$2^o$ All the other elements $\Phi(e_i, e_j)$ where $i \leq j$ are zero.

The ideals $A \alpha_i$ ($i = 1, \ldots, r$) are uniquely determined by the preceding conditions. The submodule $E^0$ of $E$ orthogonal to $E$ is generated by $e_{2r+1}, \ldots, e_n$.

We shall proceed by induction on the dimension $n$ of $E$. The theorem is evident for $n = 0$. If $\Phi = 0$, the theorem is evident also; we may therefore suppose $\Phi \neq 0$. Let $f$ be the linear mapping $d_\Phi$ of $E$ into $E^*$ associated on the right with $\Phi$ ($§ 1$, no 1); then $f(E)$ is a non-zero submodule of the module $E^*$, which is a free module of dimension $n$. Let $A\alpha_1$ be the largest invariant factor of $f(E)$ with respect to $E^*$ (chap. VII, $§ 4$, no 2, th. 1); we know (loc. cit.) that there exists a basis $(e'_1, a'_2, \ldots, a'_n)$ of $E^*$ and an element $f(e_2) \in f(E)$ such that $f(e_2) = \alpha_1 e'_1$. Let $(e_1, a_2, \ldots, a_n)$ be the basis of $E$ (identified with the bidual $E^{**}$) dual to $(e'_1, a'_2, \ldots, a'_n)$; we have

Let $P$ be the submodule $Ae_1 + Ae_2$ of $E$. We shall see that $E$ is direct sum of $Ae_1, Ae_2$ and the orthogonal submodule $P^0$ of $P$. For this, it is enough to prove that, for every $x \in E$, there exist elements $\xi_1, \xi_2$ of $A$, uniquely determined, such that $x - \xi_1 e_1 - \xi_2 e_2 \in P^0$, that is to say such that

According to (1) these conditions are written

But we know (loc. cit.) that the image of $f(E)$ by every linear form on $E^*$ is contained in the ideal $A\alpha_1$, in other words all the values $\Phi(x, y) = \langle x, f(y) \rangle$ belong to $A\alpha_1$; hence the existence and uniqueness of $\xi_1$ and $\xi_2$. Thus $P^0$ is a free module of rank $n-2$; there exists therefore, in $P^0$, according to the induction hypothesis, a basis $(e_3, e_4, \ldots, e_n)$ satisfying the conditions of the statement. To prove that the basis $(e_1, \ldots, e_n)$ of $E$ thus obtained also satisfies these conditions, it is enough to prove that $\alpha_1$ divides $\alpha_2$; now this follows from the fact that all the values $\Phi(x, y)$ are multiples of $\alpha_1$. It is then clear that $e_{2r+1}, \ldots, e_n$ generate $E^0$. Finally, if $(e'_i)$ is the dual basis of $(e_i)$, we have $f(e_{2j-1}) = -\alpha_j e'_{2j}$ and $f(e_{2j}) = \alpha_j e'_{2j}$ for $j = 1, \ldots, r$ and $f(e_k) = 0$ for $k = 2r+1, \ldots, n$; the ideals $A\alpha_1, A\alpha_1, A\alpha_2, A\alpha_2, \ldots, A\alpha_r, A\alpha_r$ are therefore the invariant factors of $f(E)$ with respect to $E^*$, which proves their uniqueness (Chapter VII, $§ 4$, No. 2, Theorem 1).

In particular $\Phi$ is of pair rank $2r$.

A basis satisfying (2) is called a symplectic basis for $\Phi$.

Indeed, it is enough to note that z identifies canonically with an alternating bilinear form on $E^*$ (Chapter III, § 8, No. 2), and to apply Corollary 1 to this form.

By translating Corollary 1 into matrix language, one obtains:

(7) $\mathrm{Pf}({}^t P . R . P) = (\det P) \mathrm{Pf}(R).$

Indeed, suppose first that $A$ is a field of characteristic 0 and put $R = (\alpha_{ij}), P = (\beta_{st})$. Associate with $R$ the bivector

