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nLab > Latest Changes: invariant set

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    • CommentRowNumber1.
    • CommentAuthorPaoloPerrone
    • CommentTimeJul 12th 2024

    Creating page for now, adding more content soon.

    v1, current

    • CommentRowNumber2.
    • CommentAuthorPaoloPerrone
    • CommentTimeJul 12th 2024

    Added most of the planned content.

    diff, v2, current

    • CommentRowNumber3.
    • CommentAuthorPaoloPerrone
    • CommentTimeJul 12th 2024

    Changed page name to singular.

    diff, v2, current

    • CommentRowNumber4.
    • CommentAuthorUrs
    • CommentTimeJul 12th 2024

    where it said that points need not be invariant, I made it read:

    the elements (“points”) of an invariant set need not themselves be fixed points

    diff, v3, current

    • CommentRowNumber5.
    • CommentAuthorzskoda
    • CommentTimeJul 12th 2024

    An analogue of an invariant open subset in a G-space in noncommutative algebraic geometry, where an action of a group is replaced by coaction of a Hopf algebra on an algebra “of functions” is the notion of coaction compatible localization and the functions on the space of orbits within the invariant subset are in that case replaced by localized coinvariants.

    • CommentRowNumber6.
    • CommentAuthorRodMcGuire
    • CommentTimeJul 12th 2024

    A subset SS of XX is called invariant if and only if for each m∈Mm\in M, m −1(S)=Sm^{-1}(S)=S. Equivalently, if for every x∈Xx\in X we have that x∈Sx\in S if and only if f(x)∈Sf(x)\in S.

    What is this unexplained ff in “f(x)∈Sf(x)\in S”? I guess that it is quantified to mean any mm but why don’t you just use mm?

    variants of the same definition which differ for the case of arbitrary monoids. In particular, sometimes one calls a subset S⊆XS\subseteq X invariant if and only if m −1(S)⊆Sm^{-1}(S)\subseteq S, i.e. if for every x∈Sx\in S, f(x)∈Sf(x)\in S as well. (Without requiring that if f(x)∈Sf(x)\in S, then x∈Sx\in S.) In other words, it is a set “which we cannot leave” under the specified action.

    this seems confused. For a subset SS closed under the action of a monoid the requirement is that for every s∈Ss\in S then f(s)∈Sf(s)\in S, which means that S⊆m −1(S)S\subseteq m^{-1}(S). There may be x∈Xx\in X with f(x)∈Sf(x)\in S without x∈Sx\in S.

    • CommentRowNumber7.
    • CommentAuthorPaoloPerrone
    • CommentTimeJul 12th 2024

    Oh, thank you, let me correct.

    • CommentRowNumber8.
    • CommentAuthorRodMcGuire
    • CommentTimeJul 12th 2024

    For a fixed monoid MM or group GG isn’t an “invariant set” just a subobject in the corresponding category of M-sets or category of G-sets?

    (I know little about those categories and not much about the different types of G-sets.)

    • CommentRowNumber9.
    • CommentAuthorPaoloPerrone
    • CommentTimeJul 13th 2024

    I believe that for monoids it should be exactly those sets SS for which if x∈Sx\in S, then m(x)∈Sm(x)\in S, which in the article are under “Variations on the definition”. (But I’m not an expert either.)

    • CommentRowNumber10.
    • CommentAuthorUrs
    • CommentTimeJul 13th 2024

    re #8: that certainly sounds right!

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