# Normal forms and the shift of a submanifold transform A submanifold transform can turn a hypersurface into a submanifold of higher codimension. Its cohomological shift records which fibre directions are integrated with compact support. We first obtain a local coordinate model from a clean conormal intersection, then compute every part of that shift. Positive, negative and null quadratic directions play different roles. The sheaf-operation inputs are the formal kernel-composition comparison, the direct-germ formula (MC.25), and proper-support base change. The fibre proof below uses the actual localization and trace maps specified at each step. We prove the clean cotangent reduction and bind the exact programme Morse lemma with parameters before applying them to the hypersurface. The smooth inverse, implicit, submersion and constant-rank theorems, differentiation of smooth parameter integrals, and finite-dimensional symmetric-form diagonalization are calculus prerequisites. We work over a commutative ring \(k\) of finite global dimension, and \(L\in D^b(k)\) may have arbitrary coefficient modules. *Written by GPT-6.1 Sol and GPT-6 Astra (OpenAI), Ultra, September–October 2026. Independently written exposition is public domain (CC0); cited human works retain their own terms.* ## Clean intersection supplies the coordinate model Let \(f:Y\to X\) be a submersion, \(N\subset Y\) a closed smooth hypersurface, and \(M\subset X\) a closed smooth submanifold. Embed \[ V=Y\times_X T^*X\subset T^*Y \] by the cotangent pullback \(f_d(y;\xi)=(y;df_y^t\xi)\), and put \(A=T_N^*Y\). Fix a nonzero \(p\in A\cap V\). Assume the intersection is clean near \(p\): it is a submanifold with tangent space \(T_pA\cap T_pV\). Assume also that the image under \(f_\pi:V\to T^*X\) is \(T_M^*X\) as a germ at \(f_\pi(p)\). The image hypothesis is part of the theorem; it does not follow from smoothness of \(f\) alone. Cleanliness will imply that the restricted map to this image is a submersion. Here is the rank calculation, including the possible excess directions. ## Clean cotangent reduction has constant rank Put \(m=\dim X\) and \(d=\dim(Y/X)\), so \(\dim Y=m+d\). In submersion coordinates \(f(x,t)=x\), write a covector on \(Y\) as \((\xi,\tau)\). Then \[ V=\{\tau=0\},\qquad f_\pi(x,t;\xi,0)=(x;\xi),\qquad \omega_Y|_V=f_\pi^*\omega_X. \] At a point \(z\in C=A\cap V\), let \(E=T_zT^*Y\), \(W=T_zV\), and \(\ell=T_zA\). The plane \(W\) is coisotropic: in these coordinates its symplectic orthogonal is the \(d\)-plane \[ K=W^\omega=\ker d(f_\pi|_V) =\operatorname{span}\{\partial_{t_j}\}. \] The plane \(\ell\) is Lagrangian. The map \(\ell\to K^*\), \(v\mapsto\omega_Y(v,\cdot)|_K\), has kernel \(\ell\cap K^\omega=\ell\cap W\). Its transpose has kernel \(K\cap\ell^\omega=K\cap\ell\). Writing \(e_z=\dim(K\cap\ell)\), equal ranks of a linear map and its transpose give \[ \dim(\ell\cap W)=m+d-(d-e_z)=m+e_z. \] Cleanliness says \(T_zC=\ell\cap W\). Its dimension is constant near the selected point, so \(e_z=\dim C-m\) is constant as well. The kernel of \(d(f_\pi|_C)\) is \(\ell\cap K\), and hence \[ \operatorname{rank}d(f_\pi|_C)=\dim C-e_z=m. \] The image tangent is isotropic: lift two of its vectors to \(T_zC\subset\ell\), use the displayed pullback identity, and use \(\omega_Y|_\ell=0\). Its dimension \(m\) makes it Lagrangian in \(T_{f_\pi(z)}T^*X\). Under the stated image-germ hypothesis the map has values in \(T_M^*X\), another \(m\)-dimensional manifold, and its derivative is onto that tangent space. The submersion theorem therefore gives a local submersion onto an open part of this conormal germ. This proves the required clean-reduction statement with excess fibre dimension \(\dim C-m\). It makes no assertion about a global image or a transverse intersection. \(\square\) This is the cotangent instance of Lagrangian reduction. The programme's Phase space and generating families, Proposition 2.3 also proves the corresponding linear statement for every Lagrangian and isotropic constraint. The argument above supplies the local rank and submersion conclusions used in this lesson. ## A nondegenerate critical family has fixed quadratic coordinates **Morse lemma with parameters.