Draft lesson.

Determinant signs and subset Koszul complexes

Original text public domain (CC0).

Two paths around a square of subsets must contribute opposite signs. Determinant lines supply those signs without choosing a permanent ordering of the labels. This construction works for diagrams in any abelian category. It also explains how splitting the indexing family into two ideals produces a cone, and how adjoining one endomorphism controls the cohomology of a Koszul complex.

Let \(\mathcal A\) be an abelian category, with the size and Choice conventions of Universes and small categories. We use the ordinary signed shifts and ordered cones of Graded complexes, contravariant signs and stupid truncations, Section 1. Their positive projection, full rotation rule and distinguished triangles are proved in Differential objects, curvature and cone homotopies, Sections 3–5. Cohomology in abelian categories with translation, Sections 4–5 proves cohomological exactness and naturality, including the difference between the cone and snake connecting maps.

1. Integral determinant coefficients in an abelian category

If \(L\) is a free abelian group of finite rank \(r\), its determinant line is \(\det L=\bigwedge^rL\). Here exterior powers are the alternating quotients of tensor powers; in particular \(v\wedge v=0\). Expanding \((v+w)\wedge(v+w)=0\) gives \(v\wedge w=-w\wedge v\), so adjacent interchanges change the sign. If \(l_1,\ldots,l_r\) is a basis, every top wedge expands into an integer multiple of \(l_1\wedge\cdots\wedge l_r\). The alternating multilinear determinant of the coordinate matrix descends to an integer-valued map on the exterior power and sends this wedge to \(1\). Consequently this wedge freely generates \(\det L\); the spanning generator has no nonzero integer relation. For rank zero the determinant is \(\mathbb Z\), with empty wedge \(1\).

For a finite set \(I\), put \(D_I=\det\mathbb Z^I\), with basis labels \(e_s\). An ordering of \(I\) supplies a generator of \(D_I\); a permutation multiplies that generator by its sign, since every permutation is a product of adjacent interchanges. The same argument proves that a bijection of labels induces the corresponding determinant isomorphism. For \(s\notin I\), define

This is an isomorphism of free rank-one groups: in any ordered bases its coefficient is \(1\) or \(-1\). If \(s,t\notin I\) are distinct, the two maps into \(D_{I\cup\{s,t\}}\) satisfy

These equations are statements about free integer groups, not about elements of an object of \(\mathcal A\).

We next define the coefficient object \(X\otimes L\) for \(X\in\mathcal A\) and finite free \(L\). Choose a basis of \(L\) and take the corresponding finite biproduct of copies of \(X\). A group map between finite free groups acts by its integer matrix, with entry \(a\) represented by \(a\,1_X\). Matrix multiplication and addition in the Hom groups of \(\mathcal A\) give identities, composition and sums. For a map \(u:X\to Y\), the same matrices with entries \(a u\) commute with these coefficient maps, giving maps \(u\otimes v\) for a group map \(v\). A change of basis has an invertible integral matrix, whose inverse gives the inverse coefficient isomorphism. The comparison matrices satisfy the composition law for successive changes, so all these objects and arrows agree up to coherent natural isomorphism. Equivalently their basis-independent universal property is

Indeed a map out of the finite biproduct is a map from each copy of \(X\), hence exactly a homomorphism assigning such maps to the basis of \(L\). Extending linearly proves (1.3); the matrix description verifies its naturality and independence of the basis. This also proves the uniqueness of the coefficient comparisons. A rank-one line gives a functor naturally isomorphic, after choosing its generator, to the identity on \(\mathcal A\). In particular tensoring with a determinant line is exact. No tensor product internal to \(\mathcal A\), and no interpretation of a formal coefficient as an element of \(X\), is required.

