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A symplectic blowup replaces a small symplectic ball around the point by an exceptional divisor ${\mathbb{CP}}^{n-1}$, for ${\dim }_{\mathbb{R}}X=2n$. One can construct its form by symplectic reduction. In a Darboux chart, consider ${\mathbb{C}}^n\times \mathbb{C}$ with the circle actionOn ${\mu }^{-1}(R^2/2)$ the action is free because $z\ne 0$. Its quotient therefore carries a smooth symplectic form. Where $w\ne 0$, choose the representative with $w$ positive real; the reduced form is the original form on $|z|>R$. Where $w=0$, the quotient is the Hopf fibration quotient of $|z|=R$, namely ${\mathbb{CP}}^{n-1}$. The local quotient is the tautological complex line bundle near its zero section, hence the local blowup. Glue this model to the exterior of the ball to obtain a symplectic form on the blowup.

The area of a projective line in the exceptional divisor is $\epsilon =\pi R^2$. For a closed $X$, changing this size changes the symplectic volume:Consequently the blowup is not determined up to symplectomorphism without a size choice. The underlying smooth blowup is fixed, but the symplectic construction has a parameter, as in symplectic blowup size changes volume.

For the blowdown map $\pi$ and exceptional divisor $E$, the First Chern class isIn real dimension four this becomes ${\pi }^{\ast }{c}_1(TX)-\mathrm{PD}[E]$, the first Chern class formula for a symplectic blowup.

The tangent bundle of the standard four-torus is a trivial complex rank-two bundle, so its symplectic canonical class is zero. For a generic fiber $F$ of a symplectic Lefschetz pencil, the symplectic adjunction formula gives $2g-2=F^2$. Two generic fibers intersect precisely at the base points, each with local intersection number one in the pencil model $[z_1:z_2]$. HenceThis is base-point count for a Lefschetz pencil on a symplectic four-torus.

A symplectic capacity assigns $c(X,\omega )\in [0,\infty ]$ to symplectic manifolds in a fixed dimension, is monotone under symplectic embeddings, satisfies $c(X,a\omega )=ac(X,\omega )$ for $a>0$, and is nontrivial in the sense thatThe second domain is the standard symplectic cylinder. A normalized symplectic capacity assigns $\pi$ to both unit domains, hence $\pi R^2$ to both radius-$R$ domains. The general axioms already give invariance under symplectomorphisms and quadratic scaling under coordinate dilation. For the rigidity argument, use the assumed existence of capacities in each dimension, including after adding a symplectic plane; normalization is not needed.

Let ${\varphi }_j$ be symplectomorphisms converging locally uniformly to a smooth diffeomorphism $\varphi$. Work in Darboux charts. For a sufficiently small solid ellipsoid $E$ centered at a chosen point and $0<a<1$, uniform convergence and degree of a continuous mapping give, for large $j$,For the second inclusion, ${\varphi }^{-1}{\varphi }_j$ is uniformly close to the identity on the larger ellipsoid. Its boundary stays outside $\overline{E}$ and is homotopic there to the identity, so it has degree one about every point of $E$ and must cover $E$. Monotonicity and scaling now give $a^{2}c(E)\le c(\varphi (E))\le a^{-2}c(E)$. Letting $a$ tend to one proves capacity preservation under uniform limits.

At the chosen point, write $T=D\varphi$. The rescaled maps $(\varphi (tz)-\varphi (0))/t$ preserve capacities of ellipsoids and converge uniformly on compact sets to $T$. Applying the same sandwich argument proves that $T$ preserves the capacities of all solid ellipsoids; so does $T^{-1}$.

We prove the required linear capacity rigidity explicitly. Write the standard symplectic form as $\Omega$. Suppose $\Omega (u,v)=1$ andComplete $u,v$ to a symplectic basis, and complete $T^{T}u/\lambda$, with the appropriately signed $T^{T}v/\lambda$, to another symplectic basis. If their basis matrices are $P,P^{\prime }$, then $A=(P^{\prime }{)}^{-1}{T}^{T}P$ has first two columns $\lambda e_1,\pm \lambda f_1$. Hence $A^T$ sends the unit ball into the symplectic cylinder of radius $\lambda$. This matrix differs from $T$ by symplectic linear maps, so it preserves capacities of ellipsoids. Its first coordinate pair is multiplied by $\lambda$, with a possible sign, and therefore $(A^T{)}^k$ sends the unit ball into the cylinder of radius ${\lambda }^k$. Monotonicity givesfor every $k$, contradicting finiteness of the right-hand unit-cylinder capacity as $k$ tends to infinity. For a normalized capacity, the first iterate already gives the contradiction.

