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Any $d$ quadratic forms on ${\mathbb{C}}^n$ have a common nonzero zero when $n\ge 2^d$. Choose a totally isotropic subspace for the first form, of dimension at least $\lfloor n/2\rfloor$. Restrict the remaining forms to it and repeat by induction. After $d$ stages at least one dimension remains. This elementary sufficient bound does not assert that $2^d$ is the best possible threshold.

A quadratic form of rank $r$ on ${\mathbb{C}}^n$ is congruent to $x_1^2+\cdots +x_r^2$. Pair its nonzero coordinates into vectors $e_{2j-1}+i{e}_{2j}$, and include the radical of a bilinear form. Their span is a totally isotropic subspace and has dimension $\lfloor r/2\rfloor +n-r=n-\lceil r/2\rceil \ge \lfloor n/2\rfloor$. The form vanishes on every vector of this subspace, not merely on its basis vectors.

On a connected compact Kähler manifold of complex dimension two, the intersection form restricted to real $(1,1)$ cohomology has signature $(1,h^{1,1}-1)$. The Kähler class spans a positive line, and its orthogonal complement consists of primitive classes, which are negative definite. Thus a totally isotropic subspace of real $(1,1)$ cohomology has dimension at most one.

Real compactified Minkowski points are the two-dimensional totally isotropic subspaces for the signature-$(2,2)$ twistor Hermitian form. In a graph chart this is the usual Hermitian reality condition on $x^{A{A}^{\prime }}$. Their twistor lines lie in projective null twistor space. On the Klein quadric the same real form is the projectivized null cone of a real quadratic form of signature $(2,4)$.

For a real quadratic form on an $n$-dimensional vector space, with rank $r$ and signature $s$ defined as positive minus negative index, the maximum dimension of a totally isotropic subspace is $n-(r+|s|)/2$. Quotienting by the radical of a bilinear form leaves a nondegenerate form of indices $p,q$. Projection of an isotropic subspace to each sign-coordinate space is injective, bounding its dimension by $\min (p,q)$. The radical and vectors pairing one positive with one negative unit coordinate attain the bound $n-r+\min (p,q)$.

First work over a field of characteristic different from two. If a nonzero symmetric bilinear form had $\varphi (v,v)=0$ for every vector, polarization would givea contradiction. Hence choose $e$ with $\varphi (e,e)\ne 0$. Every $v$ decomposes uniquely aswith the second summand orthogonal to $e$. Induct on the orthogonal complement to obtain a diagonal basis. If the restricted form becomes zero, any basis of that remaining space completes the diagonalization. In this basis the rank of a quadratic form is exactly the number of nonzero diagonal coefficients; the remaining basis vectors span its radical of a bilinear form.

Over the reals, rescale the nonzero basis vectors so the diagonal entries are $p$ copies of $+1$, $q$ copies of $-1$ and $z=n-p-q$ zeros. Define the signature of a quadratic form by $\sigma =p-q$, so $r=p+q$. To prove it is basis independent, note that $p$ is the largest dimension of a linear subspace on which the form is positive-definite. The positive coordinate linear subspace attains dimension $p$; on any other positive linear subspace, projection to those positive coordinates is injective, since its kernel has nonpositive quadratic value. Thus its dimension is at most $p$. The same argument for the negative form characterizes $q$. This proves Sylvester's law of inertia and that the signature is well-defined.

Let $e_i$ and $f_i$ be positive and negative normalized basis vectors, and let $Z$ be the radical of a bilinear form. The span of $Z$ and $e_i+f_i$ for $1\le i\le \min (p,q)$ is a totally isotropic subspace: all mutual bilinear pairings vanish. Its dimension is $z+\min (p,q)$. Conversely, project any totally isotropic subspace $U$ to the nondegenerate positive-plus-negative quotient. The projected linear subspace is still null, and its projections to both the positive and negative coordinate spaces are injective: a vector with one component zero would have strictly signed quadratic value unless it vanished. Its dimension is therefore at most $\min (p,q)$, while the kernel of the quotient map on $U$ has dimension at most $z$. HenceThis proves both construction and sharpness of the maximum dimension of a totally isotropic subspace.

For the five-variable example, the symmetric coefficient matrix isIts cyclic structure makes the vectors $(1,{\zeta }^k,{\zeta }^{2k},{\zeta }^{3k},{\zeta }^{4k})$, $\zeta =e^{2\pi i/5}$, eigenvectors with eigenvalues ${\zeta }^k+{\zeta }^{-k}=2\cos (2\pi k/5)$. Thus the eigenvalues areTheir product is $2$, and three are positive and two negative. ThereforeThe real and imaginary parts of each conjugate pair of eigenvectors give the corresponding real two-dimensional eigenspaces, consistent with the real inertia calculation.

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.

A quadratic form over $\mathbb{R}$ or $\mathbb{C}$ is a function $q(x)=B(x,x)$ for a symmetric bilinear form $B$, equivalently a homogeneous polynomial of degree two. Its polarization is $B(x,y)=[q(x+y)-q(x)-q(y)]/2$. If $q$ is not identically zero, choose $v$ with $q(v)\ne 0$. Every $x$ decomposes aswhere the second term lies in $v^{\perp }$ and $kv\cap v^{\perp }=\{0\}$. The quadratic form is consequently a direct sum of its one-dimensional value on $v$ and its restriction to $v^{\perp }$. Induction on dimension diagonalizes it. If $q=0$, any basis suffices. Rescale every nonzero diagonal coefficient: over the reals it becomes $1$ or $-1$, while over the complexes a square root makes it $1$. HenceOver $\mathbb{C}$ the negative coefficients can all be avoided.

