The Architecture Behind the Hom Differential I: Bifunctors and Totalization

The Architecture Behind the Hom Differential I: Bifunctors and TotalizationBifunctors Produce Signed Double ComplexesTotalization Is a FunctorThe Tensor Product of Chain ComplexesThe Hom Double ComplexChain Maps and Chain HomotopiesWhat Has Been Built

 

The Architecture Behind the Hom Differential I: Bifunctors and Totalization

The elementary formulas of homological algebra often look like a collection of sign conventions:

$$d(x\otimes y)=d_{X}x\otimes y+(-1{)}^{|x|}x\otimes d_{Y}y,$$

$$(df{)}_i=d_D{f}_i-(-1{)}^{|f|}{f}_{i-1}{d}_C.$$

But these formulas are not arbitrary. They come from a single structural mechanism:

$$\text{bifunctors}⟶\text{signed double complexes}⟶\text{total complexes}.$$

The tensor product of chain complexes is obtained this way. The Hom complex is obtained this way. The signs are not repairs inserted after the fact. They are already part of the passage from bifunctors to double complexes.

Throughout, we use homological grading. Thus differentials lower degree:

$$d_X:X_i\to X_{i-1}.$$

---

Bifunctors Produce Signed Double Complexes

Let

$$F:\mathcal{A}\times \mathcal{B}\to \mathcal{E}$$

be a biadditive functor between additive categories.

Given chain complexes

$$X\in \mathrm{Ch}(\mathcal{A}),Y\in \mathrm{Ch}(\mathcal{B}),$$

define

$$B_{p,q}=F(X_p,Y_q).$$

The first differential comes from $X$$X$:

$${\partial }^h=F(d_X,\mathrm{id}):B_{p,q}\to B_{p-1,q}.$$

The second differential comes from $Y$$Y$, with the Koszul sign inserted at the moment of construction:

$${\partial }^v=(-1{)}^{p}F(\mathrm{id},d_Y):B_{p,q}\to B_{p,q-1}.$$

Then

$$({\partial }^h{)}^2=0,({\partial }^v{)}^2=0,$$

because

$$d_X^2=0,d_Y^2=0.$$

Moreover,

$${\partial }^h{\partial }^v+{\partial }^v{\partial }^h=0.$$

Indeed, on $B_{p,q}$$B_{p,q}$,

$${\partial }^h{\partial }^v=(-1{)}^{p}F(d_X,d_Y),$$

whereas

$${\partial }^v{\partial }^h=(-1{)}^{p-1}F(d_X,d_Y).$$

These two terms cancel.

Thus a biadditive functor does not merely produce a double-indexed grid. With the sign

$$(-1{)}^p$$

built into the second direction, it produces an anticommuting double complex.

This is the correct place for the sign. It is part of the passage from a bifunctor to a double complex, not a later repair.

---

Totalization Is a Functor

Let

$$\mathrm{Dbl}(\mathcal{E})$$

denote the category of anticommuting double complexes in $\mathcal{E}$$\mathcal E$.

An object is a double-indexed family $B_{p,q}$$B_{p,q}$ with maps

$${\partial }^h:B_{p,q}\to B_{p-1,q},{\partial }^v:B_{p,q}\to B_{p,q-1},$$

such that

$$({\partial }^h{)}^2=0,({\partial }^v{)}^2=0,{\partial }^h{\partial }^v+{\partial }^v{\partial }^h=0.$$

A morphism of double complexes is a degreewise family of maps commuting with both differentials.

If the relevant coproducts exist, define the direct-sum totalization by

$${\mathrm{Tot}}^{\oplus }(B{)}_n=\underset{p+q=n}{⨁}{B}_{p,q}.$$

If the relevant products exist, define the product totalization by

$${\mathrm{Tot}}^{\mathrm{\Pi }}(B{)}_n=\prod _{p+q=n}{B}_{p,q}.$$

In either case, the differential is

$$d_{\mathrm{Tot}}={\partial }^h+{\partial }^v.$$

Because the two differentials anticommute,

$$d_{\mathrm{Tot}}^2=({\partial }^h{)}^2+{\partial }^h{\partial }^v+{\partial }^v{\partial }^h+({\partial }^v{)}^2=0.$$

Therefore totalization defines functors

$${\mathrm{Tot}}^{\oplus }:\mathrm{Dbl}(\mathcal{E})\to \mathrm{Ch}(\mathcal{E}),$$

and

$${\mathrm{Tot}}^{\mathrm{\Pi }}:\mathrm{Dbl}(\mathcal{E})\to \mathrm{Ch}(\mathcal{E}),$$

whenever the relevant sums or products exist.

