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Given $M$ and $N$, two connected orientable manifolds of the same dimension, when is $M$ # $N$ diffeomorphic to $M$ # $\overline{N}$, where $\overline{N}$ is $N$ with the orientation reversed? If $N$ has an orientation-reversing automorphism, is this a necessary or sufficient condition for $M$#$N$ to be diffeomorphic to $M$#$\overline{N}$? If it isn't a necessary condition, what invariants can be used to distinguish $M$ # $N$ from $M$ # $\overline{N}$? (As a baby example, how does one show that $CP^2$ # $CP^2\ncong CP^2 $#$ \overline{CP ^2}$ ?)


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up vote 8 down vote accepted

What kind of object do you want to consider $M \# N$ to be, an oriented manifold, or unoriented? Presumably you're taking the connect sum up to some kind of equivalence. You'll also need for $M$ and $N$ to be connected if you want connect-sum to be well-defined in any sense.

In the oriented sense, $M \# N$ is well-defined up to orientation-preserving diffeomorphism, and contains both the punctured $M$ and the punctured $N$ as oriented submanifolds, so it depends on the orientations of $M$ and $N$ respectively.

As an unoriented object taken up to diffeomorphism, a connect-sum is well defined provided either input manifold has an orientation-reversing diffeomorphism.

Explicit examples where you can see there is or is not diffeomorphisms between such are connect sums of complex projective spaces (and/or their orientation reverses). There's also examples with $3$-dimensional lens spaces but working out which ones of those admit orientation-reversing diffeomorphisms is more work. All $1$ and $2$-manifolds admit orientation-reversing diffeomorphisms so there's no good examples there.

edit: in detail for $\mathbb CP^2$, the intersection form on $H_2(\mathbb CP^2 \# \mathbb CP^2)$ is definite, regardless of what orientation you give the connect-sum. But if you take the connect sum with one factor orientation-reversed $H_2(\mathbb CP^2 \# \overline{\mathbb CP^2})$, the intersection form is indefinite. For $3$-dimensional lens spaces, the torsion linking form is a good analogous invariant.

Have you looked at a book like Kosinski's Differential Topology? It covers these kinds of operations on manifolds.

If you want a lower-dimensional example that's easier to see, you could take the analogous connect-sum of knots in $S^3$. This is well-defined for oriented knots, but for unoriented knots you only get a well-defined operation when the knots are invertible, which means there's an orientation-preserving diffeo of $S^3$ that preserves the knot and reverses the orientation of the knot.

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In the oriented category it's not true that if $M\#N$ is oriented diffeomorphic to $M\#(-N)$ then $N$ admits an orientation reversing diffeomorphism. I suspect there are easier counterexamples but the following works. Take $M^{4k+3}=S^3\times \mathbb{CP}^{2k}$ with $k\ge 1$. Then there exists an exotic sphere $\Sigma^{4k+3}$ such that $\Sigma$ does not admit an orientation reversing diffeomorphism but $M\#\Sigma$ is oriented diffeomorphic to $M$ (and hence also to $M\#(-\Sigma)$). Note that $M$ obviously admits an orientation reversing diffeomorphism because $S^3$ does.

I read about this fact in a paper by Belegradek, Kwasik and Schultz "Codimension two souls and cancellation phenomena". Not sure if this is the earliest reference, perhaps Igor can clarify this - he visits MO regularly.

More specifically they show that if $I(M)$ is the inertia group of $M$ (the group of oriented exotic $4k+3$-spheres $\Sigma$ such that the standard homeomorphis $M\#\Sigma\to M$ is homotopic to a diffeomorphism) then $I(S^3\times \mathbb{CP}^{2k})\cap bP_{4(k+1)}$ has index 2 in $bP_{4(k+1)}$ where $bP_{4(k+1)}$ is the group of exotic $(4k+3)$-spheres bounding parallelizable manifolds. It's known that $bP_{4(k+1)}$ is cyclic or order exponentially growing in $k$ so any nontrivial element $\Sigma$ of $I(S^3\times \mathbb{CP}^{2k})\cap bP_{4(k+1)}$ of order different from 2 works.

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You can play the same game with other $M$, e.g. you can realise all subgroups of the group $\mathbb{Z}_{28}$ of exotic smooth 7-spheres as inertia groups by looking at $S^3$-bundles over $S^4$ with Euler number 0; according On the inertia group of certain manifolds by Wilkens the inertia group of $M^7$ is determined by the greatest divisor of the obstruction in $H^4(M;\mathbb{Z})$ to stable parallelisability of $M$ (half of $p_1$), which we can prescribe for the sphere bundle by choosing the right clutching function. This way you can also get $M$ without orientation-reversing diffeomorphism. – Johannes Nordström Apr 8 '12 at 17:41
Vitali, your response triggered this question:… – Ryan Budney Apr 8 '12 at 18:33
@Johannes Nordström, Thanks. note however, that the OP specifically wanted an example where $M\#N$ does admit an orientation reversing diffeomorphism. Can one get such examples among $S^3$-bundles over $S^4$? @Ryan yes, I saw it. It's a nice question. – Vitali Kapovitch Apr 8 '12 at 19:14
@Johannes sorry, I was being very dense. Of course all such bundles with zero Euler class admit orientation reversing self diffeomorphisms induced by an orientation reversing diffeomorphism of $S^4$. – Vitali Kapovitch Apr 8 '12 at 21:10

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