Constructible traces and local Euler indices

An Euler index is a signed count of finite cohomology groups. To connect that count to geometry, we turn the identity of a constructible complex into a class with values in the dualizing complex. The diagonal supplies the comparison between an endomorphism and an evaluated tensor. Its exceptional restriction, ordinary restriction and comparison map must all remain visible: they measure different kinds of information.

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

The supported trace below is defined by its actual maps: product evaluation, exceptional restriction to the diagonal, the closed-embedding counit, graded interchange and evaluation. Its normalization is checked directly at a point by the chain-level supertrace. Use Constructible costalks and Verdier duality for the actual local dual pairings and perfection, Perfect coefficients on compact fibres for finiteness on compact subanalytic sets, and Perfect operations and finite microlocal coefficients for bounded tensor and internal Hom. The normalized maps come from the product evaluation theorem, SH02-CB-EXTERNAL-HOM, exceptional inverse image of internal Hom, SH02-EX-HOM, and exceptional composition, SH02-EX-COMPOSITION, with their stated hypotheses. The present construction uses their formal neighborhood systems, proper-support soft, fibre and composition results, and derived resolution and duality prerequisites. Proper trace transport, the global index theorem and characteristic cycles require the further arguments described below.

Two finite local measurements

Throughout this lesson, $k$ is a commutative field of characteristic zero. Let $X$ be a real analytic manifold with the standing finite uniform dimension bound, and let

$$F\in D^b_{\mathbb R\text{-c}}(k_X). \qquad\text{(1)}$$

Constructibility in (1) includes perfect stalks. The characteristic-zero field hypothesis makes the point supertrace determine an integer Euler index. No statement here extends the global trace or cycle construction to an arbitrary coefficient ring.

For a bounded complex $P$ with finite-dimensional cohomology, define

$$\chi(P)=\sum_q(-1)^q\dim_k H^q(P)\in\mathbb Z. \qquad\text{(2)}$$

The sum is finite. Over a field, such a complex is perfect. A bounded finite-dimensional representative is therefore available for calculations, although replacing a complex by a representative does not change its morphisms in the derived category.

For the point inclusion $i_x:\{x\}\hookrightarrow X$, write

$$A_x(F)=i_x^{-1}F=F_x,\qquad C_x(F)=i_x^!F\simeq R\Gamma_{\{x\}}(X;F). \qquad\text{(3)}$$

The second expression is cohomology supported at the point, equivalently the point costalk. The compact-support symbol in the notation below does not replace it with the global complex $R\Gamma_c(X;F)$.

Both complexes in (3) are perfect by the constructible-costalk theorem. We can consequently form the integer-valued functions

$$\chi(F)(x)=\chi(A_x(F)),\qquad \chi_c(F)(x)=\chi(C_x(F)). \qquad\text{(4)}$$

The natural local dual pairing identifies

$$(D_XF)_x\simeq R\operatorname{Hom}_k(C_x(F),k),\qquad D_XF=R\mathcal Hom(F,\omega_X). \qquad\text{(5)}$$

For a finite coefficient complex $P$, duality sends $H^q(P)$ to its vector-space dual in degree $-q$. Thus $\chi(P^\vee)=\chi(P)$: the signs $(-1)^{-q}$ and $(-1)^q$ agree. Equation (5) proves the dual-sections identity

$$\chi_c(F)=\chi(D_XF). \qquad\text{(6)}$$

This proof uses the actual costalk–dual-stalk comparison. It does not identify a costalk with a stalk. Since $F$ and $D_XF$ are constructible, choose a common locally finite subanalytic stratification for their cohomology sheaves. On each stratum all their cohomology ranks are locally constant. Boundedness makes (4) finite sums of those ranks, so both functions are constructible.

On an $n$-dimensional component, the manifold normalization is

$$\omega_X=\operatorname{or}_X[n]. \qquad\text{(7)}$$

For the constant sheaf $k_X$, a local coordinate ball gives $A_x(k_X)=k$ and $C_x(k_X)=\operatorname{or}_{X,x}[-n]$. Hence its ordinary local index is $1$, while its costalk index is $(-1)^n$. No global orientation is needed to count the dimension of the orientation line.

Global indices require a separate finiteness check

Define

$$\chi(X;F)=\chi(R\Gamma(X;F)),\qquad \chi_c(X;F)=\chi(R\Gamma_c(X;F)) \qquad\text{(8)}$$

only when the corresponding complex has bounded finite-dimensional cohomology. Local constructibility alone does not supply this global condition on a noncompact space.