This is why one should not reprove $d^2=0$$d^2=0$ separately in every concrete example. Once the correct signed double complex has been constructed, totalization automatically produces a chain complex.

---

The Tensor Product of Chain Complexes

Let $R$$R$ be a commutative ring. The tensor product bifunctor

$$-{\otimes }_R-:R\mathrm{-Mod}\times R\mathrm{-Mod}\to R\mathrm{-Mod}$$

sends two chain complexes $X$$X$ and $Y$$Y$ to the double complex

$$B_{p,q}=X_p{\otimes }_R{Y}_q.$$

The two differentials are

$${\partial }^h=d_X\otimes \mathrm{id},$$

and

$${\partial }^v=(-1{)}^p\mathrm{id}\otimes d_Y.$$

The tensor product complex is the direct-sum totalization:

$$X{\otimes }_{R}Y={\mathrm{Tot}}^{\oplus }(B).$$

Hence

$$(X{\otimes }_{R}Y{)}_n=\underset{p+q=n}{⨁}{X}_p{\otimes }_R{Y}_q.$$

For homogeneous elements

$$x\in X_p,y\in Y_q,$$

the total differential is

$$d(x\otimes y)=d_{X}x\otimes y+(-1{)}^{p}x\otimes d_{Y}y.$$

This is the usual tensor product differential.

The sign is not a convention chosen after the fact. It is the sign that makes the two directions of the tensor double complex anticommute.

---

The Hom Double Complex

Now let $\mathcal{C}$$\mathcal C$ be a $K$$K$-linear additive category.

For objects $A,B\in \mathcal{C}$$A,B\in\mathcal C$, the Hom object

$${\mathrm{Hom}}_{\mathcal{C}}(A,B)$$

is a $K$$K$-module. Thus we have a biadditive functor

$${\mathrm{Hom}}_{\mathcal{C}}(-,-):{\mathcal{C}}^{op}\times \mathcal{C}\to K\mathrm{-Mod}.$$

Given chain complexes

$$C,D\in \mathrm{Ch}(\mathcal{C}),$$

define

$$B_{p,q}={\mathrm{Hom}}_{\mathcal{C}}(C_{-p},D_q).$$

The reversed index $C_{-p}$$C_{-p}$ is forced by contravariance.

The differential of $C$$C$ is

$$d_C:C_i\to C_{i-1}.$$

But Hom is contravariant in the first variable. Precomposition with $d_C$$d_C$ sends a morphism

$$C_{i-1}\to D_q$$

to a morphism

$$C_i\to D_q.$$

Thus the source direction reverses after entering Hom. Writing

$$i=-p$$

absorbs this reversal. It makes the source direction lower the index $p$$p$.

There are two differentials.

The target differential is postcomposition:

$${\partial }^t(f)=d_{D}f.$$

Thus

$${\partial }^t:B_{p,q}\to B_{p,q-1}.$$

The source differential is precomposition. Since

$$d_C:C_{-p+1}\to C_{-p},$$

a morphism

$$f:C_{-p}\to D_q$$

gives

$$f{d}_C:C_{-p+1}\to D_q.$$

Thus the source direction is

$$B_{p,q}\to B_{p-1,q}.$$

To recover the usual Hom differential after totalization, we use the signed source differential

$${\partial }^s(f)=(-1{)}^{p+q+1}f{d}_C,$$

and the target differential

$${\partial }^t(f)=d_{D}f.$$

The raw operations of precomposition and postcomposition commute:

$$d_D(f{d}_C)=(d_{D}f)d_C.$$

The signs force anticommutation:

$${\partial }^s{\partial }^t+{\partial }^t{\partial }^s=0.$$

Therefore

$$B_{p,q}={\mathrm{Hom}}_{\mathcal{C}}(C_{-p},D_q)$$

is an anticommuting double complex.

