solifine

We can see that O(2) and O(0) are both just ∗, since O(2)×O(0) ≃ O(2)×O(1)×O(0) → O(1) ≃ ∗ is a bijection. Inductively, we see that O(n) ≃ ∗ for all n, since O(n-1)×O(2)×O(1)×O(1)×⋯×O(1) → O(n) is a bijection. [3/17]

OK, so what about colored operads? Here, we have a fixed set of colors with names like a,b,c,aᵢ and so on, and we replace O(r) by O(a₁,a₂,…,aᵣ;a₀). The structure maps are adjusted to be "typed". [4/17]

This is like the passage from "monoids" to "categories" (maybe someday @koszuldude will give more details). In particular, the units take the shape ∗ → O(a;a), and part of "invertibility" tells us that O(a;a) ≃ ∗ for all colors a. [5/17]

If you're like me, you might guess that "invertibility" for colored operads means that structure morphisms like O(a,b;c)×O(d;a)×O( ;b) → O(d;c) (where a,b,c,d are some colors) are bijections. [6/17]

If that were the case, you could play the same games as before to see that O(a₁,a₂,…,aᵣ;a₀) ≃ ∗ for every list of colors. That's boring! [7/17]

Instead you let O(r) be the sum over all of the sets O(a₁,a₂,…,aᵣ;a₀). You don't have a function O(r)×O(m₁)×O(m₂)×⋯×O(mᵣ) → O(Σmᵢ), but you do have a function from a subset T where all of the colors match up correctly. We're really asking if T ≃ O(Σmᵢ). [8/17]

As an example, when r = 1 and m₁ = 1, this is asking that ∐_{a,b,c} O(b;c)×O(a;b) → ∐_{a,c} O(a;c) is a bijection. Can you see why this implies that O(a;c) = ∅ whenever a≠c? [9/17]

What's a unital 2-Segal simplicial set? Well, a simplicial object X is called 2-Segal "when the commuting squares that express the simplicial identities between inner and outer face maps of X are pullback squares." (Feller et al. https://arxiv.org/abs/1905.09580 ) [11/17]

I'm going to pull the diagrams out of the paper from the previous tweet, but first I want to say that this paper is short and cool as hell and has only a single theorem "Every 2-Segal space is unital." [12/17]

Let's look directly at the case of unital 2-Segal simplicial sets with X₁ = ∗. In light of the Dyckerhoff–Kapranov theorem, these are meant to correspond to monochrome invertible operads (so there should only be one!). [15/17]

Examine the unitality diagrams in the case i=n=0. When X₁ = ∗, the pullback condition devolves into a product condition, and we are asking X₀×X₂ ≃ X₁ ≃ ∗. Thus X₀ and X₂ are both points. [16/17]