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\begin{document}

\title{Growth Rates of Finite Algebras}
\author{Keith A. Kearnes}
\address[Keith Kearnes]{Department of Mathematics\\
University of Colorado\\
Boulder, CO 80309-0395\\
USA}
\email{Keith.Kearnes@Colorado.EDU}
\author{Emil W. Kiss}
\address[Emil W. Kiss]{E{\"o}tv{\"o}s University\\
Department of Algebra and Number Theory\\
1088 Budapest, M{\'u}zeum krt. 6--8\\
Hungary}
\email{ewkiss@cs.elte.hu}
\author{\'Agnes Szendrei}
\address[\'Agnes Szendrei]{Department of Mathematics\\
University of Colorado\\
Boulder, CO 80309-0395\\
USA}
\email{Agnes.Szendrei@Colorado.EDU}
\subjclass{}
\keywords{}

\begin{abstract}
We investigate the growth sequences of finite algebras.

***
\end{abstract}

\maketitle

\section{Introduction}\label{intro_sec}
For a finite algebra $\m a$, write $d_{\m a}(n)=g$
if $g$ is the least size of a generating set
for $\m a^n$, and write $h_{\m a}(g)=n$ if the
largest power of $\m a$ that is $g$-generated is $\m a^n$.
These functions map positive integers
to positive integers and satisfy the bi-implication
\[
d_{\m a}(n) \leq g \wrel{\Longleftrightarrow} n \leq h_{\m a}(g),
\]
which asserts that $d_{\m a}$ is the lower adjoint of $h_{\m a}$
and $h_{\m a}$ is the upper adjoint of $d_{\m a}$.
It follows that $d_{\m a}, h_{\m a}\colon \mathbb Z^+\to\mathbb Z^+$ 
are monotone functions,
which are inverse bijections between their images:
\[
\im(d_{\m a}) \xrightleftharpoons[d]{h} \im(h_{\m a});
\]
and, moreover, each determines the other.

In the literature, the notation is slightly different.
The least size of a generating
set for $\m a$ is denoted $d(\m a)$. 
The \emph{growth sequence} for $\m a$
is then defined to be the sequence
\[
\dd(\m a) := (d(\m a), d(\m a^2), d(\m a^3), \ldots).
\]
One defines $h(g,\m a)$ to be the largest $n$ such that 
$\m a^n$ is $g$-generated. 
The relationship between the two notations
is $\dd(\m a) = (d_{\m a}(1),d_{\m a}(2),\ldots)$ and 
$h(g,\m a) = h_{\m a}(g)$. 

The $h$ function was the first to be introduced into mathematics,
by Philip Hall in his 1936 paper \cite{hall}. Hall proves that if $A$
is a finite, simple, nonabelian group, then 
\[
h_{A}(g) = \frac{1}{|\Aut(A)|}\sum_{H\leq A} \mu(H)|H|^g,
\]
where $\mu$ is the M\"obius function of the subgroup
lattice of $A$. 
A formula for the $d$ function
of a finite, simple, nonabelian group was later given
by James Wiegold in his 1974 paper***CHECK*** \cite{wiegold1}: for large $n$,
$d_{\m a}(n)$ is one of the two or three integers
closest to 
\[
\log_{|A|}(n) + \log_{|A|}(|\Aut(A)|).
\]
Wiegold's paper initiated a program of research
into the growth sequences (or $d$ functions) of groups
\cite{erfanian1,erfanian2,erfanian3,erfanian4,erfanian5,
erfanian_rezaei, erfanian_wiegold, 
kimmerle, lennox_wiegold, meier_wiegold, 
obraztsov, stewart_wiegold, wiegold2, wiegold3, wiegold4, 
wiegold_wilson,wise}.
\marginpar{When paper \\ is done\\ replace some\\ with range a--b}
The program later expanded to include the 
investigation of growth sequences of semigroups, in
\cite{pollak, wiegold_sem}, and later to 
include the investigation of the
growth sequences of arbitrary algebraic structures,
in \cite{glass_riedel, riedel}. Our own investigations
into growth sequences of finite algebras,
on which we are reporting here,
was stimulated
by the recent appearance of \cite{quick_ruskuc}
and by the related preprint \cite{hyde_etal}.

Some of the key open questions about growth rates of
finite algebras are related to the following
theorems of Wiegold:
\begin{enumerate}
\item[(I)] 
A finite perfect group\footnote{$G$ is perfect
if $[G,G]=G$.}
 has a growth sequence 
that is logarithmic ($d_{\m a}(n)\in \Theta(\log(n))$),
while a finite imperfect group has 
a growth sequence that is linear ($d_{\m a}(n)\in \Theta(n)$),
\cite{wiegold3}.
\item[(II)] 
A finite semigroup with identity
 has a growth sequence that is logarithmic or linear, while a finite
semigroup without identity has
a growth sequence that is exponential 
($d_{\m a}(n)\in 2^{\Theta(n)}$), \cite{wiegold_sem}.
% Note: Quick and Ruskuc omit the logarithmic case when citing 
% (II) (page 2 of their preprint). 
\end{enumerate}

Herbert Riedel partially extended item (I) to 
congruence uniform varieties in \cite{riedel}
by proving that finite perfect algebras
in such varieties have logarithmic growth sequences.
He adds that \emph{``Unfortunately, Wiegold's results 
for the imperfect case do not hold even for
modules.''} It is not clear what this sentence means.
It is not very difficult to show
that any finite imperfect algebra in a congruence uniform 
variety has linear growth, so either Riedel's remark is a mistake
or he means that Wiegold's sharp estimate on the
leading coefficient of the linear growth function
is not valid for every finite module. 

The recent paper \cite{quick_ruskuc} extends item (I)
to any variety of rings, modules, $k$-algebras
or Lie algebras, but also falls short of extending the result
to arbitrary congruence uniform varieties.

Riedel asks in \cite{riedel} 
whether his result about perfect algebras can be extended 
from congruence uniform varieties to 
congruence permutable varieties. 
Our work began in response to this question,
and more generally in response to the question of
whether ``Wiegold dichotomy'' (item (I) from above)
holds for arbitrary congruence permutable varieties. 
We shall see that it does, and in fact that
Wiegold dichotomy holds more generally
for finite algebras having
``few subpowers'', a concept from \cite{bimmvw}.

Concerning item (II), Martyn Quick and Nikola Ru\v skuc
write in \cite{quick_ruskuc} that 
\emph{``At present no finite algebraic structure is known
for which the $d$--sequence does not have 
one of logarithmic, linear or exponential growth.''}
In this paper we shall exhibit such algebras, namely
we shall construct finite algebras having 
$d_{\m a}(n)\in\Theta(n^m)$ for any given $m\in \mathbb Z^+$.

Our main results are most easily phrased
in terms of linear Maltsev conditions.
%Let $\mathcal F$ be a set of functions 
%from $\mathbb Z^+$ to $\mathbb Z^+$.
If $\mathcal L$ is a set of linear Maltsev conditions,
then we say that $\mathcal L$ \emph{enforces} growth sequences 
%in $\mathcal F$ 
of some particular type
if any finite
algebra generating a variety satisfying the Maltsev conditions
in $\mathcal L$ has growth sequence 
%in $\mathcal F$.
of that particular type.
For example, $\mathcal L$
enforces logarithmic growth sequences if any finite
algebra generating a variety satisfying 
$\mathcal L$ has a logarithmic growth sequence.

