Lorentz cones and their vertex characteristic cycles

A smooth cone boundary has an inward and an outward conormal. At its vertex, those conormals acquire additional fibre pieces. The distinction between a closed cone, its open interior and its punctured boundary changes both the pieces and their coefficients. We calculate these cycles from actual local support tests, duality and support triangles, including the dimension-dependent vertex signs.

Use Directional tests at a constructible boundary for the neighborhood definition of microsupport, Constructible costalks and Verdier duality for the natural costalk-to-dual-stalk pairing, and Antipodal duality and half-line characteristic cycles for the supported antipodal operation and its actual unit/evaluation compatibility. Integer coefficients and additive characteristic cycles supplies integral dense-piece determination and additivity. The precise topological inputs are constant-interval acyclicity, constant coefficients on locally closed convex sets, closed-cover cohomology, and proper-support base change. We derive the contraction and the uniform convex-vertex tests from those maps.

Let \(k\) be a field of characteristic zero. Set \(d=n+1\), with \(n\geq0\), and work on the real analytic manifold \(X=\mathbb R_t\times\mathbb R_x^n\), with its standard orientation. Every sheaf below is the indicated constant coefficient extended by zero; it lies in \(D^b_{\mathbb R\text{-}c}(k_X)\) with finite perfect stalks. Cycle normalizations and orientation coefficients are those of the linked lessons. In particular, the graph-normalized conormal formula is \(CC(k_M)=(-1)^{\dim M}[T_M^*X]\) for a closed smooth \(M\), and the same equality holds locally for a locally closed smooth piece. The point coefficient is \(+1\). We retain this distinction between the normalized geometric bracket and the characteristic cycle of the constant sheaf throughout.

Original lesson text and solutions: CC0 1.0 Universal. Human mathematical sources retain their named credit and their own terms.

Open pieces and closed carriers

Put

Thus \(U_0=\{|t|<|x|\}\). For \(n=0\), this middle interior and both punctured boundaries are empty. For \(n\geq1\), each \(S_\pm\) is an analytic hypersurface of dimension \(n\): its points have \(x\ne0\), where the norm is analytic. All these sets are semialgebraic and hence subanalytic. The cone vertex is the single point \(o=(0,0)\).

Write \((t,x;\tau,\xi)\) for cotangent coordinates. Define the trimmed normalized chains

The first sign of \(\tau\) labels the base hypersurface; its second sign labels the time component of the covector. The \(\gamma_\pm\) restrictions are strict timelike regions. Their closed carriers include the light boundary \(\pm\tau=|\xi|\). A restriction in (2) is a chain with its frontier germs retained, not an independently closed cycle. The sums proved below have the required frontier cancellation.

These pieces partition the dense top-dimensional parts of the relevant conormals. Missing zero-section cone boundaries and missing light directions in the point fibre have dimension at most \(d-1\). These are statements about geometric brackets, before multiplying by the constant-sheaf dimension factor. The integral top-chain injection therefore gives

This reasoning includes \(n=0\): the omitted origin in a line has dimension zero, while the chains have dimension one. It does not discard frontier boundary terms.

Contractions identify the actual constant-section map

We need constant cohomology on some nonconvex test cuts. The following argument derives it from proper interval projection.

Let \(Y\) be a nonempty contractible locally compact Hausdorff subspace of a finite-dimensional Euclidean space. The projection \(p:Y\times[0,1]\to Y\) is proper: the preimage of a compact set is its product with a compact interval. Proper base change and constant-interval acyclicity identify the stalks of the actual unit

with \(k\to R\Gamma([0,1];k)\), the constant-section isomorphism. Thus (4) is an isomorphism. For either endpoint inclusion \(i_\epsilon:Y\to Y\times[0,1]\), the equality \(p i_\epsilon=\operatorname{id}_Y\) makes its restriction a left inverse of the projection pullback on derived global sections. Since that pullback is an isomorphism by (4), both endpoint restrictions are its inverse. For any homotopy \(H:Y\times[0,1]\to Y\), the two end maps therefore induce the same map on \(R\Gamma(Y;k)\), by composing their endpoint restrictions with \(H^*\).

Apply this to a contraction from the identity to the constant map through a chosen point. The latter map factors through \(R\Gamma(\{\mathrm{pt}\};k)=k\). Point restriction and the constant-section map have both composites equal to the identity, the second by the homotopy argument. Consequently

For every inclusion between two such spaces, the restriction in these identifications is the identity on \(k\), because it restricts the same constant section. No nonproper fibrewise base-change claim was used. The argument also explains exactly which map occurs in the localization calculation below.