of $\bigwedge^2 \mathbf{A}^{2m}$, where $(e_i)$ denotes the canonical basis of $\mathbf{A}^{2m}$; consider ${}^t P$ as the matrix, with respect to the basis $(e_i)$, of an endomorphism $f$ of $\mathbf{A}^{2m}$. Then the bivector $(\bigwedge f)(u)$ is associated with the matrix ${}^t P . R . P$ since it is equal to $\frac{1}{2} \sum_{i, j, s, t} \beta_{is} \alpha_{ij} \beta_{jt} e_s \wedge e_t$. Since the extension $\bigwedge f$ of $f$ to the exterior algebra $\bigwedge \mathbf{A}^{2m}$ is an endomorphism of this algebra (chap. III, § 5, no 9), we have $\bigwedge ((\bigwedge f)(u)) = (\bigwedge f)(\bigwedge u)$; since $\bigwedge f$ is the homothety of ratio $\det f$, it follows therefore from (6) and def. 1, that we have $m! \mathrm{Pf}({}^t P . R . P) = m! (\det P) \mathrm{Pf}(R)$, whence (7) in the case under consideration. The general case follows by observing that the two members of (7) are polynomials with integer coefficients in the elements of the matrices $R$ and $P$ (chap. IV, § 2, no 5, Scholium).

If $A$ is a commutative field and $\Phi$ an alternating non-degenerate bilinear form on a vector space $E$ of even dimension $2m$ over $A$, the symplectic group associated with $\Phi$ is isomorphic to $\mathbf{Sp}(2m, A)$, according to cor. 1 of th. 1.

Exercises. — ¶ 1) Let $A$ be a commutative principal ring, $E$ a free $A$-module of finite dimension $n$, $\Phi$ an alternating bilinear form on $E$; the ideals $A \alpha_i$ ($1 \leq i \leq r$) defined in Theorem 1 of No. 1 are called the invariant factors of $\Phi$.

a) Let $F$ be a submodule of $E$, and let $\Phi_F$ be the restriction of $\Phi$ to $F \times F$. Show that if $A \beta_i$ ($1 \leq i \leq s$) are the invariant factors of $\Phi_F$ (where $\beta_i$ divides $\beta_{i+1}$), then $s \leq r$ and $\beta_i$ is a multiple of $\alpha_i$ for $1 \leq i \leq s$. (Reduce to the case where $r = s = n/2$, and use Exercises 9 b) and 9 c) of Chapter VII, § 4).

b) Let $E_1$ be a second free $A$-module of finite dimension, $\Phi_1$ an alternating bilinear form on $E_1$, $A \gamma_1, \ldots, A \gamma_s$ its invariant factors ($\gamma_i$ dividing $\gamma_{i+1}$). In order that $\Phi_1$ be the inverse image of $\Phi$ by a linear mapping of $E_1$ into $E$, it is necessary and sufficient that $s \leq r$ and that $\gamma_i$ be a multiple of $\alpha_i$ for $1 \leq i \leq s$. (Use a) and Proposition 4 of Chapter VII, § 4, No. 5).

c) Let $F, G$ be two submodules of $E$, such that $F^0$ (resp. $G^0$) is supplementary to $F$ (resp. $G$) in $E$. If the restrictions of $\Phi$ to $F$ and $G$ are equivalent, show that the same is true of the restrictions of $\Phi$ to $F^0$ and to $G^0$, and that there exists an automorphism of $E$ leaving $\Phi$ invariant and transforming $F$ into $G$.

d) Give an example of two submodules $F, G$ of $E$, of dimension 2, such that $F$ and $G$ admit complements in $E$ and such that the restrictions $\Phi_F$ and $\Phi_G$ are equivalent, but that there is no automorphism of $E$ leaving $\Phi$ invariant and transforming $F$ into $G$ (take $n = 4$).

2) Let $\Phi$ be an alternating bilinear form on a finite-dimensional vector space $E$. Show that for every vector subspace $M$ of $E$, the difference dim M – dim (M ∩ M^0) is even. (Consider first the case where $\Phi$ is non-degenerate).

¶ 3) Let E be a vector space of even dimension n = 2m over a commutative field A; let $\Phi$ and $\Psi$ be two alternating bilinear forms on E; suppose that $\Psi$ is non-degenerate. Let u and v be the linear mappings from E into E* associated on the right to $\Phi$ and $\Psi$ respectively; v is an isomorphism of E onto E*; put $\omega = v^{-1} \circ u$, so that $\omega$ is an endomorphism of E.

a) Put M_0 = E, and, recursively, M_{k+1} = ω(M_k) for k > 0. Show that if M'_k is the orthogonal subspace to M_k for $\Phi$, M_{k+1} is the orthogonal subspace to M'_k for $\Psi$.

b) Let n_0 be the dimension of M'_0; put m_0 = 0, and for k ≥ 1, denote by m_k the dimension of M_k ∩ M'_0. Show that for k ≥ 1 the dimension of M_k is n − n_0 − (m_1 + ... + m_{k−1}) and that of M'_k is n_0 + (m_1 + ... + m_k).

