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Is there any notion in the literature which captures the idea of an $\infty$-ary tensor product on a category $C$? This should include tensor products of $\alpha$-indexed families of objects in $C$ for every ordinal $\alpha$, as well as some coherence isomorphisms and compatiblity properties. Here are three motivating examples:

1) The tensor product of modules

2) A category with choosen arbitrary (co)products

3) Let $C$ be a cocomplete monoidal category such that every object $x$ is equipped with a natural morphism $1 \to x$. If this data isn't available, just pass to this slice category over $C$. For an ordinal $\alpha$ and a familiy $(x_{\beta})_{\beta < \alpha}$ in $C$ define $\bigotimes_{\beta < \alpha} x_{\beta}$ by recursion on $\alpha$. It's clear what to do for $\alpha=0$ and when $\alpha$ is a successor. When $\alpha$ is a limit ordinal, take the colimit of all the preceding tensor products. The transition maps are induced by the data above.

There are some applications for the construction in 3), for example the largest Hausdorff quotient, the associated sheaf and the coproduct of algebras. In the context of orthogonal classes in presentable categories it is also called "transfinite composition".

If there is no such notion yet, what axioms should we choose?

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A first idea is generalizing the associativity axioms to (finite) partitions on a set of (ordered) indexes, classically you have the finite set {1, 2, 3, 4} indexes and associate respect to partitions of it... –  Buschi Sergio Sep 8 '12 at 14:58
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There is a notion of "unbiased monoidal category" in which all finite monoidal products are defined at once. This can be found in Leinster's book, Higher operads, higher categories. This seems easy to generalise in comparison to the traditional one involving only a monoidal unit and a binary monoidal product. The coherence axioms are based on partitions, as Buschi Sergio indicated. –  Zhen Lin Sep 8 '12 at 15:40
    
@Zhen: Yes, that's right. So is this spelled out somewhere in that generality? Or can you do this in an answer? Thanks. –  Martin Brandenburg Sep 8 '12 at 16:25
    
Construction (3) uses a transfinite composition, and so do the largest Hausdorff quotient, the associated sheaf, and colimits of algebras. Transfinite composition is of course a widely applicable technique. But are you claiming that those three other constructions are actually infinitary tensor products? –  Mike Shulman Sep 8 '12 at 19:51
    
Perhaps the crucial thing is that there is no single concept of infinitary monoidal category. I think, the right approach to define infinitary monoidal categories is along Mike's line. If we take the grupoid of finite ordinals $\mathit{fin}$ then the Grothendieck construction over the functor $\mathbb{C}^{(-)} \colon \mathit{fin}^{op} \rightarrow \mathbf{Cat}$ gives a kind of the "free" symmetric (here permutations define symmetry) monoidal category on $\mathbb{C}$. Then monoidal categories are just algebras of the induced monad $\mathbb{C} \mapsto \int \mathbb{C}^{(-)}$. (cont) –  Michal R. Przybylek Sep 12 '12 at 21:49
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Here is a tentative definition based on finitary unbiased monoidal categories – I make no claims of usefulness!

An infinitary unbiased monoidal category consists of the following data:

  • An ordinary category $\mathcal{C}$.

  • For each small ordinal $\alpha$, a functor $T_\alpha : \mathcal{C}^\alpha \to \mathcal{C}$, which maps a $\alpha$-sequence $(A_0, A_1, \ldots)$ to $A_0 \otimes A_1 \otimes \cdots$.

  • A natural isomorphism $\textrm{id}_\mathcal{C} \Rightarrow T_1$.

