# a basic question about connect sum of two manifolds

How to prove connect sum of two manifolds doesn't depend on the choices of balls(which would be cutted) and different gluing of boundary spheres? Is it still true in differential category? Thanks!

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The connected sum of what are you talking about? –  Rasmus Oct 1 '10 at 14:01
It sounds like he's talking about surfaces. In that case the theorem is called the "Tubular Neighbourhood Theorem". –  Ryan Budney Oct 1 '10 at 14:09
The only proof I know can be found in Kosinski's book "Differential Manifolds". It's now published by dover, so quite cheap. He argues that the answer to your second question is "yes". –  Jason DeVito Oct 1 '10 at 14:11
And of course, the answer does depend on the "choice of balls" provided your manifolds are orientable and do not admit an orientation-reversing diffeomorphism. If you restrict your balls to be orientation-preserving embeddings, this is not an issue. –  Ryan Budney Oct 1 '10 at 15:20
For connected compact surfaces this is a consequence of the classification theorem for those surfaces. –  a.r. Oct 1 '10 at 15:47

First of all, you need your surfaces to be connected, otherwise the result is not true: let $X = \mathbb{T}^2 \sqcup \mathbb{S}^2$ and $Y = \mathbb{T}^2$. ($\mathbb{T}^2$ a torus, $\mathbb{S}^2$ a sphere.)

Then the connected sum $X \oplus Y$ does depend on the choice of the disks. It doesn't matter which disk you choose on $Y$, of course, but if you choose your disk on $X$ belonging to $\mathbb{T}^2$, then you get $2\mathbb{T}^2 \sqcup \mathbb{S}^2$ and if you choose your disk belonging to $\mathbb{S}^2$, you obtain $\mathbb{T}^2 \sqcup \mathbb{T}^2$.

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Now, I think I've got a possible real answer.

Once connection assumed, I'm going to develop my previous comment in order to prove that, at least for compact connected surfaces, connected sums do NOT depend on anything at all. Feel free to point out if I'm missing something or any mistakes in my reasonings.

The problem of the independence of the disks and "gluing" homeomorphisms can be stated as follows: Let $X$ and $Y$ be two topological surfaces, $D$ a disc in $X$ and $D'$ a disc in $Y$. Let $h: \partial D \longrightarrow \partial D'$ be a homeomorphism. Is it necessarily true that $X \oplus_h Y$ does not depend on $h$?

To see this, you could proceed as follows. First of all, you need the theorem of classification of surfaces, which says:

Classification theorem. Let $X$ be a connected compact surface. Then $X$ is homeomorphic to one, and only one, of the following standard surfaces:

1. The sphere $\mathbb{S}^2$.
2. A connected sum of $g$ toruses $g\mathbb{T}^2 = \mathbb{T}^2 \oplus \dots \oplus \mathbb{T}^2$.
3. A connected sum of $g$ projective planes $g\mathbb{RP}^2 = \mathbb{RP}^2 \oplus \dots \oplus \mathbb{RP}^2$.

Warning. There is no circular argument here when talking about "the" connected sum of toruses or projective planes, provided you make the following convention. For the time being, this connected sum just means that we are representing spheres, connected sums of toruses, or projective planes by means of the standard polygonal presentations

(a) $\mathbb{S}^2 = aa^{-1}$ .

(b) $g \mathbb{T}^2 =a_1b_1a_1^{-1}b_1^{-1} \dots a_gb_ga_g^{-1}b_g^{-1}$ .

(c) $g \mathbb{RP}^2 = a_1a_1 \dots a_ga_g$ .

Then, define the connected sum of the standard surfaces by choosing the same "standard" small (such that its boundary doesn't touch the polygon) $D = D'$ disk centered at the barycenters of the polygons and with $h = \mathrm{id} : \partial D \longrightarrow \partial D'$ as homeormorphism.

Of course there could be still a little problem here with orientations: you could "glue", for instance, two spheres with this convention in two apparently different ways: "side by side", or one inside the other. Fortunately, for surfaces at least, this doesn't matter (you can "extract" the inner sphere without changing the homeomorphism type).

Now, with this definition, you can prove the following:

Proposition 1. For the standard surfaces, you have

(1) $\mathbb{S}^2 \oplus X \cong X$, for any standard surface $X$.

(2) $g\mathbb{T}^2 \oplus g'\mathbb{T}^2 \cong (g+g')\mathbb{T}^2$.

(3) $g\mathbb{RP}^2 \oplus g'\mathbb{RP}^2 \cong (g+g')\mathbb{RP}^2$.

