Embedded 2-tori in $S^1\times S^4$

Question

I am interested in understanding the smooth isotopy class of embedded 2-tori in $S^1\times S^4$. Is it true that every two homotopic embedded 2-tori in $S^1\times S^4$ are smoothly isotopic? It would be great if there is a certain analogue of Haefliger's theorem in these settings...

Answer 1

I think this is true (homotopy implies isotopy). Consider a generic smooth torus embedding $f: T^2 \to S^1\times S^4$, then the projection to $S^1$ gives a circle-valued Morse function on $T^2$. Because of the ambient dimension (codimension 3), I think one may isotope $f$ to cancel critical points and obtain an isotopic embedding $f': T^2\to S^1\times S^4$ such that the projection to $S^1$ has no critical points when restricted to $T^2$ if the map $f_{|T^2}:T^2\to S^1$ is homotopically non-trivial. Thus $f'^{-1}(z)$ is a union of $k$ circles in $S^4$ for all $z\in S^1$, where $k$ is the index of $\pi_1(T^2)\to \pi_1(S^1)$. The homotopy type of $f'$ should be determined by $k$

For the homotopically trivial case, I think the technique of the Whitney-Wu unknotting theorem implies that any two homotopically trivial tori embedded in $S^1\times S^4$ are isotopic. The point is that we may use Morse theory to shrink the torus down into an $I\times S^4$ slice, then apply Whitney-Wu.

Now consider the homotopically non-trivial case, where $f’^{-1}(z)$ consists of $k$ circles for every $z$. We may think of this as a map $g: S^1 \to Emb(\sqcup^k S^1, S^4)$, where the monodromy permutes the components of $\sqcup^k S^1$ cyclically so that the mapping torus is a torus. Then $g(1)$ is an embedding of $\sqcup^k S^1$ to $S^4$. Let $g(1)=\gamma_1 \sqcup \cdots \sqcup \gamma_k$ a disjoint union of circles. By general position, this is the boundary of an embedding $D^2_1 \sqcup \cdots \sqcup D^2_k$. Assume that the monodromy sends $\gamma_i$ to $\gamma_{i+1}$, indices taken $(\mod k)$. We may extend $g$ to an ambient isotopy of $S^4$, and hence get a path $G: [0,1] \to Emb( \sqcup^k D^2 , S^4)$ with boundary restricting to $g$. We may also assume that $D_i$ is taken to $D_{i+1}$ by the isotopy, $1\leq i< k$, and $D_k$ is taken to $D_1’$ with $\partial D_1’=\gamma_1$.

By the 4-dimensional light-bulb theorem (Theorem 1.10), $D_1$ is isotopic to $D_1'$ rel $\gamma_1$. By general position, we may assume that this isotopy misses $D_2, \ldots, D_k$ (since these all sit inside of disjoint balls).

So this isotopy may be achieved by a 1-parameter isotopy of disks $G’:S^1 \to Emb(\sqcup^k D^2,S^4)$. But any such isotopy is homotopic to a cyclic rotation of order $k$. Consider a point on the disk $D_1$, the point gives a loop in $S^4$ by iterating the isotopy $k$ times, which is contractible, and hence one may assume that the isotopy the points on the disks $D_1, \dots, D_k$ arranged around a round circle. Then one may also assume that the tangent space to the disks at the point are permuted cyclically, and then that the disks are permuted by a rotation by shrinking them down to the tangent space. Hence there should be only one isotopy class of embedded tori for each $k$.

It’s possible that the application of the 4D lightbulb trick here is overkill, one might look at the classical isotopy theory linked to above to see if it can be applied directly to show that homotopy implies isotopy in this case (things tend to be more flexible in higher dimensions).