c) Show that, for every k ≥ 0, the dimension of M_k ∩ M'_k is m_k + m_{k+1} + ... + m_{2k}, and that of M'_k ∩ M_{k+1} is m_{k+1} + m_{k+2} + ... + m_{2k+1} (apply Exercise 2 b) of § 3 to calculate the dimensions of M_h ∩ M'_{2k−h} and of M'_h ∩ M_{2k+1−h} by induction on h).

d) Deduce from c) that the numbers m_k are even (use Exercise 2).

e) Conclude from d) that the number of elementary divisors of the endomorphism ω, corresponding to the characteristic root λ = 0, and having a given degree, is even (cf. Chapter VII, § 5, Exercise 20).

4) Let E be a vector space over a commutative field A, admitting a countable basis (e_n)_{n≥1}, and let $\Phi$ be a non-degenerate alternating form on E. Show that there exists in E a basis (a_n) such that Φ(a_{2n−1}, a_{2n}) = 1 for every n ≥ 1, and Φ(a_i, a_j) = 0 for every other pair of indices such that i < j (reason as in Exercise 13 of § 4).

5) For every alternating matrix X = (x_{ij}) of even order n = 2m over a commutative ring, and for every index i, show that one has

6) Let M be a square matrix of order m over a commutative ring. Show that if one puts

7) Let A be a commutative field, $\mathfrak{A}(A)$ the set of alternating matrices of order 2m over A; let I be a mapping of $\mathfrak{A}(A)$ into A such that for every matrix R ∈ $\mathfrak{A}(A)$ and every matrix P of order 2m over A, one has I(^tPRP) = (det P)^h I(R), where h is a rational integer. Show that I(R) = c(PfR)^h, where c ∈ A (use formula (7) and Theorem 1 of No. 1).

8) Let P, Q be two square alternating matrices of even order 2m over a commutative ring A. Let φ(X) = Pf(P − XQ); show that, if Q is invertible, one has $\varphi(Q^{-1}P) = 0$. (Consider first the case where A is a field; deduce then from Exercise 3 that the minimal polynomial of the matrix $Q^{-1}P$ divides $\varphi(X)$, on passing to an algebraically closed extension of A, and observing that $\varphi^2$ is the characteristic polynomial of $Q^{-1}P$ up to a scalar factor).

§ 9) Let E be a vector space of dimension $n$ over a commutative field A, $\Phi$ an alternating bilinear form on E, $\Psi$ a hermitian sesquilinear form (for an involutory automorphism of A) on E, P and Q the respective matrices of $\Phi$ and $\Psi$ with respect to the same basis of E.

a) Suppose that $\Psi$ is non-degenerate. Show that the number of elementary divisors of $Q^{-1}P$ corresponding to the characteristic root 0 and having the same even degree is an even number (method of Exercise 3).

b) Suppose that $\Phi$ is non-degenerate (which implies that $n$ is even). Show that the number of elementary divisors of $P^{-1}Q$ corresponding to the characteristic root 0, and having the same odd degree, is an even number (same method).

10) Let $\omega_{2m}(\mathbf{F}_q)$ be the order of the symplectic group $\mathbf{Sp}(2m, \mathbf{F}_q)$ over the finite field $\mathbf{F}_q$. Show that if $h_{2m}$ is the number of pairs of vectors $(x, y)$ of $\mathbf{F}_q^{2m}$ such that $\Phi_0(x, y) = 1$ (notations of No. 3), one has $\omega_{2m}(\mathbf{F}_q) = h_{2m}\omega_{2m-2}(\mathbf{F}_q)$; deduce

11) Let A be a commutative field. Show that every transformation $u$ belonging to the symplectic group $\mathbf{Sp}(2m, A)$ is a product of transvections belonging to this group (called symplectic transvections; cf. § 4, Exercise 6). (Reason by induction on $m$, showing that if $x, y$ are two non-orthogonal vectors of $E = A^{2m}$, there exists a product $v$ of symplectic transvections such that $vu$ leaves $x$ and $y$ invariant.) Deduce a new proof of Proposition 3 of No. 3.

12) Let A be a commutative field, E a vector space of even dimension $n = 2m$ over A, $\Phi$ a non-degenerate alternating bilinear form on E. Now prove that, for every similitude $u$ for the form $\Phi$ ($§ 6$, no 5), with multiplier $\alpha$, one has $\det u = \alpha^m$ (use formula (7)).