** Let \(h(a,t)\) be smooth near \((a_0,0)\), with \(a\) in a finite-dimensional parameter space and \(t\in\mathbb R^b\). Assume, for every nearby \(a\), \[ h(a,0)=0,\qquad d_t h(a,0)=0, \] and assume the Hessian \(d_t^2h(a_0,0)\) is nondegenerate. There is a parameter-preserving local coordinate change \((a,t)\mapsto(a,v)\), taking \(t=0\) to \(v=0\), for which \[ h(a,t)=\sum_{j=1}^{b_+}v_j^2 -\sum_{j=b_++1}^{b}v_j^2. \] The positive and negative dimensions are those of the original Hessian. If the critical point is instead a smooth section \(t=c(a)\), translate by that section; if its critical value is \(h(a,c(a))\), subtract that value before applying the lemma. **Proof from the exact programme prerequisite.** Lemma 4.1 of the written AN-04 lesson Stationary phase and critical manifolds proves a stronger parameter Morse statement: a smoothly varying nondegenerate critical point has coordinates satisfying \(h(F(z,a),a)=c(a)+z^TJz/2\), with fixed signs, a smooth inverse and the exact determinant normalization. Its proof constructs the critical section by the implicit function theorem, completes one square using the integrated second derivative, and proves the remaining Hessian is the invertible Schur complement before inducting. All parameter dependence and signs are retained. Apply that result with the fibre variables \(t\) and parameters \(a\). The critical-point uniqueness in its proof makes its critical section exactly our given \(t=0\); the assumed value makes \(c(a)=0\). Set \(v=z/\sqrt2\) in its final coordinates to obtain the displayed normalization without the factor \(1/2\). The same provider covers a moving critical section and a varying critical value, giving the final variant. For \(b=0\), the hypotheses themselves give \(h=0\). Thus every part of the stated lemma is supplied by an exact, written internal proof. The smooth-calculus prerequisites stated by that lesson remain prerequisites here. \(\square\) ## Coordinates adapted to the clean conormal image **Normal-form theorem.** There are coordinates \[ x=(x_1,x',x''),\qquad (x,t,u)\text{ on }Y, \quad x',u\in\mathbb R^r, \qquad\text{(1)} \] in which the selected base points are zero and \[ \begin{split} f(x,t,u)&=x, &p&=(0;dx_1),\\ M&=\{x_1=x'=0\}, &N&=\{x_1=q(t)+\langle x',u\rangle\}. \end{split} \qquad\text{(2)} \] Here \(q\) is a quadratic form, possibly degenerate, and \(\operatorname{codim}_XM=r+1\). If the conormal intersection with \(V\) is transverse, \(q\) is nondegenerate. **Proof.** Near the nonzero conormal \(p\), projection from \(A\cap V\) to \(Y\) has a one-dimensional kernel: the radial conormal direction. Cleanliness makes its rank constant. Its image is therefore a submanifold \(D\subset N\), and projectivizing the positive conormal ray identifies \(A\cap V\) locally with \(D\) times that ray. The clean-reduction prerequisite makes \(D\) a submersion over the corresponding projectivized piece of \(T_M^*X\). Choose coordinates on \(X\) with \(M=\{x_1=x'=0\}\) and the selected covector \(dx_1\). On the projectivized conormal chart use \(u=-\xi'/\xi_1\), together with \(x''\), as coordinates. Complete their pullbacks to coordinates \((x'',u,t'')\) on \(D\). The minus sign is fixed by the