2. Covariant subset diagrams and product differentials

Let \(S\) be a small set and let \(\Sigma\) be a family of its finite subsets satisfying interval convexity:

Use inclusions as the arrows of \(\Sigma\). A covariant diagram \(F:\Sigma\to\mathcal A\) supplies objects \(X_I\) and maps \(f_{JI}:X_I\to X_J\) for \(I\subseteq J\), with \(f_{II}=1\) and \(f_{KJ}f_{JI}=f_{KI}\). Assume the products in the following display exist. Finite families need only finite biproducts; for an infinite family, the indicated small products suffice.

An empty product in this abelian category is a zero object. For \(J\in\Sigma\) of size \(n+1\), let \(\pi_I\) denote the source projections and define its output component by

This sum has at most \(n+1\) terms. The universal property of the target product assembles those output components into a unique \(d^n\). A single input can affect infinitely many outputs; that does not require an infinite sum in a Hom group, because each output in (2.3) is defined separately.

Proposition 2.1. These maps make \(C^\bullet(F)\) a complex, naturally in the diagram.

Proof. Fix an output \(K\) of size \(n+2\). Expand its component of \(d^{n+1}d^n\) using (2.3) twice. This is a finite sum: both the intermediate subsets of \(K\) and their predecessors inside \(K\) are finite in number. For an endpoint \(I\in\Sigma\) with \(K\setminus I=\{s,t\}\), interval convexity ensures that both \(I\cup\{s\}\) and \(I\cup\{t\}\) occur. Functoriality gives the same coefficient arrow \(f_{KI}\) on both paths, and (1.2) gives opposite determinant maps. The two terms, each composed with the same projection \(\pi_I\), cancel. Thus the entire output component is zero. All output projections of a product jointly determine a map, so \(d^{n+1}d^n=0\). This argument uses finite sums of source projections; it does not assume that maps out of an infinite product are determined by restrictions to individual coordinate injections.

A natural transformation \(\alpha:F\to F'\) induces the degreewise product of \(\alpha_I\otimes1_{D_I}\). Its chain equation follows by comparing the finite sum (2.3) in each output, using \(\alpha_Jf_{JI}=f'_{JI}\alpha_I\). Product uniqueness gives identities and composition. The coefficient comparisons of Section 1 and bijections of labels commute with (2.3), by the naturality of the wedge maps. This proves the claimed naturality, including independence of any orders used to calculate the signs. \(\square\)

Interval convexity is needed. A family containing only \(\varnothing,\{a\},\{a,b\}\) misses the second path between its extreme subsets; a constant identity diagram on it has a nonzero square of its differential, as Exercise 1 shows.

3. Lower ideals, the full cone and its last arrow

Let \(\Sigma_0\subseteq\Sigma\) be a lower ideal: if \(J\in\Sigma_0\) and \(I\in\Sigma\) satisfies \(I\subseteq J\), then \(I\in\Sigma_0\). Its complement \(\Sigma_1\) is an upper ideal, because an inclusion from \(I\in\Sigma_1\) to \(J\in\Sigma_0\) would violate the lower-ideal property. Both families inherit (2.1): an intermediate subset belongs to \(\Sigma\), then to the appropriate ideal using its lower or upper endpoint.

Write \(A=C^\bullet(F|_{\Sigma_0})\), \(B=C^\bullet(F|_{\Sigma_1})\) and \(K=C^\bullet(F)\). Partitioning the product factors gives a canonical isomorphism \(K^n\simeq A^n\oplus B^n\). It follows from the product universal property, then the equality of binary products and biproducts; it remains valid for infinitely many factors in either group. There are no differential components from the upper ideal into the lower ideal. Consequently

Each \(c^n\) is the upper-output part of (2.3) applied to the lower-input factors. The lower-left entry of \(d_K^{n+1}d_K^n=0\) is

Define \(\phi:A[-1]\to B\) by \(\phi^n=c^{n-1}\). Since the differential of \(A[-1]\) is \(-a\) in the reindexed degrees, (3.2) is exactly its chain equation. With our ordered cone convention,

The product partition is therefore an actual chain isomorphism \(K\simeq C_\phi\), preserving the inclusion \(j:B\to K\) and the positive projection \(p:K\to A\). Rotating the entire cone triangle gives

