MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $\Theta_n$ be the set of orientation-preserving diffeomorphism classes of homotopy spheres, with abelian group structure given by #. Then for any smooth manifold $M^n$ one defines the "inertia subgroup" as

$I(M)=${$\Sigma\in\Theta_n$ | $M$#$\Sigma\cong M$}

In other words, the inertia group is the set of homotopy spheres which fix the smooth structure on $M$.

Does anyone know an example of a diffeomorphism (as explicit as possible) between an $M$ and $M$#$\Sigma$, where $\Sigma$ is as exotic sphere? Moreover, if the dimension were sufficiently large can we always find a diffeomorphism which induces the identity on $\pi_1$?

(The big question I'm trying to think about is "If $\Sigma$ is in the inertia group of $M$, how does its smooth structure get 'unraveled' in $M$?")

share|cite|improve this question
May be I'm being dense but when you say "the identity on $\pi_1$" what exactly do you mean? Fundamental group is assigned to a pointed space. Abstractly, $\pi_1(M)$ and $\pi_1(M\# \Sigma)$ will be isomorphic but how does the identity map between these abstract groups make sense? – Somnath Basu Nov 4 '11 at 2:13
It seems we asked the same question.… – Xiaolei Wu Nov 4 '11 at 2:31
You get some non-trivial examples with things like $\mathbb RP^{18}$, and $\mathbb RP^{20}$. This is because the group of exotic spheres has elements of even order larger than $2$, and these projective spaces aren't orientable. So when you connect-sum with an exotic sphere, you can slide the summation point around the projective space to turn the manifold into $\mathbb RP^n$ connect sum with the inverse exotic sphere. – Ryan Budney Nov 4 '11 at 6:28

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.