Grothendieck Spectral Sequences and Computations — Examples

For $r=2$ the only possible nonzero arrows are ${\tau }_p$ from row one to row zero. Hence $E_3^{p,0}=\mathrm{coker}{\tau }_{p-2}$ and $E_3^{p,1}=\ker {\tau }_p$. For $r\ge 3$, every outgoing arrow from either row has negative second coordinate and every incoming arrow starts above row one. These positions stay zero on successive pages, so $E_3=E_{\mathrm{\infty }}$.

In total degree $n$, the only possible filtration quotients are at $p=n$ and $p=n-1$. F1's zero/full endpoints identify the first as a subobject of the target and the second as its quotient, giving the displayed exact sequence. At $n=0$ it reduces to ${\mathbb{R}}^{0}F(K)=F(A)$; at $n=1$ its subobject is $R^{1}F(A)$ and its quotient is $\ker {\tau }_0$. Below zero there are no surviving quotients, so the finite target filtration forces vanishing. This proves convergence and identifies the unresolved extension, rather than assuming a splitting.

A fully numerical specialization takes abelian groups, $F$ the identity, and $K^0=\mathbb{Z}\mathrm{/}2$, $K^1=\mathbb{Z}\mathrm{/}3$ with zero differential and supplied replacements. Exactness of identity means its positive derived objects vanish, since applying it preserves the exact resolution. Thus $E_2^{0,0}=\mathbb{Z}\mathrm{/}2$, $E_2^{0,1}=\mathbb{Z}\mathrm{/}3$, all other entries are zero, and every ${\tau }_p$ is zero. Each total degree has one nonzero quotient: the target is $\mathbb{Z}\mathrm{/}2$ in degree zero, $\mathbb{Z}\mathrm{/}3$ in degree one and zero otherwise. The upper edges are the identity under the augmentation identification. This specialization has no extension ambiguity and uses no choice beyond the supplied-data convention.

Applying an exact $G$ to any augmented injective resolution preserves its positive exactness, so $R^{p}G(V)=0$ for $p>0$ and every $V$ is $G$-acyclic. The Grothendieck injective-image condition is therefore automatic. The page has only column zero; for every $r\ge 2$ outgoing differentials land in a zero positive column and incoming ones start in a negative column. Consequently $E_2=E_{\mathrm{\infty }}$, and $F^1{H}^n=0,F^0{H}^n=H^n$ reconstruct the target through the upper edge in F1.

For the displayed specialization resolve $\mathbb{Z}\mathrm{/}2$ by $0\to \mathbb{Z}\stackrel{2}{\to }\mathbb{Z}\to \mathbb{Z}\mathrm{/}2\to 0$. The free rank-one terms are projective by lifting the image of $1$. Applying $\mathrm{Hom}(-,\mathbb{Z})$ gives $\mathbb{Z}\stackrel{2}{\to }\mathbb{Z}$, with kernel zero and cokernel $\mathbb{Z}\mathrm{/}2$. F2 thus gives $R^{0}F(\mathbb{Z})=0$, $R^{1}F(\mathbb{Z})=\mathbb{Z}\mathrm{/}2$ and all higher terms zero. The target is zero outside degree one and is $\mathbb{Z}\mathrm{/}2$ in degree one; its upper edge is the identity under these common Hom-cohomology identifications. The lower edge is zero in positive degrees because its source is a positive derived identity functor. No extension or additional splitting choice remains.

Substitute $F=(-{)}^N$ and $T=(-{)}^Q$ into F1. The five terms become, in order, $H^1(Q,M^N)$, $H^1(G,M)$, $H^1(N,M{)}^Q$, $H^2(Q,M^N)$ and $H^2(G,M)$. F2 identifies the first and last maps with its bottom-cycle inflation, the second with restriction of invariant cocycles, and the middle with $d_2$ from $(0,1)$ to $(2,0)$. Hence every term and arrow agrees with F2, without importing a later cocycle-classification theorem.

To compute the specialization let $C_2=⟨s⟩$. Its trivial $\mathbb{Z}[C_2]$-module has the free resolution with augmentation and alternating differentials $d_1=s-1$, $d_2=1+s$, $d_3=s-1$, continuing periodically. Indeed for $a+bs$, the kernel of $s-1$ is $\mathbb{Z}(1+s)$ and the kernel of $1+s$ is $\mathbb{Z}(1-s)$; these are the preceding images. The augmentation kernel is also $\mathbb{Z}(1-s)$. Rank-one free terms are projective by lifting the image of their generator. Hom into trivial $\mathbb{Z}$ has successive differentials $0,2,0,2,\dots$. Thus F3 gives $H^1(C_2,\mathbb{Z})=0$ and $H^2(C_2,\mathbb{Z})=\mathbb{Z}\mathrm{/}2$.