If $\mathcal L$
enforces no restrictions on growth sequences,
we call $\mathcal L$ \emph{generic}.
That is, $\mathcal L$ is generic if it is the case
that for every finite algebra $\m a$ there is
a finite algebra $\m b$ that generates a variety
satisfying the Maltsev  conditions
in $\mathcal L$ such that $d_{\m b}=d_{\m a}$.

If any finite
algebra generating a variety satisfying
$\mathcal L$ has growth sequence that is logarithmic
or linear we say that $\mathcal L$ enforces the 
\emph{logarithmic/linear dichotomy}.
If any finite
algebra generating a variety satisfying
$\mathcal L$ has growth sequence that is logarithmic, then
we say that $\mathcal L$ enforces the 
\emph{logarithmic monochotomy}.\footnote{%
The terms ``dichotomy'' and ``monochotomy''
are meant to apply to the division of growth sequences
into distinct asymptotic growth classes, but 
deliberately ignores the fact that the $1$-element
algebra in a variety has a constant growth sequence, which
represents yet another asymptotic growth class.}
If $\mathcal L$ can be satisfied only by 
varieties of 1-element algebras, then we call $\mathcal L$
\emph{inconsistent}, otherwise it is \emph{consistent}.
Our main results are:
\begin{enumerate}
\item[(1)] Any consistent set of linear Maltsev conditions
can be strengthened to a consistent set of linear Maltsev 
conditions that enforces the logarithmic monochotomy.
\item[(2)]
A consistent set of linear Maltsev conditions enforces the
logarithmic monochotomy if and only if it entails the existence
of a near unanimity term.
\item[(3)] Any set of linear Maltsev conditions
that does not enforce the logarithmic monochotomy
can be strengthened to a set of linear Maltsev 
conditions that enforces the logarithmic/linear dichotomy.
\item[(4)]
A consistent set $\mathcal L$ of linear Maltsev conditions enforces
the logarithmic/linear dichotomy if and only if it entails the existence
of a cube term (or edge term or parallelogram term).
In this situation, a finite algebra generating a 
variety satisfying $\mathcal L$ will have logarithmic growth rate
if it is perfect and linear growth rate if it is imperfect.
\item[(5)] Any consistent set of linear Maltsev conditions
that does not enforce the logarithmic/linear dichotomy
is generic.
\end{enumerate}

Although (5) is easy to prove, it is the most striking
of these statements. It implies that any function
that is the growth rate of some finite algebra must arise
as the growth rate of a finite algebra in 
a variety that is (for example)
congruence distributive
and congruence 3-permutable.
But (5) does \emph{not} imply that if $\mathcal V$
is a variety whose finite algebras fail the 
logarithmic/linear dichotomy, then every
growth rate of a finite algebra
arises as the growth rate of a finite algebra
in $\mathcal V$. (The variety of semigroups
refutes this, according to (II) above.) Instead, what (5) means is that 
if $\mathcal V$ fails to have a cube term
and $\m a$ is a finite algebra of any type,
then there is another variety $\mathcal W$
that satisfies all the linear Maltsev conditions true in $\mathcal V$,
which has a finite algebra $\m b$ such that 
$d_{\m b} = d_{\m a}$.

Our results require a careful analysis of 
the structure of maximal subalgebras
of direct powers. Some of those results are interesting
for their own sake. (REFERENCES???)


\section{Preliminaries}\label{prelim_sec}

\subsection{Notation}\label{notation}

In this paper, $[n]$ denotes the set $\{1,\ldots,n\}$.
Structures are denoted in bold face font,
e.g. $\m a$, while the universe of a structure
is denoted by the same character 
in italic font, e.g., $A$. 
The subuniverse of $\m a$ generated by a subset $G\subseteq A$
is denoted $\lb G\rb$.

We will use Big Oh notation.
If $f$ and $g$ are real-valued functions defined
on some subset of the real numbers, then 
$f\in O(g)$ and $f=O(g)$ both mean that there are constants
$M$ and $N$ such that $|f(x)|\leq M|g(x)|$ for all $x>N$.
We write $f\in \Omega(g)$ and $f=\Omega(g)$ to mean that there are constants
$M$ and $N$ such that $|f(x)|\geq M|g(x)|$ for all $x>N$.
Finally, $f\in \Theta(g)$ and $f=\Theta(g)$ mean that both
$f\in O(g)$ and $f\in \Omega(g)$ hold.

\subsection{Easy estimates}\label{estimates}

\begin{thm}\label{basic_estimates}
Let $\m a$ be a finite algebra.
\begin{enumerate}
\item[(1)] $d_{\m a^k}(n) = d_{\m a}(kn)$.
\item[(2)] If $\m b$ is a homomorphic image of 
$\m a$, then $d_{\m b}(n) \leq d_{\m a}(n)$.
\item[(3)] If $\m b$ is an expansion of 
$\m a$ (equivalently, if $\m a$ is a reduct of $\m b$), 
then $d_{\m b}(n) \leq d_{\m a}(n)$.
\item[(4)] {\rm (From \cite{quick_ruskuc}.)}
If $\m b$ is the polynomial expansion
of $\m a$, then 
\[
d_{\m a}(n) - d_{\m a}(1)\leq d_{\m b}(n) \leq d_{\m a}(n).
\]
\end{enumerate}
\end{thm}

\begin{proof}
For $(1)$, both 
$d_{\m a^k}(n)$ and $d_{\m a}(kn)$
count the number of elements in a smallest
generating set for $(\m a^k)^n\cong \m a^{kn}$.

For $(2)$, if $\varphi\colon \m a\to \m b$ is surjective and
$G\subseteq A^n$ is a smallest generating set for $\m a^n$,
then $\varphi(G)$ is a generating set for $\m b^n$.
Hence $d_{\m b}(n)\leq |G| = d_{\m a}(n)$.

For $(3)$, if 
$G\subseteq A^n$ is a smallest generating set for $\m a^n$,
then $G$ is also a generating set for $\m b^n$.
Hence $d_{\m b}(n)\leq |G| = d_{\m a}(n)$.

For $(4)$, the right-hand
inequality $d_{\m b}(n) \leq d_{\m a}(n)$ follows from $(3)$.
Now let $G\subseteq A^n$ be a smallest generating set
for $\m b^n$ and let $H\subseteq A$ be a smallest generating set for $\m a$.
For each $a\in H$ let $\hat{a} = (a,a,\ldots,a)\in A^n$ be the associated
constant tuple, and let $\widehat{H}$ be the set of these.
Every tuple of $A^n$ is generated from $G$ by polynomials
of $\m a$ acting coordinatewise, hence is generated from 
$G\cup \widehat{H}$ by term operations of $\m a$ acting
coordinatewise. This proves
$
d_{\m a}(n) \leq |G| + |H| = d_{\m b}(n) + d_{\m a}(1),
$ 
from which the left-hand inequality follows.
\end{proof}

\begin{thm}\label{first_bounds}
If $\m a$ is a finite algebra of
more than one element, then 
\[
\lceil \log_{|A|}(n) \rceil\leq d_{\m a}(n) \leq |A|^n
\]
and
\[
\lfloor \log_{|A|}(n) \rfloor\leq h_{\m a}(n) \leq |A|^n.
\]

Moreover,
\begin{enumerate}
\item[(1)] $d_{\m a}(n) \in O(\log(n))$ iff 
$h_{\m a}(n) \in 2^{\Omega(n)}$.
\item[(2)] $d_{\m a}(n) \in O(n)$ iff 
$h_{\m a}(n) \in \Omega(n)$, and 
$d_{\m a}(n) \in \Omega(n)$ iff 
$h_{\m a}(n) \in O(n)$.
\item[(3)] $d_{\m a}(n) \in 2^{\Omega(n)}$ iff
$h_{\m a}(n) \in O(\log(n))$.
\end{enumerate}
\end{thm}

\begin{proof}
It follows from Theorem~\ref{basic_estimates}~(3) that,
among all algebras with universe $A$, the algebra equipped with
no operations has the smallest $d$-function and the
algebra equipped with all finitary operations has
the largest $d$-function. These two algebras are
also extremes for the $h$-function.

