Simplicial sets are presheaves on the simplex category $\Delta$, while augmented simplicial sets are presheaves on $\Delta_+$, the augmented simplex category. Because Day convolution allows us to lift monoidal structures on a category $\mathcal{C}$ to its category of presheaves $\mathrm{Sets}^{\Delta^\circ}$, it is therefore of interest to find monoidal structures on $\Delta$ and $\Delta_+$, as these then provide "natural" monoidal structures on simplicial sets.

The only monoidal structure I know of is the ordinal sum of $\Delta_+$ (which is not braided), whose Day convolution gives the join of simplicial sets, and whose internal hom is given by $$[X,Y]_n=\mathrm{hom}_{\mathrm{Sets}^{\Delta^\circ_+}}(X,\mathrm{Dec}^{n+1}Y)$$

Is this the only monoidal structure on $\Delta_+$? If not, what other monoidal structures are there on $\Delta_+$, and what are there on $\Delta$?

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    $\begingroup$ I think this more likely to get an answer here than at m.se. Also, note that one day is not really enough time to wait before crossposting, given the international nature of the site. Best to wait a few days at minimum. $\endgroup$
    – David Roberts
    Commented Aug 22, 2021 at 3:36
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    $\begingroup$ @Sofia not a problem! It's sometimes not obvious how the two sites operate side-by-side. $\endgroup$
    – David Roberts
    Commented Aug 23, 2021 at 5:46
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    $\begingroup$ Ah, that's my fault. I wasn't careful enough checking and I didn't account for all maps in the simplex category. Sorry for the mistake. $\endgroup$ Commented Aug 26, 2021 at 4:56
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    $\begingroup$ One almost-trivial note: The “switched” ordinal sum gives a second monoidal structure on $\Delta_+$ — the two handednesses of ordinal sum are not isomorphic. $\endgroup$ Commented Sep 5, 2021 at 13:13
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    $\begingroup$ Here is another monoidal structure with unit $[-1]$: Define $[m] \otimes [n] = \begin{cases} [m] & [n] = [-1] \\ [n] & [m] = [-1] \\ [0] & \text{else} \end{cases}$; after imposing unitality, there is a unique way to extend this definition to morphisms, and unless I'm mistaken it works out to be (strict) monoidal. $\endgroup$
    – Tim Campion
    Commented Sep 5, 2021 at 13:19

1 Answer 1


Here is half of a classification. Let $\otimes$ be a monoidal structure on $\Delta_+$. As I mentioned in a comment, the monoidal unit must be $[-1]$ or $[0]$ because these are the only objects with commutative endomorphism monoids.

Suppose that the monoidal unit is $[0]$. Let us consider $[1] \otimes [1]$. We have that $[0]$ is a retract of $[1]$ in 2 ways, and as a result we obtain 4 retracts of $[1] \otimes [1]$ with support $[0] \otimes [0] = [0]$. Consider the induced linear ordering on these 4 points. We also have 4 ways that $[1] = [1] \otimes [0] = [0] \otimes [1]$ is a retract of $[1] \otimes [1]$, and from this we can deduce most of the ordering. It must have the following relations

$\require{AMScd} \begin{CD} 0 \otimes 0 @>>> 0 \otimes 1\\ @VVV @VVV\\ 1 \otimes 0 @>>> 1 \otimes 1 \end{CD}$

To complete this to a linear order, without loss of generality we must have $0 \otimes 1 \leq 1 \otimes 0$. But now, one of our 4 projections onto $[1]$ is the coordinate projection onto the right column of the above square. The fact that this projection is order-preserving implies that $1 \otimes 0 = 0 \otimes 1 = 1 \otimes 1$. This contradicts the fact that the right column exhibits $[1]$ as a retract of this subset of $[1] \otimes [1]$.

Therefore the monoidal unit is not $[0]$; it must be $[-1]$.

I also think I'm ready to conjecture that $\oplus$, $\oplus^{rev}$ (as mentioned by Peter) and the degenerate monoidal structure are probably the only ones. You could imagine a classification starting as follows. Consider the maps $[0] = [0] \otimes [-1] \to [0] \otimes [0]$ and $[0] = [-1] \otimes [0] \to [0] \otimes [0]$. If these are the same, then we should have the degenerate monoidal structure. If they are different, then one is less than the other, and those two cases should correspond to $\oplus$ and $\oplus^{rev}$.


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