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Draft lesson.

Hard-Lefschetz maps and cohomology splittings

Written by GPT-6.1 Sol (OpenAI), October 2026. Public domain (CC0).

A bounded complex can carry extension information that its cohomology objects do not show. A degree-two derived map satisfying hard-Lefschetz isomorphisms forces this extension information to split. The proof constructs the two outer summands first, corrects their overlap, and repeats on an object with smaller cohomological support. The resulting splitting concerns the underlying derived object; the degree-two map can still have components invisible on cohomology.

Conventions and prerequisites

Let \(\mathcal A\) be an arbitrary abelian category. Work in an ambient Grothendieck universe containing its objects, arrows and the integers as sets. All finite sums and derived Hom sets below are taken there. Put \(D=D^b(\mathcal A)\), identified with its full image in the unbounded derived category. We require neither enough injectives nor enough projectives for the splitting theorem. A supplied injective resolution is used only in the concrete module exercises.

Use cochain complexes and the conventions

\[ X[r]^n=X^{n+r},\qquad d_{X[r]}=(-1)^r d_X, \qquad H^n(X[r])=H^{n+r}(X). \tag{0.1} \]

Thus \(A[-q]\), for \(A\in\mathcal A\), is a stalk in degree \(q\). Shifts on derived arrows have the coherent additive comparisons in Triangles after quasi-isomorphism localization, Sections 1–2. That proof also supplies both exact Hom sequences for every distinguished triangle. Our cone has target first, with its positive last projection. Positively shifting the first two arrows of a triangle by \(r\) multiplies its last arrow by \((-1)^r\).

Cohomological bounds, denominators and truncation adjunctions, Sections 1–5 proves cohomology detection, the full bounded embedding, smart truncations and their adjunctions, and the following actual Hom formulas:

\[ \begin{aligned} \operatorname{Hom}_D(A[q],B[q])&=\operatorname{Hom}_{\mathcal A}(A,B),\\ \operatorname{Hom}_D(X,Y)&=0 \quad\text{if }H^{>a}X=0\text{ and }H^{<a+1}Y=0. \end{aligned} \tag{0.2} \]

The first identification is the cohomology map in degree \(-q\). In particular negative Ext vanishes, where \(\operatorname{Ext}^r(A,B)=\operatorname{Hom}_D(A,B[r])\). The same proof identifies an object supported in a single degree with its cohomology stalk, naturally through truncation. We use the explicit complement argument from Null systems, roofs and the retract kernel, Section 7, and reproduce it below. For the exercises, Derived morphisms through quasi-isomorphism roofs, Section 5 proves that a supplied bounded-below complex of injectives computes every derived Hom into it.

1. The hard-Lefschetz hypothesis and the splitting statement

Let \(K\in D\) and let \(\ell:K\to K[2]\) be a derived arrow. Its powers mean the actual shifted composites

\[ \ell^0=1_K,\qquad \ell^{j+1}=\ell[2j]\ell^j:K\longrightarrow K[2j+2]. \tag{1.1} \]

The hard-Lefschetz condition, centered at degree zero, is

\[ H^{-j}(\ell^j):H^{-j}K\xrightarrow{\sim}H^jK \qquad(j\geq0). \tag{1.2} \]

For \(j=0\) this is the identity; when both objects are zero, the unique map is an isomorphism. No characteristic, semisimplicity or dimension assumption is part of (1.2).

Theorem 1.1. Under (1.2) there is an isomorphism

\[ \sigma:\bigoplus_{q\in\mathbb Z}H^qK[-q] \xrightarrow{\sim}K \tag{1.3} \]

whose cohomology map in every degree is the identity of \(H^qK\). Only finitely many summands are nonzero. In particular the underlying derived object is formal: it is isomorphic to the zero-differential complex formed from its cohomology.

We will construct (1.3) by induction. The statement does not require an isomorphism commuting with the pure degree-two cohomology map. Exercise 2 proves that such a stronger conclusion can fail even when all of (1.2) holds.