A zero pairing is handled by a small perturbation of $u,v$. A pairing of absolute value greater than one gives the same contradiction for $T^{-1}$, after normalizing the transformed pair. Thusfor every pair. The squared bilinear forms agree, so their difference times their sum is the zero polynomial. The real polynomial ring is an integral domain; consequently one factor vanishes identically. Therefore $T$ is either symplectic or anti-symplectic.

To eliminate the negative sign, repeat the argument for ${\varphi }_j\times {\mathrm{id}}_{{\mathbb{R}}^2}$, which are symplectomorphisms for $\omega \oplus {\omega }_{{\mathbb{R}}^2}$. If ${\varphi }^{\ast }\omega =-\omega$ at the chosen point, the limit pulls this product form back to $-\omega \oplus {\omega }_{{\mathbb{R}}^2}$, which is neither the product form nor its negative. This contradicts the preceding derivative classification. Thus ${\varphi }^{\ast }\omega =\omega$ everywhere. We have proved the Eliashberg–Gromov rigidity theorem: the symplectomorphism group is closed in the $C^0$ topology on diffeomorphisms.

Finally put $E=W^{\perp }$, with perpendicularity taken for the standard Euclidean inner product. Its dimension is two, so being nonisotropic means that the restricted symplectic form is nondegenerate. For the standard compatible complex structure $J$, $W^{\omega }=JE$. If $Jv\in W\cap JE$, then $v\in E$ and $\omega (v,e)=0$ for every $e\in E$, forcing $v=0$. Thus $W$ is a symplectic subspace and admits a symplectic basis extended to the whole space.

In the resulting linear symplectic coordinates, $W$ is $\{q_1=p_1=0\}$. The projection of the bounded set $U$ to the $(q_1,p_1)$ plane is bounded, so $U+W$ lies in a symplectic cylinder of some finite radius $R$. As $U$ is nonempty and open, it contains a ball of some radius $r>0$, which is also contained in $U+W$. Monotonicity and scaling giveas in capacity of a bounded set thickened by a symplectic subspace. For a normalized symplectic capacity, the two bounds simplify to $\pi r^2$ and $\pi R^2$.

The Symplectic Darboux theorem says that every point of a $2n$-dimensional symplectic manifold has local coordinates $(q^1,\dots ,q^n,p_1,\dots ,p_n)$ in whichFirst choose linear coordinates at the point so the form there is standard. The required symplectic basis can be constructed inductively: choose $e,f$ with $\omega (e,f)=1$, split off their span, and repeat on its nondegenerate symplectic orthogonal complement. Extend these coordinates to a chart centered at zero.

Let ${\omega }_0$ be the constant standard form in this chart and $\eta =\omega -{\omega }_0$. The closed form $\eta$ vanishes at zero. On a small star-shaped ball the radial Poincaré lemma supplies a primitiveSince ${\eta }_0=0$, this primitive is $O(|x{|}^2)$. The interpolating forms ${\omega }_t={\omega }_0+t\eta$ are nondegenerate on a common smaller ball, because they all agree with ${\omega }_0$ at zero and $t$ ranges over a compact interval.

Apply the local version of Moser's trick: solve ${\iota }_{X_t}{\omega }_t=-\alpha$. The vector fields are $O(|x{|}^2)$ and fix zero. On a sufficiently small ball their flows exist for $0\le t\le 1$ and remain inside the coordinate chart; the quadratic bound makes their displacement smaller than the available margin. The same pullback calculation gives $f_1^{\ast }\omega ={\omega }_0$. Thus $f_1$ is a local symplectomorphism from the standard ball into $M$, and its inverse supplies the desired Darboux chart. There are no local symplectic invariants beyond dimension.

Fix $p\in X$. A linear change of coordinates first identifies ${\omega }_p$ with the standard symplectic form ${\omega }_0$. After shrinking to a star-shaped neighborhood, everyis nondegenerate. The Poincaré lemma gives $\omega -{\omega }_0=d\sigma$, with $\sigma (p)=0$. Define $X_t$ by ${\iota }_{X_t}{\omega }_t=-\sigma$ and let ${\varphi }_t$ be its local flow. Cartan's magic formula givesThus ${\varphi }_1^{\ast }\omega ={\omega }_0$, proving the Darboux theorem.

