Lefschetz Splitting In Bounded Triangulated Categories

Let $A\stackrel{a}{\to }X\stackrel{b}{\to }B\to A[1]$ be distinguished with $A\in {\mathcal{T}}^{\le n}$ and $B\in {\mathcal{T}}^{\ge n+1}$. For $T\in {\mathcal{T}}^{\le n}$, the exact segment $\mathrm{Hom}(T,B[-1])\to \mathrm{Hom}(T,A)\stackrel{a_{\ast }}{\to }\mathrm{Hom}(T,X)\to \mathrm{Hom}(T,B)$ has zero outer terms by [F2], since $B[-1]\in {\mathcal{T}}^{\ge n+2}$. Thus $a_{\ast }$ is bijective. Dually, for $T\in {\mathcal{T}}^{\ge n+1}$ the segment $\mathrm{Hom}(A[1],T)\to \mathrm{Hom}(B,T)\stackrel{b^{\ast }}{\to }\mathrm{Hom}(X,T)\to \mathrm{Hom}(A,T)$ has zero outer terms, since $A[1]\in {\mathcal{T}}^{\le n-1}$, so $b^{\ast }$ is bijective.

If $A\stackrel{f}{\to }B\to 0\to A[1]$ is a distinguished triangle, then $f$ is an isomorphism. Indeed, the triangle $B\stackrel{1}{\to }B\to 0\to B[1]$ is distinguished by [F4], and $(f,1_B,0)$ is a morphism of triangles from the first to the second, commuting with all three arrows; the components $1_B$ and $0\to 0$ are adjacent isomorphisms, so $f$ is an isomorphism by [F3].

Claim (i). Let $A\stackrel{a}{\to }X\stackrel{b}{\to }B\to A[1]$ be distinguished with $A\in {\mathcal{T}}^{\le n}$, $B\in {\mathcal{T}}^{\ge n+1}$, and let $P\stackrel{p}{\to }X\stackrel{q}{\to }Q\to P[1]$ be the truncation triangle. By step 1.1 applied to the truncation triangle with $T=A$ there is a unique $f:A\to P$ with $pf=a$, and applied to the given triangle with $T=P$ there is a unique $f^{\mathrm{\prime }}:P\to A$ with $a{f}^{\mathrm{\prime }}=p$. Then $p(f{f}^{\mathrm{\prime }})=a{f}^{\mathrm{\prime }}=p=p{1}_P$, and $f{f}^{\mathrm{\prime }}=1_P$ because $\mathrm{Hom}(P,P)\to \mathrm{Hom}(P,X)$ is injective by step 1.1 applied to the truncation triangle with $T=P$; symmetrically $f^{\mathrm{\prime }}f=1_A$ using the given triangle with $T=A$. So $f$ is an isomorphism. By TR3 [F3] the commutative square $(f,1_X)$ extends to a morphism of triangles $(f,1_X,g)$, which is an isomorphism of triangles because $f$ and $1_X$ are adjacent isomorphism components, and its middle component is the identity of $X$. For uniqueness, let $(f^{\mathrm{\prime }\mathrm{\prime }},1_X,g^{\mathrm{\prime }\mathrm{\prime }})$ be another such isomorphism; then $p{f}^{\mathrm{\prime }\mathrm{\prime }}=a$, so $f^{\mathrm{\prime }\mathrm{\prime }}=f$ by the uniqueness in step 1.1 applied to the truncation triangle with $T=A$. Hence the composite of $(f^{\mathrm{\prime }\mathrm{\prime }},1_X,g^{\mathrm{\prime }\mathrm{\prime }})$ with the inverse of $(f,1_X,g)$ is an automorphism $(1_A,1_X,d)$ of the given triangle; it satisfies $db=b$, and $\mathrm{Hom}(B,B)\to \mathrm{Hom}(X,B)$, $v\mapsto vb$, is injective by step 1.1 applied to the given triangle with $T=B\in {\mathcal{T}}^{\ge n+1}$, so $d=1_B$ and the isomorphism over $X$ is unique.

