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What can we say about the connected sum of a manifold $M^n$ with an Exotic sphere? Is is possible some of them are still diffemorphic to $M^n$. Is it possible to classifying all the exotic smooth structures for a given $M^n$?

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When $n \neq 4$, $\mathbb R^n \# M \simeq \mathbb R^n$ for all homotopy $n$-spheres $M$. The group of homotopy spheres acts on the smooth structures of a given manifold, and some subgroup acts trivially. It's generally an interesting question as to what that subgroup is precisely but it can also be difficult to compute. – Ryan Budney Dec 8 '10 at 23:49
These matters are classically studied by surgery theory but I do not know any elementary exposition of your question in the literature. I struggled with these very issues while writing section 3 of which you may find useful. – Igor Belegradek Dec 9 '10 at 0:55
Farrell-Jones [JAMS 1989] showed that for certain closed hyperbolic manifolds taking connected sum with an exotic sphere always changes the smooth structure. On the other hand, they showed in [JDG 1993] that for a non-compact manifold of dimension >4 connected sum with an exotic sphere never changes the smooth structure; I am somewhat confused about their proof but this is what I think is claimed there. – Igor Belegradek Dec 9 '10 at 2:09
Google "inertia group" of M: this is the name given in surgery theory to the subgroup of the group of homotopy spheres whose connected sum doesn't change M. – Paul Dec 9 '10 at 3:15
Ryan Budney's and Paul's answers lead me to wonder when the group of smooth structures on the sphere acts transitively on the set of smooth structures of $M$. Also, when a group acts transitively on a set abstractly, the set generally has no distinguished points, but this action is not abstract. So I wonder if there are topological criteria that single out certain smooth structures. In particular, do some manifolds possess smooth structures that deserve to be called "standard" because they share some topological property that characterizes the standard smooth structure on the sphere? – David Feldman Dec 9 '10 at 5:27
up vote 10 down vote accepted

Surgery theory provides a framework for classifying closed higher-dimensional manifolds, but unfortunately, a definitive classification is known only for a very few homotopy types. Here is how surgery attempts to classify smooth structures on a given manifold $M$.

A basic object is a smooth structure set $S(M)$, which is the set of equivalence classes of simple homotopy equivalences $f: N\to M$ where $f_1: N_1\to M$, $f_2: N_2\to M$ are considered equivalent if there is a diffeomorphism $d: N_1\to N_2$ such that $f_2\circ d$ is homotopic to $f_1$. For example, every homeomorphism is a simple homotopy equivalence, so $S(M)$ contains all manifolds homeomorphic to $M$. The set $S(M)$ fits into the sugery exact sequence. Roughly, to every $f: N\to M$ one associates the so called normal invariant which lives in $[M, G/O]$, the homotopy classes of maps from $M$ to the classifying space $G/O$. In a sense, the normal invariant $n(f)$ records tangential data of $f$, but it is more compicalted than that, e.g. $n(f)$ need not be trivial even when $f$ is a homeomorphism that preserves the tangent bundles.

If $n(f)$ is trivial, then by exactness $f$ lies in the orbit of the action of the surgery $L$-group. If $M$ is simply-connected, this action is given by connected sums of $f$ with (the identity maps of) homotopy spheres bounding parallelizable manifolds; if a homotopy sphere does not bound a parallelizable manifold the connected sum may (will?) change the normal invariant. In the non-simply-connected case undertsanding $L$-groups and their action may involve heavy algebra; of course, in this case the group of homotopy spheres bounding parallelizable manifolds still acts on $S(M)$, but there is much more stuff in the $L$-group than this.

Even we are lucky to compute $S(M)$, we are not done yet because $S(M)$ could contain "repetitions", e.g. there could be homeomorphism $f_1: N\to M$, $f_2: N\to M$ that are different in $S(M)$ even though their domain $N$ is the same smooth manifold. Thus if we really want to have a list of manifolds homeomorphic to $M$, we should not count $f_1$, $f_2$ as different elements. Sorting out these repetitions has a strong homotopy theoretic flavor, and is notoriously hard.

I mentioned some papers in comments where the above classification scheme was made work, but again this is quite rare, to my knowledge. For example, even for product of several (maybe even two) spheres or complex projective spaces the classification seems to be unknown.

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This is more a long comment on a special case of the first two questions than a complete answer, but I am surprised no one else mentioned this.

One can, even without a proof or refutation of the smooth 4-dimensional Poincare conjecture, state something about the case $M^4 = \sharp m S^2 \times S^2$, where $m$ is greater than or equal to a particular positive integer $k$.

Assume there exists an exotic 4-dimensional sphere $\textbf{S}^4$ homeomorphic but not diffeomorphic to a standard $S^4$. Then, by a theorem of Wall, there exists a positive integer $k$ such that $S^4 \sharp k S^2 \times S^2$ is diffeomorphic to $\textbf{S}^4 \sharp k S^2 \times S^2$. Thus, $S^4 \sharp m S^2 \times S^2$ is diffeomorphic to $\textbf{S}^4 \sharp m S^2 \times S^2$ for $m \ge k$. Trivially, $S^4 \sharp m S^2 \times S^2$ is diffeomorphic to $\sharp m S^2 \times S^2$. Thus, $\textbf{S}^4 \sharp m S^2 \times S^2$ is diffeomorphic to $\sharp m S^2 \times S^2$ for $m \ge k$.

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Can we use a argument which is not rely on 4-dimensioan smooth poincare conjecture? – Jialong Deng Apr 25 at 15:31

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