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I am sorry that this question is probably too basic - I could not seem to find the answer though. I know the following - let $S$ be a closed orientable surface, an element of $H_1(S;\mathbb{Z})$ is represented by an embedded curve if and only if it is primitive, and the question of whether or not an element of $\pi_1(S)$ can be represented by an embedded curve is at least algorithmically decidable. Also - the sphere theorem implies that if $\pi_2(M) \neq 0$ then there will be some nonzero element represented by an embedded sphere.

Let $M$ be a (closed, orientable if you like) 3-manifold. What can be said about elements of $\pi_2(M)$ being represented by embedded spheres? Are there some simple obstructions? Are there some necessary/sufficient conditions?

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  • $\begingroup$ Let me understand a bit better. Your question is, given a (closed, oriented) 3-manifold $M$ and a map $\alpha:S^2\rightarrow M$, when is $\alpha$ homotopic to an embedding? $\endgroup$ – Tyrone May 23 '18 at 10:01
  • $\begingroup$ @Tyrone Yes I would like to know that - and I might guess that that can be algorithmically determined. But I was also fishing for what else might be known - for example if there are some simple algebraic obstructions. $\endgroup$ – user101010 May 24 '18 at 9:36
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This should be a comment, but was a bit too long.

I'll take you up on your offer to assume that $M$ is orientable, since then it is parallelizable and it follows from the Smale-Hirsch theory of immersions that any map $f:S^2\to M$ is homotopic to an immersion (see the accepted answer to Realizing a homology by a smooth immersion). The same theory shows that two immersions are regularly homotopic if and only if they're homotopic.

So your question can be reduced to: given an immersion $f:S^2 \looparrowright M^3$ in general position, is it regularly homotopic to an embedding?

Questions such as this come up in surgery theory, when trying to perform surgery just above the middle dimension. For immersions $f: S^n\looparrowright M^{2n-1}$ with $n\ge 3$, there are complete obstructions described by Hatcher and Quinn. But for $n=2$ the Whitney trick breaks down, which makes me think that sufficient conditions may be hard to come by.

Here is an idea for finding necessary conditions, i.e. obstructions to $f:S^2 \looparrowright M^3$ being regularly homotopic to an embedding. Such an immersion will have finitely many double circles $C_1,\ldots , C_n$ in $M$, and each circle $C_i$ is doubly covered by its pre-image $\bar{C_i}$ in $S^2$. If the cover $\bar{C_i}\to C_i$ is non-trivial, then the class in $\pi_1(M)$ represented by $C_i$ is $2$-torsion, since twice round the loop comes from $\pi_1(S^2)$ which is trivial.

We can choose a smooth map $h:S^2\to \mathbb{R}$ such that $g:=(f,h):S^2\looparrowright M\times\mathbb{R}$ is an immersion in general position, which therefore has finitely many double points. In fact the double points of $g$ correspond to the non-trivially covered double circles of $f$, see

Szűcs, András, Note on double points of immersions, Manuscr. Math. 76, No. 3-4, 251-256 (1992). ZBL0776.57017.

It follows that if $f$ is regularly homotopic to an embedding, then the number of double circles in each $2$-torsion class in $\pi_1(M)$ must be even.

This is difficult to turn into an algebraic condition, hence why this should be a comment rather than an answer. It's possible that someone more familiar with the literature on $3$-manifolds can come up with something more conclusive.

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If one has a connect sum of two non-trivial irreducible 3-manifolds $M=M_1\#M_2$, then there is a unique embedded essential 2-sphere $S^2$ up to isotopy (forgetting orientation). Hence the elements of $\pi_2(M)$ realized by an embedded 2-sphere will just be the orbit of this 2-sphere under the action of $\pi_1(M)=\pi_1(M_1)\ast\pi_1(M_2)$. With the identification $\pi_2(M) \cong \pi_2(\tilde{M})\cong H_2(\tilde{M})$ via Hurewicz, where $\tilde{M}$ is the universal cover, we can see that most classes (even primitive) will not be realized by an embedded 2-sphere.