Claim (ii), the $\le$ side. Fix the truncation triangle $P_X\stackrel{p_X}{\to }X\stackrel{q_X}{\to }{Q}_X\to P_X[1]$ for every $X$. For $f:X\to Y$, the composite $f{p}_X$ lies in $\mathrm{Hom}(P_X,Y)$, and step 1.1 applied to the truncation triangle of $Y$ with $T=P_X\in {\mathcal{T}}^{\le n}$ gives a unique map ${\tau }_{\le n}f:P_X\to P_Y$ with $p_Y({\tau }_{\le n}f)=f{p}_X$. This makes ${\tau }_{\le n}$ a functor: ${\tau }_{\le n}(1_X)=1_{P_X}$ by uniqueness, and for $X\stackrel{f}{\to }Y\stackrel{h}{\to }Z$ both ${\tau }_{\le n}(hf)$ and ${\tau }_{\le n}h\circ {\tau }_{\le n}f$ have the same composite with $p_Z$, so they are equal by the uniqueness. The map $\mathrm{Hom}(T,P_X)\to \mathrm{Hom}(T,X)$, $u\mapsto p_{X}u$, is a bijection for $T\in {\mathcal{T}}^{\le n}$ by step 1.1 applied to the truncation triangle of $X$, and it is natural in $X$ because both $p_Y({\tau }_{\le n}fu)$ and $f(p_{X}u)$ equal $(f{p}_X)u$. Applying the same argument with arrows reversed to the truncation triangle at $n-1$, whose third vertex is ${\tau }_{\ge n}X$, makes $X\mapsto {\tau }_{\ge n}X$ a functor and gives a bijection $\mathrm{Hom}({\tau }_{\ge n}X,T)\to \mathrm{Hom}(X,T)$ for $T\in {\mathcal{T}}^{\ge n}$. The final vanishing $\mathrm{Hom}({\mathcal{T}}^{\le n},{\mathcal{T}}^{\ge n+1})=0$ is [F2].

Claim (iii), the $\le$ side. Let $P\stackrel{p}{\to }X\stackrel{q}{\to }Q\to P[1]$ be the truncation triangle. The following are equivalent: (a) $X\in {\mathcal{T}}^{\le n}$; (b) $p$ is an isomorphism; (c) $Q=0$; (d) $\mathrm{Hom}(X,{\mathcal{T}}^{\ge n+1})=0$. (a)$\Rightarrow$(d) is [F2]. (a)$\Rightarrow$(c): if $X\in {\mathcal{T}}^{\le n}$ then $q\in \mathrm{Hom}(X,Q)=0$ by [F2], so $q^{\ast }(1_Q)=q=0=q^{\ast }(0)$, and $q^{\ast }:\mathrm{Hom}(Q,Q)\to \mathrm{Hom}(X,Q)$ is injective by step 1.1 applied to the truncation triangle with $T=Q\in {\mathcal{T}}^{\ge n+1}$, hence $1_Q=0$ and $Q$ is a zero object. (c)$\Rightarrow$(b): if $Q=0$, step 1.2 applied to the triangle $P\to X\to 0\to P[1]$ makes $p$ an isomorphism. (b)$\Rightarrow$(c): if $p$ is an isomorphism, then in the exact sequence $\mathrm{Hom}(Q,Q)\stackrel{q^{\ast }}{\to }\mathrm{Hom}(X,Q)\stackrel{p^{\ast }}{\to }\mathrm{Hom}(P,Q)$ of [F1] the map $p^{\ast }$ is injective (a map $\phi$ with $\phi p=0$ vanishes because $p$ is an isomorphism), so $\mathrm{im}(q^{\ast })=0$; since $q^{\ast }$ is a bijection by step 1.1, $1_Q=0$ and $Q=0$. (d)$\Rightarrow$(c): if $\mathrm{Hom}(X,{\mathcal{T}}^{\ge n+1})=0$ then $q=0$, so $q^{\ast }(1_Q)=0=q^{\ast }(0)$ and injectivity of $q^{\ast }$ gives $1_Q=0$, i.e. $Q=0$. Finally (b)$\Rightarrow$(a): if $p$ is an isomorphism then $X\cong P$ and $P\in {\mathcal{T}}^{\le n}$, so $X\in {\mathcal{T}}^{\le n}$.

