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Let $\mathcal{V_1}$ and $\mathcal{V_2}$ be cocomplete symmetric monoidal categories, each endowed with a cosimplicial object $\Delta^\bullet=\Delta^\bullet_{\mathcal{V}_i}:\Delta \to \mathcal{V}_i$. Denote by $|-|=|-|_{\mathcal{V}_i}:s\mathcal{V}_i\to \mathcal{V}_i$ the functor tensor product $-\otimes_\Delta \Delta^\bullet_{\mathcal{V}_i}$, i.e. the coend

$$|X_\bullet|=X_\bullet \otimes_\Delta \Delta^\bullet = \int^n X_n \otimes \Delta^n$$ for $X_\bullet\in s\mathcal{V}_i$ a simplicial object in $\mathcal{V}_i$.

Let $F:\mathcal{V}_1\to \mathcal{V}_2$ be a lax symmetric monoidal functor which is a left adjoint and such that $F(\Delta^\bullet_{\mathcal{V}_1})\cong \Delta^\bullet_{\mathcal{V}_2}$.

Then there is an induced natural transformation $$\tau:|F-|\Rightarrow F|-|$$ of functors $s\mathcal{V}_1\to \mathcal{V}_2$, since $F$ preserves coends and the chosen cosimplicial objects.

Now, the category $s\mathcal{V}_i$ has a symmetric monoidal structure given by the pointwise formula $(X \otimes Y)_n = X_n \otimes Y_n$, and it is such that the induced functor $F:s\mathcal{V}_1\to s\mathcal{V}_2$ is lax symmetric monoidal. Suppose further that $|-|:s\mathcal{V}_i\to \mathcal{V}_i$ is lax symmetric monoidal. Therefore $\tau$ is a natural transformation between lax symmetric monoidal functors, so it makes sense to ask:

Is $\tau$ a lax symmetric monoidal transformation? i.e., do the following diagrams commute?

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I am under the impression that it won't be the case in general... Are there reasonable extra conditions to impose such that it will hold?

Here is an example of the above scenario, where I'm hoping the answer is affirmative (and it may thus serve as an inspiration to extract additional conditions to ensure a general $\tau$ above to be lax symmetric monoidal).

Consider $\mathcal{V}_1$ to be simplicial sets and $\mathcal{V}_2$ to be topological spaces. Endow the first one with the Yoneda embedding as a cosimplicial simplicial set, and endow the second one with the standard cosimplicial topological space. These objects yield internal geometric realizations: of a bisimplicial set into a simplicial set (which one can prove to be the diagonal functor), and of a simplicial space into a space (the standard one).

As a functor $F$, we will consider the standard "extrinsic" geometric realization of a simplicial set into a topological space, which satisfies all the hypotheses above. So the question in this case is whether the natural isomorphism $\tau_{X_{\bullet,\bullet}}: |\mathrm{diag} X_{\bullet,\bullet}|\cong |[n]\mapsto |X_{n,\bullet}||$ in $X_{\bullet, \bullet}\in ss\mathrm{Set}$ is (cartesian) symmetric monoidal.

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The answer might depend on how one defines the monoidal structure on the realizations $|-|:s\mathcal{V}_i\to \mathcal{V}_i$. In the topological and the simplicial cases the strong monoidal structures are determined by the preservation of products.

Observe that in both of these cases the canonical morphism $|X\times Y| \rightarrow |X|\times|Y|$ is induced through the diagonals $\Delta^n \rightarrow \Delta^n\times\Delta^n$ when one thinks of it as a map between the coends. One can generalize to the monoidal situation by requiring the cosimplicial object $\Delta \to \mathcal{V}$ to be a comonoid in $[\Delta, \mathcal{V}]$, so that we have $\delta_{\Delta^n} : \Delta^n \rightarrow \Delta^n\otimes\Delta^n$. This will define a colax monoidal structure on $|-|$

$$\int^n X_n\otimes Y_n \otimes \Delta^n \rightarrow \int^{n,m} X_n\otimes Y_m \otimes \Delta^n\otimes \Delta^m$$ $$\cong (\int^n X_n \otimes\Delta^n) \otimes (\int^m Y_m \otimes \Delta^m)$$

We also need $\otimes$ to commute with coends.

