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A coisotropic submanifold $N$ satisfies $(T_{p}N{)}^{\omega }\subset T_{p}N$, where the superscript denotes the symplectic orthogonal complement. Every hypersurface of a symplectic manifold is coisotropic.

Let $V=C\oplus D$ be a symplectic orthogonal complement splitting into two symplectic subspaces of dimension two. In symplectic bases, the graph of a function $A:D\to C$ pulls back the symplectic form to ${\omega }_D+A^{\ast }{\omega }_C=(1+\det A){\omega }_D$. It is a symplectic subspace exactly when $\det A\ne -1$. Its symplectic orientation agrees with the orientation transported from $D$ exactly when $\det A>-1$. When $\det A<-1$, the transverse oriented pair $(C,{\Gamma }_A)$ has negative intersection sign, whereas two distinct complex lines in a compatible complex two-space have positive intersection sign.

Choose representatives ${\lambda }_1,\dots ,{\lambda }_r$ as in part (b). We use the symplectic centralizer decompositionTo prove it, a linear map commuting with every $a\in \mathfrak{a}$ must preserve every weight space: if $v\in V_{\lambda }$, then $a(Tv)=T(av)=\lambda (a)Tv$. Conversely, preserving the weight spaces makes a linear map commute with the scalar action of every $a$ on each space.

On $V_0$, the symplectic Lie algebra condition says exactly that the restriction belongs to $\mathfrak{sp}(V_0)$. On $V_{{\lambda }_i}\oplus V_{-{\lambda }_i}$, use the perfect pairing from part (b) to identify $V_{-{\lambda }_i}$ with the dual space $V_{{\lambda }_i}^{\ast }$. The action $A_i$ on $V_{{\lambda }_i}$ can be arbitrary, but its action on $V_{-{\lambda }_i}$ must be $-A_i^{\ast }$, where $A_i^{\ast }$ is the dual map. Indeed,determines $B_i=-A_i^{\ast }$ uniquely. There are no additional conditions between different pairs because $\omega$ vanishes between different pairs. Every choice of these blocks therefore gives an element of the centralizer of a Lie subalgebra. The block correspondence preserves Lie brackets, proving the claimed direct sum of Lie algebras.

The general linear Lie algebra ${\mathfrak{gl}}_m(k)$ is an abelian Lie algebra exactly when $m\le 1$: for $m\ge 2$, the matrix units satisfy $[E_{11},E_{12}]=E_{12}\ne 0$. A nonzero symplectic vector space has a symplectic basis and contains a two-dimensional symplectic subspace. The symplectic Lie algebra acting on that plane, extended by zero on its symplectic orthogonal complement, contains the matrices $H=\mathrm{diag}(1,-1)$ and $E=E_{12}$ with $[H,E]=2E\ne 0$. Thus $\mathfrak{sp}(V_0)$ is an abelian Lie algebra exactly when $V_0=0$.

Since different summands in a direct sum of Lie algebras commute, we obtain the precise criterionEquivalently, there are $2n$ distinct nonzero weights of a representation, each with a one-dimensional weight space. The condition concerns multiplicities in the defining representation; it does not assert that $\dim \mathfrak{a}=n$ for every such $\mathfrak{a}$.

For a symplectic vector space $(V,\omega )$ and a vector subspace $W$, the symplectic orthogonal complement isNondegenerate bilinear form identifies $V$ with its dual space, so $\dim W^{\omega }=\dim V-\dim W$. If $W$ is a symplectic subspace, $W\cap W^{\omega }=0$, and hence $V=W\oplus W^{\omega }$; the restriction to $W^{\omega }$ is also nondegenerate.

