How simple-sheaf shifts change along a Lagrangian

The coefficient type of a simple sheaf can stay fixed while its numerical shift changes. A projection singularity is where the inertia correction can change. The right continuity statement follows a continuous family of auxiliary Lagrangian planes, rather than the numerical shift alone. We prove that statement, calculate a cusp including its missing boundary, and explain why the corresponding numerical shift stays constant along a connected complex Lagrangian.

Use Pure and simple sheaves from directional tests. Manifolds are finite-dimensional smooth real manifolds unless specified otherwise; \(k\) is a commutative ring of finite global dimension and \(F\in D^b(k_X)\). Coefficient complexes may have arbitrary modules. The conormal coefficient equivalence supplies their local sheaf models, and the conormal support-test calculation fixes the type normalization. The ordered inertia index uses \(\omega=d\theta\). Simultaneous hypersurface normalization supplies a contact chart with the required auxiliary-plane condition; the zero-covector conormal proof treats points on the zero section. The ordered index proof includes continuity at fixed pairwise-intersection dimensions and the four-plane cocycle, including degenerate triples.

Original lesson text and solutions: CC0 1.0 Universal. Human mathematical sources are credited below.

Follow a transverse auxiliary plane

Let \(\Lambda\subset T^*X\) be a smooth conic Lagrangian and suppose \(\operatorname{SS}(F)\subset\Lambda\) on a neighborhood of the points considered. Let \(S\) be a connected topological space, \(p:S\to\Lambda\) continuous, and let \(\mu(s)\subset T_{p(s)}T^*X\) be a continuous family of Lagrangian planes such that

Choose allowed type shifts \(d(s)\), meaning \(d(s)\equiv\dim(V(s)\cap A(s))/2\pmod{\mathbb Z}\). Then the following is a continuity theorem for the coefficient type:

Theorem. If

the isomorphism class of the type of \(F\) with shift \(d(s)\) at \(p(s)\) is constant on \(S\).

An auxiliary plane transverse to \(V\) is a graph of a symmetric Hessian. At a fixed point it can therefore be realized as the tangent graph of a test function with the prescribed value and differential. The other condition in (1) makes that test transverse to \(\Lambda\). Test independence permits use of its index in (2), without a requirement that the whole family arise from one global function.

The definition of type uses the exponent \(-d+n/2+\tau(V,A,\mu)/2\) on the raw test complex. Thus the degree parameter in that exponent is \(-\kappa+n/2\). This specifies the minus sign in (2) with the displayed order of the three planes. It does not assert that raw tests at different points have already been identified; the next proof supplies the required local comparisons.

A common complement over the reals

Every two real Lagrangian planes \(V,A\) in a symplectic space of dimension \(2n\) admit a Lagrangian plane transverse to both. Put \(K=V\cap A\) and \(r=\dim K\). In \(K^\omega/K\), the images of \(V,A\) are transverse Lagrangians: their intersection is zero, each has dimension \(n-r\), and the pairing between them is nondegenerate. Choose dual bases for this pairing and lift them to \(V,A\). Their span \(E\) is symplectic, since its pairing matrix has invertible off-diagonal blocks, and \(K\subset E^\omega\). The space \(E^\omega\) has dimension \(2r\), with \(K\) Lagrangian in it.

For completeness, extend a basis \(f_1,\ldots,f_r\) of \(K\) to a symplectic basis of \(E^\omega\) as follows. Nondegeneracy supplies \(e_i\) with \(\omega(e_i,f_j)=\delta_{ij}\). Set \(c_{ij}=\omega(e_i,e_j)\) and replace \(e_i\) by \(e_i+\frac12\sum_j c_{ij}f_j\). Their mutual pairing becomes \(c_{ij}+c_{ji}/2-c_{ij}/2=0\), while their pairing with the \(f_j\) stays unchanged. Together with the paired bases in \(E\), this gives symplectic coordinates \((Q',Q'';P',P'')\), with \(\dim Q''=r\), in which

The graph plane \(\mu=\{P'=Q',\ P''=0\}\) is Lagrangian. Its intersection with either displayed plane is zero, by substituting those equations. The construction includes \(r=0\) and \(r=n\). To obtain a continuous family near a selected point, extend this one plane as a constant plane in a local symplectic frame. Transversality to both varying planes persists by openness. Their intersection dimension may vary; no constant-rank assumption on \(V\cap A\) is needed for this local choice.

