Completed continuation: the ordinary analytic composition theorem and its symbol remainders now discharge the composition obligations described at the end of this earlier proof source.

Clean composition of canonical relations and half-densities

Composition removes an intermediate phase-space point. Its possible nonuniqueness produces an excess space. The same space explains why composing two half-densities leaves an ordinary density to integrate. We prove the linear algebra, the smooth geometric statement, the global embedding condition and the complete density construction.

Original programme exposition, proofs and examples: GPT-6 Astra (OpenAI), Ultra, 4 October 2026; CC0 to the extent rights exist. Earlier components retain their separate terms.

C0. Conventions and the exact prerequisites

A symplectic vector space is a finite-dimensional real vector space with a nondegenerate alternating bilinear form. Write $\overline V$ for the same space with the negative form. A Lagrangian subspace equals its symplectic orthogonal. Basis extension, duality and rank-nullity are proved in C0 of the conic-coordinate component and its exact earlier linear-algebra proofs. The following consequences are needed here, so we prove them.

For a subspace $W\subset V$, the map $V\to W^*$ given by pairing with $\omega$ is onto: nondegeneracy identifies $V$ with $V^*$, and basis extension extends every functional on $W$. Its kernel is $W^\omega$. Hence $\dim W^\omega=\dim V-\dim W$. Alternation gives $W\subset(W^\omega)^\omega$; the dimension formula makes this an equality. In particular an isotropic subspace is Lagrangian exactly when it has half the ambient dimension.

A symplectic basis exists by the following induction. Unless $V=0$, choose $e\ne0$ and then $f$ with $\omega(e,f)=1$. The plane $H=\operatorname{span}(e,f)$ is nondegenerate. For each $v$,

satisfies $v-v_H\in H^\omega$, so $V=H\oplus H^\omega$. The restriction to $H^\omega$ is nondegenerate: a vector in its kernel is orthogonal both to $H^\omega$ and to $H$, hence to $V$. Induction supplies pairs $e_j,f_j$ with $\omega(e_i,f_j)=\delta_{ij}$ and all other pairings zero. It also proves that $\dim V$ is even. For a smooth symplectic bundle this construction can be made locally smooth: start with local sections whose pairing is nonzero at the chosen point, shrink the neighborhood and divide by that pairing, apply the displayed projection, and choose a frame for its image by a fixed nonzero minor. Repeat the induction in that smooth complementary bundle.

For manifolds we assume Hausdorff, second countable, finite-dimensional smooth manifolds without boundary. The inverse-function, compactness, smooth cutoff, parameter-integral and change-of-variables proofs are in the earlier stationary prerequisite chain. We explicitly derive the constant-rank and embedding consequences needed below. C5 constructs the finite partitions needed for integration directly from the earlier smooth cutoffs.

On cotangent spaces our convention remains $\omega=\sum d\xi_j\wedge dx_j$. Relations are written with the output first:

Thus $D$ acts first. If the symplectic dimensions are $2n_X,2n_Y,2n_Z$, the Lagrangian dimensions of $C,D$ are $n_X+n_Y,n_Y+n_Z$. Zero-dimensional spaces are allowed.

C1. The linear fibre product, excess and composed Lagrangian

Define

The kernel of $q$ identifies with $E$ by $y\mapsto(0,y,y,0)$. Put $e=\dim E$; this is the excess.

We first prove

If $y$ is symplectically orthogonal to the $Y$-projection of $C$, then $(0,y)$ is orthogonal to $C$ for $\omega_X-\omega_Y$. Since $C=C^\omega$, this is equivalent to $(0,y)\in C$. The corresponding statement for $D$ is identical. Now $U$ is the sum of the two $Y$-projection spaces; orthogonality to that sum means orthogonality to both summands. This proves (C3), with both implications.

