Smooth structures on the connected sum of a manifold with an Exotic sphere

Question

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 $M^n$?

Answer 1

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.

Answer 2

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 $m \ge k$.