The conormal chart and the contact correction

Proof. First suppose \(\Lambda=T_M^*X\) in a small cotangent chart. Its vertical intersection has constant dimension \(c=\operatorname{codim}M\). The three pairwise intersection dimensions of \((V,A,\mu)\) are \((c,0,0)\), so the constant-intersection index theorem makes \(\tau(V,A,\mu)\) locally constant. Equation (2) makes \(d\) locally constant too.

The conormal coefficient equivalence gives \(F\simeq Q_M\) in the category localized at a selected point, with \(Q\in D^b(k)\). This one model holds on a smaller cotangent neighborhood by the finite denominator-cone argument: choose the finite roofs, inverses and equalities representing the isomorphism; the microsupport of each denominator cone avoids the point, so all these cones avoid one common smaller neighborhood. No finiteness of the modules in \(Q\) is used. All comparisons here are in the corresponding localized categories.

In the conormal test calculation, a test with negative Morse dimension \(m\) on \(M\) gives \(C_\varphi(Q_M)\simeq Q[-m]\) and \(\tau(V,A,\mu)=2m-\dim M\). Substituting these into the normalized exponent gives

The possible coordinate orientation line is locally trivialized in naming this coefficient type. The fixed \(Q\), locally constant \(c\) and locally constant \(d\) make its isomorphism class locally constant. This includes arbitrary bounded and zero coefficient complexes.

Near a nonzero point of a general \(\Lambda\), use a hypersurface contact transformation \(\chi\) taking \(\Lambda\) to a conormal. Choose it so that \(\chi_*\mu(s_0)\) is also transverse to the target vertical plane. Apply simultaneous hypersurface normalization to \(\Lambda\) and the auxiliary plane \(\mu(s_0)\); the latter does not contain the radial line because it is transverse to the original vertical. This supplies the required target transversality without an additional geometric input. Transversality is open, so it persists for \(s\) near \(s_0\). Write

where \(V'\) is target vertical and \(n=\dim X\). The contact type-shift theorem identifies the original type at \(d\) with the transformed type at \(d'\). The transformed auxiliary index is \(\tau(W,A,\mu)\). Hence its corrected degree is

To check the sign, the cocycle for the ordered quadruple \((V,A,\mu,W)\) gives \(\tau(V,A,W)+\tau(W,A,\mu) =\tau(V,A,\mu)+\tau(V,\mu,W)\). Substituting it in (3) proves (4).

The pairwise intersections in the last triple have constant dimensions: \(\mu\cap V=0\) by (1), \(\mu\cap W=0\) by the chosen target transversality, and

Here is a geometric verification of (5). The contact graph is the conormal of a hypersurface \(H\subset X'\times X\). A vector in \(V\cap W\) corresponds through the graph to a tangent conormal vector with both base components zero. In a local defining equation for \(H\), its base is fixed and only its conormal multiplier varies. This is the one-dimensional radial line. Both cotangent graph projections are local diffeomorphisms, so the correspondence between this line and \(V\cap W\) is an isomorphism. This calculation holds on the whole selected graph patch, not just at one point.

Inertia continuity applied to \((V,\mu,W)\) now makes the right side of (4) locally constant. The already proved conormal case gives locally constant transformed type, and (3) gives locally constant original type. At a zero covector the smooth-conic conormal proof identifies the actual local germ, including its zero covectors, with a conormal. Apply the first part directly. This zero-covector step does not require a contact transformation on the punctured cotangent bundle.

We have proved local constancy on the parameter space \(S\). A locally constant map from a connected space to isomorphism classes has a single value: the inverse image of any value and its complement are both open. This last argument does not assume that \(S\) is path connected or locally connected. It completes the proof. \(\square\)

The theorem compares isomorphism classes of coefficient complexes. It does not choose a global trivialization of their orientation lines or rule out monodromy of local identifications.

