The four-Lagrangian Maslov cocycle identity

The assignment $[a]\mapsto [b]$ is a well-defined isomorphism from $U_{123}$ to $U_{231}=\frac{L_2\cap (L_3+L_1)}{(L_2\cap L_3)+(L_2\cap L_1)}\mathrm{.}$ Indeed, changing $b$ in a fixed decomposition changes it by an element of $L_2\cap L_3$; adding an element of $L_1\cap L_2$ to $a$ changes $b$ by an element of $L_1\cap L_2$, and adding an element of $L_1\cap L_3$ can be absorbed into $c$. Conversely, for $b\in L_2\cap (L_3+L_1)$ write $b=a_0+c_0$ with $a_0\in L_1,c_0\in L_3$ and take $a=-a_0,c=-c_0$, which gives $a+b+c=0$. If $a=u_{12}+u_{13}$ with $u_{12}\in L_1\cap L_2,u_{13}\in L_1\cap L_3$, then $b=-a-c$ differs from $-u_{12}$ by an element of $L_2\cap L_3$, so its class is zero. Thus the denominator is exactly the kernel and the map is an isomorphism.

For a cyclic list $L_1,\dots ,L_n$ of Lagrangians, set $K=\ker ({\sum }_i:L_1\oplus \cdots \oplus L_n\to V)$. For $w\in K$, choose ${\widehat{w}}_i\in V$ with $w_i={\widehat{w}}_i-{\widehat{w}}_{i-1}$ cyclically; such vectors exist since ${\sum }_i{w}_i=0$, and any two choices differ by a common constant. Define $q(v,w)={\sum }_i\omega (v_i,{\widehat{w}}_i)$. Changing the common constant changes the sum by $\omega ({\sum }_i{v}_i,t)=0$. Summation by parts gives $q(v,w)={\sum }_i\omega (w_i,{\widehat{v}}_{i-1})$, which differs from $q(w,v)$ by ${\sum }_i\omega (w_i,v_i)=0$, since both entries in each pair lie in $L_i$. Thus $q$ is symmetric. Choosing ${\widehat{w}}_0=0$ and ${\widehat{w}}_i={\sum }_{j=1}^i{w}_j$ gives $q(v,w)={\sum }_i{\sum }_{j\le i}\omega (v_i,w_j)={\sum }_{i>j}\omega (v_i,w_j),$ because the diagonal terms vanish. Define $\tau (L_1,\dots ,L_n)$ as the signature of the induced nondegenerate form on $K\mathrm{/}\mathrm{rad}(q)$.

The pairing $\omega (a,b)$ descends to $U_{123}\times U_{231}$. If $a$ changes by $u_{12}+u_{13}$ with $u_{rs}\in L_r\cap L_s$, then $\omega (u_{12},b)=0$ since both lie in $L_2$, and $\omega (u_{13},b)=0$ since $b=-a-c$ and each of $u_{13},a$ lies in $L_1$ while $u_{13},c$ lie in $L_3$. Changing $b$ by $L_2\cap L_3$ also pairs to zero because $a=-b-c$. For subspaces $P,Q$, $P^{\perp }+Q^{\perp }=(P\cap Q{)}^{\perp }$: inclusion follows from the definition of orthogonality, and the dimensions agree by rank-nullity and the formula $\dim U^{\perp }=\dim V-\dim U$. The latter follows because $V\to U^{\ast }$, $v\mapsto \omega (v,-){\mathrm{\mid }}_U$, is onto (extend a functional on a basis of $U$ and use nondegeneracy of $\omega$), with kernel $U^{\perp }$. Thus $(L_2\cap (L_3+L_1){)}^{\perp }=L_2+(L_3+L_1{)}^{\perp }=L_2+(L_1\cap L_3)\mathrm{.}$ The induced pairing is nondegenerate: if $a\in L_1\cap (L_2+L_3)$ pairs to zero with $L_2\cap (L_3+L_1)$, then $a\in L_2+(L_1\cap L_3)\mathrm{.}$ Writing $a=b+t$ with $b\in L_2$ and $t\in L_1\cap L_3$, the facts $a,t\in L_1$ imply $b\in L_1\cap L_2$, so $[a]=0$. In the other variable, if $b\in L_2\cap (L_3+L_1)$ pairs to zero with every $a\in L_1\cap (L_2+L_3)$, then $b\in (L_1\cap (L_2+L_3){)}^{\perp }=L_1+(L_2\cap L_3)$. Since $b\in L_2$, its class is zero in $U_{231}$.

