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An anti-symplectic involution is an anti-symplectic map whose square is the identity. Its nonempty fixed-point set is a Lagrangian submanifold; the linear version has complementary $+1$ and $-1$ Lagrangian subspaces.

A subspace $W$ of a symplectic vector space $(V,\Omega )$ is isotropic when $\Omega {|}_{W\times W}=0$. It has at most half the dimension of $V$; equality makes it a Lagrangian subspace. This use of isotropic concerns an alternating bilinear form, rather than an isotropic quadratic form.

A Lagrangian complement to $L$ in a symplectic vector space is a Lagrangian subspace $K$ with $E=L\oplus K$. Such complements exist smoothly for Lagrangian subbundles. Identify any initial complement with $L^{\ast }$ using the symplectic cross-pairing. If its internal pairing is $b(\eta ,\zeta )$, correct its lift by a vector $a\eta \in L$ with $\zeta (a\eta )=-b(\eta ,\zeta )/2$. The corrected complement has zero internal pairing.

The Lagrangian Grassmannian is the space of all Lagrangian subspaces of a symplectic vector space. The unitary group acts transitively on $\mathrm{LGr}({\mathbb{R}}^{2n})$, and the stabilizer of ${\mathbb{R}}^n$ is the orthogonal group $O(n)$, giving

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.

Relative to a fixed Lagrangian subspace $L_0$, the Maslov cycle consists of Lagrangian subspaces $L$ with $L\cap L_0\ne 0$. In a two-dimensional symplectic space all lines are Lagrangian, and the cycle is the single line $L_0$ in the projective line. Fibrewise versions yield sections of a projective-line bundle and record passage through a prescribed vertical direction.

The Maslov index assigns an integer to a loop of Lagrangian subspaces. Under $\mathrm{LGr}({\mathbb{R}}^{2n})\cong U(n)/O(n)$, it is the degree of the map induced by $A\mapsto \det (A{)}^2$.

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.

Let $B$ be a nondegenerate alternating bilinear form on $V$ of dimension $2m$ over ${\mathbb{F}}_q$. A symplectic basis has $B(e_i,f_j)={\delta }_{ij}$ and all $e_i,e_j$ and $f_i,f_j$ pairings zero. Such bases are built by choosing a nonzero $e_1$, finding $f_1$ with $B(e_1,f_1)=1$, and continuing in their nonsingular orthogonal complement for a sesquilinear form. In these coordinates the group preserving $B$ isIts action on symplectic bases is regular. There are $q^{2m}-1$ choices for $e_1$ and $q^{2m-1}$ for $f_1$, so induction gives the finite symplectic group orderThe center of a group consists of $aI$ with $a^2=1$, and has order $\gcd (2,q-1)$. To see there are no other central elements, a central element commutes with every symplectic transvection and hence fixes every symplectic transvection center line; a linear map fixing every line is scalar. Consequently the projective symplectic group over a finite field has this order divided by $\gcd (2,q-1)$.

Here is a proof of the main ingredients in simplicity. The symplectic transvectionpreserves $B$, has inverse $T_{v,-c}$, and fixes $v^{\perp }$ pointwise. These transformations generate $S{p}_{2m}(q)$. Indeed if $B(u,w)\ne 0$, then $T_{w-u,1/B(u,w)}$ sends $u$ to $w$. If the pairing is zero, choose an intermediate vector having nonzero pairing with both, so two symplectic transvections suffice. This aligns the first vector of any symplectic basis. For partners $f,f^{\prime }$ of the same $e$, their difference lies in $e^{\perp }$. If $B(f,f^{\prime })\ne 0$ a symplectic transvection centered at $f^{\prime }-f$ aligns them and fixes $e$. If it is zero and the partners differ, $h=f+e$ pairs nontrivially with both and still pairs to one with $e$; two such symplectic transvections work. After fixing the pair, induction in its orthogonal complement for a sesquilinear form aligns the rest of the basis.