  • For each partition $\alpha = \sum_{i < \gamma} \beta_i$, a natural isomorphism $$T_\gamma \circ (T_{\beta_0} \times T_{\beta_1} \times \cdots ) \Rightarrow T_\alpha$$ such that for each double partition $\alpha = \sum_{i < \gamma} \sum_{j < \delta_i} \beta_{i,j}$, where $\beta_i = \sum_{j < \delta_i} \beta_{i,j}$ and $\delta = \sum_{i < \gamma} \delta_i$, the two composites \begin{multline} T_\gamma \circ (T_{\delta_0} \circ (T_{\beta_{0,0}} \times T_{\beta_{0,1}} \times \cdots) \times T_{\delta_1} \circ (T_{\beta_{1,0}} \times T_{\beta_{1,1}} \times \cdots) \times \cdots) \newline \Rightarrow T_\gamma \circ (T_{\beta_0} \times T_{\beta_1} \times \cdots) \Rightarrow T_\alpha \end{multline} and \begin{multline} T_\gamma \circ (T_{\delta_0} \circ (T_{\beta_{0,0}} \times T_{\beta_{0,1}} \times \cdots) \times T_{\delta_1} \circ (T_{\beta_{1,0}} \times T_{\beta_{1,1}} \times \cdots) \times \cdots) \newline \Rightarrow T_\delta \circ ((T_{\beta_{0,0}} \times T_{\beta_{0,1}} \times \cdots) \times (T_{\beta_{1,0}} \times T_{\beta_{1,1}} \times \cdots) \times \cdots) \Rightarrow T_\alpha \end{multline} are equal, and the two composites $$T_\alpha \Rightarrow T_\alpha \circ (T_1 \times T_1 \times \cdots ) \Rightarrow T_\alpha$$ $$T_\alpha \Rightarrow T_1 \circ T_\alpha \Rightarrow T_\alpha$$ are the identity natural transformation on $T_\alpha$.

An infinitary lax monoidal category is what we get if we replace "natural isomorphism" by "natural transformation" in the above. An infinitary strict monoidal category is what we get if we replace "natural isomorphism" by "identity".


This is just a straightforward generalisation of the definition appearing in [Leinster, Higher operads, higher categories, §3.1]. The reason why this even makes sense is that ordinal addition is associative – if it weren't, we'd be stuck. This leads us to our first easy example:

Example. The category of small ordinals (and monotone maps) is a infinitary strict monoidal category under ordinal addition.

Less tautologically:

Example. Any (co)complete category is an infinitary unbiased monoidal category under (co)products.

(Though, in some sense we assumed this in our definition...)

Conjecture. Martin's third example probably works if we assume that $X \otimes (-)$ preserves sequential colimits, and that the chosen morphisms $I \to X$ are sufficiently nice. First, by the Mac Lane–Kelly coherence theorem, we can make a finitary biased monoidal category into an unbiased one, and we can use the colimit construction to define the infinitary monoidal products. We use the assumption on $X \otimes (-)$ to prove that, e.g. $$A_0 \otimes (A_1 \otimes A_2 \otimes \cdots) \cong A_0 \otimes A_1 \otimes A_2 \otimes \cdots$$ The universal property of colimits should be enough to ensure the coherence of the infinitary associators – but I haven't checked. We need to assume something about the choice of morphisms $I \to X$ so that $$I \otimes I \otimes I \otimes \cdots \cong I$$ holds. (For example, choosing a non-isomorphism for $I \to I$ is a bad idea!)

Non-example. Take $\mathcal{C}$ to be the category of small ordinals, and take $T_\alpha$ to be ordinal addition for all finite ordinals $\alpha$, and $T_\alpha = 0$ for all infinite ordinals $\alpha$. This is not an infinitary unbiased monoidal category, because $$T_\omega \circ (T_1 \times T_1 \times T_1 \times T_0 \times T_0 \times \cdots ) \ncong T_3$$ and so we see that there is some form of continuity required.

Also, a reminder about a famous trick:

Remark. Suppose we define infinitary monoids to be discrete infinitary strict monoidal categories. There are non-trivial small infinitary monoids: for example, take any complete small semilattice with $\sup$ as the monoid operation. However, any infinitary monoid that is also a group must be the trivial monoid: after all, for any element $x$, $$x^{-1} \cdot (x \cdot x \cdot x \cdot \cdots) = (x^{-1} \cdot x) \cdot (x \cdot x \cdot x \cdot \cdots) = x \cdot x \cdot x \cdot \cdots$$ and then cancel $x \cdot x \cdot x \cdot \cdots$ on both sides of the equation to obtain $x^{-1} = \textrm{id}$.

(EDIT) Non-example. The tensor product of modules doesn't give an infinitary monoidal product. To be precise, if we define $$\textrm{Hom}_R (A_0 \otimes_R A_1 \otimes_R A_2 \otimes_R \cdots, B) \cong \textrm{Multi}_R(A_0, A_1, A_2, \ldots ; B)$$ where by $R$-multilinear we mean that $f(\ldots, r a, \ldots) = r f (\ldots, a, \ldots)$, then I don't see how $R \otimes_R R \otimes_R R \otimes_R \cdots$ can be isomorphic to $R$. (A multilinear map $R \times R \times R \times \cdots \to B$ is not necessarily determined by just what it does to $(1, 1, 1, \ldots)$, unlike the finite case.)


