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A linear bijection between spaces carrying compatible bilinear forms, Hermitian forms or quadratic forms is an isometry of a space with a form when it preserves the specified form. Preserving the polar form alone need not preserve a quadratic form in characteristic two. Subspaces may have degenerate restricted forms even when the ambient form is nonsingular.

Here is Witt's lemma with the hypotheses appropriate to the classical groups. Let $V$ be finite-dimensional with a nonsingular alternating bilinear form, a nonsingular symmetric bilinear form in characteristic different from two, or a nonsingular Hermitian form with nontrivial involution. Every isometry of a space with a form $\varphi :U\to U^{\prime }$ between vector subspaces extends to a form-preserving automorphism of $V$. For quadratic forms, the same statement holds in every characteristic when the polar bilinear form of a quadratic form is nonsingular and the isometry of a space with a form preserves the quadratic form itself. The restricted forms on $U,U^{\prime }$ may be degenerate. We prove these versions, including the characteristic-two quadratic case needed later.

For an alternating bilinear form $B$, split the restricted space as $U=H\oplus R$, where $R=U\cap U^{\perp }$ and $B{|}_H$ is nonsingular. Choose a symplectic basis for $H$ and a basis $r_1,\dots ,r_t$ of $R$. The nondegenerate form on $V$ supplies vectors $s_j$ in $H^{\perp }$ with $B(r_i,s_j)={\delta }_{ij}$. Adding linear combinations of the $r_i$ to the $s_j$ makes $B(s_i,s_j)=0$: for each pair $i<j$ one may cancel its pairing by adding an appropriate multiple of $r_i$ to $s_j$, without disturbing the dual pairings or previously treated pairs. This uses no division by two. Thus $H$ together with the pairs $(r_i,s_i)$ is nonsingular and its orthogonal complement for a sesquilinear form has a symplectic basis.

Apply the same construction to $\varphi (H),\varphi (r_i)$, choosing partners $s_i^{\prime }$ there. The two completed bases have identical alternating Gram matrices and equal remaining dimensions. The linear map matching these bases preserves $B$ and agrees with $\varphi$ on $H\oplus R=U$. This proves the alternating version over an arbitrary field, including characteristic two.

For a quadratic form $Q$, define its polar bilinear form of a quadratic form by $B(x,y)=Q(x+y)-Q(x)-Q(y)$. Its symmetrypreserves $Q$. The zero-dimensional case is immediate. Induct on $\dim U$: choose $Q(t)\ne 0$ and a hyperplane $K\subseteq U\cap t^{\perp }$. Extending $\varphi {|}_K$ first reduces to $\varphi {|}_K=I$. Put $U=K\oplus Fu$, $v=\varphi (u)$, $d=u-v\in S=K^{\perp }$. If $Q(d)\ne 0$, $s_d$ completes the extension. Otherwise $B(u,d)=B(v,d)=0$. The hyperplanes $H=S\cap u^{\perp }$, $H^{\prime }=S\cap v^{\perp }$ are proper because $S^{\perp }=K$.

For $a\in S\setminus (H\cup H^{\prime })$ with $Q(a)\ne 0$, set $c=u-s_a(v)$. Thenand $s_a{s}_c$ fixes $K$ and sends $u$ to $v$. If no such $a$ exists and $|F|>2$, the identityforces $Q(z)=B(a,z)=0$. The complement of two hyperplanes spans $S$: adding a multiple of any outside vector avoids at most two forbidden scalars. Hence $d\perp S$, so $H=H^{\prime }$, forcing $Q{|}_S=0$, contrary to $t\in S$.

Over ${\mathbb{F}}_2$, choose $a$ outside both hyperplanes. In the remaining case $Q(a)=Q(d)=B(a,d)=0$, and $B(a,u)=B(a,v)=1$. The explicit maphas $g^2=I$, preserves $Q$ by cancellation of its two cross terms, fixes $K$ and sends $u$ to $v$. This closes the induction. The characteristic-independent symmetry argument and its binary exceptional step are also treated in Casselman's quadratic-form notes and Elman, Karpenko and Merkurjev, section 8.

In odd characteristic a symmetric bilinear form is recovered by the polarization identity for $Q(x)=B(x,x)/2$, so the quadratic proof also proves its extension theorem. For completeness, the Hermitian form case has a similarly explicit induction. Take $h$ linear in the first variable, with conjugation $a\mapsto \bar{a}$. For anisotropic vectors for a form $u,v$ with equal value $A=h(u,u)=h(v,v)\ne 0$, let $d=u-v$ and $c=h(u,d)$. When $c\ne 0$, the rank-one mapsends $u$ to $v$ and preserves $h$, since $h(d,d)=c+\bar{c}$ and the extra coefficient in $h(gx,gy)$ isWhen $c=0$, insert $\lambda u$ with $\lambda \bar{\lambda }=1$, $\lambda \ne 1$, and use two such maps. Such a scalar exists for a nontrivial involution: $\lambda =a/\bar{a}$ with $a\ne \bar{a}$ works. If the source vector subspace contains an anisotropic vector for a form, align it this way and induct in its nonsingular orthogonal complement for a sesquilinear form.

If instead every source vector is isotropic, the Hermitian polarization identity makes its whole restricted form zero. Choose $e$ in its basis and a partner $f$ pairing to one with $e$ and to zero with the remaining source basis. Make $h(f,f)=0$ by replacing $f$ by $f-te$, where $t+\bar{t}=h(f,f)$. The trace map is onto the fixed field: in odd characteristic divide by two; in characteristic two scale a nonzero trace. Choose $a+\bar{a}\ne 0$. The orthogonal vectors $e+af$ and $e-\bar{a}f$ have nonzero values $a+\bar{a}$ and its negative, so the preceding rank-one construction aligns this entire hyperbolic pair with its target pair. The remaining source basis lies in the pair's orthogonal complement for a sesquilinear form; induction finishes the Hermitian proof.

