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The standard definition of a (left) simplicial module $V$ over some simplicial algebra $A$ is the map of simplicial vector spaces $A\otimes V\to V$ that gives the usual modules component-wise. Here $\otimes$ denotes the component-wise tensor product of simplicial vector spaces (it gives to the category $sVect$ of simplicial vector spaces a symmetric monoidal structure).

But there is a natural structure of simplicial algebra on $sHom(V,V)$ (the internal $Hom$ induced by the product $\otimes$). So we can also define structure of a simplicial module over $A$ on $V$ by fixing a morphism of simplicial algebras $A\to sHom(V,V)$.

I think we can do it in more generality, for any symmetric monoidal category $C$ with internal $Hom$. If we have a monoid $A$ in $C$ and some object $V\in C$, then we can define the structure of $A$-module on $V$ in two different ways: via $A\otimes V\to V$ and $A\to \underline{Hom}(V,V)$, where $\underline{Hom}$ is the internal $Hom$.

The question is: are these two definitions equivalent for any symmetric monoidal category with internal $\underline{Hom}$? If not, is it true for simplicial vector spaces?

Thank you very much!

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    $\begingroup$ If your $\otimes$ and $\underline{\mathrm{Hom}}$ are related by the standard adjunction, then there is a canonical bijection between morphisms $A \otimes V \to V$ and morphisms $A \to \underline{\mathrm{Hom}}(V, V)$. So it just amounts to checking that the axioms for one version translate to the axioms for the other version. $\endgroup$
    – Zhen Lin
    Commented Oct 31, 2013 at 21:41
  • $\begingroup$ Yes, I know that. But this is far from being obvious (to me) that the axioms actually match up. Maybe I am just being stupid, and there is easy to see. I tried to write it down, but I didn't get anywhere. So I thought maybe there is some more high-powered way to say that everything works. $\endgroup$
    – Sasha Patotski
    Commented Oct 31, 2013 at 21:47
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    $\begingroup$ I'd say this is better for math.stackexchange $\endgroup$
    – Fernando Muro
    Commented Oct 31, 2013 at 21:47
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    $\begingroup$ Sasha, you may find useful first to check it for abelian groups, where objects are sets and morphisms are maps, and then to rephrase everything in a categorical language $\endgroup$
    – Fernando Muro
    Commented Oct 31, 2013 at 21:56

1 Answer 1

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I will write $[B, C]$ instead of $\underline{\mathrm{Hom}}(B, C)$. Recall the tensor–hom adjunction: $$\mathrm{Hom}(A \otimes B, C) \cong \mathrm{Hom}(A, [B, C])$$ Thus there is a canonical bijection between morphisms $\alpha : A \otimes V \to V$ and $\tilde{\alpha} : A \to [V, V]$. Let us show that $\alpha$ is an $A$-action on $V$ if and only if $\tilde{\alpha}$ is a monoid homomorphism. For simplicity I will work in a strict monoidal category.

  • The unit axiom for $\alpha$ says, $\alpha \circ (e \otimes \mathrm{id}_V) = \mathrm{id}_V$; and the the unit axiom for $\tilde{\alpha}$ says, $\tilde{\alpha} \circ e = \eta_V$, where $\eta_V : I \to [V, V]$ is the right adjoint transpose of $\mathrm{id}_V : V \to V$. The naturality of the tensor–hom adjunction in the first variable implies that these two conditions are equivalent.
  • The compatibility axiom for $\alpha$ says, $\alpha \circ (m \otimes \mathrm{id}_V) = \alpha \circ (\mathrm{id}_A \otimes \alpha)$; and the compatibility axiom for $\tilde{\alpha}$ says, $\tilde{\alpha} \circ m = \mu_V \circ (\tilde{\alpha} \otimes \tilde{\alpha})$, where $\mu_V : [V, V] \otimes [V, V] \to [V, V]$ is the right adjoint transpose of the composite $$[V, V] \otimes [V, V] \otimes V \xrightarrow{\mathrm{id}_{[V, V]} \otimes \epsilon_{V,V}} [V, V] \otimes V \xrightarrow{\epsilon_{V,V}} V$$ where $\epsilon_{B,C} : [B, C] \otimes B \to C$ is the left adjoint transpose of $\mathrm{id} : [B, C] \to [B, C]$. Now, naturality in the first variable implies the right adjoint transpose of $\alpha \circ (m \otimes \mathrm{id}_V)$ is $\tilde{\alpha} \circ m$, and naturality in the first variable implies the left adjoint transpose of $\mu_V \circ (\tilde{\alpha} \otimes \tilde{\alpha})$ is the following composite, $$A \otimes A \otimes V \xrightarrow{\tilde{\alpha} \otimes \tilde{\alpha} \otimes \mathrm{id}_V} [V, V] \otimes [V, V] \otimes V \xrightarrow{\mathrm{id}_{[V, V]} \otimes \epsilon_{V,V}} [V, V] \otimes V \xrightarrow{\epsilon_{V,V}} V$$ but $\alpha = \epsilon_{V,V} \circ (\tilde{\alpha} \otimes \mathrm{id}_V)$, so the above reduces to $\alpha \circ (\mathrm{id}_A \otimes \alpha)$.

Thus the claim is proved.

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  • $\begingroup$ There must be a cleaner diagrammatic language to phrase such proofs in. I know there's a nice diagrammatic language when we can write $[B, C] \cong B^{\ast} \otimes C$ but I don't know about in general. I ran into this issue while trying to find a diagrammatic language for cartesian closed categories awhile ago. $\endgroup$
    – Qiaochu Yuan
    Commented Oct 31, 2013 at 22:27
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    $\begingroup$ There are various extensions of the string diagram calculus to handle biclosed monoidal categories, or at least that's what I hear. For cartesian closed categories one could use simply typed $\lambda$-calculus though. $\endgroup$
    – Zhen Lin
    Commented Oct 31, 2013 at 23:06
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    $\begingroup$ And in the general closed monoidal case one can use linear type theory. $\endgroup$
    – Mike Shulman
    Commented Nov 1, 2013 at 4:07
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    $\begingroup$ Of course this correspondence is well-known. Does someone know a reference in the literature? $\endgroup$
    – Martin Brandenburg
    Commented Nov 2, 2013 at 20:46

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