Lefschetz traces of constructible correspondences

A correspondence can move information from one point to another before taking a trace. Its local class lives where the two maps coincide. Compactness of their common support makes the trace finite even when the original sheaf has noncompact support and infinite-dimensional global cohomology.

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

The identity and evaluated tensor fixes the diagonal identity and its graded contraction. The evaluated proper-Hom map identifies the dual of the return map through its actual pairing. The Hom comparison before contraction and its evaluation square transport the identity through a proper image on its closed coefficient support.

Perfect operations and compact supported cohomology provide the finite intermediate complex. The supported intersection construction retains the order of the graph and diagonal factors and their orientation coefficients. The proof below specifies the two support comparisons, verifies the dual return map by evaluation, and uses a compact cutoff to allow noncompact sheaf support. No properness of either correspondence map is imposed.

The two maps determine three supports

Let $k$ be a characteristic-zero field. Manifolds are real analytic, Hausdorff and countable at infinity, with the standing finite uniform dimension bounds. Let $f,g$ be real analytic maps and take

f,g:Y\longrightarrow X,\qquad F\in D^b_{\mathbb R\text{-c}}(k_X),\qquad \phi:f^{-1}F\longrightarrow g^!F. \qquad\text{(1)}

Constructibility includes perfect stalks. Set

\begin{aligned} S&=\operatorname{supp}(F),\\ T&=f^{-1}S\cap g^{-1}S,\\ W&=\{y:f(y)=g(y)\},\qquad Z=T\cap W. \end{aligned} \qquad\text{(2)}

Support means closed support: the complement of the largest open set where the complex vanishes. Thus $S,T,Z$ are closed subanalytic sets. The equality set $W$ is the inverse image of the closed diagonal under $h=(f,g)$.

Put $A=f^{-1}F$ and $B=g^!F$. The perfect-operation theorem makes both constructible. Their supports lie in $f^{-1}S$ and $g^{-1}S$, respectively. Since the source $A$ is supported on $f^{-1}S$, supported adjunction gives a unique lift

\widehat\phi:A\longrightarrow R\Gamma_TB \longrightarrow B \qquad\text{(3)}

of $\phi$. Indeed $R\Gamma_{f^{-1}S}A=A$, and $R\Gamma_{f^{-1}S}B=R\Gamma_TB$: $B$ already vanishes outside $g^{-1}S$. The equality uses the composition law for closed support functors. The lift comes from a supported source, rather than from an arbitrary map whose restriction happens to vanish.

Assume for the trace theorem that

T\text{ is compact}. \qquad\text{(4)}

Then

C=R\Gamma(X;F),\qquad P=R\Gamma_T(Y;B) \qquad\text{(5)}

have different roles. To see that $P$ is perfect, write $B_T=R\Gamma_TB=R\mathcal Hom(k_T,B)$. Perfect internal Hom makes $B_T$ bounded constructible, and its closed support lies in compact $T$. Proper direct-image finiteness for its map to a point gives $P=R\Gamma(Y;B_T)$ perfect. The cohomology of $C$ need not be finite-dimensional.

Ordinary pullback followed by (3) gives $a:C\to P$. For the return map use

\begin{aligned} b:P&=R\Gamma_c(Y;R\Gamma_TB) \longrightarrow R\Gamma_c(Y;g^!F)\\ &\longrightarrow R\Gamma_c(X;F) \longrightarrow R\Gamma(X;F)=C. \end{aligned} \qquad\text{(6)}

The middle arrow is the exceptional counit $Rg_!g^!F\to F$; the last forgets compact support. The first equality follows from compactness of $T$. This constructs the correspondence endomorphism

U_\phi=ba:C\longrightarrow C. \qquad\text{(7)}

No ordinary direct-image counit $Rg_*g^!F\to F$ has been inserted. It would not be supplied by the adjunction being used when $g$ is nonproper.

A finite-rank trace does not require a finite vector space

For a finite-rank endomorphism $u:V\to V$, choose a finite-dimensional subspace $E$ containing its image. It is invariant under $u$, and define

\operatorname{tr}_{\mathrm{fin}}(u)=\operatorname{tr}(u|_E). \qquad\text{(8)}

Lemma. This definition is independent of $E$. If $p:V\to Q$ and $q:Q\to V$, with $Q$ finite-dimensional, then

\operatorname{tr}_{\mathrm{fin}}(qp)=\operatorname{tr}(pq). \qquad\text{(9)}

Proof. For two choices $E_1,E_2$, work in $E_1+E_2$. Its quotient by either $E_i$ has zero induced endomorphism because the full image lies in $E_i$. A basis adapted to an invariant subspace gives a block triangular matrix, so the trace on the sum equals the trace on either $E_i$.

