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Let $V$ be a finite-dimensional symplectic vector space over an algebraically closed field of characteristic zero. For $a\in \mathfrak{sp}(V)$, write its Jordan–Chevalley decomposition as $a=S+N$. Applying the symplectic adjoint to this decomposition and using $a^{\ast }=-a$, uniqueness gives $S^{\ast }=-S$ and $N^{\ast }=-N$. Thus both parts belong to the symplectic Lie algebra. The operators $\mathrm{ad}S$ and $\mathrm{ad}N$ commute. The former is diagonalizable on $\mathrm{End}(V)$ in a matrix unit basis for an eigenbasis of $S$, and the latter is a nilpotent linear map, as follows fromThe restrictions to the invariant symplectic Lie algebra retain these properties. Hence these restrictions form the Jordan–Chevalley decomposition of $\mathrm{ad}a$.

If $\mathrm{ad}a$ is diagonalizable, $\mathrm{ad}N=0$. For every $v\in V$, the rank-one map $T_v(w)=\omega (v,w)v$ lies in $\mathfrak{sp}(V)$, since the two terms in $\omega (T_{v}w,u)+\omega (w,T_{v}u)$ cancel by alternation. For $v\ne 0$ its image is $kv$, so commuting with $T_v$ forces $N$ to preserve $kv$. A linear map preserving every line is scalar: compare its action on two linearly independent vectors and on their sum. Since $N$ is a nilpotent linear map, that scalar is zero. Thus $a=S$ is diagonalizable. Conversely, if $a$ is diagonalizable on $V$, its operator commutator action is diagonalizable on $\mathrm{End}(V)$ and therefore on its invariant symplectic Lie algebra. This proves both directions, including $V=0$.

A bilinear form is alternating if $B(v,v)=0$ for every vector $v$. Expanding $B(v+w,v+w)=0$ gives $B(v,w)+B(w,v)=0$, hence $B(v,w)=-B(w,v)$. If the characteristic is not two, the converse follows from $B(v,v)=-B(v,v)$ by dividing by two. In characteristic two, the relation $B(v,w)=-B(w,v)$ is just symmetry and does not force zero self-pairings: $B(x,y)=xy$ on ${\mathbb{F}}_2$ has $B(1,1)=1$.^[ref]

The matrixgives an alternating bilinear form over every field. Over the real numbers, its value is the signed area of the parallelogram with sides $v,w$: $B((2,1),(1,3))=6-1=5$, and swapping the sides gives $-5$. Every self-pairing is zero, yet $\det J=1$, so this bilinear form is nondegenerate. Directly, pairing $(x,y)$ with $(1,0)$ gives $-y$, and pairing it with $(0,1)$ gives $x$; both vanish only when $(x,y)=0$.

In characteristic different from two, a skew-symmetric matrix of odd integer size has zero determinant, since $\det A=\det (-A^T)=(-1{)}^n\det A$. Thus an alternating bilinear form on a finite odd-dimensional vector space is a degenerate bilinear form. The stronger statement that the rank of an alternating bilinear form is even holds in all characteristics: splitting off a two-dimensional block with matrix $J$ repeatedly proves it. A nondegenerate alternating bilinear form defines a symplectic vector space, the linear setting for signed area rather than squared length.

An anti-symplectic map between symplectic manifolds is a diffeomorphism $f$ satisfying $f^{\ast }{\omega }^{\prime }=-\omega$. For symplectic vector spaces, the analogous definition uses an invertible linear map.

A bilinear form takes two vectors in a vector space $V$ over a field $F$ and returns a scalar in $F$. Its defining property is separate linearity: holding either input fixed gives a linear functional of the other. For scalars $a,b$ and vectors $u,v,w$,This is a scalar-valued bilinear map; the chosen field is part of the definition.^[ref]

Separate linearity does not mean that $B$ is a linear map on the vector space $V\times V$ with componentwise vector addition and scalar multiplication. Over the real numbers, $B(2v,2w)=4B(v,w)$, whereas a linear map on the pair would give $2B(v,w)$. The simplest example is $B(x,y)=xy$ on the one-dimensional vector space $\mathbb{R}$: fixing $y=3$ gives $x\mapsto 3x$, but doubling both $x$ and $y$ quadruples their product.

