Dubious tensor-hom adjunction for chain complexes in Weibel. The differentials are wrong; can we make them right?

Question

$\newcommand{\hom}{\mathsf{Hom}}\newcommand{\tot}{\mathsf{Tot}}$The exercise $2.7.3$ from Weibel's "an introduction to homological algebra" is known to be incorrectly stated. However, even for the revised version given in that link, I think there may be an error. In a comment under that post it is said this isomorphism is useful for proving some results such as a Kunneth formula so I would quite like everything to actually work!

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For any chain complexes $A,B$ of right and left $R$-modules and any cochain complex $G$ of Abelian groups, it is claimed that: $$\tag{$\ast$}\tot(\hom_\Bbb Z(\tot^{\oplus}(A\otimes_R B),G))\cong\tot(\hom_R(A,\tot(\hom_\Bbb Z(B,G)))$$As a natural isomorphism of cochain complexes. On objects I can see this:

We want a map, for each $n\in\Bbb Z$, from: $$\prod_{k,\ell\in\Bbb Z}\hom_\Bbb Z(A_\ell\otimes_RB_{k-\ell},G^{n-k})\overset{?}{\cong}\prod_{i,j\in\Bbb Z}\hom_R(A_i,\hom_\Bbb Z(B_j,G^{n-i-j}))$$

Such a map is determined by its $(i,j)$th projections, which will be: $$\prod_{k\in\Bbb Z}\prod_{\ell\in\Bbb Z}\hom_\Bbb Z(A_\ell\otimes_RB_{k-\ell},G^{n-k})\overset{\pi_{i+j}}{\twoheadrightarrow}\prod_{\ell\in\Bbb Z}\hom_\Bbb Z(A_\ell\otimes_RB_{i+j-\ell},G^{n-i-j})\\\overset{\pi_i}{\twoheadrightarrow}\hom_\Bbb Z(A_i\otimes_RB_j,G^{n-i-j})\cong\hom_R(A_i,\hom_\Bbb Z(B_j,G^{n-i-j}))$$

Where the last isomorphism is the canonical tensor-hom adjunction isomorphism.

This map is clearly an isomorphism.

However, I remain entirely unconvinced that this is an isomorphism of chain complexes. Namely, the map I described above is not a chain map (as per the sign conventions Weibel uses, detailed below).

Specifically, I have computed that for $n\in\Bbb Z$ and a family $(g_{k,\ell})_{k,\ell\in\Bbb Z}$ of homomorphisms $A_\ell\otimes_RB_{k-\ell}\to G^{n-k}$, the element $\left(\sum_{\ell\in\Bbb Z}g_{k,\ell}\right)_{k\in\Bbb Z}$ of $\tot(\hom_\Bbb Z(\tot^{\oplus}(A\otimes_RB),G))^n$ is mapped to two distinct destinations: $$\tag{1}\left(\prod_{j\in\Bbb Z}(\overline{g}_{i+j-1,i-1}\circ d_i+(-1)^id_j^\ast\overline{g}_{i+j-1,i}+(-1)^{n+1}d_\ast^{n-i-j}\overline{g}_{i+j,i}\right)_{i\in\Bbb Z}$$And: $$\tag{2}\left(\prod_{j\in\Bbb Z}(\overline{g}_{i+j-1,i-1}\circ d_i+(-1)^id_\ast^{n-i-j}\overline{g}_{i+j,i}+(-1)^{n+1}d^\ast_j\overline{g}_{i+j-1,i}\right)_{i\in\Bbb Z}$$Both describing elements of $\tot(\hom_R(A,\tot(\hom_\Bbb Z(B,G))))^{n+1}$.

Where $(1)$ is what you get when you apply first the differential of the left hand cocomplex and then apply the isomorphism $(\ast)$, and $(2)$ is what you get if you first apply the isomorphism of $(\ast)$ and then apply the differential of the right hand cocomplex. That is, the commutative square for $(\ast)$ to be a cochain map is not actually commutative!

I've run through that calculation twice being very attentive to which sign conventions apply when, so I'm confident - but if I still somehow screwed it up and everything is fine as-is, please let me know.