For (9), take $E=\operatorname{im}q$. Regard $q$ as a map $Q\to E$ and restrict $p$ to $E$. In finite bases the two traces are both the sum $\sum_{i,j}q_{ij}p_{ji}$. This proves (9), including nonsurjective $q$ and noninjective $p$. $\square$

Each $H^j(U_\phi)$ factors through the finite-dimensional $H^j(P)$. It has finite rank, and it is zero whenever $H^j(P)=0$. Consequently

L(\phi)=\sum_j(-1)^j \operatorname{tr}_{\mathrm{fin}}H^j(U_\phi) =\operatorname{str}(ab:P\to P) \qquad\text{(10)}

is a finite sum. The point trace proof in the first prerequisite identifies the supertrace of a perfect complex endomorphism with this alternating cohomology trace. Formula (9) proves the equality in (10) degree by degree.

This is a trace in $k$. It need not be an integer: the correspondence may act by arbitrary coefficient endomorphisms. If $F$ has compact closed support, $C$ is perfect, and (10) is its usual supertrace.

In the self-map case $Y=X$, $g=\mathrm{id}_X$, compact $S$ already makes $T=f^{-1}S\cap S$ compact. Here the exceptional counit in (6) is the identity, and $b$ is exactly the support-forgetting map $R\Gamma_T(X;F)\to R\Gamma(X;F)$. The supported lift (3) composes to $\phi$, so $U_\phi$ is the ordinary composite $R\Gamma(X;F)\to R\Gamma(X;f^{-1}F)\xrightarrow{R\Gamma(\phi)}R\Gamma(X;F)$. This holds for every analytic $f$, with no properness assumption on $f$.

Pull the diagonal identity to the coincidence set

Write $D_XF=R\mathcal Hom(F,\omega_X)$ and $K=F\boxtimes D_XF$. The normalized diagonal comparison from the trace prerequisite is

R\mathcal Hom(F,F)\simeq\delta^!K,\qquad \delta:X\hookrightarrow X\times X. \qquad\text{(11)}

For clarity, the external-Hom map used in (11) is the evaluated isomorphism $F\boxtimes D_XF\simeq R\mathcal Hom(p_2^{-1}F,p_1^!F)$. Exceptional restriction along $\delta$, followed by $p_i\delta=\mathrm{id}_X$, recovers $R\mathcal Hom(F,F)$ with its actual evaluation order. Under this inverse comparison and closed exceptional adjunction, $\mathrm{id}_F$ becomes a particular morphism $u_\Delta:k_\Delta\to K$. It represents the supported diagonal class

e_\Delta(F)\in H^0_\Delta(X\times X;K). \qquad\text{(12)}

Its underlying construction is the actual exceptional diagonal restriction, with the product evaluation and its graded tensor order.

For any continuous map $h:Y\to X\times X$ in the standing sheaf-theoretic setting, the ordinary unit $K\to Rh_*h^{-1}K$, followed by supported sections, supplies

R\Gamma_\Delta(X\times X;K) \longrightarrow R\Gamma_{h^{-1}\Delta}(Y;h^{-1}K). \qquad\text{(13)}

