2
$\begingroup$

Let $\mathcal C$ be a stably monoidal $\infty$-category, and let $I \xrightarrow f X \to Y$ be a fiber sequence where $I$ is the unit. Then for each $k \in \mathbb N$, we can form a canonical cubical diagram $C_k(f) : [1]^k \to \mathcal C$, $(\epsilon_1, \dotsc, \epsilon_k) \mapsto X^{\epsilon_1} \otimes \dotsb \otimes X^{\epsilon_k}$. [1]

Let $D_k(f) = C_k(f)\rvert_{[1]^k \setminus \{(0,\dotsc,0)\}} : [1]^k \setminus \{(1,\dotsc,1)\} \to \mathcal C$ be the restriction to the full sub-poset of the cube $[1]^k$ which throws out the terminal vertex $(1,\dotsc,1)$. I'm interested in computing the colimit $\varinjlim D_k(f)$. [2]

Question: For arbitrary $k \in \mathbb N$, does $\varinjlim D_k(f)$ decompose as a direct sum of terms of the form $X^i \otimes Y^j$? If so, what is the formula?

The case $k = 2$:

When $k = 2$, we have $\varinjlim D_2(f) = X \cup_I X$. The identity map $X \to X$ induces a splitting of the inclusion $X \to X \cup_I X$, which identifies

$$\varinjlim D_2(f) = X \oplus Y.$$

I strongly suspect that there is some generalization of this formula to higher $k$ — hence the question.


[1]: Here $[1] = \{0 < 1\}$ is the arrow category, and $\epsilon_i \in \{0,1\} = Ob [1]$; we have $X^0 = I$ and $X^1 = X$.

[2]: Note that the fiber of $I \to \varprojlim D_k(f)$ is the total fiber of the cube $C_k(f)$.

$\endgroup$
1
  • $\begingroup$ Isn't the total cofiber of your cube exactly $Y^{\otimes k}$, so that you're looking for a description of $fib(X^{\otimes k}\to Y^{\otimes k})$ ? In general, this is filtered by terms like the ones you're looking for, but this filtration does not split. $\endgroup$ Commented Sep 11, 2023 at 14:43

0

You must log in to answer this question.