Fix $2\le k<n$. Let $K_A$ be the sum-zero subspace for the list $(L_1,\dots ,L_k)$ and $K_B$ that for $(L_1,L_k,L_{k+1},\dots ,L_n)$. For $a=(a_1,\dots ,a_k)$ and $b=(b_1,b_k,b_{k+1},\dots ,b_n)$ define $s(a,b)=(a_1+b_1,a_2,\dots ,a_{k-1},a_k+b_k,b_{k+1},\dots ,b_n)\in K\mathrm{.}$ Using $q(v,w)={\sum }_{i>j}\omega (v_i,w_j)$, the terms involving only $a$ give $q_A$, those involving only $b$ give $q_B$, and the cross terms reduce, by ${\sum }_i{a}_i=0=\sum b_i$, to $-\omega (a_1,b_1^{\mathrm{\prime }})+\omega (a_k^{\mathrm{\prime }},b_k)\mathrm{.}$ Both terms vanish because $a_1,b_1^{\mathrm{\prime }}\in L_1$ and $a_k^{\mathrm{\prime }},b_k\in L_k$. Thus $s$ preserves the orthogonal-sum form. Its kernel is the set of pairs with all interior coordinates zero and shared endpoint coordinates $(-u,u)$ in $K_A$ and $(u,-u)$ in $K_B$, for $u\in E:=L_1\cap L_k$. Its image is $\ker \delta$, where $\delta :K\to V\mathrm{/}(L_1+L_k),\delta (v)={\sum }_{i=1}^k{v}_i<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>(L_1+L_k)\mathrm{.}$ Indeed, the prefix sum of $s(a,b)$ is $b_1+b_k\in L_1+L_k$, because ${\sum }_i{a}_i=0$. Conversely, if the prefix $p={\sum }_{i=1}^k{v}_i$ equals $u_1+u_k$ with $u_1\in L_1,u_k\in L_k$, set $b_1=u_1,b_k=u_k$, $a_1=v_1-b_1$, $a_k=v_k-b_k$, and use $a_i=v_i$ for $1<i<k$, $b_i=v_i$ for $i>k$; then both shorter lists sum to zero and $s(a,b)=v$.

Symmetry follows from $a+b+c=a^{\mathrm{\prime }}+b^{\mathrm{\prime }}+c^{\mathrm{\prime }}=0$ and isotropy: expanding $0=\omega (a,a^{\mathrm{\prime }})=\omega (-b-c,-b^{\mathrm{\prime }}-c^{\mathrm{\prime }})$ gives $\omega (c,b^{\mathrm{\prime }})=\omega (c^{\mathrm{\prime }},b)$, and then $\omega (a,b^{\mathrm{\prime }})=\omega (a^{\mathrm{\prime }},b)$. Thus ${\psi }_{123}$ is a nondegenerate symmetric form and its signature is defined.