For example, take $X$ to be a countable discrete manifold and $F=k_X$. Every stalk and costalk is the finite complex $k$, and both local functions in (4) are $1$. Nevertheless,

$$\Gamma(X;F)=\prod_{m\geq1}k,\qquad \Gamma_c(X;F)=\bigoplus_{m\geq1}k$$

are infinite-dimensional. Neither global index in (8) is defined by (2).

If $F$ has compact closed support, both global complexes are perfect by compact constructible finiteness. Here closed support means the complement of the largest open set on which $F$ vanishes; it is the closure of the set of points with a nonzero cohomology stalk. In particular, extension by zero from an open interval has the closed interval as its closed support, even though its endpoint stalks vanish.

The support is a compact subanalytic set. Restricting to it and using the closed-embedding equivalence reduces ordinary sections to the compact-set finiteness theorem. The same theorem applies to compact sections, and the canonical map

$$R\Gamma_c(X;F)\longrightarrow R\Gamma(X;F) \qquad\text{(9)}$$

is an isomorphism because the complex is supported on that compact set. Indeed, for its closed inclusion $i:Z\hookrightarrow X$, the ordinary localization equivalence gives $F\simeq i_*i^{-1}F$. The embedding is proper, so composition identifies the two sides of (9) with $R\Gamma_c(Z;i^{-1}F)$ and $R\Gamma(Z;i^{-1}F)$. These section functors agree on the compact space $Z$, and their comparison is the identity. Consequently $\chi_c(X;F)=\chi(X;F)$ in this case. Relating this integer to the geometric characteristic class requires the proper trace compatibility proved in the next stage.

The identity and the evaluated tensor

The internal Hom adjunction gives

$$\operatorname{Hom}(F,F) \simeq\operatorname{Hom}(k_X,R\mathcal Hom(F,F)).$$

Let $e_F:k_X\to R\mathcal Hom(F,F)$ be the morphism corresponding to $\mathrm{id}_F$. Let $\mathrm{ev}_F:D_XF\otimes^L F\to\omega_X$ be evaluation. We define the contraction with the following explicit graded order as

$$t_F:F\otimes^LD_XF \xrightarrow{\,\tau\,}D_XF\otimes^LF \xrightarrow{\,\mathrm{ev}_F\,}\omega_X. \qquad\text{(10)}$$

The symmetry $\tau$ is the graded symmetry: homogeneous elements of degrees $a,b$ acquire $(-1)^{ab}$ when interchanged. This sign is part of (10).

Let $q_1,q_2:X\times X\to X$ be the projections, and let $\delta:X\to X\times X$ be the closed diagonal. Put

$$K_F=F\boxtimes^LD_XF.$$

The product evaluation theorem, SH02-CB-EXTERNAL-HOM, applied with the cohomologically constructible factor on the second copy of $X$, gives the canonical isomorphism

$$K_F\xrightarrow{\sim} R\mathcal Hom(q_2^{-1}F,q_1^!F). \qquad\text{(11)}$$

It includes the graded permutation placing the first factor $F$ before the second factor $D_XF$. This is the evaluation map of that theorem with its actual normalization. Constructibility and the perfect local section representatives establish its invertibility; an abstract isomorphism of its source and target would not suffice for the trace construction.

Apply exceptional restriction along $\delta$. The exceptional-Hom comparison, SH02-EX-HOM, is an isomorphism for a bounded first Hom input and a bounded-below second input. Here $q_2^{-1}F$ is bounded and $q_1^!F$ is bounded below; the finite manifold dimension makes the exceptional functors available. It gives

$$\begin{aligned} \delta^!K_F &\simeq\delta^!R\mathcal Hom(q_2^{-1}F,q_1^!F)\\ &\simeq R\mathcal Hom(\delta^{-1}q_2^{-1}F,\delta^!q_1^!F)\\ &\simeq R\mathcal Hom(F,F). \end{aligned} \qquad\text{(12)}$$

The last step uses $q_2\delta=\mathrm{id}_X$ and the normalized exceptional composition, SH02-EX-COMPOSITION, for $q_1\delta=\mathrm{id}_X$. Denote the inverse of (12) by

$$\theta_F:R\mathcal Hom(F,F)\xrightarrow{\sim}\delta^!K_F. \qquad\text{(13)}$$

There is no unexplained dimension shift in (12): the exceptional composition has already accounted for the relative dualizing factor in $q_1^!$. We keep $\delta^!$ until the next map.