Here is an inductive construction of a symplectic basis. Choose $e_1\ne 0$. Nondegenerate bilinear form supplies $f_1$ with $\omega (e_1,f_1)=1$. The plane $P=\mathrm{span}(e_1,f_1)$ is a symplectic subspace, and its symplectic orthogonal complement gives $V=P\oplus P^{\omega }$. Repeat on $P^{\omega }$. The process ends in dimension zero, since each step removes two dimensions and a one-dimensional alternating form is degenerate. We obtain $e_1,f_1,\dots ,e_n,f_n$ withThe linear map taking the coordinate vectors of ${\mathbb{R}}^{2n}$ to this basis pulls $\omega$ back to ${\sum }_{j}d{x}_j\wedge d{y}_j$, proving the asserted normal form.

Now split $V=C\oplus D$ with $D=C^{\omega }$, and choose symplectic bases on both planes. Parametrize ${\Gamma }_A$ by $v\mapsto (Av,v)$. Orthogonality removes the cross terms, and a two-dimensional alternating form transforms by the determinant, soConsequently the correct nondegenerate bilinear form criterion isThe strict inequality printed in the PDF requires an extra orientation condition: it says that the symplectic orientation on the graph agrees with the orientation transported from $D$. It is not equivalent to being a symplectic subspace. For example, $A=\mathrm{diag}(-2,1)$ has determinant $-2$ and gives a nondegenerate restriction $-{\omega }_D$. This is the orientation criterion for a graph of symplectic planes.

This same map supplies the requested obstruction. Orient $C$ and ${\Gamma }_A$ by their restricted symplectic forms. If $(d_1,d_2)$ is a positive basis of $D$, then $(A{d}_1+d_1,A{d}_2+d_2)$ is negatively oriented in ${\Gamma }_A$. The coordinate transformation from a basis of $C$ followed by these graph vectors to a basis of $C\oplus D$ has determinant $1$. Reversing the graph basis to give its positive orientation therefore makes the oriented direct sum $C\oplus {\Gamma }_A$ negative relative to the ambient symplectic orientation.

Two distinct complex lines in standard ${\mathbb{C}}^2$ have the opposite behaviour: complex bases of the two lines give a complex-linear isomorphism $\mathbb{C}\oplus \mathbb{C}\to {\mathbb{C}}^2$, whose real determinant is the positive number $|{\det }_{\mathbb{C}}{|}^2$. Their complex orientations are their symplectic orientations. A symplectic linear map preserves both the ambient and each plane's symplectic orientation, so it cannot change this sign. Thus the pair with $A=\mathrm{diag}(-2,1)$ cannot be the image of a pair of complex lines. Interchanging the planes does not change the sign, since both dimensions are even.

Fix the sign convention ${\iota }_{{\xi }_M}\omega =-d{\mu }^{\xi }$, where ${\xi }_M$ is the fundamental vector field and ${\mu }^{\xi }=⟨\mu ,\xi ⟩$. A Hamiltonian group action is a symplectic Lie group action together with an equivariant moment map $\mu :M\to {\mathfrak{g}}^{\ast }$ satisfying this identity for every $\xi \in \mathfrak{g}$. Equivariance means $\mu (gx)={\mathrm{Ad}}_{g^{-1}}^{\ast }\mu (x)$. Merely requiring each fundamental field to have some Hamiltonian is the weaker condition of a weakly Hamiltonian action.

For a circle group action, identify its Lie algebra with $\mathbb{R}$, and let $Y$ generate the action. Put $Z={\mu }^{-1}(t)$ and let $i:Z\hookrightarrow M$. Since $t$ is a regular value,The action is free, so $Y_x\ne 0$. The symplectic orthogonal complement of this hyperplane is exactly $\mathbb{R}{Y}_x$, which is contained in $T_{x}Z$. ThusThe free action of the compact circle group is proper, and the quotient $Q=Z/S^1$ is a smooth manifold with projection $\pi :Z\to Q$ a principal bundle. The form $i^{\ast }\omega$ is invariant and kills the orbit directions, hence is basic and descends to a unique two-form ${\omega }_Q$ satisfying ${\pi }^{\ast }{\omega }_Q=i^{\ast }\omega$. It is closed because pullback by a submersion detects differential forms. If a quotient tangent vector annihilates ${\omega }_Q$, a lift lies in the displayed kernel and hence projects to zero. Therefore ${\omega }_Q$ is nondegenerate. This proves symplectic reduction directly.

