## free algebras with "transitive" automorphisms

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• 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
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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.

Luis Sequeira
• ... 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
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> 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
• ... 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
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> > 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@...
University of Colorado Direct Phone: (303) 492-5203
395 UCB Dept Phone: (303) 492-3613
Boulder, CO 80309-0395 Fax: (303) 492-7707
• 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
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Dear Luis-

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@...