Distances between surfaces in $4$-manifolds

We direct the reader to Figure 6. We first consider a small tubular neighbourhood $\nu(\Gamma_{i}\cup f(\beta))$  in $X$, which is diffeomorphic to $S^{1}\times B^{3}$. We pick a diffeomorphism of $\nu(\Gamma_{i})$ to $B^{4}$, parametrising so that at $t=0$ we see $f(\nu(\gamma_{i}))$ (the sheet of the immersed surface containing $\Gamma_{i}$) and the ends of the arc $f(\beta)$). The rest of the sheet containing the ends of $f(\beta)$ extends into the past and future. We extend our parametrisation to one of $S^{1}\times B^{3}$ so that at each $t$ we see a copy of $S^{1}\times B^{2}$, and so $f(\beta)$ is in the $t=0$ frame and the sheet containing it extends into the past and future; see Figure 6.

The associated tubed surface $\widetilde{S}$ in this parametrisation is depicted in Figure 7.A. We perform a stabilisation between the top of the tube, and the part of surface $f(\nu(\gamma_{i}))$ to obtain Figure 7.B (provided the stabilisation pictured respects orientations; we deal below with the case in which it does not).

To obtain Figure 8.C from Figure 7.B we perform an isotopy supported in the $t=0$ frame by ‘inflating’ the tube from the stabilisation. To obtain Figure 8.D we flatten the resulting bulge, which slightly deforms the surface in other time frames.

Next, we push the sides of the tube into the $t=0$ frame to obtain Figure 9.E. Then we perform an isotopy, flattening the resulting dip to obtain Figure 9.F. To obtain Figure 9.G from Figure 9.F, we thicken the red arc $f(\beta)$ by pushing some of the surface from the future into the present. We then perform an isotopy to create a tube with a band removed, depicted in Figure 9.H.

To obtain Figure 10.I from Figure 9.H we perform an isotopy. In the same picture, we depict an embedded disc intersecting the surface on its boundary in Figure 10.J. We then remove an annulus given by a neighbourhood of the boundary of this disc, and replace it with two parallel copies of the disc in the $t=0$ frame, thus performing a destabilisation, to obtain Figure 10.K. We then push part of the band containing the arc $f(\beta)$ into the future and perform a small further isotopy to obtain Figure 10.L. Note that Figure 10.L is precisely $\widetilde{S^{\prime}}$.

In the case that the above stabilisation was not compatible with orientations, we instead perform a stabilisation on the underside of the surface to obtain Figure 11.A. We then perform the sequence of isotopies in Figure 11, which are analogous to those in the previous case. Note that in Figure 11 only the $t=0$ slice is depicted, the past and future pictures are as in the previous case, until the final picture where the past and future images are swapped (i.e. the past images become the future images and vice versa). Figure 11.F is identical to Figure 9.F, except that the past and future images are swapped. We now proceed as above (with the past and future images swapped). Again we obtain $\widetilde{S^{\prime}}$, completing the proof of the tube swap lemma. ∎

We direct the reader to Figure 13. In Figure 13.A, we see the associated tubed surface $\widetilde{S}$ in a neighbourhood of a point $p\in\Gamma_{i}$ which intersects the linking annulus of $\Gamma_{i}$ on a smaller annulus. First, we perform a stabilisation between the tube and the surface to obtain Figure 13.B (note that if this is not compatible with orientations, we instead perform the stabilisation on the underside and proceed in the same way, taking the mirror image of each figure). We then perform an isotopy to obtain Figure 13.C, then push part of the sides of the tube into the present to obtain Figure 13.D. We then perform a further isotopy to obtain Figure 13.E. Here we see two disconnected tubes, whose ends join the surface as depicted in Figure 13.F.

We can now move these ends around freely on the surface, provided during the isotopy they are disjoint from the other tubes and double points. We depict how we drag the ends diagrammatically in Figure 14. First, we retract the two tubes along $\Gamma_{i}$ dragging the ends towards $z_{i}^{+}$ and $z_{i}^{-}$. We then perform an isotopy that pushes these ends along $\alpha$ so that the two tubes now run along $\alpha$ and the ends of the tubes lie in a neighbourhood of some point $q\in f(\alpha)$. In this neighbourhood, we see the final picture of Figure 13. We then rejoin the tubes, by performing an isotopy and destabilisation which can be seen by reading the pictures in Figure 13 in reverse order. After rejoining the tubes they form the linking annulus of $\alpha$. The resulting surface is $\widetilde{S^{\prime}}$ as required. This completes the proof of the tube move lemma. ∎