2. Actual maps at the two ends

Choose an integer \(N\geq1\) for which \(H^qK=0\) outside \([-N,N]\), and write

\[ A=H^{-N}K,\qquad B=H^NK, \qquad \phi=H^{-N}(\ell^N):A\xrightarrow{\sim}B. \tag{2.1} \]

Smart truncation gives actual derived maps

\[ a:A[N]\longrightarrow K, \qquad b:K\longrightarrow B[-N], \tag{2.2} \]

inducing the identity at the lower and upper degree, respectively. More explicitly, \(\tau_{\leq-N}K\) has cohomology only in degree \(-N\); identify it with \(A[N]\) by the canonical single-degree identification, then use its truncation map into \(K\). Similarly identify \(\tau_{\geq N}K\) with \(B[-N]\) and use the truncation map out of \(K\). These are derived arrows, not hypothetical termwise inclusions and projections in an arbitrary chosen representative of \(K\).

Their essential composite is

\[ b[2N]\ell^N a=\phi[N]:A[N]\longrightarrow B[N]. \tag{2.3} \]

Both sides have cohomology map \(\phi\) in degree \(-N\). The first formula of (0.2) makes that cohomology map a bijective test on this Hom group, so proves equality of the actual derived arrows.

Define

\[ \begin{aligned} r&=\phi[N]^{-1}b[2N]\ell^N:K\longrightarrow A[N],\\ s&=\ell^N[-2N]a[-2N]\phi[-N]^{-1}:B[-N]\longrightarrow K. \end{aligned} \tag{2.4} \]

All source and target shifts in these formulas are literal. Equation (2.3) and its shift by \(-2N\) give

\[ ra=1_{A[N]},\qquad bs=1_{B[-N]}. \tag{2.5} \]

Also

\[ ba=0. \tag{2.6} \]

Indeed its source is a stalk in degree \(-N\), its target a stalk in degree \(N\), and \(N>0\); apply the strict orthogonality in (0.2). Equivalently this Hom group is \(\operatorname{Ext}^{-2N}(A,B)=0\).

There is no analogous negative-Ext argument for \(rs\). Its group is

\[ \operatorname{Hom}_D(B[-N],A[N]) =\operatorname{Ext}^{2N}(B,A), \tag{2.7} \]

which can be nonzero. The two one-sided splittings (2.5) therefore need one more operation before they split a biproduct.

3. Correcting the triangular composite

Set \(E=A[N]\oplus B[-N]\), and consider

\[ i=(a,s):E\longrightarrow K, \qquad p_0=\binom r b:K\longrightarrow E. \tag{3.1} \]

Equations (2.5)–(2.6) give the complete matrix

\[ p_0i=\begin{pmatrix}1&rs\\0&1\end{pmatrix}, \qquad (p_0i)^{-1}=\begin{pmatrix}1&-rs\\0&1\end{pmatrix}. \tag{3.2} \]

Multiplying in either order proves the displayed inverse. Only addition and composition in an additive category are used; there is no division by a scalar or restriction on characteristic.

Put

\[ r'=r-(rs)b,\qquad p=(p_0i)^{-1}p_0=\binom{r'}b. \tag{3.3} \]

Then

\[ r'a=1,\quad r's=0,\quad ba=0,\quad bs=1, \qquad pi=1_E. \tag{3.4} \]

For example \(r's=rs-(rs)bs=0\), whereas \(r'a=ra-(rs)ba=1\). Thus the overlap is removed without changing either end inclusion. The maps \(ip\) and \(1_K-ip\) are complementary idempotents, but we still need an actual object carrying the second part.

4. Constructing the smaller-support complement

Lemma 4.1 (a complement for a specified retraction). In any triangulated category, suppose \(i:E\to K\), \(p:K\to E\), and \(pi=1_E\). There are an object \(C\) and maps \(u:C\to K\), \(j:K\to C\) with

\[ ju=1_C,\qquad pu=0,\qquad ji=0, \qquad ip+uj=1_K. \tag{4.1} \]

Consequently \((i,u):E\oplus C\to K\) and \(\binom p j:K\to E\oplus C\) are inverse isomorphisms.

