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I think we can argue that Sg --the genus-g surface -- has only a trivial embedding in S4 , since Sg is topologically a wedge of g S1's, and there are no knotted S1's in S4 (meaning that any two embeddings of S1 in S4 are isotopic.) But I am not clear on why there are no non-trivial embeddings of Hg ----a 3D handlebody; a 3-sphere with g handles----in S4.

If the first argument about Sg works, can I use it somehow; specifically using the fact that Sg is the boundary of Hg, to argue that there are no non-trivial embeddings of an Hg in S4? Or do I need an additional assumption for this last to be true? Or can I use some sort of surgery argument ?

Thanks in Advance.

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    Please see my comment on today's post on the Torelli group. I am posting from a public computer, and I lost track of my login, so I have been having trouble logging in, so that I cannot vote an answer as the correct one. So for the meantime, I have been posting comments instead of answers, since it would not be fair to receive points when I do not-- for the time being --vote others up.2011-06-01

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There are lots of knotted embeddings of surfaces in $\mathbb R^4$. One of the easiest examples is a spun knot. This is a knotted sphere in $\mathbb R^4$ which is obtained by rotating a knotted arc in $\mathbb R^3$ around a plane in $\mathbb R^4$. There are several results in the literature that show that these are not trivial embeddings.

Now to get an embedding of any higher genus surface, you can take a knotted embedding of a sphere and connect it by a thin tube to a standard embedding of a surface to obtain a knotted embedding of any oriented surface.

The problem with your argument is that you can't necessarily extend an embedding of $S_g$ to an embedding of a handlebody.

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    @$g$ary: okay. That's a very unusual definition of "embedded."2011-06-12