Quadratic Hamilton maps and positive complex planes

A quadratic Hamiltonian gives a linear flow. A symplectic coordinate change conjugates that flow, preserving information that an arbitrary change of basis can lose. We classify real nonnegative forms and forms with exactly one negative square, including their degenerate blocks. We then develop the positive complex planes associated with a quadratic form whose real part is positive.

We retain the symplectic conventions of Phase space and generating families and the linear Lagrangian geometry developed in Clean Lagrangian pairs and common transversals. The spectral algebra companion proves the polynomial-root, characteristic-identity, generalized-eigenspace, Jordan-chain and contour-projection arguments used here. Its A7–A9 give semidefinite Hermitian Cauchy–Schwarz, positive square roots, orthonormal bases and complete linear flows. Real symmetric diagonalization and singular inertia are proved in U001 Q5 and M0a. Symplectic complements, bases and quotient dimensions are G21–G23. These finite algebra proofs also apply over the complex field; A0 supplies complex rank and basis constructions. The arguments use generalized eigenspaces and include repeated eigenvalues. The source is Hörmander III, Section 21.5, Definition 21.5.1 through Corollary 21.5.11 and its concluding decomposition, printed pages 321–328 / PDF pages 336–343 of the 1994 reprint. The proof map binds every prerequisite to its current proof.

The coordinate changes here are ordinary real linear symplectic changes on a vector space. The quadratic forms have degree two under simultaneous scaling of all vector coordinates; no fiber-homogeneous coordinate assertion is being made.

1. The invariant Hamilton map

Let $S$ be a real symplectic vector space of dimension $2n$, with form $\omega$. In ordered coordinates $z=(x,\xi)$, put

Our Hamilton convention is $\iota_{H_f}\omega=-df$.

First consider a vector field $v$ vanishing at a point $c$ of a manifold. Its derivative there is an intrinsic endomorphism of $T_cS$. Indeed, if $y=\chi(z)$, the transformed field is $D\chi(z)v(z)$. Differentiating at $c$ gives $D\chi(c)Dv(c)$: the term involving $D^2\chi$ vanishes because $v(c)=0$. Thus its derivative changes by conjugacy. Equivalently, the adjoint endomorphism sends $d\phi(c)$ to $d(v\phi)(c)$; this depends only on $d\phi(c)$.

Definition 1.1 (Hamilton map). For a real $C^2$ function $f$ with $df(c)=0$, its Hamilton map is the derivative of $H_f/2$ at $c$. It depends only on the Hessian of $f$ there. For a quadratic form

write its polarized form as $Q(Y,X)=Y^TBX$. Its Hamilton map $F$ is characterized by

For a general $f$ at the critical point, $B=f''(c)/2$. In block notation,

The factor $1/2$ removes the factor two arising from differentiating a quadratic form.

Symmetry of $B$ gives

Conversely an endomorphism satisfying (1.4) has symmetric matrix $JF$, so it arises from exactly one quadratic form by (1.2). If $z=Cw$ and $C^TJC=J$, then

To verify the second identity, $JC^{-1}=C^TJ$, so both $JF'$ and $B'$ equal $C^TB C$. Consequently the spectrum, generalized eigenspace dimensions and Jordan lengths of $F$ are symplectic invariants of $Q$.

2. Spectral orthogonality, including Jordan chains

Extend $S$, $\omega$, $Q$ and $F$ complex linearly to $S_{\mathbb C}$. The form $\omega$ remains bilinear. Write $V_\lambda$ for the full generalized eigenspace of $F$ at $\lambda$. The actual direct-sum decomposition, its invariant projections and the finite inverse on each other summand are proved in A1–A3 of the companion; A4 proves the full chain decomposition and its multiplicities.

Lemma 2.1 (opposite eigenvalues pair). If $\lambda+\mu\ne0$, then

The pairing between $V_\lambda$ and $V_{-\lambda}$ is nondegenerate, and their dimensions are equal. The zero generalized eigenspace is symplectic. Each space $V_\lambda\oplus V_{-\lambda}$, $\lambda\ne0$, is symplectic; distinct such pairs are symplectically and quadratically orthogonal.

Proof. From (1.4), for every integer $N\ge0$,

On $V_\lambda$, $F+\mu$ is invertible when $\lambda+\mu\ne0$. On $V_\mu$, the right-hand operator vanishes for sufficiently large $N$. As the left-hand arguments run through all of $V_\lambda$, (2.1) follows.

