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@MartinEscardo@mathstodon.xyz

Post #1816228

2026-02-23 21:15 UTC

Paraphrasing @dwarn , "there are no higher semi-lattices". I found this rather surprising and interesting, and so I wanted to fully understand it. Here is my elaboration of David's Agda file, in which I add some diagrams and some explanation in English prose, and rename some things that I hope are correct and what David intended. https://martinescardo.github.io/TypeTopology/gist.ThereAreNoHigherSemilattices.html If you wondered about David's result, this may help. I had to spend a couple of hours this evening to understand it, and the above link is the result of my underdstanding.

Replies (4)

  • @MartinEscardo @dwarn I like this, because it does not seem clear to me that once algebra is done within hott, every basic algebraic structure should a priori be a set. Maybe it's because higher groups are much more complicated than classical groups, but still.

    Open ##1816229

  • @ToucanIan@mathstodon.xyz 2026-02-25 02:11

    @MartinEscardo thanks for this! I tried and failed to do the very same thing. @dwarn

    Open ##1816233

  • @de_Jong_Tom@mathstodon.xyz 2026-02-25 10:47

    @MartinEscardo @dwarn To really understand it, I've been working on my own retelling. If I manage to complete it, I'll share it here. Thanks to both of you for sharing!

    Open ##1816234

  • @jakub_et_al@mathstodon.xyz 2026-02-28 15:56

    @MartinEscardo @dwarn This thread forced me to finally learn some agda, and try to write some proofs. I am attempting to reprove another statement from Taylor's paper. If a topological space admits a majority operation ๐‘š satisfying ๐‘š(๐‘ฅ,๐‘ฅ,๐‘ฆ)=๐‘š(๐‘ฅ,๐‘ฆ,๐‘ฅ)=๐‘š(๐‘ฆ,๐‘ฆ,๐‘ฅ)=๐‘ฅ, then it is weakly-equivalent to a discrete set. Naturally, for homotopy types this would translate to three equations like ๐‘’โ‚ : ฮ (๐‘ฅ,๐‘ฆ : ๐ด) ๐‘š(๐‘ฆ,๐‘ฅ,๐‘ฅ)=๐‘ฅ ๐‘’โ‚‚ : ฮ (๐‘ฅ,๐‘ฆ : ๐ด) ๐‘š(๐‘ฅ,๐‘ฆ,๐‘ฅ)=๐‘ฅ ๐‘’โ‚ƒ : ฮ (๐‘ฅ,๐‘ฆ : ๐ด) ๐‘š(๐‘ฅ,๐‘ฅ,๐‘ฆ)=๐‘ฅ I managed to make ๐‘š act idempotently on the loop space, but encountered an interesting problem: There are a few ways how to prove that ๐‘š is idempotent, and it is not clear which choice to make! I would need something like ฮ (๐‘ฅ:๐ด) ๐‘’โ‚(๐‘ฅ,๐‘ฅ) = ๐‘’โ‚‚(๐‘ฅ,๐‘ฅ). But maybe this is a consequence of what we mean when we write the above majority identities.

    Open ##1816240