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Hard Lefschetz splits a bounded t-structure complex

Statement

Let $T$ be a triangulated category with a bounded t-structure, with truncation functors and t-cohomology as in Truncation functors, t-cohomology, amplitude and bounded t-structures and t-cohomology is functorial, additive, shift-compatible and detects amplitude. Let $K \in T$ and let $L : K \to K [2]$ be a morphism such that for every integer $j \geq 0$ the map $H_{t}^{- j} (L^{j}) : H_{t}^{- j} (K) ⟶ H_{t}^{- j} (K [2 j]) ≅ H_{t}^{j} (K)$ is an isomorphism, where $L^{j} : = L [2 j - 2] \circ \dots \circ L [2] \circ L$ is the $j$ -fold iterate (so $L^{0} = 1_{K}$ ) and the identification is the shift isomorphism $H_{t}^{j} (K) ≅ H_{t}^{- j} (K [2 j])$ of that lemma. Then there is an isomorphism $K ≅ ⨁_{j} H_{t}^{j} (K) [- j]$ in $T$ ; the sum is finite because the t-structure is bounded. The isomorphism is not asserted to be canonical, and no compatibility with primitive lifts is claimed: this is the abstract splitting criterion, not the stronger primitive-compatible decomposition used in the Hodge theory of projective maps. The degenerate case in which $H_{t}^{- n} (K)$ and $H_{t}^{n} (K)$ are both zero (they vanish together because $H_{t}^{- n} (L^{n})$ is an isomorphism) is covered by the same formulas, all of which are then the zero maps between zero objects.

Facts & Assumptions

Given: A t-structure $(T^{\leq 0}, T^{\geq 0})$ on a triangulated category $T$ , bounded in the sense of Truncation functors, t-cohomology, amplitude and bounded t-structures, with truncation functors $τ_{\leq n}, τ_{\geq n}$ and t-cohomology $H_{t}^{n}$ ; an object $K \in T$ and a morphism $L : K \to K [2]$ whose iterates $L^{j} : K \to K [2 j]$ satisfy the hard Lefschetz hypothesis of the Statement.

[F1]

Boundedness and amplitude detection: there is $n \geq 0$ with $K \in T^{\leq n} \cap T^{\geq - n}$ , and then $H_{t}^{j} (K) = 0$ for every $∣ j ∣ > n$ ; conversely, if $H_{t}^{j} (K) = 0$ for every $∣ j ∣ > n$ then $K \in T^{\leq n} \cap T^{\geq - n}$ (Truncation functors, t-cohomology, amplitude and bounded t-structures, (iv) of t-cohomology is functorial, additive, shift-compatible and detects amplitude).

[F2]

Truncation package (Truncation triangles are essentially unique and give adjoint truncation functors, t-cohomology is functorial, additive, shift-compatible and detects amplitude): $Hom (T^{\leq m}, T^{\geq m + 1}) = 0$ for every $m$ , and $T^{\leq m} \subseteq T^{\leq m^{'}}$ , $T^{\geq m^{'}} \subseteq T^{\geq m}$ whenever $m \leq m^{'}$ ; $X \in T^{\leq m}$ if and only if $τ_{\geq m + 1} X = 0$ , and $X \in T^{\geq m}$ if and only if $τ_{\leq m - 1} X = 0$ ; the aisles are closed under extensions and finite direct sums, and a finite direct sum lies in an aisle exactly when each summand does. If $X \in T^{\geq m}$ then $τ_{\leq m} X ≅ H_{t}^{m} (X) [- m]$ , and if $X \in T^{\leq m}$ then $τ_{\geq m} X ≅ H_{t}^{m} (X) [- m]$ , the identification using the truncation maps of $X$ ; in particular an object lying in $T^{\leq m} \cap T^{\geq m}$ is canonically isomorphic to $H_{t}^{m} (X) [- m]$ . The assignments $H_{t}^{m}$ are functors, $H_{t}^{m} (X [r]) ≅ H_{t}^{m + r} (X)$ naturally, and for a finite biproduct $X \oplus Y$ the structure maps give mutually inverse isomorphisms $(H_{t}^{m} (i_{X}), H_{t}^{m} (i_{Y})) : H_{t}^{m} (X) \oplus H_{t}^{m} (Y) \to H_{t}^{m} (X \oplus Y)$ and $(H_{t}^{m} (p_{X}), H_{t}^{m} (p_{Y}))$ . If $X, Y$ lie in $T^{\leq m} \cap T^{\geq m}$ , then under the canonical identifications $X ≅ H_{t}^{m} (X) [- m]$ , $Y ≅ H_{t}^{m} (Y) [- m]$ the map $Hom (X, Y) \to Hom (H_{t}^{m} (X), H_{t}^{m} (Y))$ , $f \mapsto H_{t}^{m} (f)$ , is a bijection; in particular a morphism $X \to Y$ is an isomorphism exactly when $H_{t}^{m} (f)$ is.