Proof. Complete \(i\) to a distinguished triangle

\[ E\xrightarrow iK\xrightarrow jC\xrightarrow hE[1]. \tag{4.2} \]

Consecutive arrows give \(i[1]h=0\). Multiplication by \(p[1]\) and \(p[1]i[1]=1\) gives \(h=0\). Exactness of \(\operatorname{Hom}(C,-)\) then provides \(u_0:C\to K\) with \(ju_0=1_C\). Put \(u=(1_K-ip)u_0\). We still have \(ju=1_C\), because \(ji=0\), and now \(pu=0\).

For the remaining equation, let \(v=1_K-ip-uj\). We have \(jv=0\) and \(pv=0\). Exactness of \(\operatorname{Hom}(K,-)\) for (4.2) writes \(v=iw\) for some \(w:K\to E\). Thus \(w=piw=pv=0\), proving \(v=0\). All the block equations for the asserted inverse maps are now (4.1) and \(pi=1\). This constructs a complement for the supplied section and retraction; it does not assume that arbitrary idempotents already split. \(\square\)

Apply this to (3.4), taking the completion inside \(D^b(\mathcal A)\). The resulting object \(C\) stays bounded. In degree \(-N\), \(H^{-N}(i)\) is the identity on \(A\); in degree \(N\), it is the identity on \(B\), since \(bs=1\) and \(H^N(b)=1_B\). All other cohomology of \(E\) is zero. The biproduct isomorphism therefore gives

\[ H^qC=0\quad(q\notin[-N+1,N-1]), \qquad \alpha_q=H^q(u):H^qC\xrightarrow{\sim}H^qK \quad(-N<q<N). \tag{4.3} \]

For the second assertion, \(H^q(j)\) is the inverse of \(\alpha_q\) by (4.1). At the end degrees, the isomorphism from \(H^qE\oplus H^qC\) to \(H^qK\) has an invertible first component; its inverse equations force the second summand to be zero. Outside \([-N,N]\) both \(H^qE\) and \(H^qK\) are zero, so the same biproduct equation forces \(H^qC=0\). This proves every bound in (4.3), including the case of a vanished end object.

For any section \(u_0\) in the proof, the corrected \(u\) is uniquely characterized by \(ju=1_C\), \(pu=0\). Indeed a difference killed by \(j\) factors through \(i\), and applying \(p\) then kills its factor. If a different completion is chosen, its cone comparison with identity first two components is invertible by exact Hom sequences. The equations \(j'i=0\), \(j'u= v\), \(u'v=u\), \(vj=j'\) give the resulting compatible isomorphism \(v:C\to C'\). One can verify the last equation from \(uj=1-ip\), and the preceding one from \(u'j'=1-ip\). These comparisons suffice for existence of the splitting; no functorial choice of all completions is assumed.

5. Inheriting every Lefschetz isomorphism

On the constructed complement define the actual arrow

\[ \ell_C=j[2]\ell u:C\longrightarrow C[2]. \tag{5.1} \]

Define its powers by shifted composition just as in (1.1). For \(q\) and \(q+2\) strictly between \(-N\) and \(N\), apply cohomology to (5.1). The inverse description after (4.3) gives

\[ \alpha_{q+2}H^q(\ell_C)=H^q(\ell)\alpha_q. \tag{5.2} \]

This identity compares degree \(q\) to degree \(q+2\). It is valid because \(H^{q+2}(u)H^{q+2}(j)=1_{H^{q+2}K}\) at that interior degree. There is no such identity at a removed end degree.

Now let \(m\geq0\) and suppose that all the integers

\[ q,q+2,\ldots,q+2m \quad\text{are strictly between }-N\text{ and }N. \tag{5.3} \]

Repeatedly composing (5.2), with its successive degree indices, yields

\[ \alpha_{q+2m}H^q(\ell_C^m) =H^q(\ell^m)\alpha_q. \tag{5.4} \]

Here the comparison \(H^q(\ell[2r])=H^{q+2r}(\ell)\) follows from (0.1), so this calculation uses exactly the powers defined in (1.1). For \(m=0\), it is the identity. It proves a statement about cohomology, rather than the usually unjustified derived equality \(\ell_C^m=j[2m]\ell^m u\); the inserted projections between powers need not disappear as derived arrows.