For $Q=1$, invariants are identity and the positive cohomology is zero by exactness of any supplied resolution. Consequently step 1.1 has four zero non-initial terms before its final $\mathbb{Z}\mathrm{/}2$, as claimed. All maps in that displayed finite portion are zero, its transgression is zero and exactness holds at every required position. There is no final surjectivity. The rank-one periodic calculation itself is choice-free; only the common resolution-independent comparison convention is inherited.

Present any abelian group by the free group on its underlying set. F2 makes the kernel free, and AC lifts arbitrary basis images to prove these free groups projective. Thus every such group has a projective resolution of length at most one and Ext vanishes for $p>1$. In F1 every $d_r$ for $r\ge 2$ changes $p$ by at least two, so its source or target vanishes. Therefore $E_2=E_{\mathrm{\infty }}$, and $F^1{H}^n$ is the Ext-one term while $H^n\mathrm{/}{F}^1{H}^n$ is the Hom term. The edge is restriction of a Hom cocycle to cycles, hence evaluation as in F3; the boundary quotient defining its kernel is the same free presentation of $H_{n-1}C$ used in F3. This proves the displayed exact sequence with its actual arrows.

For the specified $C$, its homology is $H_{0}C=\mathbb{Z}\mathrm{/}2$, $H_{1}C=\mathbb{Z}b$, zero elsewhere. Hom of the presentation $\mathbb{Z}\stackrel{2}{\to }\mathbb{Z}$ into $M$ has zero differential, so Hom and Ext-one of $\mathbb{Z}\mathrm{/}2$ into $M$ are both $M$. Hom of $\mathbb{Z}$ is $M$ and its positive Ext is zero, using its one-term projective resolution. This gives exactly the three asserted page entries. Directly, $\mathrm{Hom}(C,M)$ is $M\stackrel{0}{\to }{M}^2$ in degrees zero and one, confirming both targets and zero in all other degrees.

In degree zero $F^0{H}^0=M,F^1{H}^0=0$. In degree one, identify a cochain by $(x,y)=(f(a),f(b))$; then $F^0{H}^1=M^2$, $F^1{H}^1=M\oplus 0$, $F^2{H}^1=0$. Evaluation is $(x,y)\mapsto y$ and the Ext injection is $x\mapsto (x,0)$. The section $y\mapsto (0,y)$ splits this particular extension. But $a\mapsto a+b$ fixes homology and induces $(x,y)\mapsto (x+y,y)$, which moves every lift of $1$; hence no section is natural in $C$, in agreement with F4. The finite endpoints prove there is no unresolved convergence issue. AC is used for the general free-kernel argument and optional general sections; this displayed finite calculation itself makes no infinite choices.

F1 gives zero in every column $p>1$, because free presentations have free kernels under AC. A later differential changes $p$ by $-r$ for $r\ge 2$, so no two surviving columns can be joined. Hence $E^2=E^{\mathrm{\infty }}$. The increasing target filtration is $F_{-1}=0$, $F_0=T_n$ and $F_1=H_n$, with $F_1\mathrm{/}{F}_0=U_n$. Its maps are the cross product and Tor quotient of F1. AC is used in free-submodule/projectivity and section choices, and a splitting is additional to this natural exact sequence.

In the numerical case $H_{0}C=\mathbb{Z}\mathrm{/}2$, $H_{1}C=\mathbb{Z}$, $H_{0}D=\mathbb{Z}\mathrm{/}2$ and $H_{1}D=0$. Tensoring the rank-one resolution $\mathbb{Z}\stackrel{2}{\to }\mathbb{Z}$ of $\mathbb{Z}\mathrm{/}2$ with $\mathbb{Z}\mathrm{/}2$ gives a zero differential, so its Tor-zero and Tor-one groups are both $\mathbb{Z}\mathrm{/}2$. The one-term resolution of $\mathbb{Z}$ gives tensor $\mathbb{Z}\mathrm{/}2$ and zero positive Tor. Thus the three listed page positions are precisely the surviving ones.

