Wall nonadditivity, RT cap references, and the Walker weight sign

For any symmetric finite-dimensional real form $q$ on $E$ and subspace $V$, one has $\mathrm{sign}(E)=\mathrm{sign}(q{\mathrm{\mid }}_V)+\mathrm{sign}(q{\mathrm{\mid }}_{V^{\perp }})$, allowing all restrictions to be degenerate. Split off a nondegenerate complement to the radical of $q{\mathrm{\mid }}_V$, reducing to an isotropic $V$, and discard the ambient radical without changing signatures. For an isotropic basis $v_i$, choose dual vectors $u_j$ with $q(v_i,u_j)={\delta }_{ij}$ and replace $u_i$ by $u_i-\frac{1}{2}{\sum }_{j}q(u_i,u_j)v_j$. The resulting pair span has matrix $\left[\begin{array}{cc}0& I\\ I& 0\end{array}\right]$ and signature0. Its orthogonal complement is the nondegenerate reduction of $V^{\perp }\mathrm{/}V$, so its signature is that of $V^{\perp }$. If $V$ met the discarded ambient radical, first discard those basis vectors. Reintroducing the first nondegenerate orthogonal summand proves the asserted formula.

A surgery trace $W_L$ is $B^4$ with framed two-handles. A handle core capped by a Seifert surface has self-intersection equal to its framing; distinct capped cores intersect by the oriented linking number, counted by pushing their Seifert surfaces normally. Thus its intersection matrix is the actual linking matrix $B_L$, and $\sigma (W_L)=\sigma (B_L)$ even when singular. To identify the stacked trace cut, let the separating sphere bound a three-ball $P\subset B^4$ and take its collar $P\times I_t$. The old $L_1,L_2$ handles lie on their respective side hemispheres. Each new joined-handle surgery circle crosses the equator twice; parametrize its new handle as $h_r=D^2\times D^2\cong I_s\times I_t\times D_z^2$. Only its two opposite attaching patches $\{s=\pm 1\}\times I_t\times D_z^2$ meet the middle collar, along two disjoint $D^2\times I_t$ disk products. Its other attaching patches $I_s\times \{t=-1\}\times D_z^2$ and $I_s\times \{t=+1\}\times D_z^2$ lie in the two side hemispheres. The zero framing is exactly the matching of these $z$ coordinates without a rotation; it is available from the explicit paired disk-product presentation in [F4]. Round only the small product corners. This specifies which portions of each original attaching region lie in each piece.

Put $E=H_2(W)$ with its absolute intersection form and $V=\mathrm{im}[H_2(W_1)\oplus H_2(W_2)\to H_2(W)]$. Interior cycles in different pieces are disjoint, and the restriction within each is its original intersection form. Thus the pullback of the form on $V$ is the direct sum of the forms of $W_1,W_2$. Its inclusion kernel is radical because its image intersects every cycle trivially; quotienting by it changes no signature. Hence $\mathrm{sign}(q_W{\mathrm{\mid }}_V)=\sigma (W_1)+\sigma (W_2)$, and step 1.1 reduces the entire defect to the form on $V^{\perp }$.

Let $C$ be the union of $P\times I_t$ and those new handles with only their $s=\pm 1$ patches attached to that collar. Relative to $C$, each handle is $I_s\times (I_t\times D_z^2)=D^1\times D^3$, attached along $S^0\times D^3$. It is therefore a four-dimensional one-handle relative to the middle collar, although it was a two-handle relative to the original whole attaching circle. All disjoint disk products match, giving the literal product $C=[P\cup ({⨆}_r{I}_s\times D_z^2)]\times I_t=U_G\times I_t$. In particular $H_1(C)={\mathbb{R}}^G$, $H_2(C)=0$, $\sigma (C)=0$. The two side intersections are exactly $(P\times \{\pm 1\})\cup ({⨆}_r{I}_s\times \{\pm 1\}\times D_z^2)\cong U_G$. The complementary side pieces are the original four-ball halves with their old $L_i$ handles: rounding the product indentations changes only boundary collars; no whole new handle remains on either side. Thus the actual decomposition is $W_{L_{\mathrm{c}\mathrm{o}\mathrm{m}\mathrm{p}}}=W_{L_1}{\cup }_{U_G}(U_G\times I){\cup }_{U_G}{W}_{L_2}$ with these exact faces. This avoids the false ordinary-handle claim: $B^4$ with $G$ whole attaching-circle two-handles instead has $H_2$ of rank$G$ and is not this middle product.

