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When define the connected sum of two oriented manifods, one gluing along the reversed orientation of the boundary spheres. I am wondering what is the connected sum of $S^2$ connected sum with $S^2$ with the reversed orientation? (suppose to be the same $S^2$ as the regular connected sume of two $S^2$'s) But why connected sume between different oriented $CP^2$ matters?

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up vote 5 down vote accepted

The key fact is that $S^2$ has an orientation reversing diffeomorphism. That is, there is a diffeomorphism $f:S^2\rightarrow S^2$ with the property that if $z\in H^2(S^2;\mathbb{Z})\cong \mathbb{Z}$ generates, then $f^*(z) = -z$. For example, $f$ could be a reflection through the $xy$ plane, or the antipodal map. (The reflection case generalizes to all $n$, while the antipodal case only generalizes for even $n$).

$\mathbb{C}P^2$ (or more generally, $\mathbb{C}P^{2n}$) doesn't have such a diffeomorphism. To see this, recall that $H^*(\mathbb{C}P^n; \mathbb{Z})\cong \mathbb{Z}[x]/x^{n+1}$ where $x$ lives in $H^2(\mathbb{C}P^n;\mathbb{Z})\cong \mathbb{Z}$. If $f:\mathbb{C}P^n\rightarrow\mathbb{C}P^n$ is a diffeomorphism, then we must either have $f^*(x) = x$ or $f^*(x) = -x$.

If $n$ is even, then we have $f^*(x^n) = f^*(x)^n = (\pm x)^n = x^n$. Recalling that $x^n$ generates $H^{2n}(\mathbb{C}P^n;\mathbb{Z})$, this implies that $f$ must preserve orientation.

Finally, it's not too hard to show that if $M$ has an orientation reversing diffeomorphism, then it doesn't matter what orientation you pick on it when connect summing with another space - you'll get diffeomorphic manifolds either way. The details would be in Kosinski's Differential Manifolds book, for example.

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