Directly the degree-one tensor differential sends $\alpha (a\otimes y)+\beta (b\otimes y)+\gamma (c\otimes x)$ to $2(\alpha +\gamma )c\otimes y$. Its kernel is generated by $u=b\otimes y$ and $t=a\otimes y-c\otimes x$. Degree-two differentials send $a\otimes x$ to $-2t$ and $b\otimes x$ to $-2u$, an injective map onto $2\mathbb{Z}u\oplus 2\mathbb{Z}t$. Thus $H_2=0$, $H_1=(\mathbb{Z}\mathrm{/}2)[u]\oplus (\mathbb{Z}\mathrm{/}2)[t]$, and $H_0=\mathbb{Z}(c\otimes y)\mathrm{/}2=\mathbb{Z}\mathrm{/}2$. In degree one the tensor subobject is $⟨[u]⟩$, and the Tor quotient is generated by the image of $[t]$, in agreement with F1.

Sending the quotient generator to $[t]$ gives a section. The automorphism $a\mapsto a+b$ fixes both end terms but sends $[t]\mapsto [t]+[u]$ and fixes $[u]$, so neither lift of the quotient generator is invariant. This verifies nonnaturality in the actual computed extension. Degree zero has one quotient and filtration $F_{-1}=0,F_0=H_0$; all other degrees except one are zero. These finite filtrations solve reconstruction and convergence for the example. The matrices and their kernels require no additional choice.

The displayed operators satisfy $T^2=S^2=TS=ST=0$. In characteristic two this gives $(1+T{)}^2=(1+S{)}^2=1$ and commuting actions, so $M$ is a well-defined $G$-module. The map $s\mapsto s$ is a group section, and its conjugation action on $N$ is trivial by commutativity, as in F3. It need not act trivially on the coefficient module.

For either cyclic factor use the rank-one free integral group-ring resolution with alternating differentials $t-1,1+t,t-1,1+t,\dots$ (replace $t$ by $s$ for $Q$). It is exact: in $\mathbb{Z}[C_2]$, the kernels are respectively $\mathbb{Z}(1+t)$ and $\mathbb{Z}(1-t)$, equal to the preceding images, and the augmentation kernel is $\mathbb{Z}(1-t)$. Hom into a characteristic-two module replaces every differential by $T$, or by $S$. Thus its positive cohomology is $\ker T\mathrm{/}\mathrm{im}T$, or $\ker S\mathrm{/}\mathrm{im}S$. F4 licenses this computation; each rank-one free module is projective by a single generator lift.

Here $W=M^N=⟨b,c,d⟩$, and $H^1(N,M)=⟨[b],[d]⟩$. The quotient action is trivial on these two classes, since $Sb=c$ is a $T$-boundary and $Sd=0$. To see the action agrees with F1's resolution convention, let $Q$ act trivially on the cyclic $N$-resolution and by its given action on $M$; the Hom-to-injective-total comparison in F4 commutes with these actions and its augmentation. On $W$, $Sb=c$ and $Sc=Sd=0$, so $H^1(Q,W)=H^2(Q,W)=⟨[d]⟩$. Consequently the three page entries in F2 have dimensions $1,2,1$.

Tensor the two cyclic resolutions over $\mathbb{Z}$ and take the signed total. This is a free $\mathbb{Z}[G]$-resolution of $\mathbb{Z}$: each bidegree is rank one over that ring. For exactness, each augmented factor, as an abelian complex, splits into its degree-zero copy of $\mathbb{Z}$ and contractible two-term complexes. Indeed its successive boundary groups have the single displayed generator in step 1.2, and each surjection to that generator has the explicit lift $1$ or $-1$; the augmentation also has lift $1$. These splittings decompose the differentials into identity maps on adjacent summands. Tensoring such a contractible summand with the other complex stays contractible: the homotopy $h\otimes 1$ has cross terms cancelling under the tensor sign. There are finitely many summands in each degree. Thus the total has homology $\mathbb{Z}$ in degree zero and zero above it, proving the resolution claim.