Uniform tests at a closed convex vertex

Take \(C=Z_+\), and write \(q(t,x)=\tau t+\xi\cdot x\). Its positive polar is

Indeed, if \(\tau\geq|\xi|\), then \(q(t,x)\geq(\tau-|\xi|)t\geq0\) on \(C\). If \(\tau<|\xi|\) and \(\xi\ne0\), evaluation on \((1,-\xi/|\xi|)\) is negative. Moving its spatial component slightly towards zero keeps that evaluation negative and gives a vector in \(\operatorname{Int}C\). If \(\xi=0\), necessarily \(\tau<0\), and \((1,0)\) serves. Thus outside the polar there is an interior vector \(v\) with \(q(v)<0\).

Fix that \(v\). Choose a complement \(H\) to its span. In coordinates \(z=y+sv\), the cone is a Lipschitz epigraph

Here is the elementary geometry behind this assertion. Choose \(\eta>0\) with \(v+B_\eta\subset C\). For each \(y\), sufficiently large positive \(s\) puts \(y+sv\) in \(C\); sufficiently negative \(s\) makes its time coordinate negative. Closedness, convexity and stability under addition by \(v\) make the allowed parameters a closed ray. Its first value is \(g(y)\). For \(\delta\in H\), the vector \(\delta+(|\delta|/\eta)v\) lies in \(C\). Adding it to a point of \(C\) and then interchanging \(y,y+\delta\) proves

Now choose a cotangent neighborhood of \((o,q)\) on which the evaluation of every covector on \(v\) is negative, bounded away from zero. For any allowed \(C^1\) test \(\varphi\), at any testing point \(a\in C\) in that neighborhood, continuity lets us use a small box centered at \(a\) on which \(\partial_s\varphi<-c<0\). Subtract \(\varphi(a)\). Strict monotonicity and the intermediate value theorem give a unique continuous graph \(s=h(y)\) on a smaller base ball, with \(h(0)=0\), such that

Continuity is not an unproved implicit-function step: the derivative bound makes a change of root at most the change of the function at the old root divided by \(c\). Uniform continuity on a smaller closed box therefore gives continuity of \(h\).

Write the translated epigraph threshold as \(g_a\); because \(a\in C\), \(g_a(0)\leq0\). Choose the base ball so small that \(g_a(y),h(y)<\varepsilon/4\), with vertical coordinate in \((-\varepsilon,\varepsilon)\). Both

are nonempty and contractible. Move every vertical coordinate linearly to \(\varepsilon/2\); the entire path retains \(s\geq g_a(y)\), and for \(A_<\) retains \(s>h(y)\). Then contract the base ball at that common height. The sets are locally compact: they are the intersection of a closed cone with an open set. Formula (5) makes the actual restriction from the constant stalk to sections on \(A_<\) the identity \(k\to k\), with no higher cohomology. These boxes can be taken arbitrarily small. Localization thus gives

At points outside \(C\), shrink to its open complement. The fixed cotangent neighborhood, chosen before \(a\) and \(\varphi\), works for all the tests just described. Hence every vertex covector outside (6) is absent from microsupport. This proves a neighborhood statement, not merely vanishing for one linear test.

Conversely, if \(q\in C^\circ\), the linear function \(q\) is nonnegative on all of \(C\). Its supported sheaf equals \(k_C\) near the vertex, so its support-test stalk is \(k\ne0\). Every such covector belongs to microsupport. Together,

Away from the vertex, the constant interior has only zero covectors. On \(S_+\), the analytic coordinate \(u=t-|x|\) straightens the support to \(u\geq0\). The closed-half-line test with the constant tangential coordinates gives precisely the positive conormal multiples of \(du\). For completeness, negative normal covectors are excluded by the same monotone epigraph boxes as above. A covector with a nonzero tangential component is excluded by using a tangent coordinate on which the test decreases; the half-space is invariant in that direction, and the cut and uncut boxes contract to the same slice over a contractible half-ball. Positive pure normal multiples have the nonzero linear support test. Thus the only nonzero smooth-boundary part is \(\tau_{+,+}\). Repeating with the coordinate \(-t\) proves the corresponding result for \(Z_-\), whose inward conormal is \(\tau_{-,-}\).