My questions:

• Can we alter the choice of levelwise isomorphism in $(\ast)$ so that this really is a bona fide isomorphism of cochain complexes? I suspect yes because it's only slightly wrong (the final two summands differ by $(-1)^{n+i+1})$ but I can't see it myself (the indexing scares me)

• Alternatively, is my choice of isomorphism in $(\ast)$ correct as written if you use different sign conventions to Weibel? Note these different sign conventions should still have the nice properties such as satisfying the double complex condition. I would be unhappy if a different sign convention is truly necessary since that somewhat invalidates the material of the book.

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Definitions, conventions: Given a chain complex $C$ and a cochain complex $K$ in an Abelian category, Weibel defines the hom double cochain complex $\hom(C_\bullet,K^\bullet)$ to have horizontal differentials: $$d_h^{k,\ell}:=(d^C_{k+1})^\ast:\hom(C_k,D^\ell)\to\hom(C_{k+1},D^\ell)$$ and vertical differentials: $$d_v^{k,\ell}:=(-1)^{k+\ell+1}\cdot(d_D^\ell)_\ast:\hom(C_k,D^\ell)\to\hom(C_k,D^{\ell+1})$$

Given chain complexes of right and left $R$-modules $A$ and $B$, he defines $A\otimes_RB$ to be the double chain complex of Abelian groups $A_\bullet\otimes_R B_\bullet$ where again the differentials are twisted by a sign: the horizontals: $$d^h_{k,\ell}:=d^A_k\otimes1:A_k\otimes_RB_\ell\to A_{k-1}\otimes_RB_\ell$$And verticals: $$d^v_{k,\ell}:=(-1)^k\otimes d^B_\ell:A_k\otimes_RB_\ell\to A_k\otimes_RB_{\ell-1}$$

Answer 1

Signs always seem to cause problems, and everyone seems to have their own conventions. Brian Conrad spends 11 pages at the beginning of his book 'Grothendieck duality and base change' discussing such issues.

When I recently had cause to use triple complexes, I felt the easiest approach was to define them having commutative squares, and only introduce signs when taking total complexes. The stacks project seems to take the same view. In that respect this answer may not be that helpful to you, as it uses a different convention to Weibel.

We take a triple complex $T_{ijk}$ (with commuting squares), and with differentials $s,t,u$, so $$ s_{ijk} : T_{ijk} \to T_{i-1jk}, \quad t_{ijk} : T_{ijk} \to T_{ij-1k}, \quad u_{ijk} : T_{ijk} \to T_{ijk-1}. $$ We can then form the double complex $\mathrm{Tot}_{12}T$, having terms $\prod_{i+j=d}T_{ijk}$ and differentials $(s_{ijk}+(-1)^it_{ijk})$ and $(u_{ijk})$. More precisely, it sends $(x_{ijk})\in\prod_{i+j=d}T_{ijk}$ to the element of $\prod_{i+j=d-1}T_{ijk}$ having term $s_{i+1jk}(x_{i+1jk})+(-1)^it_{ij+1k}(x_{ij+1k})$ in position $ijk$.

We can also form the double complex $\mathrm{Tot}_{23}T$, having terms $\prod_{j+k=e}T_{ijk}$, and differentials $(s_{ijk})$ and $(t_{ijk}+(-1)^ju_{ijk})$.

Finally we can directly form the complex $\mathrm{Tot}T$, having terms $\prod_{i+j+k=d}T_{ijk}$ in position $d$, and differential $(s_{ijk}+(-1)^it_{ijk}+(-1)^{i+j}u_{ijk})$.

It is then easy to check that $$ \mathrm{Tot}\mathrm{Tot}_{12}T = \mathrm{Tot}T = \mathrm{Tot}\mathrm{Tot}_{23}T, $$ where on the left and right we have taken the total complexes of the double complexes.

So far so good.