Recall that $\Sigma$ and $\Sigma^{\prime}$ are immersed surfaces in $X$ of genus $g$. In the case the surfaces are not regularly homotopic $d_{\operatorname{sing}}(\Sigma,\Sigma^{\prime})=\infty$ and we are done. Hence assume that $\Sigma$ and $\Sigma^{\prime}$ are regularly homotopic and suppose $d_{\operatorname{sing}}(\Sigma,\Sigma^{\prime})=n$. Given distance minimising regular homotopy from $\Sigma$ to $\Sigma^{\prime}$, let $P_{1},\ldots P_{k}$ be the sequence of immersed surfaces describing this homotopy, each differing from the previous either a finger move, a Whitney move, or an ambient isotopy.

We shall describe a sequence of surface tubing diagrams with associated tubed surfaces that differ by stabilisations, destabilisations, and ambient isotopy, such that the genus of any intermediate surface never exceeds $g+n+1$.

Fix the abstract surface $S$, and immersions $f_{i}\colon S\rightarrow X$ with image $P_{i}$ for each $i$. The immersion $f_{1}\colon S\rightarrow X$ gives a surface diagram $\mathcal{S}_{1}$ (the empty diagram in $S$) with associated tubed surface $\widetilde{S}_{1}=S_{1}^{\operatorname{im}}=P_{1}$.

We now suppose for induction that we have a surface tubing diagram $\mathcal{S}_{i}$ with immersion data $f_{i}\colon S\rightarrow X$, and associated tubed surface $\widetilde{S}_{i}$. We will construct a surface tubing diagram $\mathcal{S}_{i+1}$ with immersion data $f_{i+1}\colon S\rightarrow X$, such that $\widetilde{S}_{i+1}$ differs from $\widetilde{S}_{i}$ by a series of stabilisations, destabilisations, and ambient isotopy, such that the genus of any intermediate surface does not exceed $g+n+1$. There are three cases; $P_{i+1}$ is obtained from $P_{i}$ by ambient isotopy, a finger move, or a Whitney move.

Ambient Isotopy: If $P_{i+1}$ just differs from $P_{i}$ by ambient isotopy, by Remark 2.12(1) we have a new surface tubing diagram $\mathcal{S}_{i+1}$ with immersion data $f_{i+1}\colon S\rightarrow X$. Furthermore, $\widetilde{S_{i}}$ is ambiently isotopic to $\widetilde{S}_{i+1}$.

Finger Move: If $P_{i+1}$ differs from $P_{i}$ by a finger move, let $\alpha$ and $\beta$ be the Whitney arcs of the Whitney disc that undoes the finger move. Note these arcs are in $S$, the abstract surface. Before we perform the finger move, we perform an isotopy of the arcs, to push any arcs off $\alpha$ and $\beta$ to form $\mathcal{S}_{i}^{\prime}=(f_{i},\{\gamma_{j}^{\prime}\})$. We then add the new double points and $\beta$ to the diagram to obtain diagram $\mathcal{S}_{i+1}=(f_{i+1},\{\gamma_{j}^{\prime}\}\cup\{\beta\})$ which we note uses the new map $f_{i+1}$, which differs from $f_{i}$ by the finger move; see Figure 15. The associated tubed surfaces $\widetilde{S_{i}}$ and $\widetilde{S^{\prime}_{i}}$ are ambiently isotopic by Remark 2.12(2), and $\widetilde{S^{\prime}_{i}}$ and $\widetilde{S}_{i+1}$ differ by isotopy and a stabilisation, as in Figure 16.

Whitney Move: The Whitney moves present the main difficulty. If $P_{i+1}$ differs from $P_{i}$ by a Whitney move, there are several cases to consider.

The endpoints of the Whitney arcs form the set $\{x_{i}^{+},y_{i}^{+},x_{j}^{-},y_{j}^{-}\}$ for some $i$ and $j$. In the case that $i=j$ then either $\partial\alpha=\{x_{i}^{+},x_{i}^{-}\}$ and $\partial\beta=\{y_{i}^{+},y_{i}^{-}\}$, or $\partial\alpha=\{x_{i}^{+},y_{i}^{-}\}$ and $\partial\beta=\{y_{i}^{+},x_{i}^{-}\}$ (up to relabeling of $\alpha$ and $\beta$). We call the former situation Case 1 and the latter situation Case 2. Case 2 is the crossed Whitney disc .

