Draft lesson.

Acyclic lines, Weil staircases and corner removal

Original text public domain (CC0).

When a double complex is exact away from one line, its total can be compared with that line by two actual truncation maps. In a module quadrant, these maps give a finite staircase calculation. We prove its independence of every choice, its inverse and its sign. We also determine the degree shift when both coordinate axes of an exact quadrant are removed.

Use Finite diagonal totals and signed double truncations, especially Theorem 5.1 and its transposed version. Our double complexes commute, and their total differential on bidegree \((i,j)\) is \(d'+(-1)^id''\). Ordinary shifts have differential \(d_{E[r]}=(-1)^rd_E\). These conventions and the natural transpose multiplier are proved in Signed bifunctors and sum and product total complexes, Sections 3–5. All categories in Sections 1–2 and 5 are abelian; Sections 3–4 specialize to left modules over a ring \(R\).

1. Acyclicity in either direction

Write a row as \(X^{i,\bullet}\), with differential \(d''\), and a column as \(X^{\bullet,j}\), with differential \(d'\). Assume finite diagonal support throughout this section.

Proposition 1.1. If every column is acyclic, the total is acyclic. The same holds if every row is acyclic.

Proof. When every column is acyclic, all complexes \(H_I^p(X)\) are zero. Thus \(X\to0\) induces an isomorphism on \(H_{II}H_I\), and Theorem 5.1 of the prerequisite makes \(\operatorname{Tot}X\to0\) a quasi-isomorphism. When every row is acyclic, apply the same argument to the transposed double complex. It retains finite diagonal support. The natural total isomorphism from its total to the original total acts on \(X^{i,j}\) by \((-1)^{ij}\), so the acyclicity conclusion transports to the original total. \(\square\)

2. An isolated line and its two comparison maps

Let \(\sigma^I_{\ge p}X\) keep the terms in first degrees \(i\ge p\) and discard the rest, and let \(\sigma^I_{\le p}\) keep the terms in first degrees \(i\le p\). Retained differentials are original; differentials incident to a discarded term are zero. These are the stupid truncations in the outer complex of \(F_I(X)\). Both double-complex equations survive, and the support remains a subset of the original support. Put \(A=\sigma^I_{\ge p}X\) and \(E=\sigma^I_{\le p}A\). There are natural double-complex maps

the inclusion and the projection. The inclusion commutes with the first differential because no arrow leaves the retained tail for a discarded first degree; the projection commutes because all arrows from the retained first-degree-\(p\) term to higher first degrees have zero projected image. Their second equations hold termwise, as \(d''\) preserves the first degree.

Proposition 2.1. If \(X^{i,\bullet}\) is acyclic for every \(i\ne p\), both total maps in (2.1) are quasi-isomorphisms, and

Proof. Second-direction cohomology of \(X\), \(A\) and \(E\) is supported only in first degree \(p\). In that degree both maps induce the identity on the cohomology of \(X^{p,\bullet}\). In all other first degrees they induce the isomorphism \(0\to0\). Thus both maps induce isomorphisms on the second-direction cohomology complexes, including their first differentials: those differentials are zero because their adjacent terms vanish. In particular they induce isomorphisms on \(H_IH_{II}\). The transposed version of Theorem 5.1 makes their totals quasi-isomorphisms. The isolated total has term \(X^{p,n-p}\) in total degree \(n\) and differential \((-1)^pd''\). This is precisely the ordinary shifted complex in (2.2), with the identity comparison on each term. All maps and identifications are natural in double-complex maps satisfying the hypotheses. \(\square\)

For columns, use the direction-II stupid truncations to obtain

If every column \(X^{\bullet,j}\) with \(j\ne p\) is acyclic, both total maps are quasi-isomorphisms by the original direction of Theorem 5.1: first-direction cohomology is unchanged at second degree \(p\) and zero elsewhere. The isolated column total has differential \(d'\), whereas \(X^{\bullet,p}[-p]\) has differential \((-1)^pd'\). Its actual comparison with this shifted column multiplies the term \(X^{i,p}\) by

Indeed the multiplier at \(i+1\) is \((-1)^p\) times the multiplier at \(i\), which is exactly the chain equation. Its inverse has the same multipliers, and it is natural under component maps. This also follows by transposing (2.2) and using the multiplier \((-1)^{ip}\) on the isolated line. Thus both variances include their full shift comparisons.