Hom of this total into $M$ has degree zero $M$ and degree one $M\oplus M$, with coboundary $w\mapsto (Tw,Sw)$. Its degree-one cycles $(u,v)$ satisfy $Tu=0$, $Sv=0$ and $Su=Tv$; the three equations come from bidegrees $(2,0),(0,2),(1,1)$ and signs disappear over $k$. Here $u\in ⟨b,c,d⟩$ and $v\in ⟨c,d⟩$, so the third equation forces the $b$-coefficient of $u$ to vanish. Cycles are therefore $⟨c,d⟩\oplus ⟨c,d⟩$, of dimension four. Boundaries are generated by $(c,d)$ and $(0,c)$, of dimension two. F4 gives $H^1(G,M)\cong k^2$.

Restriction to $N$ sends $[(u,v)]$ to $[u]\in \ker T\mathrm{/}\mathrm{im}T$. Indeed inclusion of the $N$-resolution at degree zero of the other factor lifts the identity augmentation, so its Hom map is this projection; the canonical comparison of F4 identifies it with F2's restriction. Its image is exactly $⟨[d]⟩$: all allowable $u$ lie in $⟨c,d⟩$, and $(d,0)$ is a cycle. Thus F2 forces the kernel of transgression to be $⟨[d]⟩$ in $⟨[b],[d]⟩$. Its target is the one-dimensional $⟨[d]⟩$ from step 2.1, so $d_2([b])=[d]\mathrm{\ne }0$. The five-term portion is $0\to k\to k^2\to k^2\to k\to H^2(G,M)$, with middle restriction of rank one, transgression of rank one, and the last inflation zero. This proves noncollapse despite the group section.

Restriction to the section subgroup projects a cycle to $[v]\in H^1(Q,M)$. Here $\ker S=\mathrm{im}S=⟨c,d⟩$, so this target is zero. It cannot retract the nonzero injection $H^1(Q,M^N)\hookrightarrow H^1(G,M)$: the coefficient modules in those two quotient-group cohomologies differ. With trivial coefficients $k$ instead, $T=S=0$, the same resolution gives $H^1(G,k)=k^2$ and $H^1(N,k)=H^1(Q,k)=k$. Inflation is $z\mapsto (0,z)$, restriction is $(u,v)\mapsto u$, and restriction to the section is $(u,v)\mapsto v$. To verify the inflation formula, project the tensor resolution onto the $Q$ factor by augmentation of the $N$ factor; its Hom map is the displayed inclusion and lifts the quotient fixed-point map in F2. This is a valid split degree-one sequence and has zero transgression by exactness.

F1 gives finite strong convergence for both coefficient modules. The calculations establish only the stated low-degree portion; other page differentials and the full degree-two target in the first example are not claimed computed. A group section imposes no bidegree vanishing on those uncomputed arrows. All displayed resolutions, bases and linear equations are explicit and require no AC; resolution independence retains the supplied-data/DC convention.

Put $H=\mathbb{Z}\mathrm{/}4$ in cochain degree one, with zero differential, $F^{0}H=H$, $F^{1}H=2\mathbb{Z}\mathrm{/}4$ and $F^{2}H=0$. The associated graded is $\mathbb{Z}\mathrm{/}2$ at $(0,1)$ and $(1,0)$, zero elsewhere. Every differential is zero, so all pages equal this graded object, and actual cohomology is $H$ with exactly the displayed finite filtration. A section of $H\to H\mathrm{/}(2\mathbb{Z}\mathrm{/}4)$ would send the element of order two to an element killed by two lifting the odd coset. Its only lifts are $1$ and $3$, both of order four. Thus no section exists; equivalently $H$ has order-four elements while $(\mathbb{Z}\mathrm{/}2{)}^2$ does not. This is F1's extension obstruction with no choice assumption.

Similarly place $k^2$ in degree one, $F^1=k{e}_1$, with zero differential. Its stationary entries are $k$ at the same two positions, and its finite image filtration gives the actual abutment. Each line $L_a=k(a{e}_1+e_2)$ is a complement, since its intersection with $k{e}_1$ is zero and every vector is the sum of elements in those two lines. Conversely every complement has this form by normalizing the second coordinate of a nonzero vector in it. The shear $e_1\mapsto e_1$, $e_2\mapsto e_2+e_1$ sends $L_a$ to $L_{a+1}$ and fixes no complement, while inducing identity on both graded pieces. Hence existence and even an explicit choice $L_0$ do not give naturality.

