Draft lesson.

Differential objects, curvature and cone homotopies

Original text public domain (CC0).

A map called a differential need not have square zero. Its square is then a curvature map, and a proposed homotopy need not even produce a differential morphism. We determine the exact condition, construct the cone triangles for all differential objects, and show how their homotopies factor through actual contractible objects.

Let \(\mathcal A\) be a locally small additive category with an additive translation equivalence \(T\). No abelian structure, kernels, infinite sums or exactness hypothesis is assumed. Use the universe and Choice conventions of Universes and small categories. Formal linear combinations and finite sums, Section 1 supplies finite biproduct calculus. Translations, coherent powers and universal triangle comparisons, Sections 1–2 proves the coherent inverse and integer-power comparisons used below. All coordinate formulas mean biproduct matrices.

1. Making the translation calculations precise

Choose the coherent powers \(P_n\) and comparisons \(\mu_{r,s}:P_rP_s\to P_{r+s}\) constructed in the translation prerequisite. Here \(P_0=1\), \(P_1=T\), and

There is an equivalent category with a strictly invertible translation. Its objects are pairs \((X,n)\), with

Identities and composition are those in \(\mathcal A\). Evaluation \(E(X,n)=P_nX\) is fully faithful and essentially surjective, since every \(X\) occurs as \(E(X,0)\). Give these Hom groups their existing addition. A biproduct of \((X,n),(Y,m)\) is \((P_nX\oplus P_mY,0)\), with the evaluated biproduct legs; the zero object is \((0,0)\). Thus evaluation is an additive equivalence.

Put \(a_{n,X}=\mu_{1,n,X}:TP_nX\to P_{n+1}X\). Define

This preserves identities, composition and addition. For each pair of objects its map on Hom groups is bijective, by full faithfulness of \(T\). Consequently the inverse translation decreases the labels by one and sends a morphism to its unique inverse image under (1.3). These two functors compose to the identity on both objects and morphisms. Evaluation has the natural translation comparison

Naturality is exactly (1.3). All inverse comparisons follow from (1.1). For example, if \(S\) is the chosen inverse of \(T\) and \(\eta:1\to ST\) its unit, the comparison

is invertible. Transporting an arrow \(d:X\to UX\) gives the arrow \(\xi_X E(d):EX\to TEX\). Transporting a degree-minus-one arrow \(h:X\to U^{-1}Y\) gives \(\lambda_YE(h):EX\to SEY\). Naturality of \(\xi\) and the two unit/counit cancellation equations identify the transported boundary

Indeed the first term uses \(T\lambda_Y\,\xi_{U^{-1}Y}=1_{EY}\) after the counit identification, and the second uses naturality of (1.5). This also identifies the square-zero condition and all finite cone matrices. Full faithfulness lifts every differential equation, morphism and homotopy, and essential surjectivity lifts every object with its differential. Thus these constructions give equivalences on differential objects and on their homotopy quotients, preserving the entire cone triangles.

We may therefore calculate with a strictly invertible \(T\). In the original category every cancellation of \(TT^{-1}\) or \(T^{-1}T\) means the specified coherent comparison. This reduction proves the formulas for translation equivalences; it does not add a strictness hypothesis.

2. Curvature controls which homotopies are allowed

A differential object is a pair \((X,d_X:X\to TX)\). A differential morphism \(f:X\to Y\) satisfies

Identities satisfy this equation; composition follows by substituting it twice. The sum of two solutions is a solution by bilinearity. Zero objects and finite biproducts, with the block diagonal differentials, satisfy the same equation on every injection and projection. Hence these objects form an additive category \(\mathcal A_d\). A complex is a differential object whose curvature

vanishes. Complexes form a full additive subcategory \(\mathcal A_c\). Equation (2.1) gives

Define the shifted differential by \(d_{TX}=-T(d_X)\), and shift morphisms by \(T\). Substitution in (2.1) proves that this defines a functor on \(\mathcal A_d\). Its strict inverse has differential \(-T^{-1}(d_X)\). For every integer \(r\), induction upwards and downwards gives

In particular both shifts preserve complexes.