The last map in degree \(n\) is \(-c^n\); (3.2) verifies its chain equation into the signed shift \(B[1]\). Thus the upper-ideal complex is a subcomplex and the lower-ideal complex is the quotient, and

is degreewise split exact. The full cohomological theorem in the prerequisite applies to (3.4), yielding the natural exact sequence

Here \(H(c)\) means the cohomology map into the shifted target, using its natural degree identification. It is defined by (3.2), not by treating \(c\) as a degree-zero map into \(B\). Naturality follows either from the cone comparison for a commuting arrow square or directly from the product projections in (3.1); those maps preserve \(j,p,c\), hence every arrow of (3.4). With the snake connecting convention the connecting arrow is the negative of the one in (3.6), by the complete comparison proved in the cohomology prerequisite.

4. Commuting endomorphisms, top cohomology and regularity

Let \(S\) be finite, let \(X\in\mathcal A\), and let \((f_s)_{s\in S}\) be pairwise commuting endomorphisms. On all subsets of \(S\), take \(X_I=X\) and

The empty product is the identity. Commutativity makes every order of the factors equal, since adjacent factors may be interchanged. For \(I\subseteq J\subseteq K\), the two disjoint difference sets concatenate to \(K\setminus I\), so (4.1) has the diagram composition law. Define the cochain Koszul complex \(K^\bullet(X;(f_s))\) to be the complex of Section 2 for this diagram. Its terms have degrees \(0,\ldots,m\), where \(m=|S|\), and its differential adds a label on the left with its endomorphism as coefficient. This is an intrinsic labeled construction; different orders used to present it are related by their actual determinant isomorphisms.

Its top cohomology has the natural description

Indeed the top term is \(X\otimes D_S\), with no outgoing differential. Each incoming component is \(f_s\otimes w_s^{S\setminus\{s\}}\). The determinant map is invertible, so that component has image \((\operatorname{im}f_s)\otimes D_S\). The image of their finite biproduct map is their sum, by the universal image factorization, and tensoring with the free line is exact by Section 1. Taking the cokernel gives (4.2). An ordering trivializes \(D_S\) and identifies this with the ordinary quotient; a change of orientation changes that trivialization by its determinant sign. The empty family gives \(X\) in degree zero, and (4.2) gives \(X\otimes\mathbb Z\simeq X\).

Call the family quasi-regular when its cohomology is zero in every degree except \(m\). This definition includes a zero top quotient and makes the empty family quasi-regular.

Now choose \(t\in S\), put \(S'=S\setminus\{t\}\), and let \(A=K^\bullet(X;(f_s)_{s\in S'})\). Partition all subsets according to whether they contain \(t\). The subsets not containing \(t\) form the lower ideal. For the upper-ideal complex \(B\), adjoining \(t\) gives maps

They are degreewise isomorphisms. For every other added label \(s\), (1.2) gives \(b^{n+1}\epsilon^n=-\epsilon^{n+1}a^n\). Thus \(\epsilon:A\to B[1]\) is a chain isomorphism with the signed differential on \(B[1]\). The cross component is \(c^n=\epsilon^nf_t^n\), where \(f_t^n\) acts by \(f_t\otimes1\) on every factor of \(A^n\). Commutativity with the other endomorphisms makes this a chain endomorphism of \(A\).

These equations give the entire cone identification

Explicitly the cone has terms \(A^n\oplus A^{n-1}\) and differential \(\begin{pmatrix}a^n&0\\f_t^n&-a^{n-1}\end{pmatrix}\). The isomorphism to the lower/upper decomposition of \(K\) is \(\operatorname{diag}(1_{A^n},\epsilon^{n-1})\). Its lower-left equation is \(c^n=\epsilon^nf_t^n\); its lower-right equation is precisely the anti-commutation just proved. This verifies the source shift, both diagonal signs and every off-diagonal map in (4.4).