[F3]

TR1 and TR3: every morphism extends to a distinguished triangle, and a commutative square formed by the first maps of two distinguished triangles extends to a morphism of triangles (Triangulated-category axiom TR1, Triangulated-category axiom TR3, Morphism and isomorphism of triangles).

[F4]

Rotations and split triangles: with the signs of Rotation of a triangle, the inverse rotation $Z [- 1] \overset{- h [- 1]}{\to} X \overset{f}{\to} Y \overset{g}{\to} Z$ of a distinguished triangle $X \overset{f}{\to} Y \overset{g}{\to} Z \overset{h}{\to} X [1]$ is distinguished (Distinguished triangles are closed under shifts and both rotations); the triangle $X \overset{1_{X}}{\to} X \to 0 \to X [1]$ is distinguished, and a distinguished triangle $X \overset{0}{\to} Y \overset{g}{\to} Z \overset{h}{\to} X [1]$ is isomorphic to $X \overset{0}{\to} Y \overset{(\binom{1}{0})}{\to} Y \oplus X [1] \overset{(0 1)}{\to} X [1]$ (Zero and split triangles are distinguished, A distinguished triangle with zero first map is split).

[F5]

The triangulated category is additive and has finite biproducts (Triangulated category, Additive category, Biproduct): for a biproduct diagram $(X \oplus Y, i_{X}, i_{Y}, p_{X}, p_{Y})$ one has $p_{X} i_{X} = 1_{X}$ , $p_{Y} i_{Y} = 1_{Y}$ , $p_{X} i_{Y} = 0$ , $p_{Y} i_{X} = 0$ and $i_{X} p_{X} + i_{Y} p_{Y} = 1_{X \oplus Y}$ , and a pair of morphisms into or out of a biproduct is determined by its components, composition being matrix multiplication.

Proof

technique · induction on an amplitude bound

1.1basegivenF1F2

Base case. Let $n = 0$ and let $K \in T^{\leq 0} \cap T^{\geq 0}$ admit such an $L$ . Then $K$ lies in both aisles, so the concentration identity of (F2) applied with $m = 0$ gives a canonical isomorphism $K ≅ H_{t}^{0} (K) [0] = H_{t}^{0} (K)$ , while amplitude detection gives $H_{t}^{j} (K) = 0$ for every $j \neq 0$ ; hence the asserted direct sum is the single term $H_{t}^{0} (K)$ and the conclusion holds.

1.2ihgivenF1F2

Induction step. Fix $n \geq 1$ and assume as induction hypothesis that the assertion holds for every object of $T^{\leq n - 1} \cap T^{\geq - (n - 1)}$ admitting a morphism $L^{'}$ with $H_{t}^{- j} (L^{' j})$ an isomorphism for every $j \geq 0$ . Let $K \in T^{\leq n} \cap T^{\geq - n}$ and let $L : K \to K [2]$ satisfy the hard Lefschetz hypothesis, and put $A : = H_{t}^{- n} (K)$ , $B : = H_{t}^{n} (K)$ and $h : = H_{t}^{- n} (L^{n}) : A \to B$ ; the map $h$ is an isomorphism by hypothesis and $A, B$ lie in the heart. Since $K \in T^{\geq - n}$ and $K \in T^{\leq n}$ , concentration (F2) identifies the truncation maps of $K$ at $- n$ and $n$ with morphisms $a : A [n] \to K$ and $b : K \to B [- n]$ for which $H_{t}^{- n} (a) : A \to H_{t}^{- n} (K) = A$ and $H_{t}^{n} (b) : B \to H_{t}^{n} (B [- n]) = B$ are the canonical identifications. Since $A [n] \in T^{\leq - n}$ , $B [- n] \in T^{\geq n}$ and $T^{\geq n} \subseteq T^{\geq - n + 1}$ as $n \geq 1$ , the orthogonality in (F2) gives $b a = 0$ .