The projective action on one-spaces is faithful. For $m\ge 2$, a line stabilizer has three orbits: the line itself, the other lines perpendicular to it, and the lines not perpendicular to it. Witt's lemma proves transitivity within the latter two orbits by matching the two-dimensional restricted form. This action is primitive. A block containing distinct perpendicular lines $L,M$ is invariant under both stabilizers and therefore contains a line perpendicular to $M$ but not to $L$. A block containing distinct nonperpendicular lines likewise contains a line perpendicular to $L$ but not to $M$, by using the $M$ stabilizer. Such lines exist by nonsingularity in dimension at least four. Either possibility makes the block contain both nontrivial stabilizer orbits, hence every line. For $m=1$ the action on the projective line is two-transitive and thus primitive.

Symplectic transvections with one fixed center line form an Abelian normal subgroup of its stabilizer, and their conjugates generate the whole group. The Iwasawa simplicity lemma now reduces simplicity to being perfect. Its short proof is useful: any nontrivial normal subgroup of a faithful primitive group is transitive, since its orbits are blocks. Writing $G=NH$, where $H$ is a point stabilizer, shows that all conjugates of the specified Abelian subgroup have the same image modulo $N$. Thus $G/N$ is Abelian. If $G$ is perfect, $N=G$.

The groups can also be shown to be perfect concretely. For $q>3$ choose $a$ with $a^2\ne 1$ and let $D$ scale $e$ by $a$ and $f$ by $a^{-1}$. Thenso every symplectic transvection is a group commutator. For $q=3,m\ge 2$, symplectic transvections of coefficient one are conjugate, have order three, and the four with centers $e_1,e_2,e_1+e_2,e_1-e_2$ in a two-dimensional totally isotropic subspace multiply to the identity. They commute, and the sum of their rank-one terms is zero modulo three. Their common class in the abelianization is therefore annihilated by both three and four, hence is zero. For $q=2,m\ge 3$, use the seven nonzero vectors of a three-dimensional totally isotropic subspace. The symplectic transvections commute and their product is the identity, because each diagonal coefficient occurs four times and each cross coefficient twice. Their common abelianized class is annihilated by two and seven, hence is zero. Generation now proves these groups are perfect.

It follows thatThe exceptions are $S_3$, $A_4$ and $S_6$ respectively. For odd $q$ the full symplectic group over a finite field has nontrivial center of a group, so simplicity refers to its quotient group; for even $q$ the center of a group is trivial.

A symplectic parabolic subgroup stabilizes a flag of totally isotropic subspaces. The maximal such parabolic subgroups $P_r$ stabilize a single $r$-space, $1\le r\le m$. In an adapted basis their Levi subgroup is $G{L}_r(q)\times S{p}_{2m-2r}(q)$, with an arbitrary block of $r(2m-2r)$ parameters and a remaining symmetric block of $r(r+1)/2$ parameters. ThusThese are the single-member flag stabilizers, equivalently the standard maximal parabolic subgroups obtained by omitting one node of the symplectic Dynkin diagram. Two particular cases have transparent descriptions.

For the symplectic point stabilizer, write $V=⟨e,f⟩\perp W$. Its unipotent radical consists ofAll pairings are preserved by direct substitution. The product law is $(b,t)(b^{\prime },t^{\prime })=(b+b^{\prime },t+t^{\prime }+B(b,b^{\prime }))$, so this radical has order $q^{2m-1}$. With group commutator convention $uv{u}^{-1}{v}^{-1}$, its commutator has central parameter $2B(b,b^{\prime })$: it is Heisenberg-type in odd characteristic, but Abelian in characteristic two. A Levi subgroup scales $e,f$ inversely and acts as $Sp(W)$, givingThis index counts the one-spaces, all of which are isotropic for an alternating bilinear form.