If $X\in V_\lambda$ pairs to zero also with $V_{-\lambda}$, it pairs to zero with every generalized eigenspace. Their direct sum is $S_{\mathbb C}$, so nondegeneracy of $\omega$ gives $X=0$. The same argument with the two spaces exchanged proves the nondegenerate dual pairing and equality of dimensions. For $\lambda\ne0$, each individual space is isotropic by (2.1), and the pairing makes their sum symplectic. The argument for $V_0$ is identical. Finally $Q(X,Y)=\omega(X,FY)$; an invariant symplectically orthogonal splitting is therefore $Q$-orthogonal. □

For real $Q$, complex conjugation sends $V_\lambda$ onto $V_{\bar\lambda}$. The associated real invariant subspaces inherit the splitting. In particular, non-real eigenvalues ordinarily occur in the set $\lambda,-\lambda,\bar\lambda,-\bar\lambda$; the signature restrictions below will rule out some of these possibilities.

3. Real blocks and the four-step nilpotent mechanism

For this section assume that $Q$ is real and its negative index is at most one. Its radical is

Lemma 3.1 (a signature obstruction). If a real subspace $E$ satisfies $Q\le0$ on $E$ and $E\cap\operatorname{Rad}Q=0$, then $\dim E\le1$. If $Q\ge0$ on all of $S$, such an $E$ is zero.

Proof. Diagonalize the real symmetric form by an arbitrary real basis, only for this argument. It is a sum of positive squares, at most one negative square, and zero coordinates. The projection from $E$ to the negative coordinate is injective: if that coordinate vanishes, $Q\le0$ forces every positive coordinate to vanish too, placing the vector in the radical. Thus $\dim E$ is at most the number of negative squares. □

We now split off invariant symplectic blocks; their orthogonal complements are invariant by (1.4).

Real eigenvalues. Suppose $\operatorname{Re}\lambda\ne0$. The real subspace $\operatorname{Re}V_\lambda$ is $Q$-null. For non-real $\lambda$, all pairings among $V_\lambda$ and $V_{\bar\lambda}$ vanish by Lemma 2.1; for real $\lambda$, the same assertion uses just $V_\lambda$. Also this real subspace has no intersection with $\ker F$. If $\lambda$ is non-real, its real dimension is twice $\dim_{\mathbb C}V_\lambda$, contradicting Lemma 3.1. Hence $\lambda$ is real and $V_\lambda$ is one-dimensional. Choose real eigenvectors $e,\varepsilon$ in the dual opposite spaces with $\omega(\varepsilon,e)=1$. Then

Exchanging the canonical pair by $(e,\varepsilon)\mapsto(\varepsilon,-e)$ changes the sign of $\lambda$, so take $\lambda>0$. This block has one positive and one negative square. It consumes the available negative square, and the orthogonal complement is nonnegative. For a nonnegative $Q$, no real nonzero eigenvalue occurs.

Imaginary eigenvalues. Let $FX=i\mu X$, $\mu>0$, and write $X=X_1+iX_2$ with real vectors. The two vectors are independent and their span has no intersection with $\ker F$. The eigenvector equation gives $FX_1=-\mu X_2$, $FX_2=\mu X_1$, and

If $\omega(X_1,X_2)\ge0$, this is a nonpositive two-plane disjoint from the radical, contradicting Lemma 3.1. Normalize $\omega(X_1,X_2)=-1$. In the canonical basis $e=X_1,\varepsilon=X_2$, the block is

The plane is invariant and symplectic. Split it off and repeat on its invariant complement. At each stage a nonzero imaginary eigenvalue supplies an eigenvector there. This induction exhausts the nonzero imaginary spectral spaces, including repeated frequencies, and proves that there are no nontrivial Jordan blocks at these eigenvalues. It does not assume their semisimplicity in advance.

Rank-one and zero blocks. The residual real generalized zero space is symplectic, and $F$ is nilpotent on it. If $F^2X=0$ and $a=Q(X)\ne0$, then $X,FX$ span an invariant symplectic two-plane, since $\omega(X,FX)=a$. Set

Then $\omega(\varepsilon,e)=1$, $Fe=-\operatorname{sgn}(a)\varepsilon$, $F\varepsilon=0$, and $Q=\operatorname{sgn}(a)x^2$ on the plane. If instead $\ker F$ contains a symplectic pair, split off its plane; $F$ and $Q$ are zero there. Repeat these operations on invariant complements.

