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Consider the space of smoothly embedded genus $g$ surfaces in the 3-sphere under the $C^\infty$ topology: $$\mathcal E_g:=\operatorname{Emb}(\Sigma_g,S^3)/{\operatorname{Diff}(\Sigma_g)}$$ where $\Sigma_g$ is a genus $g$ surface. Note that Hatcher proved that $\mathcal E_0$ has the homotopy type of $\mathbb{RP}^3$.

What is known about the topology, e.g. homology groups, cohomology ring, or homotopy groups, of $\mathcal E_g$?

Edit: For example, if we consider a genus $g$ surface which looks like one big sphere with $g$ small handles, and we move the handles around, then it is like moving points in $\mathbb R^2$. Thus we have a map from the unordered configuration space $\mathrm{UConf}_g(\mathbb R^2)\to\mathcal E_g$. Since the cohomology of the former is well-understood, what would it tell us about that of the latter?

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  • $\begingroup$ The $g=1$ case has a homotopy-type close to that of $Emb(S^1,S^3) / Diff(S^1)$, the map is given by taking tubular neighbourhoods. For $g \geq 2$ not much is known about the homotopy-type of these embedding spaces. $\endgroup$
    – Ryan Budney
    Commented Nov 18, 2022 at 16:12
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    $\begingroup$ Moreover, the "map" you describe, I do not think it is well-defined. $\endgroup$
    – Ryan Budney
    Commented Nov 19, 2022 at 2:02
  • $\begingroup$ One way to get a well-defined map is to use "spaces of graphs" rather than configuration spaces. $\endgroup$
    – Ryan Budney
    Commented Nov 19, 2022 at 2:15
  • $\begingroup$ @Ryan Budney: If we allow varying the size of the handles, then the map should be well-defined? Alternatively, we can also consider $\mathrm{UConf}_g(S^2)$ instead, but then the "rotation" of individual small handles matter, so we have a map from some $(S^1)^g$-bundle over $\mathrm{UConf}_g(S^2)$ to $\mathcal E_g$. $\endgroup$
    – Adrian Chu
    Commented Nov 19, 2022 at 2:58
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    $\begingroup$ No, I don't think such a map exists. The problem comes when you try to extend your map to the part of the configuration space where you have collinear triples (or more) points. You have to deform the handles to avoid collisions, and your domain does not have enough information to define such maps. $\endgroup$
    – Ryan Budney
    Commented Nov 19, 2022 at 3:05

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The first observation is that $\mathcal{E}_g$ is not connected when $g > 0$. This is due to the existence of "knotting". For example, in genus one, let $K$ and $K'$ be smooth knots. Let $T$ and $T'$ be boundaries of small tubular neighbourhoods of $K$ and $K'$ respectively. Then $T$ is isotopic to $T'$ if and only if $K$ is isotopic to $K'$.

To avoid this issue one may assume that the surface $\Sigma_g$ divides $S^3$ into a pair of handlebodies. In this case it is a result of Waldhausen that the embedding space $\mathrm{Emb}$ is connected. It will simplify the discussion to assume that $\Sigma_g$ is oriented (and use $\mathrm{Diff}^+$).

For $g > 1$ I then believe that the fundamental group of $\mathrm{Emb}$ is the "Goeritz group". If $g = 2$ this known to be finitely generated. In higher genus finite generation is open.

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  • $\begingroup$ By the way, is there a known example of two connected components of Emb$\Sigma$ which are not homotopy equivalent? $\endgroup$
    – Denis T
    Commented Nov 18, 2022 at 15:47
  • $\begingroup$ I don't know. However, there is a natural place to look. Suppose that $T$ is a two-torus. So $T$ bounds a solid torus on at least one side. If it bounds a solid torus on both sides call it unknotted. In this case, the component of $\mathrm{Emb}(T)$ should have the homotopy type of the Grassmaniann $\mathrm{Gr}(2, 4)$ (modulo an involution depending on some orientation choices). $\endgroup$
    – Sam Nead
    Commented Nov 18, 2022 at 19:23
  • $\begingroup$ If $T$ bounds a solid torus on one side, and a Seifert fibered space on the other, then I'll guess that the homotopy type becomes $\mathrm{SO(4)}/S^1$. And in the general case, I'll guess that the homotopy type is $\mathrm{SO(4)}$. $\endgroup$
    – Sam Nead
    Commented Nov 18, 2022 at 19:28

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