Use smooth singular chains subordinate to product collars, or a finite smooth subdivision respecting all faces. Excision gives $H_2(W,M_0)=H_2(W_1,M_0)\oplus H_2(W_2,M_0)$. For $z\in H_2(W)$ let $z_i$ be its relative image. Its intersection with any $y\in H_2(W_i)$ is the evaluation of the image of $z_i$ in $H_2(W_i,\mathrm{\partial }{W}_i)$ on $y$: all transverse intersections are in the interior of $W_i$, so enlarging the allowed relative boundary from $M_0$ to $\mathrm{\partial }{W}_i$ does not alter them. Perfection in [F1] therefore gives $z\in V^{\perp }$ exactly when both those relative images vanish. The triple sequence and excision contain $H_2(M_i,\mathrm{\Sigma })\to H_2(W_i,M_0)\to H_2(W_i,\mathrm{\partial }{W}_i)$. Hence for such a $z$ choose $s_i\in H_2(M_i,\mathrm{\Sigma })$ mapping to $z_i$, and write $a_i=\mathrm{\partial }{s}_i\in K_i$. The boundary of $z$ in the sequence of $(W,M_0)$ is zero, so $a_1+a_2$ maps to zero in $H_1(M_0)$. Thus $a_0=-a_1-a_2\in K_0$ also bounds a relative chain in $M_0$.

Conversely take a triple $(a_1,a_0,a_2)\in K_1\oplus K_0\oplus K_2$ with sum0. Choose actual smooth singular one-cycle representatives for $a_1,a_2$, define $a_0=-a_1-a_2$ at the chain level, and choose bounding two-chains $S_i\subset M_i$. Their boundaries may be made those exact one-cycles by adding their homology corrections in face collars. Thus their sum is a closed two-cycle with product collars and no unaccounted central two-chain in $\mathrm{\Sigma }$. Push its faces slightly into $W$, obtaining $z(a)\in V^{\perp }$ by step 3.1. Different bounding choices in $M_1,M_2$ change it by boundary classes and hence globally radical classes; choices in $M_0$ change it by classes in $V\cap V^{\perp }$, radical on $V^{\perp }$. For an arbitrary original $z$ of step 3.1, the constructed $z(a)$ has the same image in $H_2(W,M_0)$, so $z-z(a)$ is in $\mathrm{im}{H}_2(M_0)\subset V$ and in $V^{\perp }$, again radical. The triple construction therefore surjects onto the nondegenerate reduction of $V^{\perp }$.

Near $\mathrm{\Sigma }$ choose $u\ge 0$, $v\in \mathbb{R}$ and $p\in \mathrm{\Sigma }$, with $\mathrm{or}(W)=du\wedge dv\wedge \mathrm{or}(\mathrm{\Sigma })$, $W_1=\{v\le 0\}$, $W_2=\{v\ge 0\}$, $M_0=\{v=0\}$, $M_1=\{u=0,v\le 0\}$ and $M_2=\{u=0,v\ge 0\}$. The $M_0$ face of $W_1$ has orientation $-du\wedge \mathrm{or}\mathrm{\Sigma }$; its outward boundary vector at $u=0$ is $-{\mathrm{\partial }}_u$ and gives precisely $\mathrm{or}\mathrm{\Sigma }$. The outer face of $W_1$ is $-dv\wedge \mathrm{or}\mathrm{\Sigma }$, so the convention $\mathrm{\partial }{W}_1=M_0-M_1$ gives $\mathrm{or}(M_1)=dv\wedge \mathrm{or}\mathrm{\Sigma }$ and its boundary has that same $\mathrm{\Sigma }$ orientation. Similarly $\mathrm{or}(M_2)=-dv\wedge \mathrm{or}\mathrm{\Sigma }$ and its boundary agrees. These coordinates realize exactly the Statement's signs. A bounding strip along a normal ray coordinate $t$ pointing away from the corner has orientation $-dt\wedge a_i$, giving boundary $a_i$ at $t=0$.

