Post #2516084

2026-04-17 16:00 UTC

Before that, here's an example of two other composites that might appear different but are actually equal to each other and to the figure eight. We call one Cₓ, the *figure C*, and the other Hₓ, the *figure H*, since the string diagrams sort of look like those letters. So, Cₓ and Hₓ also have *odd* parity, because they're each equal to one instance of 8ₓ. Not pictured: There is also a "reverse C" that uses the braiding of x' with itself, and a figure eight on x' that reverses the roles of x and x'. Both of these are also equal to 8ₓ = Cₓ = Hₓ, and therefore have odd parity. (4/9)

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@nilesjohnson@mathstodon.xyz 2026-04-17 16:00

As often happens, our main theorem is a little easier to state with some additional background. I'll add that now. In Pₓ, the free symmetric monoidal category on the invertible object x, the morphisms are generated as sums and composites of six basic morphisms, with the following parities: There are two de/cancel morphisms, and they have even parity: ηₓ: 0 → x'+x εₓ: x+x' → 0 Then, there are four basic braiding morphisms βₚ,ₛ: p+s → s+p where s and p are each either x or x'; each of these has odd parity. These parities follow from the parities of the "figure morphisms", 8, C, and H above. Then, parity for any other morphisms in Pₓ are computed from these: parity is additive on sums or composites of morphisms. Our main theorem says that any parallel morphisms with the same parity are equal. (5/9)

Open ##2516085