The positive vertex germ is the actual point sheaf

Let \(A=C\setminus\{o\}\). The support triangle is

For \(q\) strictly inside (6), there is \(b>0\) with \(q(z)\geq b|z|\) on \(C\). One may take any \(b<(\tau-|\xi|)/\sqrt2\). This strict inequality persists for nearby covectors. For every \(C^1\) test with such differential at \(o\), the mean-value estimate along the segment from \(o\) to \(z\in C\) gives \(\varphi(z)-\varphi(o)>0\) for all sufficiently small nonzero \(z\in C\). On a sufficiently small neighborhood the restriction of \(k_A\) to the open lower test cut is zero. The localization triangle therefore identifies its sections with closed test support with \(k_A\) itself on that neighborhood. Its stalk at \(o\) is zero, proving the required test vanishing there.

At nearby nonvertex points, \(k_A=k_C\), whose nonzero boundary normals have \(|\tau|=|\xi|\), and whose interiors are constant. They avoid the same strict timelike cotangent neighborhood. Thus \(SS(k_A)\) misses that neighborhood and (13) makes the actual restriction \(r\) a microlocal isomorphism there:

This is an isomorphism in the point-localized derived category, not equality of ordinary stalk ranks. At a positive conormal over \(S_+\), the support triangle for the local closed half-space identifies its sheaf with \(k_{S_+}\), since the open half-space is invisible in the inward direction. The smooth support \(S_+\) has dimension \(n\), so its constant sheaf has coefficient \((-1)^n\) in the normalized bracket \(\tau_{+,+}\). At a positive timelike vertex covector, (14) instead gives the point model and coefficient \(+1\) on \(\gamma_+\). In the interior the constant sheaf lives on dimension \(d\) and gives coefficient \((-1)^d\) on \(\sigma_+\). These are coefficients of the actual localized objects, not ordinary vertex stalk ranks. The microsupport calculation excludes other top-dimensional pieces. Integral dense-piece determination, including frontier germs, proves

For the minus cone the same restriction argument uses \(-\tau>|\xi|\). These are equalities of cycles through the vertex: the sum agrees with the existing characteristic cycle as a top chain, so its frontier cancels. No individual term is asserted to be a cycle on its own.

Costalks fix the duality shift

We compute the dual rather than inferring it from a picture. At an interior point of \(C\), the point costalk of \(k_C\) is the ambient orientation coefficient in degree \(d\), namely \(k[-d]\) with the chosen orientation. At a smooth boundary point it is zero: a small closed half-ball and that half-ball with its boundary center removed are contractible, and their constant-section restriction is the identity by (5).

The same statement holds at the vertex. The link

is homeomorphic to the closed unit \(n\)-ball by \(u=x/t\); its inverse is \((t,x)=(1,u)/\sqrt{1+|u|^2}\). The punctured cone in a small ball is \((0,\varepsilon)\times L\), hence contractible. The unpunctured intersection is convex. Their restriction is again \(k\to k\), so its fibre, the point costalk at \(o\), is zero. The link description includes \(n=0\), where \(L\) is one point.

Natural constructible duality identifies the dual stalk with the coefficient dual of that costalk. The dual therefore has zero stalks outside \(U_+\), and its restriction to \(U_+\) is the orientation sheaf in shift \([d]\). The actual open-extension counit is a stalkwise isomorphism and yields

The second equality follows from constructible biduality and reversal of the shift under the contravariant dual. No differentiation of a nonsmooth homeomorphism, or cotangent transport by one, is needed.

The supported antipode acts on the trimmed normalized pieces by

These signs use the base dimensions \(d,n,0\) of the corresponding conormals and their orientation coefficients. Combining (15), (17), actual antipodal duality and the shift rule gives the formula below. Indeed, the antipode changes the first coefficient by \((-1)^d\), the second by \((-1)^n\), and exchanges the point pieces with no sign. Applied to (15), its three resulting coefficients are all \(+1\). To recover the open sheaf from its shift by \([d]\), multiply the entire result by \((-1)^d\). Thus

The open smooth-boundary coefficient in the normalized bracket is \((-1)^d=-(-1)^n\). It is the negative of the closed smooth-boundary coefficient, after reversing the ray. That relative minus sign is dimension independent; the absolute coefficient is not. The open vertex coefficient is \((-1)^d\), obtained from the ambient duality shift acting on the positive point coefficient.