Identifying \(B\) with \(A[-1]\) by \(-\epsilon[-1]^{-1}\), the triangle (3.4) becomes

The target of the last arrow is \(A[-1][1]=A\). Thus a positive-\(f_t\) form of its natural long exact sequence is

The negative on the inclusion is part of this particular identification; (4.5) specifies every structural map.

Proposition 4.1 (one-endomorphism criterion). Suppose the smaller family on \(S'\) is quasi-regular, with \(r=|S'|\). The full family is quasi-regular if and only if \(f_t\) induces a monomorphism on \(X/\sum_{s\in S'}\operatorname{im}f_s\).

Proof. Each \(f_s\) commutes with \(f_t\), so \(f_t\) preserves its image subobject: the commutative square with \(f_s\) induces the corresponding image map. The finite sum of these images is preserved as well, using its image presentation by a finite biproduct. The cokernel therefore gives the asserted induced endomorphism on the quotient. Tensoring it with \(D_{S'}\) is its endomorphism on \(H^r(A)\), by (4.2), and preserves and reflects monomorphisms because it is an equivalence.

All other cohomology objects of \(A\) vanish. Sequence (4.6) then gives \(H^j(K)=0\) outside \(j=r,r+1\), identifies \(H^r(K)\) with the kernel of that quotient endomorphism, and identifies \(H^{r+1}(K)\) with its cokernel after the determinant comparison (4.3). These assertions follow at the indicated exact positions, with zero objects on their other sides. Vanishing of the kernel is exactly monicity. The top cokernel agrees with (4.2) for the full family, since quotienting by the previous sum and then by the image of \(f_t\) is the quotient by the sum of all images; this follows from the two successive cokernel universal properties. This proves both directions, including \(r=0\). \(\square\)

For an ordered list \(f_1,\ldots,f_m\), define regularity by requiring the endomorphism induced by \(f_i\) on \(X/\sum_{j<i}\operatorname{im}f_j\) to be monic for every \(i\). The preceding image argument proves each of these induced maps exists. The empty prefix is quasi-regular. Apply Proposition 4.1 successively to each prefix to prove that every regular ordered list is quasi-regular, with top cohomology (4.2). At the first step the condition is simply that \(f_1:X\to X\) is monic; no special starting assumption or properness of a quotient is needed.

5. Contravariant diagrams and the homological decomposition

For a contravariant diagram \(G:\Sigma^{\mathrm{op}}\to\mathcal A\), write \(X_I=G(I)\) and \(g_{IJ}:X_J\to X_I\) for \(I\subseteq J\). Assume the required small coproducts exist and put

For \(s\in J\), let \(\iota_s^J:D_J\to D_{J\setminus\{s\}}\) be the inverse of \(w_s^{J\setminus\{s\}}\). Define the boundary on the source injection belonging to \(J\) by

Each source has finitely many outgoing terms, so this is a map into the actual target coproduct. The source coproduct universal property assembles these maps uniquely even if a degree has infinitely many summands. For distinct removed labels \(s,t\), inverting (1.2) gives the opposite-sign identity between the two successive inverse-wedge maps into \(D_{J\setminus\{s,t\}}\). Interval convexity supplies both intermediate subsets, and contravariant functoriality gives the same coefficient arrow on their two paths. Expanding (5.2) twice on each source injection is a finite sum whose terms cancel in pairs. The source injections jointly determine the map, proving \(\partial_{n-1}\partial_n=0\). In an ordered wedge the inverse-wedge coefficient for removing the label in position \(j\), starting at position zero, is \((-1)^j\), because (1.1) inserts it on the left. Natural transformations give chain maps by checking (5.2) on each injection; the uniqueness proves identity and composition laws.