Claim (iii), the $\ge$ side, by the same arguments with the roles of the two maps interchanged. Let $P\stackrel{p}{\to }X\stackrel{q}{\to }Q\to P[1]$ be the truncation triangle at $n-1$, so that $P={\tau }_{\le n-1}X$ and $Q={\tau }_{\ge n}X$. Then: (a${}^{\mathrm{\prime }}$) $X\in {\mathcal{T}}^{\ge n}$ implies $\mathrm{Hom}({\mathcal{T}}^{\le n-1},X)=0$ by [F2], hence $p=0$ and $p_{\ast }(1_P)=p=0=p_{\ast }(0)$ with $p_{\ast }$ injective by step 1.1, so $P=0$; (b${}^{\mathrm{\prime }}$) $P=0$ makes the triangle $X\to Q\to 0\to X[1]$ distinguished by [F4], so $X\to Q$ is an isomorphism by step 1.2 and $X\cong Q\in {\mathcal{T}}^{\ge n}$; (c${}^{\mathrm{\prime }}$) if $X\to Q$ is an isomorphism then $qp=0$ forces $p=0$ and hence $1_P=0$ as above, so $P=0$; (d${}^{\mathrm{\prime }}$) if $\mathrm{Hom}({\mathcal{T}}^{\le n-1},X)=0$ then $p=0$ and $1_P=0$ as in (a${}^{\mathrm{\prime }}$), so $P=0$ and $X\in {\mathcal{T}}^{\ge n}$ by (b${}^{\mathrm{\prime }}$); and $X\in {\mathcal{T}}^{\ge n}$ implies $\mathrm{Hom}({\mathcal{T}}^{\le n-1},X)=0$ by [F2]. This proves $X\in {\mathcal{T}}^{\ge n}$ if and only if $X\to {\tau }_{\ge n}X$ is an isomorphism, if and only if ${\tau }_{\le n-1}X=0$, if and only if $\mathrm{Hom}({\mathcal{T}}^{\le n-1},X)=0$.

Claim (iv), extensions: let $X\to Y\to Z\to X[1]$ be distinguished. If $X,Z\in {\mathcal{T}}^{\ge n}$, then for $W\in {\mathcal{T}}^{\le n-1}$ exactness of $\mathrm{Hom}(W,X)\to \mathrm{Hom}(W,Y)\to \mathrm{Hom}(W,Z)$ and orthogonality force $\mathrm{Hom}(W,Y)=0$, hence $Y\in {\mathcal{T}}^{\ge n}$ by step 2.4. Dually, $X,Z\in {\mathcal{T}}^{\le n}$ force $Y\in {\mathcal{T}}^{\le n}$ using $\mathrm{Hom}(Z,W)\to \mathrm{Hom}(Y,W)\to \mathrm{Hom}(X,W)$ and step 2.3. Applying these extension statements to the rotations $Y\to Z\to X[1]\to$ and $Z[-1]\to X\to Y\to$, and using $X[1]\in {\mathcal{T}}^{\le n-1}$ when $X\in {\mathcal{T}}^{\le n}$ and $Z[-1]\in {\mathcal{T}}^{\ge n+1}$ when $Z\in {\mathcal{T}}^{\ge n}$, gives the four other bounds in (iv).

Claim (iv), direct sums: for objects $X,Y$ one has $\mathrm{Hom}(X\oplus Y,W)\cong \mathrm{Hom}(X,W)\oplus \mathrm{Hom}(Y,W)$ and $\mathrm{Hom}(W,X\oplus Y)\cong \mathrm{Hom}(W,X)\oplus \mathrm{Hom}(W,Y)$ by [F5]. Hence $X\oplus Y\in {\mathcal{T}}^{\le n}$ implies $\mathrm{Hom}(X,W)=0$ for every $W\in {\mathcal{T}}^{\ge n+1}$, as a direct summand of the vanishing group $\mathrm{Hom}(X\oplus Y,W)$, so $X\in {\mathcal{T}}^{\le n}$ and likewise $Y\in {\mathcal{T}}^{\le n}$ by step 2.3; conversely if $X,Y\in {\mathcal{T}}^{\le n}$ then every such $\mathrm{Hom}(X\oplus Y,W)$ vanishes and $X\oplus Y\in {\mathcal{T}}^{\le n}$ by step 2.3. The same argument with the second displayed bijection and step 2.4 gives the statement for ${\mathcal{T}}^{\ge n}$; the empty direct sum is the zero object and the claims are vacuous for it.

