Fundamental group of a generalized connected sum

Question

Let $M$ and $N$ be two $n$ dimensional connected closed manifolds with $n \ge 3$, and let $S$ be a $(n-1)$ dimensional closed submanifold common to $M$ and $N$. Consider the connected sum of $M$ and $N$ along the submanifold $S$, denoted $M\#_{S} N$ (described in https://en.wikipedia.org/wiki/Connected_sum#Connected_sum_along_a_submanifold). Roughly speaking, this is obtained by deleting $S$ from both the manifolds and then gluing along the boundary. This generalizes the canonical idea of connected sum, denoted $M \#N$, in which $S$ is taken to be a singleton.

I want to calculate the fundamental group of $M\#_{S} N$. I know how to calculate $\pi_1(M\# N)$ using the Seifert–Van Kampen theorem. I believe the same idea can be applied to this general case of connecting along $S$ but I'm running into problems since $S$ is arbitrary. Is there a way I can see how $\pi_1(M\#_S N)$ looks like in general?

I understand that it might depend on $S$. Just like $\pi_1(M\# N) = \pi_1(M) \ast \pi_1(N)$, can $\pi_1(M\#_S N)$ be described in a similar, simple way in terms of $\pi_1(M)$, $\pi_1(N)$, and $\pi_1(\partial S)$ only? Thanks in advance for your help.

Edit: two corrections made in light of the comments.

Answer 1

This basic question is unfortunately not well explained anywhere in the literature that I know of, although the answer is well known to lots of people. When $\pi_1(S)$ embeds into $\pi_1(M)$ and $\pi_1(N)$, then the answer can be expressed in terms of graphs of groups, explained in Serre's book Trees. Without the $\pi_1$-injectivity assumption, the answer is no longer a graph of groups, but the presentations that Serre gives you still work. Since you are motivated by topology, you might also be interested in Scott and Wall's article Topological methods in group theory.

In any case, I will try to sketch the answer, with an emphasis on giving actual presentations that you can use.

I will assume that $S$ is 2-sided, so the boundary of the normal bundle of $S$ consists of two copies of $S$; let's denote them $S_+$ and $S_-$. Let $M_0$ be the result of cutting $M$ along $S$ and $N_0$ the result of cutting $N$ along $S$. The generalised connect sum is now the glued manifold

$M\#_S N=M_0\cup_{S_+\cup S_-} N_0$.

I will also assume that $S$ is connected; if not, you can deduce the final answer by iterating the calculation below. Finally, let's use $i:S_+\cup S_-\to M_0$ and $j:S_+\cup S_-\to N_0$ for the inclusion maps.

There are several cases depending on whether or not $M_0$ and $N_0$ are connected. If $M_0$ is disconnected, write $M_0=M_+\sqcup M_-$, where $S_\pm\subseteq M_\pm$, and similarly for $N_0$.

If $M_0$ and $N_0$ are both disconnected, then the glued manifold is also disconnected, and so you need to compute the fundamental groups of the components separately, which can be done using the Seifert--van Kampen theorem. You get:

$\pi_1(M_\pm\cup_{S_\pm}N_\pm)\cong \pi_1(M_\pm)*\pi_1(N_\pm)/\langle\langle i(g)^{-1}j(g)\mid g\in\pi_1(S_\pm)\rangle\rangle$.

When the inclusion maps are injective on $\pi_1$, this is the amalgamated free product

$\pi_1(M_\pm)*_{\pi_1(S_\pm)}\pi_1(N_\pm)$.

If $M_0$ is connected but $N_0$ is disconnected, then

$M\#_S N= N_+\cup_{S_+} M_0\cup_{S_-}N_-$

and the fundamental group can be computed by iterating the calculation from the previous case. In particular, if the inclusion maps are $\pi_1$-injective then you get the result

$\pi_1(M\#_S N)=\pi_1(N_+)*_{\pi_1(S_+)} \pi_1(M_0)*_{\pi_1(S_-)}\pi_1(N_-)$.

Finally, the most interesting case is when both $M_0$ and $N_0$ are connected. Many people will start to talk about groupoids at this point, but this is an unnecessary complication. You should read about HNN extensions in the sources I mentioned above. In any case, the answer is

$\pi_1(M\#_S N)=\pi_1(M_0)*\pi_1(N_0)*\langle t\rangle/\langle\langle\{ i(g_+)^{-1}j(g_+)\mid g_+\in\pi_1(S_+)\},\{ i(g_-)^{-1}tj(g_-)t^{-1}\mid g_-\in\pi_1(S_-)\}\rangle\rangle$.

Note the conjugation by $t$. Topologically, $t$ represents a loop that starts in $M_0$ (say), traverses $S_+$ to enter $N_0$ and exits $N_0$ through $S_-$.

When the inclusion maps are $\pi_1$-injective, this decomposes $\pi_1(M\#_S N)$ as a graph of groups, with underlying graph a circle.