The residual chain. If a nonzero remainder survives, it has

A complement of $\ker F$ in $\ker F^2$ is $Q$-null and disjoint from the radical, so its dimension is at most one by Lemma 3.1. The identity

follows from (1.4) and equality of dimensions. Isotropy gives $\ker F\subset\operatorname{im}F$. Hence $F:\ker F^2\to\ker F$ is onto, with kernel $\ker F$, and

is exact. Thus $\dim\ker F\le1$. Nilpotence on a nonzero space forces a nonzero kernel. The invariant chain projection (A7) proves the chain decomposition without assuming it: each chain contributes exactly one kernel vector. Hence there is precisely one Jordan chain

The dimension $N$ is even because the remainder is symplectic. It exceeds two: otherwise $F^2=0$, and (3.6) would make $Q=0$ on the entire remainder, hence $F=0$, incompatible with a nonzero symplectic space having isotropic kernel.

Moving powers of $F$ across $\omega$ gives the exact formula

If $N\ge6$, this vanishes on the two-plane spanned by $F^{N-3}X,F^{N-2}X$, because the smallest exponent is $2N-5\ge N$. That plane is disjoint from $\ker F=\mathbb RF^{N-1}X$, contradicting Lemma 3.1. Therefore $N=4$.

Put $b=\omega(X,F^3X)$. On $\operatorname{span}(FX,F^2X)$, the form is $\operatorname{diag}(-b,0)$ by (3.10), and that plane again misses the radical. Thus $b<0$. Scale $X$ to make $b=-1$, and set

Here $\omega(Y,F^3Y)=-1$ and $\omega(Y,FY)=\omega(X,FX)-2t=0$. Define

The two diagonal pairings $\omega(\varepsilon_j,e_j)$ equal one. The same-type pairings vanish by (3.11) or $F^4=0$. The off-diagonal mixed pairings vanish because $\omega(Y,F^2Y)=0$, which follows from (1.4) and alternation. Thus this is a canonical basis. Its Hamilton map is

Using $Q(Z)=\omega(Z,FZ)$, the block is exactly

It has two positive squares, one negative square and one radical direction. Its single Jordan chain has length four. This completes the exhaustion of all residual blocks under the stated signature hypothesis.

4. The complete real classification

Theorem 4.1 (nonnegative and index-one forms). A real nonnegative quadratic form on a $2n$-dimensional symplectic space has canonical coordinates in which

The unused pairs are zero blocks. If the negative index is exactly one, its canonical form is the nonnegative portion in (4.1) plus precisely one of

The three types are inequivalent. Within a type, the dimension, the listed positive frequencies with multiplicities, the number of positive rank-one blocks, and $\lambda$ when present determine the symplectic equivalence class, up to reordering equal types of pairs.

Proof. Section 3 constructs invariant symplectic blocks until the dimension is exhausted. Each complement remains real, symplectic and of negative index at most one, so its argument applies repeatedly. In the nonnegative case only positive oscillators, positive rank-one blocks and zero pairs can occur. In the index-one case exactly one block uses the negative square. These are precisely the three choices in (4.2); the others must be nonnegative. Their sizes give the displayed dimension constraints.

The eigenvalues on the positive imaginary axis are exactly $i\mu_j$, counted with multiplicity. The hyperbolic block adds the real eigenvalues $\pm\lambda$. The four-dimensional block adds a Jordan chain of length four at zero. In the remaining negative rank-one case there is no such real pair or length-four chain; its negative square is detected by inertia. These invariants separate the three types. Each positive rank-one block has a length-two zero chain, and each unused zero pair has two length-one chains. After the type is fixed, the zero Jordan data and dimension determine their counts. In the negative rank-one type, one of the length-two chains belongs to that negative block. Conversely, if these data agree, reorder the canonical blocks and map their displayed canonical bases to one another. This is a symplectic change and preserves the displayed form. □

The theorem is a complete classification under these two signature hypotheses. Forms with two or more negative squares can have longer real Jordan blocks and complex quartets; Lemma 3.1 is the step that fails in that wider setting.

