Let $k$ be a field and let $\mathcal{C}=\mathbf{StLin}_k$ be the $\infty$-category of stable infinity categories enriched over the $\infty$-category $\mathbf{Vect}_k$, regarded as a symmetric monoidal $\infty$-category with unit object $\mathbf{Vect}_k$.

The forgetful functor $\operatorname{CAlg}(\mathcal{C}) \to \mathcal{C}$ admits a left adjoint $Sym^*: \mathcal{C} \to \operatorname{CAlg}(\mathcal{C})$. Now let $\mathcal{C}[z]=Sym^*(\mathbf{Vect}_k)$ - i.e. $\mathcal{C}[z]$ is the free stable symmetric monoidal category generated by $\mathbf{Vect}_k$.

My question is: Is the $\infty$-category $\mathcal{C}[z]$ compactly generated in $\mathcal{C}$? (Recall that a category $\mathcal{A}$ is compactly generated if there exists a subcategory $\mathcal{A}_0 \subseteq \mathcal{A}$ and an equivalence $\mathcal{A} \simeq Ind(\mathcal{A}_0)$.)

More generally: If $\mathcal{D} \in \mathcal{C}$ is compactly generated, is $Sym^*(\mathcal{D})$ also compactly generated?

My idea was to show that the map $\mathbf{Vect}_k \to \mathcal{C}[z]$ corresponding to the identity:$ \mathcal{C}[z] \to \mathcal{C}[z]$ under the adjunction above identifies the image of $k$ with a compact generator of $\mathcal{C}[z]$, but I'm not sure if this would work.

  • 1
    $\begingroup$ I don't see why you expect $k$ to be a generator ? It seems more reasonable to expect that $(k, k \otimes k, k \otimes k \otimes k, \dot... )$ would be a set of compact generator, isn't it ? More generally I would expect that if $D$ has a sets of compact generator $E$ then $Sym^*(D)$ has a set of compact generator given by all the tensor product of finite families of elements in $E$. $\endgroup$ – Simon Henry Nov 16 '18 at 8:13
  • 1
    $\begingroup$ Do you want $\mathcal C$ to consist of cocomplete ∞-categories? If so the answer is yes, because $Ind(Sym^*(\mathcal D^c))$ has the correct universal property. If not, then no as $Sym^*(\mathcal D)$ is not cocomplete. $\endgroup$ – Marc Hoyois Nov 16 '18 at 15:40
  • $\begingroup$ @MarcHoyois Right, if all the categories $X$ are cocomplete then $Map_{CAlg}(Ind(Sym(D^c)), X)=Map_{CAlg}(Sym(D^c),X)=Map_{StLin}(D^c, X)=Map(D,X)$. Is this correct? $\endgroup$ – leibnewtz Nov 16 '18 at 18:56
  • $\begingroup$ Yes. A reference for this symmetric monoidal universal property of Ind is HA $\endgroup$ – Marc Hoyois Nov 16 '18 at 22:15

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.