Otherwise, if $i\neq j$ then either $\partial\alpha=\{x_{i}^{+},x_{j}^{-}\}$ and $\partial\beta=\{y_{i}^{+},y_{j}^{-}\}$, or $\partial\alpha=\{x_{i}^{+},y_{j}^{-}\}$ and $\partial\beta=\{y_{j}^{+},x_{i}^{-}\}$ (again up to relabeling of $\alpha$ and $\beta$). We call the former situation Case 3 and the latter Case 4.

Note that $\alpha$ and $\beta$ may intersect other arcs, including $\gamma_{i}$ and $\gamma_{j}$. In all cases, we perform a small isotopy of all the arcs so that they intersect $\alpha$ and $\beta$ transversely.

Case 1: The Whitney arcs are $\alpha$ between $x_{i}^{+}$ and $x_{i}^{-}$, and $\beta$ between $y_{i}^{+}$ and $y_{i}^{-}$ for some $i$. We wish to run in reverse what we did in the case of a finger move, however, we need $\gamma_{i}$ to run over either $\alpha$ or $\beta$, and we also need $\alpha$ and $\beta$ to be disjoint from other arcs, so that the surface tubing diagram looks like the diagram at the end of a finger move, as in Figure 15.

We arrange this by first using the tube move lemma to move any arcs off $\beta$. To do so we remove intersection points with $\beta$ one by one. We consider the arc $\gamma_{r}$ which has the intersection with $\beta$ closest to $y_{i}^{+}$ (in the sense of distance along $\beta$), at $p\in\beta$. We form the arc $\gamma_{r}^{\prime}$, by removing an arc neighbourhood of $p$ from $\gamma_{r}$, and replacing it with an arc that runs along the boundary of a small neighbourhood of $\beta\subset S$; see Figure 17. Provided the neighbourhood is taken to be sufficiently small, $\gamma_{r}^{\prime}$ is disjoint from $\{\gamma_{p}\}_{p\neq r}$ and from marked points other than $x_{r}^{\pm}$, and so the corresponding associated surfaces differ by a stabilisation and destabilisation by the tube move lemma. We do this for every intersection point of arcs with $\beta$, until all arcs are disjoint from $\beta$. Note that we may move $\gamma_{i}$ during this process.

Next, we use the tube swap lemma to swap $\gamma_{i}$ to $\beta$. We then use the tube move lemma to move any arcs off $\alpha$ as we did for $\beta$; see the third to fourth picture of Figure 17. Finally, we perform the destabilisation and isotopy coming from reading Figure 16 in reverse. This has the effect of removing the points $x_{i}^{\pm}$ and $y_{i}^{\pm}$, and the arc $\gamma_{i}$ (which now runs along $\beta$) from the diagram.

Case 2: The Whitney arcs are $\alpha$ between $x_{i}^{+}$ and $y_{i}^{-}$, and $\beta$ between $y_{i}^{+}$ and $x_{i}^{-}$ for some $i$. This is a crossed Whitney disc. As in Case 1, we first remove any arcs running over $\alpha$ and $\beta$ using the tube move lemma, which may move $\gamma_{i}$; see Figure 18. Let $\mathcal{S}_{i}^{\prime}$ be the resulting surface diagram, and $\mathcal{S}_{i+1}$ be the diagram with the same points and arcs but with the points $x_{i}^{\pm}$ $y_{i}^{\pm}$, the arc $\gamma_{i}$ deleted, and immersion data $f_{i+1}$; See Figure 18.

We show $\widetilde{S_{i}^{\prime}}$ and $\widetilde{S}_{i+1}$ are related by a stabilisation and two destabilisations (and isotopy). We depict the immersed surface $S_{i}^{\prime\operatorname{im}}$ in a neighbourhood of the Whitney disc in Figure 20. The associated tubed surface in this neighbourhood is shown in Figure 20. To obtain the second picture of Figure 20 we perform an isotopy which pulls apart the surface (taking the sheets of the surface to where they need to be after the Whitney move), at the expense of creating a double tube again using the terminology of Gabai [Gab17], a Hopf link$\times[0,1]$ running through $X$ between two disc neighbourhoods of the surface. The Hopf link$\times[0,1]$ has four boundary components, two of which are joined to the surface (the top of the green tube and bottom of the pink), while the other two join to the linking annulus of $\Gamma_{i}$ (the bottom of the green and top of the pink).

We now perform a stabilisation inside the Hopf link$\times[0,1]$, which can be seen in the movie picture as attaching two bands (provided this is compatible with orientation, if not see below); see Figure 22. The middle picture of this movie is then an unknot, along which we perform a destabilisation to cut the double tube; again see Figure 22.