2.1step 1.2F2

Define $c : = L^{n} [- 2 n] \circ a [- 2 n] \circ h^{- 1} [- n] : B [- n] \to K$ and $r : = h^{- 1} [n] \circ b [2 n] \circ L^{n} : K \to A [n]$ , so that $b c = 1_{B [- n]}$ and $r a = 1_{A [n]}$ . Indeed, on $H_{t}^{n}$ the composite $b c$ induces $h \circ 1_{A} \circ h^{- 1} = 1_{B}$ : the shift isomorphisms of (F2) convert the induced maps of $h^{- 1} [- n]$ , $a [- 2 n]$ and $L^{n} [- 2 n]$ into $h^{- 1}$ , the canonical identification of $A ≅ H_{t}^{- n} (K)$ and $h$ respectively, and $b$ induces the canonical identification of $B$ ; since $B [- n]$ lies in $T^{\leq n} \cap T^{\geq n}$ and therefore has $H_{t}^{m} = 0$ for $m \neq n$ , map detection (F2) upgrades the identity on $H_{t}^{n}$ to the identity of morphisms $b c = 1_{B [- n]}$ . Symmetrically $r a = 1_{A [n]}$ : the shift isomorphisms convert the induced maps of $L^{n}$ , $b [2 n]$ and $h^{- 1} [n]$ into $h$ , the canonical identification of $B$ and $h^{- 1}$ , so $r a$ induces $h^{- 1} \circ 1_{B} \circ h = 1_{A}$ on $H_{t}^{- n}$ , and $A [n]$ lies in $T^{\leq - n} \cap T^{\geq - n}$ with t-cohomology concentrated in degree $- n$ .

3.1step 1.2step 2.1F5

Put $r^{'} : = r - (r c) b$ , where $r c : = r \circ c : B [- n] \to A [n]$ and $(r c) b : = (r c) \circ b : K \to A [n]$ . Then $r^{'} a = r a - (r c) (b a) = 1 - 0 = 1$ and $r^{'} c = r c - (r c) (b c) = r c - r c = 0$ , while $b a = 0$ and $b c = 1$ by steps 1.2 and 2.1; hence with respect to the biproduct structure of $A [n] \oplus B [- n]$ the composite $(r^{'}, b) \circ (a, c)$ has matrix $(\begin{array}{cc} r^{'} a & r^{'} c \\ b a & b c \end{array}) = (\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array})$ and equals the identity, so $(a, c)$ has the left inverse $(r^{'}, b)$ .

4.1step 3.1F3F4

By TR1 complete the map $u : = (a, c) : A [n] \oplus B [- n] \to K$ to a distinguished triangle $A [n] \oplus B [- n] \overset{u}{\to} K \overset{v}{\to} K_{0} \overset{w}{\to} (A [n] \oplus B [- n]) [1]$ . The square formed by $u$ and by $(r^{'}, b) : K \to A [n] \oplus B [- n]$ commutes, since $(r^{'}, b) u = 1$ by step 3.1; TR3 applied to this square and to the distinguished triangle $A [n] \oplus B [- n] \overset{1}{\to} A [n] \oplus B [- n] \to 0 \to \dots$ of (F4) yields a morphism of triangles whose first two components are the identity and $(r^{'}, b)$ and whose third component $h^{'}$ satisfies $w = δ h^{'}$ , where $δ : 0 \to (A [n] \oplus B [- n]) [1]$ is the zero morphism; hence $w = 0$ .

5.1step 4.1F4F5

The inverse rotation is $K_{0} [- 1] \overset{0}{\to} X \overset{u}{\to} K \overset{v}{\to} K_{0}$ , where $X = A [n] \oplus B [- n]$ , by step 4.1 and (F4). Applying the zero-first-map split criterion (F4) supplies an isomorphism $g : K \to X \oplus K_{0}$ with $g u = (\binom{1}{0})$ and $(0 1) g = v$ . Here the criterion can be taken to fix its first two vertices: its proof constructs the third-vertex isomorphism from a section of the final map. Put $R : = (1 0) g$ , $i : = g^{- 1} (\binom{0}{1})$ and $p : = (0 1) g = v$ . Then $p i = 1_{K_{0}}$ , $R u = 1_{X}$ , $p u = 0$ , $R i = 0$ and $u R + i p = 1_{K}$ . Thus $(u, i, R, p)$ is a biproduct diagram and $K ≅ A [n] \oplus K_{0} \oplus B [- n]$ .