For \(h:X\to T^{-1}Y\), its proposed boundary is

Lemma 2.1. The arrow (2.5) is differential exactly when

Proof. Expanding \(d_Y\partial h-T(\partial h)d_X\) cancels the two terms \(d_YT(h)d_X\). The remainder is

This vanishes exactly under (2.6). For complexes both sides vanish. \(\square\)

Thus two differential maps are homotopic if their difference equals (2.5) for some \(h\). Such an \(h\) automatically satisfies (2.6). Zero, negation and addition of the homotopies prove reflexivity, symmetry and transitivity. If \(a:W\to X\) and \(b:Y\to Z\) are differential maps, the homotopies for \((\partial h)a\) and \(b(\partial h)\) are respectively

Substitution uses (2.1). Their boundaries are differential because the original composites are. Null-homotopic maps therefore form an additive two-sided ideal. The quotient by this ideal is the additive category \(K_d(\mathcal A)\); the full subcategory on complexes is \(K_c(\mathcal A)\). Shifting a homotopy uses \(-T(h)\), whose boundary is \(T(\partial h)\) with the shifted differentials. The inverse shift works in the same way. Thus the translation descends as an equivalence. An object is contractible exactly when its identity is zero in this quotient.

For perspective, on \(\mathsf{Ab}\) with \(T=1\), the object \((\mathbb Z,d=1)\) has curvature one. Its differential endomorphisms are all integers; their null homotopies are precisely the even integers. Its endomorphism group in \(K_d\) is \(\mathbb Z/2\). This object is not a complex. Between \((\mathbb Z,0)\) and \((\mathbb Z,1)\), the proposed boundary of \(h=1\) is the underlying identity, which is not differential. Equation (2.6) detects that failure.

3. Cones retain all the differential data

For any underlying map \(f:X\to Y\), define the differential object

Its curvature is

If \(X,Y\) are complexes, this matrix vanishes exactly when \(f\) is differential: the off-diagonal entry vanishes exactly when its untranslated entry does, by faithfulness of \(T\). If \(f\) is differential, no restriction on the two diagonal curvatures is required.

For a differential \(f\), the maps

are differential. Their composite is zero. The composite \(j_ff\) is the boundary of \(X\to T^{-1}C_f=X\oplus T^{-1}Y\) with components \((1_X,0)\). The composite \(T(f)p_f\) is the boundary of the projection \(C_f\to Y=T^{-1}(TY)\), using the shifted differential \(-T(d_Y)\) on \(TY\). These calculations prove that all consecutive maps of the cone triangle vanish in \(K_d\).

A strict commutative arrow square \(bf=f'a\) induces the differential cone map \(\operatorname{diag}(T(a),b)\). These matrices preserve identities and composition. For an additive translation-compatible functor \((F,\xi:FT\to UF)\), give \(FX\) the differential \(\xi_XF(d_X)\). Naturality of \(\xi\) proves (2.1) for \(F(f)\). Its cone comparison is

For the top-left component the required equality is naturality of \(\xi\) at \(d_X\), followed by \(U(\xi_X)\); the bottom-left component is naturality at \(f\), and the bottom-right component is the definition of the transported differential. The transported curvature is \(U(\xi_X)\xi_{TX}F(\kappa_X)\), so complexes remain complexes. The inclusion square and the projection square, with comparison \(\xi_X\) on the last vertex, commute. The compatible inverse-translation comparison transports (2.5), so null homotopies remain null homotopies. Consequently this construction descends to \(K_d\) and \(K_c\), and preserves the whole cone triangles.

Call a triangle distinguished if it is isomorphic in \(K_d\) to

The last projection is positive. In the ordinary complex case this is the same convention as Chain homotopies, mapping cones and triangles, Section 2, after interchanging the two cone summands.

4. The triangulated axioms before the octahedron

Lemma 4.1. Distinguished triangles have the identity, completion, rotation and triangle-morphism properties for all differential objects.

Proof. Isomorphism closure is part of the definition. Every quotient morphism has a differential representative and hence a cone triangle. The identity cone has contraction

Matrix multiplication gives \(\partial H=1\), with the two terms containing \(d_X\) cancelling. This proves the identity triangle.