If one is given a lax monoidal functor $F$, then the composites $F|-|$ and $|F-|$ are neither lax nor colax, since $|-|$ is colax. However, one can still form the diagrams for $\tau$ by inverting in them the monoidality stucture maps for $|-|$. If $\delta_{F\Delta^n}$ equals to $\delta_{\Delta^n}$ up to the isomorphism $F(\Delta^n) \cong \Delta^n$, then these diagram will commute. The first diagram will commutive since (writing $.$ for the tensor, and omitting an index $n$)

$$FX.FY.F\Delta \rightarrow{} FX.FY.F\Delta.F\Delta \rightarrow F(X.\Delta).F(Y.\Delta) \rightarrow F(X.\Delta.Y.\Delta)$$ $$=$$ $$FX.FY.F\Delta \rightarrow F(X.Y).F\Delta \rightarrow F(X.Y.\Delta) \rightarrow F(X.\Delta.Y.\Delta),$$ which holds because of the naturality and coherence of the lax structure of $F$. The second diagram commutes because of a similar argument.

Now, if the colax monoidality morphisms are invertible, then $|-|$ will become strong monoidal, and the original diagrams in the question will commute. Hence $\tau$ will become a monoidal transformation. This is the case in the simplicial-topological example.

For the purpose of defining a lax structure on $|-|$ we could again look at our examples. In these examples the inverses of the lax structure of $|-|$ are defined as follows. In the topological case the lax monoidal structure $|X|\times|Y| \rightarrow |X\times Y|$ is constructed (-> Theorem 11.5 "The geometry of iterated loops") using certain maps

$$X_n\times X_m \times \Delta^n\times \Delta^m \rightarrow X_{n+m}\times Y_{n+m} \times \Delta^{n+m}$$

which depend on certain isomorphisms $\Delta^n\times\Delta^m \cong \Delta^{n+m}$. In the bisimplicial situation $|X|\times|Y| \rightarrow |X\times Y|$ is explicitly given via

$$(X_n\times Y_m \times \Delta^n\times \Delta^m)_r \rightarrow (X_r\times Y_r \times \Delta^r)_r$$ $$(x, y, u : r \rightarrow n, v : r \rightarrow m) \mapsto (u^\ast(x), v^\ast(y), 1 : r \rightarrow r).$$

However, these do not have straightforward generalization to the monoidal context.

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  • $\begingroup$ Thanks for taking the time! Sorry, but when you say "these diagrams will commute" -- could you elaborate? (also, I think there's an $m$ which should be an $n$ in your displayed coends) $\endgroup$ – Bruno Stonek Jul 1 '16 at 10:51
  • $\begingroup$ Ok, added some details. $\endgroup$ – Dimitri Chikhladze Jul 1 '16 at 12:28
  • $\begingroup$ I'm confused... How do you get the morphism in your displayed coends? $\endgroup$ – Bruno Stonek Jul 1 '16 at 13:22
  • $\begingroup$ The coend is defined using the natural family $1\otimes 1\otimes \delta_{\Delta^n} : X_n\otimes Y_n\otimes\Delta^n \rightarrow X_n\otimes Y_n\otimes\Delta^n\otimes \Delta^n$ $\endgroup$ – Dimitri Chikhladze Jul 1 '16 at 13:34
  • $\begingroup$ Thanks for the clarification. However, I don't see why you end up in a coend over $\Delta\times \Delta$. Shouldn't the arrow point to $\int^n X_n \otimes Y_n \otimes \Delta^n \otimes \Delta^n$? $\endgroup$ – Bruno Stonek Jul 1 '16 at 13:44

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