There is a functor from bicomplexes to chain complexes sending a bicomplex to its associated total chain complex. Does this functor have a left adjoint, and if so, what is it?

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My guess would be yes, but I can't recall off the top of my head what it is. – David Roberts Jun 11 '12 at 10:09

It's not to hard to work out by hand what such an object would need to be if it existed. By considering the case that $D$ is a bicomplex supported in degree $(m,n)$, the adjunction condition $\operatorname{Hom}_{bicomplex}(L(C), D) = \operatorname{Hom}_{complex}(C, \operatorname{Tot}(D))$ imposes the condition that $L(C)$ have copies of $C_k$ in all degrees $(m,k-m)$. Similarly, by examining the case where $D$ has a single nonzero arrow, we find that up to signs, the differential $L(C)_{m,n} \to L(C)_{m-1,n}$ is given by $C_{m+n} \to C_{m+n-1}$, and the same holds for the vertical arrows.

As I understand it, there is a direct sum totalization, and a direct product totalization, and they have different adjunction properties. In particular, if $D$ has large support, then the left side of the adjunction condition involves a countably infinite product in each degree. While the direct product totalization yields an adjunction, the direct sum totalization will give you a dimension mismatch.

If you work with finite periodic complexes, the totalizations coincide, and the adjunction holds.

Since chain complexes are (up to some details about signs) representations of the $x \mapsto ax+b$ group $\mathbb{G}_a \rtimes \mathbb{G}_m$, the adjunction amounts to Frobenius reciprocity, and the problem here is quite similar to the problem of finding a well-behaved induction functor for passing from representations of this group to representations of its direct product with itself. Since the group has positive dimension, the usual induction ends up being some kind of direct integral. The direct product totalization seems to be some kind of completed restriction, but it is rather unclear to me.

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The direct sum totalization does not preserve limits (namely, infinite producs) - so there won't be a left adjoint. – Martin Brandenburg Jun 12 '12 at 7:05
Thank you, that is a much more concise way to say it. – S. Carnahan Jun 12 '12 at 7:16
Though there is a right adjoint, which is exactly the functor that S. Carnahan described. I think it's fairly straightforward to check that his functor is both left adjoint to direct product totalization and right adjoint to direct sum totalization. – Jeremy Rickard Jun 12 '12 at 13:01
@JeremyRickard I was just trying to use this, and ran into something weird: consider two (homological) double complexes $C,D$, both with copies of the integers at the origin, one with a $\mathbf{Z}$ at $(-1,0)$ and the other with a $\mathbf{Z}$ at $(0,-1)$, and the nontrivial differentials the identity. Writing $R$ for the apparent right adjoint to $\oplus$ totalization, $R\text{Tot}D$ has integers along the diagonal and subdiagonal, with identities everywhere up to some signs. A double complex map $C\to R\text{Tot}D$ must be zero at the origin, to make the vertical square commute... – Kevin Carlson Oct 7 '15 at 16:19
@KevinCarlson Ah, I don't think S.Carnahan's method of identifying the adjoint quite works, as considering maps to a double complex concentrated in one degree only tells you about the terms modulo the images of the differentials. But it can be adapted. By considering maps to and from complexes supported on a $2\times 2$ square with isomorphisms as differentials, the right adjoint of direct sum totalization must satisfy $(RC)_{p,q}=C_{p+q}\oplus C_{p+q-1}$, and for the left adjoint of direct product totalization $(LC)_{p,q}=C_{p+q+1}\oplus C_{p+q}$. – Jeremy Rickard Oct 8 '15 at 9:13