If every row except row zero and every column except column zero is acyclic, (2.1) and (2.3), with \(p=0\), give natural isomorphisms in every degree

Here the isomorphisms are the specified cohomology maps of the two zigzags and their inverses. The isolated-line shifts and multipliers are identity at \(p=0\); no sign is hidden in (2.5). Negative degrees are included, and no quadrant assumption is required for this assertion.

3. The actual comparison in a module quadrant

Now suppose \(X^{i,j}=0\) if \(i<0\) or \(j<0\). Each total diagonal is finite. Assume that rows \(i>0\) and columns \(j>0\) are acyclic. Their zeroth lines may have cohomology. Because there are no negative first or second degrees, the tail inclusions in (2.1) and (2.3) at zero are identities. The remaining maps are actual projections

The first retains the summands with second degree zero, and the second those with first degree zero. They are chain maps: outgoing arrows into a discarded coordinate have zero image, and the retained arrows have exactly the axis differential. Proposition 2.1 and its column version show that both are quasi-isomorphisms. Consequently the canonical comparison from column cohomology to row cohomology is

This specifies an actual isomorphism, natural in a morphism of such double complexes. The inverse maps in (3.2) are inverses of induced isomorphisms of modules; no choice of a chain inverse is assumed.

For \(n<0\) both axes have zero terms and zero cohomology. For \(n\ge0\), take a cycle \(x_0\in X^{n,0}\), so \(d'x_0=0\). Inductively choose

These are the unsigned forward staircase equations. To prove existence at each step, the first step has \(d'd''x_0=d''d'x_0=0\). After a step, \(d'd''x_r=d''d'x_r=(d'')^2x_{r-1}=0\). Thus the next right-hand side is a first-direction cycle in column \(r+1>0\). Exactness of that column supplies its preimage under \(d'\). For left modules, an epimorphism onto the image is surjective on elements, so this use of a preimage is valid. There are only \(n\) choices, not an infinite lifting process.

At the end, \(d'd''x_n=0\), also when \(n=0\), by the same calculation. In column \(n+1>0\), exactness in first degree zero makes \(d':X^{0,n+1}\to X^{1,n+1}\) injective: its preceding term is zero. Hence \(d''x_n=0\). The endpoint is a row-zero cycle.

4. Choices, representatives, inverse and the total sign

Define signs by

The product of the \(n\) recurrence factors gives \(e_n=(-1)^{n(n+3)/2}=\varepsilon_n\), because the two integer exponents differ by \(2n\). The finite diagonal element

is a total cycle. At the coordinate \((n-r,r+1)\), its differential is

by (3.3) and the sign recurrence. Its two extreme outgoing terms are \(d'x_0=0\) and \(d''x_n=0\). These list all possible outputs of its summands, so \(Dz=0\). Projecting gives \(\pi_cz=x_0\) and \(\pi_rz=\varepsilon_nx_n\). Therefore

In particular the unsigned endpoint map is

This proves independence of all choices: any permitted sequence (3.3) produces a cycle (4.2) with the same projected column class, and \(H^n(\pi_c)\) is invertible. All those total classes therefore coincide; their projected row classes coincide too, and multiplication by \(\varepsilon_n\) is invertible. It also proves independence of the initial representative. If \(x_0\) is replaced by a homologous cycle, its column class remains unchanged in (4.4); hence its row endpoint class remains unchanged. If the original cycle is a boundary, the total class in (4.2) is zero by injectivity of \(H^n(\pi_c)\), and its endpoint class is zero. Thus the construction descends to cohomology, including boundaries.