conormal of a graph \(x_1=g\): its covector has \(\xi'=-\xi_1\partial_{x'}g\). Because \(f\) is a submersion, the independent coordinates \(x_1,x'\) vanish on \(D\) and can be completed by coordinates \(t'\) vanishing there. Thus \((x_1,x',x'',t',t'',u)\) are coordinates on \(Y\), with \(D=\{x_1=x'=t'=0\}\). The selected conormal has nonzero \(dx_1\) component, so the hypersurface can be written \[ x_1=g(x',x'',t',t'',u) =\langle x',u\rangle+h(x',x'',t',t'',u). \qquad\text{(3)} \] On \(\{x'=t'=0\}\) we have \[ h=0,\qquad \partial_{x'}h=0, \qquad \partial_{t'}h=0. \qquad\text{(4)} \] The first comes from \(D\subset N\), the second from the definition of \(u\), and the third from the requirement that the conormal over \(D\) be a pulled-back covector. Derivatives along \(x'',t'',u\) of these identities vanish there as well. We claim the Hessian \(H=\partial_{t't'}^2h(0)\) is nondegenerate. Write fibre coordinates \((\xi,\tau',\tau'',v)\) dual to \((x,t',t'',u)\). The conormal of (3), with multiplier \(\xi_1\), has \(\tau'=-\xi_1\partial_{t'}h\), \(\tau''=-\xi_1\partial_{t''}h\), and \(v=-\xi_1(x'+\partial_uh)\). The tangent equations at the selected point, using (4), imply that membership in \(T_pA\cap T_pV\) forces \[ \delta x_1=\delta x'=0,\qquad \delta\tau'=\delta\tau''=\delta v=0, \qquad H\,\delta t'=0. \qquad\text{(5)} \] Conversely a vector in \(\ker H\) can be completed in the free \(\delta\xi'\) components to a tangent vector with these properties. Cleanliness says this intersection is the tangent of \(A\cap V\). Its projection lies in \(T_0D\), where \(\delta t'=0\). Thus \(\ker H=0\). Apply the Morse lemma with parameters \((x'',t'',u)\) to \(h(0,x'',t',t'',u)\). Its critical value is zero by (4), its critical point is \(t'=0\), and its Hessian is nondegenerate. A parameter-dependent change of \(t'\), extended independently of \(x'\), makes that restriction a fixed nondegenerate quadratic form \(q(t')\). Hadamard's lemma then gives \[ h=q(t')+\langle x',\phi(x',x'',t',t'',u)\rangle, \qquad \phi=0\text{ on }\{x'=t'=0\}. \qquad\text{(6)} \] Replace the old coordinate \(u\) by \(u_{\mathrm{new}}=u+\phi\). Its derivative in \(u\) on the displayed submanifold is the identity, so this is a coordinate change near the origin. It preserves \(f\) and turns (3) into (2), with \(t=(t',t'')\) and \(q\) zero on \(t''\). The dimension of \(A\cap V\) is \(\dim X+\dim t''\): projectivized conormal coordinates contribute \(\dim X-1\), the ray contributes one, and \(t''\) is the remaining fibre. A transverse intersection has the expected dimension \(\dim X\). Hence in the transverse case there are no \(t''\) directions, and \(q\) is nondegenerate. \(\square\) ## The closed side and the three quadratic dimensions Choose the closed upper side \[ N^+=\{x_1\geq q(t)+\langle x',u\rangle\}. \qquad\text{(7)} \] Diagonalize the form and write \[ t=(s_+,s_-,w),\qquad q(t)=|s_+|^2-|s_-|^2. \qquad\text{(8)} \] Let their dimensions be \(n_+,n_-,n_0\), respectively. The last is the dimension of the radical \(\ker q\). It is not the dimension of a maximal isotropic subspace of an indefinite nondegenerate form. Set \(d=\dim(Y/X)\) and \(c=\operatorname{codim}M=r+1\). Counting the fibre variables gives \[ d=n_++n_-+n_0+r, \qquad n_++n_-+n_0=d-c+1. \qquad\text{(9)} \] Fix the coordinate orientations for the \(s_-,w,u\) spaces. The compact-support generator, (M5)–(M6) starts with the increasing-interval boundary difference and takes ordered products. It gives \(R\Gamma_c(\mathbb R^a;L)=L[-a]\), with its actual normalization for arbitrary bounded \(L\). Here is the extension to the open convex fibres used below. Translate an interior point of a nonempty open convex \(O\subset\mathbb