# nontrivial $\pi_2(Diff(M))$

Homotopy groups of Lie groups I asked it also there, and I still don't know the answer, so I try again.

I would like to know a closed manifold (possibly of low dimension) such that $\pi_2(Diff(M))\neq 0$.

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My guess is that there exists an $n$ such that the diffeomorphism group of the $n$-sphere has non-trivial $\pi_2$. E.g. the orientation preserving diffeomorphism group of $S^6$ has non-trivial $\pi_0$, as witnessed by the existence of exotic 7-spheres. So I don't see any reason for $\pi_2$ to always vanish. –  André Henriques May 24 '12 at 11:08
Thanks! so what is a guess of a suitable n? –  Mircea May 24 '12 at 12:03
See Hatcher's expanded answer and the comments to it. –  Tom Goodwillie May 25 '12 at 20:39
By the way, since $\pi_2$ of a Lie group is always trivial the fact that $\pi_2Diff(M)$ can be nontrivial is especially striking. –  Tom Goodwillie May 25 '12 at 20:41

Probably the simplest such manifold is $S^1 \times S^2$, whose diffeomorphism group has the homotopy type of $O(2) \times O(3) \times \Omega SO(3)$. This has $\pi_2$ equal to $\pi_2\Omega SO(3)=\pi_3 SO(3) = {\mathbb Z}$. The $\Omega SO(3)$ term is realized by rotating the $S^2$ slices of $S^1\times S^2$ by an element of $SO(3)$ that varies as one goes around the $S^1$ factor of $S^1 \times S^2$. This calculation was originally done in a paper of mine in the 1981 AMS Proceedings. An updated version of this paper is posted on my webpage.

In higher dimensions, spheres provide interesting examples. By an elementary argument there is a homotopy equivalence $Diff(S^n) \simeq O(n+1) \times Diff_\partial(D^n)$ for all $n$, where the subscript $\partial$ denotes diffeomorphisms that restrict to the identity on the boundary of $D^n$. The question is whether $Diff_\partial(D^n)$ is contractible. The status of this is: true for $n\le3$, unknown for $n=4$, and false for each $n\ge 5$. The noncontractibility can be deduced from the sequence $$\cdots\ \to \pi_2 Diff_\partial(D^{n-2})\to \pi_1 Diff_\partial(D^{n-1})\to \pi_0Diff_\partial(D^n) = \Theta_{n+1}$$ where $\Theta_{n+1}$ is the group of exotic (n+1)-spheres and the equality $\pi_0Diff_\partial(D^n) = \Theta_{n+1}$ is assuming $n\ge 5$. Usually $\pi_0Diff_\partial(D^n)$ is nonzero since exotic spheres exist in most dimensions greater than $6$, and in the rare dimensions in which they don't exist one can appeal to known results about how far some elements of $\Theta_{n+1}$ pull back in the sequence above. For example, Cerf's pseudoisotopy theorem says the map from $\pi_1 Diff_\partial(D^{n-1})$ to $\pi_0Diff_\partial(D^n)$ is surjective for all $n\ge 6$, so in particular $\pi_1 Diff_\partial(D^5)$ is nonzero since it maps onto $\Theta_7={\mathbb Z}/28$. A paper of Crowley and Schick posted on the arXiv last month shows there is a nontrivial element of $\Theta_{8k+2}$ that pulls all the way back to $\pi_{8k-6}Diff_\partial (D^7)$, hence $Diff_\partial(D^n)$ has infinitely many nontrivial homotopy groups for all $n\ge 7$.

Since the groups $\Theta_{n+1}$ are finite (apart perhaps from the unknown $\Theta_4$), these constructions don't give nontrivial rational homotopy groups of $Diff_\partial(D^n)$, but there is another construction that does, coming from algebraic K-theory. Using Waldhausen's big machine, Farrell and Hsiang in 1978 computed $\pi_iDiff_\partial (D^n) \otimes{\mathbb Q}$ in a stable range $n>>i$ to be ${\mathbb Q}$ for $i\equiv 3$ mod $4$ and $n$ odd, and $0$ otherwise. (This is the result mentioned in Vitali Kapovitch's answer.)

