# How to embed 3-manifold being homology sphere to $S^4$?

Can someone explain for me following sentence "Mike Freedman has proven all homology 3-spheres admit tame topological embeddings into $S^4$.", which I found on "open problem garden" http://www.openproblemgarden.org/op/which_compact_boundaryless_3_manifolds_embed_smoothly_in_the_4_sphere ?

On wikipedia I read that "Surgery on a knot in the 3-sphere with framing +1 or −1 gives homology sphere". From the topology books I read long time ago I remember that $H_1(X)$ is abelianization of $\pi_1(X)$.

My question has two parts 1) understand which 3-manifolds are "homology spheres" 2) how the embedding in $S^4$ for "homology sphere" is constructed ?

I have also asked question on mathoverflow whether exists method to determine which 3-manifold embeds in $R^4$ or $S^4$. I understand that on mathoverflow I should ask advanced questions while on stackexchange I should aks the basic ones $:)$

2017-11-28 Clarification

I add this clarification, because someone asked what is my question here. My original interest was which closed 3-manifolds embeds in $R^4$. I found out on mathoverflow from Ryan Budney that "Poincare homology sphere has tame topological embedding in $S^4$ but it does not have smooth embedding.". This another mistery for me. My intuition is that 3-manifolds can be divided to two families: first one can be embedded to $R^4$. Second one can only be embedded in $R^5$. Next we could make order in these families by some invariants like fundamental group.

To gain some intuition I would like to see some examples of 3-manifolds which can be embedded in $R^4$. One of the examples are "homology spheres". They have trivial homology groups, but they can have finite or infinite fundamental group. Homology spheres embed in $S^4$ with tame topological embedding. According to Ryan Budney explanations: "This is a result of Mike Freedman's. The embedding construction involves a type of infinite handle adjunctions. So the fact that you have a tubular neighbourhood is not obvious at all".

I would like to see also examples of 3-manifolds which cannot be embedded in $R^4$.

• A homology 3-sphere is a compact, connected, oriented 3-manifold $M$ which has the same homology as $S^3$; equivalently, $H_1(M;\mathbb{Z})$ is trivial (which implies, by Poincare duality, that $H_2(M;\mathbb{Z})$ is also trivial). – Lee Mosher Mar 1 '16 at 13:33
• Also, the "..." that you left out of that wikipedia quote yields a false statement: surgery on a knot in the 3-sphere does not always give a homology sphere. – Lee Mosher Mar 1 '16 at 13:34
• Thank you ! I corrected the sentence about "surgery on knot". It means that fundamental group of such manifold is perfect. Wikipedia says that the only finite fundamental group is binary icosahedral group of size 120. Otherwise fundamental group is infinite. I am not good in understanding infinite groups. Can we say that 3-manifold is "homology sphere" if and only if it's fundamental group is perfect ? – Marek Mitros Mar 1 '16 at 16:49
• Yes. For a group $G$ to be perfect is equivalent to saying that the abelianization of $G$ is trivial, and the abelianization of $\pi_1(M)$ is isomorphic to $H_1(M;\mathbb{Z})$. – Lee Mosher Mar 1 '16 at 17:56
• Can you be more precise about what your question is? – user98602 Mar 2 '16 at 7:21