The middle-boundary Maslov correction composes for Lagrangian relations

The Wall form in [F1] is unchanged by a symplectic isomorphism: the isomorphism carries its numerator and denominator to the corresponding subspaces and preserves $\omega (a,b^{\mathrm{\prime }})$. The cyclic map $[a]\mapsto [b]$, with $a+b+c=0$, is a well-defined isomorphism between the cyclic Wall quotients: changing a decomposition adds an adjacent intersection, and its inverse is obtained by solving the same sum-zero equation. It preserves the form because $\omega (b,c^{\mathrm{\prime }})=-\omega (b,a^{\mathrm{\prime }})=\omega (a^{\mathrm{\prime }},b)=\omega (a,b^{\mathrm{\prime }})$. Hence $\mu$ is cyclically invariant. If the first two Lagrangians agree, its numerator and denominator are both that Lagrangian, so the quotient is zero. Cyclic invariance therefore shows that $\mu$ vanishes whenever any two arguments agree. Orthogonal direct sums give orthogonal direct sums of Wall quotients and forms, so their signatures add.

Set $K=\{v:(v,0)\in B\}$ and $K^{\mathrm{\prime }}=\{w:(0,w)\in B\}$. Both are isotropic. A vector $x$ annihilates ${\pi }_{V}B$ exactly when $(x,0)$ annihilates $B$, hence belongs to $B=B^{\perp }$. Thus $({\pi }_{V}B{)}^{\perp }=K$ and ${\pi }_{V}B=K^{\perp }$; likewise ${\pi }_{W}B=K^{\mathrm{\prime }\perp }$. A pair $(v,w)\in B$ defines an isomorphism $\varphi :K^{\perp }\mathrm{/}K\to K^{\mathrm{\prime }\perp }\mathrm{/}{K}^{\mathrm{\prime }}$. Changing $w$ adds precisely $K^{\mathrm{\prime }}$, and changing $v$ adds $K$; the two projections are onto and these are their kernels. Isotropy of $B$ gives ${\omega }_W(w,w^{\mathrm{\prime }})={\omega }_V(v,v^{\mathrm{\prime }})$, so $\varphi$ is symplectic.

Choose a Lagrangian $N_0$ containing $K$. This finite construction needs no transversality: extend a basis $e_i$ of $K$ to hyperbolic pairs by choosing vectors $f_j$ with $\omega (e_i,f_j)={\delta }_{ij}$, possible by nondegeneracy. Replace $f_i$ by $f_i-\frac{1}{2}{\sum }_j\omega (f_i,f_j)e_j$ to make their pairings zero. Their span is symplectic. In its symplectic orthogonal complement, repeatedly choose a nonzero vector and a vector pairing to $1$, splitting off a symplectic plane each time. Adjoining one vector from each such plane to the $e_i$ spans the required Lagrangian. The construction ends by finite dimension.

For a Lagrangian $L\subseteq V$, put $L_K=((L\cap K^{\perp })+K)\mathrm{/}K$. It is Lagrangian in $K^{\perp }\mathrm{/}K$: the inverse image $P_L=(L\cap K^{\perp })+K$ is isotropic, and $P_L^{\perp }=(L+K)\cap K^{\perp }=(L\cap K^{\perp })+K=P_L$. The last equality follows by writing $x=l+k$ and observing that $x,k\in K^{\perp }$ force $l\in K^{\perp }$. Here $(U\cap T{)}^{\perp }=U^{\perp }+T^{\perp }$ follows from inclusion and equal dimensions by [F3]. The relation normal form in step 1.2 gives $B(L)\mathrm{/}{K}^{\mathrm{\prime }}=\varphi (L_K)$ and $B^t(Q)\mathrm{/}K={\varphi }^{-1}(Q_{K^{\mathrm{\prime }}})$. These are equalities of the actual subspaces containing $K^{\mathrm{\prime }}$ or $K$, not an assumption that relation images are injective.