All outside-corner intersections can be removed explicitly. The three faces have disjoint tubular neighborhoods away from a fixed corner collar. For the first and second surfaces supported on the same face choose two distinct constant small normal levels into $W$: distinct positive $u$ levels for $M_1,M_2$, and distinct $v$ levels for the seam $M_0$. Even if their underlying chains in a three-face meet along curves, their normal levels separate them. Different faces remain in disjoint neighborhoods. In the annular transition to the corner normal fan, interpolate the two same-face levels to two ordered endpoints on the boundary of that face's angular sector. Linear interpolation preserves their order; the three sectors are disjoint there. Hence no intersections occur in the transition annulus either. For both triple chains use the exact chain equality $a_1+a_0+a_2=0$ of step 4.1, so no extra central $\mathrm{\Sigma }$ surface is hidden in this construction. The perturbations are realized on the finitely many smooth chain pieces with product collars; bilinear intersection counts add over these pieces, so the argument applies even when a representing singular chain has multiple components or self-intersections.

Move the second triple's junction in the normal half-plane to $(ϵ,ϵ)$, leaving its negative-$v$ ray directed from that point toward the negative-$v$ boundary ray, its positive-$v$ ray toward the positive boundary ray, and its seam ray toward increasing $u$. Choose the already ordered transition paths of step 6.1. In the central disk the only crossing is then between the first seam ray and the second negative-$v$ ray. The first tangent is $+{\mathrm{\partial }}_u$ and the second points toward decreasing $v$, so their determinant in the ordered normal plane $({\mathrm{\partial }}_u,{\mathrm{\partial }}_v)$ is negative. The two oriented surface tangent bases are $(-t_0,a_0,-t_1^{\mathrm{\prime }},a_1^{\mathrm{\prime }})$. Moving the first surface's $\mathrm{\Sigma }$ tangent past the second surface's normal tangent gives one minus; the two displayed prefixed minus signs cancel. Thus the four-dimensional sign is minus the normal crossing sign times the $\mathrm{\Sigma }$ crossing sign, yielding $q_W(z(a),z(a^{\mathrm{\prime }}))=\omega (a_0,a_1^{\mathrm{\prime }})=-\omega (a_1,a_0^{\mathrm{\prime }})$. The last equality is symmetry of the Walker form for order $(K_1,K_0,K_2)$, using isotropy and the sum-zero equations. step 6.1 excluded every other intersection. Therefore the pullback remaining form is $-{\psi }_W$. The surjection of step 4.1 carries radicals to radicals, and its kernel is radical because the target pairing pulls back to this form. Its nondegenerate reduction is consequently isometric to that of $V^{\perp }$. Combining this with step 1.1, step 2.1 proves $\sigma (W)=\sigma (W_1)+\sigma (W_2)-{\mu }_W(K_1,K_0,K_2)$. Disconnected faces give orthogonal sums throughout.

Absorb $U_G\times I$ into the first trace by a boundary collar extension, preserving its other cap faces and signature. Apply step 7.1 to its remaining gluing face with the second trace. For $r>1$ middle components, first cut $P$ along the $r-1$ disjoint separating disk products of [F4]'s family-meridian model. They separate precisely the specified arc families; their thickened compressions split the middle product into the disjoint products $U_{g_q}\times I_q$, with all other side pieces unchanged. The redundant final family-meridian in the alternative all-$r$ graph is a separated zero-framed trace handle with matrix$[0]$ and signature0; omit it for the actual cut. Thus the same argument applies componentwise, and the incidence-component connected-sum reduction in [F4] adds only zero-signature trace blocks.

The corner surface orientation in step 5.1 is opposite the canonical middle object orientation: it is induced from the incoming face of $M_2$, and $\mathrm{\partial }{M}_2$ contains the negative of that object. The corner kernels identified with its canonical $H_1$ are $K_1=R_1({\alpha }_0)$, $K_0={\alpha }_1$, $K_2=R_2^t({\alpha }_2)$. Indeed Mayer–Vietoris after capping the incoming boundary of $M_1$ says that a middle class dies exactly when it is homologous in $M_1$ to an element of the cap kernel ${\alpha }_0$; this is precisely $R_1{\alpha }_0$. The identical outgoing-cap argument gives $K_2$, and the common actual handlebody has kernel ${\alpha }_1$ by its definition. These use the individual bordisms, not inclusion maps into their final composite. Negating the surface orientation negates the entire Walker form. step 7.1 with the actual trace cut of step 1.2, step 2.2, step 8.1 therefore gives $\mathrm{\Delta }\sigma =+{\mu }_W(R_1{\alpha }_0,{\alpha }_1,R_2^t{\alpha }_2)={\mu }_{\alpha }$. Substituting into [F4] gives $Z_{\alpha }(M_2{M}_1)={\kappa }^{-{\mu }_{\alpha }}{Z}_{\alpha }(M_2)Z_{\alpha }(M_1)$.