For a Lagrangian subspace $E=⟨e_1,\dots ,e_m⟩$, the symplectic Lagrangian stabilizer has block formSubstituting into $g^{\mathsf{T}}Jg=J$ proves both the inverse-transpose lower block and symmetry of $S$. The unipotent radical is the additive group of symmetric matrices, and $A$ acts by $S\mapsto AS{A}^{\mathsf{T}}$. ThereforeThe last identity follows by canceling $q$-powers and $(q^j-1)$ factors in the two order formulas; it counts maximal totally isotropic subspaces.

For the Johnson graph, an edge exchanges one member of a $k$-set. Along one edge, intersection size with a fixed target can increase by at most one, so any path in a graph from $A$ to $B$ has length at least $k-|A\cap B|$. Replacing the elements of $A\setminus B$ successively by elements of $B\setminus A$ gives a path in a graph attaining that bound. Hence the distance in a Johnson graph isIf two ordered pairs have the same distance $r$, their four regions (intersection, first-only, second-only and outside the union) have sizes $k-r,r,r,n-k-r$. Choose bijections between corresponding regions. Their union is a ground-set permutation sending the ordered pair to the other pair and preserving graph adjacency. The symmetric group is distance-transitive. For $n\ge 2k$ the graph diameter is $k$.

For the Grassmann graph, a codimension-one exchange changes intersection dimension with a fixed target by at most one. More explicitly, if $A,A^{\prime }$ are adjacent, their common $(k-1)$-space shows $\dim (A^{\prime }\cap B)\ge \dim (A\cap B)-1$, and exchanging the two gives the reverse bound. Thus distance is at least $r=k-\dim (A\cap B)$.

Choose a basis $c_1,\dots ,c_{k-r}$ of $I=A\cap B$, complement it in $A$ by $a_1,\dots ,a_r$, and in $B$ by $b_1,\dots ,b_r$. These combined vectors are independent because $A\cap B=I$. The spacesform a path in a graph of length $r$. ConsequentlyEqual-distance ordered pairs have the same dimensions in this adapted basis construction. Extend both combined bases to bases of $V$, and match them by a invertible linear map. It maps the first ordered pair to the second and preserves dimensions of intersections, hence graph adjacency. $G{L}_n(F)$ is distance-transitive, over an arbitrary field, with graph diameter $k$ under $n\ge 2k$.

For the symplectic dual polar graph, vertices are Lagrangian $m$-spaces. Let $I=A\cap B$ have dimension $m-r$. The quotient $I^{\perp }/I$ is a nonsingular symplectic vector space of dimension $2r$, and $A/I,B/I$ are complementary Lagrangian subspaces. Their mutual pairing is nonsingular: a vector orthogonal to both lies in $(A+B{)}^{\perp }=I$, so vanishes in the quotient. Choose dual bases $e_1,\dots ,e_r$ in $A/I$ and $f_1,\dots ,f_r$ in $B/I$, and lift them into $A,B$. Their span is a nonsingular symplectic $2r$-space. Complete it by a basis $e_{r+1},\dots ,e_m$ of $I$ and suitable dual partners. This gives the pair normal formThe spaces $A_j=⟨f_1,\dots ,f_j,e_{j+1},\dots ,e_m⟩$ remain totally isotropic subspace: a present $f_i$ pairs only with the absent $e_i$. They give an adjacent-step path in a graph of length $r$. The previous intersection-dimension lower bound still applies, provingFor two ordered pairs at the same distance, match their adapted symplectic bases. The resulting map preserves the alternating bilinear form and sends the pairs to each other, so lies in $S{p}_{2m}(F)$. Equivalently, construct the isometry of a space with a form on $A+B$ from the displayed normal form and extend it by Witt's lemma. The action of the symplectic group over a field is distance-transitive, over every field, and its graph has graph diameter $m$. The zero-dimensional boundary cases give a single vertex and trivial transitivity.

An $N$-dimensional stratified pseudomanifold has a closed filtration $\varnothing =X_{-1}\subseteq X_0\subseteq \cdots \subseteq X_N=X$ with each nonempty $X_s\setminus X_{s-1}$ an $s$-dimensional manifold, dense top stratum, and $X_{N-1}=X_{N-2}$. Each point of an $s$-dimensional stratum has a stratum-preserving conical neighbourhood ${\mathbb{R}}^s\times cL$, where $L$ is a compact $(N-s-1)$-dimensional stratified pseudomanifold. The definition is inductive, starting with zero-dimensional spaces. There are no singular codimension-one strata; connectedness of links is not required. One may work in the usual triangulable category used for Witt bordism.