For the same lower/upper partition, now put \(A=C_\bullet(G|_{\Sigma_0})\), \(B=C_\bullet(G|_{\Sigma_1})\) and \(K=C_\bullet(G)\). The lower ideal is a subcomplex: removing a label from a lower-ideal subset, when the result remains in \(\Sigma\), stays in the lower ideal. The coproduct partition gives \(K_n\simeq A_n\oplus B_n\), with

We specify the shifts and cone to translate the entire triangle. Our homological shift is \(N[r]_n=N_{n-r}\), with differential \((-1)^r\partial_{N,n-r}\). Reindexing by \(\widehat N^k=N_{-k}\), with the same differential, identifies this shift with the cohomological signed shift \(\widehat N[r]\). Thus the cochain cone and triangle theorems apply under this reindexing.

Define \(\psi:B[-1]\to A\) by \(\psi_n=c_{n+1}\). Its chain equation is \(a_nc_{n+1}=-c_nb_{n+1}\), exactly (5.3). The homological ordered cone of a map \(f:P\to Q\) has terms \(P_{n-1}\oplus Q_n\) and differential \(\begin{pmatrix}-\partial_{P,n-1}&0\\f_{n-1}&\partial_{Q,n}\end{pmatrix}\), by reindexing the cochain formula. For \(\psi\) these terms are \(B_n\oplus A_n\) and its differential is \(\begin{pmatrix}b_n&0\\c_n&a_n\end{pmatrix}\). Swapping the two summands identifies it with the full complex in (5.3). It preserves the inclusion of \(A\) and the positive projection onto \(B\). Consequently the entire distinguished triangle is

Here distinguished means that its reindexed cochain triangle is distinguished in \(K(\mathcal A)\). The natural long exact homology sequence, by that exact reindexing, is

All signs and the last arrow come from the positive cone projection and its rotation. The degreewise split exact sequence here is \(0\to A\to K\to B\to0\); this is the reversed ideal role from the covariant product construction. The proof uses only the required coproducts, finite incidence sums and the checked cone/cohomology theorems, with no exactness assumption on arbitrary coproducts.

6. Four graded exercises with complete solutions

Exercise 1 (introductory: a missing face). Give \(\Sigma=\{\varnothing,\{a\},\{a,b\}\}\) the constant diagram \(\mathbb Z\) with identity transition maps. Order \(a<b\). Compute the two consecutive differentials and their square. Explain why an infinite full subset family would not cause this particular difficulty in the covariant product construction.

Solution. The three determinant terms are \(\mathbb Z\) in degrees \(0,1,2\), with generators \(1,e_a,e_a\wedge e_b\). The first differential adds \(a\), so has coefficient \(1\). The second adds \(b\) on the left: \(e_b\wedge e_a=-e_a\wedge e_b\), so has coefficient \(-1\). Their composite is \(-1\), which is nonzero. The missing intermediate subset \(\{b\}\) prevents the opposite contribution from appearing. The extreme subsets are in \(\Sigma\), yet an intermediate one is absent, so (2.1) fails.

In the full finite-subset family of an infinite small set, both intermediates for every two-label addition occur. For each fixed output \(J\), only its finitely many immediate predecessors contribute to (2.3). For example the degree-zero map of the constant identity diagram is the diagonal \(\mathbb Z\to\prod_{s\in S}\mathbb Z\), whose every output is the identity. It is a valid product map even though its column has infinitely many nonzero entries. The pair cancellation is proved separately in each finite output row, so no finite-support condition on that column is needed.

Exercise 2 (intermediate: an integral contraction). Compute \(K^\bullet(\mathbb Z;(2,3))\) with labels ordered \(1<2\). Give an actual contraction and relate the result to ordered regularity.

Solution. Its complex is

in degrees \(0,1,2\): adding label \(2\) to \(e_1\) contributes \(-3\), while adding label \(1\) to \(e_2\) contributes \(2\). The first map is injective, and the kernel of the second consists of \((2t,3t)\), since \(3a=2b\) forces \(a\) even and then \(b=3a/2\). This is exactly the incoming image. The top cokernel is \(\mathbb Z/(2,3)=0\), since \(-2+3=1\).