From the proof of the geometrization conjecture, one may conclude that $\tilde{M} \cong \mathbb{S^3}-K$, where $K$ is either 2 points (if $M_1\cong M_2\cong \mathbb{RP}^3$, or $K$ is a tamely embedded Cantor set. I have some notes describing how this goes. In the first case, $H_2(\tilde{M})\cong \mathbb{Z}$, and the element is represented by an embedded sphere iff it is primitive.

In the second case, one may represent any element of $H_2(\tilde{M})$ by a disjoint union of oriented 2-spheres $\Sigma$. Moreover, we may assume this is minimal, so that if components of $\Sigma$ are adjacent, then the orientations agree (otherwise, we may tube them together to get fewer spheres). If $\Sigma$ is disconnected, the the corresponding class of $\pi_2(\tilde{M})$ cannot be represented by an embedded 2-sphere, and hence the same for $\pi_2(M)$. But $\Sigma$ may be disconnected without being non-primitive, since there may be components of $K$ between each of the components of $\Sigma$ which prevent them from being parallel. I believe figuring out if such a class is connected should be algorithmic (but not practically), by constructing enough of the universal cover, and then performing 3-manifold algorithms to find the embedded spheres.

However, embeddedness of spheres in $\pi_2(\tilde{M})$ is not equivalent to embeddedness in $\pi_2(M)$. In the case at hand, where there is only one class of embedded 2-sphere (up to the action of $\pi_1(M)$), I believe that there ought to be an algorithm. Pull back the embedded 2-sphere $S$ from $M$ to $\tilde{M}$ to get a surface $\tilde{S}$ with infinitely many components. Then we want to be able to determine whether an embedded 2-sphere $\Sigma$ in $\pi_2(\tilde{M})$ is isotopic to one of the components of $\tilde{S}$. As described in my notes, $\tilde{M}$ is a union of thrice-punctured spheres. We may assume that the components of $\tilde{S}$ are contained in this union of 3-punctured spheres. Then take enough of these 3-punctured spheres to contain $\Sigma$ and be connected, giving a submanifold $N\subset \tilde{M}$. The by Hurewicz, $\pi_2(N)=H_2(N)$. Hence $\Sigma$ will be parallel to a sphere of $\tilde{S}$ iff it is homologically equivalent to a sphere of $\tilde{S}$ in $N$ (up to orientation).

I believe that this approach could be made algorithmic, although not in a practical way.

For the general case, I think there ought to be an abstract criterion that implies embeddedness. By the previous consideration, a 2-sphere in $\pi_2(M)$ lifts to a class of $H_2(\tilde{M})$ represented by an embedded sphere. This sphere separates the ends $E=\partial_{\infty}\tilde{M}$ into two pieces $E_+, E_-$. Then one can associate to this partition an action on a CAT(0) cube complex by Sageev's theory. There is a way to do this canonically in this case using a wall space on $E$, so that the sphere is embedded iff the complex is a tree. I won't delve into the theory, but suffice it to say that the sphere will be embedded iff the partition $\{E_+,E_-\}$ has the property that for each element $g\in \pi_1(M)$ acting on $\tilde{M}$ by covering translations and hence on $E$, either $g(E_+)\subset E_+$ or $g(E_+)\subset E_-$.

Again, it should be possible to make this criterion algorithmic by building "enough" of $\tilde{M}$ to see if a lift of $S$ crosses any of its covering translates in an essential way. One might even be able to make this computable using PL minimal surface theory.

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    $\begingroup$ In the first sentence it looks like you forgot to include the hypothesis that $M_1$ and $M_2$ are irreducible, which is needed in order to have uniqueness up to isotopy of an essential separating sphere. $\endgroup$ – Allen Hatcher Jul 29 '18 at 20:55

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