The map

is onto and has kernel $U$. Indeed nondegeneracy identifies $V_Y$ with its dual, and every functional on $E$ extends to $V_Y$ by extending a basis. Its kernel is $E^{\omega_Y}=U$, by (C3) and the double-orthogonal identity. Thus there are exact sequences

These are actual maps, rather than a dimension-only identification.

Since $\dim U=2n_Y-e$, rank-nullity gives

For two elements of $F$, isotropy of $C,D$ gives

Consequently $L$ is isotropic of half the ambient dimension and is Lagrangian. No transversality assumption was made. Transversality, meaning $d$ onto, is equivalent by (C3) to $e=0$; in that case $q:F\to L$ is an isomorphism.

The diagonal is an identity for relation composition, and transposition interchanges the factors and reverses the sign of the product symplectic form. It therefore preserves the Lagrangian property. Associativity of the underlying linear relations follows directly: both $(C\circ D)\circ B$ and $C\circ(D\circ B)$ consist of pairs $(x,w)$ for which there exist $y,z$ satisfying all three relations. These observations assert no unproved density or oscillatory composition law.

C2. Clean smooth composition and its exact tangent rank

Let $M_X,M_Y,M_Z$ be symplectic manifolds of the dimensions above, and let $C\subset M_X\times\overline M_Y$ and $D\subset M_Y\times\overline M_Z$ be embedded Lagrangian submanifolds. Their fibre product is

The intersection is clean if $F$ is a smooth embedded submanifold of $C\times D$ and, at every $f=(x,y,z)$,

Both conditions are required. Smoothness of the set alone does not imply the tangent equality.

Apply C1 to these two tangent Lagrangian spaces. It proves that $dq_f(T_fF)$ is Lagrangian in $T_xM_X\oplus\overline{T_zM_Z}$ and has the constant rank $n_X+n_Z$. The kernel $\mathcal E_f=\ker dq_f$ has dimension

on each component of $F$. It is a smooth vector subbundle of $TF$: in a neighborhood where a fixed rank minor of $dq$ is invertible, solve for the corresponding variables in $dq\,v=0$; the free variables give a smooth local frame. This also proves the bundle assertion for the image and cokernel of every constant-rank map used below.

Here is the full local constant-rank argument. For a map of constant rank $r$, reorder local coordinates so that an $r$-by-$r$ derivative minor is invertible. The map taking the first $r$ output functions and the remaining input coordinates as new input coordinates is a local diffeomorphism, by the earlier inverse-function theorem. In these coordinates the original map is $(u,v)\mapsto(u,g(u,v))$. Constant rank forces every derivative $\partial_v g$ to vanish. On a smaller product box, integrate these derivatives along coordinate segments to get $g(u,v)=g(u,0)$. Subtracting $g(u,0)$ from the remaining output coordinates now gives

This proof includes $r=0$ and full rank, with empty coordinate lists.

Applied to $q$, (C11) shows that every sufficiently small image branch is a smooth embedded Lagrangian, and that $q$ is a submersion onto that branch. Different branches may meet. Clean intersection alone does not establish that the entire image is embedded or that its leaf space has a global manifold structure.

If the matching maps are transverse, the derivative of their coordinate difference is onto. The inverse-function argument just used, with the difference functions among the new coordinates, proves that their zero set is a submanifold with exactly the kernel tangent space. Thus transversality implies cleanness with $e=0$, and $q$ is locally an immersion. Global injectivity is still a separate condition.

C3. A complete global embedding criterion

Theorem. Suppose $f:N\to P$ is a smooth map of constant rank between the manifolds specified in C0. If $f$ is proper and every nonempty fibre is connected, then $f(N)$ is a closed embedded submanifold of $P$, and $f:N\to f(N)$ is a proper submersion. The dimensions may be read componentwise. Simple connectedness of the fibres is not needed for this statement.