The same equality proves that \(W_n\) is \(R\)-linear and natural: (3.2) is a natural isomorphism of left modules, and \(\varepsilon_n\) is the central scalar \(1\) or \(-1\). One may also add or scalar-multiply chosen staircases, obtaining permitted staircases for the resulting cycles; a double-complex map sends every equation (3.3) to the corresponding equation. Choice independence makes these direct checks agree with (4.5).

For an explicit reverse procedure, begin with a row cycle \(y_0\in X^{0,n}\) and choose

The right-hand side is a second-direction cycle: initially \(d''d'y_0=d'd''y_0=0\), and subsequently \(d''d'y_r=d'd''y_r=(d')^2y_{r-1}=0\). Exactness of row \(r>0\) supplies each lift. At the final endpoint, the same square equation gives \(d''d'y_n=0\). Exactness of row \(n+1>0\) at second degree zero makes \(d''\) injective there, so \(d'y_n=0\). This includes \(n=0\), when there are no lifts.

Set \(a_0=1\), \(a_{r+1}=(-1)^ra_r\). The endpoint coefficient is \(a_n=\varepsilon_n\). For the total element \(w=\sum_{r=0}^na_ry_r\), the output at \((r+1,n-r)\) is

Its extreme terms vanish, so it too is a total cycle. Its row projection is \(y_0\) and its column projection is \(\varepsilon_ny_n\). Consequently the unsigned reverse endpoint map is well-defined, independent of choices and representatives, linear and natural, with

Equations (4.5) and (4.8), and \(\varepsilon_n^2=1\), give \(V_nW_n=1\) and \(W_nV_n=1\). They are actual inverse maps. For \(n=0\), both staircases have their single corner term; the injectivity arguments above show it is a cycle in both axes, and the coefficient is one. For negative degrees define both maps as the unique map between the zero cohomology modules. This proves the entire staircase comparison in every degree and accounts for its relation to the signed total comparison.

5. Removing both axes gives a two-degree shift

Return to any abelian category \(\mathcal A\). Assume that \(X\) is first quadrant and that all its rows and columns are acyclic, including the zero ones. Define \(Y\) by keeping exactly the terms \(X^{i,j}\) with \(i,j\ge1\), using the retained differentials and zero elsewhere. They still commute, and every total diagonal is finite.

Proposition 5.1. There is a natural quasi-isomorphism

Here \(X^{0,0}\) on the left denotes the point complex in degree zero before shifting. All other components of \(\gamma\) are zero.

Proof. For each \(j\ge1\), deleting first degree zero from the exact column leaves first-direction cohomology only in degree one. Exactness and the zero term in degree minus one make \(d'^{0,j}\) monic and identify it isomorphically with the cycles in degree one. All first-direction degrees above one retain the incoming image and outgoing kernel of the original exact column, while degrees below one are zero. Therefore \(H_I^1(Y)\) is naturally identified with the complex

and every other \(H_I^i(Y)\) is zero. Commutation with \(d''\) gives the differential of this identification: \(d''d'^{0,j}=d'^{0,j+1}d''\), so it is the original second differential of row zero, rather than its negative.

Exactness of row zero now shows that \(H^j(E)=0\) for \(j\ne1\), and that \(d''^{0,0}\) identifies \(X^{0,0}\) isomorphically with \(H^1(E)=\ker d''^{0,1}\). Indeed \(d''^{0,0}\) is monic because the preceding term is zero, and its image is that kernel. In degree one there is no incoming term in \(E\); in higher degrees its original incoming image equals the outgoing kernel. Thus \(H_{II}H_I(Y)\) is supported only at \((1,1)\), where the indicated object is \(X^{0,0}\).