For rotation compute the cone of \(j_f\), ordered as \(TY,TX,Y\). Its differential is

Define \(r(a,b,c)=b\) to \(TX\), and \(i(b)=(-T(f)b,b,0)\) in the other direction. Both are differential, by (2.1), and \(ri=1\). The homotopy \(H(a,b,c)=(c,0,0)\) into \(T^{-1}C_{j_f}\) gives

Thus \(r\) is invertible in \(K_d\). Composing with the cone inclusion gives \(p_f\), and the last cone projection composed with \(i\) is \(-T(f)\). This proves rotation with precisely the required last sign.

For every integer \(r\), the differential isomorphism \(T^rC_f\to C_{T^rf}\) has matrix \(\operatorname{diag}((-1)^r,1)\). It preserves the shifted inclusion and identifies the new projection with \((-1)^rT^r(p_f)\). Three rotations have arrows \(-T(f),-T(j_f),-T(p_f)\); changing the sign of the middle vertex identifies them with that shifted triangle. Two more rotations and an inverse shift therefore prove the reverse implication in the rotation axiom.

Finally suppose \(bf-f'a=\partial h\), for \(h:X\to T^{-1}Y'\). Set

The diagonal differential equations are those for \(a,b\). Equality of the lower-left entries is exactly the translate of \(bf-f'a=T(h)d_X+T^{-1}(d_{Y'})h\). Thus \(c\) is differential. The inclusion and projection squares commute strictly; the first square commutes in the quotient by the given homotopy. Transport through the cone presentations completes every triangle square. None of these calculations used \(\kappa=0\). \(\square\)

At this point Hom exactness in both variances is already available. The proof of Triangles, finite universal maps and the limits of cohomology, Lemma 1.1 uses only identity triangles, rotation, full faithfulness of translation and completion of a square. All four have just been proved; it does not use the octahedron. Consequently its argument applies here without circularity.

We will also need the following consequence: a completed triangle morphism with invertible first two vertices has invertible third vertex. Applying \(\operatorname{Hom}(W,-)\) gives five exact group positions with that vertex in the middle and four outside isomorphisms. Here is the required group argument. For injectivity, a middle element killed by the vertical map has zero outgoing image by injectivity at the next position. Lift it from the preceding position. The image of that lift is in the kernel of the lower incoming map, so it comes from the first position; lift again through its isomorphism and subtract. The original element is then zero. For surjectivity, lift the outgoing image of a desired lower-middle element through the next isomorphism. Its further image is zero by injectivity at the last position, so lift it from the upper-middle position. The remaining lower-middle difference has zero outgoing image and therefore comes from the preceding position; lift through its isomorphism and add. This proves the middle group map bijective. Taking \(W\) equal to the target vertex supplies a right inverse; injectivity with \(W\) the source vertex makes it a left inverse.

In particular every distinguished completion of a quotient arrow is isomorphic to the cone of any chosen differential representative with identity maps on its first two vertices. Apply (4.4) to the square comparing its cone presentation with that representative, and use the preceding result.

5. The entire octahedron for nonzero curvature

Take differential maps \(X\xrightarrow fY\xrightarrow gZ\), and put \(h=gf\). Define

The first two are the strict-square cone maps of Section 3, and \(w\) is a composite of differential maps. The cone of \(u\), ordered as \(T^2X,TY,TX,Z\), has differential

Define differential maps to and from \(C_g=TY\oplus Z\) by

For \(T(r)D\), the first component is \[ -T^2(f)a-T(d_Y)b+T^2(f)(a-T(d_X)c)=-T(d_Y)(b+T(f)c). \] The second is \(T(g)(b+T(f)c)+d_Zz\). These are exactly the components of \(D_gr\), using the translated equation for \(f\). For \(i\), (5.2) gives \((0,-T(d_Y)b,0,T(g)b+d_Zz)\), namely \(T(i)D_g\). Thus both maps are differential and \(ri=1\).

For \(H:C_u\to T^{-1}C_u\) put \(H(a,b,c,z)=(c,0,0,0)\). Multiplying the two matrices in (2.5) gives

The two terms involving \(T(d_X)c\) cancel. The boundary is an actual differential morphism because it equals the difference of two differential maps; thus this is an admissible homotopy even when curvature is nonzero.