For a smooth function $H$ on a closed symplectic manifold, its Hamiltonian vector field is defined by ${\iota }_{X_H}\omega =-dH$. The same formula gives ${\mathcal{L}}_{X_H}\omega =-d^{2}H=0$, so the Hamiltonian flow preserves the symplectic form. To move one point to another in a connected $X$, join them by a path, cover the path by finitely many Darboux charts, and in each chart use a cutoff linear Hamiltonian to perform a small translation. Composing these compactly supported Hamiltonian diffeomorphisms proves that symplectomorphisms act transitively on each connected component.

In ${\mathbb{R}}^2$, every embedded curve is a Lagrangian submanifold. Let $L_1,L_2$ be circles enclosing different Euclidean areas. A plane symplectomorphism preserves area and carries the bounded complementary component of one circle to that of its image, so no symplectomorphism maps $L_1$ to $L_2$.

The same phenomenon exists in every ${\mathbb{R}}^{2n}$. Withtake the product toriThe Liouville class of a Lagrangian submanifold has respective period vectors $\pi (1,\dots ,1)$ and $2\pi (1,\dots ,1)$ on these tori. Every symplectomorphism of ${\mathbb{R}}^{2n}$ preserves the Liouville class up to the induced integral change of basis on $H_1(T^n)$, because its pullback changes $\lambda$ only by an exact form. An integral automorphism sends a primitive vector to a primitive vector and therefore cannot send the first period vector to the second. Hence

With the conventionthe time-dependent Hamiltonian vector field is uniquely determined because $\omega$ is nondegenerate. If ${\varphi }_t$ is its local flow, Cartan's magic formula givesThus ${\varphi }_t^{\ast }\omega =\omega$ wherever the flow is defined, which is the Hamiltonian flow preserves the symplectic form property.

Write $v=(a,b)$ and ${\omega }_0=dx\wedge dy$. The linear Hamiltonianhas Hamiltonian vector field $X_{H_0}=v$. Choose a smooth cutoff $\rho$ that is one on $B(r+|v|)$ and zero outside $B(r+|v|+ϵ)$, and put $H=\rho H_0$. Every trajectory beginning in $B(r)$ and following $v$ remains in $B(r+|v|)$ for time $0\le t\le 1$, so the time-one map translates that ball by $v$. Outside $B(r+|v|+ϵ)$ its vector field vanishes, so the map is the identity. This is a compactly supported Hamiltonian translation and hence a compactly supported symplectomorphism.

For the connected-sum construction, choose a Darboux chart about the unique transverse intersection and straighten the two Lagrangian sheets to ${\mathbb{R}}^2$ and $i{\mathbb{R}}^2$ in ${\mathbb{C}}^2$. Remove small disks from the two sheets and join their boundary circles by the standard Lagrangian neckin ${\mathbb{R}}_x^2\oplus {\mathbb{R}}_y^2$, where $(a(t),b(t))$ follows a smooth arc from one positive coordinate ray to the other and agrees with those rays near its ends. Its pullback of ${\sum }_{j}d{x}_j\wedge d{y}_j$ vanishes because $u\cdot u^{\prime }=0$. Gluing this neck to the unchanged surfaces performs Lagrangian surgery. Topologically it is their connected sum, so it gives a Lagrangian connected sum

Finally, choose an immersed circle ${\gamma }_l:S^1\to {\mathbb{R}}^2$ with exactly $l$ transverse double points and no other multiple points; one may add $l$ small figure-eight kinks to an embedded circle. Let $c:S^1\to {\mathbb{R}}^2$ be an embedded circle and defineThis Product Lagrangian immersion satisfies ${\iota }^{\ast }{\omega }_0=0$. If $p$ is a double point of ${\gamma }_l$, its two local branches times $c(S^1)$ meet along $\{p\}\times c(S^1)$. Their tangent spaces intersect precisely in the tangent line to that circle, so the intersection is clean. The $l$ double points therefore give exactly $l$ disjoint clean self-intersection circles.

Near each clean circle, perturb one Lagrangian sheet by the graph of $\delta dh$ in its Weinstein neighborhood, where $h:S^1\to \mathbb{R}$ is a Morse function with one minimum and one maximum. The clean circle is replaced by two transverse double points. Resolve both by Lagrangian surgery. Each resolution attaches one one-handle and lowers the Euler characteristic by two, so resolving all $l$ clean circles changes the Euler characteristic of the original torus from zero toChoose the orientation-reversing neck at one double point; the resulting connected surface is nonorientable, while all the surgeries remove their double points and leave an embedding. Thus the Givental construction of nonorientable Lagrangian surfaces gives a closed connected nonorientable surface of Euler characteristic $-4l$ Lagrangian embedded in $({\mathbb{R}}^4,{\omega }_0)$.