This is a follow-up to this question, in which Denis Nardin nicely explained that $$ \operatorname{Map}_{\operatorname{Fun}(X, \operatorname{Sp})}(E_X, E'_X) \simeq \operatorname{Map}(X, \operatorname{Map}_{\operatorname{Sp}}(E, E')). $$ where $E, E'$ are spectra and $E_X, E'_X$ are their "constant" parametrized versions over a base space $X$.

Now when $X$ is an $E_\infty$-space and $E$ is a commutative ring spectrum, then unless I am wrong think so should be $E_X$ a commutative algebra object in parametrized spectra for the Day product monoidal structure. (please correct me if I'm wrong on this!)

Question: Can we in that case obtain a version of the above equivalence which takes the monoidal structure into account, i.e. with morphisms of parametrized ring spectra on the left and morphisms of $E_\infty$-spaces on the right (potentially also replacing morphisms of spectra with morphisms of commutative ring spectra, if that makes more sense)?

I am not sure the same sort of proof as in the previous question can be used because it relied on expressing natural transformation with an end, but here we only want to consider morphisms of parametrized ring spectra, that is to say monoidal functors. Though all of this is quite new to me and I may be missing something quite easy.

P.S.: I was initially going to ask this in a comment to the previous question but I found it is both too long and potentially interesting on its own.


So the answer is a bit surprising (maybe I have a mistake). You have an adjunction between $Fun(X,\mathrm{Sp})$ and $\mathrm{Sp}$ which in one direction sends a functor to its (homotopy) colimit and on the other hand sends a spectrum to the constant functor $X \to \mathrm{Sp}$ with that value. If $X$ has an $E_\infty$-structure then $Fun(X,\mathrm{Sp})$ inherits the Day convolution symmetric monoidal structure and the functor $colim: Fun(X,\mathrm{Sp}) \to \mathrm{Sp}$ is symmetric monoidal. As a result the right adjoint $const: \mathrm{Sp} \to Fun(X,\mathrm{Sp})$ carries a canonical lax-monoidal structure, and the adjunction $colim \dashv const$ can be promoted to an adjunction $$Alg_{E_{\infty}}(Fun(X,\mathrm{Sp})) \rightleftarrows Alg_{E_{\infty}}(\mathrm{Sp})$$ on the level of $E_\infty$-algebras. In particular, if $E$ is an $E_\infty$-ring spectrum then $E_X = const(E)$ is naturally an $E_\infty$-algebra with respsect to the day convolution product. By adjunction one gets that $$ Map_{Alg_{E_{\infty}}(Fun(X,\mathrm{Sp}))}(E_X,E'_X) \simeq Map_{Alg_{E_{\infty}}(\mathrm{Sp})}(colim_XE_X,E') \simeq Map_{Alg_{E_{\infty}}(\mathrm{Sp})}(\Sigma^{\infty}_+(X) \wedge E,E') $$ On the other hand, smash of $E_\infty$-ring spectra is just the $\infty$-categorical coproduct of ring spectra! We then get that $$ Map_{Alg_{E_{\infty}}(\mathrm{Sp})}(\Sigma^{\infty}_+(X) \wedge E,E') \simeq Map_{Alg_{E_{\infty}}(\mathrm{Sp})}(\Sigma^{\infty}_+(X),E') \times Map_{Alg_{E_{\infty}}(\mathrm{Sp})}(E,E') $$ In other words, to describe a map of $E_\infty$-algebras $E_X \to E'_X$ one needs to give a map of $E_\infty$-rings spectra $f:E \to E'$ together with a map of $E_\infty$-ring spectra $a:\Sigma^{\infty}_+(X) \to E'$. Alternatively, we may write a map of $E_\infty$-ring spectra $a:\Sigma^{\infty}_+(X) \to E'$ as a map of $E_{\infty}$-monoids $a:X \to Map(S^0,E')$. Given a pair $(f,a)$ the associated map $E_X \to E_X'$ is presumably given at the point $x \in X$ by the map $a_x\cdot f \in Map_{\mathrm{Sp}}(E,E')$.