A bilinear form measures how two inputs interact. The dot product measures alignment; a symmetric bilinear form can describe a quadratic form; an alternating bilinear form can measure signed area. A nondegenerate bilinear form turns each vector into a unique linear functional and so connects geometry with the dual space. The same short definition therefore supports lengths, bilinear orthogonal complements, and symplectic vector spaces, depending on the additional properties imposed.

Over the complex numbers, a bilinear form is linear in both inputs without conjugating either scalar. A Hermitian form instead is a sesquilinear form and conjugates when its inputs are exchanged. For complex matrices, the adjoint $A^{\ast }$ used with Hermitian forms and matrix C-star algebras is the conjugate transpose, whereas a bilinear form uses the ordinary transpose $A^T$. For example, $B(z,w)=zw$ has $B(i,i)=-1$, while the Hermitian form $H(z,w)=z\overline{w}$ has $H(i,i)=1$.^[ref]

For odd $r\ge 1$ and $W_g={\#}_{i=1}^g(S^r\times S^r)$, the integral cohomology has one copy of $\mathbb{Z}$ in degrees zero and $2r$, ${\mathbb{Z}}^{2g}$ in degree $r$, and zero elsewhere. Choose the top orientation class $\omega$ and degree-$r$ classes ${\alpha }_i,{\beta }_i$ from the two factors of summand $i$. Their nonzero positive-degree basis products areAll ${\alpha }_i{\alpha }_j$ and ${\beta }_i{\beta }_j$ vanish, and $\omega$ annihilates positive-degree classes. The Künneth theorem computes each sphere product; the Mayer–Vietoris sequence computes the connected sum groups, and its degree-one pinch maps determine the products. Thus its middle-degree Poincare duality pairing is a symplectic vector space after changing coefficients to a field of characteristic different from two. The case $r=1$ is the usual cohomology ring of a closed oriented surface.

Over $\mathbb{C}$, the symplectic Lie algebra on a four-dimensional symplectic vector space acts faithfully on its five-dimensional primitive exterior square, preserving its symmetric bilinear form. Faithfulness follows because an operator killing all wedges $e_i\wedge e_j$ has all off-diagonal coefficients zero and satisfies $A_{ii}+A_{jj}=0$ for every pair, hence is zero. The source and target Lie algebras both have dimension ten, so this faithful map is an isomorphism to the Special orthogonal Lie algebra ${\mathfrak{so}}_5$.

Fix a volume form on a four-dimensional real symplectic vector space. Wedge pairing on its exterior square has signature $(3,3)$. The symplectic group fixes the bivector inverse to the symplectic form. Its orthogonal complement is five-dimensional with signature $(3,2)$ after choosing the appropriate overall sign. The induced representation has kernel $\{\pm I\}$: identity action on the entire exterior square preserves every two-plane and thus every line, forcing a scalar transformation with square one. Its derivative is injective and both group dimensions are ten, so its image is the identity component of $SO(3,2)$.

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.

A Lagrangian subspace of a $2n$-dimensional symplectic vector space $(V,\omega )$ is an $n$-dimensional subspace $L$ on which $\omega$ vanishes.

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 $\mathfrak{a}$.

Writeon $V=W\times W^{\prime }$, where $W^{\prime }$ is the dual space. The dual pairing is linear separately in the vector and the linear functional. For example, for real numbers $a,b$,Thus $\omega$ is a bilinear form. It is also an alternating bilinear form, since

To prove that this is a nondegenerate bilinear form, suppose $(v,f)$ pairs to zero with every $(w,g)$. Setting $g=0$ gives $f(w)=0$ for every $w\in W$, hence $f=0$. Setting $w=0$ gives $g(v)=0$ for every $g\in W^{\prime }$. If $v\ne 0$, basis extension supplies a coordinate functional with $g(v)=1$, a contradiction. Thus $v=0$. The left radical is zero, and the displayed change of sign implies the right radical is zero as well. Hence $\omega$ is a nondegenerate bilinear form, making $V$ a symplectic vector space.

For the required example take the coordinate subspaceUsing the definition of the bilinear orthogonal complement in part (b),ThereforeThis is a Lagrangian subspace. Its dimension is $\dim W$, exactly half of $\dim V=2\dim W$, so part (b)(i) is satisfied.

There is no contradiction with part (b)(ii): that part asks whether a direct sum is guaranteed, and the answer is already no. Moreover, when $W\ne 0$, the present alternating bilinear form is not a symmetric bilinear form. Choose $v\ne 0$ and $f\in W^{\prime }$ with $f(v)=1$; thenIt therefore does not satisfy the symmetry hypothesis of part (b)(ii). If $W=0$, then $V=U=0$, and both $U=U^{\perp }$ and $V=U\oplus U^{\perp }$ hold trivially.

