Lagrangian surgery Assume that $L_1$ and $L_2$ are connected Lagrangian submanifolds (of dimension at least 2) which intersect transversally. Do we always get a connected Lagrangian after performing Lagrangian surgeries at the points of intersections?
If not, what is an example?
 A: Actually, one can always do a (local) surgery to 'resolve' transverse intersection points of Lagrangian submanifolds, and this works in all dimensions, not just dimension $2$.  Here is a sketch of an argument:
First, it's a local problem, so you can assume that $L_1$ and $L_2$ are $n$-dimensional Lagrangian embedded submanifolds of $\mathbb{R}^{2n}$ that meet transversely at the origin, i.e., their tangent planes at the origin are Lagrangian subspaces that have trivial intersection.  Using Darboux' Theorem, we can choose local coordinates $x^1,\ldots, x^n,p_1,\ldots, p_n$ centered on the origin so that the symplectic form is $\Omega = \mathrm{d}p_i\wedge\mathrm{d}x^i$ (sum on repeated indices in opposition is assumed) and so that $L_1$ is defined in these coordinates by $p_1 = p_2 = \cdots = p_n = 0$.  Because $L_2$ is transverse to $L_1$ at $x=p=0$, it follows that the differentials $\mathrm{d}p_i$ are linearly independent on $L_2$ near the origin and so, on $L_2$ we can write $\mathrm{d}x^i = h^{ij}\,\mathrm{d}p_j$ for some functions $h^{ij} = h^{ij}(p)$.  The assumption that $L_2$ be Lagrangian is equivalent to the condition that $h^{ij}=h^{ji}$, and then the closure condition $0=\mathrm{d}(\mathrm{d}x^i) = \mathrm{d}h^{ij}\wedge\mathrm{d}p_j$ implies that there exist a function $f = f(p_1,\ldots,p_n)$ such that $h^{ij} = \partial^2f/\partial p_i\partial p_j$, where we can assume that $f$ and its first partials vanish at $p=0$.  Then, replacing $x^i$ by $x^i - \partial f/\partial p_i$, we get new Darboux coordinates in which $L_1$ is defined by $p=0$ and $L_2$ is defined by $x=0$.  Thus, we are reduced to the linear case, locally.
Now, let $u:S^{n-1}\to\mathbb{R}^n$ be the inclusion of the unit sphere, let $\epsilon>0$ be chosen (as small as you like), and let $f(t)$ and $g(t)$ be two functions on the real line such that $f(t) = t$ for $t<-\epsilon$, g(t) = t for $t>\epsilon$ and $f(t) = 0$ for $t>\epsilon$ while $g(t) = 0$ for $t<-\epsilon$.  Assume also that $f(t)$ and $g(t)$ never vanish simultaneously and that $f'(t)$ and $g'(t)$ never vanish simultaneously.  (It is easy to construct such functions.)  Let $a:\mathbb{R}\times S^{n-1}\to\mathbb{R}^{2n}$ be given by 
$$
a(t,u) = \bigl(x(t,u),p(t,u)\bigr) = \bigl(\ f(t)\,u,\ g(t)\,u\ \bigr).
$$
It is easy to veryify that $a$ is a Lagrangian immersion and that, when $t<-\epsilon$, its image is $L_1$ with a small ball about the origin cut out while, when $t>\epsilon$, its image is $L_2$ with a small ball about the origin cut out.  Thus, this Lagrangian 'surgery' joins the two Lagrangian submanifolds smoothly. 
N.B.: Note that, if we also arrange that the curve $\bigl(f(t),g(t)\bigr)$ be a smooth embedding of $\mathbb{R}$ into $\mathbb{R}^2$, then $a$ will be a smooth embedding as well.