From exceptional to ordinary restriction along a closed embedding

For any closed embedding $i:Z\hookrightarrow W$, proper and ordinary direct image agree, and ordinary restriction satisfies $i^{-1}i_*\simeq\mathrm{id}$. Apply $i^{-1}$ to the exceptional counit:

$$i_*i^!A\xrightarrow{\epsilon_A}A.$$

This defines a natural comparison

$$\beta_{i,A}:i^!A \simeq i^{-1}i_*i^!A \xrightarrow{\,i^{-1}\epsilon_A\,}i^{-1}A. \qquad\text{(14)}$$

Its adjunction characterization also gives uniqueness. Compose a candidate $b:i^!A\to i^{-1}A$ with the ordinary unit $A\to i_*i^{-1}A$ and the exceptional counit. Requiring

$$i_*b: i_*i^!A\longrightarrow i_*i^{-1}A \quad=\quad \bigl(i_*i^!A\xrightarrow{\epsilon_A}A \longrightarrow i_*i^{-1}A\bigr) \qquad\text{(15)}$$

forces $b$ to be (14), because $i_*$ is fully faithful. Applying $i^{-1}$ to the right side of (15) recovers exactly (14): the ordinary unit restricts to the identity. This is the unit–counit characterization of the closed-diagonal comparison used here.

For $i=\delta$, ordinary restriction gives $\delta^{-1}K_F\simeq F\otimes^LD_XF$. The complete evaluated endomorphism map is therefore

$$R\mathcal Hom(F,F) \xrightarrow{\theta_F}\delta^!K_F \xrightarrow{\beta_{\delta,K_F}}\delta^{-1}K_F \simeq F\otimes^LD_XF \xrightarrow{t_F}\omega_X. \qquad\text{(16)}$$

The comparison (14) is generally not an isomorphism. For example, if $i$ includes a point in a positive-dimensional manifold and $A=k_W$, its two restrictions are $\operatorname{or}_{W,x}[-n]$ and $k$. Their degrees differ. Replacing $i^!$ by $i^{-1}$ would erase precisely the local support information retained in (16).

The characteristic class has closed support

Set $Z=\operatorname{supp}(F)$, with the closed-support convention above, and abbreviate $E_F=R\mathcal Hom(F,F)$. Outside $Z$ the restriction of $F$ is zero, so $E_F$ is zero there as well. Thus $E_F$ is supported on $Z$.

If $i:Z\hookrightarrow X$ is the closed embedding, support localization gives $R\Gamma_ZA=i_*i^!A$ and an isomorphism

$$R\Gamma_ZE_F\xrightarrow{\sim}E_F. \qquad\text{(17)}$$

Indeed the complementary open restriction of $E_F$ vanishes, so its localization triangle has zero open term. For any map $v:E_F\to\omega_X$, apply $R\Gamma_Z$ to $v$ and use the inverse of (17). This gives its unique supported lift

$$E_F\longrightarrow R\Gamma_Z\omega_X. \qquad\text{(18)}$$

Uniqueness follows from the adjunction between the inclusion of complexes supported on $Z$ and $R\Gamma_Z$. It is the supported source $E_F$ that gives this uniqueness. A map from $k_X$ whose open restriction happens to vanish would not by itself justify a unique lift.

Use (16) for $v$ in (18), then precompose with $e_F$. The image of the global unit is

$$C(F)\in H_Z^0(X;\omega_X),\qquad k_X\xrightarrow{e_F}E_F \longrightarrow R\Gamma_Z\omega_X. \qquad\text{(19)}$$

Here $H_Z^0(X;\omega_X)=H^0R\Gamma(X;R\Gamma_Z\omega_X)$. Formula (19) defines the supported trace class used in this lesson. If $Z\subset S$ with $S$ closed, the inclusion of support conditions gives a natural map $R\Gamma_Z\omega_X\to R\Gamma_S\omega_X$. The image of (19) is the class with support condition $S$.

All maps are natural under an isomorphism of $F$ in the derived category: its identity conjugates to the new identity, evaluation pairs the conjugate morphisms, and the diagonal comparisons are natural. This proves that (19) depends on the derived object, not on a chosen representative. They are also compatible with open restriction, since the exceptional comparisons, closed-diagonal counit and supported localization all restrict to their counterparts on an open subset.

