Tilt derived-dual truncation and canonical finite saturation

Let β be real and $E\in B_{\beta }^{\mathrm{f}\mathrm{i}\mathrm{n}}$. Write $F=H^{-1}(E)$, $T=H^0(E)$. Then F is reflexive and ${\mu }_{\max }(F)<\beta$ when F≠0. In fact $Q=F\ast \ast \mathrm{/}F$ is supported in dimension at most one; $F\ast \ast ,F\in F_{\beta }$ and the exact coherent sequence gives a monomorphism Q→F[1]→E in $B_{\beta }$. Since $Q\in C_{\beta }$, it vanishes. Similarly a nonzero HN boundary part of F of slope β gives a monomorphism of its shift into E from $C_{\beta }$, impossible. This proves both assertions.

We claim that D(E) has standard cohomology only in degrees −1,0,1, with $H^1$ zero-dimensional. Initially it has no cohomology below −1 or above 2: dualize F[1]→E→T, use regular-local dimension three and torsion-freeness of F (depth at least one, projective dimension at most two). Its $H^2$ is supported at closed points by local global dimension. If $H^2$ were nonzero, its quotient to a residue-field sheaf would give a nonzero composite $D(E)\to H^2(D(E))[-2]\to k(x)[-2]$. Applying the involutive D produces a nonzero map k(x)^D→E, contradicting $Hom(C_{\beta },E)=0$. Hence $H^2=0$.

Let $G=H^1(D(E))$. At a point of codimension at most one, F is free and T has projective dimension at most one, so the same triangle gives no $H^1$. Therefore G is supported in dimension at most one. If its one-dimensional part were nonzero, consider the canonical map D(E)→G[−1] and dualize. The source D(G[−1]) has cohomology $K=Ex{t}^2(G,O_X)$ in degree zero and $Ex{t}^3(G,O_X)$ in degree one. Thus the canonical map K→D(G[−1])→E exists. At a generic codimension-two point where G is nonzero, the local ring is regular of dimension two and G has finite length; Koszul duality and finite-length dévissage show that D(G[−1]) there is exactly its nonzero Ext² module. The map is nonzero there, since the original truncation map induces an isomorphism on its highest cohomology and derived duality is faithful. This yields a nonzero map from K, a dimension-at-most-one sheaf, to E, again impossible. Hence G is zero-dimensional.

Define

$E^d=\tau \le 0$ D(E), $T_E=H^1$ D(E).

This gives a canonical triangle $E^d\to D(E)\to T_E[-1]$. We now prove $E^d\in B_{-\beta }$, not just that it is a two-term complex. Dualizing the cohomology triangle gives $H^{-1}$ $D(E)=T\ast$. Its maximal slope is less than −β by ordinary slope duality on T's torsion-free quotient, so it belongs to $F_{-\beta }$. Also $H^{0}D(E)$ fits into an exact sequence with a torsion image of $Ex{t}^1(T,O_X)$ as its subobject and an image K inside F* as its quotient; F*/K is supported in codimension at least two, being a subsheaf of $Ex{t}^2(T,O_X)$. The HN slopes of K equal those of F*, whose minimum exceeds −β. Thus $H^{0}D(E)\in T_{-\beta }$. This proves the asserted heart membership. The statement remains valid when F or T is zero.

Dualize the last triangle. Since $D(T_E[-1])=T_E^D[-1]$, it gives a canonical exact sequence in $B_{\beta }$

$0\to E\to E^{\mathrm{s}\mathrm{a}\mathrm{t}}:=D(E^d)\to T_E^D\to 0$.

In particular $E^{\mathrm{s}\mathrm{a}\mathrm{t}}$ has standard cohomology only in degrees −1,0. For any K of dimension at most one,

$Hom(K,E^{\mathrm{s}\mathrm{a}\mathrm{t}})=Hom(E^d,D(K))=0$,

because $E^d$ has cohomology in degrees at most zero and D(K) in degrees at least one. For a torsion-free slope-semistable G of slope β, a map $G[1]\to E^{\mathrm{s}\mathrm{a}\mathrm{t}}$ factors through E: $Hom(G[1],T_E^D)=0$ by negative ordinary Ext. It is therefore zero. Every object of $C_{\beta }$ is an extension of such a G[1] by a dimension-at-most-one sheaf. Thus $E^{\mathrm{s}\mathrm{a}\mathrm{t}}\in B_{\beta }^{\mathrm{f}\mathrm{i}\mathrm{n}}$.

The dual truncation itself lies in $B_{-\beta }^{\mathrm{f}\mathrm{i}\mathrm{n}}$. For K of dimension at most one, use $Hom(K,E^d)=Hom(E^{\mathrm{s}\mathrm{a}\mathrm{t}},D(K))=0$ by the same standard degree argument. For a slope-semistable G of slope −β,

$Hom(G[1],E^d)=Hom(E^{\mathrm{s}\mathrm{a}\mathrm{t}},RHom(G,O_X))$.

The standard degree-zero term G* has slope β, so $G\ast [1]\in B_{\beta }$ and $Hom(E^{\mathrm{s}\mathrm{a}\mathrm{t}},G\ast )=0$ by negative-heart orthogonality. The remaining terms of $RHom(G,O_X)$ have standard cohomology in degrees at least one, giving no maps from $E^{\mathrm{s}\mathrm{a}\mathrm{t}}$. This proves the other necessary Hom vanishing, and hence the finite-part assertion.

The saturation is finite and universal. Given 0→E→U→Z→0 in $B_{\beta }$ with Z zero-dimensional and with no nonzero zero-dimensional subobject of U, every $C_{\beta }-subobject$ of U has intersection zero with E and hence injects into Z; it would be zero-dimensional, so must vanish. Thus $U\in B_{\beta }^{\mathrm{f}\mathrm{i}\mathrm{n}}$. Since $D(Z)=Z^D[-2]$, the dual map induces a canonical isomorphism $\tau \le 0D(U)\cong \tau \le 0D(E)=E^d$. Apply the preceding saturation construction to U to obtain the canonical injection $U\to D(E^d)=E^{\mathrm{s}\mathrm{a}\mathrm{t}}$ over E. It is unique: two such maps differ by a map $Z\to E^{\mathrm{s}\mathrm{a}\mathrm{t}}$, which is zero. Consequently $length(U\mathrm{/}E)\le length(T_E)$, uniformly for every such extension. No existence of a maximal extension is assumed; $E^{\mathrm{s}\mathrm{a}\mathrm{t}}$ is constructed by two actual derived duals and one standard truncation.

Functoriality also gives the exact dual sequence. From an exact 0→A→E→B→0 of finite-part objects, take tilted cohomology of D(B)→D(E)→D(A). Because each D object is its dual truncation plus a zero-dimensional degree-one term, the result is

$0\to B^d\to E^d\to A^d\to T_B\to T_E\to T_A\to 0$

in $B_{-\beta }$. In particular the injection of the dual of a finite-part quotient is genuine; no dual of an arbitrary heart exact sequence is claimed exact without the defect terms. The first three numerical coordinates transform as (C0,C1,C2)↦(−C0,C1,−C2); $ch3(E^d)=ch3(E)+length(T_E)$.