The cone inclusion followed by \(r\) is \(v\). The last cone projection preceded by \(i\) is \(w\). It follows that

is a distinguished triangle. Every required face of the octahedron is fixed by

Each equality follows on the two biproduct components. The other three distinguished triangles are the cones of \(f,g,h\). This supplies all four triangles, their specified arrows and the last-arrow face. For three prescribed distinguished completions of the same quotient arrows, use the identity-on-endpoints comparisons established at the end of Section 4 and transport (5.1) and (5.6) through them. The fourth triangle remains distinguished by isomorphism closure. This proves the octahedral axiom for the prescribed triangles as well.

Theorem 5.1. The category \(K_d(\mathcal A)\), its signed translation and cone triangles (3.5) form a triangulated category. Its full subcategory \(K_c(\mathcal A)\) is triangulated. Every additive translation-compatible functor induces triangulated functors on these categories.

Proof. Sections 4–5 prove every axiom. Complexes are closed under both shifts, biproducts and cones by (2.4) and (3.2). Hence the cone completion of a morphism between complexes stays among complexes, and the full subcategory inherits the axioms and comparisons. Section 3 supplies the entire functorial cone and last-arrow comparison; it therefore preserves distinguished triangles. The equivalence of Section 1 transfers these assertions from strict translations to arbitrary translation equivalences. \(\square\)

Corollary 5.2. A differential map is invertible in \(K_d\) exactly when its cone is contractible.

Proof. A contractible cone is zero in the quotient. Hom exactness of its triangle then makes \(\operatorname{Hom}(W,f)\) bijective for every \(W\), which gives a two-sided inverse as in Section 4. Conversely, if \(f\) is invertible, Hom exactness makes \(\operatorname{Hom}(W,C_f)=0\) for every \(W\). Take \(W=C_f\) to obtain a zero identity. \(\square\)

6. Fixed curvature and the category of curvature labels

Suppose \(\eta:1_{\mathcal A}\to T^2\) is natural and satisfies \(\eta_{TX}=T(\eta_X)\). Let \(\mathcal A_\eta\) consist of differential objects with \(\kappa_X=\eta_X\).

Naturality at \(h:X\to T^{-1}Y\) says \(T^2(h)\eta_X=\eta_{T^{-1}Y}h\), and the translation condition says \(\eta_{T^{-1}Y}=T^{-1}(\eta_Y)\). Thus every proposed homotopy between these objects satisfies (2.6). Formula (3.2) shows that a differential cone has curvature \(\operatorname{diag}(T\eta_X,\eta_Y)=\eta_{TX\oplus Y}\). The last equality follows from naturality on the biproduct injections. Formula (2.4) gives stability under both shifts. Hence the full subcategory \(K_\eta\) of \(K_d\) is triangulated. The case \(\eta=0\) is \(K_c\).

Even arbitrary curvature can be expressed this way after enlarging the category of objects. Let \(\mathcal B\) have objects \((X,e:X\to T^2X)\) and maps

Identities, composition and addition satisfy this equation. The zero label and block diagonal labels give zero objects and biproducts. Set

Its inverse is induced by \(T^{-1}\). Forgetting the labels is additive and translation-compatible. The arrow \(e\) is itself a morphism \((X,e)\to T'^2(X,e)\): its required equation is the identity \(T^2(e)e=T^2(e)e\). Equations (6.1)–(6.2) show that these arrows form a natural transformation \(\eta:1_{\mathcal B}\to T'^2\), with \(\eta T'=T'\eta\).

For a differential object \((X,d)\) in \(\mathcal A\), give it the label \(e=T(d)d\). The arrow \(d:(X,e)\to(TX,T(e))\) is a morphism of \(\mathcal B\), since

Its curvature equals its label. Conversely, an \(\eta\)-curved differential object in \(\mathcal B\) necessarily has precisely this label. Differential maps in \(\mathcal A\) preserve these labels by (2.3), so forgetting is an equivalence on these differential objects and morphisms. A homotopy \(h:X\to T^{-1}Y\) is a morphism of their labelled objects exactly under (2.6), which is exactly the condition for its boundary to be differential. Thus the null-homotopy ideals agree as well. Forgetting preserves the shifts and entire cone triangles, so it induces an equivalence of triangulated categories

This argument establishes (6.4) for intrinsic curvature; it does not assume that all original objects already have one common natural curvature in \(\mathcal A\).