Here is the derived adjunction check of the support comparison. Let $i:W=h^{-1}\Delta\hookrightarrow Y$ and let $v:W\to X$ be the map to the diagonal. Closed proper base change gives $h^{-1}\delta_*Q=i_*v^{-1}Q$. For a bounded test object $Q$, the adjunctions consequently give $\begin{aligned} \operatorname{Hom}(Q,\delta^!Rh_*H) &\simeq\operatorname{Hom}(\delta_*Q,Rh_*H)\\ &\simeq\operatorname{Hom}(h^{-1}\delta_*Q,H)\\ &\simeq\operatorname{Hom}(v^{-1}Q,i^!H) \simeq\operatorname{Hom}(Q,Rv_*i^!H). \end{aligned} \qquad\text{(13a)}$ Thus $\delta^!Rh_*H\simeq Rv_*i^!H$, with the isomorphism fixed by these adjunctions. Apply it to the ordinary unit $K\to Rh_*h^{-1}K$, then take global sections. The result is (13), since $R\Gamma(W;i^!h^{-1}K)=R\Gamma_W(Y;h^{-1}K)$. This proves its map and support, without imposing properness on $h$ or choosing a pullback of dualizing complexes. Equivalently, apply exact ordinary inverse image to $u_\Delta$; closed base change identifies its source $h^{-1}k_\Delta$ with $k_W$. This morphism $k_W\to h^{-1}K$ represents precisely (13), since its adjunction transpose is the same ordinary unit followed by (13a). This description proves compatibility with further ordinary pullback, open restriction, and support-forgetting maps. Analyticity of $h$ is not used for this comparison.

For $h=(f,g)$, constructible duality gives the evaluated comparison

h^{-1}K =f^{-1}F\otimes g^{-1}D_XF \simeq A\otimes D_YB. \qquad\text{(14)}

Apply exceptional-Hom duality to $D_XF$: $g^!D_XD_XF\simeq D_Y(g^{-1}D_XF)$. Constructible biduality identifies its left side with $B$; dualizing again gives the actual comparison $D_YB\simeq g^{-1}D_XF$. This proves (14) through evaluation and biduality, without identifying an arbitrary exceptional inverse image with a shifted ordinary inverse image.

The tensor in (14) vanishes off $T$, since its two factors vanish off $f^{-1}S$ and $g^{-1}S$. Thus the closed-support composition law identifies $R\Gamma_W(h^{-1}K)$ with $R\Gamma_Z(h^{-1}K)$. Equivalently, closed adjunction factors the pulled-back morphism $k_W\to h^{-1}K$ uniquely through $k_Z$. This is the supported refinement used below; it does not infer a supported lift merely from a class whose restriction happens to vanish.

Apply $\phi$ to the first factor and then the ordered contraction $B\otimes D_YB\to\omega_Y$. This defines

C(\phi)\in H^0_Z(Y;\omega_Y). \qquad\text{(15)}

Explicitly, (15) is the composite $k_Z\to A\otimes D_YB\xrightarrow{\phi\otimes1}B\otimes D_YB\to\omega_Y$, with the first arrow just constructed. The last arrow is graded symmetry followed by dual-first evaluation; its point value is the supertrace with signs $(-1)^j$. All support maps and both coefficient arrows are specified before taking cohomology. Construction (11)–(15) requires no compactness of $T$. Compactness is used for the global integration and finite intermediate trace.

The class is linear in $\phi$, invariant under an isomorphism of the coefficient complex compatible with $\phi$, and compatible with open restriction. These follow from the actual units, counits, support comparison and evaluation: conjugating the coefficient object conjugates the identity in (12), and evaluation cancels the two conjugate factors.

The compact-support trace square

Theorem. Under (1) and (4), with the finite-rank meaning (10),

L(\phi)=\int_Y C(\phi). \qquad\text{(16)}

The integral is the map from support $Z$ to compact support, followed by the proper point trace $R\Gamma_c(Y;\omega_Y)\to k$.

First suppose that $S$ is compact. Then $C$ is perfect and its ordinary and compact section complexes agree. Evaluated duality for the map to a point gives $R\Gamma(X;D_XF)\simeq C^\vee$. Its product comparison identifies $R\Gamma(X\times X;K)$ with $C\otimes C^\vee$: both coefficient supports are compact, so the proper-image product and projection maps apply to these factors. The map forgetting the diagonal support in (12) is therefore a morphism $k\to C\otimes C^\vee$.

The endomorphism comparison in the proper characteristic-class proof identifies this particular morphism with $\mathrm{coev}_C$. Indeed its Hom route takes $\mathrm{id}_F$ first to the exceptional unit of its supported direct image and then to the identity of $C$, by the exceptional triangular identity. The product evaluation comparison identifies that route with the tensor route before contraction. Consequently the tensor is the identity tensor itself, not merely some tensor with the same numerical trace.