A rational Witt space is such a stratified pseudomanifold for which every stratum of odd codimension $2r+1$ has link $L$ satisfying $I{H}_r^{\bar{m}}(L;\mathbb{Q})=0$. No extra condition is imposed at even codimension. A Witt space with collared boundary has Witt interior and Witt boundary, and a stratum-preserving collar $\partial X\times [0,\epsilon )$ with the product stratification. An orientation means an orientation of the regular stratum, inducing the boundary orientation on its collar.

A three-pronged graph, with its central vertex and open edges as strata, is a stratified space but not a one-dimensional pseudomanifold: the central vertex is a singular codimension-one stratum, and deleting it leaves three local branches instead of the two branches of a one-manifold. For a non-Witt pseudomanifold, take the suspension of $T^2$. Its two vertices have codimension three and link $T^2$, whose middle group is $I{H}_1(T^2;\mathbb{Q})=H_1(T^2;\mathbb{Q})={\mathbb{Q}}^2$. It violates the Witt space condition.

The rational Witt bordism classification in dimensions divisible by four states that, for every $k\ge 1$, the middle intersection form induces an isomorphismHere $W(\mathbb{Q})$ is the Witt group of a field, so a nonsingular symmetric form represents zero precisely when it is metabolic, that is, when it has a half-dimensional totally isotropic subspace. The Complex projective plane gives a nonzero example: its middle group is generated by the degree-two class $h$, with $⟨h^2,[{\mathbb{CP}}^2]⟩=1$. Its form is $⟨1⟩$, of signature one; a metabolic form has signature zero. Thus $[{\mathbb{CP}}^2]\ne 0$ in ${\Omega }_4^{\mathrm{Witt}}$.

For a closed oriented Witt space $X$ of dimension $2k+1$, take the closed coneAway from its vertex it is a stratified product with an interval, so all old Witt space conditions persist. At the new vertex the codimension is $2k+2$, which is even; hence its odd-dimensional link $X$ creates no new Witt condition. The collar $X\times (1-\epsilon ,1]$ is explicit, and the product orientation may be chosen to induce the given orientation on $X$. This is a compact oriented Witt space with collared boundary, proving

Now let $\dim X=2$. Its singular strata are zero-dimensional. Their links are compact one-dimensional pseudomanifolds, which have no singular strata and are therefore nonempty finite disjoint unions of circles. A single circle is allowed, since a chosen stratum need not be intrinsically singular. The Witt condition is automatic in dimension two, because singular strata have even codimension.

Construct the normalization of a two-dimensional pseudomanifold concretely. Remove disjoint small conical neighbourhoods of the finitely many zero-stratum points. The remainder is a compact oriented surface with boundary, with one boundary circle for each local branch. Cap every such circle with a separate disk. The result $\tilde{X}$ is a closed compact oriented topological surface. Map each new disk homeomorphically onto its branch of the original cone, sending all the disk centres belonging to one old singular point to that point. Along the disk boundaries this agrees with the identification with the remainder, so it defines a continuous surjection $\pi :\tilde{X}\to X$. It is a homeomorphism off these finitely many centres, in particular over the prescribed open stratum. Every fibre is a point except over a zero-stratum point, where it is the finite set of its branch centres. Thus every fibre is zero-dimensional, as required.

Each component of $\tilde{X}$ bounds an oriented handlebody: a surface of genus $g$ is the boundary of a three-ball with $g$ one-handles attached. Let $H$ be the disjoint union of these handlebodies, oriented so that $\partial H=\tilde{X}$. Formgluing $x\in \partial H$ to $(\pi (x),0)$. The quotient is compact and Hausdorff, and its only remaining boundary is $X\times \{1\}$, with an explicit product collar. Near regular points of the gluing locus, two half-ball neighbourhoods glue to a three-ball; the induced orientations match after choosing the collar orientation in the usual boundary-gluing convention.

