Homotopy groups of the blow-ups and monotone symplectic manifolds I want to calculate the second homotopy group of the blow-up of the unit cotangent disk bundle of a closed surface $\Sigma$, i.e $\pi_2\left(D^*T\Sigma\#\overline{\mathbb{C}P^2}\right).$ Actually, I want to understand whether these symplectic manifolds are monotone. If yes, I want to find a closed monotone Lagrangian submanifold in it.   
 A: Write $X$ for the manifold in question (i.e. the symplectic blowup of the unit disc bundle in the cotangent bundle of a surface $\Sigma$).
If you're interested just in monotonicity, what's relevant isn't so much $\pi_2(X)$ as the image of $\pi_2(X)$ under the Hurewicz map to $H_2(X;\mathbb{Z})$, which is quite a bit simpler--it has rank either one or two (generated at most by the zero section and the exceptional sphere, depending on what $\Sigma$ is).  Now $\omega$ evaluates as zero on the zero section and is positive on the exceptional sphere, while the first Chern class is zero on the zero section and $1$ on the exceptional sphere.  Thus $[\omega]$ and $c_1$ are positively proportional as elements of $H^2\left(X;R\right)$,
so   $X$ is always monotone, regardless of $\Sigma$.
Regarding monotone Lagrangian submanifolds, the obvious candidate is the zero section $\Sigma$.  Since $\pi_1(\Sigma)\to \pi_1(X)$ is an isomorphism, all classes in the relative $\pi_2$ come from the absolute $\pi_2$, so since the ambient manifold is monotone it directly follows that $\Sigma$ is a monotone Lagrangian submanifold.
When $\Sigma$ is $S^2$, there's another interesting example of a monotone Lagrangian $L\subset D^*TS^2$ discussed in this paper of Albers and Frauenfelder.  If you carefully choose the size of the blowup (so that the proportionality constant on the exceptional sphere is the same as the monotonicity constant for $L$), then it should be possible to arrange for $L$ to lift to a monotone Lagrangian torus in the blowup.  (However for a generic size blowup this probably won't work.)
A: I don't know how to comment, so here is a comment on Dmitri's answer:
I think that if $\Sigma$ is not $S^2$ $\pi_2$ is bigger than $\mathbb{Z}$. For $\mathbb{R}P^2$ it is $\mathbb{Z}^3$ and for others it is a free abelian group generated by elements in $\pi_1(\Sigma)$ (each of those elements gives a copy of $\overline{\mathbb{C}P^2}$ in the universal cover).
