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• CommentRowNumber1.
• CommentAuthorUrs
• CommentTimeAug 26th 2012

I started putting some content into internal hom of chain complexes. But am being interrupted now…

• CommentRowNumber2.
• CommentAuthorTodd_Trimble
• CommentTimeAug 26th 2012

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.)

• CommentRowNumber3.
• CommentAuthorUrs
• CommentTimeAug 26th 2012

Sure. I have changed it.

• CommentRowNumber4.
• CommentAuthorMike Shulman
• CommentTimeJun 7th 2014

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?

• CommentRowNumber5.
• CommentAuthorTodd_Trimble
• CommentTimeJun 7th 2014
• (edited Jun 7th 2014)

For the positively graded case, can’t we just use $[X, Y]_n = \prod_{p \geq 0} Hom(X_p, Y_{n+p})$?

• CommentRowNumber6.
• CommentAuthorZhen Lin
• CommentTimeJun 8th 2014
• (edited Jun 8th 2014)

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…

• CommentRowNumber7.
• CommentAuthorMike Shulman
• CommentTimeJun 8th 2014

That’s about the conclusion I had come to as well. It’s curious that the two cases seem so different.

• CommentRowNumber8.
• CommentAuthorMike Shulman
• CommentTimeJun 8th 2014

(@Zhen, why did you put “inclusion” in quotes?)

• CommentRowNumber9.
• CommentAuthorMike Shulman
• CommentTimeJun 8th 2014

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.

• CommentRowNumber10.
• CommentAuthorZhen Lin
• CommentTimeJun 8th 2014

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?

• CommentRowNumber11.
• CommentAuthorMike Shulman
• CommentTimeJun 8th 2014

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 $\Sigma\Omega=Id$.

• CommentRowNumber12.
• CommentAuthorZhen Lin
• CommentTimeJun 9th 2014

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.

• CommentRowNumber13.
• CommentAuthorMike Shulman
• CommentTimeJun 9th 2014

Yes, that makes sense.