If $\m a$ has no operations, then every element of $A^n$
is a required generator, so $d_{\m a}(n) = |A|^n$.
In this case, $h_{\m a}(n)= \lfloor \log_{|A|}(n)\rfloor$,
since $h$ is the upper adjoint of $d$.

Now assume that $\m a$ is equipped with all finitary operations,
i.e., $\m a$ is primal.
The $n$-generated free algebra in the variety generated by $\m a$
is isomorphic to $\m a^{|A|^n}$ (Theorem 3 of \cite{foster}).
Since the largest $n$-generated algebra in this variety
is a power of $\m a$, it is also the largest $n$-generated 
power of $\m a$ in the variety; we obtain that
$h_{\m a}(n) = |A|^n$. 
In this case, $d_{\m a}(n)= \lceil \log_{|A|}(n)\rceil$,
since $d$ is the lower adjoint of $h$.

The fact that $d_{\m a}$ is the lower adjoint 
of $h_{\m a}$ suggests an asymmetry, in that
\begin{equation}\label{adjoint1}
d_{\m a}(n)\leq k\Longleftrightarrow n\leq h_{\m a}(k),
\end{equation}
relates an upper bound of $d_{\m a}$ to a lower bound
of $h_{\m a}$. But the fact that these functions
are defined between totally ordered sets allows us to
rewrite (\ref{adjoint1}) as
\begin{equation}\label{adjoint2}
h_{\m a}(k) < n\Longleftrightarrow k < d_{\m a}(n),
\end{equation}
which almost exactly reverses condition (\ref{adjoint1}) on 
$d_{\m a}$ and $h_{\m a}$.
Using this fact and the following claim, 
one easily verifies items (1)--(3).

\begin{clm}
If $f, g\colon [a,\infty)\to \mathbb R$ are increasing
functions that tend to infinity as $x$ tends to infinity,
then 
$\lfloor f(n)\rfloor < d_{\m a}(n)\leq \lceil g(n)\rceil$ 
holds for all large $n$
iff $\lfloor g^{-1}(n)\rfloor \leq h_{\m a}(n) < \lceil f^{-1}(n)\rceil$ 
holds for all large $n$.
\end{clm}

Allow ``$\forall N$'' to stand for ``for all large $n$''. We have
\[
\begin{array}{rl}
\forall N(d_{\m a}(n)\leq \lceil g(n)\rceil) 
&\Longrightarrow\;\;
\forall N(n\leq h_{\m a}(\lceil g(n)\rceil))\\
&\Longrightarrow\;\;
\forall N(\lfloor g^{-1}(n)\rfloor \leq 
h_{\m a}(\lceil g(\lfloor g^{-1}(n)\rfloor)\rceil))\\
&\Longrightarrow\;\;
\forall N(\lfloor g^{-1}(n)\rfloor \leq h_{\m a}(n)),
\end{array}
\]
because the monotonicity of $g$ guarantees that 
$\lceil g(\lfloor g^{-1}(n)\rfloor)\rceil))\leq n$.
The reverse implication is proved the same way, as are 
both implications in 
$\lfloor f\rfloor < d \Leftrightarrow h < \lceil f^{-1}\rceil$.
\end{proof}

%The claims of Theorem~\ref{first_bounds} can be found in 
%\cite{quick_ruskuc}, but they are not justified correctly there.
%(Specifically, their formula (2.4) is false.)

Recall that the \emph{free spectrum} of a variety
$\mathcal V$ is the function $f_{\mathcal V}(n):=|F_{\mathcal V}(n)|$
whose value at $n$ is the cardinality of the
$n$-generated free algebra in $\mathcal V$.

\begin{thm}\label{spec1}
If $\m a$ is a finite algebra and $f_{\mathcal V}$
is the free spectrum of the variety $\mathcal V = {\mathcal V}(\m a)$,
then 
$
%|A|^{h_{\m a}(n)}\leq f_{\mathcal V}(n).
h_{\m a}(n)\leq \log_{|A|}(f_{\mathcal V}(n)).
$
In particular,
\begin{enumerate}
\item[(1)] 
if $f_{\mathcal V}(n)\in O(n^m)$ for some fixed $m\in\mathbb Z^+$, 
then $d_{\m a}(n)\in 2^{\Theta(n)}$;
\item[(2)] 
if $f_{\mathcal V}(n)\in 2^{O(n)}$,
then $d_{\m a}(n)\in \Omega(n)$.
\end{enumerate}
\end{thm}

\begin{proof}
The algebra $\m a^{h_{\m a}(n)}$ is $n$-generated, hence
a quotient of the $n$-generated free algebra
$\m f_{\mathcal V}(n)$. This proves that 
$|A|^{h_{\m a}(n)}\leq f_{\mathcal V}(n)$, or 
$h_{\m a}(n)\leq \log_{|A|}(f_{\mathcal V}(n))$.

If $f_{\mathcal V}(n)\in O(n^m)$ for some fixed $m\in\mathbb Z^+$, 
then $\log(f_{\mathcal V}(n))\in O(\log(n))$,
hence $h_{\m a}(n)\in O(\log(n))$. It follows from 
Theorem~\ref{first_bounds}~(3) that 
$d_{\m a}(n)\in 2^{\Theta(n)}$.

If $f_{\mathcal V}(n)\in 2^{O(n)}$, then
$\log(f_{\mathcal V}(n))\in O(n)$,
hence $h_{\m a}(n)\in O(n)$. It follows from 
Theorem~\ref{first_bounds}~(2) that 
$d_{\m a}(n)\in \Omega(n)$.
\end{proof}

\begin{cor}
Let $\m a$ be a finite algebra and let $\m b$
be a homomorphic image of $\m a^k$ for some $k$.
\begin{enumerate}
\item[(1)] 
If $\m b$ is strongly abelian (or even just strongly rectangular), 
then $d_{\m a}(n)\in 2^{\Theta(n)}$.
\item[(2)] 
If $\m b$ is abelian, then $d_{\m a}(n)\in \Omega(n)$.
\end{enumerate}
\end{cor}

\begin{proof}
For (1), Theorem~5.3
of \cite{strnil} proves that a finite strongly rectangular
algebra generates a variety with free spectrum bounded
above by a polynomial.
By Theorem~\ref{spec1}, $d_{\m a}(n)\in 2^{\Theta(n)}$
in this case. The strong abelian property 
is more restrictive than the strong rectangular
property by Lemma~2.2~(11) of \cite{strnil}.