Define \(h^1(a,b)=-a+b\) and \(h^2(c)=(-c,-c)\), with other components zero. Then \(h^1d^0(x)=x\), \(d^1h^2(c)=c\), and

Thus \(1=dh+hd\) in every degree, so the complex is contractible. The ordered pair is regular: multiplication by \(2\) is monic on \(\mathbb Z\), and multiplication by \(3\) on \(\mathbb Z/2\mathbb Z\) is the identity. The vanishing top quotient is permitted by the definition in Section 4.

Exercise 3 (advanced: a nonzero top quotient). For a field \(k\), apply the regularity criterion to multiplication by \(x,y\) on \(R=k[x,y]\). Compute every cohomology group and its top comparison map.

Solution. Multiplication by \(x\) is injective, because shifting the exponent of \(x\) in the unique finite monomial expansion cannot annihilate a nonzero coefficient. The quotient \(R/xR\) is naturally \(k[y]\), by retaining the monomials with \(x\)-exponent zero. Multiplication by \(y\) on that quotient is injective for the same coefficient reason. These endomorphisms commute, so the ordered list is regular and therefore quasi-regular. Its terms are \(R\to R^2\to R\) in degrees \(0,1,2\), with maps \((x,y)^{\mathrm t}\) and \((-y,x)\). We obtain \(H^0=H^1=0\), \(H^2=R/(x,y)\otimes D_{\{1,2\}}\simeq k\) after choosing the generator \(e_1\wedge e_2\), and \(H^j=0\) for all other degrees. The top map takes the class of a coefficient multiplying \(e_1\wedge e_2\) to its constant term. It is well-defined and bijective because the polynomials of zero constant term form the ideal \((x,y)\). Reversing the determinant generator negates this last trivialization; the intrinsic quotient-with-line in (4.2) is unchanged.

Exercise 4 (advanced: the reversed ideal and its connecting sign). Take the constant contravariant identity diagram \(\mathbb Q\) on all subsets of \(\{1,2\}\), and let the lower ideal contain only \(\varnothing\). Compute the lower and upper complexes and the full homological connecting map in (5.5).

Solution. The lower complex \(A\) is \(\mathbb Q\) in degree zero. The upper complex \(B\) is \(\mathbb Q\xrightarrow{(-1,1)^{\mathrm t}}\mathbb Q^2\) in homological degrees \(2,1\); its degree-zero term is zero. This follows from \(\iota_1(e_1\wedge e_2)=e_2\) and \(\iota_2(e_1\wedge e_2)=-e_1\). It has \(H_2(B)=0\), and the map \((a,b)\mapsto a+b\) identifies \(H_1(B)\) with \(\mathbb Q\), because its kernel is the line spanned by \((-1,1)\).

The entire complex is \(\mathbb Q\xrightarrow{(-1,1)^{\mathrm t}}\mathbb Q^2\xrightarrow{(1,1)}\mathbb Q\) in degrees \(2,1,0\). Its homology is zero in every degree: the left map is monic, its image is the kernel of the right map, and the right map is epic. The cross component \(c_1:B_1\to A_0\) is \((a,b)\mapsto a+b\). Thus the last arrow of the triangle is \(-c_1\), and the connecting map \(H_1(B)\to H_0(A)\) is \(-1\) under the displayed identifications. All its other connecting groups are zero. The nonzero part of the long exact sequence is \(0\to H_1(B)\xrightarrow{-1}H_0(A)\to0\). This verifies both the lower ideal's subcomplex role and the last-arrow sign in a case where that sign acts on a nonzero homology group.

References

• The Stacks Project, Section 15.29, The Koszul complex, Definitions 15.29.1–2 and Lemmas 15.29.7–8, for the module construction and the adjoining-element cone in its homological convention. The general abelian-category subset diagrams, both directions and all shift comparisons used here are proved above.
• Differential objects, curvature and cone homotopies, Sections 3–5, for full ordered cone triangles and their rotations.
• Cohomology in abelian categories with translation, Sections 4–5, for natural cohomological exactness and the actual cone/snake connecting comparison.