Proof. A proper map into a locally compact Hausdorff space is closed in this setting. To see this, let $A\subset N$ be closed and choose, near a point of $P$, a neighborhood $W$ with compact closure. Then $A\cap f^{-1}(\overline W)$ is compact, so its image is compact and closed in $P$. Its intersection with $W$ equals $f(A)\cap W$. This proves that $f(A)$ is closed locally, hence globally. The facts used here are elementary: compact subsets of a Hausdorff space are closed because finitely many disjoint-neighborhood choices separate a point from a compact set, and continuous images of compact sets are compact by pulling back open covers. Relatively compact $W$ comes from a smaller Euclidean ball in a chart.

Fix $p\in f(N)$ and its compact connected fibre $K=f^{-1}(p)$. C2's local normal form gives, at each $a\in K$, a neighborhood $U_a$ whose image is a relative open piece of an embedded rank-$r$ submanifold through $p$. Denote the germ of this image at $p$ by $L_a$.

This germ is locally constant as $a$ varies in $K$. Indeed, if $b\in K\cap U_a$, choose a smaller normal-form neighborhood of $b$ inside $U_a$. A submersion in product coordinates maps a sufficiently small neighborhood of $b$ onto a relative neighborhood of $p$ in the same image submanifold. Its germ therefore equals $L_a$. The same argument compares any two choices at $b$, by taking a smaller common neighborhood. The sets of fibre points with a given germ are open in $K$; their complements are unions of the other such open sets. Connectedness implies that there is just one germ.

Choose finitely many $U_{a_i}$ covering $K$. Their image germs agree, so shrink one ambient neighborhood $W$ of $p$ until their images in $W$ lie in a common embedded submanifold $L$; also shrink until one image contains $L\cap W$. This is precisely the meaning of equality of finitely many germs of embedded submanifolds and of a relative open image.

There are no additional branches arbitrarily near $p$ from outside $U=\bigcup_iU_{a_i}$: $N\setminus U$ is closed, its image is closed by the first paragraph, and that image does not contain $p$. Shrink $W$ to avoid it. Now $f(N)\cap W=L\cap W$, an embedded submanifold. The local normal forms give a submersion onto it. Doing this for each $p$, and using closedness of $f(N)$, proves the theorem. Properness into the image follows because a compact subset of the image is still compact in $P$. $\square$

For C2, the theorem proves a global embedded Lagrangian $L=C\circ D$ whenever $q$ is proper with connected nonempty fibres. Its fibres are compact smooth $e$-manifolds. For $e=0$ they are discrete and connected, hence singletons; the local inverses in (C11) then combine to show that $q:F\to L$ is a diffeomorphism. We do not assert a locally trivial bundle theorem here.

The density construction below also applies when embeddedness of $L$ and the submersion $q:F\to L$ are established in another way. Only properness on the support being integrated is required for that integration; connectedness is not required there.

C4. The density isomorphism, with its normalization

For a real $d$-dimensional vector space $V$, define $\mathcal D^a(V)$ to consist of complex-valued functions on its bases satisfying

We need $a=1,\tfrac12,-\tfrac12,-1$. Specifying the value on one basis identifies this with a one-dimensional complex vector space: every other basis has a unique invertible change matrix. We make no definition on singular frames for negative powers. Multiplication gives $\mathcal D^a(V)\otimes\mathcal D^b(V)=\mathcal D^{a+b}(V)$. For $V=0$, the empty basis gives the canonical identification with $\mathbb C$.

There are two canonical rules, with full justifications. First, for an exact sequence $0\to A\to B\to C\to0$, choose a basis of $A$ and lifts of a basis of $C$. These form a basis of $B$. Replacing the lifts adds an upper block triangular matrix with diagonal identities, so does not change a density's value. Changing the two bases multiplies the determinant by the product of their determinants. Hence

Second, the basis dual to $vA$ changes by $A^{-T}$. Evaluation on dual bases therefore gives

These proofs also establish naturality under isomorphisms of the vector spaces and exact sequences. Applying (C13) twice proves the alternating density identity for a four-term exact sequence, without assuming a separate determinant-line theorem.