R^a\) to zero and choose a ball of radius \(\rho>0\) inside \(O\). Its gauge \[ p(v)=\inf\{t>0:v\in tO\} \] is finite, nonnegative and positively homogeneous. Convexity gives subadditivity, and the interior ball gives \(p(v)\leq\lVert v\rVert/\rho\). Applying these bounds to \(v-w\) and \(w-v\) proves \(|p(v)-p(w)|\leq\lVert v-w\rVert/\rho\). Thus \(p\) is continuous, and openness and convexity give \(O=\{p<1\}\). The maps \[ h(v)=\frac{v}{1+p(v)},\qquad h^{-1}(u)=\frac{u}{1-p(u)} \] are inverse homeomorphisms \(\mathbb R^a\leftrightarrow O\), by homogeneity. This includes unbounded directions where \(p=0\). Near zero, \(h(v)=v+O(\lVert v\rVert^2)\), so it preserves the positive local orientation class. Transporting the compact generator by \(h\) gives \(R\Gamma_c(O;L)=L[-a]\) with that orientation. The composition and open-extension identity for trace, (M27) says that extending a compact class from one such open convex set to another preserves its integral. Both oriented integrals are isomorphisms, so their extension map is the identity under the same \(L[-a]\) identification. The dual-generator normalization, (O9)–(O10) fixes its sign, and projection with arbitrary bounded coefficients carries the generator and the trace to \(L\). Closed balls instead have ordinary cohomology \(L\) by constant-coefficient convex acyclicity and are compact. The zero-dimensional ball is a point. No flatness or finite-generation condition on \(L\), or global orientability of \(Y\), is assumed. ## Four fibre integrations determine the shift For small \(\epsilon>0\), choose \(\eta>0\) so that \(\eta(1+\epsilon)<\epsilon\). Consider the open neighborhood \[ U_{\eta,\epsilon}= \{|x|<\eta,\ |w|<\epsilon,\ |u|<\epsilon, \ |s_+|^2<\epsilon,\ |s_-|^2<\epsilon\}. \qquad\text{(10)} \] Shrinking these parameters gives a neighborhood system at zero. Factor \(f\) into projections which forget \(s_-\), then \(s_+\), then \(w\), then \(u\). At each step the derived proper-image fibre formula and composition of proper images calculate the fibre. We also identify the extension across its parameter boundaries, so that a stalk calculation alone is not asked to determine a sheaf. For the first projection put \(a=|s_+|^2+\langle x',u\rangle-x_1\). Its fibre is \[ \{s_-:a\leq|s_-|^2<\epsilon\}. \qquad\text{(11)} \] For \(a\leq0\) this is an open ball, with compact cohomology \(L[-n_-]\). For \(a\geq\epsilon\) it is empty. When \(0x_1\}} \longrightarrow L_{\{|u|<\epsilon\}} \longrightarrow L_{\{|u|<\epsilon,\ \langle x',u\rangle\leq x_1\}} \xrightarrow{+1}. \qquad\text{(12)} \] The first fibre is empty when \(x_1\geq\epsilon|x'|\). Otherwise it is a nonempty open convex set; its compact-cohomology map to the whole open ball is the orientation trace isomorphism. Thus the third term has compact cohomology zero outside that closed parameter condition, and \(L[-r]\) on it. The first term's convex-set trace identifies its direct image with the open-extension coefficient on \(\{x_1<\epsilon|x'|\}\), shifted by \([-r]\). Triangle (12) identifies the third image with the closed-extension coefficient on its complement. Therefore, as a sheaf calculation on \(|x|<\eta\), all four integrations give \[ Rf_!L_{N^+\cap U_{\eta,\epsilon}} \simeq L_{\{x_1\geq\epsilon|x'|,\ |x|<\eta\}}[\delta], \qquad \delta=-n_--n_0-r=n_+-d. \qquad\text{(13)} \] The support and the degree can be checked independently at each stage. In this table, the base restriction \(|x|<\eta\) and the cutoffs on variables not yet integrated are understood. | Variable forgotten | Fibre and comparison which select the support | Surviving closed condition | Additional shift | |---|---|---|---| | \(s_-\) | The open ball survives when \(a\leq0\); for \(00\}\), and let \(B_\epsilon\) be the product of the four open fibre balls in (10). Work over the interior target ball \(|x|<\eta\). There, \[ L_{O\cap U_{\eta,\epsilon}} =L_O\otimes^L k_{B_\epsilon}. \] By the open-extension boundary estimate, the nonzero boundary covectors of \(L_O\) have the form \(a\,dF\) with \(a<0\), hence negative \(dx_1\) coefficient. All covectors of the independent fibre cutoff \(k_{B_\epsilon}\) have zero \(dx_1\) coefficient. The tensor is noncharacteristic: a cancellation \(a\,dF+b=0\) forces \(a=0\) from that component and then \(b=0\). The tensor estimate (MO21) consequently puts every microsupport covector of \(L_{O\cap U_{\eta,\epsilon}}\) in the half-space with nonpositive \(dx_1\) coefficient, including all fibre faces and corners. Choose the closures of the fibre balls inside the coordinate chart. The closed support is then proper over the retained base ball. The proper direct-image estimate (MO8) gives the same nonpositive coefficient bound for \(Rf_!L_{O\cap U_{\eta,\epsilon}}\), so it is invisible at \((0;dx_1)\). Applying \(Rf_!\) to the ordinary localization triangle \[ L_{O\cap U_{\eta,\epsilon}} \longrightarrow L_{N^+\cap U_{\eta,\epsilon}} \longrightarrow L_{N\cap U_{\eta,\epsilon}} \xrightarrow{+1} \] makes this actual boundary restriction an isomorphism at the selected positive covector. These arrows commute with compact-support extension between nested neighborhoods, so they give the formal comparison with the same shift as (15). This verifies the comparison directly for every clean excess dimension. ## Expressing the shift by three Lagrangian planes Use the cotangent symplectic form \(\omega=d\theta\), so in linear coordinates \(\omega((x;\xi),(y;\eta))=\langle\xi,y\rangle-\langle\eta,x\rangle\). For three Lagrangian planes define their inertia index as the signature of \[ Q(v_0,v_N,v_1)= \omega(v_0,v_N)+\omega(v_N,v_1)+\omega(v_1,v_0). \qquad\text{(16)} \] ### The ordered index: degeneracy, parity and the cocycle Here is a proof of the structural identities needed in the next lessons. Let \(E\) be a real symplectic vector space of dimension \(2n\), with the convention above, and let \(\lambda_1,\lambda_2,\lambda_3\) be Lagrangian planes. There is a Lagrangian complement transverse to any prescribed finite collection of Lagrangian planes: AN-04, Gaussian lines and invariant symbols, Proposition 1.2 proves this using the positive-codimension intersection strata in symmetric graph charts. Choose this common complement as the vertical plane and a complementary Lagrangian as horizontal. Then every \(\lambda_i\) is \(p=A_iq\), where \(A_i\) is symmetric. Formula (16) becomes \[ Q=q_1^T(A_1-A_2)q_2+q_2^T(A_2-A_3)q_3 +q_3^T(A_3-A_1)q_1. \] Make the invertible change \[ q_1=x+z,\quad q_2=x+y,\quad q_3=y+z; \qquad x=\tfrac12(q_1+q_2-q_3),\quad y=\tfrac12(-q_1+q_2+q_3),\quad z=\tfrac12(q_1-q_2+q_3). \] Expanding and using \((A_1-A_2)+(A_2-A_3)+(A_3-A_1)=0\) cancels all mixed terms. Consequently \[ Q=x^T(A_1-A_2)x+y^T(A_2-A_3)y+z^T(A_3-A_1)z, \qquad \tau(\lambda_1,\lambda_2,\lambda_3) =\operatorname{sgn}(A_1-A_2)+\operatorname{sgn}(A_2-A_3) +\operatorname{sgn}(A_3-A_1). \] No difference matrix was inverted, so this proof includes degenerate intersections. A cyclic permutation preserves \(Q\); an odd permutation negates it. Reversing \(\omega\) negates \(\tau\), and symplectic direct sums add the signatures. For four planes choose one common chart. Cancelling the six ordered matrix differences in the displayed signature formula gives the full four-plane identity \[ \tau(\lambda_1,\lambda_2,\lambda_3) +\tau(\lambda_1,\lambda_3,\lambda_4) =\tau(\lambda_1,\lambda_2,\lambda_4) +\tau(\lambda_2,\lambda_3,\lambda_4). \] For