Let \(M\) be the double complex with sole term \(X^{0,0}\) at \((1,1)\), and zero differentials. Define \(g:M\to Y\) there by \(d'^{0,1}d''^{0,0}\). Its outgoing first equation is zero by \((d')^2=0\), and its outgoing second equation is zero by commutation of the two differentials and \((d'')^2=0\). Its other equations have zero source or target. Thus it is a double-complex map. Under the identification (5.2), its first-direction cohomology map is precisely \(d''^{0,0}:X^{0,0}[\text{degree }1]\to E\). The preceding kernel/image argument shows this is a quasi-isomorphism. Hence \(g\) induces an isomorphism on \(H_{II}H_I\) at \((1,1)\) and the isomorphism \(0\to0\) everywhere else. Theorem 5.1 of the prerequisite makes \(\operatorname{Tot}g\) a quasi-isomorphism. Its source has sole term \(X^{0,0}\) in total degree two, exactly \(X^{0,0}[-2]\), and its component is the positive map in (5.1). This proves the formula and every cohomology degree. The construction uses actual differential composites and their natural kernel/image identifications, so it is natural in \(X\). \(\square\)

The unshifted point complex is not the source of this comparison. For example, over a nonzero field \(k\), take \(X^{i,j}=k\) at the four positions \(i,j\in\{0,1\}\), with both arrows along each direction the identity, and zero elsewhere. Every row and column is an exact two-term complex. Removing the axes leaves only \(Y^{1,1}=k\), whose total is concentrated in degree two. Every degree-preserving map from the unshifted point complex \(k\) to this total is zero, and it cannot be a quasi-isomorphism. The map from \(k[-2]\) in (5.1) is instead the identity in degree two. This detects the shift with an explicit exact quadrant.

6. Four graded exercises with complete solutions

Exercise 1 (both isolated-line signs). Put the complex \(E=(\mathbb Z\xrightarrow{2}\mathbb Z)\), in degrees zero and one, first in the isolated row \(i=-1\), and then in the isolated column \(j=-1\). Compute both totals and their comparisons with \(E[1]\).

Solution. The row total has \(\mathbb Z\) in degrees \(-1,0\) with differential \(-2\), since \((-1)^{-1}=-1\). This is \(E[1]\) with the identity comparison. The column total has the same degrees but differential \(2\), because the first-direction total differential has coefficient one. Its comparison with \(E[1]\) is identity in degree \(-1\), and minus identity in degree zero, by (2.4) at first degrees \(0,1\). The chain equation is \((-2)\cdot1=(-1)\cdot2\). Both totals have zero cohomology in degree \(-1\) and \(\mathbb Z/2\mathbb Z\) in degree zero. The induced minus sign happens to act trivially on this quotient, but the nonzero integer chain equation still distinguishes the two comparisons.

Exercise 2 (a degree-two staircase sign). Over \(\mathbb Z\), put one free rank-one term with labels \(x,u,v,w,y\) at \((2,0),(2,1),(1,1),(1,2),(0,2)\), respectively. Let \(d''x=u\), \(d'v=u\), \(d''v=w\), and \(d'y=w\), with other arrows zero. Determine the unsigned staircase, the total cycle representing the column class, and the canonical comparison (3.2).

Solution. Both same-direction squares vanish. The mixed square starting at \(v\) has both routes zero at \((2,2)\); every other mixed square also has both routes zero. Rows one and two are exact identity pairs \(v\to w\) and \(x\to u\). Columns one and two are exact identity pairs \(v\to u\) and \(y\to w\). All other positive rows and columns are zero, so the hypotheses of Section 3 hold. Column zero has only \(x\) in degree two, and row zero has only \(y\) in degree two. The unsigned equations take \(x\) to \(v\), then to \(y\), so \(W_2\) sends \([x]\) to \([y]\).

The sign recurrence gives \(e_0=1,e_1=-1,e_2=-1\). Hence \(z=x-v-y\) is a total cycle. Directly, \(Dz=u-(u-w)-w=0\), where the second arrow on \(v\), of first degree one, has a minus sign. The column projection is \(x\), and the row projection is \(-y\). Thus \(\Psi_2([x])=-[y]\) whereas the unsigned endpoint is \([y]\), in agreement with \(\varepsilon_2=-1\). The reverse unsigned procedure takes \(y\) to \(v\) to \(x\), giving the inverse of \(W_2\).