The resulting ‘end’ of the double tube can be made by taking an open double tube, adding a band between the two tubes, then capping off the resulting boundary circle with a disc. It can also be thought of as the standard annulus whose boundary is the Hopf link whose interior is pushed into the 4-ball. We now suck back these ends to be close to the surface. The resulting surface differs from $\widetilde{S}_{i+1}$ only in a 4-ball neighbourhood of the arc $\Gamma_{i}$. The resulting capped off double tube in this 4-ball neighbourhood is pictured on the right in Figure 22. In this neighbourhood, we see a genus 1 slice surface for the unknot. We call this slice surface $K$.

If the above stabilisation was not compatible with the orientation on the double tube, we instead perform a different stabilisation and destabilisation which are compatible with the alternative possible orientations; see Figure 23. Again the result is a cut open double tube, and we again suck the ends back to lie in a 4-ball neighbourhood of $\Gamma_{i}$. We call the resulting surface in this neighbourhood $K^{\prime}\subset B^{4}$ and note the boundary of $K^{\prime}$ is the unknot in $S^{3}$ as before.

We now wish to construct a sequence of stabilisations and destabilisations taking $K$ and $K^{\prime}$ to the trivial disc. Such a sequence would remove the mess of tubes, and take our surface to $\widetilde{S}_{i+1}$ as required. We exhibit such a sequence in Lemma 6.2. Note that results in [BS15] imply that some sequence of stabilisations and destabilisations exists but we show that in fact, one destabilisation is sufficient, and so the genus of any intermediate surface does not exceed $g+n+1$.

Case 3: The Whitney arcs consist of an arc $\alpha$ between $x_{i}^{+}$ and $x_{j}^{-}$, and an arc $\beta$ between $y_{i}^{+}$ and $y_{j}^{-}$ for $i\neq j$. This case presents the most diagrammatic difficulty. We wish to remove the intersections of arcs with $\alpha$ and $\beta$, however this is made difficult by the fact that $\gamma_{i}$ and $\gamma_{j}$ may intersect $\alpha$ and $\beta$ in a complicated way; see Figure 24.

To overcome this, we first we pick an arc $a_{i}$ from $y_{i}^{+}$ to $y_{i}^{-}$ which is disjoint from $\alpha\cup\mathring{\beta}$ and all other marked points on the surface; clearly, we may do so since the complement of $\alpha\cup\mathring{\beta}$ and all points is connected. We then similarly pick an arc $a_{j}$ from $y_{j}^{+}$ to $y_{j}^{-}$ which is disjoint from $\alpha\cup\mathring{\beta}\cup a_{i}$ and all marked points other than $y_{j}^{+}$ and $y_{j}^{-}$; again the complement of these arcs and points is connected since $a_{i}$ does not intersect $\alpha$ or $\beta$. We remove the intersections of all arcs with $a_{i}\cup\beta\cup a_{j}$, which is one long embedded arc, one by one using the tube move lemma as in previous cases; see Figure 24. Note that this may move $\gamma_{i}$ and $\gamma_{j}$.

We now use the tube swap lemma to swap $\gamma_{i}$ to $a_{i}$ and $\gamma_{j}$ to $a_{j}$. We then move any arcs off $\alpha$ using the tube move lemma.

Finally we remove the intersection points $y_{i}^{+}$, $x_{i}^{+}$, $y_{j}^{-}$, $x_{j}^{-}$, join the arcs $\gamma_{i}$ and $\gamma_{j}$ using $\beta$, and change the index of the points labelled $j$ to $i$; see Figure 24. The corresponding isotopy and destabilisation of the associated surface is given by Figure 25.

Case 4: The Whitney arcs are an arc $\alpha$ between $x_{i}^{+}$ and $y_{j}^{-}$, and an arc $\beta$ between $y_{i}^{+}$ and $x_{j}^{-}$ for $i\neq j$. In this case, we use the tube swap lemma to swap $\gamma_{i}$ to any arc between $y_{i}^{+}$ and $y_{i}^{-}$ disjoint from other tube arcs, but which may intersect $\alpha$ and $\beta$; some such arc exists since the complement of $\cup_{r}\gamma_{r}\cup_{k}y_{k}^{\pm}$ is a punctured surface, so is connected. We are then in Case 3 and proceed as before.

At stage $P_{k}$, we have a surface diagram $\mathcal{S}_{k}$ with immersion data $f_{k}\rightarrow X$. Since $f_{k}$ is an embedding it has no intersection points so must be the empty diagram, hence the associated tubed surface $\widetilde{S}_{k}$ is $f_{k}(S)=P_{k}$ as required. This completes the proof of Theorem 1.4 modulo Lemma 6.2. ∎