Support triangles calculate both middle regions and the boundaries

The open complement of \(Z_0\) is the disjoint union \(U_+\sqcup U_-\). Its localization triangle is

Let \(D=Z_+\cup Z_-\), a closed set with intersection \(Z_+\cap Z_-=\{o\}\). Actual restrictions, diagonal and difference give a stalkwise exact sequence

At the vertex its maps are \(k\to k^2\to k\), diagonal then difference; elsewhere they are identities on the single present cone. Hence no open-cover interpretation is being substituted for this closed-cover sequence. The open complement of \(D\) is \(U_0\), with triangle \(k_{U_0}\to k_X\to k_D\xrightarrow{+1}\).

Finally let \(B_\pm=Z_\pm\setminus U_\pm=S_\pm\cup\{o\}\). Two support triangles give \([k_{S_\pm}]=[k_{Z_\pm}]-[k_{U_\pm}]-[k_o]\). Apply the proved additive characteristic-cycle map to these actual triangles, and substitute (3), (15), (19). This proves every row of the following table.

For example, (21) and the complement triangle give \(CC(k_{U_0})=CC(k_X)-CC(k_{Z_+})-CC(k_{Z_-})+CC(k_o)\). The zero-section contributions leave \((-1)^d\sigma_0\), the two same-sign boundary contributions have coefficient \(-(-1)^n=(-1)^d\), and the point contributions leave \(\gamma_0\). For \(S_+\), the \(\gamma_+\) coefficients cancel, whereas the remaining opposite timelike coefficient is \(-(-1)^{n+1}-1=(-1)^n-1\). Both smooth conormal rays have coefficient \((-1)^n\). These computations explain the displayed covector signs and the coefficient \(-2\) that occurs in odd spatial dimension. All lower-dimensional frontier terms cancel because the equations were obtained by the characteristic-cycle map on genuine sheaf triangles.

Exercises with complete solutions

The light boundary is not the timelike interior

Difficulty: Introductory.

For \(\xi\ne0\), find a nonzero \(z\in Z_+\) on which \(q=(|\xi|,\xi)\) vanishes. Explain why the strict vertex-chain region in (2) differs from the closed microsupport fibre in (12).

Solution. Take \(z=(1,-\xi/|\xi|)\). It lies on \(S_+\) and \(q(z)=0\). Every strictly timelike positive covector is uniformly positive on nonzero future-cone directions, which is the margin used in (14). A light covector has no such margin. Microsupport is closed and includes all \(\tau\geq|\xi|\); the top-dimensional point-conormal chain is trimmed to \(\tau>|\xi|\), and its carrier closure includes that boundary. The light directions have dimension \(d-1\) in the point fibre and meet closures of the smooth-boundary conormals. They are retained by the chain frontier, not assigned a separate top-dimensional \(\gamma_+\) coefficient.

A nonlinear test outside the polar

Difficulty: Intermediate.

Assume \(n\geq1\) and test \(k_{Z_+}\) at \(o\) with \(\varphi(t,x)=x_1+t^2\). Determine its local support complex, and compare the test \(\psi=t\). Give an interior cone direction controlling the first test.

Solution. The first differential is \((\tau,\xi)=(0,e_1)\), outside (6). Take \(v=(1,-e_1/2)\): it is strictly inside \(Z_+\), and \(d\varphi_o(v)=-1/2\). Nearby differentials retain negative evaluation on \(v\); along this direction the nonlinear derivative remains negative on a sufficiently small box. The epigraph construction (7)–(11) makes the smaller-value cut contractible and identifies the actual restriction with \(k\to k\). Its fibre is zero. For \(\psi=t\), the smaller-value cut misses \(Z_+\) entirely, so the support complex is \(k\) in degree zero. The nonlinear first calculation uses a neighborhood of the differential and monotonicity, rather than replacing all \(C^1\) tests by a single linear half-space.

Two dimensions change the vertex parity

Difficulty: Intermediate.

Compare \(n=1\) and \(n=2\) in the open future cone, closed middle and punctured future boundary formulas. Which comparison of the open and closed smooth-boundary coefficients is independent of this parity?

Solution. For \(n=1\), the open future vertex coefficient is \(+\gamma_-\); the closed middle has \(-\gamma_+-\gamma_-\); and the punctured future boundary has opposite timelike coefficient \(-2\gamma_-\), besides \(-\gamma_0\) and its two boundary conormals. For \(n=2\), those three coefficients become \(-\gamma_-\), \(+\gamma_++\gamma_-\), and zero on \(\gamma_-\), respectively. The open smooth-boundary terms are \(+\tau_{+,-}\) when \(n=1\) and \(-\tau_{+,-}\) when \(n=2\); the corresponding interior terms are \(+\sigma_+\) and \(-\sigma_+\). The closed smooth-boundary coefficient is \((-1)^n\), so in both dimensions the open coefficient is its negative on the opposite ray. The ambient shift, the antipodal conormal sign, and the original closed coefficient all enter: \((-1)^{n+1}(-1)^n(-1)^n=(-1)^{n+1}\). Omitting the original coefficient or using the ambient dimension twice loses the stated convention.