The ordinary $f$-pullback and the supported lift of $\phi$ send the first tensor factor to $P$ by $a$. Put $B_T=R\Gamma_TB$. The internal-Hom duality formula gives $B_T\simeq D_Y(k_T\otimes D_YB)$; constructible biduality therefore gives $D_YB_T\simeq k_T\otimes D_YB$. Both sides have compact support. Evaluated global duality identifies $P^\vee$ with $R\Gamma(T;(D_YB)|_T)$, pairing a supported $B$-section with the ordinary restriction of a $D_YB$-section to $T$.

The candidate second-factor map is $R\Gamma(X;D_XF)\to R\Gamma(Y;g^{-1}D_XF)\simeq R\Gamma(Y;D_YB)\to R\Gamma(T;(D_YB)|_T)$. The first arrow is ordinary pullback and the last is ordinary closed restriction. Under the two evaluated duality identifications this map is $b^\vee:C^\vee\to P^\vee$, as the following pairing check shows.

Pair a section of $B_T$ with the ordinary pullback of a dual section on $X$. Projection formula moves the latter through $Rg_!$; the ordinary inverse-image counit evaluates its pulled-back germ. The coefficient comparison (14) then contracts it with $g^!F$. Evaluated internal adjunction identifies this contraction followed by the $g$-trace with evaluation after $Rg_!g^!F\to F$. Composing with the point trace gives the pairing of the original dual section with (6). These are equalities of derived pairings, not just pointwise section values: they are the curry/evaluation identity in the closed-support prerequisite, followed by the support-forgetting identity (EX.41). The properly supported factor is $B_T$, whose support lies in compact $T$; no properness of $g$ on all of $Y$, or on all of $\operatorname{supp}(B)$, is needed. Currying this equality of pairings proves the asserted formula for $b^\vee$, with the same graded order.

The resulting trace square can be written

\begin{array}{ccc} k&\xrightarrow{\mathrm{coev}_C}&C\otimes C^\vee\\ &&\downarrow\,a\otimes b^\vee\\ &&P\otimes P^\vee\xrightarrow{\mathrm{contr}_P}k. \end{array} \qquad\text{(17)}

For a homogeneous basis $e_{j,r}$ of a finite representative of $C$, its value is $\sum_{j,r}(-1)^j e^*_{j,r}(ba(e_{j,r}))$. Boundary traces cancel by the point trace proof, leaving $\operatorname{str}(ba)$.

To identify the sheaf path with this tensor path, first enlarge the class support from $Z$ to $T$. Ordinary pullback of the representing diagonal morphism, as checked after (13a), commutes with forgetting diagonal support. Thus its image is the ordinary pullback of the actual identity tensor just computed. Naturality of tensor pullback and the supported lift (3) takes its first factor by $a$; the second factor is the ordinary restriction of the pulled-back dual section to $T$, the map just proved to be $b^\vee$. Their contraction and point trace are exactly $\mathrm{contr}_P$, by the evaluated duality pairing of $B_T$.

Integration of the original $Z$-supported class factors through that same enlargement to $T$, since $Z\subset T$ and $T$ is compact. Hence this comparison proves equality of the two integrals without inferring equality of $Z$-supported classes from their images after support is forgotten. The identity/Hom calculation supplies the top cell of (17), ordinary support pullback supplies its middle cell, and the verified pairing supplies its bottom cell. This proves (16) for compact $S$.

A compact cutoff removes the extra support assumption

Keep compact $T$, but allow $S$ to be noncompact. Choose a compact subanalytic $K_0\subset X$ whose interior contains $f(T)\cup g(T)$. Such a set is obtained from finitely many closed coordinate balls covering this compact image, with closures inside their charts.

Set

G=R\Gamma_{K_0}F,\qquad c:G\longrightarrow F. \qquad\text{(18)}

The perfect-operation theorem makes $G$ constructible; its support is contained in $S\cap K_0$, hence compact. On the interior of $K_0$, $c$ is the identity comparison. It therefore induces an isomorphism

R\Gamma_T(g^!G)\xrightarrow{\sim}R\Gamma_T(g^!F). \qquad\text{(19)}

Indeed $g^!c$ is an isomorphism on the open neighbourhood $g^{-1}(\operatorname{int}K_0)$ of $T$; supported excision proves (19).