§ 13) We suppose that A is a commutative field of characteristic 0, E a vector space of dimension $2m$ over A, $\Phi$ a non-degenerate alternating bilinear form on E. We identify the inverse form $\widehat{\Phi}$ of $\Phi$ with a bivector $\Gamma \in \bigwedge^2 E$, in such a way that for every symplectic basis $(e_i)_{1 \leq i \leq 2m}$ of E (for $\Phi$), indexed so that $\Phi(e_i, e_j) = \Phi(e_{m+i}, e_{m+j}) = 0$, $\Phi(e_i, e_{m+j}) = \delta_{ij}$ ($1 \leq i \leq m, 1 \leq j \leq m$), one has

We say that a decomposable and nonzero $p$-vector $z \in \bigwedge^p E$ is isotropic (resp. totally isotropic) for $\Phi$, if the corresponding vector subspace $V_z$ (chap. III, § 7, no 3) is isotropic (resp. totally isotropic).

a) If $z$ is a decomposable $p$-vector $\neq 0$, $2r$ the dimension of a supplement of $V_z \cap V_z^0$ relative to $V_z$, now prove that $m - p + r$ is the largest of the integers $h$ such that $z \wedge \Gamma^h \neq 0$, denoting by $\Gamma^h$ the $h$-th power of $\Gamma$ in the exterior algebra $\wedge E$ (use prop. 2 of § 4, n° 2).

b) Now prove that every bivector $x \in \overset{2}{\wedge} E$ can be written in the form $\lambda \Gamma + x_1$, where $\lambda$ is a scalar and $x_1$ a linear combination of decomposable totally isotropic bivectors. (Reduce to the case where $x$ is decomposable, and observe that if $(e_i)$ is a symplectic basis, $(e_1 + e_2) \wedge (e_{m+1} - e_{m+2})$ is totally isotropic).

c) If $p \leq m$, now prove that every $p$-vector $z \in \overset{p}{\wedge} E$ can be written $z = x \wedge \Gamma + z_1$, where $x$ is a $(p-2)$-vector and $z_1$ a linear combination of decomposable totally isotropic $p$-vectors. (Reduce to the case where $z$ is decomposable, and reason by induction on $p$. Thus reduce to the case where, $(e_i)$ being a symplectic basis, one has $z = e_1 \wedge e_2 \wedge \ldots \wedge e_{p-1} \wedge e_{m+p-1}$, and consider the bivectors $(e_{p-1} + e_{p+i}) \wedge (e_{m+p-1} - e_{m+p+i})$ for $0 \leq i \leq m-p$.

d) For $1 \leq p \leq m$, now prove that every $(m+p)$-vector $z \in \overset{m+p}{\wedge} E$ can be written $z = y \wedge \Gamma^p$, where $y$ is an $(m-p)$-vector. (Reduce to the case where $z$ is decomposable; if $2r$ is the dimension of a supplement of $V_z \cap V_z^0$ with respect to $V_z$, distinguish two cases, according as $r = p$ or $r > p$; in the second case, reason by induction on $r$, in the same way as in c)). Deduce that the mapping $y \to y \wedge \Gamma^p$ of $\overset{m+p}{\wedge} E$ into $\overset{m+p}{\wedge} E$ is bijective (observe that the two spaces have the same dimension). If $y$ is a decomposable $(m-p)$-vector, show that $z = y \wedge \Gamma^p$ is decomposable and that $V_z = V_y^0$.

e) Deduce from d) that, for $p \leq m$, the subspace $\overset{p}{\wedge} E$ is the direct sum of the homogeneous component of degree $p$ of the two-sided ideal $c$ generated by $\Gamma$ in $\wedge E$, and of the subspace $R_p$ of the $p$-vectors $z_1$ such that $z_1 \wedge \Gamma^{m-p+1} = 0$ (for $z \in \overset{p}{\wedge} E$, apply d) to the $(2m-p+2)$-vector $z \wedge \Gamma^{m-p+1}$). Using c), proving that $R_p$ is generated by the totally isotropic decomposable $p$-vectors, and using d), show that the mapping $x \to x \wedge \Gamma$ of $\overset{p-2}{\wedge} E$ into $\overset{p}{\wedge} E$ is injective, and that $R_p$ has dimension $\binom{2m}{p} - \binom{2m}{p-2}$.