For (2), any finite abelian algebra generates a variety
$\mathcal V$ 
whose free spectrum satisfies 
$f_{\mathcal V}(n)\in 2^{O(n)}$, according to \cite{berman_mck},
so Theorem~\ref{spec1}~(2) completes the argument.
\end{proof}

\begin{thm}
Let $\m a$ be a finite algebra of more than one element.
\begin{enumerate}
\item[(1)] If $\m a$ is an affine algebra, then 
$d_{\m a}(n)\in \Theta(n)$.
\item[(2)] If $\m a$ is a \emph{simple} affine algebra, then 
$r(n-1)\leq h_{\m a}(n) \leq  rn + 1$, where 
$r = \log_{|A|}(R)$ and $R$ is the associated ring.
\end{enumerate}
\end{thm}

\begin{proof}
Is this really needed?
\end{proof}

\subsection{Maltsev conditions}\label{maltsev_subsection}

Say that a set $\Sigma$ of identities is \emph{realizable} 
in a variety $\mathcal V$ of $\mathcal L$-algebras 
if $\Sigma$ holds in the algebras of $\mathcal V$ 
when the function symbols occurring in $\Sigma$ are 
interpreted as $\mathcal L$-terms. 
A class ${\mathcal K}$ of varieties is 
\emph{definable by a strong Maltsev condition}
if there is a finite set $\Sigma$ such that 
a variety $\mathcal V$ belongs to 
$\mathcal K$ if and only if $\Sigma$
is realizable in $\mathcal V$.
(A \emph{strong Maltsev condition }
is an assertion that a given finite $\Sigma$
is realizable, and the strong Maltsev condition
is \emph{satisfied}
by those varieties which realize it.)

For example, it is a result of 
A.~I.~Maltsev from \cite{maltsev} that  the class of
varieties whose members have permuting congruences
is definable by the realizability of the set 
$\Sigma$ consisting of the following two identities:
\[
M(x,y,y) \approx x\qquad\textrm{and}\qquad 
M(y,y,x) \approx x, 
\]
which may be written succinctly as
\begin{equation}\label{m}
{\mathcal V}\models 
{M}\left(\begin{matrix}
x&y&y\\
y&y&x
\end{matrix}
\right) \approx
\left(\begin{matrix}
x\\
x
\end{matrix}
\right).\footnote{Here and below a matrix identity
is to be read along rows, and represents a sequence
of row identities.}
\end{equation}
A realization of (\ref{m}) in a variety $\mathcal V$
is called a \emph{Maltsev term} for $\mathcal V$.

Bjarni J\'onsson in \cite{jonsson} 
proved that $\mathcal V$ has 
the property that all members have distributive 
congruence lattices if and only if, for some $n\geq 1$, $\mathcal V$ has a 
a sequence of terms $d_1,\ldots,d_{n}$
such that the following identities hold in $\mathcal V$:
\begin{enumerate}
\item[(1)] $d_1(x,x,y)\approx x$; 
\item[(2)] $d_i(x,y,x)\approx x$ for all $i$;
\item[(3)] $d_i(x,x,y)\approx d_{i+1}(x,x,y)$ for $i$ even; 
\item[(4)] $d_i(x,y,y)\approx d_{i+1}(x,y,y)$ for $i$ odd; and
\item[(5)] $d_{n}(x,x,y)\approx y$ if $i$ is even and 
$d_{n}(x,y,y)\approx y$ if $i$ is odd.
\end{enumerate}
These terms are now called \emph{J\'onsson terms}.
(To simplify Example~\ref{example1}, we have written
these identities using only two variables.)

A class $\mathcal K$ of varieties is definable by an ordinary
\emph{Maltsev condition} if there is 
a sequence of sets of identities, $\Sigma_n$, which get
weaker as $n$ increases, such that 
a variety $\mathcal V$ belongs to $\mathcal K$
if and only if some $\Sigma_n$ is realizable in $\mathcal V$.
The class of varieties whose members
have distributive congruence lattices 
is definable by a Maltsev condition,
according to J\'{o}nsson's result. (In fact,
J\'onsson's result was the model for the definition
of ``Maltsev condition''.)

A term is \emph{linear} (or \emph{simple})
if it is a variable or a function symbol applied to variables.
An identity $s\approx t$ is linear if the terms
on both sides are linear. A (strong) Maltsev condition is 
linear if all identities in the condition are linear.
As examples, the assertion that a variety has a 
Maltsev term is a linear
strong Maltsev condition, and the assertion that a variety
has J\'onsson terms is a linear Maltsev condition.

There are two other linear Maltsev conditions 
that play an essential role in this paper.
A $k$-ary \emph{near unanimity} for a variety
$\mathcal V$ is a $k$-ary term ${\sf U}$ such that:
\[
{\mathcal V}\models 
{\sf U}\left(\begin{matrix}
y&x&x&\cdots&x&x\\
x&y&x&&x&x\\
x&x&y&&x&x\\
\vdots&&&\ddots&&\vdots\\
x&x&x&&y&x\\
x&x&x&\cdots&x&y
\end{matrix}
\right) \approx
\left(\begin{matrix}
x\\
x\\
x\\
\vdots\\
x\\
x
\end{matrix}
\right).
\]


Next we define cube terms.
For any subset $U\subseteq [k]$
let $\chi_U\colon k\to \{x,y\}$ denote 
the function satisfying $\chi_U(i)=y$ if $i\in U$
and $\chi_U(i)=x$ if $i\notin U$.

\begin{df}\label{cube} {\rm \cite{bimmvw}}
Let $\mathcal V$ be a variety, and let 
$\m F = \m F_{\mathcal V}(x,y)$ be the
free $\mathcal V$-algebra over $\{x,y\}$.
A {\em $k$-cube term} for 
$\mathcal V$ for $k>1$ is a
%$(2^k-1)$-ary 
$\mathcal V$-term ${\sf C}$ for which 
$ 
{\sf C}^{\m F^k}\left( \langle \chi_U\rangle_{U\neq \emptyset} \right) 
= \chi_{\emptyset}.
$
%holds in $\m F^k$.
\end{df}
In words, $\sf C^{\m F^k}$ applied to the characteristic functions
of nonempty subsets of $k$ produces the characteristic
function of the empty set. We do not specify the
order of the characteristic functions,
so we consider any term obtained from $\sf C$ by permutation
of variables to be a $k$-cube term also.

For example, a $2$-cube term is a term $\sf C$ such that 
\begin{equation}\label{2-cube}
{\sf C}^{\m F^2}\left(\chi_{\{1\}},  
\chi_{\{1,2\}},  \chi_{\{2\}}\right) = 
\chi_{\emptyset}.
\end{equation}
%holds in $\m F^2$.
If we write characteristic functions as column vectors,
$\left(\chi_U(1),\chi_U(2),
\ldots,\chi_U(k)\right)^\transpose$, 
then (\ref{2-cube}) assumes a more familiar form
(compare with (\ref{m})):
\begin{equation}\label{2-cube2}
{\sf C}^{\m F^2}
\left(
\begin{matrix}
x & y & y\\
y & y & x
\end{matrix}
\right) = 
\left(
\begin{matrix}
x\\
x
\end{matrix}
\right).
\end{equation}
Since the rows of (\ref{2-cube2})
express equations relating the free generators
of $\m F_{\mathcal V}(x,y)$, these equations
hold as identities throughout $\mathcal V$.
Hence, a $2$-cube term for $\mathcal V$ is 
just a Maltsev term.
A $3$-cube term is something more general:
$${\sf C}^{\m F^3}
\left(
\chi_{\{1\}},
\chi_{\{2\}},
\chi_{\{3\}},
\chi_{\{1,2\}},
\chi_{\{1,3\}},
\chi_{\{2,3\}},
\chi_{\{1,2,3\}}
\right) = 
\chi_{\emptyset},$$ 
or equivalently
\begin{equation}\label{3-cube}
{\mathcal V}\models {\sf C}\left(
\begin{matrix}
y&x&x&y&y&x&y\\
x&y&x&y&x&y&y\\
x&x&y&x&y&y&y
\end{matrix}
\right) = 
\left(
\begin{matrix}
x\\
x\\
x
\end{matrix}
\right).
\end{equation}

The statement that a variety has a $k$-cube term for some
$k$ is a linear Maltsev condition. It is shown in \cite{bimmvw}
that a finite algebra $\m a$ has a cube term if and only
if the enumeration function for the
subalgebras of powers of $\m a$
grows slower than a doubly exponential function of $n$.