Claim (iv), iteration: let $m\le n$. Since ${\tau }_{\le m}X\in {\mathcal{T}}^{\le n}$, its truncation map ${\tau }_{\le n}{\tau }_{\le m}X\to {\tau }_{\le m}X$ is an isomorphism by step 2.3. Put $P={\tau }_{\le m}{\tau }_{\le n}X$ and $Q={\tau }_{\le m}X$, and let $\alpha :P\to X$ and $\beta :Q\to X$ be the composites of their truncation maps. For every $T\in {\mathcal{T}}^{\le m}$, composition with either $\alpha$ or $\beta$ is a bijection onto $\mathrm{Hom}(T,X)$ by step 2.2 and ${\mathcal{T}}^{\le m}\subseteq {\mathcal{T}}^{\le n}$. Thus there are unique $d:P\to Q$ and $v:Q\to P$ with $\beta d=\alpha$ and $\alpha v=\beta$, and injectivity gives $dv=1_Q$ and $vd=1_P$. These isomorphisms commute with the maps to $X$ and are natural by uniqueness. Dually, since ${\tau }_{\ge n}X\in {\mathcal{T}}^{\ge m}$, the map ${\tau }_{\ge n}X\to {\tau }_{\ge m}{\tau }_{\ge n}X$ is an isomorphism by step 2.4; the dual factorization argument identifies ${\tau }_{\ge n}{\tau }_{\ge m}X$ with ${\tau }_{\ge n}X$, compatibly with the maps from $X$. Finally, if $m<n$, then ${\tau }_{\ge n}X\in {\mathcal{T}}^{\ge m+1}$, so ${\tau }_{\le m}{\tau }_{\ge n}X=0$ by step 2.4.

Steps 2.1, 2.2, 2.3, 2.4, 3.1, 3.2 and 3.3 prove claims (i), (ii), (iii) and (iv) respectively, including the uniqueness of the isomorphism in (i), the naturality and the dual statements in (ii), all four equivalent formulations in (iii), and the closure, direct-sum and iteration statements in (iv).

Claim (i). The formula $H_t^n(X)=({\tau }_{\le n}{\tau }_{\ge n}X)[n]$ is a composite of functors by [F1] and [F2], and the induced maps are uniquely compatible with the truncation maps by [F1]. Put $Y={\tau }_{\ge n}X$ and $Z={\tau }_{\le n}Y$. For $T\in {\mathcal{T}}^{\le n-1}$, adjunction gives $\mathrm{Hom}(T,Z)\cong \mathrm{Hom}(T,Y)=0$, since $Y\in {\mathcal{T}}^{\ge n}$. The aisle criterion therefore puts $Z$ in ${\mathcal{T}}^{\ge n}$; it is in ${\mathcal{T}}^{\le n}$ by construction. Consequently $Z[n]=H_t^n(X)$ lies in the heart by the shift conventions.

Claim (ii). For all $a,b$ there is a canonical isomorphism ${\tau }_{\le a}(X[b])\cong ({\tau }_{\le a+b}X)[b]$ over $X[b]$, and dually for ${\tau }_{\ge a}$: for $T\in {\mathcal{T}}^{\le a}$ both objects represent the same functor, $\mathrm{Hom}(T,{\tau }_{\le a}(X[b]))\cong \mathrm{Hom}(T,X[b])$ by the adjunction [F1], and $\mathrm{Hom}(T,({\tau }_{\le a+b}X)[b])\cong \mathrm{Hom}(T[-b],{\tau }_{\le a+b}X)\cong \mathrm{Hom}(T[-b],X)\cong \mathrm{Hom}(T,X[b])$ because $T[-b]\in {\mathcal{T}}^{\le a+b}$, so the essential uniqueness in [F1] identifies them. Hence $H_t^n(X[m])=({\tau }_{\le n}{\tau }_{\ge n}(X[m]))[n]\cong ({\tau }_{\le n}(({\tau }_{\ge n+m}X)[m]))[n]\cong (({\tau }_{\le n+m}{\tau }_{\ge n+m}X)[m])[n]\cong H_t^{n+m}(X)$, using the coherence isomorphisms of [F2], and the isomorphisms are natural in $X$ by functoriality of the truncations.

Claim (iv), the forward implications. If $X\in {\mathcal{T}}^{\le m}$ and $n>m$, then $X\in {\mathcal{T}}^{\le n-1}$ by [F1], so ${\tau }_{\ge n}X=0$ by the aisle criterion [F1] and $H_t^n(X)=0$. If $X\in {\mathcal{T}}^{\ge m}$ and $n<m$, then $X\in {\mathcal{T}}^{\ge n+1}$ by [F1], so ${\tau }_{\ge n}X\cong X$ and ${\tau }_{\le n}{\tau }_{\ge n}X={\tau }_{\le n}X=0$ by the dual aisle criterion, whence $H_t^n(X)=0$.