Take \(q=-m\), \(0\leq m\leq N-1\). Every degree in the path (5.3) lies in \([-m,m]\subseteq[-N+1,N-1]\), so (5.4) and (1.2) prove

\[ H^{-m}(\ell_C^m):H^{-m}C\xrightarrow{\sim}H^mC. \tag{5.5} \]

For \(m\geq N\), both its end objects are zero by (4.3), so the same condition holds there too. Thus \((C,\ell_C)\) satisfies the entire hard-Lefschetz condition, with a smaller support bound. Showing just the outer isomorphism for \(K\) would not justify this step: Exercise 3 exhibits that failure.

6. Full induction and cohomology normalization

Proof of Theorem 1.1. Because \(K\) is bounded, choose some \(N\geq0\) with cohomology supported in \([-N,N]\). We induct on this integer; it need not be the least possible bound.

If \(N=0\), the single-degree truncation identification gives an isomorphism \(H^0K\to K\) inducing the identity on \(H^0\). No use of the two-end construction is appropriate here: the lower and upper end are the same object. If \(H^0K=0\), cohomology detection identifies \(K\) with zero, and the same conclusion uses the empty sum.

For \(N\geq1\), Sections 2–4 give \(K\simeq A[N]\oplus C\oplus B[-N]\), with the specified maps \(a,u,s\). Section 5 supplies \(\ell_C\) satisfying (1.2) and support in \([-N+1,N-1]\). By the induction hypothesis choose

\[ \sigma_C: \bigoplus_{-N<q<N}H^qC[-q]\xrightarrow{\sim}C \tag{6.1} \]

inducing each cohomology identity. Use the inverse isomorphisms \(\alpha_q^{-1}:H^qK\to H^qC\) from (4.3). The middle inclusion in the final splitting is

\[ \bigoplus_{-N<q<N}H^qK[-q] \xrightarrow{\ \bigoplus\alpha_q^{-1}[-q]\ } \bigoplus_{-N<q<N}H^qC[-q] \xrightarrow{\sigma_C}C\xrightarrow uK. \tag{6.2} \]

Combine it with \(a\) on \(A[N]\) and \(s\) on \(B[-N]\), arranging summands in increasing cohomological degree. Each step is a biproduct of isomorphisms followed by the actual complement isomorphism, so yields (1.3). At an interior degree, its cohomology map is \(\alpha_q 1\alpha_q^{-1}=1\); at the ends it is \(H^{-N}(a)=1\) and \(H^N(s)=1\). This proves the required normalization.

The support shrinks at every step even when \(A=B=0\): then \(E=0\), \(C\simeq K\), but the stricter bound is already true. Condition (1.2) makes one vanished end imply that the other vanishes. At \(N=1\) the complement is supported only in degree zero and the next step is the base case. Thus there is no exception for a zero middle object, a missing outer pair, an acyclic object or a nonminimal initial bound. The induction terminates after finitely many steps. \(\square\)

If \(\pi=\sigma^{-1}\), denote the components by \(\sigma_q:H^qK[-q]\to K\), \(\pi_q:K\to H^qK[-q]\). They satisfy the actual derived equations

\[ \pi_p\sigma_q=\begin{cases}1,&p=q,\\0,&p\ne q,\end{cases} \qquad \sum_q\sigma_q\pi_q=1_K. \tag{6.3} \]

These are stronger than an equality of cohomology groups. They are the section and projection equations needed to apply functors to the splitting.

7. Scope, invisible maps and changed centers

The construction splits the underlying derived object and provides its cohomology-normalized inclusions and projections. It does not discard higher Ext morphisms between those stalks. After any splitting, an endomorphism of \(K\) is a matrix whose \((p,q)\) entry lies in \(\operatorname{Ext}^{q-p}(H^qK,H^pK)\); a map to \(K[2]\) has entries in \(\operatorname{Ext}^{q-p+2}(H^qK,H^pK)\). Only the zero-Ext entries of the latter are the prescribed cohomology degree-two maps. Negative Ext entries vanish, while positive Ext entries can remain.

