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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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    A genus $g$ surface is not topologically a wedge of $S^1$s: look at their fundamental groups or their homology.2011-04-02
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    you're right, I think it should be the Hg, not the Sg; we need a solid ball. Then we can continuously shrink each solid torus into an S12011-04-02
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    If my last statement is valid, then it would seem to easily follow that Sg has only trivial embeddings, seeing Sg as a subspace/restriction of Hg.2011-04-02
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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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    Thanks, Jim: does the extra point in S^4 (i.e., S^4 as R^4 \/{pt.} using stereo. proj.) give me some leeway in this respect? Also, what if I restrict my choice of S^g's to those who are bdries. of Hg's (I think this would imply the S^g's are closed)? I am mostly interested in extending the embedding from the known boundary Sg of some Hg, than from any2011-04-03
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    No the extra point doesn't help! Any isotopy in $S^4$ can be deformed slightly to miss a point, so can be realized in $\mathbb R^4$. As for your second question, if your $S_g$ extends to an $H_g$ in $S^4$ (or $\mathbb R^4$), then indeed it is standardly embedded by your argument. But an arbitrary embedding of $S_g$ (which is closed) may not extend to an embedding of $H_g$. (Just like an arbitrary embedding of a circle in $S^3$ doesn't extend to an embedding of a disk in $S^3$.)2011-04-03
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    One more, Jim, please, just to clarify: an Sg is then standardly-embedded if/when the embedding of the Sg can be extended into an embedding of an Hg? I did a search for 'standard embeddings', and did not find any clear definition. I would appreciate a def. and/or some refs. Thanks Again.2011-04-03
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    @gary: One definition of a standard embedding of a surface $S_g$ is that it extends to an embedding of $H_g$. All such embeddings are isotopic to the one where you take a neighborhood of a wedge of circles in $\mathbb R^3\times\{0\}\subset\mathbb R^4$. I hope that helps!2011-04-04
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    It did help, Jim. How do I give you the points?2011-04-04
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    Dammit Jim, I am a doctor, not a Mathema....or, wait, I am a Mathematician--in training :) . Anyway:Just in case, I realized I misunderstood--or misunderestimated--the issue: the surface Sg is described/defined as being embedded, which means (at least for this author), that Sg bounds a g-handlebody. The pair (S^,K); ambient space, embedding--as a knot-- is then defined to be trivial, since any other embedded Sg is isotopic to K. Nice explanation, though.2011-06-12
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    @gary: okay. That's a very unusual definition of "embedded."2011-06-12