\subsection{A decision procedure for cube terms, I}\label{decision1}

In this subsection we present a procedure to decide whether
an arbitrary linear strong Maltsev condition entails
the existence of a cube term.
We will prove that the procedure is adequate in Subsection~\ref{decision2},
after the necessary construction becomes available,
but we find it useful to introduce the terminology here.

Before getting to the decision procedure, we state 
Kelly's Completeness
Theorem for linear identities, \cite{kelly}.\footnote{Kelly proved
the theorem for \emph{basic identities} in place of linear identities.
This is more general. Refer to the discussion in Section~\ref{discussion}.}
For this theorem, 
let $\mathcal L$ be an algebraic language,
let $\Sigma$ be a set of linear 
$\mathcal L$-identities, and
let $X$ be a set of variables 
containing all variables 
occurring in $\Sigma$.
%Let $\mathcal T$ be the set of linear $\mathcal L$-terms
%in the variables $X$. 
The \emph{Kelly closure} of $\Sigma$ 
with respect to $X$ is defined to be 
the smallest set $\overline{\Sigma}$ 
of linear identities obtained from $\Sigma$ by the following rules:
\begin{enumerate}
\item[(i)]
$\Sigma\subseteq \overline{\Sigma}$;
\item[(ii)] $(t\approx t)\in \overline{\Sigma}$ for all linear 
$\mathcal L$-terms $t$ in the variables $X$;
\item[(iii)] if $(s\approx t)\in \overline{\Sigma}$, then 
$(t\approx s)\in \overline{\Sigma}$;
\item[(iv)] if $(r\approx s)\in \overline{\Sigma}$ and
$(s\approx t)\in \overline{\Sigma}$, then 
$(r\approx t)\in \overline{\Sigma}$; and
\item[(v)] if $(s\approx t)\in \overline{\Sigma}$
and $f\colon X\to X$ is a function, then 
$(s[f]\approx t[f])\in \overline{\Sigma}$, where 
$s[f]$ denotes the linear term obtained from 
$s$ by replacing each variable $x\in X$ with $f(x)\in X$.
\end{enumerate}

Observe that the Kelly closure of $\Sigma$
may be constructed by first applying variable
replacement (items (i) and (v)) to $\Sigma$ in all possible ways,
and then generating an equivalence relation on 
the set of linear $\mathcal L$-terms in $X$ (items (ii)--(iv))
with the resulting pairs. The generated
equivalence relation will be closed under (i)--(v).
\marginpar{Should we say \\
``v.\ replacement''\\
or ``v.\ substitution''?}

\begin{thm}
Let notation be as explained above.
\begin{enumerate}
\item[(1)] Let $X$ be any set of variables 
containing all variables occurring in $\Sigma$.
$\Sigma$ is inconsistent (i.e., has only
$1$-element models) if and only if 
there exist distinct $x, y\in X$ such that
$(x\approx y)\in \overline{\Sigma}$.
\item[(2)] 
Suppose $\Sigma\cup\{\varphi\}$ consists of linear 
$\mathcal L$-identities, and $X$ is any set
of variables containing all variables
occurring in these identities.
Then $\Sigma\models \varphi$
(meaning that every model of $\Sigma$ is a model of $\varphi$)
if and only if
\begin{enumerate}
\item[(i)] $\Sigma$ is inconsistent; or
\item[(ii)] $\varphi\in \overline{\Sigma}$.
\end{enumerate}
\end{enumerate}
\end{thm}

\marginpar{Use provability\\
instead of consequence?}
\begin{df}
Let $\mathcal L$ be an algebraic language, and let 
$\Sigma$ be a set of $\mathcal L$-identities.
Let $t$ be an $\mathcal L$-term.
Call $t$ \emph{idempotent relative to $\Sigma$} if 
\[
\Sigma\models t(x,x,\ldots,x)\approx x.
\]
Call $t$ a \emph{$k$th-variable pseudoprojection relative
to $\Sigma$} if 
\[
\Sigma\models t(x_{i_1},x_{i_2},\ldots,x_{i_n})\approx x_j \qquad
\textrm{implies} \qquad j=i_k.
\]
\end{df}

\begin{thm}\label{decide_cube}
Let $\mathcal L$ be an algebraic language and 
let $\Sigma$ be a set of linear $\mathcal L$-identities.
\begin{enumerate}
\item[(1)] 
Suppose that, for every function symbol $F$ of $\mathcal L$,
if $F(x_1,\ldots,x_n)$ is idempotent relative to $\Sigma$, then 
$F(x_1,\ldots,x_n)$ is a $k$th-variable pseudoprojection for some $k$.
Then the variety axiomatized by $\Sigma$ has no cube term.
\item[(2)]
Suppose that, for some function symbol $F$ of $\mathcal L$,
$F(x_1,\ldots,x_n)$ is idempotent relative to $\Sigma$, but
$F(x_1,\ldots,x_n)$ is not a $k$th-variable pseudoprojection for any $k$.
Then $F(x_1,\ldots,x_n)$ is a cube term for the variety
axiomatized by $\Sigma$.
\end{enumerate}
\end{thm}

This theorem provides a procedure for deciding
if a linear strong Maltsev condition entails the 
existence of a cube term, namely, if $\Sigma$
is a finite set of linear identities 
defining a strong Maltsev condition
and some
function symbol in $\Sigma$ is not a pseudoprojection
relative to $\Sigma$, then it is a cube term;
otherwise $\Sigma$ does not entail the existence of a cube term.
In the latter case, a witness for the fact that
$\Sigma$ does not entail the existence of a cube term
is the variety axiomatized by $\Sigma$;
it satisfies the associated Maltsev condition,
but has no cube term.

\begin{exmp}\label{example1}
Let's apply the decision procedure to
the sequence of J\'onsson terms for $n=2$
to show that the associated Maltsev condition
does not entail the existence of a cube term.

Here our language has two ternary function
symbols, $d_1$ and $d_2$, and
our set of variables is $X=\{x,y\}$.
There are 18 linear terms using these symbols.
The equivalence classes under Kelly provability are:
\begin{enumerate}
\item[(i)] 
$\{x, d_1(x,x,x), d_1(x,y,x), d_1(x,x,y), 
d_2(x,x,x), d_2(x,y,x), d_2(y,y,x)\}$; 
\item[(ii)]
$\{y, d_1(y,y,y), d_1(y,x,y), d_1(y,y,x), 
d_2(y,y,y), d_2(y,x,y), d_2(x,x,y)\}$; 
\item[(iii)] 
$\{d_1(x,y,y), d_2(x,y,y)\}$; 
\item[(iv)] 
$\{d_1(y,x,x), d_2(y,x,x)\}.$
\end{enumerate}

The fact that $d_1(x,x,x)$ and $d_2(x,x,x)$ both appear in 
the equivalence class of $x$ implies that 
both $d_1$ and $d_2$ are idempotent relative to $\Sigma$.
By examining the equivalence class of $x$ further, we see that
$d_1$ is a first variable pseudoprojection
relative to $\Sigma$ (since $x_{i_1}=x$ for every 
$d_1(x_{i_1}, x_{i_2}, x_{i_3})$ in the equivalence class of $x$). 
Similarly, 
$d_2$ is a third variable pseudoprojection
relative to $\Sigma$. Theorem~\ref{decide_cube} asserts
that this is enough to conclude
that the variety axiomatized by the identities
of a sequence of two J\'{o}nsson terms does not
have a cube term.
\end{exmp}