Even the base case illustrates this. If \(K=A[0]\) and \(\operatorname{Ext}^2(A,A)\ne0\), choose a nonzero \(\ell:A\to A[2]\). All the hard-Lefschetz conditions with \(j>0\) are isomorphisms between zero objects. The theorem's underlying splitting is the identity, but the degree-two map on the cohomology stalk complex is zero. Hence an isomorphism intertwining those two maps cannot exist. Exercise 2 supplies an explicit abelian category with this nonzero Ext group.

Once a splitting is chosen, equation (6.3) gives functorial Hom decompositions in a separate test object. It gives no automatic statement that the chosen splitting is natural in every arrow of \(K\), unique among all normalized splittings, or compatible with \(\ell\). For instance any nonzero positive-Ext entry between distinct summands gives a nontrivial triangular automorphism inducing the identity on cohomology; composing \(\sigma\) with it produces another normalized splitting. Such entries are concretely nonzero in Exercise 2.

The same theorem has an arbitrary integer center \(c\). If

\[ H^{c-j}(\ell^j):H^{c-j}K\xrightarrow{\sim}H^{c+j}K \quad(j\geq0), \tag{7.1} \]

apply Theorem 1.1 to \(K[c]\) and \(\ell[c]\). Its powers are \(\ell^j[c]\) by shifted composition. Shifting the resulting splitting back by \(-c\), and putting \(q=c+r\), gives

\[ K\simeq\bigoplus_r H^{c+r}K[-r-c] =\bigoplus_q H^qK[-q], \tag{7.2} \]

again inducing the identity on each cohomology object. This reindexing uses a finite sum and requires no existence of infinite coproducts.

8. Four graded exercises with complete solutions

Exercise 1 — a complete computation with a contractible middle (introductory)

Over a field \(k\), put

\[ K^{-2}=k,\quad K^{-1}=k^2,\quad K^0=k^3, \quad K^1=k^2,\quad K^2=k. \]

The only nonzero differentials are \(d^{-1}(u,v)=(u,0,0)\) and \(d^0(a,b,c)=(b,0)\). Define a degree-two chain map by

\[ \ell^{-2}(z)=(0,0,z),\qquad \ell^{-1}(u,v)=(u,v),\qquad \ell^0(a,b,c)=c, \]

with all other components zero. Check the chain-map and hard-Lefschetz conditions, construct the two outer splittings and the complement, and give a full cohomology splitting with a chain homotopy for its inverse equations.

Solution. We have \(d^0d^{-1}=0\). Since the shift by two has positive differential, the chain equations are \(d^{n+2}\ell^n=\ell^{n+1}d^n\). In degree \(-2\), both sides are zero because \(d^0(0,0,z)=0\). In degree \(-1\), both sides are zero because \(d^1=0\) and \(\ell^0(u,0,0)=0\). In all remaining degrees an outgoing differential or a further component is zero. Thus \(\ell\) is a chain map.

All five cohomology objects, in degrees \(-2,-1,0,1,2\), are \(k\). At \(-1\) use the second coordinate \(v\) in the kernel, at zero use \(c\) modulo the first-coordinate image, and at one use the second coordinate modulo the first-coordinate image. The other two are the unchanged end terms. The map from \(H^{-1}\) to \(H^1\) is the identity. The composite \(\ell^2\) sends \(z\) in degree \(-2\) to \(z\) in degree two, so is the identity on those cohomology objects. These prove (1.2), including \(j=0\) and the zero objects for \(j>2\).

For \(N=2\), take \(a\) to be the inclusion of \(K^{-2}\) and \(b\) the projection onto \(K^2\). We have \(\phi=1_k\). Formula (2.4) gives \(r\) as the projection onto \(K^{-2}\) and \(s\) as the inclusion of \(K^2\), directly from the displayed composite \(\ell^2\). Hence \(rs=0\) and the correction changes nothing. The complement is the complex with the same three middle terms and differentials, in degrees \(-1,0,1\). Its projected map has the one component \(\ell_C^{-1}=1_{k^2}:k^2\to k^2\), and induces the required identity on \(H^{-1}\to H^1\).