If $L\subset Q$ is a Lagrangian submanifold, its inverse image $\tilde{L}={\pi }^{-1}(L)$ is a smooth circle bundle over $L$. The restricted symplectic form is ${\pi }^{\ast }({\omega }_Q{|}_L)=0$. If $\dim M=2N$, then $\dim Q=2N-2$, $\dim L=N-1$, and $\dim \tilde{L}=N$, proving that $\tilde{L}$ is Lagrangian. This is the Lagrangian lift through circle reduction.

For scalar multiplication on ${\mathbb{C}}^{n+1}$, the generator is $Y={\sum }_j(-y_j{\partial }_{x_j}+x_j{\partial }_{y_j})$. Direct contraction givesChoose $t>0$. Its level is the sphere of radius $R=\sqrt{2t}$ and its quotient is ${\mathbb{CP}}^n$. Complex conjugation on the sphere reverses the ambient symplectic form and descends to an anti-symplectic involution of the quotient. Its fixed set is ${\mathbb{RP}}^n$: a projective point fixed by conjugation has a real representative after a phase change. For two tangent vectors to this fixed set, anti-symplecticity makes the reduced form equal to its negative, so it vanishes. The real dimension is $n$, half the quotient dimension, proving the Lagrangian fixed locus of an anti-symplectic involution property.

The lifted Lagrangian submanifold can be written explicitly asThe only redundancy in this parametrization is the displayed free involution: two nonzero real representatives on the same complex ray differ by a real sign. The quotient embeds smoothly in the sphere, and projection to $[x]\in {\mathbb{RP}}^n$ exhibits its circle bundle structure. Its Lagrangian property follows from the reduction argument, including the case $n=0$.

Replace $F^n$ by a symplectic vector space $(V,B)$ of dimension $2m$, and elementary row operations by symplectic transvectionsThe cross terms in $B(Tx,Ty)$ cancel, and $B(v,v)=0$, so these maps preserve $B$. Their inverses are $T_{v,-c}$.

They generate $Sp(V)$ by the following symplectic basis elimination. If $B(u,v)\ne 0$, a single transvection with direction $v-u$ and parameter $1/B(u,v)$ sends $u$ to $v$. If $B(u,v)=0$ and $u\ne v$, choose $w$ with both $B(u,w)$ and $B(w,v)$ nonzero: the union of two proper hyperplanes does not exhaust a vector space, even over ${\mathbb{F}}_2$. Two transvections then suffice. Thus one can make a form-preserving linear map fix the first basis vector $e$. Next align its partner $f$, while keeping $e$ fixed. Two partners have difference in $e^{\perp }$, so the preceding single-transvection construction uses a direction in $e^{\perp }$ whenever their mutual pairing is nonzero. If it is zero, the intermediate partner $f+ae$, $a\ne 0$, has nonzero pairing with both. All these transvections fix $e$. Once both $e,f$ are fixed, recurse on their nondegenerate symplectic orthogonal complement. This proves generation over every field.

Use the action on the points of projective space, whose kernel is the scalar center $Z=\{\lambda I:{\lambda }^2=1\}$. For $m=1$ this is the $S{L}_2$ case. For $m\ge 2$ the action is primitive, although it is generally not two-transitive. The stabilizer of a line $L$ has two orbits on other lines: those orthogonal to $L$ and those not orthogonal to $L$. This follows by extending a hyperbolic pair, or an independent isotropic pair, to a symplectic basis. A block containing $L$ and a second line $M$ contains the entire corresponding orbit of the stabilizer of $L$. The stabilizer of $M$ then supplies a line of the other type: choose a vector orthogonal to one of $L,M$ and not to the other. Nondegeneracy and independence guarantee such a vector, also in dimension four. Hence the block contains both orbits and is the whole set.