The Hom complex is its product totalization:

$$\underset{―}{\mathrm{Hom}}(C,D)={\mathrm{Tot}}^{\mathrm{\Pi }}(B).$$

Hence

$$\underset{―}{\mathrm{Hom}}(C,D{)}_n=\prod _{p+q=n}{\mathrm{Hom}}_{\mathcal{C}}(C_{-p},D_q).$$

Put

$$i=-p.$$

Since

$$p+q=n,$$

we get

$$q=i+n.$$

Therefore

$$\underset{―}{\mathrm{Hom}}(C,D{)}_n=\prod _i{\mathrm{Hom}}_{\mathcal{C}}(C_i,D_{i+n}).$$

So a degree $n$$n$ element is a family

$$f=(f_i:C_i\to D_{i+n}{)}_i.$$

Its differential is the total differential.

The target contribution is

$$d_D{f}_i.$$

The source contribution comes from

$$f_{i-1}:C_{i-1}\to D_{i+n-1}$$

precomposed with

$$d_C:C_i\to C_{i-1}.$$

The sign at that bidegree is

$$(-1{)}^{n+1}.$$

Thus

$$(df{)}_i=d_D{f}_i+(-1{)}^{n+1}{f}_{i-1}{d}_C.$$

Equivalently,

$$(df{)}_i=d_D{f}_i-(-1{)}^n{f}_{i-1}{d}_C.$$

This is the usual Hom differential.

Again, $d^2=0$$d^2=0$ is not a separate miracle. It follows because the Hom complex is the totalization of an anticommuting double complex.

---

Chain Maps and Chain Homotopies

The Hom complex explains where chain maps come from.

A degree $0$$0$ element of

$$\underset{―}{\mathrm{Hom}}(C,D)$$

is a family

$$f=(f_i:C_i\to D_i{)}_i.$$

Its differential is

$$(df{)}_i=d_D{f}_i-f_{i-1}{d}_C.$$

Thus $df=0$$df=0$ if and only if

$$d_D{f}_i=f_{i-1}{d}_C$$

for every $i$$i$.

This is exactly the condition that $f$$f$ is a chain map. Hence

$$Z_0\underset{―}{\mathrm{Hom}}(C,D)={\mathrm{Hom}}_{\mathrm{Ch}(\mathcal{C})}(C,D).$$

A degree $1$$1$ element is a family

$$h=(h_i:C_i\to D_{i+1}{)}_i.$$

Its boundary is

$$(dh{)}_i=d_D{h}_i+h_{i-1}{d}_C.$$

This is precisely the usual formula for a null-homotopic chain map.

Therefore

$$H_0\underset{―}{\mathrm{Hom}}(C,D)$$

is the module of chain maps modulo chain homotopy.

So the Hom complex contains three layers:

$$\underset{―}{\mathrm{Hom}}(C,D{)}_0=\text{graded maps},$$

$$Z_0\underset{―}{\mathrm{Hom}}(C,D)=\text{chain maps},$$

$$H_0\underset{―}{\mathrm{Hom}}(C,D)=\text{chain maps modulo chain homotopy}.$$

The ordinary category of chain complexes sees only $Z_0$$Z_0$.

The homotopy category sees $H_0$$H_0$.

The dg structure remembers the whole Hom complex.

---

What Has Been Built

We have now built the two fundamental complexes associated to chain complexes.

The tensor product complex comes from the tensor bifunctor and direct-sum totalization:

$$X{\otimes }_{R}Y={\mathrm{Tot}}^{\oplus }(X_p{\otimes }_R{Y}_q).$$

The Hom complex comes from the Hom bifunctor and product totalization:

$$\underset{―}{\mathrm{Hom}}(C,D)={\mathrm{Tot}}^{\mathrm{\Pi }}({\mathrm{Hom}}_{\mathcal{C}}(C_{-p},D_q)).$$

The reversed source index in Hom comes from contravariance.

The Hom differential

$$(df{)}_i=d_D{f}_i-(-1{)}^n{f}_{i-1}{d}_C$$

comes from totalization.

The next step is to explain how tensor and Hom interact, and why the Hom complex upgrades the whole category of chain complexes into a dg category.