- 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 - Luis Sequeira asked:

> Is there a description of the varieties V (or a known sufficient, or

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

> 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.

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 > > For a free algebra F in V and x,y in F\constants, there is an

I think Luis means "x, y in F MINUS the constants".

> > 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?

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- 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