Let \(G\) be the direct sum of the five cohomology stalks. Define \(\sigma:G\to K\) by the identity at the two ends, by \(v\mapsto(0,v)\) in degrees \(-1,1\), and by \(c\mapsto(0,0,c)\) in degree zero. Define \(\pi:K\to G\) by the identity at the ends and by the coordinates \(v,c,v\) in the middle. Each is a chain map, because \(\sigma\) lands in cycles and \(\pi\) kills every displayed boundary. Clearly \(\pi\sigma=1_G\).

Define the only nonzero homotopy components by

\[ h^0(a,b,c)=(a,0),\qquad h^1(u,v)=(0,u,0). \]

In degree \(-1\), \(dh+hd\) sends \((u,v)\) to \((u,0)\); in degree zero it sends \((a,b,c)\) to \((a,b,0)\); in degree one it sends \((u,v)\) to \((u,0)\); at the ends it is zero. These are exactly the components of \(1_K-\sigma\pi\). Thus \(\sigma\pi=1_K\) in the homotopy category, and \(\sigma\) is the desired cohomology-normalized derived isomorphism. The two removed middle pairs are visibly contractible, with the actual contraction just given.

Exercise 2 — a nonzero overlap and failure of equivariance (intermediate)

Let \(\Lambda=k[t]/(t^2)\), and work in the abelian category of finite-dimensional left \(\Lambda\)-modules. Construct a nonzero \(\eta\in\operatorname{Ext}^2_\Lambda(k,k)\), where \(k=\Lambda/(t)\), and put \(\beta=\eta[-1]:k[-1]\to k[1]\). For

\[ K=k[1]\oplus k[-1],\qquad \ell:K\longrightarrow K[2],\qquad \ell=\begin{pmatrix}0&0\\1&\beta\end{pmatrix}, \tag{8.1} \]

compute \(r,s,rs,r'\) for \(N=1\). Prove that (1.2) holds. Let \(L\) be the same displayed matrix with \(\beta\) replaced by zero. Prove that no isomorphism of the underlying objects intertwines \(L\) and \(\ell\).

Solution. First \(\Lambda\) is injective in this finite module category. For a \(k\)-linear form \(f:V\to k\), set \(e_f(v)=f(tv)+f(v)t\). This is \(\Lambda\)-linear, since \(e_f(tv)=f(tv)t=t e_f(v)\). The coefficient of \(t\) gives the inverse identification \(\operatorname{Hom}_\Lambda(V,\Lambda)=\operatorname{Hom}_k(V,k)\): if \(g(v)=a(v)+b(v)t\), linearity imposes \(a(v)=b(tv)\). A linear form on a subspace extends to a whole finite-dimensional vector space by extending a basis. Under this identification, every map into \(\Lambda\) therefore extends across a module monomorphism, proving injectivity.

Use the explicit resolution

\[ 0\longrightarrow k\xrightarrow{1\mapsto t}\Lambda \xrightarrow t\Lambda\xrightarrow t\Lambda\xrightarrow t\cdots, \tag{8.2} \]

with its injective terms \(I^n=\Lambda\) for \(n\geq0\). It is exact after its first term because the kernel and image of multiplication by \(t\) both equal \(kt\). The inclusion makes \(k[0]\to I\) a quasi-isomorphism. The supplied bounded-below injective comparison proves that \(\operatorname{Ext}^r(k,k)\) is \(\operatorname{Hom}_{K(\mathcal A)}(k,I[r])\); the full bounded embedding allows this computation in the unbounded category.

For \(r\geq0\), a chain map \(k[0]\to I[r]\) has its sole component \(1\mapsto c t\) in \(I^r\), with \(c\in k\). Its outgoing differential kills that component. Every possible homotopy has degree-zero source component into \(I^{r-1}\), also landing in \(kt\); its composition with the possibly signed multiplication-by-\(t\) differential is zero. Thus no distinct values of \(c\) are homotopic. At \(r=0\) the same statement uses the missing \(I^{-1}\) term. Consequently

\[ \operatorname{Ext}^r(k,k)=k\quad(r\geq0),\qquad \operatorname{Ext}^r(k,k)=0\quad(r<0). \tag{8.3} \]

Choose \(\eta\) as the class represented by \(1\mapsto t\) in \(I^2\). It is nonzero by this homotopy calculation, and its shift \(\beta\) is nonzero too.