Now, some closing remarks:

  • How do we define an "infinitary braided/symmetric monoidal category" as extra structure on top of an infinitary unbiased monoidal category? The trouble is that a permutation of an infinite ordinal can change its order type (e.g. $\omega + \omega_1 = \omega_1 \ne \omega_1 + \omega$)... but this probably isn't a big problem. It seems to me that the morally correct way of defining an "infinitary unbiased symmetric monoidal category" would define a functor $S_\kappa : \mathcal{C}^\kappa / \textrm{Sym}(\kappa) \to \mathcal{C}$ for each cardinal $\kappa$ and some natural isomorphisms to the various $T_\alpha$.

  • Is there a sensible notion of an "infinitary biased monoidal category"? We would need to define $T_\alpha$ for at least $\alpha = 0$, $\alpha = 2$, and every infinite regular ordinal $\alpha$. This is still a large amount of data! For the finitary fragment we can adopt the same coherence axioms, but I don't know what the coherence axioms for the infinitary fragment ought to be. There will have to be some new ones that don't show up in the finitary fragment: For example, we would need an axiom like $$I \otimes I \otimes I \otimes \cdots \cong I$$ for the monoidal unit $I$ – because inductively applying the left unitor can never delete infinitely many copies of $I$; or we could instead make $$A_0 \otimes A_1 \otimes \cdots \otimes A_n \otimes I \otimes I \otimes I \otimes \cdots \cong A_0 \otimes A_1 \otimes \cdots \otimes A_n$$ into an axiom, because we can't even apply the right unitor here!

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+1: nice answer! –  Todd Trimble Sep 8 '12 at 19:21
    
Thank you Zhen. But this does not include the tensor product of modules, as you say. If $K$ is a field, then $\bigotimes_{i < \alpha} K$ is a twisted group algebra of the giant group ${K^*}^{\alpha} / {K^*}^{({\alpha})}$ (see the link in my question). So this contradicts the definition with the trivial partition $0 = \sum_{i < \alpha} 0$. How can we adjust the definition so that it works? For example, we may treat the zero case separatedly. –  Martin Brandenburg Sep 12 '12 at 7:40
    
I'm not sure how to resolve that problem. Perhaps zero should be excluded everywhere – leading to some kind of non-unital infinitary monoidal category – and then we adjoin the axioms for the unit from the theory for finitary monoidal categories. Just excluding partitions containing infinitely many copies of $0$ is troublesome for the statement of the associative law: if we do that, then $1 + 1 + 1 + \cdots = \omega$ and $0 + 1 = 1$ are both valid partitions, but the substituted partition $(0 + 1) + (0 + 1) + (0 + 1) + \cdots = \omega$ is not. –  Zhen Lin Sep 12 '12 at 9:05
    
I want to publicly thank you for answering to (part of) this related question mathoverflow.net/questions/138518/… –  tetrapharmakon Aug 21 '13 at 22:50
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Here's a sketch of a direct way to define a symmetric infinitary-monoidal category. For a category $C$, let $T C$ be the Grothendieck construction (lax colimit) of the functor $\mathrm{Core}(\mathrm{Set}) \to \mathrm{Cat}$ sending $X$ to $C^X$. Thus an object of $T C$ is a set-indexed family of objects of $C$, and a morphism is a bijection between indexing sets which indexes a family of morphims in $C$. There is an obvious functor $C \to T C$ sending an object to the corresponding singleton family. And there is a functor $T T C \to T C$ induced by coproducts in Set. This makes $T$ into a pseudomonad on Cat; then a symmetric infinitary-monoidal category is a pseudo $T$-algebra.

If you use the (discrete) category $\mathrm{Core}(\mathrm{Ord})$ instead of the groupoid $\mathrm{Core}(\mathrm{Set})$, then I think you get Zhen's notion of non-symmetric infinitary-monoidal category. I don't have any suggestions for what a "braided" version would mean, though.

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So how does this definition look like in more "down-to-earth" terms? –  Martin Brandenburg Sep 20 '12 at 15:05
    
That's left as an exercise for the reader. (-: –  Mike Shulman Sep 21 '12 at 2:37
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