The transvections along $L$ form an abelian normal subgroup of its stabilizer; their conjugates generate the group. Thus the Iwasawa simplicity lemma applies as soon as perfectness is established. For $|F|>3$, every transvection lies in an embedded $S{L}_2(F)$ acting on a hyperbolic plane, and 4(c) proves that embedded group perfect.

For the remaining fields, the perfectness of finite symplectic groups has a short direct proof. Over ${\mathbb{F}}_3$ with $m\ge 2$, all parameter-one transvections have the same class $t$ in the abelianization, and $3t=0$. In an isotropic plane with basis $u,v$, the four directions $u,v,u+v,u-v$ give commuting transvections whose product is $1$: their rank-one terms add to zero, sinceover ${\mathbb{F}}_3$. Thus $4t=0$, giving $t=0$. Parameter-two transvections are inverses, so all generators vanish in the abelianization. Over ${\mathbb{F}}_2$ with $m\ge 3$, use the seven nonzero vectors of an isotropic three-space. Their commuting transvections multiply to $1$, since each diagonal coordinate of ${\sum }_{0\ne v}v{v}^{\mathsf{T}}$ occurs four times and each off-diagonal coordinate twice. Their common abelianized class satisfies $2t=7t=0$, and again the group is perfect.

Consequently the projective symplectic group is simple, with the necessary exceptionsThe first two are the linear exceptions; the last is $S{p}_4(2)\cong S_6$, proved in 6(c). The word “projective” matters: the unquotiented symplectic group may have the nontrivial scalar center $\{I,-I\}$.

Let $V$ have dimension $2m$ over the finite field ${\mathbb{F}}_q$, with a nondegenerate alternating bilinear form $B$. The symplectic group over a finite field isIt acts simply transitively on ordered symplectic bases: a linear map between two such bases is uniquely specified and preserves all pairings, hence the whole bilinear form.

Count a first hyperbolic pair $(e_1,f_1)$. There are $q^{2m}-1$ choices for $e_1\ne 0$. Nondegeneracy makes $B(e_1,-)$ a nonzero linear functional, so there are $q^{2m-1}$ choices of $f_1$ with $B(e_1,f_1)=1$. Their span is nondegenerate, and its symplectic orthogonal complement is a symplectic vector space of dimension $2m-2$. Recursing, with $|S{p}_0(q)|=1$, yieldsThe construction also proves the existence of the required symplectic bases, in characteristic two as well as odd characteristic.

Choose an orientation for each curve and let $c_i=[C_i]\in H_1(X;\mathbb{Q})$. The algebraic intersection number of curves on an oriented surface gives a nondegenerate alternating bilinear form on the $2g$-dimensional space $H_1(X;\mathbb{Q})$, by Poincare duality. Distinct curves are disjoint, so their intersection numbers are zero; each self-pairing is zero by alternation. Thus the span $W=\mathrm{span}\{c_1,\dots ,c_n\}$ is an isotropic subspace of a symplectic vector space. The assumed linear independence gives $\dim W=n$.

Nondegeneracy identifies $H_1(X;\mathbb{Q})$ with its dual. Restricting functionals to $W$ is surjective, so the symplectic orthogonal complement has dimension $\dim W^{\perp }=2g-n$. Isotropy says $W\subseteq W^{\perp }$, giving $n\le 2g-n$. HenceThe bound is sharp: one meridian on each handle of a genus-$g$ surface gives $g$ pairwise disjoint curves with independent homology classes. This is an algebraic bound on independent classes, not on the number of disjoint separating or mutually homologous curves.

Replace the determinant condition by preservation of a nondegenerate alternating bilinear form $B$ on $F^{2m}$. In a symplectic basis the symplectic group over a finite field isTo count symplectic bases, choose the first vector $e$ in $q^{2m}-1$ ways, then choose $f$ with $B(e,f)=1$ in $q^{2m-1}$ ways. Their span is nondegenerate; its symplectic orthogonal complement has dimension $2m-2$ and supplies the rest of the basis. The resulting recurrence gives the finite symplectic group order

The elementary replacements are symplectic transvectionsExpanding $B(Tx,Ty)$ shows that the two cross terms cancel and the last term vanishes because $B(v,v)=0$. Thus these maps preserve $B$; their inverses are $T_{v,-c}$.