The zero-spatial-dimension specialization

Difficulty: Intermediate.

Set \(n=0\). Reduce all seven rows to the line formulas, including the middle regions and the punctured boundaries.

Solution. There are no \(\tau\) chains, \(\sigma_0=\gamma_0=0\), \(Z_0=\{o\}\), and \(U_0=S_+=S_-=\varnothing\). The two \(\sigma\) chains are the positive and negative zero-section rays, while \(\gamma_\pm\) are the corresponding point-fibre rays. The constant line has cycle \(-\sigma_+-\sigma_-\), whereas the point has cycle \(\gamma_++\gamma_-\). Closed half-lines give \(-\sigma_\pm+\gamma_\pm\), and open half-lines give \(-\sigma_\pm-\gamma_\mp\), because \(d=1\). These are exactly the four-ray formulas with the graph-normalized zero-section bracket; replacing that bracket by the positive fibre Thom unit would change the meaning of each \(\sigma\). The closed-middle row is \(\gamma_++\gamma_-\), the point cycle. The open-middle row is zero. In the boundary row, \((-1)^n-1=0\) and all other terms vanish, as required for the empty sheaves. No connectedness assumption on a spacelike region was needed.

A zero vertex stalk can carry a nonzero vertex cycle

Difficulty: Advanced.

Assume \(n\geq1\) and take a nonzero spacelike covector \(p=(o;\tau,\xi)\) with \(|\tau|<|\xi|\). Determine the actual localized objects \((k_{U_0})_p\) and \((k_{S_\pm})_p\) in terms of \((k_o)_p\). Compare their shifts with the \(\gamma_0\) coefficients.

Solution. The closed cone objects are invisible at \(p\) by (12) and its minus version. Their open interiors are invisible too, by (17) and dual microsupport. Localizing the triangle of (21) therefore gives \((k_D)_p\simeq(k_o)_p[-1]\). The complement triangle, with \((k_X)_p=0\), gives \((k_{U_0})_p\simeq(k_o)_p[-2]\). For \(B_\pm\), the triangle of open interior and closed cone makes \((k_{B_\pm})_p=0\). Its point-removal triangle gives \((k_{S_\pm})_p\simeq(k_o)_p[-1]\). Thus the two coefficients are \(+1\) and \(-1\), respectively, by the even and odd shift rules. All these sheaves have ordinary stalk zero at \(o\); that stalk cannot recover their microlocal vertex complexes or cycles. The two-step shift in the open middle is actual derived information behind its positive coefficient.

Graded coefficients retain the same geometric signs

Difficulty: Advanced.

Take \(n=2\), and let \(P=k\oplus k^2[1]\). Calculate the cycles of \(P_{Z_+}\) and its Verdier dual, and verify antipodal compatibility. What changes for \(Q=k\oplus k[1]\)?

Solution. Here \(d=3\) and \(n=2\), so the unweighted closed-cone cycle is \(-\sigma_++\tau_{+,+}+\gamma_+\). The Euler weight is \(\chi(P)=1-2=-1\), making \(CC(P_{Z_+})=\sigma_+-\tau_{+,+}-\gamma_+\). Its dual is \(P^\vee_{U_+}[3]\), with Euler weight \((-1)^3\chi(P^\vee)=1\). The open-cone formula gives \(-\sigma_+-\tau_{+,-}-\gamma_-\). Applying (18) to the first cycle gives exactly that result: the zero-section sign is \(-1\), the smooth hypersurface sign is \(+1\), and the point pieces are exchanged with sign \(+1\). For \(Q\), the Euler weight is zero, so both cycles vanish. The coefficient complexes and their sheaves remain nonzero, and their microsupport still records the cone directions: direct summands with nonzero tests cannot disappear by an Euler cancellation. Characteristic cycles forget that cancellation's graded constituents.

A distant cap changes the compact index

Difficulty: Advanced.