7. Actual factorizations and split underlying sequences

Proposition 7.1. For a differential map \(f:X\to Y\), the following are equivalent: it is null-homotopic; it factors through \(j_1:X\to C_{1_X}\); it factors through \(T^{-1}p_1:T^{-1}C_{1_Y}\to Y\); it factors through some contractible differential object.

Proof. If \(f=\partial h\), the differential map

satisfies \(bj_1=f\). Its first differential equation is \(d_YT(h)=-T^2(h)T(d_X)+T(f)\), the translate of the boundary equation. Its second is (2.1) for \(f\).

The object \(T^{-1}C_{1_Y}=Y\oplus T^{-1}Y\) has differential \(\left(\begin{smallmatrix}d_Y&0\\-1&-T^{-1}(d_Y)\end{smallmatrix}\right)\). The map \(a=(f,-h):X\to T^{-1}C_{1_Y}\) is differential: its second equation is \(-f+T^{-1}(d_Y)h=-T(h)d_X\), and its first is (2.1). Projection onto \(Y\) gives \(f\). Both identity cones are contractible by (4.1), and their shifts are contractible.

If \(f=ba\) through an object with contraction \(s\), (2.7) gives \(f=\partial(T^{-1}(b)sa)\). This proves all reverse implications and the equivalences. \(\square\)

Maps factoring through contractible objects form an additive ideal: sums factor through the biproduct of two contractible objects, whose two contractions assemble, and compositions retain the factorization. Proposition 7.1 identifies this ideal with the null-homotopy ideal. Therefore the additive quotient by these factorizations is exactly \(K_d\), with unchanged objects and composition. No Verdier localization is asserted here.

Now consider complexes and a sequence of differential maps \(X\xrightarrow fY\xrightarrow gZ\) with \(gf=0\). The map \(u=(0,g):C_f\to Z\) is differential: its equation on the first summand is \(T(g)T(f)=0\), and on the second is the equation for \(g\). Suppose the underlying sequence splits, with

These are underlying maps; \(k,h\) need not be differential. Multiplying the last equation by \(k\) gives \(khg=0\), and then \(kh=khgh=0\). The maps \((f,h):X\oplus Z\to Y\) and \((k,g)\) are inverse. In these coordinates the differential on \(Y\) has matrix

Since \(Y\) is a complex, \(T(d_X)b+T(b)d_Z=0\). In the cone coordinates \((TX,X,Z)\), the differential is

Define \(v:Z\to C_f\) by \(v(z)=(-bz,0,z)\). Equation (7.3) makes its first differential equation hold; its second is \(-b+b=0\), and its third is the equation for \(1_Z\). Thus \(v\) is differential, and \(uv=1\). The homotopy \(H(x,y,z)=(y,0,0)\) into \(T^{-1}C_f\) gives

Consequently \(u\) is a homotopy equivalence. In the original coordinates \(v=(-T(k)d_Yh,h)\), and the resulting entire distinguished triangle has last map \(p_fv=-b\). This also fixes the connecting sign for a split underlying sequence.

One further useful contraction does not require the same homotopy in both terms. For a complex \(X\), suppose

Composing on the left with \(d_X\), and composing the translated equation on the right with \(d_X\), gives respectively \(d_XT(s)d_X=d_X\) and \(d_XT(t)d_X=d_X\); the omitted terms contain \(T(d_X)d_X=0\). Apply \(T^{-1}\) to the first identity. For \(h=sT^{-1}(d_X)t\) the two terms of its boundary are then

Their sum is the identity. Thus (7.6) gives a genuine contraction. The square-zero hypothesis was used in obtaining those two intermediate identities.

8. Four graded exercises with complete solutions

Exercise 1 (foundation: curvature and a two-torsion identity). Take \(T=1\) on abelian groups. Compare the differential objects \((\mathbb Z,0)\) and \((\mathbb Z,1)\). Compute their endomorphism groups in \(K_d\), and explain why an arbitrary underlying map cannot be used as a homotopy between them.

Solution. For differential zero, every integer endomorphism is differential and every boundary is zero, so the endomorphism group is \(\mathbb Z\). For differential one, every endomorphism still commutes with the differential, while \(\partial h=h+h=2h\); thus the quotient is \(\mathbb Z/2\). The latter identity survives and the object is not contractible. Its curvature is one, so it is not a complex. A differential map from the first object to the second must satisfy \(f=0\). An underlying homotopy \(h\) would have boundary \(h\); it is differential only when \(h=0\). Equivalently (2.6) says \(0=h\). This checks both the boundary and its admissibility.