Example 4.2 (frequencies survive). For $a,c>0$ and $ac-b^2>0$,

has frequency $\mu=\sqrt{ac-b^2}$, because $F=\begin{pmatrix}b&c\\-a&-b\end{pmatrix}$ has eigenvalues $\pm i\mu$. The canonical shear $\eta=\xi+(b/c)x$, followed by $y=rx,\nu=\eta/r$ with $r^4=(ac-b^2)/c^2$, gives $Q=\mu(y^2+\nu^2)$. Its one-form satisfies $\nu\,dy=\xi\,dx+d(bx^2/(2c))$, so the whole change is symplectic. Different $\mu$'s cannot be removed by a symplectic change.

Example 4.3 (the actual nilpotent orbit). For (3.14),

Starting at $\varepsilon_1=(0,0,1,0)$, its half-Hamiltonian flow is

Substitution gives $Q=0$ for every real $t$. The full Hamiltonian flow uses $2F$, so the parameter in (4.3) would be replaced by $2t$.

Open the full-size figure.

Figure 4.1. The first two panels show exact orbits of $F=H_Q/2$ for the unit oscillator and unit real hyperbolic block, with arrows in increasing flow time. The third shows only the $(x_2,\xi_2)$ projection of the exact four-coordinate orbit (4.3); the omitted coordinates are given explicitly. A projected parabola alone does not encode the full length-four chain. The formulas (3.12)–(3.14) identify that chain and its symplectic form. All curves are original plots of the stated formulas.

5. Sectorial complex forms, kernels and reduction

Now allow $Q=P+iR$ to be complex valued, with real quadratic forms $P,R$. Assume $P\ge0$ and that for some finite $C\ge0$,

The nonnegative-real-part hypothesis is retained also when $C=0$; the inequality alone would then impose no sign on a real form. The closed convex sector $\Gamma$ contains all real-vector values of $Q$. The Hamilton map $F=-JB$ is complex linear; it need not preserve the real space $S$.

Theorem 5.1 (sectorial kernel and spectrum). Under (5.1), for $X\in S_{\mathbb C}$,

Thus $\ker F$ is spanned by its real elements. On the zero generalized eigenspace, $F^2V_0=0$, and

Every eigenvalue satisfies

Proof. We first identify the real radical. If $P(U)=0$, then (5.1) gives $R(U)=0$. For a real nonnegative quadratic form $A$, $A(U)=0$ implies $A(U,Y)=0$ for all real $Y$: the quadratic polynomial $A(U+tY)$ would otherwise have a nonzero linear term and take negative values for one sign of small $t$. Apply this to $P$. To get the same conclusion for $R$, choose any small nonzero real $\varepsilon$ with $|\varepsilon|C<1$. The form

is nonnegative and vanishes at $U$; its polarized row there is zero. Together with the row of $P$, this gives $R(U,Y)=0$. Therefore

Here $\ker P$ means the radical of the polarized real form, equivalently its zero set because it is nonnegative.

For $X=U+iV$, symmetry and bilinearity give

If this is zero, its real part is $P(U)+P(V)=0$. Both vectors belong to (5.5), hence $FU=FV=0$, and therefore $FX=0$. Conversely $FX=0$ implies $Q(\bar X,X)=\omega(\bar X,FX)=0$. This also shows that $FX=0$ forces its two real components separately into the kernel. The real vectors in that kernel are in the common radical of $P,R$, so they are also killed by $\bar F$. Applying the same reasoning to $\bar Q$, which satisfies the same bound, proves the reverse implication from $\bar F X=0$. This establishes every equivalence in (5.2).

If $F^3X=0$, then $F^2X\in\ker F$, and (5.2) gives $\bar F F^2X=0$. Using (1.4) for $\bar F$,

Apply (5.2) to $FX$ to obtain $F^2X=0$. Thus no Jordan chain at zero has length three or more: such a chain would contain a vector with $F^3X=0$, $F^2X\ne0$. Hence $V_0=\ker F^2$ and $F^2V_0=0$.

Identity (3.7) remains valid over the complex numbers. For $Y\in(\ker F)^\omega$, write $Y=FZ$. Then

This vanishes for all $Z$ exactly when $F^2X=0$, proving (5.3).