We need both generation and primitivity to use the argument from part (i). For nonzero vectors $v,w$ with $B(v,w)\ne 0$, the symplectic transvection with direction $w-v$ and parameter $1/B(v,w)$ sends $v$ to $w$. When $B(v,w)=0$, choose $z$ nonorthogonal to both and use two such steps. Such $z$ exists because two proper hyperplanes cannot cover a vector space: if the two linear forms are independent prescribe both values to be one, and if dependent prescribe either a nonzero value. Hence symplectic transvections act transitively on nonzero vectors.

They generate the whole symplectic group over a finite field. Indeed first use them to make a given symplectic transformation fix $e_1$. Its image $f^{\prime }$ of $f_1$ still satisfies $B(e_1,f^{\prime })=1$. A symplectic transvection with direction $f_1-f^{\prime }$ fixes $e_1$ and sends $f^{\prime }$ to $f_1$ if the pairing is nonzero. If that pairing is zero, pass through $f^{\prime }+c{e}_1$ for any $c\ne 0$: it has nonzero pairing with both partners, and both required directions lie in $e_1^{\perp }$. We have now fixed the first hyperbolic pair, and induction applies in its nondegenerate orthogonal complement. The final two-dimensional case is exactly the elementary generation of $S{L}_2(q)=S{p}_2(q)$ proved in part (i).

For a fixed projective point $⟨e⟩$, the symplectic transvections $T_{e,c}$ form an abelian subgroup normal in its stabilizer subgroup; conjugation scales $c$ by the square of the scalar multiplying $e$. Their conjugates supply all symplectic transvections. A central element commutes with every symplectic transvection, so it preserves every direction line and must be scalar. Conversely scalars commute with the whole group. The kernel of the projective action consists of scalar matrices, and preservation of $B$ restricts the scalar to ${\lambda }^2=1$. Thus the center of a group has size $\gcd (2,q-1)$ and the projective symplectic group over a finite field acts faithfully.

For $m\ge 2$ the stabilizer subgroup has precisely three orbits: its fixed point, the other points in $e^{\perp }$, and the points outside $e^{\perp }$. Transitivity on each follows by extending, respectively, an independent isotropic pair or a hyperbolic pair to a symplectic basis. WriteA nontrivial proper block through the fixed point would be a union of these stabilizer subgroup orbits, hence would have size either $h$ or $q^{2m-1}+1$. A block size divides $N$. But $N-qh=1$, so $h>1$ cannot divide $N$, and $N/2<q^{2m-1}+1<N$, excluding the other size. This proves primitivity. For $m=1$ the projective action is the two-transitive action of $PS{L}_2(q)$.

Here is a direct check of perfectness of finite symplectic groups, including the small fields where the diagonal group commutator needs replacement. If $q>3$, every symplectic transvection lies in an embedded $S{L}_2(q)$ on a hyperbolic plane containing its direction, and that subgroup is perfect by part (i). Since the symplectic transvections generate, the whole symplectic group over a finite field is perfect.

For $q=3$ and $m\ge 2$, all $T_{v,1}$ are conjugate, so their common image $z$ generates the abelianization and satisfies $3z=0$. Choose independent perpendicular vectors $e,f$. The four commuting symplectic transvections with directions $e,f,e+f,e-f$ and parameter one multiply to the identity: the sum of their rank-one matrices is zero in characteristic 3. Thus also $4z=0$, giving $z=0$.