Suppose now that we have complexes $(X,s)$, $(Y,t)$ and $(Z,u)$, so $s_i:X_i\to X_{i-1}$ etc. The idea is then that we can form the triple complexes $T=\mathrm{Hom}(X\otimes Y,Z)$ and $U=\mathrm{Hom}(X,\mathrm{Hom}(Y,Z))$, and the standard tensor-hom adjunction quickly shows (there are no sign problems at this stage) that $T\cong U$ as triple complexes. It follows that their total complexes are isomorphic, and using the above factorisations we see that $$ \mathrm{Tot}\mathrm{Tot}_{12}\mathrm{Hom}(X\otimes Y,Z) \cong \mathrm{Tot}\mathrm{Tot}_{23}\mathrm{Hom}(X,\mathrm{Hom}(Y,Z)). $$

We then show that we have isomorphisms of double complexes $$ \mathrm{Tot}_{12}\mathrm{Hom}(X\otimes Y,Z) \cong \mathrm{Hom}(\mathrm{tot}(X\otimes Y),Z) $$ and $$ \mathrm{Tot}_{23}\mathrm{Hom}(X,\mathrm{Hom}(Y,Z)) \cong \mathrm{Hom}(X,\mathrm{Tot}\mathrm{Hom}(Y,Z)). $$

Here we have used the small total complex $\mathrm{tot}(X\otimes Y)$, having terms $\bigoplus_{i+j=d}X_i\otimes Y_j$ and differential $p_d=\sum s_i\otimes1+(-1)^i\otimes t_j$.

That's the plan.

The triple complex $T=\mathrm{Hom}(X\otimes Y,Z)$ has terms $\mathrm{Hom}(X_{-i}\otimes Y_{-j},Z_k)$, and differentials $(s_{1-i}\otimes1)^\ast$, $(1\otimes t_{1-j})^\ast$, and $(u_k)_\ast$.

The triple complex $U=\mathrm{Hom}(X,\mathrm{Hom}(Y,Z))$ has terms $\mathrm{Hom}(X_{-i},\mathrm{Hom}(Y_{-j},Z_k)$ and differentials $(s_{1-i})^\ast$, $((t_{1-j})_\ast)^\ast$ and $((u_k)_\ast)_\ast$ (sorry about the notation!)

The isomorphism $T\cong U$ of triple complexes just comes from the tensor-hom adjunctions $\mathrm{Hom}(X_{-i}\otimes Y_{-j},Z_k)\cong\mathrm{Hom}(X_{-i},\mathrm{Hom}(Y_{-j},Z_k))$. That's the first part out of the way.

The double complex $\mathrm{Hom}(\mathrm{tot}(X\otimes Y),Z)$ has terms $\mathrm{Hom}(\bigoplus_{i+j=-d}X_i\otimes Y_j,Z_k)$, and differentials $(p_{1-d})^\ast$ and $(u_k)_\ast$. Using the standard isomorphism $$ \prod_{i+j=-d}\mathrm{Hom}(X_i\otimes Y_j,Z_k) \cong \mathrm{Hom}(\bigoplus_{i+j=-d}X_i\otimes Y_j,Z_k) $$ the first differential sends a tuple $f_{ij}:X_i\otimes Y_j\to Z_k$ for $i+j=-d$ to the tuple $f_{i-1j}(s_i\otimes1)+(-1)^if_{ij-1}(1\otimes t_j):X_i\otimes Y_j\to Z_k$ for $i+j=1-d$. This is precisely what the first differential for $\mathrm{Tot}_{12}\mathrm{Hom}(X\otimes Y,Z)$ does, though. Thus the standard isomorphisms above determine an isomorphism of double complexes $$ \mathrm{Tot}_{12}\mathrm{Hom}(X\otimes Y,Z) \cong \mathrm{Hom}(\mathrm{tot}(X\otimes Y),Z) $$ as required.