In arbitrary dimension F2 uses AC for bases and complements, followed by only finitely many filtration splittings. In the displayed two-dimensional example the formula already supplies complements in ZF. In both cases the full target has zero/full filtration endpoints $F^2=0,F^0=H$, so convergence has no hidden issue; the unresolved question from page data alone is the extension or its choice of section. With just one nonzero quotient this particular extension obstruction disappears.

For $t=0,1$, let $C_t^1=kx$, $C_t^2=ky$ and $dx=ty$, zero elsewhere. Let $x$ have filtration degree zero and $y$ degree two in the decreasing filtration. This means $F^p{C}_t^1=kx$ exactly for $p\le 0$ and $F^p{C}_t^2=ky$ exactly for $p\le 2$. Both filtrations are finite in each degree and preserved by $d$. Their graded terms are $kx$ at $(0,1)$ and $ky$ at $(2,0)$, zero elsewhere. As in F1, negate chain degree and filtration index to apply F2's homological formulas.

Through page two the numerator at each of these positions is the full displayed line and its denominator is zero: $dx$ already lies in $F^2{C}_t^2$, and no boundary from filtration zero enters the denominator at filtration two until page three. Thus $d_0=d_1=0$ and both $E_2$ pages have the same two lines. At $r=2$, F2's representative rule gives $d_{2}x=ty$. If $t=1$, the page-three cycle numerator at $x$ is zero because $dx\mathrm{\notin }{F}^3{C}_t^2=0$, and the page-three denominator at $y$ is all $ky=d(kx)$. Hence $E_3=0$. If $t=0$, both numerators remain full and both denominators remain zero on every page, so $E_3=E_{\mathrm{\infty }}=E_2$.

Direct total cohomology is zero for $t=1$ and is $kx$ in degree one and $ky$ in degree two for $t=0$. In the latter case the induced filtration has $F^0{H}^1=H^1,F^1{H}^1=0$ and $F^0{H}^2=F^1{H}^2=F^2{H}^2=H^2,F^3{H}^2=0$. These finite normalized filtrations have exactly the calculated stationary graded pieces; all later incident arrows are zero. Thus the difference persists in the actual filtered abutments, with strong convergence verified directly. No infinite choices or splitting assumptions occur.

The input homology is $H_{0}C=\mathbb{Z}\mathrm{/}2$, $H_{1}C=\mathbb{Z}$, zero elsewhere. Applying Hom into $M$ to $\mathbb{Z}\stackrel{2}{\to }\mathbb{Z}$ gives $M\stackrel{0}{\to }M$, so the degree-zero and degree-one Ext entries of $H_{0}C$ are $M$. The one-term projective resolution of $H_{1}C$ contributes only $M$ at $(0,1)$. The AC free-kernel argument in F2 kills every $p>1$ entry, giving exactly the stated page.

No later arrow can join columns zero and one: its first-coordinate change is $r\ge 2$. Every possible incoming arrow also starts in a zero column, unless its target has first coordinate at least two, in which case that target is zero. Thus every later differential vanishes at every bidegree, not just those displayed, and $E_2=E_{\mathrm{\infty }}$. F2 supplies first-quadrant finite-filtration convergence to Hom cohomology. Directly this Hom complex is $M\stackrel{0}{\to }{M}^2$, confirming the target and vanishing in all other degrees.

In degree zero the only quotient has filtration index zero, so $F^0{H}^0=M,F^1{H}^0=0$. In degree one, evaluation on $b$ gives $(x,y)\mapsto y$, with kernel $M\oplus 0$; these are the two graded pieces at $(0,1)$ and $(1,0)$. Thus the exact extension is $0\to M\stackrel{x\mapsto (x,0)}{\to }{M}^2\stackrel{(x,y)\mapsto y}{\to }M\to 0$. The lower edge in degree one is the displayed injection; the upper edge is evaluation. In degree zero both edges are the identity under the kernel identification. All edges in degrees at least two have zero target and zero source here. This fixes every endpoint and reconstructs the actual extension.

The section $y\mapsto (0,y)$ exists explicitly. Under the chain automorphism $a\mapsto a+b$, Hom cohomology transforms as $(x,y)\mapsto (x+y,y)$ while the graded endpoints are fixed; no lift of $1$ is fixed. Consequently the section is not natural, and no alternative section restores naturality for all chain maps. AC has been used only through the general UCT free-submodule/projective and replacement conventions of F2; all computations for these specified finite free complexes are explicit. Every obligation of F1 is now resolved in all degrees, with no unknown differential, extension or convergence qualification.