It remains to check the singular links, rather than merely assert that this quotient is a Witt space. If an old point $p$ has $r$ circle branches, the new one-dimensional stratum $\{p\}\times (0,1)$ has link the disjoint union of those $r$ circles. This has codimension two and imposes no Witt condition. At $(p,0)$, each branch supplies a half-three-ball from $H$ and another from the collar. The halves glue to a full three-ball; the different branches share only the positive collar axis. Intersecting with a small link therefore gives $r$ copies of $S^2$ identified at one common point. This is a two-dimensional pseudomanifold with zero middle intersection homology: an allowable one-cycle avoids the common point, so decomposes into cycles in the $r$ punctured spheres, each homeomorphic to ${\mathbb{R}}^2$; each such cycle bounds there. Consequently $I{H}_1({\bigvee }_{a=1}^r{S}^2;\mathbb{Q})=0$. This is exactly the Witt condition at the new codimension-three stratum. The boundary has the required product stratification. Hence $Z$ is a three-dimensional Witt space with collared boundary with $\partial Z\cong X$, proving

For the final product assertion, “product” means a nontrivial product with at least two positive-dimensional factors; a single four-dimensional factor would allow the preceding ${\mathbb{CP}}^2$ counterexample. State the standard rational intersection-homology product fact used here: products of Witt spaces are Witt, their intersection cohomology satisfies the Künneth theorem, and the complementary-degree intersection pairing on elementary tensors is the product of the factor pairings, with the sign $(-1{)}^{\mathrm{deg}\beta \mathrm{deg}{\alpha }^{\prime }}$ for $\alpha \otimes \beta$ and ${\alpha }^{\prime }\otimes {\beta }^{\prime }$. Terms whose degrees in a factor do not add to that factor's dimension pair to zero. Regrouping factors reduces total dimension four to cases $1+3$ or $2+2$.

In the $1+3$ case, write $X=A\times B$. The middle intersection cohomology isEach summand is totally isotropic, because the degrees in $A$ are respectively $0+0$ and $1+1$, neither equal to one. They pair perfectly with one another by factor Poincare duality. Thus either summand has half the total dimension and is a Lagrangian for the middle symmetric form.

For dimensions $2+2$, the middle group is the direct sumThe outside summands are isotropic and pair perfectly with one another. Each factor's degree-one pairing is a nonsingular skew-symmetric form, so choose a Lagrangian subspace $L_A\subset I{H}^1(A)$. Inside the central summand, $L_A\otimes I{H}^1(B)$ is isotropic and has half its dimension. It is orthogonal to both outside summands by their degrees. Thereforeis an isotropic subspace of half the full middle dimension. In both cases the form is metabolic. The Witt bordism classification now proves the desired $[X]=0$ in ${\Omega }_4^{\mathrm{Witt}}$. This also proves the assertion for more factors, using the regrouping just described.

In weight one, the Hodge numbers are $h^{1,0}=h^{0,1}=r$, and the Hodge filtration is determined by the $r$-plane $H=F^1$. The bilinear orthogonality condition is $\psi (H,H)=0$. An isotropic $r$-plane in a symplectic $2r$-space is a Lagrangian subspace, soThe ordinary Grassmannian has dimension $r^2$. Restriction of the alternating form to an $r$-plane imposes $r(r-1)/2$ independent conditions. Equivalently, near a fixed Lagrangian plane, transverse Lagrangians are graphs of symmetric matrices, with $r(r+1)/2$ independent entries. Therefore