Define the cutoff correspondence by

f^{-1}G\longrightarrow f^{-1}F \xrightarrow{\widehat\phi}R\Gamma_Tg^!F \xrightarrow{(19)^{-1}}R\Gamma_Tg^!G \longrightarrow g^!G. \qquad\text{(20)}

Its common closed support $T_G$ equals $T$. Support containment gives $T_G\subset T$, while at every $y\in T$, $G$ agrees with $F$ near both $f(y)$ and $g(y)$, proving the reverse inclusion. On a neighbourhood of $T$, the maps and objects in (20) identify with those of $\phi$. In particular $Z_G=Z$. Excision and the naturality of (11)–(15) consequently give

C(\phi_G)=C(\phi)\quad\text{in }H_Z^0(Y;\omega_Y). \qquad\text{(21)}

For the traces, identify the intermediate supported complex by (19). Put $C_G=R\Gamma(X;G)$, and let $c_C:C_G\to C$ be induced by $c$. Naturality of ordinary pullback and of the exceptional counit gives

a_G=a\,c_C,\qquad b=c_C\,b_G. \qquad\text{(22)}

Therefore $U_\phi=c_Cb_Ga$ and $U_{\phi_G}=b_Gac_C$. Each degree factors through finite-dimensional $H^j(C_G)$, so (9) proves equality of their alternating traces. Applying the compact-$S$ theorem to $G$ and using (21) proves (16) in full.

This proves the theorem for noncompact $S$ with compact common support $T$. The cutoff introduces no new choice in the answer: its trace is the intrinsic finite-rank trace of (7), and (21) identifies its class with the intrinsic supported class (15).

Local contributions keep their own trace

For an isolated $y\in Z$, support excision separates $y$ from $Z\setminus\{y\}$. Project (15) to $H^0_{\{y\}}(Y;\omega_Y)=k$, using the normalized point trace, and call the result $C_y(\phi)$. If $Z$ is finite, disjoint neighbourhoods give the direct sum of its point-supported complexes. Proper trace additivity and (16) then yield

L(\phi)=\sum_{y\in Z}C_y(\phi). \qquad\text{(23)}

The point identification retains the orientation line and dimension shift in $i_y^!\omega_Y=k$; it introduces no additional ambient sign.

Now specialize to $Y=X$, $g=\mathrm{id}$, and $\phi:f^{-1}F\to F$. At a fixed point $x$, ordinary restriction gives a stalk endomorphism $\phi_x$. If $x$ is also isolated in $f^{-1}(x)\cap T$, there is a point-support action on the complexes of sections

\begin{aligned} R\Gamma_{\{x\}}(X;F) &\longrightarrow R\Gamma_{f^{-1}(x)}(X;f^{-1}F)\\ &\longrightarrow R\Gamma_{f^{-1}(x)}(X;F) \longrightarrow R\Gamma_{\{x\}}(X;F). \end{aligned} \qquad\text{(24)}

The first arrow is the ordinary supported pullback (13) for the closed set $\{x\}$. For the second arrow use the supported lift (3): after composition of support functors its output has support in $E=f^{-1}(x)\cap T$. Since $x$ is isolated in the closed set $E$, both $\{x\}$ and $E\setminus\{x\}$ are closed, disjoint components of this support. The support complex splits over these two sets, and its projection to the first is the last arrow of (24). No compactness of the second component is required for this finite splitting. The point-supported complex is perfect by compact supported-cohomology finiteness. Thus (24) has a defined supertrace, but its projection is not an ordinary stalk restriction. The traces of $\phi_x$, of (24), and the number $C_x(\phi)$ can differ.

The local number depends only on the germ of the pair: if $F$ and $F'$ are isomorphic on a neighbourhood $U$ of an isolated fixed point, and the isomorphism intertwines their morphisms on $U\cap f^{-1}U$, then $C_x(\phi)=C_x(\phi'). \qquad\text{(25)}$ To prove this, choose a smaller neighbourhood whose image under $f$ remains in $U$. Open restriction of (11)–(15) identifies the two classes there; excision and the same normalized point trace identify their local numbers. No global isomorphism is needed.