Note that, in all cases, the answer isn't just a function of $\pi_1(M)$ and $\pi_1(N)$; you also need to understand the complementary manifolds $M_0$ and $N_0$, and how the surface $S$ sits inside them.

Answer 2

You want to look up the idea of “free product with amalgamation”. Also, you need side-conditions on how $S$ lies in the manifolds $M$ and $N$ - namely, the normal bundles need to be isomorphic. Finally, the eventual answer may (and generally will) depend on the choice of isomorphism.

Answer 3

Disclaimer: I wrote this answer before realizing the question was specifically about codimension 1. This answer only works in codimension 1 for the case that the normal bundle is not orientable.

First, before we compute the answer, let's fix the data needed to make the question make sense:

We take two closed $n$-manifolds $M, N$, and two submanifolds $S_M, S_N$ together with an isomorphism $\varphi$ of their respective normal bundles $\nu S_M, \nu S_N$ in $M, N$ respectively (in particular $S_M, S_N$ have to be isomorphic, but this is more data than that). When I don't care about embeddings, I'll call the manifold $S$. I'll assume $S$ is connected, though you can try to see how much carries through if it's not.

Now we can form what you call the generalized connected sum: After appealing to compactness, we may identify $\nu S_M, \nu S_N$ with small tubular neighborhoods of $S_M, S_N$, pick some metric on the normal bundles compatible via $\varphi$, and define $$M\#_S N = (M\backslash S_M \sqcup N\backslash S_N )/ (v\sim\frac{\varphi(v)}{|v|^2}) \text{ for $v\in \nu S_M\backslash S_M$}$$

To compute the fundamental group of $M \#_S N$, we apply Seifert-van Kampen to the natural cover of this space by the open sets $M\backslash S_M$ and $N\backslash S_N$, where we consider these as subsets through the above definition of $M \#_S N$. (Note we should assume at this point that $\nu(S)\backslash S$ is connected, i.e. that either the codimension is at least 2 or that the codimension is 1 and the normal bundle is non-orientable)

By looking at the equivalence relation we divided out in the construction, the intersection of these sets is given by $\nu S_M \backslash S_M$ ($=\nu S_N \backslash S_N$ in the quotient), so is a deformation retract of the (unit) sphere bundle $S(\nu S_M)$ of the normal bundle.

Similarly, $M\backslash S_M$ deformation retracts onto the compact manifold with boundary $(M\backslash D_1(\nu S_M), S(\nu S_M))$, where $D_1(\nu S_M)$ denotes the bundle of vectors of length $< 1$ in $\nu S_M,$ and similarly for $N$. Thus we get the general answer

$$\pi_1(M\#_S N) = \pi_1(M\backslash S) \star_{\pi_1(S(\nu S))}\pi_1(N\backslash S),$$

where $\pi_1(S(\nu S))$ is mapped into the two factors through the identification of the normal bundle with a tubular neighborhood of $S$ in each manifold.

If your submanifold has sufficiently high codimension, you can do a bit better than this: Assume $\dim(S)\le n-3$. Then we can look at the long exact sequence of homotopy groups for the bundle $S^{codim-1} \to S(\nu S) \to S,$ $$\ldots \to \pi_1(S^{codim-1}) = 0 \to \pi_1(S(\nu(S)) \to \pi_1(S) \to 0$$ to find that the projection of the sphere bundle induces an isomorphism $\pi_1(S(\nu(S)) \to \pi_1(S).$

Using this on the decomposition $M = M\backslash S_M \cup \nu(S_M)$ and invoking Seifert-van Kampen again, we get $$\pi_1(M)=\pi_1(M\backslash S_M)\star_{\pi_1(S(\nu S_M))}\pi_1(\nu S_M) = \pi_1(M\backslash S_M),$$ where the last equality holds because $\pi_1(S(\nu S_M))\to \pi_1(\nu S_M) = \pi_1(S_M)$ is an isomorphism.

Since we can do the same for $N$, in this case in fact the general solution becomes isomorphic to $$\pi_1(M\#_S N) = \pi_1(M) \star_{\pi_1(S)} \pi_1(N)$$ with the natural maps. However, for codimension $2$ and below, care has to be taken. Look for example at the inclusion of two points into $S^2$, connecting two copies of that gives the two-torus, but $\pi_1(S_2)=0$, so this is certainly not obtained from that by amalgamated product.

Ok, after writing all this I rechecked the question and you only asked about codimension one to begin with. In this case the homotopy exact sequence does give that $\pi_1(S(\nu S)) \to \pi_1(S)$ is an isomorphism, so things do work out if $S(\nu(S))$ is connected, i.e. if the normal bundle is non-orientable. If it is orientable, one has to look at the version of Seifert-van Kampen with disconnected intersection. Things presumably become subtle and one has to track whether/which of $M$ and $N$ is disconnected by removing $S$. (It's late and I don't want to spend my night figuring out what groupoid pushouts do)