\section{Examples}\label{examples}

This section houses our results concerning
the diversity of possible growth rates.
In Subsection \ref{partial} we prove that any function
that arises as the growth rate of a partial algebra
also arises as the growth rate of a total algebra.
In Subsection \ref{poly_growth} we construct the
first examples of finite algebras with polynomial
growth rates. In Subsection \ref{pseudoprojections}
we show that if $\Sigma$ defines a linear strong Maltsev condition
that does not imply the existence of a cube term,
then the class of varieties satisfying $\Sigma$
is generic, which, the reader will remember (from the Introduction),
means that every function that occurs as the growth rate
of a finite algebra occurs as the growth rate of a 
finite algebra in a variety satisfying the Maltsev
condition associated to $\Sigma$.
In Subsection \ref{decision2} we prove 
Theorem~\ref{decide_cube}, which describes
the procedure for deciding if a linear 
strong Maltsev condition entails the existence of a cube term. 
In Subsection \ref{exp} we identify some reasons
for exponential growth.

\subsection{Growth rates of partial algebras}\label{partial}

By a partial algebra we mean, of course, a set
equipped with a set of partial operations.
Our definitions of the functions 
$d_{\m a}$ and $h_{\m a}$ make sense when $\m a$
is a partial algebra,
as does the problem of determining growth rates 
of partial algebras. We will learn in this subsection 
that a function arises as the growth function
of a partial algebra if and only if it arises 
as the growth function of a total algebra.

Let $\m a = \lb A; P\rb$ be a partial algebra
with universe $A$ and a set $P$ of partial operations on $A$,
and let $0\notin A$ be a new element.
$\m a_{0}$ will be the total algebra with universe
$A_{0}:=A\cup \{0\}$ whose operations 
$P_0 = \{p_0\;|\;p\in P\}\cup \{\wedge\}$ are defined
as follows.

\begin{enumerate}
\item[(1)] If $p\in P$ is a partial 
$m$-ary operation on $A$ with domain $D\subseteq A^m$,
then the total operation $p_{0}\colon (A_{0})^m\to A_{0}$
is defined by 
\[
p_{0}(\wec{a}) = 
\begin{cases}
p(\wec{a}) & \textrm{if $\wec{a}\in D$; or}\\
0 & \textrm{otherwise.}
\end{cases}
\]
\item[(2)]
A meet operation $\wedge$ on $A_{0}$ defined by
\[
a\wedge b = 
\begin{cases}
a & \textrm{if $a=b$;}\\
0 & \textrm{otherwise.}
\end{cases}
\]
\end{enumerate}
Note that the meet operation in (2) is the same as the operation
$b_0(x,y)$ where $b$ is the partial binary operation on $A$
defined on pairs of the form $(a,a)$ by $b(a,a) = a$ and is undefined
on all other pairs. 

\begin{thm}\label{partial}
If $\m a$ is a partial algebra of more than one element, 
then $\m a^n$ and the total algebra
$\m a_0^n$ have the same minimal generating
sets. Hence $d_{\m a_0} = d_{\m a}$.
\end{thm}

\begin{proof}
It suffices to prove the following two statements:
(i) any generating set for $\m a^n$ is a generating set for 
$\m a_0^n$, and (ii) any \emph{minimal} generating set for
$\m a_0^n$ is a subset of $A^n$ and a generating set for $\m a^n$.

In this paragraph we prove (i).
If $G\subseteq A^n$ is a generating set for $\m a^n$, then as a subset
of $\m a_0^n$ it will generate 
(in exactly the same manner) all tuples in $A_0^n$
which have no $0$'s.
If $\wec{z}\in A_0^n$ is a tuple and $a, b\in A$ are distinct, let
$\wec{z}_a$ and $\wec{z}_b$
be the tuples obtained from $\wec{z}$ by replacing
all $0$'s with $a$ and $b$, respectively. Then $\wec{z}_a, \wec{z}_b
\in A^n$, so they are generated by $G$, and 
$\wec{z} = \wec{z}_a\wedge \wec{z}_b$, so $\wec{z}$ is also generated
by $G$. Hence $G$ generates all of $\m a_0^n$. 

Now we prove (ii). Assume that 
$H\subseteq A_0^n$ is a minimal generating set for
$\m a_0^n$. If $\wec{a}\in A_0^n$, let $Z(\wec{a})\subseteq [n]$
be the \emph{zero set} of $\wec{a}$, by which we mean the set
of coordinates where $\wec{a}$ is $0$. It is easy to see
that for any basic operation $F$ of $\m a_0$ 
it is the case that 
\begin{equation}\label{generation}
Z(\wec{a}_1)\cup \cdots \cup Z(\wec{a}_m)
\subseteq Z(F(\wec{a}_1,\ldots,\wec{a}_m)),
\end{equation}
since $0$ is absorbing for every basic operation. 
If the right-hand side is empty, then the left-hand
side is empty as well; i.e., tuples with empty zero sets
can only be generated by tuples with empty zero sets.
Said a different way, 
if $H\subseteq A_0^n$ generates $\m a_0^n$,
then $H\cap A^n$ suffices to generate all tuples in $A^n$,
and, by (i), these tuples suffice
to generate $\m a_0^n$; hence $H\cap A^n$ generates
$\m A_0^n$. Therefore, if $H\subseteq A_0^n$
is a minimal generating set for $\m a_0^n$, then 
$H\subseteq A^n$. 
If you consider how $H$ generates elements of $A^n$
in the algebra $\m a_0^n$, it is clear that $H$ generates
those elements in the algebra $\m a^n$ in exactly the same way, so
$H$ is a generating set for $\m a^n$.
\end{proof}

\subsection{Finite algebras with polynomial growth}\label{poly_growth}

\begin{thm}\label{example}
If $m$ is an integer greater than $1$, 
then there is a partial algebra $\m a$
with $m+1$ elements for which
$d_{\m a}(n)
%=1+(m-1)\binom{n}{1} + (m-2)\binom{n}{2} + \cdots + 
%2\binom{n}{m-2} + \binom{n}{m-1} 
\in\Theta(n^{m-1})$.
\end{thm}

\begin{proof}
The universe of the partial
algebra is $A = \{a_1, \ldots, a_m, 1\}$.
We equip this set with a single partial
$m$-ary operation $F$ defined by
\[
F(x_1,\ldots,x_m) =
\begin{cases}
y & \textrm{if $\{x_1,\ldots,x_m,y\}=A$ and $y\neq 1$;}\\
1 & \textrm{if $x_1=x_2=\cdots=x_m=1$; and}\\
%\textrm{undefined} 
& \textrm{else undefined.}
\end{cases}
\]
Set $\m a = \langle A; F\rangle$. 

It is worth remembering that 
the only way to generate the element $1$
using $F$ 
is by taking $x_1=\cdots =x_n = 1$, and the only
way to generate any $a_i\in A\setminus\{1\}$
is to use each element of $A\setminus \{a_i\}$
exactly once in some order in $F$.

We shall prove the exact formula
\begin{equation}\label{d_formula}
d_{\m a}(n)=
\binom{n}{m-1} + 2\binom{n}{m-2} + \cdots + (m-2)\binom{n}{2} + 
(m-1)\binom{n}{1} + 1,
\end{equation}
which is a polynomial in $n$ of degree $m-1$.

