Connected sum in an ambient space

Let $M$ be a smooth c connected $m$-manifold, $N_1$, $N_2$ two smooth disjoint connected $n$-submanifolds with boundary, both contained in the interior of $M$ (if necessary). Assume that $M\setminus (N_1\cup N_2)$ is still connected. I would like to do a "boundary connected sum of $N_1$ and $N_2$ in $M$"; that is, cat out small $(n-1)$-discs in $\partial N_i$ and attach $B^{n-1}\times B^1$ that connects $N_1$ and $N_2$ and get a smooth connected submanifold. While such a construction is described in, for example, Kosinski's Differential Manifolds for the case of abstract $N_1, N_2$, how can I do it if I want the resulting connected manifold to be a smooth submanifold of $M$?

If it helps, I can assume that I have a framing of the normal bundles of $N_i$ in $M$. • The proof of existence of embedded connected sum is not hard, you just need to know how to "smooth corners". The trouble is that this connected sum is ill-defined: the main culprit is the relative isotopy class of the arc connecting the manifolds. You need codimension of $N_i$ to be at least 4 for this problem to go away. Jun 19 '14 at 1:53
• @Studious: this is funny: I have an independent assumption (from completely different reasons) that the codimension is at least 4. However, I thought it can be proved if the codimension is at least 2. Please, could you give me more hints? Jun 19 '14 at 6:18
• @studiosus: do I understand it right that if the codimension is smaller, you still can do such construction, but the resulting submanifold will depend on the choice of the arc connecting the manifolds? Jun 19 '14 at 8:36
• @DanielRust Thanks, they are connected -- corrected. Their boundary might be disconnected but I just want to connect one particular component of $\partial N_1$ and one particular component of $\partial N_2$. Jun 19 '14 at 12:35

As pointed out in the comments, the resulting manifold will depend in general on the path chosen to connect $N_1$ to $N_2$.

Choose a point $x_i\in N_i$ for $i=1,2$ and an embedded arc $\alpha$ connecting $x_1$ to $x_2$ in $M\setminus(N_1\cup N_2)$. The arc $\alpha$ has a regular product neighborhood of the form $\alpha\times D^{m-1}$ whose endpopint-disks, that i denote by $D_1^{m-1}$ and $D_2^{m-1}$, contains smaller disks $D_1^{n-1}$ and $D_2^{n-1}$ on $\partial N_1$ and $\partial N_2$. You can now chose a strip-section $\gamma$ of the form $\alpha\times D^{n-1}$ and attach it to $N_1\cup N_2$ along $D_1^{n-1}$ and $D_2^{n-1}$.

This is the connected sum of $N_1$ and $N_2$ along $\gamma$. As said it depends on $\gamma$.

To make the construction smooth, is enough to do in in local coordinate charts $X_1,\dots, X_m$ near $x_i, i=1,2$.

There you have that locally $N_i$ correspond to $\{X_{n+1}=\dots=X_m=0, X_n\leq 0\}$ and you can arrange things so that $\alpha$ corresponds to the $X_{n}$-axis.

In such coordinates the smoothing is easy: just do it for $n=2$ and then extend in a rotationally symmetric way.

As for the in/dependence on $\gamma$, note that if you are in codimension at least $3$ then any $\alpha$ and $\alpha'$ as above are isotopic (provided the $N_i$ are connected, otherwise will depend on the connected compoents where the $x_i$ belong). This is because there is enouch room in $\mathbb R^{m+3}$ for separating $R^m$ from a line, so one can resolve easily evetual crossings.

EDIT: The above sentence is not formally correct. To be more precise, once you chose an arc $\alpha$ then you can move the points $x_i$ where you want, but the dependence on the homotopy class of $\alpha$ in $M$ still is in play.

Finally, if the codimension is at least $2$ then any two strip-section $D^{n-1}\times [0,1]$ in $D^{m-1}\times [0,1]$ are isotopic. Indeed a strip-section is given by $n-1$ linearly independent sections, that we may ortonormalize. Given two such sections $(s_1,\dots,s_{n-1})$ and $(s_1',\dots,s_{n-1}')$ we can isotope $s_1$ to $s_1$ because $S^{m-2}$ is simply connected. The isotopy of $s_2$ now take place in $s_1^\perp$, thus in $S^{m-3}$. And so on. For the last isotopy we need $S^{m-1-(n-1)}=S^{m-2}$ simply connected, hence $m-n\geq 2$.

In the comments I see that the right codimension is at least $4$. So I may have been confusing with indices and notations, but this is easily checked.

• The arc $\alpha$ is probably embedded to $M\setminus \mathrm{interior} (N_1\cup N_2)$, right? Why are the end-point discs $D_1^{m-1}$ and $D_2^{m-1}$ "on $N_{12}$"? Can it not happen that you have, for example a unit ball in $R^2$ and the arc attached to a boundary point $(1,0)$, and the disc $D_1^{m-1}$ a vertical streight-line segment, for example? Jun 19 '14 at 13:07
• first quesiton: yes. second question: yes the disks that are in $N_i$ are the $D^{n-1}$ I'll correct. Third quesstion: I don't understand the question. Jun 19 '14 at 13:29
• The fact that a product neighborhood exists is a fact, that need a proof, but it is true: you can find such a neighborhood. Here I'm assuming that "submanifold" means always "tame submanifold" topospaces.subwiki.org/wiki/Tame_submanifold Jun 19 '14 at 13:34
• I'm familiar with the product neighborhood theorem and agree that we can find a neighborhood diffeomorphic to $\alpha×D^{m−1}$ whose end-point discs are $\simeq D^{m−1}$ and they contain $x_1$, resp. $x_2$. I just can't see why the intersection of these discs cannot be "only" $x_i$, or why it needs to contain a neighborhood of $x_i$ in $\partial N_i$. Jun 19 '14 at 15:32
• @PeterFranek well, suppose you have any product neighborhood. Now adjust it locally by a small perturbation in local coordinates so to get the desired condition. Locally everything happens in $\mathbb R^n\subset \mathbb R^m$ Jun 19 '14 at 16:41