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.

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.

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.

Let $U$ be a four-dimensional symplectic vector space, with a symplectic basis $e_1,f_1,e_2,f_2$. Its invariant inverse-form bivector may be chosen as $\eta =e_1\wedge f_1+e_2\wedge f_2$. Put $\mathrm{vol}=e_1\wedge f_1\wedge e_2\wedge f_2$, so $\eta \wedge \eta =2\mathrm{vol}$. Define a symmetric bilinear form on the six-dimensional exterior square byIt is symmetric because both factors have degree two. It is nondegenerate: in the six-element wedge basis, every $e_a\wedge e_b$ pairs nontrivially with the wedge of the complementary pair and with no other basis vector. Since $B(\eta ,\eta )=2$, the orthogonal complementhas dimension five and inherits a nondegenerate bilinear form. It is the primitive exterior square, equivalently the kernel of the symplectic contraction of an exterior square.

The symplectic Lie algebra preserves $\eta$. It also preserves $\mathrm{vol}$, since its elements have trace zero and the induced action on ${\Lambda }^{4}U$ is multiplication by that trace. The exterior-power Lie algebra representation therefore preserves $B$ and $W$, giving a Lie algebra homomorphismTo show injectivity, suppose $A$ acts as zero on $W$. It already acts as zero on the invariant line $\mathbb{C}\eta$, so it kills all of ${\Lambda }^{2}U=W\oplus \mathbb{C}\eta$. For any basis $u_1,\dots ,u_4$, this meansIf $k\ne i$, choose $j$ distinct from $i,k$; the coefficient of $u_k\wedge u_j$ forces $A_{ki}=0$. Thus $A$ is diagonal. The same identity gives $A_{ii}+A_{jj}=0$ for every pair. Applying this to three distinct indices forces each diagonal entry to vanish. Hence $A=0$ and $\Phi$ is injective.

Finally $\dim {\mathfrak{sp}}_4=2(2\cdot 2+1)=10$ and $\dim \mathfrak{so}(W,B)=5\cdot 4/2=10$. Injectivity between these equal-dimensional vector spaces gives surjectivity. Every nondegenerate complex symmetric bilinear form on a five-dimensional space is equivalent to the standard one. We have therefore constructed the exceptional isomorphism between sp4 and so5:

Let $U$ be a four-dimensional symplectic vector space. The symplectic contraction of an exterior square splits ${\Lambda }^{2}U$ into its five-dimensional primitive exterior square $W$ and the invariant line spanned by the inverse symplectic bivector $\Omega$. Choose $\mathrm{vol}=\Omega \wedge \Omega /2$. Exterior multiplication defines a symmetric bilinear form on ${\Lambda }^{2}U$ by $\xi \wedge \eta =B(\xi ,\eta )\mathrm{vol}$; it is symmetric because both degrees are two. In a symplectic basis, $\Omega \wedge \Omega =2\mathrm{vol}\ne 0$, and $W={\Omega }^{\perp }$, so the restriction to $W$ is nondegenerate.

The symplectic Lie algebra ${\mathfrak{sp}}_4$ preserves this form, giving a homomorphism ${\mathfrak{sp}}_4\to \mathfrak{so}(W)$. It is injective: an element acting trivially on $W$ also acts trivially on $\Omega$, hence on all of ${\Lambda }^{2}U$. In a four-vector basis, the identities $A{e}_i\wedge e_j+e_i\wedge A{e}_j=0$ first force every off-diagonal coefficient of $A$ to vanish, then force $A_{ii}+A_{jj}=0$ for every pair. Over $\mathbb{C}$ these equalities imply $A=0$.