completeness, the radical has an intrinsic description. Polarizing (16) says that a triple lies in the radical precisely when \(v_2-v_3\in\lambda_1\), \(v_3-v_1\in\lambda_2\), and \(v_1-v_2\in\lambda_3\). The assignments \[ a_{12}=\tfrac12(v_1+v_2-v_3),\quad a_{23}=\tfrac12(-v_1+v_2+v_3),\quad a_{31}=\tfrac12(v_1-v_2+v_3) \] put \(a_{ij}\) in \(\lambda_i\cap\lambda_j\). Their inverse is \(v_1=a_{12}+a_{31}\), \(v_2=a_{12}+a_{23}\), \(v_3=a_{23}+a_{31}\). Thus the radical is isomorphic to the direct sum of the three pairwise intersections. The rank is \(3n-\sum_{i0\); alternation gives the same rule when \(I\) is contained in either other pair. This proves reduction for a constraint contained in one pair of arguments. The following argument proves the stronger rule for the sum of the pairwise intersections. **Reduction by a subspace of the sum of the pairwise intersections.** Set \(J_{ij}=\lambda_i\cap\lambda_j\) and \(J=J_{12}+J_{23}+J_{31}\). The sum \(J\) is isotropic: each summand is isotropic, and any two summands lie in one common Lagrangian. We claim that every \(I\subset J\) can be reduced without changing \(\tau\). First record reduction in stages. If \(I\subset J\) are isotropic and \(E_I=I^\omega/I\), then \[ (J/I)^{\omega_{E_I}}=J^\omega/I, \qquad (E_I)_{J/I}=J^\omega/J=E_J. \] The reduced planes agree as well. Since \(I\subset J^\omega\subset I^\omega\), \[ \bigl((\lambda\cap I^\omega)+I\bigr)\cap J^\omega =(\lambda\cap J^\omega)+I. \] To check this equality, write a vector on the left as \(l+i\). Both \(l+i\) and \(i\) belong to \(J^\omega\), hence \(l\) does too. The reverse inclusion is immediate. Adding \(J\) and quotienting by \(J\) gives the plane \(((\lambda\cap J^\omega)+J)/J\). Thus these are canonical identifications of the quotient spaces and their three reduced planes. Now reduce successively by \(J_{12}\), the image of \(J_{23}\), and the image of \(J_{31}\). Each constraint is still in the indicated pairwise intersection: all the original \(J_{ij}\) lie in the isotropic \(J\), so they are orthogonal to the constraints already removed. The common-pair proof applies at every stage, and reduction in stages identifies the final triple with its reduction in \(E_J\). Its index equals the original one. For an arbitrary \(I\subset J\), the images \((J_{ij}+I)/I\) in \(E_I\) belong to the corresponding pairwise intersections and span \(J/I\). Apply the same three common-pair reductions to this triple. Its final quotient and planes are again those in \(E_J\). Comparing the two routes proves \[ \tau_E(\lambda_1,\lambda_2,\lambda_3) =\tau_{E_I}\bigl((\lambda_1)_I,(\lambda_2)_I,(\lambda_3)_I\bigr), \qquad I\subset (\lambda_1\cap\lambda_2)+(\lambda_2\cap\lambda_3)+(\lambda_3\cap\lambda_1). \] No complement to the pairwise intersections and no transversality among the three planes has been assumed. The restriction \(I\subset J\) is essential to this argument. \(\square\) ### The submanifold transform's ordered triple The order matters. In \(E=T_pT^*Y\) take \(\lambda_0=T_p\pi_Y^{-1}(0)\), \(\lambda_N=T_pT_N^*Y\), and \(\lambda_1=T_p f_\pi^{-1}(\pi_X^{-1}(0))\), with the last inverse image taken inside \(V\). Set \(\tau=\tau_E(\lambda_0,\lambda_N,\lambda_1)\). Write \(q(t)=\tfrac12\langle At,t\rangle\), with \(A\) symmetric, in the normal form. Its tangent conormal equations are \(x_1=\xi''=0\), \(\xi'=-u\), \(\tau_t=-At\), \(v=-x'\). The vertical plane has \(x=t=u=0\), while \(\lambda_1\) has \(x=\tau_t=v=0\). We calculate the signature directly in these coordinates. The fibre-variable contribution is the triple in \((t;\tau_t)\) \[ \{t=0\},\qquad \{\tau_t=-At\}, \qquad \{\tau_t=0\}. \qquad\text{(17)} \] For the \(x_1\) coordinate all