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thanks! this is very clear. –  Mircea May 24 '12 at 12:03
is there a similar reference for questions on the higher $\pi_k(Diff(M))$? –  Mircea May 24 '12 at 12:21
Allen, I'm embarrassed to say that I can't remember whether $\pi_1$ of stable smooth pseudoisotopies of a point is trivial or of order two. If the former, then of course every element of $\Theta_{n+1}$ comes from an element of $\pi_2 Diff_\partial(D^{n-2}$ if $n$ is not too small, giving lots of $n$ such that $\pi_2Diff(S^{n-2})$ is nontrivial. –  Tom Goodwillie May 25 '12 at 20:38
Tom, it's of order two. So one gets $\pi_2 Diff (S^{n-2})$ nonzero when $\Theta_{n+1}$ has more than two elements, which is most of the time. –  Allen Hatcher May 25 '12 at 20:57
Thanks, Allen. More than two, of course! –  Tom Goodwillie May 26 '12 at 11:08

Regarding the question about nontrivial $\pi_k(Diff(M))$ for $k>2$ there are plenty of such examples too. For example Farrell and Hsiang in "On the rational homotopy groups of the diffeomorphism groups of discs, spheres and aspherical manifolds." computed the rational homotopy groups of $Diff(S^n)$ in the stability range ($i\lt n/6-7$ ) which in particular gives that in the stability range $\pi_{4i-1}(Diff(S^n))\otimes \mathbb Q\ne 0$. They have some other computations there too. Those are hard results however and if you are not interested in computing the homotopy groups exactly and only want to prove that they are not trivial then much more elementary considerations are sufficient.

For example it's well known that for any odd $n$ the space $Aut(S^{n})$ (the identity component of space of self homotopy equivalences of $S^n$) is rationally equivalent to $S^{n}$ and moreover the obvious map $SO(n+1)\to Aut(S^n)$ is an epimorphism on $\pi_n \otimes \mathbb Q$. Since the map $SO(n+1)\to Aut(S^n)$ factors through $SO(n+1)\to Diff(S^n)\to Aut(S^n)$ it follows that $SO(n+1)\to Diff(S^n)$ is not zero on $\pi_n\otimes \mathbb Q$. This gives you lots of examples with nontrivial odd $\pi_i(Diff(M))$. If you want even ones too then one can use the same trick as in Allen Hatcher's example.

Pick an element in $\pi_n(SO(n+1))\otimes \mathbb Q$ which maps to the generator of $\pi_n(S^n)\otimes\mathbb Q$ under the evaluation map and let $\alpha: S^{n-1}\to \Omega SO(n+1)$ be the corresponding spheroid in $\pi_{n-1}(\Omega SO(n+1))\cong \pi_n(SO(n+1))$ . This gives you a map $\Phi: S^{n-1}\times S^n\times S^1\to S^n\times S^1$ given by $\Phi(x,y,t)=(\alpha(x)(t)(y),t)$. By construction, this is an (n-1)-spheroid in $Diff(S^n\times S^1)$. And it's clearly nontrivial because of the action of $\Phi$ on the cohomology. Therefore, $\pi_{n-1}(Diff(S^1\times S^n))\otimes \mathbb Q\ne 0$ for any odd $n>1$.

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Thanks! So one can say that for any even $n$ there exists $M_n$ such that $\pi_n(Diff(M_n))\otimes Q\neq 0$, and for odd $n>1$ we can achieve $\pi_n(Diff(M_n))\neq 0$. In the realm of elementary statements just examples of $\pi_{2n+1}(Diff(M))\otimes Q\neq 0$ seem to be missing. Or did I miss something? –  Mircea May 26 '12 at 8:41
@Mircea All the computations I gave work with rational coefficients. In particular the argument in my answer shows that $\pi_n(Diff(S^n))\otimes Q\ne 0$ for any odd n. This also easily follows from Allen Hatcher's answer because $Diff(S^n)$ splits $O(n+1)$ homotopically for any $n$ and $O(n+1)$ has lots of nontrivial odd rational homotopy groups (in dimensions $4i-1$ for $i\le n/2$ and also in dimension $n$ for odd $n$). –  Vitali Kapovitch May 26 '12 at 19:55