Finally let $FX=\lambda X$, $X=U+iV\ne0$, with $\lambda\ne0$. Then

The left side belongs to $\Gamma$, by convexity and (5.1). The real scalar $\omega(U,V)$ is nonzero: if it were zero, (5.2) would give $FX=0$, a contradiction. Dividing (5.9) by this scalar gives (5.4), with the sign depending on the scalar. For $\lambda=0$, (5.4) holds trivially. □

In particular a nonzero eigenvalue never lies on the real axis: a purely imaginary number in $\Gamma\cup-\Gamma$ must be zero. The theorem permits nontrivial Jordan blocks at nonzero complex eigenvalues. The real semidefinite conclusion in Section 3 does not assert their absence here.

Proposition 5.2 (remove the real radical symplectically). Set

This is a real symplectic space. The form $Q$ descends to $Q'$ there, satisfying (5.1), and $\operatorname{Re}Q'$ is positive definite. Its Hamilton map is isomorphic, after complexification, to the restriction of $F$ to the sum of its nonzero generalized eigenspaces.

Proof. The radical of $\omega|_{W^\omega}$ is $W^\omega\cap(W^\omega)^\omega=W^\omega\cap W$. Quotienting by precisely this radical gives a nondegenerate alternating form. Every vector of $W$ is in the quadratic radical (5.5), so adding a vector of $W\cap W^\omega$ changes neither $Q$ nor its polarized values. Thus $Q'$ is well-defined, as is the bound (5.1). If a real representative $U\in W^\omega$ has $P(U)=0$, (5.5) puts $U$ in $W\cap W^\omega$; its quotient class is zero. This proves strict real positivity.

By (5.2), $\ker F=W_{\mathbb C}$. Hence $(W^\omega)_{\mathbb C}=\operatorname{im}F$, and

To verify the last isomorphism, split into $V_0$ and the nonzero spectral sum. On the latter, $F$ is invertible. On $V_0$, $F^2=0$, so its image is exactly the part removed by the intersection with the kernel. Both the symplectic form and the quadratic form agree under this identification by spectral orthogonality. The defining identity (1.2) therefore identifies the induced Hamilton map with the nonzero restriction of $F$. □

The quotient in (5.10) uses the symplectic orthogonal space. The quotient $S/W$ need not carry a symplectic form and does not give this construction.

6. A strictly positive spectral plane

For a complex subspace of $S_{\mathbb C}$, introduce the Hermitian form

It is linear in its first argument and conjugate linear in its second; alternation of $\omega$ gives $h(Y,X)=\overline{h(X,Y)}$. A complex Lagrangian plane $\Lambda$ is positive when $h(X,X)\ge0$ for $X\in\Lambda$, and strictly positive when equality forces $X=0$.

Theorem 6.1 (positive spectral plane). If $\operatorname{Re}Q$ is positive definite, then

is an $F$-invariant, strictly positive Lagrangian plane. For its Hermitian form, the numerical range satisfies

Some finite $C$ as in (5.1) always exists under this strict hypothesis.

Proof. On the real unit sphere, $P$ has a positive minimum and $|R|$ a finite maximum; their ratio gives such a $C$. Theorem 5.1 excludes zero and real eigenvalues. Lemma 2.1 pairs the upper spectral spaces with the lower ones with equal dimensions. The sum in (6.2) thus has dimension $n$ and is isotropic, since two upper eigenvalues cannot sum to zero. It is Lagrangian and invariant.

When $Q$ is real positive definite, Theorem 4.1 gives only oscillator blocks. On each such block the upper eigenvector has $\xi=ix$. Thus

For the general form use $Q_t=P+itR$, $0\le t\le1$, with Hamilton map $F_t$. Every $Q_t$ has the same positive real part and a uniform sector bound. Its spectrum avoids the real axis. Here is the common-domain spectral argument. Bound $\|F_t\|\le M$ by continuity on the compact parameter interval. Every eigenvalue obeys $|\lambda|\le M$. If no uniform positive imaginary-part bound existed, a convergent subsequence of pairs $(t_j,\lambda_j)$ would have a real limiting $\lambda$ and $\det(\lambda I-F_t)=0$ by determinant continuity, contradicting the spectral exclusion. Thus $|\operatorname{Im}\lambda|\ge\delta>0$. Companion A6 constructs an actual common circle: with $L=M^2/\delta+\delta$ and radius $L-\delta/2$, the circle centered at $iL$ lies in the upper half-plane and encloses all upper spectra. Choose this one simple closed contour $\gamma$ in the upper half-plane, enclosing all upper spectra and none of the lower spectra for all $t$, with counterclockwise orientation. Then

is a continuous family of rank-$n$ projections onto the upper generalized spectral sum. This assertion does not require choosing individual eigenvectors: on a Jordan block $\lambda I+N$, the resolvent is $\sum_{j\ge0}N^j(z-\lambda)^{-j-1}$, and the direct circle calculation A5 proves that the integral is the identity if $\lambda$ is inside, zero if outside. The integrand is uniformly continuous in $t$ on the fixed contour. This proves continuity and the claimed range.