Choose and fix $n$. Define
the \emph{support} of a tuple $\wec{a}\in A^n$ to be the 
subset $\supp(\wec{a})\subseteq [n]$ 
consisting of indices $s$ where $a_s\neq 1$.
If $\wec{u}, \wec{v}\in A^n$ have the same support $S\subseteq [n]$,
then we say that 
$\wec{u}$ and $\wec{v}$ are \emph{adjacent} if there is some $i\in S$
such that $u_i=v_i$. A set $I \subseteq A^n$ is
\emph{independent} if $I$ consists of pairwise
nonadjacent tuples with the same support.

To outline the proof, suppose that $G\subseteq A^n$
is a generating for $\m a^n$. We will show that $G$ must
contain the tuple $[1,\ldots,1]^{\sf T}$ of empty support.
Then we show that, for any support set $S$ of size $k$ with 
$0<k<m$, $G$ must contain an independent set 
$I_S$ of size $m-k$ consisting of some of the 
tuples of support $S$.
Finally we argue that these independent sets $I_S\subseteq G$
with $0<|S|<m$ together with the tuple of empty support
already suffice to generate $\m a^n$. This will show that
any minimal generating set $G$ of $\m a^n$ has a special
structure: namely, $G$ must contain the
tuple of empty support, an independent set
$I_S$ of size exactly $m-k$ consisting of tuples of support $S$
for each subset $S\subseteq [n]$ of size $k$
for $k$ in the range $0<k<m$, and no other tuples.
This implies that any minimal generating set has size
$1+\sum_{0<|S|<m} |I_S|=1+\sum_{k=1}^{m-1} (m-k)\binom{n}{k}$.

\begin{clm}\label{support}
If $S\subseteq [n]$ and $G\subseteq A^n$, then let $G_S$ denote the tuples
in $G$ that have support contained in $S$.
If $\wec{a}\in \lb G\rb$, then $\wec{a}\in \lb G_S\rb$.
\end{clm}

In $\m a$, we have 
\[
F(x_1,\ldots,x_n) = 1\Longleftrightarrow x_1 = x_2 = \cdots = x_m = 1.
\]
Hence, in $\m a^n$, 
if $F(\wec{g}_1,\ldots,\wec{g}_m)$
is defined and equal to $\wec{b}$,
then $i\notin \supp(\wec{b})$ if and only if $i\notin \supp(\wec{g}_i)$
for any $\wec{g}_i$. Equivalently, 
\begin{equation}\label{generate}
\supp(F(\wec{g}_1,\ldots,\wec{g}_m)) = \bigcup_{i=1}^m \supp(\wec{g}_i)
\end{equation}
whenever $F(\wec{g}_1,\ldots,\wec{g}_m)$ is defined.
Now let $G(0) = G$, $G_S(0)=G_S$,
$G(k+1) = G(k)\cup F(G(k),\ldots,G(k))$, and
$G_S(k+1) = G_S(k)\cup F(G_S(k),\ldots,G_S(k))$.
By induction on $k$, using (\ref{generate}),
it can be shown that any tuple in $G(k)$ that has support
in $S$ lies in $G_S(k)$. Since $\lb G\rb = \bigcup_k G(k)$
and $\lb G_S\rb = \bigcup_k G_S(k)$, 
any tuple in $\lb G\rb$ with support in $S$ lies in $\lb G_S\rb$.
%this proves the claim.

\medskip

\begin{clm}\label{empty}
Any subset $G\subseteq A^n$ that generates the unique tuple
$[1,1,\ldots,1]^{\sf T}$
of empty support must contain this tuple.
\end{clm}

This follows immediately from Claim~\ref{support}.

\medskip

\begin{clm}\label{supp1}
Any subset $G\subseteq A^n$ that generates all tuples
of support $S\subseteq [n]$, where $S$ is a singleton, 
must contain an independent subset 
$I_S\subseteq G$ of size $m-1$
consisting of tuples of support $S$.
Conversely, any independent 
set of $m-1$ tuples of singleton support $S$,
together with the tuple of empty support, generates
all tuples of support $S$.
\end{clm}

We may assume that $S = [1] = \{1\}$.
If $G$ generates all tuples of support $[1]$,
then $G_S=G_{[1]}$ also generates these tuples. 
If $G_{[1]}$ already contains all tuples of support $[1]$,
then the first statement of the claim is true
because the set of all tuples of support
$[1]$ is an independent set of size $m$.
If $G_{[1]}$ does not contain all tuples of support $[1]$,
but does generate all such tuples, then $G_{[1]}$ must generate
some tuple of support ${[1]}$ not already in $G_{[1]}$. For this to happen, 
there must exist tuples $\wec{x}_i\in G_{[1]}$ 
and a tuple $\wec{b}\notin G_{[1]}$ of support $[1]$ such that 
\[
\small{
F(\wec{x}_1,\ldots,\wec{x}_m)=
F\left( 
\left[
\begin{array}{c}
x_{1,1}\\
1\\
\vdots\\
1\\
\end{array}
\right],
\ldots,
\left[
\begin{array}{c}
x_{1,m}\\
1\\
\vdots\\
1\\
\end{array}
\right]
\right) = 
\left[
\begin{array}{c}
b_1\\
1\\
\vdots\\
1\\
\end{array}
\right] = \wec{b}.
}
\]
In this case, $\{x_{1,1},\ldots,x_{1,m},b_1\} = 
\{a_1,\ldots,a_m,1\}$ and $b_1\neq 1$, which implies that
$\{\wec{x}_1,\ldots,\wec{x}_m\}$
consists of $m-1$ tuples of support $[1]$
along with the tuple of empty support.
Again the first statement of the claim holds.

Conversely, the equation
\[
\small{
F\left( 
\left[
\begin{array}{c}
a_1\\
1\\
\vdots\\
1\\
\end{array}
\right],
\ldots,
\left[
\begin{array}{c}
a_{m-1}\\
1\\
\vdots\\
1\\
\end{array}
\right],
\left[
\begin{array}{c}
1\\
1\\
\vdots\\
1\\
\end{array}
\right]
\right) = 
\left[
\begin{array}{c}
a_m\\
1\\
\vdots\\
1\\
\end{array}
\right]
}
\]
shows that a particular set of $m-1$ tuples
of support $[1]$ together with the tuple
of empty support generates the remaining
tuple of support $[1]$. By permuting the order of the $a_i$'s,
one sees that any independent 
set of $m-1$ tuples of support $[1]$,
together with the tuple of empty support, generates
all tuples of support $[1]$.

\medskip

\begin{clm}\label{suppk}
Fix $k$ satisfying $1 < k < m$.
Any subset $G\subseteq A^n$ that generates all tuples
of support $S$ for some subset $S\subseteq [n]$
of size $k$ must contain an independent
set $I_S$ of size $m-k$ consisting of tuples of support $S$. 
Conversely, any independent 
set of $m-k$ tuples of support $S$, $|S|=k$,
together with tuples whose support is a proper subset
of $S$, generates
all tuples of support $S$.
\end{clm}

We may assume that $S = [k]  =\{1,\ldots,k\}$.
If $G$ generates all tuples of support $S$,
then $G_S$ also generates these tuples,
according to Claim~\ref{support}.
If $G_S$ already contains all tuples of support $S$,
then it contains an independent set
of $m$ tuples of support $S$, namely 
all tuples of support $S$ that have
the form $[a_i,a_i,\ldots,a_i,1,\ldots,1]^{\sf T}$
for $i = 1,\ldots,m$.
In this case the first statement of the claim is true.