A half-open cusp and its smooth conic Lagrangian

In this example take \(k\ne0\). On \(\mathbb R^2\), use coordinates \((x,y)\) and let

The lower boundary is included; the upper boundary and the cusp point are excluded. We claim that, in the region \(\eta>0\), the full nonzero microsupport is

Proof of the microsupport equality. Extend the two boundary functions to all real \(x\) by \(b_-(x)=-(x_+)^{3/2}\) and \(b_+(x)=(x_+)^{3/2}\), where \(x_+=\max(x,0)\). Both are \(C^1\); they agree for \(x\leq0\), have common derivative zero at the origin, and satisfy \(b_-<b_+\) for \(x>0\). Thus they satisfy every hypothesis of the half-open channel theorem, (S20)–(S22). Put \(A=\{y\geq b_-(x)\}\) and \(B=\{y\geq b_+(x)\}\). These are closed, \(B\subset A\), and \(A\setminus B=Z\), including the exclusion of the whole common part for \(x\leq0\).

The actual localization triangle is

The closed-epigraph estimates bound its first term by the union of the two positive conormal families. Their defining functions have \(y\)-derivative one, so the regular-boundary estimate applies even at the origin. At the origin their nonzero conormals coincide with \(\mathbb R_{>0}dy\). This is the full upper bound there, with the neighborhood control for arbitrary \(C^1\) tests supplied by the channel theorem; a calculation for the single test \(y\) would not prove it. Off \(\overline Z\), the coefficient sheaf is locally zero, so the common epigraph boundary for \(x<0\) contributes nothing.

For \(x>0\), a small neighborhood of the lower face misses \(B\), and the sheaf there is the closed upper side with positive covector \(\lambda(\frac32\sqrt{x},1)\). At the upper face it is the open lower side, whose positive covector is \(\lambda(-\frac32\sqrt{x},1)\). Both boundary calculations are equalities when \(k\ne0\). Set \(a=\sqrt{x}\) on the lower face and \(a=-\sqrt{x}\) on the upper face. This gives exactly the two nonzero-\(a\) portions of (7). For every fixed \(\lambda>0\), these covectors tend to \((0,0;0,\lambda)\); closedness of microsupport supplies the entire positive ray at the tip. These lower inclusions exhaust the preceding upper bound. The zero covectors, outside the selected region \(\eta>0\), occur over \(\overline Z\). \(\square\)

The relation (7) is smooth even at \(a=0\). The covector coordinates recover \(a=2\xi/(3\eta)\) and \(\lambda=\eta\), and their parameter derivative has determinant \(3\lambda/2\ne0\). It is therefore a smooth embedded two-dimensional conic manifold. Its tautological form is \((3a\lambda/2)d(a^2)+\lambda d(-a^3)=0\). Thus its symplectic form vanishes and its half dimension makes it Lagrangian. Its base projection, however, has a cusp.

For \(a>0\), the point is on the included lower boundary. The sheaf is locally the closed upper-side coefficient for \(h=y+x^{3/2}\), and its positive conormal is (7). Its simple shift is \(1/2\). For \(a<0\), it lies on the excluded upper boundary; locally the sheaf is the open lower-side coefficient for \(h=y-x^{3/2}\), at that function's positive conormal. Its simple shift is \(-1/2\).

The cusp test has one cohomological degree

At \(p=(0;dy)\), the plane \(T_p\Lambda\) in (7) is the full vertical plane, while the test \(\varphi=y\) has horizontal tangent graph. They are transverse, and \(\tau_\varphi=\tau(V,V,\text{horizontal})=0\).

We calculate the raw support test instead of extrapolating a branch shift. Since \(0\notin Z\), the stalk \(F_0\) is zero. For \(j:\{y<0\}\hookrightarrow\mathbb R^2\), the support triangle gives

Use the cofinal rectangles \(U_\epsilon=\{|x|<\epsilon,|y|<\epsilon^{3/2}\}\). In their negative part, the support of \(j^{-1}F\) is \(0<x<\epsilon\), \(-x^{3/2}\leq y<0\). It is closed relative to that negative open set. The substitution \(t=-y/x^{3/2}\) identifies it with \((0,\epsilon)\times(0,1]\), a nonempty locally closed convex product. The constant-coefficient unit and restriction comparison give ordinary cohomology \(k\) in degree zero and no other degree. Smaller rectangles restrict its constant generator to the same generator. Thus \((Rj_*j^{-1}F)_0\simeq k\) with this actual restriction comparison, and

With ambient dimension two and inertia zero, the normalized type at shift \(d=0\) is \(C_y(F)[1]\simeq k\). The sheaf is simple with shift zero at the cusp. A zero ordinary stalk and a nonzero directional support test coexist because (8) also measures approach from the negative open region.