Strict positivity is open for this family of planes, by the positive compact-sphere lower bound in companion A7. For instance, near a fixed $t_0$, $\Pi_t$ applied to a basis of $\operatorname{ran}\Pi_{t_0}$ stays a basis of $\operatorname{ran}\Pi_t$; its Hermitian Gram matrix varies continuously. Suppose now $t_j\to t$ and these planes are strictly positive. For each $X\in\operatorname{ran}\Pi_t$, the vectors $\Pi_{t_j}X$ tend to $X$, so $h\ge0$ on the limiting plane. If $h(X,X)=0$, the semidefinite Cauchy–Schwarz inequality proved in A7 gives $h(F_tX,X)=0$, because $F_tX$ lies in that plane. But

Its vanishing implies $X=0$ by (5.2) and strict positivity of $P$. Thus the limiting plane is also strictly positive. The set of $t$ with the desired property is nonempty by (6.4), open and closed in $[0,1]$; it is the whole interval. This proves strict positivity at $t=1$, even through repeated eigenvalues and Jordan degeneracies.

For the numerical range, write $X=U+iV$. Equations (5.6) and (6.6) give $h(FX,X)=i(Q(U)+Q(V))$. The sum belongs to the convex sector $\Gamma$, and $h(X,X)>0$. Division by this positive scalar proves (6.3). □

Formula (6.5) states both its resolvent sign and its orientation. Using $(F_t-zI)^{-1}$ instead requires the opposite prefactor or the opposite orientation. The positivity proof uses a whole spectral plane, so it also covers a nonsemisimple example developed in Section 10.

7. The compatible complex structure and metric

The construction in this section applies to every strictly positive complex Lagrangian plane, whether it arose from a quadratic form or was given independently.

Theorem 7.1 (unique compatible Hermitian structure). If $\Lambda\subset S_{\mathbb C}$ is strictly positive Lagrangian, the real-part map $r:\Lambda\to S$ is a real-linear bijection. It induces a complex structure $I$ on $S$ by

There is a unique positive definite Hermitian form $H$, linear in its first argument with respect to $I$, such that

Its real part and its expression on $\Lambda$ are

Proof. For $X=U+iV\in\Lambda$,

If $rX=U=0$, this is zero, hence $X=0$. The real dimensions of $\Lambda$ and $S$ are both $2n$, so $r$ is a bijection. Transporting multiplication by $i$ through it gives (7.1) and $I^2=-1$. The inverse map is

Expand $\omega(Z(U),Z(V))=0$. Its real and imaginary parts yield

Consequently $g(U,V)=\omega(U,IV)$ is symmetric, and (7.4) gives $h(Z(U),Z(U))=2g(U,U)$. It is positive definite.

Define $H(U,V)=g(U,V)-i\omega(U,V)$. Symmetry of $g$ and alternation of $\omega$ give Hermitian symmetry. Identities (7.6) give $H(IU,V)=iH(U,V)$, and hence $H(U,IV)=-iH(U,V)$; thus it has the required complex linearities. It is positive definite and satisfies (7.2). Expanding $i\omega(\overline{Z(V)},Z(U))/2$ gives exactly $g(U,V)-i\omega(U,V)$, proving (7.3).