Suppose $N_0,N\subseteq V$ both contain $K$. We prove ${\mu }_V(L,N_0,N)={\mu }_{K^{\perp }\mathrm{/}K}(L_K,N_0\mathrm{/}K,N\mathrm{/}K)$ by identifying their complete Wall forms. The left numerator $L\cap (N_0+N)$ lies in $K^{\perp }$. It projects onto the right numerator: if a representative $l+k$, with $l\in L\cap K^{\perp }$, lies in the sum of the two reduced spaces, then $l\in N_0+N$, since that sum contains $K$. The image of $(L\cap N_0)+(L\cap N)$ is precisely the right denominator: $l+k\in N_0$ is equivalent to $l\in N_0$, and likewise for $N$. The projection kernel $L\cap K$ is already in this denominator. Hence the Wall quotient spaces are isomorphic. A decomposition $a+b+c=0$ projects to the corresponding decomposition, and $\omega (a,b^{\mathrm{\prime }})$ descends to the quotient form because both entries lie in $K^{\perp }$. The isomorphism is an isometry, proving equality of signatures. By step 1.1 the same reduction holds when any two arguments contain $K$.

For arbitrary $L,P$ and $N$ containing $K$, the four-Lagrangian identity gives $\mu (L,P,N)-\mu (L,P,N_0)=\mu (L,N_0,N)-\mu (P,N_0,N)$. Reduce the two right-hand terms by step 3.1 and apply that same identity in $K^{\perp }\mathrm{/}K$. Thus ${\mu }_V(L,P,N)={\mu }_{K^{\perp }\mathrm{/}K}(L_K,P_K,N\mathrm{/}K)+{\eta }_K(L,P)$, where ${\eta }_K(L,P)={\mu }_V(L,P,N_0)-{\mu }_{K^{\perp }\mathrm{/}K}(L_K,P_K,N_0\mathrm{/}K)$. In particular the defect is independent of $N$. The auxiliary $N_0$ disappears from the differences used below.

In step 4.1 take $N=B^t(Q)$. By step 2.1 and symplectic invariance, its reduced triple transports by $\varphi$ to $(B(L)\mathrm{/}{K}^{\mathrm{\prime }},B(P)\mathrm{/}{K}^{\mathrm{\prime }},Q_{K^{\mathrm{\prime }}})$. Since $B(L)$ and $B(P)$ both contain $K^{\mathrm{\prime }}$, step 3.1 applied in $W$ identifies this reduced signature with ${\mu }_W(B(L),B(P),Q)$. Therefore ${\mu }_V(L,P,B^t(Q))-{\mu }_W(B(L),B(P),Q)={\eta }_K(L,P)$, independent of $Q$. Subtract the equations for $Q$ and $R$ and use the four-Lagrangian identity on $(B(L),B(P),Q,R)$ in $W$. This yields exactly the first identity in the Statement.

Relational composition obeys $(BA)({\lambda }_0)=B(A({\lambda }_0))$ and $(CB{)}^t({\lambda }_3)=B^t(C^t({\lambda }_3))$ by the defining existential equations. Substitute $L=A({\lambda }_0)$, $P={\lambda }_1$, $Q={\lambda }_2$, $R=C^t({\lambda }_3)$ into step 5.1. All these spaces are Lagrangian by [F2], including inverse images by applying it to transposed relations. The resulting equality is precisely $m(B,A)+m(C,BA)=m(C,B)+m(CB,A)$. Thus adding these integer corrections gives associative weights.

For the identity of a specified object $(V,\lambda )$ on either side, its source and target Lagrangians are the same $\lambda$, so its image or inverse image is the specified middle Lagrangian, so two arguments of $\mu$ agree and the correction is zero by step 1.1. Direct sums add corrections by that same step, including the zero-dimensional empty space. For graphs of symplectic isomorphisms the correction is $\mu (f_1{\lambda }_0,{\lambda }_1,f_2^{-1}{\lambda }_2)$. Transporting all three arguments by $f_2$ gives $\mu (f_2{f}_1{\lambda }_0,f_2{\lambda }_1,{\lambda }_2)$ with precisely the stated sign and order. These prove the remaining conclusions.