed two non-equivalent knots in S^3, such as the trefoil and the unknot, they are certainly not isotopic even though the first homology group of S^3 is trivial.

Although this case was for a 3-manifold, I would imagine that you could find a similar example for 4-manifolds.