MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).
show/hide this revision's text 2 deleted 7 characters in body

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 $\alpha\in \pi_n(SO(n+1))\otimes \pi_n(SO(n+1))\otimes \mathbb Q$ 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 $n>1$.

show/hide this revision's text 1

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 $\alpha\in \pi_n(SO(n+1))\otimes \mathbb Q$ 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 $n>1$.