# connected sum of smooth, compact, orientable manifolds is smooth, compact, orientable

I've come across this in the book "Algebraic Topology: An Introduction" by W.S.Massey, and he simply dismissed it by stating it as some kind of fact. I got curious, and first tried it myself, and managed to proof compactness and orientability (i had to assume, that the connected sum is already smooth). I am struggeling with the proof for smoothness, and the scarce information i find in the internet is not helpful either, at least not for me.

So what i am really after is a proof, that the connected sum of two smooth, compact, orientable manifolds is smooth, compact and orientable.

## 1 Answer

The more precise statement is that the connected sum can be endowed with a smooth structure that is obtained from the given smooth structures in a "nice" way.

For the mere topological structure, we may define the connected sum of $M$ and $N$ as follows: Pick points $p\in M$, $q\in N$. Pick euklidean neighbourhoods, i.e., pick a chart $\phi\colon U\to V$ (with $V\subseteq M$, $U\subseteq \Bbb R^n$) from the atlas that take $p\mapsto 0$ and such that $V$ contains the closed ball $\overline{B_1(0)}$; remove $\phi^{-1}(B_1(0))$ from $M$; do the same construction with a similar chart $\psi\colon U'\to V'\subseteq \Bbb R^n$ of $N$; finally, identify the produced boundaries via $\phi$ and $\psi$.

However, in order to have better control of the manifold structure, the following (topologically equivalent) process is more adequate: As above, we have charts that cover the closed unit ball $\overline{B_1(0)}$. But then they also cover the open ball $B_r(0)$ for some $r>0$. Now instead of (the preimage of) $B_1(0)$, we remove (the preimage of) $\overline{B_{1/r}(0)}$, and instead of two copies of $S^{n-1}$, we identify two copies of $B_r(0)\setminus\overline{B_{1/r}(0)}$, via $x\mapsto\frac x{\|x\|^2}$. As a consequence, the transition between the specifically picked charts $\phi$ and $\psi$ (restricted to the chopped versions of $M$ and $N$) is smooth on the non-empty open overlap of their domains. Thus, if we define our atlas of the connected sum to consist precisely of the charts of the smooth atlas of $M$ restricted to the chopped $M$ and the charts of the smooth atlas of $N$ restricted to the chopped $N$, we obtain a smooth atlas of the connectd sum.

• apology, but i dont quiet see, how the transition is now smooth. can we now say that $\phi \circ \psi^{-1} = x/{\|x\|^2}$ which is a smooth map? – PlatinTato Apr 17 '17 at 16:54
• Dear god how did i write down the argument you've meant without getting to the conclusion. Thank you very much – PlatinTato Apr 17 '17 at 17:02