Loading ...
Sorry, an error occurred while loading the content.

free algebras with "transitive" automorphisms

Expand Messages
  • Luis Sequeira
    On behalf of a friend, I would like to ask the knowledgeable people on this list about the following problem: Is there a description of the varieties V (or a
    Message 1 of 4 , Dec 9, 2004
    • 0 Attachment
      On behalf of a friend, I would like to ask the knowledgeable people
      on this list about the following problem:

      Is there a description of the varieties V (or a known sufficient, or
      necessary condition for)
      which have the following property:

      For a free algebra F in V and x,y in F\constants, there is an
      automorphism of F that maps x to y.

      Thanks in advance

      Luis Sequeira
    • Fred E.J. Linton
      ... I m having trouble guessing how you mean . Are you asking that x and y in F be constants? If so, and if by constant you mean the sort of element
      Message 2 of 4 , Dec 17, 2004
      • 0 Attachment
        Luis Sequeira asked:

        > Is there a description of the varieties V (or a known sufficient, or
        > necessary condition for) which have the following property:
        >
        > For a free algebra F in V and x,y in F\constants, there is an
        > automorphism of F that maps x to y.

        I'm having trouble guessing how you mean "\".

        Are you asking that x and y in F "be" constants?

        If so, and if by "constant" you mean the sort of element that is
        the value of a nullary (or even constant unary) operation, then
        the fact that homomorphisms and operations "commute" entails that,
        no matter what the algebra F (free or otherwise), the existence of
        ANY endomorphism h: F --> F (automorphism or not) with h(x)=y
        ensures that x=y.

        If instead by "constant" you mean a free generator of F, then
        there will always be such automorphisms (just extend to a homorphism
        any permutation of the generators exchanging x with y).

        But if by "constant" you mean merely "element (of the carrier) of F"
        -- or if you're asking x and y to be in the complement in F of the constants
        in the sense described above -- then (apart from the free generator answer)
        I'm at a loss for anything to say, sorry.

        Cheers,

        -- Fred Linton
      • Keith A. Kearnes
        ... I think Luis means x, y in F MINUS the constants . For example, vector spaces are free and have the property that their automorphism groups act
        Message 3 of 4 , Dec 19, 2004
        • 0 Attachment
          > > For a free algebra F in V and x,y in F\constants, there is an
          > > automorphism of F that maps x to y.
          >
          > I'm having trouble guessing how you mean "\".
          >
          > Are you asking that x and y in F "be" constants?

          I think Luis means "x, y in F MINUS the constants".
          For example, vector spaces are free and have the property
          that their automorphism groups act transitively on the
          elements that are not named by constants.


          Yours,
          Keith

          --
          Keith A. Kearnes Email: kearnes@...
          Department of Mathematics WWW: http://spot.colorado.edu/~kearnes
          University of Colorado Direct Phone: (303) 492-5203
          395 UCB Dept Phone: (303) 492-3613
          Boulder, CO 80309-0395 Fax: (303) 492-7707
        • Keith A. Kearnes
          Dear Luis- I don t know the answer to your question, but have a few minor observations about the property: (*) For all free algebras F, Aut(F) acts
          Message 4 of 4 , Dec 19, 2004
          • 0 Attachment
            Dear Luis-

            I don't know the answer to your question, but
            have a few minor observations about the property:

            (*) For all free algebras F, Aut(F) acts transitively
            on F-{elements named by constants}.

            (I) A reformulation:
            (here automorphisms act on right)

            (*) holds for all free F iff it holds for all
            finitely generated free F. So fix an n, let F =
            F_V(x1,...,xn), and let \sigma be an automorphism
            defined by x(i)\alpha = x(i+1 mod n).
            For each automorphism \alpha, let t_{\alpha}(x1,...,xn)
            be an n-ary term that represents the element
            x1\alpha \in F. The following equations must hold in V:

            (i) t_1(x1,...,xn) = x1
            (ii) t_{\sigma^k}(x1,...,xn) = x(k+1)
            (iii) t_{\alpha\beta}(x1,...,xn) =
            t_{\alpha}(t_{\beta}(X),t_{\sigma\beta}(X),t_{\sigma^2\beta}(X),...)
            where (X)=(x1,...,xn).

            Conversely, if V has a set of terms satisfying these
            equations, where the subscripts are indexed by a group G,
            then the map \alpha |-> the endomorphism defined by
            xi |-> t_{\sigma^(k-1)\alpha}(X) is a homomorphism
            from G to Aut(F_V(X)).

            (*) is the condition that every element of F-{constants}
            lies in the Aut(F)-orbit of the element x1.
            In other words

            (**) each element of F is either named by a constant,
            or is equal to some t_{\alpha}.

            This does not eseem like an improvement at first, but it
            has the consequence that:

            (Ii) if V has property (*), then so does every subvariety of V.

            Another consequence is obtained from the above by letting n=1.
            Then the equations listed above show that

            (Iii) the nonconstant unary terms form a group under composition.
            -------------
            (II)
            The subalgebra of constants F_V(1) is a minimal
            algebra. If finite, it is isomorphic to the
            constant expansion of a vector space or G-set.

            This is because the polynomials of this algebra
            are the same as its unary terms, which are
            constant or permutations by observation (Iii).
            If finite, Palfy's Theorem forces this algebra
            of constants to be the constant expansion of a G-set,
            vector space, or a 2-element nonabelian algebra.
            The latter cases can be ruled out, because they
            don't generate varieties with (*).

            [Possibly you don't have to assume that there are only
            finitely many inequivalent constant terms to get
            this conclusion, if you make more use of (*).]
            --------------
            Observation (Ii) suggests that one first try to
            characterize the minimal varieties with (*).
            If such a minimal variety has finitely many
            distinct constant terms, then it can only have
            1 constant term up to equivalence. (Otherwise,
            observation (II) implies that the minimal variety
            must be generated by the constant expansion of
            a vector space or G-set, but such varieties
            are not minimal.)
            I haven't thought about having exactly 1 constant,
            but here is an observation about having 0 constants,
            assuming V is a locally finite minimal variety.

            (III)
            A minimal locally finite variety V with no constant terms
            has (*) iff it is term equivalent to
            (a) the variety of sets
            (b) a variety of affine spaces, or
            (c) a variety generated by a strictly simple
            quasiprimal algebra with transitive automorphism group.

            Reason: V must be generated by a strictly simple
            algebra whose operations are surjective (since
            for any operation f(x1,...,xm) the unary term
            f(x,x,...,x) is already surjective since nonconstant).
            But these strictly simple algebras have been classified.
            It can be argued that none of them generate varieties
            with (*) except those of type (a),(b) and (c).

            [Probably this argument can be modified to work for
            locally finite minimal varieties with some constants.]

            Yours,
            Keith

            ---
            Keith A. Kearnes Email: kearnes@...
            Department of Mathematics WWW: http://spot.colorado.edu/~kearnes
            University of Colorado Direct Phone: (303) 492-5203
            395 UCB Dept Phone: (303) 492-3613
            Boulder, CO 80309-0395 Fax: (303) 492-7707
          Your message has been successfully submitted and would be delivered to recipients shortly.