For a real quadratic form, let $p$ and $m$ be its positive and negative indices in a diagonalization. Its rank of a quadratic form is $p+m$ and its signature of a quadratic form is $p-m$, or the ordered pair $(p,m)$ in the alternative convention. These indices are intrinsic: the largest dimension of a subspace where the form is positive definite is $p$, since any larger subspace intersects the nonpositive coordinate subspace nontrivially; similarly for $m$. For the form here, completing squares givesThe two displayed linear coordinates, together with $x_1$, form an invertible coordinate change. ThereforeThere is one zero coordinate direction.

The PDF's final dimension condition is $n\ge 2^d$, not the fraction produced by the TeX conversion. We prove the requested common zero of complex quadratic forms by induction on $d$. A complex quadratic form of rank $r$ has canonical expression $x_1^2+\cdots +x_r^2$. Pair its nonzero coordinates into vectors $e_{2j-1}+i{e}_{2j}$ and add the $n-r$ radical coordinate vectors. Their span is a totally isotropic subspace, because the form is zero on every linear combination. This proves that complex quadratic forms have a large isotropic subspace: its dimension isFor $d=0$ there is a nonzero vector whenever $n\ge 1$. For $d\ge 1$, take such a subspace $E$ for $q_1$. If $n\ge 2^d$, then $\dim E\ge 2^{d-1}$. Apply the induction hypothesis to the restrictions of $q_2,\dots ,q_d$ to $E$. It supplies a nonzero $x\in E$ annihilated by those forms, and $q_1(x)=0$ by construction. ThusThe argument makes no nondegeneracy assumption on any of the forms.

Associate to $f$ the symmetric bilinear form $B(u,v)=\frac{1}{2}(f(u+v)-f(u)-f(v))$. If $f$ is identically zero, every basis is diagonal. Otherwise choose $v$ with $f(v)\ne 0$. Every $w$ has the decompositionwhose second term lies in $v^{\perp }$. Thus $V=\mathbb{R}v\oplus v^{\perp }$, with no cross term in $f$. Induction on dimension gives a diagonal basis. Rescaling its nonzero entries reduces the form towith $k$ additional zero coordinates. Define the rank of a quadratic form $r=p+q$ and its signature $s=p-q$. Rank is intrinsic, since it is the rank of $v\mapsto B(v,\cdot )$, equivalently $n-\dim \mathrm{rad}B$. Also $p$ is the greatest dimension of a subspace on which the restriction of $f$ is a positive-definite quadratic form: any such subspace intersects the negative-plus-zero coordinate space only at zero, so its dimension is at most $p$, while the positive coordinate space attains $p$. The corresponding negative statement identifies $q$. This proves the basis independence of $r$ and $s$, the Sylvester's law of inertia conclusion.

For a totally isotropic subspace $W$ on which $f$ vanishes, polarization identity also makes $B$ vanish on $W\times W$. Quotient by the radical of a bilinear form. In the resulting nondegenerate form, projection of an isotropic subspace to either the positive or negative coordinate space is injective, so its dimension is at most $\min (p,q)$. Consequently $\dim W\le k+\min (p,q)$. The radical of a bilinear form together with the vectors $e_i^{+}+e_i^{-}$, $1\le i\le \min (p,q)$, attains this bound. The maximum dimension of a totally isotropic subspace is thereforeThe criterion for membership in a diagonal basis follows similarly: a nonzero vector can belong to a diagonal basis if it has nonzero quadratic value, by the same orthogonal splitting, or if it lies in the radical of a bilinear form, by choosing a radical basis containing it. An isotropic vector in a diagonal basis must be orthogonal to every basis vector, and hence must lie in that radical. For the specified form, its radical is the $z$-axis, giving exactlyThe zero vector never belongs to a basis.

Use the usual connected-manifold convention for this assertion. Suppose, for contradiction, that $H^2(M,{\mathcal{O}}_M)=0$. The Dolbeault theorem gives $H^{0,2}=0$, and the Hodge decomposition theorem for compact Kähler manifolds and its conjugation symmetry give $H^{2,0}=0$. Hence every real class in $H^2(M,\mathbb{R})$ has a real harmonic differential form of type $(1,1)$ as its representative.

The intersection form is $Q([u],[v])={\int }_{M}u\wedge v$. Its value on the Kähler class is positive: $Q([\omega ],[\omega ])={\int }_M{\omega }^2>0$. Every real harmonic $(1,1)$ form splits uniquely as $c\omega +u_0$, with $u_0$ primitive and $c$ constant. Indeed, $\Lambda u$ is a harmonic function, hence constant by connectedness, and $\Lambda \omega =2$; subtracting $c\omega$ kills $\Lambda u$, and part (b) makes $u_0$ primitive.

Part (c) gives $\ast u_0=-u_0$, while $\omega \wedge u_0=0$. ThusThis proves the Hodge index theorem for compact Kähler surfaces in this case: $Q$ has signature $(1,b_2-1)$.

On the alleged two-dimensional totally isotropic subspace $V$, the linear functional $v\mapsto Q(v,[\omega ])$ has a nonzero vector in its kernel. Its harmonic differential form representative is primitive, so the preceding negative-definiteness statement gives $Q(v,v)<0$, contradicting the vanishing of $Q$ on $V$. ThereforeConnectedness is essential if one's definition of a manifold allows disconnected spaces. For $M=Y\bigsqcup Y$, with $Y={\mathbb{CP}}^1\times {\mathbb{CP}}^1$, one has $H^2(M,{\mathcal{O}}_M)=0$. Take on each component the pullback of the degree-two class from its first factor. Each class has square zero, and their mutual pairing is zero because their supports lie on different components. Their span is then a two-dimensional totally isotropic subspace, disproving the unqualified disconnected version.