6.1step 1.2step 5.1F2F5

Additivity (F2) applied to this biproduct gives, for every $m$ , an isomorphism $(H_{t}^{m} (u), H_{t}^{m} (i)) : H_{t}^{m} (X) \oplus H_{t}^{m} (K_{0}) \to H_{t}^{m} (K)$ with inverse $(H_{t}^{m} (R), H_{t}^{m} (p))$ , where $X : = A [n] \oplus B [- n]$ . On the extreme degrees the first component is an isomorphism: with respect to the biproduct $X = A [n] \oplus B [- n]$ the map $H_{t}^{- n} (u)$ has components $(H_{t}^{- n} (a), H_{t}^{- n} (c)) = (1_{A}, 0)$ , since $H_{t}^{- n} (a) = 1_{A}$ by step 1.2 and $H_{t}^{- n} (B [- n]) = H_{t}^{- 2 n} (B) = 0$ ; and $H_{t}^{n} (u)$ has components $(H_{t}^{n} (a), H_{t}^{n} (c)) = (0, H_{t}^{n} (c))$ , since $H_{t}^{n} (A [n]) = H_{t}^{2 n} (A) = 0$ while $H_{t}^{n} (c)$ is an isomorphism because $H_{t}^{n} (b) H_{t}^{n} (c) = H_{t}^{n} (b c) = 1_{B}$ and $H_{t}^{n} (b)$ is the canonical identification. An isomorphism $(f, g) : C \oplus D \to E$ with $f$ an isomorphism has $D = 0$ : if $(ψ_{1}, ψ_{2})$ is its inverse, then $ψ_{2} f = 0$ forces $ψ_{2} = ψ_{2} f f^{- 1} = 0$ , and then $1_{D} = ψ_{2} g = 0$ . Applying this to the isomorphisms $(H_{t}^{- n} (u), H_{t}^{- n} (i))$ and $(H_{t}^{n} (u), H_{t}^{n} (i))$ , whose first components are isomorphisms, yields $H_{t}^{- n} (K_{0}) = 0$ and $H_{t}^{n} (K_{0}) = 0$ .

7.1step 6.1F2

Since $K ≅ X \oplus K_{0}$ and $K \in T^{\leq n} \cap T^{\geq - n}$ , the direct-sum clause of (F2) gives $K_{0} \in T^{\leq n} \cap T^{\geq - n}$ . From $K_{0} \in T^{\geq - n}$ and $H_{t}^{- n} (K_{0}) = 0$ , concentration gives $τ_{\leq - n} K_{0} ≅ H_{t}^{- n} (K_{0}) [n] = 0$ , so $K_{0} \in T^{\geq - n + 1}$ by the aisle criterion of (F2); dually $K_{0} \in T^{\leq n}$ and $H_{t}^{n} (K_{0}) = 0$ give $τ_{\geq n} K_{0} ≅ H_{t}^{n} (K_{0}) [- n] = 0$ , so $K_{0} \in T^{\leq n - 1}$ . Moreover additivity gives morphisms $(H_{t}^{m} (u), H_{t}^{m} (i)) : H_{t}^{m} (X) \oplus H_{t}^{m} (K_{0}) \to H_{t}^{m} (K)$ with inverse $(H_{t}^{m} (R), H_{t}^{m} (p))$ ; for $- n < m < n$ both components $H_{t}^{m} (a)$ and $H_{t}^{m} (c)$ of $H_{t}^{m} (u)$ vanish, because their sources $H_{t}^{m} (A [n]) = H_{t}^{m + n} (A)$ and $H_{t}^{m} (B [- n]) = H_{t}^{m - n} (B)$ are zero when $m \neq - n$ respectively $m \neq n$ , so $H_{t}^{m} (u) = 0$ and $H_{t}^{m} (i) : H_{t}^{m} (K_{0}) \to H_{t}^{m} (K)$ is an isomorphism with inverse $H_{t}^{m} (p)$ , since $H_{t}^{m} (p) H_{t}^{m} (i) = H_{t}^{m} (p i) = 1$ . In particular $H_{t}^{m} (K_{0}) ≅ H_{t}^{m} (K)$ for every $∣ m ∣ < n$ .

