Want to take part in these discussions? Sign in if you have an account, or apply for one below
Vanilla 1.1.10 is a product of Lussumo. More Information: Documentation, Community Support.
I started putting some content into internal hom of chain complexes. But am being interrupted now…
Are these non-negatively graded chain complexes? Because if you want $\mathbb{Z}$-graded, then you need to use a product (not a sum) in the construction. (Basically, I think it would be clearer anyway to use a product, even for non-negatively graded objects.)
Sure. I have changed it.
The definition of the internal-hom of chain complexes doesn’t respect non-negative or non-positive grading. That is, even if $X$ and $Y$ are non-negatively graded, or non-positively graded, then $[X,Y]$ may not be either. Can one define a variant internal-hom that applies in either of these situations? Or are we just forced into the unbounded case if we want internal-homs?
For the positively graded case, can’t we just use $[X, Y]_n = \prod_{p \geq 0} Hom(X_p, Y_{n+p})$?
The “inclusion” $U$ of non-negatively-graded cochain complexes into unbounded complexes is fully faithful, strongly monoidal, and has a right adjoint, say $R$, so
$Hom (U (A \otimes B), C) \cong Hom (U A \otimes U B, C) \cong Hom (U A, [U B, C]) \cong Hom (A, R [U B, C])$therefore non-negatively-graded cochain complexes is also monoidal closed. The functor $R$ is the good truncation, which replaces negative degrees with the zero object and the object in degree 0 with the kernel of the differential. I suppose the story for chain complexes has a left adjoint instead, which is troublesome…
That’s about the conclusion I had come to as well. It’s curious that the two cases seem so different.
(@Zhen, why did you put “inclusion” in quotes?)
I asked because I was wondering whether you could have a closed monoidal homotopy theory which is not stable, but where $\Sigma\Omega=Id$. For a while I thought that non-positively graded chain complexes should be an answer, but then I ran into this problem. Now I think I have a proof that this is impossible: since $\otimes$ is cocontinuous in each variable, $\Sigma\Omega=Id$ gives $\Sigma I \otimes \Omega I \cong I$, where $I$ is the monoidal unit, so the functor $\Sigma \cong (\Sigma I \otimes -)$ is an equivalence and we are stable.
Oh, I suppose because I was thinking of it as being a diagram of shape $\bullet \to \bullet \to \bullet \to \cdots$ instead of a diagram of shape $\cdots \to \bullet \to \bullet \to \cdots$. But $U$ isn’t induced by a functor between diagram shapes, unfortunately.
I guess you are assuming that $\Sigma$ and $\Omega$ are strong functors or something like that? Otherwise I do not see why $\Sigma I \otimes \Omega I \cong I$. But then why should they be strong functors?
I would have thought that $U$ is induced by a functor of $Set_*$-enriched diagrams, which is where we have to be to consider chain complexes as diagrams anyway. If we include zero objects in our diagrams (i.e. we $Set_*$-Cauchy-complete them), then can’t we just send all the positive objects to $0$?
$\Sigma$ is a colimit construction, so it’s preserved by $\otimes$, so we get
$\Sigma I \otimes \Omega I \cong \Sigma(I\otimes \Omega I) \cong \Sigma\Omega I \cong I$the last step by the assumption that $\Sigma\Omega=Id$.
Yes, we would have to do that. But if $U$ is induced by a functor between diagram shapes, then it really should have both adjoints. I think the left adjoint is the functor that sends an unbounded cochain complex $A^{\bullet}$ to $coker (A^{-1} \to A^0) \to A^1 \to A^2 \to \cdots$. I guess that resolves the duality question.
Yes, that makes sense.
edit: This is wrong, see below or ignore.
I think it should be noted, that this does not make $\mathcal{Ch}$ into a closed category, but rather a category of chain complexes with morphisms all graded maps, not just chain maps. I.e., we do not have
$\mathcal{Ch}(A\otimes B,C) \cong \mathcal{Ch}(A, [B,C]),$but instead
$[A\otimes B, C]_0 \cong [A,[B,C]]_0.$The former failing is most easily seen when taking $A=I$, the unit of the tensor product.
However this category should be more useful than a potential closed structure on $\mathcal{Ch}$ as the internal hom in this case would not include homotopies, since it only contains chain maps.
Re #14: Given that Ch(X,Y) is simply the closed elements of degree 0 in [X,Y], i.e., closed elements in [X,Y]_0, the isomorphism [A⊗B,C]≅[A,[B,C]] implies both displayed formulas in #14.
Take $A = I$ the unit of the tensor product of chain complexes, then
$\mathcal{Ch}(I\otimes B,C)\cong\mathcal{Ch}(B,C) \neq [B,C]_0 \cong \mathcal{Ch}(I,[B,C])$Though the third isomorphism is wrong, as Urs wrote below. I noticed this shortly after posting myself and edited it away, but put back what I originally wrote so the discussion remains readable.
No, the chain maps out of $I$ are the closed elements of degree=0 in the codomain.
[edit: This was in reply to an earlier version of comment #16, which was subsequently edited away.]
[edit: Now it’s back with commentary.]
If you want to do the community and the next generation a favor, then you could add a remark/example to the entry to help the next reader who runs into this issue.
Thank you, also for taking the time to sort out my confusion.
added pointer to:
1 to 22 of 22