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A rank-$n$ vector subbundle of a rank-$2n$ symplectic vector bundle is Lagrangian when each fibre is a Lagrangian subspace. If it exists, a compatible almost complex structure supplies a complementary Lagrangian subbundle by applying $J$. Existence is a global condition: the area-oriented tangent bundle of the two-sphere has no real line subbundle, by triviality of real line bundles over the sphere and the Hairy ball theorem.

False. Give $E=T{S}^2\to S^2$ a fibrewise area form. This is a rank-two symplectic vector bundle. A Lagrangian subbundle would be a real line subbundle $L\subset T{S}^2$. Real line bundles are classified by their first Stiefel–Whitney class in $H^1(B;\mathbb{Z}/2)$; since this group vanishes for $S^2$, $L$ would be trivial. A nonzero section of $L$ would therefore be a nowhere-vanishing tangent vector field on $S^2$. The Hairy ball theorem rules this out: every continuous tangent vector field on an even-dimensional sphere has a zero. Hence $T{S}^2$ does not even have one Lagrangian subbundle, let alone a splitting into two. Fibrewise existence of Lagrangian subspaces does not guarantee a globally continuous choice.

The symplectic neighborhood theorem says that a symplectomorphism between compact symplectic submanifolds, together with a symplectic vector bundle isomorphism of their symplectic normal bundles, extends to a symplectomorphism of neighborhoods. To outline the proof, use tubular neighborhoods to extend the bundle identification smoothly. The two pulled-back symplectic forms agree as bilinear forms along the zero section. The relative Poincaré lemma writes their difference as $d\alpha$, with $\alpha$ vanishing there. Their linear interpolation ${\omega }_t$ remains nondegenerate on a sufficiently small neighborhood. Solvingand integrating $V_t$ gives the relative Moser theorem; the flow fixes the submanifold and carries one form to the other.

For the symplectic fiber sum along a square-zero surface, the self-intersection is the Euler number of the oriented rank-two normal bundle. Thus each normal bundle is trivial. Rescale one ambient symplectic form if necessary to equalize the areas of the two copies of $C$. The Moser theorem for the surface then supplies a base identification preserving their area forms. The symplectic neighborhood theorem identifies each neighborhood with a product carryingRemove small disk neighborhoods and glue collars with the base identification andfor a suitable positive constant $a$. Both signs reverse, so the normal two-forms match; the base forms already match. They define a closed nondegenerate two-form across the neck, agreeing with the ambient forms elsewhere. This constructs the symplectic sum. Its construction uses the area normalization and the chosen normal-bundle gluing.

For a smooth counterexample, take $X=Y={\mathbb{CP}}^2$, and in a small four-ball in each choose an unknotted sphere bounding a three-ball. These spheres are null-homologous and have self-intersection zero. Their standard smooth fiber sum isIndeed, the complement of the standard $S^2\times D^2$ in $S^4$ is $S^1\times D^3$; doubling these complements gives $S^1\times S^3$, and the two ambient projective-plane summands remain as connected sums. This is the smooth fiber sum along unknotted null-homologous spheres.

In the induced orientation, both summands in $Z={\mathbb{CP}}^2\#({\mathbb{CP}}^2\#(S^1\times S^3))$ have positive positive index of the intersection form. The symplectic connected-sum obstruction therefore rules out a symplectic form: connected-sum vanishing of the Seiberg–Witten invariant of a four-manifold contradicts Taubes nonvanishing theorem. In the opposite orientation the intersection form is negative definite, also ruling out a symplectic form, whose class has positive square. Thus the smooth fiber sum need not be symplectic.