For (8.1), the only nonzero cohomology objects are \(H^{-1}K=k\) and \(H^1K=k\). The entry \(1\) induces the identity between them, whereas \(\beta\) has zero cohomology maps. This proves (1.2); all \(j>1\) involve zero end objects. The truncation maps are the first inclusion \(a\) and second projection \(b\), with \(\phi=1\). Formula (2.4) gives

\[ r=(1,\beta),\qquad s=\binom01,\qquad rs=\beta\ne0,\qquad r'=r-\beta b=(1,0). \tag{8.4} \]

Thus the triangular inverse in (3.2) is necessary even in an object whose underlying decomposition is already displayed.

Every isomorphism \(v:K\to K\) has matrix \(\begin{pmatrix}a&\gamma\\0&d\end{pmatrix}\), with \(a,d\in k^\times\) and \(\gamma\in\operatorname{Hom}_D(k[-1],k[1])\). The lower-left group is negative Ext and hence zero. Its diagonal entries must be nonzero since they are its two cohomology maps. Conversely such a matrix is invertible by its triangular inverse. The equation \(\ell v=v[2]L\) would give

\[ \begin{pmatrix}0&0\\a&\gamma+\beta d\end{pmatrix} =\begin{pmatrix}\gamma[2]&0\\d&0\end{pmatrix}. \tag{8.5} \]

The upper-left equation forces \(\gamma[2]=0\), hence \(\gamma=0\), since shift is an equivalence. The lower-right equation then forces \(\beta d=0\), impossible because \(d\) is invertible and \(\beta\ne0\). Thus no intertwining isomorphism exists. In particular \(1+\begin{pmatrix}0&\beta\\0&0\end{pmatrix}\) is a nonidentity automorphism inducing the identity on cohomology, so normalized underlying splittings need not be unique.

Exercise 3 — why the outer power alone is insufficient (advanced)

In the module category of Exercise 2, choose the nonzero class \(\chi\in\operatorname{Ext}^3(k,k)\) represented by \(1\mapsto t\) in \(I^3\), and put \(\kappa=\chi[-1]:k[-1]\to k[2]\). Choose a distinguished triangle

\[ k[1]\longrightarrow X\longrightarrow k[-1] \xrightarrow\kappa k[2]. \tag{8.6} \]

Let \(T=k[2]\oplus k[0]\oplus k[-2]\) with the degree-two arrow \(L_T\) sending the first summand identically to the shifted middle one and the middle one identically to the shifted last one. Put \(K=X\oplus T\), \(\ell=0_X\oplus L_T\). Show that \(H^{-2}(\ell^2)\) is invertible but the cohomology splitting conclusion fails. Locate exactly where the induction loses its hypothesis.

Solution. The class \(\chi\) is nonzero by (8.3), whose calculation covers every nonnegative degree. A triangle (8.6) exists: complete \(-\kappa:k[-1]\to k[2]\), shift its cone triangle by \(-1\) with the signed last arrow, and rotate once. Its first two vertices are \(k[1]\) and \(C(-\kappa)[-1]\); its third is \(k[-1]\); its final arrow is \(\kappa\). Thus this construction can be used for \(X\) without overlooking the rotation sign.

The long cohomology sequence gives \(H^{-1}X=k\), \(H^1X=k\), and zero in every other degree. The first and third objects in (8.6) lie in \(D^{\leq0}\) and \(D^{\geq1}\). The truncation adjunctions compare this triangle to the canonical truncation triangle of \(X\). The first comparison induces the identity on its only cohomology, and is invertible by cohomology detection; the same is true of the third comparison. Consequently the boundary of that truncation triangle identifies with the nonzero \(\kappa\). It cannot become the zero boundary of \(k[1]\oplus k[-1]\): naturality under an isomorphism would multiply \(\kappa\) on either side by invertible arrows and could not kill it. Therefore \(X\) does not split into its cohomology stalks.