Edit: Let $\mathcal{C} = Fun(X,\mathrm{Sp})$ with the Day convolution symmetric monoidal structure. The OP is asking if one can write $Map_{Alg_{E_\infty}(\mathcal{C})}(E_X,E'_X)$ as a mapping space of $E_\infty$-spaces $X \to Map_{\mathrm{Sp}}(E,E')$. This statement cannot be true as is, because $Map_{\mathrm{Sp}}(E,E')$ has no natural $E_\infty$-structure. Instead, it has an $\infty$-operad structure. More precisely, there exists an $\infty$-opead $\mathcal{O}_{E,E'}$ whose colors are the spectra maps $E \to E'$, and such that for $f_1,...,f_n \in Map_{\mathrm{Sp}}(E,E')$ and a target $f_0 \in Map_{\mathrm{Sp}}(E,E')$ the space of multi-maps $(f_1,...,f_n) \to f_0$ is the space of homotopies from the composition $E^{\wedge n} \stackrel{f_1 \wedge ... \wedge f_n}{\longrightarrow} (E')^{\wedge n} \to E'$ to the composition $E^{\wedge n} \to E \stackrel{f_0}{\to} E'$. Note that $E_\infty$-algebras in $\mathcal{O}_{E,E'}$ are exactly the $E_\infty$-maps $E \to E'$. Since $X$ is an $E_\infty$-space we can also think of it as an $\infty$-operad $\mathcal{O}_X$ whose colors are the points of $X$ and such that for $x_1,...,x_n \in X$ and a target point $x_0 \in X$ the space of multimaps $(x_1,...,x_n) \to x$ is the space of paths from $x_1\cdot... \cdot x_n$ to $x_0$. One can then show that $$ Map_{Alg_{E_\infty}(\mathcal{C})}(E_X,E'_X) \simeq Map_{\infty-operads}(\mathcal{O}_X,\mathcal{O}_{E,E'}) $$ It so happens that in this particular case this space of operad maps breaks into a product $Map_{E_\infty}(X,Map_{\mathrm{Sp}}(S^0,E')) \times Map_{Alg_{E_{\infty}}(\mathrm{Sp})}(E,E')$ as described above. To understand why this happens, we note that the $\infty$-operad $\mathcal{O}_X$ has a unit, i.e., it has a color which corepresents the space of $0$-ary maps, namely, the color corresponding to the unit $1 \in X$. This color carries a canonical structure of an $E_\infty$-object in $\mathcal{O}_X$. In particular, every $\infty$-operad map $\mathcal{O}_X \to \mathcal{O}_{E,E'}$ will send $1$ to an $E_\infty$-algebra object in $\mathcal{O}_{E,E'}$, i.e., to a map of $E_\infty$-algebras $f:E \to E'$. We hence obtain a map of spaces $$ p:Map_{\infty-operads}(\mathcal{O}_X,\mathcal{O}_{E,E'}) \to Map_{Alg_{E_\infty}(\mathrm{Sp})}(E,E') $$ Now for every $E_\infty$-algebra object $f$ of $\mathcal{O}_{E,E'}$ one can form the $\infty$-operad $\mathrm{Mod}_f(\mathcal{O}_{E,E'})$ of $f$-modules. One can then show that the homotopy fiber of $p$ over $f$ is equivalent to the subspace of $\infty$-operad maps $\mathcal{O}_X \to \mathrm{Mod}_f(\mathcal{O}_{E,E'})$ which send $1$ to the $f$ (considered as an $f$-module). Now happens something which is very specific to this situation: the $\infty$-operads $\mathrm{Mod}_f(\mathcal{O}_{E,E'})$ are canonically equivalent to each other for all $f$! The reason is that, just like $E_\infty$-algebra objects in $\mathcal{O}_{E,E'}$ correspond to $E_\infty$-algebra maps $E \to E'$, modules over $f$ correspond to "affine maps" $E \to E'$ whose "linear part" is $f$, i.e., maps $g: E \to E'$ which loosely speaking satisfy $g(xy) = f(x)g(y)$. Such maps are all given by the informal formula $x \mapsto f(x)\cdot b$. More precisely, evaluation at the unit $S^0 \to E$ induces an equivalence between the space of $f$-modules and the space of maps $S^0 \to E'$. We then see that for every such $f$, the $\infty$-operad $\mathrm{Mod}_f(\mathcal{O}_{E,E'})$ is canonically equivalent to the $\infty$-operad associated with the $E_\infty$-space $Map_{\mathrm{Sp}}(S^0,E')$. We then obtain a canonical splitting $$ Map_{Alg_{E_\infty}(\mathcal{C})}(E_X,E'_X) \simeq Map_{E_\infty}(X,Map_{\mathrm{Sp}}(S^0,E')) \times Map_{Alg_{E_{\infty}}(\mathrm{Sp})}(E,E') $$

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  • $\begingroup$ but the colimit is not the category of $E_\infty$-algebras, it is in the category of spectra. Since the functor $colim$ (of spectra) is monoidal with repsect to day convolution on one side and smash on the other side it extends to a functor on algebras (but with the same underlying definition). $\endgroup$ – Yonatan Harpaz Oct 5 '16 at 20:28
  • $\begingroup$ For the mapping space to be $Map(X,Map_{E_\infty}(E,E'))$ you want to work in the category $Fun(X,Alg_{E_\infty}(\mathrm{Sp}))$, which is not the same as the category of $E_\infty$-algebras in $Fun(X,\mathrm{Sp})$ with respect to Day convolution (it is however the category of algebras in $Fun(X,\mathrm{Sp})$ with respect to the pointwise smash product). $\endgroup$ – Yonatan Harpaz Oct 5 '16 at 21:04
  • $\begingroup$ Thank you for the answer Yonatan! Do I understand your last comment correctly as saying that if we equip $Fun(X, \operatorname{Sp})$ with the monoidal structure coming from the pointwise smash product, then the mapping space is $Map_{\text{Spaces}}(X, Map_{E_\infty}(E, E'))$ while if we use the Day convolution we get $Map_{E_\infty-\text{Spaces}}(X, Map_{E_\infty}(E, E'))$? $\endgroup$ – A Rock and a Hard Place Oct 6 '16 at 20:13
  • $\begingroup$ This is not exactly true because $Map_{E_\infty}(E,E')$ is not in any natural way an $E_\infty$-space (how would do you multiply two $E_\infty$-maps?). There is a modified statement that can be made but it's too long for a comment, so I will add it as an edit to the answer. $\endgroup$ – Yonatan Harpaz Oct 7 '16 at 7:13
  • $\begingroup$ @YonatanHarpaz Sorry I had misunderstood your notation. Great answer :). $\endgroup$ – Denis Nardin Oct 7 '16 at 14:33

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