On a positive-dimensional manifold, $C(F)$ has values in a dualizing complex. It is not obtained by placing the stalk numbers $\chi(F)(x)$ in degree-zero constant coefficients. Relating constructible functions to these geometric classes is a later theorem.

A point fixes the trace sign

Take $X=\{\mathrm{pt}\}$. Its dualizing complex is $k$, both diagonal restrictions are the identity, and $\beta$ is the identity. Let $P$ be a bounded complex of finite-dimensional vector spaces. In these conventions the tensor–Hom map is

$$\Phi:P\otimes P^\vee\longrightarrow\operatorname{Hom}^\bullet(P,P), \qquad \Phi(p\otimes\varphi)(q)=p\,\varphi(q). \qquad\text{(20)}$$

For homogeneous $\varphi$ of degree $b$, the dual differential is $d\varphi=(-1)^{b+1}\varphi d$. Consequently (20) is a chain map: its tensor differential evaluates as $dp\,\varphi(q)+(-1)^{a+b+1}p\,\varphi(dq)$, which is the Hom differential for an element of degree $a+b$. This checks the normalization used in (11)–(13).

Choose homogeneous bases $e_{q,j}$ of $P^q$, with dual basis $e_{q,j}^*$ of degree $-q$. Under (20), the element representing the identity is

$$\sum_{q,j}e_{q,j}\otimes e_{q,j}^*.$$

It is closed because $\mathrm{id}_P$ is a chain map. Applying (10), the swap contributes $(-1)^{q(-q)}=(-1)^q$, and evaluation contributes $1$. Thus the trace of the identity is $\sum_q(-1)^q\dim P^q$ as an element of $k$.

We must still show that this alternating count of terms equals the alternating count of cohomology. Write $B^q=\operatorname{im}d^{q-1}$ and $Z^q=\ker d^q$. The two finite-dimensional exact sequences

$$0\longrightarrow Z^q\longrightarrow P^q\longrightarrow B^{q+1}\longrightarrow0, \qquad 0\longrightarrow B^q\longrightarrow Z^q\longrightarrow H^q(P)\longrightarrow0$$

give $\dim P^q=\dim B^q+\dim H^q(P)+\dim B^{q+1}$. The boundary contributions cancel in the finite alternating sum. Therefore

$$C(P)=\chi(P)\,1_k. \qquad\text{(21)}$$

This cancellation is a computation on a representative; it does not assert a canonical splitting of a general sheaf complex into its cohomology.

More generally a degree-zero chain endomorphism $u$ preserves $B^q$ and $Z^q$. Trace is additive on a finite invariant subspace and its quotient: a basis adapted to the subspace makes its matrix block triangular. The isomorphism $P^q/Z^q\simeq B^{q+1}$ conjugates the induced endomorphisms. The same cancellation proves

$$\operatorname{str}(u):=\sum_q(-1)^q\operatorname{tr}(u^q) =\sum_q(-1)^q\operatorname{tr}(H^q(u)). \qquad\text{(22)}$$

A chain homotopy changes neither side. One can also see the invariance directly: for a degree $-1$ map $h$, the terms in $\operatorname{str}(dh+hd)$ cancel after reindexing, using $\operatorname{tr}(AB)=\operatorname{tr}(BA)$ for maps between two finite-dimensional vector spaces. This last identity follows by writing both traces as the same sum of matrix products.

The integer in (21) embeds into $k$ because the characteristic is zero. The same graded construction over a positive-characteristic field would return the image of that integer in the field; it could lose its value. For instance, the identity of $k^2$ has Euler index $2$ but trace $0$ in characteristic two.

Shifts and triangles give local accounting rules

With the cohomological convention $H^q(P[r])=H^{q+r}(P)$, reindexing (2) gives

$$\chi(P[r])=(-1)^r\chi(P). \qquad\text{(23)}$$

If $P\to Q\to R\xrightarrow{+1}$ is a distinguished triangle of bounded finite coefficient complexes, its long exact cohomology sequence is a finite exact sequence after appending zero terms. Alternating dimensions in any finite exact sequence sum to zero: writing each term as its incoming image plus outgoing image cancels consecutive contributions. Applied in the order $H^q(P),H^q(Q),H^q(R),H^{q+1}(P)$, this yields

$$\chi(Q)=\chi(P)+\chi(R). \qquad\text{(24)}$$

Stalk and costalk functors preserve distinguished triangles. Equations (23)–(24) consequently apply pointwise to both functions in (4). They also apply to global indices whenever all three section complexes satisfy the finiteness condition in (8).