8.1step 7.1F1F2

Define $L_{0} : = p [2] \circ L \circ i : K_{0} \to K_{0} [2]$ , so that $L_{0}^{0} = 1_{K_{0}}$ . For $1 \leq j < n$ the map $H_{t}^{- j} (L_{0}^{j})$ is an isomorphism. Indeed, expanding the iterate $L_{0}^{j} = L_{0} [2 j - 2] \circ \dots \circ L_{0} [2] \circ L_{0}$ and applying $H_{t}^{- j}$ factor by factor, the factor $H_{t}^{- j} (L_{0} [2 k])$ corresponds under the shift isomorphisms of (F2) to $H_{t}^{m_{k}} (p [2] \circ L \circ i) = H_{t}^{m_{k} + 2} (p) \circ H_{t}^{m_{k}} (L) \circ H_{t}^{m_{k}} (i)$ , where $m_{k} : = - j + 2 k$ ; for the intermediate degrees $m_{k}$ with $1 \leq k \leq j - 1$ one has $∣ m_{k} ∣ < n$ , so by step 7.1 together with $H_{t}^{m_{k}} (p) H_{t}^{m_{k}} (i) = H_{t}^{m_{k}} (p i) = 1$ the maps $H_{t}^{m_{k}} (i)$ and $H_{t}^{m_{k}} (p)$ are mutually inverse isomorphisms and the inserted pair $H_{t}^{m_{k}} (i) \circ H_{t}^{m_{k}} (p) = H_{t}^{m_{k}} (i p)$ is the identity. The composite therefore corresponds to $H_{t}^{j} (p) \circ H_{t}^{- j} (L^{j}) \circ H_{t}^{- j} (i)$ , in which $H_{t}^{- j} (L^{j})$ is an isomorphism by hypothesis and the outer maps are isomorphisms because $- j$ and $j$ are interior degrees. For $j \geq n$ , both $H_{t}^{- j} (K_{0})$ and $H_{t}^{j} (K_{0})$ vanish by step 7.1 and amplitude detection, so $H_{t}^{- j} (L_{0}^{j})$ is a morphism between zero objects, hence an isomorphism. Thus $L_{0}$ satisfies the hard Lefschetz hypothesis on the object $K_{0} \in T^{\leq n - 1} \cap T^{\geq - (n - 1)}$ .

9.1step 1.2step 8.1

By the induction hypothesis of step 1.2, applied to the object $K_{0} \in T^{\leq n - 1} \cap T^{\geq - (n - 1)}$ and the morphism $L_{0}$ of step 8.1, there is an isomorphism $K_{0} ≅ ⨁_{j} H_{t}^{j} (K_{0}) [- j]$ , a finite sum because $H_{t}^{j} (K_{0}) = 0$ for $∣ j ∣ > n - 1$ .

10.1step 1.1step 5.1step 7.1step 9.1discharge-induction∎

Combining the results: step 5.1 gives an isomorphism $K ≅ A [n] \oplus K_{0} \oplus B [- n]$ ; step 7.1 identifies $H_{t}^{m} (K_{0}) ≅ H_{t}^{m} (K)$ for $∣ m ∣ < n$ ; step 9.1 decomposes $K_{0}$ . Substituting $A = H_{t}^{- n} (K)$ , $B = H_{t}^{n} (K)$ and inserting the vanishing extreme cohomology of $K_{0}$ gives the asserted isomorphism $K ≅ H_{t}^{- n} (K) [n] \oplus ⨁_{∣ j ∣ < n} H_{t}^{j} (K) [- j] \oplus H_{t}^{n} (K) [- n] = ⨁_{j} H_{t}^{j} (K) [- j]$ , a finite sum because $H_{t}^{j} (K) = 0$ for $∣ j ∣ > n$ by amplitude detection. Boundedness supplies such an $n$ for every object $K$ , the base case of step 1.1 and the induction step of steps 1.2 to 9.1 cover every $n \geq 0$ , and therefore the splitting holds for arbitrary $K$ .

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Sources

P. Deligne, «Théorème de Lefschetz et critères de dégénérescence de suites spectrales», Publ. Math. IHÉS 35 (1968) 107–126 (complete numdam PDF) (standard reference, not scraped)
M. A. A. de Cataldo and L. Migliorini, «The Hodge theory of algebraic maps», arXiv:math/0306030v2 (complete arXiv PDF) (standard reference, not scraped)
A. A. Beilinson, J. Bernstein and P. Deligne, «Faisceaux pervers», Astérisque 100 (1982), Société Mathématique de France (complete numdam PDF) (standard reference, not scraped)