Similarly, the standard isomorphisms $$ \prod_{j+k=e}\mathrm{Hom}(X_{-i},\mathrm{Hom}(Y_{-j},Z_k)) \cong \mathrm{Hom}(X_{-i},\prod_{j+k=e}\mathrm{Hom}(Y_{-j},Z_k)) $$ determine an isomorphism of double complexes $$ \mathrm{Tot}_{23}\mathrm{Hom}(X,\mathrm{Hom}(Y,Z)) \cong \mathrm{Hom}(X,\mathrm{Tot}\mathrm{Hom}(Y,Z)). $$ In particular, the second differential in both cases sends a tuple $g_{jk}:X_{-i}\to\mathrm{Hom}(Y_{-j},Z_k)$ with $j+k=e$ to the tuple $(t_{-j})^\ast g_{j+1k}+(-1)^j(u_{k+1})_\ast g_{jk+1}$ with $j+k=e-1$.

This proves the second part, and so we are done.

Answer 2

$\newcommand{\hom}{\mathsf{Hom}}\newcommand{\tot}{\mathsf{Tot}}$Time to be unhappy, I guess.

Weibel has used an unconventional definition but this choice of differential ruins the tensor hom adjunction as I suspected. Unless, we change the definition of tensor. There is nothing wrong with Weibel's definition, it just clashes with other standards.

In fairness, Weibel warns us different notions of "hom complex" arise in the literature but it seems this notion is not a "right" one. It is quite common to consider the hom complex as merely a single complex i.e. the totalisation of the double complex Weibel describes and to reindex things so that it is an ordinary complex, in the following way.

If we think of complexes as graded modules, and take two chain complexes $X,Y$ "we" (from what I've gathered, this is the most common definition) define the hom-complex $\hom_\ast(X,Y)$ to be the graded $R$-module whose $n$th degree is the set of all graded homomorphisms $X\to Y$ of degree $n$, and with differential $d(f):=d_Y\circ f-(-1)^{|f|}\cdot(f\circ d_X)$. Ok, winding back into complex notation, the $n$th degree $\hom_n(X,Y)$ is the set $\prod_{i\in\Bbb Z}\hom_R(X_i,Y_{i+n})$ or, identifying $Y$ with its dual cochain complex $Y^j:=Y_{-j}$, $\prod_{i\in\Bbb Z}\hom_R(X_i,Y^{-n-i})$. Now let's also identify $\hom_\ast(X,Y)$ with its dual cochain complex $\hom(X,Y)^n:=\hom(X,Y)_{-n}$ in which case that becomes $\hom(X,Y)^n=\prod_{i\in\Bbb Z}\hom(X_i,Y^{n-i})$. This is looking familiar - that's nothing but $\tot(\hom_R(X,Y))^n$ in the sense of Weibel.

What then should the horizontal and vertical differentials of double cochain complex $\hom_R(X,Y)^\ast$ be, so that its totalisation agrees (after all the dual indexing is taken care of) with the more standard definition? It should be the exact opposite of what Weibel wrote! He placed the sign $(-1)^{k+\ell+1}$ on the vertical differential but "actually" we "should" place it on the horizontal differential. Though, we could have always taken the differential on the graded module $\hom_\ast(X,Y)$ to be (and this is nonstandard) $d(f)=f\circ d_X-(-1)^{|f|}d_Y\circ f$ and this would have the same property of encoding chain maps and chain homotopies in its homology and then this would unwind to give Weibel's definition of the double hom complex. It just requires a nonstandard sign in the corresponding tensor product.

With this altered definition of the $\hom$ double complex, the map I wrote - with no alterations or strings attached - is a genuine isomorphism of chain complexes, the differentials commute and the final "destination" which both routes around the chain-map-commutative-square agree with can be shown to be: $$\left(\left(d_\ast^{n-i-j}\overline{g}_{i+j,i}-(-1)^{n+i}d_j^\ast\overline{g}_{i-1+j,i}-(-1)^n\overline{g}_{i-1+j,i-1}\circ d_i\right)_{j\in\Bbb Z}\right)_{i\in\Bbb Z}$$So the tensor hom adjunction really works.

As discussed elsewhere on this site, really the graded single hom-complex is the fundamental concept since its homology encodes chain homotopy classes, the double complex is an artifice motivated by this, and the tensor (double) complexes have the signs that they have in order to make this adjunction work. I'm sure there are other reasons for the sign in the tensor complex such as the sign behaviour one observes with cup products.