These rules concern Euler indices. We have not yet proved the corresponding general additivity theorem for sheaf characteristic classes. At a point it follows from (21) and (24), but the geometric statement requires its own trace argument.

Exercises with complete solutions

Cancel the boundaries in a nontrivial chain trace

Difficulty: Intermediate.

Let $P^0=k^2$, $P^1=k^3$, with differential $d(x,y)=(x,0,0)$ and all other terms zero. Let $u^0=\operatorname{diag}(a,b)$ and $u^1=\operatorname{diag}(a,c,e)$. Verify that $u$ is a chain map, compute its trace through cohomology and through terms, and calculate $C(P)$ for $u=\mathrm{id}$. Explain why adding a chain homotopy does not change the answer.

Solution. The equality $du^0=u^1d$ is $(ax,0,0)=(ax,0,0)$. Its zeroth cohomology is $\ker d=k(0,1)$, where $u$ acts by $b$. Its first cohomology is $k^3/k(1,0,0)$, with induced diagonal entries $c,e$. Thus the alternating cohomology trace is $b-c-e$.

The alternating term trace is $(a+b)-(a+c+e)=b-c-e$. The same entry $a$ occurs on the boundary and its preceding quotient, so it cancels. For the identity, $C(P)=(2-3)1_k=-1_k$, also equal to $(1-2)1_k$ from cohomology.

If $u$ changes by $dh+hd$, its cohomology action is unchanged. Explicitly, write $h:P^1\to P^0$ as a $2\times3$ matrix. The only potentially nonzero traces are $\operatorname{tr}(hd)$ on $P^0$ and $\operatorname{tr}(dh)$ on $P^1$, both the top-left entry of $h$. They have opposite signs. Their difference is zero, so the evaluated trace is homotopy invariant.

A cone accounts for a shift without choosing a splitting

Difficulty: Introductory.

Let $v:k^2\to k^3$ have rank $r$, and let $R=\operatorname{Cone}(v)$, with the input spaces placed in degree zero. Find $\chi(R)$ from its cohomology and from its distinguished triangle. What are $C(k[1])$ and $C(k\oplus k[1])$ on a point? Does vanishing of this last class imply that its complex is zero?

Solution. The cone has $H^{-1}(R)=\ker v$ of dimension $2-r$ and $H^0(R)=\operatorname{coker}v$ of dimension $3-r$. Consequently $\chi(R)=-(2-r)+(3-r)=1$. The triangle $k^2\to k^3\to R\xrightarrow{+1}$ gives the same result $3-2=1$ through (24).

The shift $k[1]$ has its nonzero cohomology in degree $-1$, so (21) gives $C(k[1])=-1_k$. Direct-sum evaluation makes the identity block diagonal, hence $C(k\oplus k[1])=1_k-1_k=0$. The complex still has two nonzero cohomology groups. Its characteristic class is a signed trace and need not detect the object. No sheaf-level splitting or general geometric additivity theorem was used.

Calculate the closed-embedding comparison rather than replacing it

Difficulty: Advanced.

Let $i:\{0\}\hookrightarrow\mathbb R$ and $A=k_{\mathbb R}$. Compute $i^!A$, $i^{-1}A$ and $\beta_{i,A}$. Then take $A=i_*P$ for a bounded finite coefficient complex $P$, and compute the same map. Prove uniqueness of the support lift (18), and indicate the hypothesis that makes the proof work.

Solution. Point-supported cohomology of a constant sheaf on a line is the fibre of $k\to k\oplus k$, with map $a\mapsto(a,a)$, obtained by deleting the point from a small interval. This fibre is $k[-1]$. Ordinary restriction is $k$. Since $\operatorname{Hom}_{D(k)}(k[-1],k)=\operatorname{Ext}^1_k(k,k)=0$, the comparison $\beta_{i,A}$ is the zero morphism. Its source is nevertheless nonzero.