This sign can also be checked directly against Turaev's printed convention: his form on $a=a_1+a_2\in L_3$ is $h(a,b)=\omega (a_2,b)=\omega (a_2,a_1^{\mathrm{\prime }})=-\omega (a_1^{\mathrm{\prime }},a_2)=-{\psi }_W(a^{\mathrm{\prime }},a)$. Symmetry gives ${\mu }_T=-{\mu }_W$. His surgery parameter is $\mathrm{\Delta }=p_{-}$, so $D\mathrm{/}\mathrm{\Delta }=\kappa$ and his anomaly ${\kappa }^{{\mu }_T}$ is exactly ${\kappa }^{-{\mu }_W}$. This is a sign comparison only; the local Wall proof above does not cite that source theorem as its proof. In particular preserving the prescribed composition $n_{\mathrm{c}\mathrm{o}\mathrm{m}\mathrm{p}}=n_1+n_2+{\mu }_W$ requires weight action ${\kappa }^{+n}$, since ${\kappa }^{-n}$ would duplicate instead of cancel the computed anomaly.

No arbitrary real $\lambda$ needs realization by a geometric handlebody. Keep the actual reference cap meridians $\alpha$ and set $w(R)={\mu }_W({\lambda }_0,{\alpha }_0,R^t{\alpha }_1)+{\mu }_W(R{\lambda }_0,{\alpha }_1,{\lambda }_1)$, $a(R)=-w(R)$. These spaces are Lagrangian by [F3], even for noninjective relations. To compute its coboundary use [F3]'s proved associative category of weighted relations. The diagonal weight-zero relation $J:(H,\lambda )\to (H,\alpha )$ has the weight-zero diagonal inverse: their composition corrections are $\mu (\lambda ,\alpha ,\lambda )=0$. For a relation $R$ between the reference objects with weight0, the arrow $J_{\mathrm{o}\mathrm{u}\mathrm{t}}^{-1}\circ R\circ J_{\mathrm{i}\mathrm{n}}$ between the $\lambda$ objects has exactly weight $w(R)$, by the two displayed middle-space corrections. For composable $A,B$, cancelling adjacent $J$ arrows calculates its weight as ${\mu }_{\alpha }+w(BA)$; composing the two $\lambda$ arrows calculates it as $w(B)+w(A)+{\mu }_{\lambda }$. Associativity equates these actual weights. Therefore $a(BA)-a(B)-a(A)={\mu }_{\alpha }-{\mu }_{\lambda }$, without defining an integer by the desired equality.

Define $Z(M,n;\lambda )={\kappa }^{n+a(R_M)}{Z}_{\alpha }(M)$. The exponent in its composite is $n_2+n_1+{\mu }_{\lambda }+a(BA)-{\mu }_{\alpha }=n_2+a(B)+n_1+a(A)$ by step 10.2, so step 9.1 proves strict weighted composition. The correction vanishes when $\lambda =\alpha$, and for the identity relation it is $-\mu (\lambda ,\alpha ,\alpha )-\mu (\lambda ,\alpha ,\lambda )=0$. The cylinder and tensor/empty laws of [F4] therefore persist; direct sums add both Maslov terms and $a$. A weight increase by1 acts by $\kappa$ with no anomaly-free assumption. Equivalently the connected block formula is ${\kappa }^{n+a(R_M)-\sigma (B)}{D}^{-m-1+\chi ({\mathrm{\Sigma }}_{\mathrm{o}\mathrm{u}\mathrm{t}})\mathrm{/}2}d(j)F_0(\mathrm{\Omega };i,j)$, with [F4]'s multiplicative disconnected version. This proves all stated signature, cap-reference and weight assertions while leaving genuine relative presentation/move completeness to its own theorem. Full AC is inherited precisely through [F1], and countable choice through [F4]; the new form reductions and bases are finite.