Fixed-boundary disk bubble compactness contract

A compact Lagrangian is totally real for tame $J$: $v,Jv\in TL$ and $v\mathrm{\ne }0$ would imply both $\omega (v,Jv)=0$ and $\omega (v,Jv)>0$. The symmetric form in [F2] is $J$-invariant, since direct expansion using $J^2=-I$ gives $g_J(Jv,Jw)=g_J(v,w)$. It is positive by tameness. Thus bounded areas give precisely bounded Dirichlet energies in this fixed metric. Pass to a subsequence with these energies converging to $E_{\mathrm{\infty }}$.

Apply [F1] to the fixed disk with $J_i(z,x)=J(x)$, boundary structure $J_b=J$ and boundary submanifold $L$. Every hypothesis holds by step 1.1. It gives a finite tree with the original principal disk, disk children only at boundary nodes and sphere children at interior nodes, the claimed rescaling convergence and matching, and the exact energy identity. Since $J$ is autonomous, all bubbles are $J$-holomorphic. All bubble constant components already satisfy stability; the principal disk is not asserted stable.

We give explicitly the marking extension for this FIXED marking set. In the original disk coordinates each marking is fixed and distinct. A marking outside the finite concentration set stays on the principal disk. For a marking at a boundary atom, use the real boundary coordinates of the actual disk rescalings in [F1]. In a component chart pass to a subsequence for the marking coordinate in the compactified real line. A finite limit away from nodes puts it on the smooth boundary of that component; a finite child-node limit follows that child. Sphere children cannot receive boundary marks, since their attaching points are interior. The component tree is finite, so this procedure can follow only finitely many children. Distinct original marks cannot enter the same first boundary atom, since that would force their distinct fixed original coordinates to converge to the same point. Thus each bubble subtree has at most one original marking to place.

The remaining case is a marking at a parent/child neck, including infinity in the child coordinate. Write the child coordinate in the parent's straight boundary chart as $c_i+{\rho }_{i}w$, and write the marking coordinate there as $a_i$. If $(a_i-c_i)\mathrm{/}{\rho }_i$ has a finite limit, step 3.1 applies inside that child, descending further if necessary. If its modulus tends to infinity while $a_i,c_i$ tend to the same parent node, insert a constant disk in the neck at scale ${\sigma }_i=\mathrm{\mid }{a}_i-c_i\mathrm{\mid }$. In the coordinate $(z-a_i)\mathrm{/}{\sigma }_i$, the mark is $0$, the child cluster is at $+1$ or $-1$ after a subsequence, and the parent is at infinity; ${\rho }_i\mathrm{/}{\sigma }_i\to 0$. This disk therefore has three distinct boundary special points. If a previously extracted component has a scale comparable to ${\sigma }_i$, place the marking on that component instead, and apply step 3.1 at its finite nodes. Finiteness of the existing component tree makes this alternative terminate. The inserted disk is constant: its compact subsets outside attachment points lie in the connecting half-annulus, whose energy tends to zero by [F1]; local derivative estimates and bootstrap used in [F1] give smooth convergence there, while the no-neck diameter assertion makes that constant the common node value. Removability fills its punctures. It lies in $L$ because its boundary values do. This inserts at most one disk per marking. No area is added. Positive real rescalings preserve the boundary order. The resulting finite marked tree is therefore exactly the prestable map of [F3], and preserves the original cyclic order, evaluations and energy.

Stabilization is a finite combinatorial operation on this tree, with matching of values through each constant component. A constant sphere with at most two special points is deleted if it is a leaf or contracted if it is a bridge. For a constant disk let $b$ count its boundary nodes and markings, and $i$ its interior nodes. An unstable disk has $b+2i\le 2$. If $i=0$, a boundary leaf without a mark can be removed, a boundary bridge can be contracted, and a boundary leaf with one mark can be removed while transferring its mark to the former attachment point on its neighbor. This point is now smooth and is distinct from all its other markings and nodes. If $i=1$, instability forces $b=0$: there are no boundary nodes or markings, hence this is the only disk, and it is attached to a sphere tree at one interior point. Remove the disk and retain that attaching point on the sphere tree as the distinguished collapsed-boundary point with value in $L$. If there is no node at all, an unstable disk is isolated with at most two markings. These alternatives exhaust all nonnegative pairs $(b,i)$ with $b+2i\le 2$. After each contraction repeat; the number of vertices decreases, so the process terminates. No nonconstant component or positive energy is removed.

When $E_{\mathrm{\infty }}>0$, the energy identity leaves at least one nonconstant component. Thus the isolated all-constant terminal disk cannot occur, and the procedure ends with a stable marked disk tree, or with the unmarked collapsed-boundary sphere tree. Marked cases cannot lose their last disk: the only disk-removal-to-spheres case in step 5.1 has no marks. Sphere contractions preserve the distinguished point as an additional special point, transferring it to the neighboring component if necessary. The finite automorphism conditions of [F3] now hold. Node values and evaluations are preserved because each removed map is constant. Concatenating finitely many zero-energy, zero-diameter necks retains those properties, so the stabilized maps have the same Gromov convergence and total area.

If $E_{\mathrm{\infty }}=0$, every component is constant by [F2]. The bubble energy quantum in [F1] excludes nonconstant bubbles, and its stable constant-bubble tree rules exclude a finite nonempty all-constant bubble tree: a terminal bubble would violate its stability rule. Hence there is just the constant principal disk, with the original distinct fixed marks. Its automorphism group is finite exactly when there are at least three boundary markings: a disk biholomorphism is determined by three distinct cyclically ordered boundary points, while the stabilizer of zero, one or two such points has positive dimension (in the upper half-plane these are respectively the full real Möbius group, its affine subgroup fixing infinity, or the dilations fixing zero and infinity). Consequently no stable limit is claimed in the remaining cases. Indeed any all-constant stable bordered tree with $K<3$ boundary marks is impossible: doubling gives a genus-zero nodal constant sphere tree with $K$ marks and $N$ vertices, whose total special-point count $2(N-1)+K$ is less than $3N$. Thus the unstable exception is necessary. This proves every branch without omitting any original input.