Compute compactly supported cohomology for the constant sheaves on \(C_1=Z_+\cap\{t\leq1\}\), \(C_1\setminus\{o\}\), \(U_+\cap\{t\leq1\}\), and \(U_+\cap\{t<1\}\). Explain why the last two have identical vertex-cycle germs despite different compact Euler characteristics.

Solution. The first set is compact and convex, so its ordinary and compact cohomology are \(k\) in degree zero. Point removal has the actual restriction \(k\to k\); its localization triangle makes the second compact cohomology zero. The third is the complement in \(C_1\) of its compact lateral boundary \(L_1=\{t=|x|,\ 0\leq t\leq1\}\). This boundary contracts to the vertex by scalar multiplication, so (5) gives its cohomology \(k\). Restriction from \(C_1\) to \(L_1\) is the constant-section identity. Its localization triangle therefore makes the third compact cohomology zero, including \(n=0\), when \(L_1\) is one point. The fourth is homeomorphic to \((0,1)\times B^n\), with \(B^n\) the open unit ball, by \((t,u)\mapsto(t,tu)\). This is an open \(d\)-ball up to homeomorphism, with compact cohomology \(k[-d]\). Their Euler characteristics are \(1,0,0,(-1)^d\). The last two sheaves agree on a neighborhood of the vertex, so their characteristic cycles agree there by locality. Their different compact indices involve the upper cap at \(t=1\); a vertex coefficient alone is not a whole-support compact index.

The closed middle has a different dual vertex

Difficulty: Advanced.

Assume \(n\geq1\). Compute the point costalk of \(k_{Z_0}\) and the vertex stalk of its Verdier dual. Can \(D_Xk_{Z_0}\) equal \(k_{U_0}[d]\), as it does for the two convex closed cones? Compare their cycles.

Solution. The unpunctured middle cone in a small ball contracts to its vertex, so its ordinary cohomology is \(k\) by (5). Its link on the unit sphere is described by \(t\in[-1/\sqrt2,1/\sqrt2]\) and \(x=\sqrt{1-t^2}\,u\), with \(u\in S^{n-1}\). Thus it is an interval times \(S^{n-1}\), and the punctured middle cone retracts to that sphere. For \(n=1\), it has two components and the actual restriction is diagonal \(k\to k^2\); its fibre is \(k[-1]\). For \(n\geq2\), the sphere has cohomology \(k\) in degree zero and \(k\) in degree \(n-1\). Here is the closed-cover induction, including the small dimension. For \(S^0\) there are two points. For \(S^1\), its two closed semicircles are acyclic and their intersection is \(S^0\); the difference of restrictions \(k^2\to k^2\) has image the diagonal, so its kernel and cokernel are each \(k\), in degrees zero and one. For \(S^r\) with \(r\geq2\), the closed hemispheres are contractible compact balls and their intersection is connected \(S^{r-1}\). The degree-zero difference \(k^2\to k\) is onto with kernel \(k\). The remaining part of the closed-cover exact sequence shifts the reduced cohomology of \(S^{r-1}\) up by one. Induction gives exactly the asserted groups. The restriction from the cone is the constant-section isomorphism in degree zero. Its fibre therefore has a single cohomology group \(k\) in degree \(n\), namely \(i_o^!k_{Z_0}\simeq k[-n]\), with the spatial orientation fixing a generator. Natural costalk duality gives \((D_Xk_{Z_0})_o\simeq k[n]\). The open-middle extension has zero stalk at \(o\), even after shift, so the proposed isomorphism fails.

Applying (18) to the closed-middle row and comparing the open-middle row gives precisely

The additional point-cycle coefficient agrees with the Euler weight of the nonzero dual vertex stalk. This cycle equation is not by itself a direct-sum decomposition of the dual sheaf. The noncontractible punctured middle link is the reason the convex-cone duality argument cannot simply be reused for this closed set.

References

M. Kashiwara, Index theorem for constructible sheaves, Astérisque 130 (1985), §§1.5–3.4, pp. 196–199, provides coefficient-valued chains, an explicit twisted conormal orientation and the generic local Euler-multiplicity construction. The normalization of a conormal generator must be compared before transferring a coefficient between conventions. The present cone calculation uses the graph-normalized brackets and the evaluated characteristic-cycle construction in the programme lessons linked above.

The polar, uniform epigraph tests, proper-interval contraction, point-localized vertex restriction, costalk calculation and seven-row support-triangle table are proved here. The worked examples distinguish the dimensions of the smooth base pieces from the ambient duality shift, and local vertex data from a compact index. The seven-row table follows from these local tests and support triangles.