Exercise 2 (intermediate: strict cone comparisons). For differential maps \(f,g:X\to Y\), prove that they are homotopic exactly when a differential map \(c:C_f\to C_g\) commutes strictly with the two inclusions from \(Y\) and the two projections to \(TX\). Prove that any such \(c\) is an actual differential isomorphism.

Solution. The inclusion equation forces the second column of \(c\) to be \((0,1_Y)\). The projection equation forces its first row to be \((1_{TX},0)\). Hence

Full faithfulness of translation supplies a unique \(h:X\to T^{-1}Y\) with \(a=T(h)\). The lower-left differential equation for (8.1) is \(T(f-g)=T^2(h)T(d_X)+d_YT(h)\), which is exactly \(f-g=\partial h\). The other entries are automatically equal. Conversely that homotopy equation proves (8.1) differential. Its underlying inverse has lower-left entry \(-a\), and either the same calculation with \(-h\), or the equation for the inverse of a differential isomorphism, proves it differential. Thus the comparison is a strict isomorphism, not merely a homotopy equivalence.

Exercise 3 (advanced: turn a homotopy retract into a strict retract). Let complexes \(X,Y\) have differential maps \(\phi:X\to Y\), \(\psi:Y\to X\) with \(\psi\phi-1_X\) null-homotopic. Construct a contractible complex \(Z\) and differential maps \(\alpha:X\to Y\oplus Z\), \(\beta:Y\oplus Z\to X\) with \(\beta\alpha=1_X\) strictly.

Solution. Put \(q=\psi\phi-1_X=\partial h\). Use Proposition 7.1 with \(Z=C_{1_X}\), \(a=j_1:X\to Z\) and \(b=(T(h),q):Z\to X\). Then \(ba=q\); both maps are differential by (7.1). The object \(Z\) is a complex by (3.2) and contractible by (4.1). Set \(\alpha=(\phi,a)\) and \(\beta=(\psi,-b)\). These are differential biproduct maps, and \(\beta\alpha=\psi\phi-ba=1_X\). The extra summand is an actual contractible object, so no splitting of an idempotent in the original additive category has been assumed.

Exercise 4 (synthesis: a non-differential underlying splitting). In graded abelian groups with the index-shift translation, let \(X\) be \(\mathbb Z\) in degree zero and \(Z\) be \(\mathbb Z\) in degree minus one, both with zero differential. Let \(Y^{-1}=Y^0=\mathbb Z\), with differential multiplication by \(m\). Take the evident inclusion \(X\to Y\) and projection \(Y\to Z\). Compute the homotopy equivalence from their first cone to \(Z\), its contraction difference and the last arrow of the resulting triangle.

Solution. The graded underlying sequence splits, but its section from \(Z\) to \(Y\) is a chain map only when \(m=0\). In (7.3), \(b^{-1}=m:\mathbb Z\to\mathbb Z=X^0\). The cone is \(\mathbb Z\oplus\mathbb Z\) in degree minus one and \(\mathbb Z\) in degree zero, with differential \((x,z)\mapsto x+mz\). The map \(u\) is projection onto \(z\) in degree minus one and zero in degree zero. Its homotopy inverse is \(v(z)=(-mz,z)\), for which \(uv=1\) and the cone differential is zero on the image. The degree-minus-one homotopy sends a degree-zero integer \(y\) to \((y,0)\). Its boundary is \((x+mz,0)\) in degree minus one and the identity in degree zero, exactly \(1-vu\). The distinguished triangle \(X\to Y\to Z\to X[1]\) has last arrow multiplication by \(-m\) in degree minus one. Thus the underlying splitting determines the cone equivalence, while the differential still contributes a nonzero connecting arrow.

References

• Pierre Schapira, An Introduction to Categories and Homological Algebra, Sections 4.2 and 6.2, for ordinary complexes and their homotopy categories.
• The Stacks Project Authors, Cones and termwise split sequences and Distinguished triangles in the homotopy category, for the square-zero case in additive categories. Their negative last-projection convention is stated explicitly there; the positive convention used throughout this lesson is (3.5).