Write ${\mathbb{C}}^{2r}={\mathbb{C}}^r\oplus {\mathbb{C}}^r$ in the block order of the given matrix. On $D$, positivity is $i\psi (v,\bar{v})>0$ for every nonzero $v\in H$. The plane $H$ cannot contain $(x,0)\ne 0$, because that vector has $\psi (v,\bar{v})=0$. Projection onto the second coordinate factor is therefore injective on $H$ and, by equal dimensions, an isomorphism. There is a unique matrix $Z$ such thatFor $v=(Zu,u)$ and $v^{\prime }=(Zw,w)$, direct evaluation givesThus isotropy is exactly $Z=Z^{\mathsf{T}}$. For such $Z$,Hence positivity is precisely positive definiteness of $\mathrm{Im}Z$. Conversely, symmetry and this inequality define a positive Lagrangian plane; it is opposed to its conjugate because $(Z-\bar{Z})u=0$ forces $u=0$. We have explicitly identifiedthe Siegel upper half-space. The choice of the graph $(Zu,u)$, rather than $(u,Zu)$, is what matches the sign of the given symplectic form.

Let an integral element of the symplectic group have block matrixActing on graph vectors givesThe image remains a positive Lagrangian plane, so its second projection is invertible. Thus $CZ+D_0$ is invertible and the induced action isThe symplectic identities give symmetry of this matrix, and preservation of $i\psi (v,\bar{v})$ givesThis is a Hermitian congruence by an invertible matrix, so it proves preservation of the positive domain as well as describing the action. These are coordinates on marked Hodge structures; passing to the integral-group quotient forgets the symplectic marking.

We prove the Weinstein neighborhood theorem with the canonical sign convention ${\omega }_0=-d\alpha$. The essential first step is to match the two symplectic forms as bilinear forms on the entire tangent space along $X$, not only after pulling them back to $TX$.

Here is the needed symplectic splitting along a Lagrangian submanifold. Write $E=TM{|}_X$ and $L=TX$. Choose a smooth vector bundle complement $N$ of $L$, for example using a Riemannian metric. Since $L$ is a Lagrangian subspace in every fiber, the pairing mapis an isomorphism: its kernel is $N\cap L^{\omega }=N\cap L=0$, and both bundles have rank $n$. Let $s:L^{\ast }\to N$ be its inverse and put $b(\eta ,\zeta )=\omega (s\eta ,s\zeta )$. Define $a:L^{\ast }\to L$ byand set $\tilde{s}=s+a$. Nondegeneracy of the dual pairing gives a unique smooth $a$. Since $b$ is alternating,Thus $K=\tilde{s}(L^{\ast })$ is a smooth Lagrangian complement to $L$. The mapis a vector bundle isomorphism, is the identity on $TX$, and satisfiesThe last equality uses the canonical splitting of the tangent bundle of $T^{\ast }X$ along its zero section.

Choose a Riemannian metric near $X$ for which $K$ is orthogonal to $TX$. The tubular neighborhood construction using its Riemannian exponential map gives a diffeomorphismafter shrinking around the zero section. It restricts to the given inclusion of $X$ and has differential $A$ along it. Consequently $\beta ={\psi }^{\ast }\omega$ and ${\omega }_0$ agree pointwise as ambient bilinear forms at every point of the zero section.

We use the following precise relative Moser theorem. If two closed symplectic forms ${\gamma }_0,{\gamma }_1$ near a compact embedded submanifold $S$ agree as ambient bilinear forms along $S$, then, on smaller neighborhoods, there is a diffeomorphism $\kappa$ fixing $S$ pointwise with ${\kappa }^{\ast }{\gamma }_1={\gamma }_0$. One sufficient version allows a symplectic path ${\gamma }_t$ with ${\partial }_t{\gamma }_t=d{\eta }_t$ and ${\eta }_t=0$ as ambient covectors along $S$: solve ${\iota }_{Z_t}{\gamma }_t=-{\eta }_t$ and use its flow. The vector field vanishes on $S$, so the flow fixes $S$, and compactness permits a common smaller neighborhood for the full time interval.