f) Now prove that if $p \leq m$, a necessary and sufficient condition for a $p$-vector $z$ to be of the form $x \wedge \Gamma$ is that, for every totally isotropic decomposable $m$-vector $u$, one have $z \wedge u = 0$. (Now prove that, if this condition is satisfied, and if $z \in R_p$, one has $z \wedge y = 0$ for every $(2m-p)$-vector $y$; put $y$ in the form $x \wedge \Gamma^{m-p}$, where $x$ is a $p$-vector, and then apply c) to $x$, and observe that, if $x_1$ is a totally isotropic decomposable $p$-vector, $x_1 \wedge \Gamma^{m-p}$ is a decomposable $(2m-p)$-vector which can be written $u_1 \wedge v_1$, where $u_1$ is a totally isotropic decomposable $m$-vector).

Let $\alpha$ be the annihilator of the ideal $c$ in $\wedge E$; $\alpha$ is the direct sum of the subspaces $R_m, R_{m-1} \wedge \Gamma, \ldots, R_1 \wedge \Gamma^{m-1}$ and $K \Gamma^m$. Now prove that the annihilator of $\alpha$ in $\wedge E$ is equal to $c$ (cf. § 2, exerc. 4 b)).

g) For every symplectic automorphism $u$ of $E$ (for $\Phi$), let $\bar{u}$ be the canonical extension of $u$ to an automorphism of the algebra $\wedge E$ (chap. III, § 5, no 9). Now prove that the only elements of $\wedge E$ invariant under all the automorphisms $\bar{u}$ are the linear combinations of $1, \Gamma, \Gamma^2, \ldots, \Gamma^m$ with coefficients in $K$. (If $(e_i)$ is a symplectic basis, write that an element of $\wedge E$ is invariant under the symplectic transvections (exerc. 11) corresponding to the hyperplanes orthogonal to the $e_i$; then consider the symplectic transformations $u_{ij}$ defined by $u_{ij}(e_i) = e_j, u_{ij}(e_j) = e_i, u_{ij}(e_{m+i}) = e_{m+j}, u_{ij}(e_{m+j}) = e_{m+i}, u_{ij}(e_k) = e_k$ for every other index $k$).

14) Let $A$ be a commutative field of characteristic 2, $E$ the vector space $A^{2m}$; we identify the symplectic group $Sp(2m, A)$ with the group of symplectic matrices $U$, satisfying therefore the relation $^tU . R . U = R$, where $R = \begin{pmatrix} 0 & I_m \\ I_m & 0 \end{pmatrix}$. Put $D = \begin{pmatrix} 0 & 0 \\ I_m & 0 \end{pmatrix}$; now prove that every symplectic matrix $U$ such that $I + U$ is invertible can be written in one and only one way in the form $(^tD + S)^{-1}(D + S)$, where $S$ is a symmetric matrix such that $D + S$ is invertible, and conversely (cf. § 4, exerc. 11).

15) Let $A$ be a commutative field, $E$ a vector space of dimension $2m$ over $A$, $\Phi$ a non-degenerate alternating bilinear form on $E$. a) Extend to the endomorphisms $u$ of $E$ such that $u^*u = uu^*$ ($u^*$ being the adjoint of $u$ for $\Phi$) the results of § 4, exerc. 12 a) and b).

b) Suppose that $u^* = u$. With the notations of exerc. 12 of § 4, show that if $M$ is a minimal element of the set $\mathfrak{M}$ of non-isotropic subspaces contained in $G(p, p)$ and stable under $u$, then either $M$ is an indecomposable submodule of $E_u$, or it is the direct sum of two such isomorphic submodules (reason as in exerc. 12 c) of § 4). Give examples (with $p(X) = X - 1$) showing that both cases can occur.

c) Suppose that $u^* = -u$, $A$ not being of characteristic 2. With the same notations, let $p^h$ be the minimal polynomial of the restriction of $u$ to $M$, and let $d$ be the degree of $p$. Show that if $d(h-1)$ is odd, $M$ is an indecomposable submodule of $E_u$; if, on the contrary, $d(h-1)$ is even, $M$ is either indecomposable or the direct sum of two isomorphic indecomposable submodules.

d) Suppose that $u^*u = 1$ (in other words, $u \in Sp(\Phi)$). With the same notations as in c), if $p(X)$ does not divide $X^{d(h-1)} - 1$, or if $p(X) = X - 1$ and $(-1)^h \neq -1$, $M$ is an indecomposable submodule of $E_u$; otherwise, $M$ is either indecomposable or the direct sum of two isomorphic indecomposable submodules.