In finite-dimensional symplectic $V$, the map $V\to U^{\ast }$ given by $v\mapsto \omega (v,-){\mathrm{\mid }}_U$ is onto for every subspace $U$: extend a functional on $U$ to a basis of $V$ and use the isomorphism $V\to V^{\ast }$ defined by $\omega$. Its kernel is $U^{\perp }$, so $\dim U^{\perp }=\dim V-\dim U$. The inclusion $A^{\perp }+B^{\perp }\subseteq (A\cap B{)}^{\perp }$ is equality because both sides then have the same dimension. In particular $(L_1\cap L_k{)}^{\perp }=L_1+L_k$. Embed $E$ into $K$ by $r(u)=(-u,0,\dots ,0,u,0,\dots ,0)$, with entries at positions $1$ and $k$, and let $J=r(E)$. The partial-sum formula gives $q(r(u),w)=\omega \text{\hspace{0.17em}⁣}(u,{\sum }_{i=1}^k{w}_i),$ since the omitted $w_k$ term pairs to zero with $u\in L_k$. Thus $J$ is isotropic and $J^{\perp }=\ker \delta =\mathrm{im}s$, because $(L_1\cap L_k{)}^{\perp }=L_1+L_k$ by [F1]–[F2]. Put $\bar{K}=K\mathrm{/}\mathrm{rad}(q)$ and let $\bar{J}$ be the image of $J$ in $\bar{K}$. The map induced by $s$ onto ${\bar{J}}^{\perp }\mathrm{/}\bar{J}$ is surjective because $\mathrm{im}s=J^{\perp }$. Its kernel is radical by preservation of the form. Conversely, a source vector in the radical maps to a vector orthogonal to every element of ${\bar{J}}^{\perp }$; nondegeneracy of $\bar{K}$ makes that orthogonal complement $\bar{J}$, so the vector is in the kernel. Hence the nondegenerate quotient of $K_A\oplus K_B$ is isometric to ${\bar{J}}^{\perp }\mathrm{/}\bar{J}$.

For $n=3$, the partial-sum formula of step 1.2 gives $q(v,w)=\omega (v_2,w_1)+\omega (v_3,w_1)+\omega (v_3,w_2)=\omega (v_3,w_2)$, since $v_2+v_3=-v_1$ and $v_1,w_1\in L_1$. Since $v_3=-v_1-v_2$ and $v_2,w_2\in L_2$, this is $-\omega (v_1,w_2)=-{\psi }_{123}([v_1],[w_1])$. The projection $K\to U_{123}$, $(v_1,v_2,v_3)\mapsto [v_1]$, is onto. Its kernel is the image of the adjacent-intersection difference map $((u_{12},u_{23},u_{31}))⟼(u_{12}-u_{31},u_{23}-u_{12},u_{31}-u_{23})\mathrm{.}$ For the reverse inclusion, if $v_1=u_{12}+u_{13}$ represents zero, set $u_{31}=-u_{13}$ and $u_{23}=v_2+u_{12}$; then $u_{23}\in L_2\cap L_3$ by $v_1+v_2+v_3=0$, and the displayed difference is $v$. The nondegeneracy in step 2.1 now shows that this kernel is exactly $\mathrm{rad}(q)$. Consequently $\tau (L_1,L_2,L_3)=-\mu (L_1,L_2,L_3)$; this is the sign translation from the cyclic convention to Walker's Wall-form convention.

If $Q$ is a nondegenerate real symmetric form and $I$ is isotropic, choose a basis $e_1,\dots ,e_r$ of $I$ and, by nondegeneracy, vectors $f_j$ with $Q(e_i,f_j)={\delta }_{ij}$. Replace $f_j$ by $f_j-\frac{1}{2}{\sum }_{\mathrm{\ell }}Q(f_{\mathrm{\ell }},f_j)e_{\mathrm{\ell }}$; symmetry makes all pairings among the modified $f_j$ vanish, so their span with $I$ is hyperbolic of signature zero. Its orthogonal complement identifies with $I^{\perp }\mathrm{/}I$. Therefore $\mathrm{sign}(I^{\perp }\mathrm{/}I)=\mathrm{sign}(Q)$. Applying this to step 3.2 proves $\tau (L_1,\dots ,L_n)=\tau (L_1,\dots ,L_k)+\tau (L_1,L_k,\dots ,L_n)\mathrm{.}$

Reindexing a cyclic list identifies its sum-zero space with the same space and changes each chosen antiderivative only by a common constant; step 1.2 shows that this does not change $q$. Hence $\tau$ is cyclically invariant. Apply the chain identity of step 4.1 to $(L_1,L_2,L_3,L_4)$ and to its cyclic rotation $(L_2,L_3,L_4,L_1)$. Thus ${\tau }_{123}+{\tau }_{134}={\tau }_{1234}={\tau }_{234}+{\tau }_{124}\mathrm{.}$ Multiplying by $-1$ and using step 3.3 gives the asserted four-Lagrangian identity for $\mu$.