For $q=2$ and $m\ge 3$, use the seven nonzero vectors of a three-dimensional isotropic subspace of a symplectic vector space. Their symplectic transvections commute, and their product is the identity. To see this explicitly, write a vector in three isotropic basis coordinates: in the sum of the outer products $v{v}^{\mathsf{T}}$ each diagonal coefficient occurs four times and each off-diagonal coefficient twice, all zero in characteristic 2. The common abelianization image $z$ therefore satisfies $2z=7z=0$. Again the group is perfect.

The simplicity argument from part (i) now applies to the projective symplectic group over a finite field. The remaining exceptional group besides $S{p}_2(2)$ and $PS{p}_2(3)$ is $S{p}_4(2)$. It is not simple: it is isomorphic to $S_6$. One can verify this identification without assuming it. On $F_2^4$ setThese are all 16 quadratic forms with polar form $B$; the Arf invariant of a quadratic form gives $\mathrm{Arf}(Q_a)=Q_0(a)$. There are six with invariant one. The symplectic group over a finite field permutes these six forms. Differences of their parameter vectors span $F_2^4$: writing $e_i,f_i$ for the symplectic basis, the six parameters areTheir pairwise differences include each of $e_1,e_2,f_1,f_2$. An element fixing all six forms therefore fixes all linear forms $B(a-b,-)$, hence is the identity. This gives a faithful action on six objects; the group order is 720, so its image is all of $S_6$. Consequently

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.

Let $\dim V=2n$ and let $V_{+}=\ker (S-I)$ and $V_{-}=\ker (S+I)$ be the two eigenspaces. The involution identity gives a direct sum, sinceFor $u,v$ in either one of these eigenspaces, the anti-symplectic involution identity impliesso both are isotropic subspaces of a symplectic vector space. Such a subspace has dimension at most $n$: $W\subset W^{\Omega }$ and the nondegenerate bilinear form $\Omega$ gives $\dim W^{\Omega }=2n-\dim W$, where $W^{\Omega }$ is the symplectic orthogonal complement. As their dimensions add to $2n$, both have dimension $n$. Therefore

On $N={\mu }^{-1}(-1/4)$, the diagonal fundamental vector field $X$ is nonzero, since the fixed-point set lies at moment map values $0$ and $-1/2$. The nondegenerate bilinear form ${\omega }_p$ and $d\mu ={\iota }_X\omega$ imply $d\mu \ne 0$, so $-1/4$ is a regular value. By the regular level set theorem, $N$ is a three-dimensional embedded submanifold of the four-dimensional Complex projective plane.

At $p\in N$, set $W=T_{p}N=\ker d{\mu }_p$. Then $\omega (X,v)=0$ for every $v\in W$, so $X\in W^{\omega }$, the symplectic orthogonal complement. This complement is one-dimensional and $d\mu (X)=\omega (X,X)=0$, givingThus the level set is a coisotropic submanifold, neither an isotropic submanifold nor a Lagrangian submanifold. Indeed an isotropic subspace of a symplectic vector space in dimension four has dimension at most two, whereas $\dim N=3$. Its characteristic line field is generated by the diagonal circle action.

Construct a symplectic basis two vectors at a time. Choose $e\ne 0$; the nondegenerate bilinear form supplies $g$ with $\omega (e,g)\ne 0$, and set $f=g/\omega (e,g)$. For any vector $x$, remove its paired components byThen $\omega (x^{\prime },e)=\omega (x^{\prime },f)=0$. Thus, for $U=\mathrm{span}\{e,f\}$, the symplectic orthogonal complement gives a direct sum $V=U\oplus U^{\omega }$; its restricted bilinear form remains nondegenerate. Repeat there until no vectors remain, then put all $e_i$ before all $f_i$. This proves even dimension and existence together. Only the nonzero pairing is divided out, so the construction also works in characteristic two. The pair-splitting argument is described in Bilinear Forms by Keith Conrad (Keith Conrad, Bilinear Forms, §5, proof of Theorem 5.1 and Theorem 5.4).

A subspace of a symplectic vector space is symplectic if the restricted symplectic form is nondegenerate. It has a symplectic orthogonal complement and the ambient space is the direct sum of those two subspaces.