For $A=i_*P$, closed-support localization is an isomorphism $i_*i^!A\to A$. The equivalences $i^!i_*P\simeq P$ and $i^{-1}i_*P\simeq P$ turn the defining counit in (14) into the identity. Thus $\beta_{i,i_*P}=\mathrm{id}_P$. The contrast records dependence on the object; it does not give an isomorphism between the two functors in general.

For the support lift, let $E$ be supported on a closed set $Z$ and let $j:X\setminus Z\hookrightarrow X$. The triangle $R\Gamma_ZE\to E\to Rj_*j^{-1}E\xrightarrow{+1}$ has zero open term. Hence $R\Gamma_ZE\to E$ is an isomorphism. The right adjunction for $R\Gamma_Z$ gives

$$\operatorname{Hom}(E,R\Gamma_ZA)\xrightarrow{\sim}\operatorname{Hom}(E,A)$$

for every $A$. This proves existence and uniqueness of the lift of $E\to A$, and applying the functor to that map produces it explicitly. The needed hypothesis is that the source $E$ is supported on $Z$. In (19) we apply it to $E=E_F$, then precompose with the unit; we do not assume $k_X$ is supported on $Z$.

Check finiteness before reading a trace as an integer

Difficulty: Intermediate.

Compare $F=k_{\mathbb R^n}$, a constant sheaf on a countable discrete manifold, and a bounded constructible complex with compact closed support. Determine which global Euler indices are defined in these examples. Why does a field-valued trace cease to recover the integer index in characteristic $p>0$?

Solution. Contractibility gives $R\Gamma(\mathbb R^n;k)=k$, while the compactly supported cohomology of an oriented real $n$-ball, or its one-point compactification, gives $R\Gamma_c(\mathbb R^n;k)=k[-n]$. The same holds for $\mathbb R^n$ by the usual exhaustion and compact-support extension maps. Thus both indices are defined, with values $1$ and $(-1)^n$. Noncompactness by itself does not prevent finiteness, and it does not force ordinary and compact indices to agree.

On the countable discrete manifold, the section groups are the infinite product and direct sum displayed above. Each contains arbitrarily large finite linearly independent sets, so its dimension is infinite. The local Euler functions remain $1$, but (8) supplies no global integer.

A bounded constructible complex with compact closed support has perfect ordinary and compact section complexes by compact finiteness. The map (9) identifies them, so both indices exist and agree. These are consequences of the support and coefficient hypotheses, not of counting local ranks alone.

Over characteristic $p$, the supertrace of the identity still computes the image of the alternating integer in $k$. The nonzero integer $p$ maps to zero, for example for a $p$-dimensional vector space in degree zero. Equality of field-valued traces therefore determines the integer only modulo $p$. Characteristic zero makes the map $\mathbb Z\to k$ injective and avoids this loss.

Keep the real orientation line and the cohomological shift

Difficulty: Intermediate.

Let $L$ be a finite-rank local system on a real $n$-manifold, and let $F=L[r]$. Compute $A_x(F)$, $C_x(F)$, $(D_XF)_x$ and both functions in (4). Which computations require an orientation of the entire manifold? Which shift would be incorrect if one treated the manifold as complex without that hypothesis?

Solution. On a small coordinate ball, $L$ is constant with fibre $L_x$. Ordinary restriction gives $A_x(F)=L_x[r]$. The local orientation calculation gives

$$C_x(F)=L_x\otimes\operatorname{or}_{X,x}[r-n].$$

Duality reverses the coefficient shift and contributes the manifold dualizing complex, so

$$(D_XF)_x=L_x^\vee\otimes\operatorname{or}_{X,x}[n-r].$$

These expressions also match (5), using the canonical self-duality of the orientation line: its transition functions are signs, whose inverse equals itself. Consequently

$$\chi(F)(x)=(-1)^r\operatorname{rank}L_x,\qquad \chi_c(F)(x)=(-1)^{r-n}\operatorname{rank}L_x =(-1)^{n-r}\operatorname{rank}L_x.$$

The final equality is equality of parities. All calculations are local and retain the orientation line, so none requires a global orientation. One may trivialize that line only after choosing an orientation. The shift is the real dimension $n$. Replacing it with $2n$ is justified only when $n$ instead denotes a complex dimension of a complex manifold, which is a different hypothesis and convention.

Compare an open interval and a closed interval at every point

Difficulty: Advanced.