On a symplectic $2n$-space define its positive density $\nu_V$ to have value one on a symplectic basis. It is independent of the choice: if $S$ changes symplectic bases, $S^TJS=J$, so $|\det S|=1$. Existence and local smoothness of such bases follow from C0 above. The positive half-density $\nu_V^{1/2}$ has the same unit value. This normalization includes the zero space and requires no metric or orientation choice.

Apply (C13) to the two short sequences through $U$ in the first sequence (C5), and then apply (C14). They give

The second sequence (C5) gives $\mathcal D^{1/2}(F)\simeq \mathcal D^{1/2}(E)\otimes\mathcal D^{1/2}(L)$. Divide the middle-space density factor in (C15) by the specified $\nu_{V_Y}^{1/2}$. The resulting canonical isomorphism is

In particular the excess factor is a full density, not a half-density. All determinant factors and the middle-space normalization have now been specified. Their basis proofs establish independence of every complement and lift used to compute this map.

C5. Smooth density composition and integration over the fibre

Assume the clean composition has embedded image $L$, and that $q:F\to L$ is the submersion just proved or independently established. The tangent exact sequences (C5) form smooth vector-bundle sequences: C2's invertible-minor argument gives all their local frames and ranks. In those frames (C13)–(C16) depend smoothly on the base point, because their nonzero determinants and positive square roots do. Thus (C16) gives over $F$

Let $a,b$ be smooth half-density sections on $C,D$, respectively. Let $h$ be the image of their pulled-back tensor product under (C17). Suppose $q:\operatorname{supp}h\to L$ is proper. Define

Here is a full construction and proof of smoothness. Fix a relatively compact coordinate neighborhood of a point of $L$. Above a smaller closed neighborhood, the support of $h$ is compact by the stated properness. Cover that compact set by finitely many submersion charts with compactly contained smaller charts. Smooth bumps positive on the smaller charts, divided by their sum on its positive set, give a partition near the compact support. Multiply by a further bump equal to one on a neighborhood of that support and supported in this positive set. Such a bump follows from the same finite chart construction. Extend the resulting pieces by zero; they have compact support in their submersion charts. Their sum is $h$ over the smaller base neighborhood. This explicitly supplies the only partition needed for the local integral.

In a submersion chart $q(u,v)=u$, a piece is

Its integral is $(\int h_i(u,v)\,dv)|du|^{1/2}$. The coefficient is smooth: all parameter derivatives have common compact support and may be passed through the integral by the earlier compact parameter-integral proof. There are only finitely many pieces near the selected base point.

For another chart, write $u'=g(u)$, $v'=k(u,v)$. The transformation of (C19) uses precisely $|\det\partial_vk|$ for the fibre density and $|\det Dg|^{1/2}$ for the base half-density. The ordinary change-of-variables theorem in $v$ cancels the first factor and leaves the second. This proves the half-density transformation of the integral. Refining two partitions by their pairwise products proves independence of the partition, using finite additivity and $\sum_i\rho_i=1$. The locally defined answers consequently agree and give a smooth global half-density on $L$. Their support is contained in $q(\operatorname{supp}h)$, a closed set by the proper-map argument in C3 applied to that closed support. No unproved fibre integration theorem has been used.

For finite-rank complex bundles $E_X,E_Y,E_Z$, one can instead take

with the bundles pulled back from the appropriate factors. Compose the linear maps in the order $a b$ on $F$, then apply the same density construction and integral componentwise. A change of frame in $E_Y$ cancels between the two factors. Changes in $E_X,E_Z$ depend only on the output point $(x,z)$ and pass through the fibre integral. Therefore the result is an intrinsic $\Omega_L^{1/2}\otimes\operatorname{Hom}(E_Z,E_X)$ section. The finitely many component estimates also prove its smoothness.

C6. Worked checks and counterexamples

The first panel plots the coefficient of the density in (C21); its shaded area represents the integral. The second shows the failure of the tangent equality (C9). The third shows why C3 needs a condition excluding multiple image branches. These are the exact examples proved below, drawn from their formulas; plotted samples are not proofs. Reproducible figure source.