For completeness, all hypotheses can be checked directly here. Put $\delta =\beta -{\omega }_0$ and ${\gamma }_t={\omega }_0+t\delta$. At the zero section, every ${\gamma }_t$ equals ${\omega }_0$. The nondegenerate property is open, and compactness of $X\times [0,1]$ allows a single smaller neighborhood on which all ${\gamma }_t$ are symplectic forms. Choose it fiberwise star-shaped. Write $H_s(q,p)=(q,sp)$ and let $E$ be the vertical radial vector field. The radial homotopy primitive near a zero sectionis smooth, is zero as an ambient covector along the zero section, and satisfiesThe radial homotopy operator gives this identity, using $d\delta =0$ and $H_0^{\ast }\delta ={\pi }^{\ast }{i}_0^{\ast }\delta =0$. Smoothness at $s=0$ follows because contraction with $E$ supplies a factor of $s$ after evaluation at $(q,sp)$; the vanishing of $\delta$ along $X$ supplies additional vanishing. Therefore solve ${\iota }_{Z_t}{\gamma }_t=-\eta$. By Cartan's magic formula, its flow maps ${\kappa }_t$ satisfyAfter shrinking the starting neighborhood, these flows exist for $0\le t\le 1$ and fix $X$. This is a relative local construction, not an assertion of completeness throughout the noncompact cotangent bundle.

Finally set $\phi =\psi \circ {\kappa }_1$, let $U_0$ be its sufficiently small domain, and let $U=\phi (U_0)$. ThenThus the map is a symplectomorphism of neighborhoods and agrees with the specified inclusion at every point of $X$.

Use the decomposition into Lagrangian subspaces $V=V_{+}\oplus V_{-}$ from part (a). The bilinear form $\Omega$ pairs $V_{+}$ and $V_{-}$ nondegenerately: a vector in either summand annihilating the other already annihilates its own summand and hence all of $V$, so is zero.

Choose a basis $e_1,\dots ,e_n$ of $V_{+}$. The linear mapis an linear isomorphism. Let $f_j$ correspond to the dual basis of the $e_i$. Then $\Omega (e_i,f_j)={\delta }_{ij}$, and the pairings within each summand vanish. Thus

For $p\in \mathrm{Fix}(\sigma )$, differentiation gives $(d{\sigma }_p{)}^2=I$ and $(d{\sigma }_p{)}^{\ast }{\omega }_p=-{\omega }_p$. Thus $d{\sigma }_p$ is an anti-symplectic involution of the symplectic vector space $T_{p}M$. By part (a), its $+1$ eigenspace is a Lagrangian subspace.

In the local manifold chart supplied in the question, $\sigma$ is linear, so its fixed-point set is the intersection of the chart with this $+1$ eigenspace. Therefore the fixed-point set is an embedded submanifold andEvery such tangent space has half the ambient dimension and the symplectic form vanishes on it. Hence every nonempty fixed-point set is a Lagrangian submanifold. The fixed-point set itself need not be connected.

Use the compatible metrics $g_{\Sigma }(\cdot ,\cdot )={\omega }_{\Sigma }(\cdot ,j\cdot )$ and $g_X(\cdot ,\cdot )={\omega }_X(\cdot ,J\cdot )$. The energy of a smooth map isIt is a J-holomorphic curve when $du\circ j=J\circ du$. Splitting $du$ into its complex-linear and complex-antilinear parts gives the Energy identity for a J-holomorphic curveThe first term depends only on the homology class. It follows that a J-holomorphic map minimizes energy among all maps in its homology class.

One Monotonicity theorem for a J-holomorphic curve says that a nonconstant J-holomorphic curve through the center of a sufficiently small radius-$\rho$ ball, with boundary outside that ball, has area at least $c{\rho }^2$; in the standard complex ball one may take the sharp value $\pi {\rho }^2$. The Gromov non-squeezing theorem says

Write ${\mathbb{R}}^4=T^{\ast }{\mathbb{R}}^2$ with coordinates $(q,p)$ and form $d{q}_1\wedge d{p}_1+d{q}_2\wedge d{p}_2$. For $c>0$, the graphis a Lagrangian subspace. Points of $L_c\cap B(R)$ have $|q|\le R/\sqrt{1+c^2}$. Choose $c$ so large thatThen no two points of the $\epsilon$-neighborhood of $L_c\cap B(R)$ differ by a nonzero vector $(m,0)$ with $m\in {\mathbb{Z}}^2$, so quotienting $q$ modulo ${\mathbb{Z}}^2$ is injective there.