The constant sheaf gives a graph intersection

Let $X$ be compact, $F=k_X$, and $\phi:f^{-1}k_X\to k_X$ the canonical map. Write $\Gamma_f=\{(f(x),x)\}$. Then

C(\phi)=[\Gamma_f]\cap[\Delta_X],\qquad \sum_j(-1)^j\operatorname{tr}H^j(f^*) =\#([\Gamma_f]\cap[\Delta_X]). \qquad\text{(26)}

Here the right-hand class is first formed on $M=X\times X$ and then identified with a class on its closed graph parameter $X$. Write $p_1,p_2:M\to X$, $h=(f,\mathrm{id})$, $o=\operatorname{or}_X$, and $n=\dim X$. The two fundamental classes have the following precise, different coefficient presentations: $[\Gamma_f]\in H^0_{\Gamma_f}(M;p_2^!k_X) =H^n_{\Gamma_f}(M;p_1^{-1}o),\qquad [\Delta_X]\in H^0_{\Delta_X}(M;p_1^!k_X) =H^n_{\Delta_X}(M;p_2^{-1}o).$ The graph class is the exceptional counit whose adjoint under $p_2h=\mathrm{id}$ is $1$; the diagonal class is the one whose adjoint under $p_1\delta=\mathrm{id}$ is $1$. They use the positive normal Thom maps fixed by these adjunctions. For the constant coefficient object, the diagonal identity (12) is exactly this second class, since $K=k_X\boxtimes\omega_X=p_2^{-1}\omega_X$.

Cup the graph first and the diagonal second. The coefficient map is the trace-normalized exceptional tensor comparison $p_2^!k_X\otimes p_2^{-1}\omega_X\to p_2^!\omega_X=\omega_M$. In codimension degrees it pairs $p_1^{-1}o\otimes p_2^{-1}o$ with the product orientation line. Thus the cup lies in $H^0_{\Gamma_f\cap\Delta_X}(M;\omega_M)$. Closed graph adjunction and $h^!\omega_M=\omega_X$ identify this group with $H^0_{\operatorname{Fix}(f)}(X;\omega_X)$, the group containing $C(\phi)$. These are the identifications in (26). No global orientation or identification of $f^{-1}o$ with $o$ is needed.

Proof. The external-Hom comparison for $k_X$ identifies the diagonal identity with the class just described: exceptional composition along $p_1\delta=\mathrm{id}$ takes its adjoint to $1$. Pull that class back ordinarily along $h$ using (13). Its coefficient becomes $h^{-1}p_2^{-1}\omega_X=\omega_X$, and its support becomes the fixed locus. In (14) the first factor is $k_X$, and contraction is the unit action $k_X\otimes\omega_X\to\omega_X$. Therefore this pulled-back class is precisely (15).

The graph-first cup gives the same class under closed graph adjunction. To verify the comparison, project the second factor through $h_*$ in the graph counit, then use the exceptional tensor map above. The resulting map into $\omega_M$ has adjoint the pulled-back second factor followed by $h^!\omega_M=\omega_X$: the first factor has adjoint $1$, and the exceptional triangular identity cancels its unit and counit. This is the supported projection formula for these particular maps. It proves equality with the ordinary diagonal pullback while keeping the fixed support, including at excess or nontransverse intersections. Composing the graph trace with the point trace is the point trace on $X$, so integration and (16) prove (26). $\square$

The factors in this base-manifold intersection are the stated normal Thom classes. Their tensor order and relative dualizing coefficients already determine their signs; no cotangent conormal normalization factor is appended. In particular, (26) remains a supported cohomology class when the fixed locus has positive dimension, rather than a literal zero-dimensional cycle.

When there are no coincidences, (15) lies in the zero support group, so the correspondence trace is zero. A positive-dimensional fixed locus still has the supported class and global integral; the finite point sum (23) is used only for a finite locus. Computing general isolated contributions from specialization and expanding or shrinking spaces is the next treatment.

Exercises with complete solutions

An infinite global space with a three-point correspondence

Difficulty: Intermediate.

Take $X=\mathbb N$ with its discrete zero-dimensional analytic structure, $F=k_X$, and $Y=\{a,b,c\}$. Let $(f(a),g(a))=(1,2),\quad (f(b),g(b))=(2,1),\quad (f(c),g(c))=(3,3),$ and let the three coefficient maps be multiplication by $4,5,7$. Compute $T,Z,P,U_\phi$, and both sides of (16). Repeat for $F[1]$.