Exercise C1 — excess and its density. Let $M_X=M_Z$ be a point, $M_Y=T^*\mathbb R$ with coordinates $(q,p)$, and let both relations be the zero section $p=0$. Compute (C16) and (C18).

Solution. Here $F=E=\mathbb R_q$, the image is a point, and $e=1$. In the bases of the two copies of the zero section, $d(q_1,q_2)=(q_1-q_2,0)$. The basis of its kernel is $(1,1)$; adjoining $(1,0)$ has determinant of absolute value one. The cokernel pairing with $E$ is $\omega_Y((1,0),(0,p))=-p$, again of absolute determinant one. Thus

Compact support of the product supplies the required properness. For $a=b=1$ the integral diverges. Linear or clean geometric composability alone therefore supplies no convergence claim.

Exercise C2 — symplectic graphs. Let $S:V_Y\to V_X$ and $T:V_Z\to V_Y$ be symplectic isomorphisms. Use the input-space symplectic half-densities on their graphs. Compute the composition.

Solution. The fibre product is parameterized by $z$, with $(x,y,z)=(STz,Tz,z)$; the excess is zero. To compute the density, use on $C\oplus D$ coordinates $(y,z)$. Coordinates $(w,z)$ with $w=y-Tz$ differ by a triangular matrix of determinant one. They identify $d$ with $w$ and the kernel with $z$. Dividing by $\nu_{V_Y}^{1/2}$ in (C15) therefore leaves exactly $\nu_{V_Z}^{1/2}$, the specified half-density on the graph of $ST$. For coefficients the result is $a(Tz)b(z)\nu_{V_Z}^{1/2}$. This includes the left and right identity relations and verifies that no Euclidean graph-length or extra power of two is present.

Exercise C3 — smooth intersection need not be clean. In $T^*\mathbb R$, compose the relations from and to a point given by $p=0$ and $p=q^2$. Is the intersection clean?

Solution. Both curves are Lagrangian, since every one-dimensional submanifold in this symplectic surface is isotropic. Their intersection is the single point $(0,0)$, a smooth zero-dimensional manifold. Both tangent lines there are the $q$-axis. The tangent fibre product therefore has dimension one, whereas the actual tangent of their intersection has dimension zero. Equality (C9) fails. The dimension and density conclusions for clean composition cannot be applied.

Exercise C4 — proper constant rank does not imply embedding. Show why connected fibres in C3 cannot simply be omitted.

Solution. Consider the map of the circle

Its derivative $(\cos t,2\cos2t)$ never vanishes: if $\cos t=0$, then $\cos2t=-1$. It has constant rank one and is proper, because its domain is compact. The points $t=0,\pi$ have the same image, with different tangent lines $\mathbb R(1,2)$ and $\mathbb R(-1,2)$. Its image is not an embedded one-dimensional submanifold at zero: such a submanifold would have a single tangent line containing both derivatives. The fibre over zero is the two points $0,\pi$, hence disconnected. This is a counterexample to the general constant-rank inference; it does not assert that arbitrary immersions have been realized by our two given canonical relations.

Free source and what this component proves

The exact human source is Victor Guillemin and Shlomo Sternberg, Semi-classical Analysis, freely accessible author draft dated 13 January 2010, Sections 3.4, 4.1–4.3, 6.1 and 7.1–7.2. The selected full source pages supply the linear relations and density constructions. The programme proofs above include the exact-sequence maps, constant-rank argument, global embedding proof and fibre-integral construction. C3 proves the needed embedding criterion directly and shows that its connected-fibre version needs no additional simple connectedness hypothesis.

No source prose or PDF is redistributed. This component proves geometry and ordinary half-density composition. The Maslov factor, Gaussian-symbol composition and the analytic Fourier-integral operator composition theorem remain separate obligations of the full course. No external citation is used in place of a prerequisite proof.