Both Lie algebras have dimension ten: $\dim {\mathfrak{sp}}_4=4\cdot 5/2$ and $\dim {\mathfrak{so}}_5=5\cdot 4/2$. The injective map is therefore an isomorphism, proving the exceptional isomorphism between sp4 and so5:

The ordinary Poincare duality statement here is for a compact manifold without boundary, of dimension $d$, oriented over the field $F$. Its fundamental class $[M{]}_F$ induces isomorphismsfor every $q$. For a manifold with boundary the appropriate statement is Poincare-Lefschetz duality with relative groups; the ordinary pairing need not be nondegenerate. Over a field, the universal coefficient theorem for cohomology identifies $H^{d-q}(M;F)$ with the full dual of $H_{d-q}(M;F)$. The cap-cup evaluation identity and duality therefore makea perfect pairing. Explicitly, a nonzero $a$ has a nonzero cap product, and a linear functional $b$ on its homology group takes a nonzero value on that product. The same argument in the other variable proves nonsingularity. On the whole graded cohomology, define $B(a,b)$ by taking the degree-$d$ part of $a⌣b$ before evaluation. For a nonzero component $a_q$ choose $b$ homogeneous of degree $d-q$ to pair nontrivially with it. All other components contribute zero in degree $d$. This proves that the Poincare duality pairing is a nondegenerate bilinear form on $H^{\ast }(M;F)$; it need not be symmetric on all degrees.

Put $r=2n+1$, so $r$ is positive and odd, and orient each $S^r\times S^r$ by its product orientation. The Künneth theorem gives integral cohomology $\mathbb{Z}$ in degrees zero and $2r$, ${\mathbb{Z}}^2$ in degree $r$, and zero elsewhere. Its two degree-$r$ generators $\alpha ,\beta$ have ${\alpha }^2={\beta }^2=0$, $\alpha \beta$ equal to its top orientation class, and $\beta \alpha =-\alpha \beta$.

For a connected sum of oriented manifolds, excision and the long exact sequence for deleting a ball show that deleting a ball removes the top homology class and leaves all lower positive homology groups unchanged. The boundary sphere represents zero in the punctured manifold: it is the boundary of its relative fundamental chain. In the Mayer–Vietoris sequence for the two punctured pieces joined along the separating sphere, the orientation class of the connected sum maps onto that sphere's class. The intermediate positive groups are consequently the direct sums of the groups of the two original manifolds. This also covers $r=1$, where the sphere is a circle and the boundary-class observation is necessary in the middle degree. Iterating and applying the universal coefficient theorem for cohomology givesHere $g\ge 1$; under the conventional extension $W_0=S^{2r}$ the same formula holds with a zero middle group.

Let $\omega$ be the top cohomological orientation class. The connected-sum pinch map to the wedge of the $g$ sphere products gives degree-one projections to each summand. Pull back the two factor classes from summand $i$ to obtain ${\alpha }_i,{\beta }_i$. Their product is $\omega$, since the projection has degree one. Classes from different wedge summands have zero positive-degree cup products, and graded commutativity of the cup product supplies the reversed sign. Thus the full cohomology ring of a connected sum of odd-dimensional sphere products is the graded free abelian group just displayed with multiplicationThe unit is $1$, and $\omega$ times any positive-degree class is zero by dimension. This describes all products, including ${\omega }^2=0$.

The smooth involution is a diffeomorphism, since it is its own inverse. Its fixed set is closed; discreteness and compactness therefore make it finite. The supplied positivity of $\det (I-D{f}_x)$ makes every fixed point nondegenerate with local index $+1$. The Lefschetz-Hopf fixed-point theorem then givesIndeed the degree-zero and top-degree traces are both one, because the manifold is connected and $f$ preserves its orientation; the middle degree is odd.

On $V=H^r(W_g;\mathbb{R})$, the form $\Omega (a,b)=⟨a⌣b,[W_g]⟩$ is a nondegenerate alternating bilinear form by Poincare duality and the oddness of $r$. Thus it is a symplectic vector space of dimension $2g$. Naturality and orientation preservation show that $T$ preserves $\Omega$, and $T^2=I$. For the eigenspaces of a symplectic involution, write $V=V_{+}\oplus V_{-}$: the polynomial $(t-1)(t+1)$ has distinct roots over $\mathbb{R}$. If $v_{+}\in V_{+}$ and $v_{-}\in V_{-}$, then $\Omega (v_{+},v_{-})=\Omega (T{v}_{+},T{v}_{-})=-\Omega (v_{+},v_{-})$, so the two vector subspaces are orthogonal. Each restricted form is nondegenerate, since a vector annihilating its own vector subspace also annihilates the other and hence all of $V$. Their dimensions are therefore even, say $\dim V_{+}=2a$, $\dim V_{-}=2b$, with $a+b=g$. ConsequentlyThis proves the fixed-point congruence for an involution on an odd-sphere connected sum:

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.