Solution. All points are in $S$, so $T=Y$ is compact and $Z=\{c\}$. Global sections are the infinite-dimensional product $C=\prod_{n\ge1}k$ in degree zero, whereas $P=k^3$. Ordinary pullback followed by the coefficient maps gives $a(v)=(4v_1,5v_2,7v_3)$. The properly supported $g$-trace sends these three values to coordinates $2,1,3$, respectively. Hence $U_\phi(v)=(5v_2,4v_1,7v_3,0,\ldots).$ Its image lies in the first three coordinates, and its finite-rank trace is $7$. Equivalently $ab$ on $P$ has a $2\times2$ off-diagonal block and diagonal entry $7$; its trace is $7$.

Only $c$ is a coincidence. At a zero-dimensional point the dualizing trace is ordinary coefficient evaluation, so $C_c(\phi)=7$, and its integral is $7$. The two off-diagonal arrows contribute no local class. Shifting the input by $[1]$ places both intermediate and global cohomology in degree $-1$, giving trace and local contribution $-7$. Infinite-dimensional global sections have not been assigned an identity trace.

Two graded loops must cancel the boundary terms

Difficulty: Intermediate.

Let $X$ be a point, $Y$ two points, and both maps constant. Let $F$ be the complex $k^2\to k^3$ in degrees zero and one, with $d(x,y)=(x,0,0)$. At the first point use the chain map with matrices $\operatorname{diag}(2,3)$ and $\operatorname{diag}(2,5,7)$; at the second use $4\,\mathrm{id}_F$. Compute the local contributions and the global trace.

Solution. The first map commutes with $d$, since both composites send $(x,y)$ to $(2x,0,0)$. Its cohomology action is multiplication by $3$ on $H^0=k$, and by $5,7$ on $H^1=k^2$. Its local supertrace is $3-5-7=-9$. On terms it is $(2+3)-(2+5+7)=-9$; the boundary entry $2$ cancels.

The second local contribution is $4(1-2)=-4$. The exceptional trace from two points adds the maps, so the global endomorphism is their sum and its supertrace is $-13$. Both points are coincidences, and the integral of the two point classes is $-9-4=-13$. Ignoring the degree-one coefficient or summing ungraded matrix traces would give a different number.

Two endpoint stalks cannot be the two local contributions

Difficulty: Advanced.

On $\mathbb R$, put $f(t)=t+\tfrac1{10}t(1-t)e^{-t^2},\qquad F=k_{[0,1]}.$ Verify that $f$ is an analytic diffeomorphism with exactly two fixed points and that it preserves the interval. For the canonical coefficient isomorphism, compare the global trace, the two stalk traces, and the two point-support traces. What follows about the local contributions without yet computing each separately?

Solution. The derivative of $t(1-t)e^{-t^2}$ is $(1-2t-2t^2+2t^3)e^{-t^2}$. For $p=1,2,3$, differentiation on $t\geq0$ gives $\max |t|^pe^{-t^2}=(p/(2e))^{p/2}<1$. Also $e^{-t^2}\leq1$. The absolute value of the displayed derivative is therefore at most $1+2+2+2=7$. Thus $f'(t)\ge3/10>0$. Also $f(t)-t\to0$ at both infinities. The map is bijective, and its nonzero derivative makes its inverse analytic. Its displacement vanishes exactly at $0,1$; monotonicity makes it a homeomorphism of $[0,1]$.

Here $T=[0,1]$, $Z=\{0,1\}$, and $R\Gamma(\mathbb R;F)=k$. Pullback of a constant interval section is the same section, so $L(\phi)=1$. Both endpoint stalk actions are the identity on $k$, with trace one.

Both endpoint costalks of the closed-interval sheaf are zero. At either endpoint, a small neighbourhood restricts $F$ to a closed half-interval; deleting the endpoint leaves a contractible half-interval, and restriction of constant sections is an isomorphism. The localization triangle therefore has zero point-support term. The actions (24) have trace zero.

Nevertheless $C_0(\phi)+C_1(\phi)=1$ by (23). They cannot both equal their stalk traces, whose sum is two, nor both equal their point-support traces, whose sum is zero. Their individual values require the later local localization calculation; the discrepancy already follows from this complete global calculation.

A reflection changes the costalk action

Difficulty: Intermediate.

Let $f(t)=1-t$, $F=k_{[0,1]}[r]$, and take the canonical coefficient map. Find the local contribution at $1/2$, the stalk trace, and the trace of (24). Explain the orientation sign in the last calculation.

Solution. The global section complex is $k[r]$, and reflection fixes its constant section. Thus $L(\phi)=(-1)^r$. The only fixed point in the support is $1/2$, so its local contribution is the same number by (23). Its stalk is $k[r]$ with identity action, giving the same trace.