Exercise 3 (the corner degree shift). Use the four-term identity square over \(\mathbb Z\): its objects are \(\mathbb Z\) at \((0,0),(1,0),(0,1),(1,1)\), with both directional arrows the identity. Compute the full total and the total after deleting the two axes. Verify (5.1) and rule out an unshifted quasi-isomorphism.

Solution. Every row and column is an identity pair, hence exact. The full total has terms \(\mathbb Z,\mathbb Z^2,\mathbb Z\) in degrees \(0,1,2\). Order the middle coordinates as \((1,0),(0,1)\). Its differentials are \(t\mapsto(t,t)\) and \((a,b)\mapsto -a+b\): the vertical arrow from first degree one carries the minus sign. The first map is injective, the second is surjective, and its kernel is the diagonal, the image of the first. Thus the full total is acyclic. Deleting the axes leaves the point \(\mathbb Z\) in total degree two. The composite \(d'^{0,1}d''^{0,0}\) is identity, so (5.1) is the identity from \(\mathbb Z[-2]\) to that total. An unshifted chain map has degree-zero component into zero and is therefore zero. It sends the nonzero \(H^0=\mathbb Z\) of its proposed source to zero, so it is not a quasi-isomorphism.

Exercise 4 (all staircase lengths). For \(n\ge0\), put \(Rx_r\) at \((n-r,r)\), \(0\le r\le n\), and \(Rt_r\) at \((n-r,r+1)\), \(0\le r<n\). Define \(d''x_r=t_r\) for \(r<n\) and \(d'x_{r+1}=t_r\), with every other arrow zero. Prove the hypotheses of Section 3, compute both endpoint maps, and verify that both projections in (3.1) induce isomorphisms directly.

Solution. Every output \(t_r\) has zero outgoing arrows, so both squares vanish and both mixed composites are zero. For each positive first degree \(i=n-r\), its row is the identity pair \(Rx_r\to Rt_r\). For each positive second degree \(j=r+1\), its column is the identity pair \(Rx_{r+1}\to Rt_r\). Other positive rows and columns are zero. The axes consist of \(Rx_0\) in first degree \(n\) and \(Rx_n\) in second degree \(n\). For \(n=0\) there are no positive lines or \(t_r\)'s, and both axes are the corner term.

The only total differential goes from the free module on \(x_0,\ldots,x_n\) in degree \(n\) to the free module on \(t_0,\ldots,t_{n-1}\) in degree \(n+1\). On coefficients it is

It is surjective over every ring: set \(b_0=0\) and, for any specified target coefficients \(c_r\), define \(b_{r+1}=c_r-(-1)^{n-r}b_r\) recursively. Its kernel is the free rank-one module of tuples \(b_r=e_rb_0\), where (4.1) gives the recurrence. Thus total cohomology is \(R\) in degree \(n\) and zero elsewhere, including degree \(n+1\). Column projection sends this generator to \(x_0\), and row projection sends it to \(\varepsilon_nx_n\). Both are isomorphisms because those coefficients are units. The unsigned staircase has every coefficient one, so \(W_n\) sends \(x_0\) to \(x_n\); the reverse does the reverse. Their inverses and the sign of the canonical comparison therefore agree with (4.5) and (4.8), also for the empty staircase \(n=0\).

References

• Finite diagonal totals and signed double truncations, Theorem 5.1 and equation (5.7), provides the complete iterated-comparison theorem in both directions, used in all the quasi-isomorphism claims here.
• Signed bifunctors and sum and product total complexes, Sections 3–5, provides the commuting and signed total conventions.
• The Stacks Project, Lemma 12.25.4, gives a related comparison from an axis complex to a one-sided exact double complex. The explicit truncation projections, staircase choices and corner map used here have their proofs above.