The five cohomology objects of \(K\) are all \(k\) in degrees \(-2,-1,0,1,2\). The square of \(L_T\) carries \(k\) in degree \(-2\) identically onto \(k\) in degree two, so \(H^{-2}(\ell^2)=1_k\). But \(H^{-1}(\ell):k\to k\) is zero, because the two odd degrees belong to \(X\). Thus the condition with \(j=1\) fails.

The outer splitting maps from \(\ell^2\) remove exactly the two end stalks of \(T\). Their complement is \(C=X\oplus k[0]\). The projected \(\ell_C\) is zero; in particular its map from \(H^{-1}C\) to \(H^1C\) is zero, so it cannot satisfy the induction hypothesis. It is nonformal as well: its truncation boundary at zero is

\[ k[-1]\longrightarrow (k[1]\oplus k[0])[1], \qquad \binom\kappa0, \tag{8.7} \]

which is nonzero by projection onto its first component. For the zero-differential cohomology complex this canonical boundary is zero, and truncation naturality rules out an isomorphism between them. Finally \(\tau_{\geq-1}\tau_{\leq1}K\simeq C\), whereas applying those functors to a formal \(K\) would give its formal middle three stalks. Hence \(K\) itself is nonformal. This proves failure of the desired conclusion from an outer isomorphism alone, and explains why every interior power was checked in Section 5.

Exercise 4 — the two derived Hom decompositions (advanced)

For a splitting in Theorem 1.1, prove the two finite-sum formulas

\[ \begin{aligned} \operatorname{Hom}_D(M,K[m]) &\simeq\bigoplus_q\operatorname{Ext}^{m-q}(M,H^qK),\\ \operatorname{Hom}_D(K,M[m]) &\simeq\bigoplus_q\operatorname{Ext}^{m+q}(H^qK,M) \end{aligned} \tag{8.8} \]

for \(M\in\mathcal A\), \(m\in\mathbb Z\). Give both directions of each map, prove naturality in \(M\), and compute both sides for \(M=k\) and the \(K\) of Exercise 2, for every integer \(m\).

Solution. Use the exact sections and projections (6.3). In the first formula send \(f:M\to K[m]\) to the family \(\pi_q[m]f:M\to H^qK[m-q]\). Send a family \((f_q)\) back to \(\sum_q\sigma_q[m]f_q\). Their two composites are identities, respectively by \(\pi_p\sigma_q=\delta_{pq}\) and \(\sum_q\sigma_q\pi_q=1_K\), shifted by \(m\). Every sum is finite.

In the second formula, send \(g:K\to M[m]\) to \((g\sigma_q)[q]:H^qK\to M[m+q]\). Send a family \((g_q:H^qK\to M[m+q])\) back to \(\sum_q g_q[-q]\pi_q\). The same identities, with these explicitly opposite shifts, prove both inverse equations. The definitions of Ext and the full bounded embedding identify the groups exactly as in (8.8); negative indices give zero by (0.2).

Precomposition with an arrow \(M'\to M\) commutes with each first-formula component and sum, so proves its contravariant naturality in \(M\). Postcomposition by the shifted arrow \(M[m]\to M'[m]\) commutes with each second-formula component and sum, so proves its covariant naturality. These maps use the chosen \(\sigma\); this assertion is not a new claim of canonical naturality in \(K\).

For Exercise 2 the two degrees are \(-1,1\), both with cohomology \(k\). The first group is \(\operatorname{Ext}^{m+1}(k,k)\oplus\operatorname{Ext}^{m-1}(k,k)\), and the second is the same two groups in the opposite order. Formula (8.3) therefore gives, for either Hom group,

\[ \begin{cases} 0,&m\leq-2,\\ k,&m=-1\text{ or }m=0,\\ k\oplus k,&m\geq1. \end{cases} \tag{8.9} \]

In particular positive Ext remains present after the underlying object splits. The calculation also covers the degree-zero and negative-degree endpoints; it does not replace a negative Ext group by a second copy of \(k\).

9. Freely accessible references