Claim (v), concentration. The identity $H_t^m(X)[-m]={\tau }_{\le m}{\tau }_{\ge m}X$ is the definition. If $X\in {\mathcal{T}}^{\ge m}$ then ${\tau }_{\ge m}X\cong X$ by [F1], giving the second identity. If $X\in {\mathcal{T}}^{\le m}$, then ${\tau }_{\ge m}X\in {\mathcal{T}}^{\le m}$: for $W\in {\mathcal{T}}^{\ge m+1}$ one has $\mathrm{Hom}({\tau }_{\ge m}X,W)\cong \mathrm{Hom}(X,W)=0$ by the adjunction and orthogonality of [F1], so the aisle criterion gives ${\tau }_{\ge m}X\in {\mathcal{T}}^{\le m}$; hence ${\tau }_{\le m}{\tau }_{\ge m}X\cong {\tau }_{\ge m}X$, giving the first identity. If $X$ lies in both aisles, then ${\tau }_{\le m}X\cong X$ and ${\tau }_{\ge m}X\cong X$, and the composite canonical map $X\cong H_t^m(X)[-m]$ is the truncation map.

Claim (iii). The inclusion ${\mathcal{T}}^{\le n}\hookrightarrow \mathcal{T}$ has the right adjoint ${\tau }_{\le n}$ by [F1], and the inclusion ${\mathcal{T}}^{\ge n}\hookrightarrow \mathcal{T}$ has the left adjoint ${\tau }_{\ge n}$; these inclusions and $\mathcal{T}$ are additive by [F3], so ${\tau }_{\le n}$ and ${\tau }_{\ge n}$ are additive functors by [F3], and so is the autoequivalence $[n]$ by [F2]. Hence $H_t^n$ is additive, and by [F3] it preserves the split biproduct diagram of $X\oplus Y$: the images of the injections and projections under $H_t^n$ are the two mutually inverse structure maps of a biproduct diagram of $H_t^n(X)$ and $H_t^n(Y)$, which is precisely the assertion, and the resulting isomorphism $H_t^n(X\oplus Y)\cong H_t^n(X)\oplus H_t^n(Y)$ is natural by functoriality.

Claim (iv), the converses for a bounded t-structure. Suppose $H_t^n(X)=0$ for all $n>m$. By boundedness choose $N\ge m$ with $X\in {\mathcal{T}}^{\le N}\cap {\mathcal{T}}^{\ge -N}$; then $X\in {\mathcal{T}}^{\le N}$, and if $X\in {\mathcal{T}}^{\le k}$ for some $k\ge m+1$, the concentration identity of step 1.4 gives ${\tau }_{\ge k}X\cong H_t^k(X)[-k]=0$, so $X\cong {\tau }_{\le k-1}X\in {\mathcal{T}}^{\le k-1}$ by the aisle criterion. Descending finitely from $N$ to $m+1$ gives $X\in {\mathcal{T}}^{\le m}$. The lower bound is dual: if $H_t^n(X)=0$ for all $n<m$, choose $N\ge -m$ with $X\in {\mathcal{T}}^{\le N}\cap {\mathcal{T}}^{\ge -N}$ and ascend from $-N$ to $m-1$, using ${\tau }_{\le k}X\cong H_t^k(X)[-k]=0$ to conclude $X\to {\tau }_{\ge k+1}X$ is an isomorphism and $X\in {\mathcal{T}}^{\ge k+1}$; the two arguments are independent.

Claim (v), map detection. Let $X,Y\in {\mathcal{T}}^{\le m}\cap {\mathcal{T}}^{\ge m}$. By step 1.4 the canonical maps give isomorphisms $X\cong H_t^m(X)[-m]$ and $Y\cong H_t^m(Y)[-m]$; under these, a morphism $f:X\to Y$ corresponds to $H_t^m(f)$: both the source and target identifications are the truncation maps, and $H_t^m$ is a functor commuting with the truncations by step 1.1, so the induced bijection is $f\mapsto H_t^m(f)$, and it is a bijection because the identifications are isomorphisms.

Claim (v), the bounded single-degree case: if the t-structure is bounded and $X,Y$ have $H_t^n=0$ for every $n\mathrm{\ne }m$, then step 2.2 (applied to the positive and negative ranges) gives $X,Y\in {\mathcal{T}}^{\le m}\cap {\mathcal{T}}^{\ge m}$, so step 2.3 applies and $\mathrm{Hom}(X,Y)\to \mathrm{Hom}(H_t^m(X),H_t^m(Y))$, $f\mapsto H_t^m(f)$, is a bijection under the canonical identifications.

Steps 1.1 to 1.4 and 2.2 to 3.1 prove all the clauses: (i) in step 1.1, (ii) in step 1.2, (iii) in step 2.1, both directions of (iv) in steps 1.3 and 2.2, and the concentration identities and map detection of (v) in steps 1.4, 2.3 and 3.1.