Relative to the Lagrangian splitting $L_c\oplus J{L}_c$, the map $(v,w)\mapsto (\lambda v,{\lambda }^{-1}w)$ is symplectic. It sends $B^4(r)$ into the indicated long thin neighborhood wheneverThese inequalities are compatible when $R>r^2/\epsilon$. Taking such an $R$ and then the quotient constructs the arbitrarily large symplectic balls in a cotangent cylinder:

On the quotient from part (b), the Maslov mapis well defined because $\det (Q{)}^2=1$ for $Q\in O(n)$. Consider the loop of Lagrangian subspacesIts endpoints agree as unoriented subspaces, and $\mu \circ \ell (t)=e^{2\pi it}$ has degree one. If $\eta$ is the generator of $H^1(S^1;\mathbb{Z})$, thenConsequently ${\mu }^{\ast }\eta \ne 0$, provingThe same integer is the Maslov index of $\ell$.

Positive complex Lagrangian subspaces for the integral symplectic form $\left(\begin{array}{cc}0& -I\\ I& 0\end{array}\right)$ are graphs $H_Z=\{(Zu,u):u\in {\mathbb{C}}^r\}$. Isotropy is the symmetry of $Z$, and $iQ(v,\bar{v})=2u^{\mathsf{T}}(\mathrm{Im}Z)\bar{u}$ gives positivity. For $g=\left(\begin{array}{cc}A& B\\ C& D\end{array}\right)$ in the symplectic group, the action is $Z\mapsto (AZ+B)(CZ+D{)}^{-1}$.

A symplectic basis organizes a nondegenerate alternating bilinear form into independent pairs. Let $(V,\omega )$ be a finite-dimensional vector space over a field $K$, with $\omega$ a nondegenerate bilinear form satisfying $\omega (v,v)=0$ for every vector $v$. Its dimension is even, say $2n$, and a symplectic basis is an ordered basis $\mathcal{B}=(e_1,\dots ,e_n,f_1,\dots ,f_n)$ such thatHere ${\delta }_{ij}$ is the Kronecker delta. Each $(e_i,f_i)$ is a hyperbolic pair; different pairs have zero mutual pairing. These conventions apply in every characteristic. In characteristic two, $-1=1$, but the condition $\omega (v,v)=0$ is still essential: a symmetric bilinear form need not be an alternating bilinear form. The existence theorem and the characteristic caveat are given in Bilinear Forms by Keith Conrad (Keith Conrad, Bilinear Forms, §1, Theorem 1.6; §5, Theorems 5.1 and 5.4 and Definition 5.3).

The benefit is a common coordinate language for every symplectic vector space of the same finite dimension over the same field. The span of all $e_i$, and separately of all $f_i$, is a Lagrangian subspace; each paired span is a symplectic subspace. In Hamiltonian mechanics, the corresponding pairing of position and momentum coordinates describes phase space and canonical transformations. On a symplectic manifold, a symplectic basis describes one tangent space; Darboux coordinates additionally describe a whole local neighborhood.^[ref]

Vertices are Lagrangian subspaces of a $2m$-dimensional symplectic vector space, joined when they intersect in codimension one. A pair with intersection dimension $m-r$ admits a symplectic basis in which $U=⟨e_1,\dots ,e_m⟩$ and $W=⟨f_1,\dots ,f_r,e_{r+1},\dots ,e_m⟩$. Replacing $e_i$ successively by $f_i$ gives an isotropic path of length $r$, while one edge can increase intersection dimension with a fixed target by at most one. Equal-distance ordered pairs have the same adapted form, so the symplectic group acts distance-transitively.