The hypotheses matter. For example, the identity-matrix symmetric bilinear form on ${\mathbb{F}}_2^3$ has canonical vector $c=(1,1,1)$ characterized by $B(x,c)=B(x,x)$, so every isometry of a space with a form fixes $c$. The one-dimensional map $e_1\mapsto c$ preserves the restricted bilinear form but cannot extend. Thus “all symmetric forms in characteristic two” is not a correct replacement for the quadratic or alternating versions proved above.

For the degree-ten partition action of S6, take unordered partitions of six letters into two triples. There are $\left(\genfrac{}{}{0ex}{}{6}{3}\right)/2=10$. Fix $A\mid A^c$. Its stabilizer is $(S_3\times S_3)⋊S_2=S_3\wr S_2$, permuting each triple and interchanging them. The subgroup generated by a 3-cycle in each triple has order nine, so is a Sylow subgroup of $S_6$. Any normalizer preserves its two three-point orbits; conversely all the indicated within-triple permutations and the swap normalize it. This proves the asserted normalizer exactly.

For any different partition $C\mid C^c$, $|A\cap C|$ is one or two after excluding the identical partition. Interchanging $C$ and $C^c$ identifies these two possibilities. The stabilizer of $A\mid A^c$ is transitive on each fixed intersection pattern, hence on all other nine partitions. The degree-ten action is two-transitive. It is also faithful: two distinct partitions have a common refinement with two singleton cells and two cells of size two. Every unordered pair of letters occurs as those singleton cells. A permutation fixing every partition therefore fixes every pair and consequently every letter.

For the subgroup factorization from coset transitivity, use cosets $Hg$ (right cosets) and right multiplication by $K$. The orbit of $H$ is precisely $\{Hk:k\in K\}$. This is every coset if and only if every $g$ can be written $hk$, provingA natural $S_5$ fixing one letter is transitive on the ten partitions: the triple containing the fixed letter is determined by any two of the other five. Taking $H=S_3\wr S_2$, $K=S_5$ gives $S_6=HK$; inversion gives the requested order as well,

Now let $B$ be the binary nondegenerate alternating bilinear form. Its binary quadratic refinements form an affine space: the difference of two such forms is linear, so for a fixed $Q_0$ every one is uniquelyIn a symplectic basis each refinement isA two-dimensional summand with $(a_i,b_i)\ne (1,1)$ admits a basis of singular vectors paired to one and is hyperbolic; the $(1,1)$ summand is anisotropic. Two planes spanned by anisotropic vectors for a form together become two hyperbolic planes of a quadratic form: with anisotropic basis pairs $(e_1,f_1),(e_2,f_2)$, the singular vectors $e_1+e_2,f_1+e_2$ pair to one; their orthogonal complement for a sesquilinear form has singular paired basis $e_1+f_1+f_2,e_1+f_1+e_2+f_2$. Consequently every refinement is equivalent to either $m$ hyperbolic planes of a quadratic form or $(m-1)$ hyperbolic planes of a quadratic form and one anisotropic plane (all three nonzero vectors are anisotropic vectors for a form).

These two types are different. Indeed the character sum of each plane is $+2$ or $-2$, soThus there are exactly two symplectic orbits $Q^{+}(B),Q^{-}(B)$, the signs denoting $\mathrm{Arf}=0,1$. Counting the three coefficient pairs with product zero and the one with product one givesThe same character sum gives the corresponding numbers of singular vectors, including zero, in a fixed form of either type.

Fix a form $Q$ of type $ϵ$. Completing the square by $x\mapsto x+v$ showsThe stabilizer of $Q$ is exactly $O(V,Q)$. It sends the parameter $v$ by its natural linear action or inverse action according to the chosen left/right convention. Other forms of the same type correspond to nonzero singular vectors $v$, and all forms of opposite type correspond to nonsingular vectors $v$. For two nonzero vectors of the same $Q$ value, the map between their spans is an isometry of a space with a form. Witt's lemma extends it, so the orthogonal stabilizer is transitive on each of these two vector sets. This proves the two-transitive binary quadratic-form actions: the symplectic group over a finite field is transitive on each type and its point stabilizer is transitive on all other members of that type.

In particular $O_{2m}^{-}(2)$ is transitive on the plus orbit. Identify that orbit with the cosets (right cosets) of a plus stabilizer, using $Q^g(x)=Q(gx)$ as a right action. The coset criterion then provesBoth genuine two-transitive actions require $m\ge 2$. At $m=1$ the plus orbit has three forms and the minus orbit is a singleton; transitivity and the factorization remain true, while two-transitive action on the singleton is only vacuous if one's convention permits degree one. No special simplicity assertion is needed for this argument. At $m=2$ the orbit sizes are ten and six, linking these actions with $S{p}_4(2)\cong S_6$ and the partition action above.

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 isometry of a space with a form between subspaces of a finite-dimensional nonsingular classical space extends to an ambient isometry. This includes alternating bilinear forms in all characteristics, symmetric bilinear forms in characteristic different from two, Hermitian forms with nontrivial involution, and quadratic forms with nonsingular polar form. No nonsingularity of the restricted subspace is required. Bare nonalternating symmetric bilinear forms in characteristic two require additional hypotheses and must not be confused with quadratic spaces.