Its point costalk is $\operatorname{or}_{\mathbb R,1/2}\otimes k[r-1]$. Reflection reverses the local orientation and acts by $-1$ on the point-support generator. One can compute this from the localization cokernel: deletion gives two components, and the quotient of $k^2$ by the diagonal $k$ is the support cohomology in degree one; exchanging the two components acts by $-1$. The action (24) consequently has supertrace $(-1)^{r-1}(-1)=(-1)^r$. Using only the costalk Euler number and ignoring its induced map would give the opposite sign.

A circle map has a negative local contribution

Difficulty: Advanced.

Let $X=S^1$, $f(e^{2\pi it})=e^{4\pi it}$, $F=k_X$, and take the canonical coefficient map. Compute the global trace and the local contribution at its fixed point. Compare with the stalk action, and with (26).

Solution. The map has exactly one fixed point: $2t=t$ modulo integers forces $t=0$ modulo integers. The circle has $H^0=k,H^1=k$. The action on $H^0$ is one. The action on $H^1$ is multiplication by two: subdivide a positively oriented loop at the two preimages of a chosen point. Each of the two arcs maps once around the target loop, so evaluation of a pulled-back Čech or cellular degree-one cocycle on the loop is twice the original evaluation. This computes the induced cohomology map, rather than merely its stalk map.

The global supertrace is $1-2=-1$. Since the fixed locus is one point, (23) gives local contribution $-1$. Its stalk map is the identity on $k$, with trace $+1$. The supported graph-diagonal intersection in (26) therefore has number $-1$, agreeing with the global trace. No transversality-based formula was needed to infer this particular number.

Compactness of the common support is a substantive hypothesis

Difficulty: Introductory.

On the same infinite discrete $X=\mathbb N$, take $Y=X$, $f=g=\mathrm{id}$, $F=k_X$, and $\phi=\mathrm{id}$. Explain exactly which construction remains valid and which trace argument fails. Does replacing $\phi$ by zero make compactness a logically necessary condition for every trace?

Solution. The supported class (15) is still defined, with $S=T=Z=X$. At every point it is the coefficient identity class. But $P=C=\prod_{n\ge1}k$ is not perfect, and the identity has infinite-dimensional image. No finite-dimensional invariant subspace can contain it, so (8) does not define its trace. The infinitely many local contributions cannot be summed by the finite support argument (23), and there is no proper point trace for this noncompactly supported class.

For the zero self-correspondence the ordinary pullback followed by the zero coefficient map is the zero endomorphism of $R\Gamma(X;F)$. It has rank zero and finite-rank trace zero even though $T$ is noncompact. This action is defined directly; the compactly supported return map (6) is not being asserted for arbitrary sections with noncompact support. The class (15) is also zero by linearity and thus has the empty compact support. Consequently (4) is a sufficient geometric hypothesis for the theorem, not a claim that every individually traceable map must have compact $T$. The counterexample concerns the identity and the failure of the theorem's general finiteness and proper integration guarantees.

References and further reading

Y. Matsui and K. Takeuchi, Microlocal study of Lefschetz fixed point formulas for higher-dimensional fixed point sets, arXiv 0812.4480v1, 24 December 2008, §2, pp. 4–5, present the self-map characteristic class in Kashiwara's Lefschetz theory. Their equations (2.4)–(2.10) pull the diagonal identity to the graph and evaluate it; Definitions 2.2 and 2.4 distinguish the supported class and its local contributions, and Theorem 2.3 identifies their sum with the global trace for compact sheaf support. This account uses complex coefficients. The correspondence theorem above treats two maps, an exceptional target and compact common support; its proof gives the actual factorization through the perfect complex $P$.

Y. Ike, Microlocal Lefschetz classes of graph trace kernels, arXiv 1504.05439v3, 15 February 2016, §3.2, pp. 9–11, packages a self-map identity and evaluation into a graph trace kernel. Lemma 3.7 constructs its two maps by diagonal and graph adjunction; equations (3.15)–(3.16) recover the graded trace over a point. The paper uses the graph $(x,f(x))$ and dual-first factors. Comparing with our $(f(x),x)$ requires the graded exchange of both factors. Its later microlocal composition theorem is a further development.