For uniqueness, a Hermitian form satisfying (7.2) must obey $H(U,IV)=-iH(U,V)$. Taking imaginary parts gives $-\operatorname{Re}H(U,V)=-\omega(U,IV)$. Its real part must therefore be $g$, while its imaginary part was prescribed. □

Corollary 7.2 (coordinates adapted to a real and a positive plane). Given a real Lagrangian plane $L\subset S$ and a strictly positive complex Lagrangian plane $\Lambda$, there are real canonical coordinates such that

Proof. Use the explicit Gram–Schmidt construction A8 to choose a $g$-orthonormal real basis $\varepsilon_1,\ldots,\varepsilon_n$ of $L$, and put $e_j=I\varepsilon_j$. Since $L$ is Lagrangian and $I$ preserves $\omega$, both $\omega(\varepsilon_j,\varepsilon_k)$ and $\omega(e_j,e_k)$ vanish. Also

Thus $(e_1,\ldots,e_n,\varepsilon_1,\ldots,\varepsilon_n)$ is a real canonical basis, and $L$ is its momentum plane. By (7.5), $\varepsilon_j-ie_j\in\Lambda$, so multiplying these vectors by $i$ gives the spanning vectors $e_j+i\varepsilon_j$. This is the graph in (7.7). □

Apply the corollary to $\Lambda=S^+$ from Theorem 6.1. Invariance and isotropy give $Q(X)=\omega(X,FX)=0$ on that plane. In the adapted coordinates, the complex polynomial $Q$ therefore belongs to the ideal

For completeness, the invertible complex linear variables $u=x+i\xi$, $v=x-i\xi$ identify the plane with $u=0$. A quadratic polynomial vanishing there has no terms involving only $v$. Every remaining monomial contains some $u_j$, so $Q=\sum_j u_j\ell_j(u,v)$, with linear $\ell_j$. This is a statement about a complex polynomial on a complex plane.

8. Positive graph matrices in arbitrary coordinates

Proposition 8.1 (graph criterion). In any real canonical coordinates, a strictly positive complex Lagrangian plane is the graph

Conversely every matrix in (8.1) defines such a plane.

Proof. A vertical vector $X=(0,\xi)$ has $h(X,X)=0$. Strict positivity excludes any nonzero such vector in $\Lambda$. Projection of the $n$-dimensional plane to the $n$-dimensional position space is thus a bijection, so it is a graph. For graph vectors,

The graph is Lagrangian precisely when $A^T=A$. Its Hermitian diagonal is

Thus strict positivity is equivalent to $A_2>0$. Both implications follow, including nondegeneracy and the exact factor two. □

The whole real symplectic coordinate change normalizing this graph is explicit. Let $D=A_2^{1/2}$ be the positive real symmetric square root constructed from the proved real diagonalization in companion A8. Set

Then

so differentiation gives $\sum d\eta_j\wedge dy_j=\sum d\xi_j\wedge dx_j$. On the graph, $\eta=iDx=iy$. This verifies the change on the entire real vector space as well as on the plane.

The compatible structure and metric can also be read directly from $A$. For a real vector $U=(u,v)$, the real part of $(z,Az)$, with $z=a+ib$, is $(a,A_1a-A_2b)$. Hence $a=u$, $b=A_2^{-1}(A_1u-v)$. Multiplication by $i$, followed by taking the real part, gives

In particular $I^2=-1$, $I^TJI=J$, and $g$ has matrix $JI$. Its diagonal is the positive expression

This also verifies the metric without having to guess a Euclidean structure first.

9. Non-strict positivity and its real part

Theorem 9.1 (reduce a positive plane). Let $\Lambda\subset S_{\mathbb C}$ be positive Lagrangian, and let

This is real isotropic. The nullspace of $h|_\Lambda$ is exactly $(L_{\mathbb R})_{\mathbb C}$. The image of $\Lambda$ in

is a strictly positive Lagrangian plane. There is also a real symplectically orthogonal splitting $S=S_1\oplus S_2$ for which $L_{\mathbb R}$ is Lagrangian in $S_1$ and

Proof. A semidefinite Hermitian form satisfies Cauchy–Schwarz. This follows, including null vectors, by applying nonnegativity to $X+tY$ for all complex $t$: if $h(X,X)=0$ and $h(Y,X)\ne0$, choose a small $t$ with its linear term negative. Thus a null vector $X\in\Lambda$ satisfies $h(Y,X)=i\omega(\bar X,Y)=0$ for all $Y\in\Lambda$. Since a Lagrangian plane equals its symplectic orthogonal, $\bar X\in\Lambda$. Hence $\operatorname{Re}X,\operatorname{Im}X\in L_{\mathbb R}$. Conversely for real $U\in\Lambda$, isotropy gives $h(Y,U)=i\omega(U,Y)=0$ for all $Y\in\Lambda$. Its complex span is the whole nullspace.