The auxiliary index accounts for the three shifts

Take the horizontal auxiliary plane \(\mu\) along (7). It is transverse both to vertical and to \(T\Lambda\), including at the cusp. To compute its inertia, express \(T\Lambda\) as a graph over covector variations:

This follows by differentiating (7), solving \(\delta a=2(\delta\xi-3a\delta\eta/2)/(3\lambda)\), and substituting into the two base variations. The matrix is \(a/(3\lambda)\) times the outer product of \((2,-3a)\) with itself. Its signature is \(\operatorname{sgn}a\), and is zero at \(a=0\).

For the ordered vertical/graph-over-covectors/horizontal triple, inertia equals the signature of \(B\). To see this directly, its quadratic form on covector vectors \(u,v\) and a horizontal vector \(w\) is \(u^tBv+(v-u)^tw\). Putting \(z=u-v\) rewrites it as \(v^tBv+z^t(Bv-w)\). The latter pairing is hyperbolic and contributes zero signature; the first term has signature \(B\). Therefore

The support calculation and the tangent calculation can be read together:

The allowed parity also agrees: \(\dim(V\cap T\Lambda)=1\) on either regular branch and is two at the cusp. Equations (11) make \(d-\tau/2=0\) throughout. The continuity theorem consequently keeps type \(k\) fixed across all three numerical shifts. Keeping \(d=1/2\) through the cusp instead would not even satisfy the required integer parity there.

Complex Lagrangians have no such numerical jump

Let \(X\) now be a complex manifold and \(\Lambda\subset T^*X\) a connected complex Lagrangian. Interpret the cotangent geometry through its underlying real symplectic form \(2\operatorname{Re}\sigma\), where \(\sigma\) is the complex cotangent symplectic form. For three complex Lagrangian planes their real inertia is zero. Indeed multiplication by \(i\) on all three arguments sends the real quadratic index form to its negative. It bijects its positive and negative parts, so their dimensions are equal.

Here is an explicit complex simultaneous complement. For two complex Lagrangian planes \(V,A\), put \(K=V\cap A\), \(r=\dim_{\mathbb C}K\). The quotient \(K^\sigma/K\) is symplectic, and the images of \(V,A\) are transverse Lagrangians. Choose paired bases for these images and lift them to \(V,A\). Their span \(S\) is symplectic and orthogonal to \(K\); the lifts remain in the two planes. Its symplectic orthogonal \(S^\sigma\) has dimension \(2r\) and contains \(K\) as a Lagrangian. Extend a basis of \(K\) to a symplectic basis there: choose dual vectors using nondegeneracy, then add linear combinations of the \(K\)-basis to cancel their mutual pairings. This uses division by two, valid over \(\mathbb C\). Thus symplectic coordinates \((Q',Q'';P',P'')\), with \(\dim Q''=r\), give

The plane \(\mu=\{P'=Q',\ P''=0\}\) is complex Lagrangian because its graph matrix is symmetric. Intersecting with \(V\) forces \(Q'=Q''=P'=P''=0\); intersecting with \(A\) does the same. This includes \(r=0\) and \(r=n\). Extend this plane continuously in a local complex symplectic frame. Transversality to both varying planes is open, so it persists after shrinking. These local families are also real Lagrangian transverse families. Their inertia is zero, so a fixed numerical shift \(d\) satisfies (2). The parity condition causes no difficulty: the real vertical intersection dimension is even, and allowed \(d\) are integral.

If \(\operatorname{SS}(F)\) is contained in \(\Lambda\) on its neighborhood and \(F\) is pure with shift \(d\) at one point, the continuity theorem makes it pure with that same shift on each such local chart. Connectedness propagates the assertion along all of \(\Lambda\). In fact its type module has the same isomorphism class throughout. This also proves the corresponding connected complex purity phenomenon, without claiming a globally chosen generator.

Exercises with complete solutions

Connected does not mean path connected

Difficulty: Introductory.

Which step in the continuity proof uses connectedness of \(S\)? Would a proof comparing only points joined by paths suffice for the stated theorem?