On $X=\mathbb R$, put $F_{c}=k_{[0,1]}$ and $F_{o}=k_{(0,1)}$, the latter extended by zero. Compute both local Euler functions at interior points, endpoints and exterior points, then both global indices. Verify (6) using the dual sheaves and explain why the zero endpoint stalks of $F_{o}$ do not remove the endpoints from its closed support.

Solution. At an interior point both sheaves are locally constant $k$, so their ordinary stalk is $k$ and their costalk is $k[-1]$. At an endpoint, use the localization fibre

$$C_x(F)\longrightarrow F_x \longrightarrow R\Gamma(B\setminus\{x\};F)\xrightarrow{+1}$$

in a sufficiently small interval $B$. For $F_{c}$, the middle term is $k$ and the punctured term is $k$ on the side inside $[0,1]$; restriction is the identity. The costalk is zero. For $F_{o}$ the middle term is zero and the punctured term is $k$ on the interior side. The fibre is $k[-1]$. At an exterior point both measurements vanish. The complete table is

For $j:(0,1)\hookrightarrow\mathbb R$, open internal-Hom adjunction gives $D_{\mathbb R}F_{o}\simeq Rj_*\omega_{(0,1)}=Rj_*k_{(0,1)}[1]$. On a small interval about either endpoint, the nonempty intersection with $(0,1)$ is a contractible interval. Its derived constant sections are $k$ and the restriction maps preserve that constant value. Thus the actual constant-section comparison gives $Rj_*k_{(0,1)}\simeq F_{c}$. This proves $D_{\mathbb R}F_{o}\simeq F_{c}[1]$ as a sheaf complex, with its maps. Constructible biduality and reversal of shifts now give $D_{\mathbb R}F_{c}\simeq F_{o}[1]$. Taking their stalk Euler indices reproduces the two costalk rows, including endpoints, and verifies (6).

For global sections, the closed interval is contractible and compact, giving $R\Gamma(\mathbb R;F_{c})=R\Gamma_c(\mathbb R;F_{c})=k$. Extension by zero identifies compact sections of $F_{o}$ with compact sections on the open interval, giving $R\Gamma_c(\mathbb R;F_{o})=k[-1]$. Its closed support is $[0,1]$, so (9) also gives $R\Gamma(\mathbb R;F_{o})=k[-1]$. Therefore both global indices of $F_{c}$ are $1$ and both global indices of $F_{o}$ are $-1$.

Every neighborhood of either endpoint contains interior points where $F_{o}$ has nonzero stalk. There is no open vanishing neighborhood of an endpoint. Hence the endpoints belong to the closed support used for properness and supported characteristic classes. This remains true despite their zero ordinary stalks.

References

Masaki Kashiwara, Index theorem for constructible sheaves, Astérisque 130 (1985), §8.3–8.4, pp. 205–206, expresses local stalk and costalk Euler indices through characteristic-cycle intersections and relates constructible functions to cycles; freely readable article. Those intersection statements belong to the subsequent geometric arguments. The normalized supported map and point trace in this lesson are constructed above using the linked product evaluation, exceptional-Hom and composition proofs.

What the construction prepares

The identity, the normalized diagonal evaluation, the closed-embedding comparison and contraction now define the supported class (19). Its point value is the Euler index with the graded sign proved in (21). Local stalk and costalk indices and their elementary accounting rules are established, with separate global finiteness conditions. The next lesson must compare this full chain with proper direct image on the closed support, retaining the unit, exceptional and ordinary exchanges and the evaluated tensor map. Only that compatibility will identify the integral of $C(F)$ with the global Euler index for compact support; cotangent characteristic cycles require further geometric constructions beyond this lesson.

Source account for the supported normalization

Schapira, An Introduction to Sheaves on Grothendieck Topologies, §§4.6–4.8, provides the exceptional-Hom, dual-sections and external-Hom framework. Its §4.8 states a Noetherian coefficient convention and explains the perfect-complex replacement; its external-Hom proposition has a bounded second input. Both restrictions hold in this field-coefficient, bounded construction. The more general neighbourhood-system proof required by the linked provider remains that provider’s explicit argument. Kashiwara, Index theorem for constructible sheaves, §8.3–8.4, concerns local Euler indices and characteristic cycles, not a substitute proof of the supported diagonal map. Here that map is derived from the cited operation contracts in (9)–(19), and its point sign is proved in (20)–(21). Additivity of Euler numbers is proved; additivity of the supported class, its proper transport, and the global index theorem are not asserted without their further proofs.