three planes are vertical, so its form is zero. In the \(x''\) block the planes are vertical, horizontal, vertical; its form is \((a-c)^Tb\). This is a hyperbolic pairing plus a radical and has signature zero. In the coupled \((x',u;\xi',v)\) block, take a vertical vector with momenta \((a,b)\), a conormal vector with bases \((c,d)\) and momenta \((-d,-c)\), and a third-plane vector with bases \((0,e)\) and momenta \((f,0)\). Formula (16) becomes \[ a^Tc+b^Td-c^Te-f^Tc-b^Te =c^T(a-f-e)+b^T(d-e). \] The change of variables \((a,d)\mapsto(a-f-e,d-e)\), retaining \((b,c,e,f)\), is invertible. The resulting two hyperbolic pairings have zero signature, and the remaining variables are radical. Finally, on (17), with a vertical vector \(a\), a graph vector \((b,-Ab)\), and a horizontal vector \(c\), the form is \[ a^T(b-c)-b^TAc =(a+Ab)^Tz-b^TAb,\qquad z=b-c. \] The change \((a,b,c)\mapsto(a+Ab,b,z)\) is invertible even when \(A\) is singular. Its hyperbolic pairing has signature zero, leaving the signature of \(-A\). The blocks are independent, so \(\tau=-\operatorname{sgn}A=-(n_+-n_-)\). This proves the complete ordered calculation, including null directions. The graph sign in (17) fixes the minus sign. Furthermore \(\dim((T_pV)^\perp\cap\lambda_N)=n_0\), since that intersection imposes \(u=0\) and \(At=0\) on the remaining fibre-base vector. Substitute these identities and (9) into (15). We obtain \[ \delta=\frac12\bigl[ 1+\dim M-\dim Y -\dim((T_pV)^\perp\cap\lambda_N)-\tau \bigr]. \qquad\text{(18)} \] Thus the half-integer-looking expression is an integer. Indeed its bracket equals \(-2(n_-+n_0+r)\). Formula (18) is a coordinate-independent description of the same local shift once the symplectic and index conventions are fixed. ## Exercises with complete solutions ### Count the integrated directions *Difficulty: Introductory.* Let \(q\) have two positive, three negative and one null direction, and let \(\operatorname{codim}M=3\). Find \(d\) and \(\delta\) in both forms of (15). **Solution.** Here \(r=2\), so \(d=2+3+1+2=8\). Counting compactly integrated negative, null and \(u\) directions gives \(\delta=-3-1-2=-6\). The alternative expression is \(n_+-d=2-8=-6\). The positive closed balls contribute no shift. ### An annulus with one boundary missing *Difficulty: Intermediate.* For \(00\), the inner open ball \(\{|v|^20\) and set \(v=t\sqrt{1+a+t}\), \(w=u\). Then \(h=v^2-w^2\) exactly. At \((a,t,u)=(0,0,0)\), the derivative of \((a,t,u)\mapsto(a,v,w)\) is invertible; more generally its fibre derivative at \(t=u=0\) is \(\operatorname{diag}(\sqrt{1+a},1)\). The inverse function theorem gives smooth local inverse coordinates, retaining \(a\). The positive and negative dimensions remain one and one. For \(g\), the fibre derivative at \(t=0\) is \(a\), so \(t=0\) is not the critical family when \(a\ne0\). The true critical point is \(t_c(a)=-a/2\), with value \(-a^2/4\). Completing the square gives \(g=(t+a/2)^2-a^2/4\); translate by that critical section and subtract that critical value. Omitting either change would falsely claim the zero-value quadratic identity. ### A changing intersection can change inertia *Difficulty: Intermediate.* In \(\mathbb R^2\) with \(\omega=dp\wedge dq\), let \(\lambda_a=\{p=aq\}\). For \(-10\). All pairwise intersections vanish off zero, so the radical is zero and its rank is three; the signature is odd, as required. At zero the first two planes coincide, with one-dimensional intersection, while both meet the third in zero. The radical is one-dimensional, its rank is two and its signature is even. At \(t=0\), the four ordered indices \((\tau_{123},\tau_{134},\tau_{124},\tau_{234})\) are \((0,-1,0,-1)\), giving \(-1=-1\). For \(0