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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 XX and YY are non-negatively graded, or non-positively graded, then [X,Y][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= p0Hom(X p,Y n+p)[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” UU of non-negatively-graded cochain complexes into unbounded complexes is fully faithful, strongly monoidal, and has a right adjoint, say RR, so

    Hom(U(AB),C)Hom(UAUB,C)Hom(UA,[UB,C])Hom(A,R[UB,C])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 RR 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 ΣΩ=Id\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, ΣΩ=Id\Sigma\Omega=Id gives ΣIΩII\Sigma I \otimes \Omega I \cong I, where II is the monoidal unit, so the functor Σ(ΣI)\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 UU isn’t induced by a functor between diagram shapes, unfortunately.

    I guess you are assuming that Σ\Sigma and Ω\Omega are strong functors or something like that? Otherwise I do not see why ΣIΩII\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 UU is induced by a functor of Set *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 *Set_*-Cauchy-complete them), then can’t we just send all the positive objects to 00?

    Σ\Sigma is a colimit construction, so it’s preserved by \otimes, so we get

    ΣIΩIΣ(IΩI)ΣΩII\Sigma I \otimes \Omega I \cong \Sigma(I\otimes \Omega I) \cong \Sigma\Omega I \cong I

    the last step by the assumption that ΣΩ=Id\Sigma\Omega=Id.

    • CommentRowNumber12.
    • CommentAuthorZhen Lin
    • CommentTimeJun 9th 2014

    Yes, we would have to do that. But if UU 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 A^{\bullet} to coker(A 1A 0)A 1A 2coker (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.

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