(4) $g\mathbb{T}^2 \oplus g'\mathbb{RP}^2 \cong (2g +g')\mathbb{RP}^2$.

So, we already know what the connected sum of the standard surfaces (with the aforementioned convention) is.

In fact, we can say more. For standard surfaces, keeping the "standard" disk fixed once for all we can choose any homeomorphism $h: \partial D \longrightarrow \partial D'$ we please. To see this we must rely on the aforementioned classification theorem and the computation of the homology of the standard surfaces, which says, among other things, that the knowledge of $H_1(X)$ suffices in order to classify our surface.

Then, when we try to compute, for instance, the homology of $X = g\mathbb{T}^2 \oplus g'\mathbb{T}^2$, we may use Mayer-Vietoris, choosing $g\mathbb{T}^2 \backslash V$ and $g'\mathbb{T}^2\backslash V'$ as a cover for $g\mathbb{T}^2 \oplus g'\mathbb{T}^2$, where $V$ and $V'$ are the interior of $D$ and $D'$, respectively. We have $\mathbb{S}^1 = (g\mathbb{T}^2 \backslash V) \cap (g'\mathbb{T}^2 \backslash V')$ and the interesting part of the Mayer-Vietoris sequence is just

$$0 \longrightarrow H_2(X) \longrightarrow H_1(S^1) \stackrel{f}{\longrightarrow} H_1( g\mathbb{T}^2 \backslash V )\oplus H_1 ( g'\mathbb{T}^2\backslash V' ) \longrightarrow H_1(X) \longrightarrow 0 \ .$$

Here, $f$ is the morphism induced by the inclusions $\mathbb{S}^1 \hookrightarrow g\mathbb{T}^2 \backslash V$ and $\mathbb{S}^1 \hookrightarrow g'\mathbb{T}^2 \backslash V'$. Now, the first morphism is just the identity on $\mathbb{S}^1 \longrightarrow \mathbb{S}^1 \subset g\mathbb{T}^2 \backslash V$ and the second one is $h:\mathbb{S}^1 \longrightarrow \mathbb{S}^1 \subset g'\mathbb{T}^2 \backslash V'$. But any homeomorphism $\mathbb{S}^1 \longrightarrow \mathbb{S}^1$ is homotopic to $\pm \mathrm{id} : \mathbb{S}^1 \longrightarrow \mathbb{S}^1$. So, an easy computation shows that $f = 0$ and the result follows.

VERY IMPORTANT REMARK. Here is the point where, for standard surfaces at least, the "gluing" homeomorphism $h: \partial D = \mathbb{S}^1 \longrightarrow \mathbb{S}^1 = \partial D'$ "disappears".

The next result we need is that we can "move" our disks wherever we please on the surface.

Proposition 2. Let $X$ be a connected surface and $D, D'$ two disks in $X$. Then there exists a homeomorphism $H: X \longrightarrow X$ such that $H(D) = D'$.

The proof of this result is more involved than the previous ones and needs Schönflies theorem.

Finally, we are prepared to prove the independence of the connected sum. Let $X$ and $Y$ be two connected compact surfaces, $D\subset X$ and $D'\subset Y$ two disks inside of them, $h: \partial D \longrightarrow \partial D'$ a homeomorphism. Let's show that $X \oplus_h Y$ doesn't depend on $h$:

(1) Because of the classification theorem $X$ and $Y$ are homeomorphic to two standard surfaces: $S_X$ and $S_Y$, respectively. Let $\phi : X \longrightarrow S_X$ and $\psi : Y \longrightarrow S_Y$ denote the homeomorphisms.

(2) These homeomorphisms $X \cong S_X$ and $Y \cong S_Y$ don't need to identify $D$ and $D'$ with our standard disks on the standard surfaces $S_X$ and $S_Y$, but for this we have proposition 2, which tells that we can "move" $D$, $D'$ to those standard disks using two homeomorphisms $H_X : X \longrightarrow X$ and $H_Y: Y \longrightarrow Y$.

(3) Now we glue our standard surfaces along their standard disks using the homeomorphism induced by $h$ on them: $h' = H_Y\psi h \phi^{-1} H_X^{-1}$.

(4) Because our "we can say more", the connected sum $S_X \oplus_{h'} S_Y$ does not depend on $h'$. Hence, neither $X\oplus_h Y \cong S_X \oplus_{h'} S_Y$ depends on $h$.

References:

V. Navarro, P. Pascual, "Topologia Algebraica" (Algebraic Topology), Edicions UB (1999).

J.M. Lee, "Introduction to topological manifolds", Springer (2000).

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