1
$\begingroup$

Suppose we have a map of $E_n$-spaces $X\to Y$. Then there is a highly structured action of $X$ on $Y$, $X\wedge Y\to Y\wedge Y\to Y$, using the multiplication of $Y$. As such, I believe that there should be a lax $E_{n-1}$-monoidal functor of quasicategories (where the domain is clearly a Kan complex): $$BX\to Top$$

which maps the unique point of $BX$ to $Y\in Top$ and each morphism of $BX$ (corresponding to the points of $X$) to an automorphism of $Y$ in $Top$ corresponding that element of $X$ acting on $Y$. Moreover, the colimit of this morphism, denote it $Y/X$, should be an $E_{n-1}$-algebra in $Top$ (by recent work of Antonlin-Camarena and Barthel), since $BX$ is an $E_{n-1}$-algebra and the morphism is lax monoidal.

My question is: how do I produce the functor $BX\to Top$? Also, is it clearly $E_{n-1}$-monoidal?

Improve this question
$\endgroup$
2
  • $\begingroup$ @QiaochuYuan I completely changed the question in an attempt to make it a little clearer what precisely I'm looking for. $\endgroup$
    – Jonathan Beardsley
    Jun 1, 2015 at 1:25
  • $\begingroup$ By the way, your source is only a Kan complex if $X$ is additionally grouplike. $\endgroup$
    – Aaron Mazel-Gee
    Jun 1, 2015 at 3:50

1 Answer 1

Reset to default
3
$\begingroup$

Your functor $BX \to \mathrm{Top}$ factors through the inclusion of the full subcategory of $\mathrm{Top}$ generated by the single object $Y$, which is just the data of $\mathrm{End}(Y)$ (an $E_1$-algebra) -- you might call this subcategory $B(\mathrm{End}(Y))$. By adjunction, the multiplication on $Y$ gives a map $Y \to \mathrm{End}(Y)$, and your map (of one-point categories, which is equivalent to a map of $E_1$-algebras) is just the composite $X \to Y \to \mathrm{End}(Y)$.

Improve this answer
$\endgroup$
6
  • $\begingroup$ Right, so for $n \ge 2$ there's no (natural, but in most cases there just won't be any) $E_{n-1}$-monoidal structure on $B \text{End}(Y)$ and so it's meaningless to ask for an $E_{n-1}$-monoidal functor into it. $\endgroup$
    – Qiaochu Yuan
    Jun 1, 2015 at 3:48
  • $\begingroup$ @QiaochuYuan in general an endomorphism space is only a monoid but is there no analogy to be made with the construction of GL_1(R) being an E_n-space when R is an E_n-ring spectrum? $\endgroup$
    – Jonathan Beardsley
    Jun 1, 2015 at 17:53
  • $\begingroup$ @Jon: well, there you start with an $E_n$-structure and attempt to transport it along a functor (namely the $GL_1(-)$ functor), so the question is how monoidal the functor is. Here there's no $E_n$-structure to start with. If you want a notion of "$E_n$-action" of an $E_n$-algebra $X$ one option is to change the source of the functor from $BX$ to $B^n X$. Of course it depends on what you want to do. $\endgroup$
    – Qiaochu Yuan
    Jun 2, 2015 at 3:08
  • $\begingroup$ Of course there's an $E_n$-structure to start with, the $E_n$-structure on $Y$ (and on $X$, and on the morphism $X\to Y$). This makes $Y$ into an $E_n$-$X$-algebra in $Top$. $\endgroup$
    – Jonathan Beardsley
    Jun 2, 2015 at 3:46
  • $\begingroup$ And so one can ask about the universal $E_n$-algebra (or $E_m$-algebra for any $m\leq n$) $Y/X$ such that any map out of $Y$ that lands on an $E_m$-algebra on which $X$ acts trivially factors through $Y/X$. $\endgroup$
    – Jonathan Beardsley
    Jun 2, 2015 at 3:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.