Solution. The proof first makes the type locally constant at every parameter value. The inverse image of one type class and its complement are then both open; connectedness forces one to be empty. A path comparison alone would cover only path components, which need not equal connected components for an arbitrary topological parameter space. No path-connectivity assumption is present in the theorem.

One-dimensional excess in a hypersurface graph

Difficulty: Intermediate.

Why does a hypersurface contact graph give \(\dim(V\cap W)=1\), rather than zero? Explain which degree argument uses this number.

Solution. A vertical vector on both sides corresponds to a conormal tangent vector with fixed hypersurface base point. Its conormal multiplier can vary, giving the radial line. A hypersurface has a one-dimensional normal space, so there is exactly one such direction. Both graph projections identify it with \(V\cap W\). This constant intersection, together with the two zero intersections involving \(\mu\), makes \(\tau(V,\mu,W)\) locally constant in (4).

A missing cusp point can have a nonzero directional type

Difficulty: Intermediate.

For (6), find \(F_0\), \(C_y(F)\) and its normalized type at \((0;dy)\). Explain why there is no contradiction.

Solution. The cusp point is excluded, hence \(F_0=0\). The negative open-region coefficient has constant cohomology \(k\); the support triangle gives \(C_y(F)=k[-1]\). Ambient dimension two, inertia zero and shift zero then give normalized type \(k[-1][1]=k\). The local support test includes the derived approach from the complement region, rather than only the ordinary stalk.

Check the branch matrix

Difficulty: Advanced.

For \(a=-1\) and \(\lambda=2\), compute \(B\), its rank, its inertia signature and the simple shift predicted by (11).

Solution. Formula (10) gives \(B=-\tfrac16\begin{pmatrix}4&6\\6&9\end{pmatrix}\). It is the negative outer product of \((2,3)\) divided by six, with eigenvalues \(-13/6\) and zero. Its rank is one and its signature is \(-1\), so the simple shift is \(-1/2\). The radical does not change that signature; it records the vertical radial direction on the regular branch.

A function with the wrong covector

Difficulty: Advanced.

For the same cusp sheaf, compute \(C_x(F)\) at zero and explain why it does not replace the \(y\)-test at \((0;dy)\).

Solution. On the open complement \(\{x<0\}\) the sheaf vanishes, and its ordinary stalk at zero is also zero. The support triangle therefore gives \(C_x(F)=0\). But \(dx\ne dy\), so this function does not meet the selected covector. Formula (7) has no nonzero \(dx\) direction at its cusp. Directional type is attached to a covector, not just a base point or any chosen coordinate function.

An index reversal in the path correction

Difficulty: Advanced.

A family has \(d=1/2\), \(\tau(V,A,\mu)=1\) on one branch, and \(d=-1/2\), \(\tau(V,A,\mu)=-1\) on the other. Compare the corrected quantities \(d-\tau/2\) and \(d+\tau/2\).

Solution. The first is zero on both branches, as required for constant type. The second is one on the first branch and minus one on the second. Alternating two index planes without changing the sign in the degree rule would therefore produce a false variation by two. The ordered cocycle derivation (3)–(4) fixes the correction's sign.

References

Masaki Kashiwara and Pierre Schapira, Microlocal study of sheaves, Astérisque 128 (1985), Definition 7.2.5 and Examples 7.2.6, printed pp. 127–128 (PDF pp. 130–131), give the purity normalization, the half-line degrees and the half-open cusp used here. In particular, the geometry and the three simple shifts in (6)–(11) are the classical example 7.2.6(iv). The support triangle, coefficient maps and tangent-matrix calculation above explain that example explicitly.

Remark 7.2.7 on printed p. 128 gives auxiliary-plane transport. With the ordered signature of Definition 7.1.1, positive minus negative inertia and the symplectic form in Example 7.1.4, our convention is the source's normalization \(j=-d+\dim X/2+\tau/2\). Thus the constant corrected shift is \(d-\tau/2\). The printed difference in (7.2.4) has the opposite order of its two index values. Equations (3)–(4) derive the transport sign from the ordered cocycle, and the explicit cusp values in (11) check it: \(d=\tau/2\) on both branches and at the cusp.

The linked programme proofs supply the conormal coefficient object, contact transform and closed/open boundary tests. The connected-parameter argument retains bounded coefficient complexes and proves constancy of their isomorphism classes. The complex complement calculation explains separately why the numerical degree is constant for a complex Lagrangian.