Otherwise $G_S$ generates
some tuple not already in $G_S$. For this to happen, 
there must exist tuples $\wec{x}_i\in G_S$ 
and a tuple $\wec{b}\notin G_S$ of support $S$ such that 
\begin{equation}\label{dividers}
\small{
F(\wec{x}_1,\ldots,\wec{x}_m)=
F\left( 
\left[
\begin{array}{c}
x_{1,1}\\
\vdots\\
\underline{\;\;x_{k,1}\;\;}\\
1\\
\vdots\\
1\\
\end{array}
\right],
\left[
\begin{array}{c}
x_{1,2}\\
\vdots\\
\underline{\;\;x_{k,2}\;\;}\\
1\\
\vdots\\
1\\
\end{array}
\right],
\ldots,
\left[
\begin{array}{c}
x_{1,m}\\
\vdots\\
\underline{\;\;x_{k,m}\;\;}\\
1\\
\vdots\\
1\\
\end{array}
\right]
\right) = 
\left[
\begin{array}{c}
b_1\\
\vdots\\
\underline{\;\;\;b_{k}\;\;\;}\\
1\\
\vdots\\
1\\
\end{array}
\right]=\wec{b}.
}
\end{equation}
We have introduced horizontal segments
as dividers separating the coordinates
in $S = [k]$ from the
remaining coordinates in order
to make the argument clearer.
Since $F(\wec{x}_1,\ldots,\wec{x}_m)$ is defined,
%no two tuples $\wec{x}_i$ and $\wec{x}_j$ can agree
%at any coordinate in $S$, and, in particular,
no row above the dividers
has more than one $1$. Hence there are at most 
$k$ $1$'s above the dividers.
This means that there are at most $k$ columns which
contain a $1$ above the dividers; equivalently, there are at least
$m-k$ columns which contain no $1$'s above the dividers.
These columns
are at least $m-k$ distinct tuples of $G_S$
of support $S$. Moreover, this set of $m-k$ tuples
must be independent if $F(\wec{x}_1,\ldots,\wec{x}_m)$ is defined.
This proves the first part of the claim.

To simplify the proof of the converse,
we prove it only in the case where $k=n$,
and therefore $S = [k] = [n]$. This special case
contains all the complexities of the general case,
but we no longer need to
refer to dividers in figures like (\ref{dividers}).
(To derive the general case from the special case,
simply place dividers after $k$ rows and fill in the
remaining rows with $1$'s.) In the case $n=k$, the statement to prove
is that any independent set of size $m-k$  in $A^k$,
together with the tuples whose support is a proper subset
of $[k]$, generates $A^k$.

It follows from the definition of $F$ that if 
$\wec{u}_1,\ldots,\wec{u}_{m-k},\wec{b}\in A^k$
have support $[k]$, then the row equations of
$F(\wec{x}_1,\ldots,\wec{x}_k,\wec{u}_1,\ldots,\wec{u}_{m-k})=\wec{b}$,
written out in (\ref{x&u}),
can be solved for $x_{i,j}$ if and only if 
$\{\wec{u}_1,\ldots,\wec{u}_{m-k},\wec{b}\}$ is an independent set.
%Moreover, when the row equations can be solved at all, 
%they can be solved 
%with $x_{i,i}=1$ for $i=1,\ldots,k$, in which case
%$\wec{x}_1, \ldots, \wec{x}_k\in G^-$.
\begin{equation}\label{x&u}
\small{
F\left( 
\left[
\begin{array}{c}
1\\
x_{2,1}\\
\vdots\\
x_{k,1}\\
\end{array}
\right],
\left[
\begin{array}{c}
x_{1,2}\\
1\\
\vdots\\
x_{k,2}\\
\end{array}
\right],
\ldots,
\left[
\begin{array}{c}
x_{1,m}\\
x_{2,m}\\
\vdots\\
1\\
\end{array}
\right],
\left[
\begin{array}{c}
u_{1,1}\\
u_{2,1}\\
\vdots\\
u_{k,1}\\
\end{array}
\right],
\ldots,
\left[
\begin{array}{c}
u_{1,m-k}\\
u_{2,m-k}\\
\vdots\\
u_{k,m-k}\\
\end{array}
\right]
\right) = 
\left[
\begin{array}{c}
b_1\\
b_2\\
\vdots\\
b_{k}\\
\end{array}
\right].
}
\end{equation}

Let $G^-$ denote the set of all tuples in $A^k$
whose support is a proper subset of $[k]$.
Let $I = \{\wec{u}_1,\ldots,\wec{u}_{m-k}\}\subseteq A^k$ 
be an arbitrary independent set of size $m-k$
consisting of tuples of support $[k]$.
(Such sets do exist, e.g., 
$I = \{\alpha_1,\ldots,\alpha_{m-k}\}$ where 
$\alpha_i = [a_i,a_i,\ldots,a_i]^{\sf T}$.)
Let $G(0) = I\cup G^-$ and let
$G(j+1) = F(G(j),G(j),\ldots,G(j))\cup G(j)$
be the set of tuples that can be constructed
in $j+1$ steps from $G(0)$.
We prove by induction on $\ell$ that if $\wec{b}\in A^k$ has support
$[k]$, and $\wec{b}$ is adjacent to 
$\ell$ tuples in $I$, then $\wec{b}\in G(\ell+1)$.
This and the fact that $G^-\subseteq G(0)$
are enough to prove that $G(0)$ is a generating
set for $A^k$.

For $\ell=0$ we must show that if $\wec{b}$ is adjacent
to none of the tuples in $\{\wec{u}_1,\ldots,\wec{u}_{m-k}\}$,
then $\wec{b}\in G(1)$; i.e., $\wec{b}$ is generated 
in one step by $I\cup G^-$. This follows from our earlier
remark that the row equations
in (\ref{x&u}) can be solved when 
when $\{\wec{u}_1,\ldots,\wec{u}_{m-k},\wec{b}\}$
is independent.

Now assume that 
whenever $\wec{b}\in A^k$ has support
$[k]$ and $\wec{b}$ is adjacent to 
$\ell$ tuples in $I$, then $\wec{b}\in G(\ell+1)$.
Tp prove the same statement with $\ell$
replaced by $\ell+1$, assume that $\wec{b}\in A^k$ 
is a tuple of support $[k]$ that is is adjacent
to $\ell+1$ tuples in $I=\{\wec{u}_1,\ldots,\wec{u}_{m-k}\}$.
Since there are fewer than $m$ tuples
in $I\cup \{\wec{b}\}$, 
there must exist a tuple $\wec{v}$ of support $[k]$
that is not adjacent to any tuple in this set.
(We may construct $\wec{v} = [v_1,\ldots,v_k]^{\sf T}$
entry by entry. The only restriction
on its $i$th coordinate $v_i$ is that it must be in
$
\{a_1, \ldots, a_m\}\setminus
\{u_{i,1}, \ldots, u_{i,m-k}, b_i\}.
$
We are subtracting fewer than $m$ elements from $m$, so
there is always a choice for $v_i$.)
Since $\wec{v}$ has support $[k]$ and $\wec{v}$
is not adjacent to any tuple in $I$,
$\wec{v}\in G(1)$, by our argument
for the basis of induction. Now let  
\[
I' = (I\setminus \{\wec{u}_h\})\cup \{\wec{v}\}
%\;(\subseteq G(1))
\]
be the set obtained from $I$ by replacing
one tuple that is adjacent to $\wec{b}$ with the tuple $\wec{v}$.
$I'$ is an independent subset of $G(1)$ of size $m-k$
of tuples of support $[k]$ which 
contains only $\ell$ tuples adjacent to $\wec{b}$.
By the inductive hypothesis, $\wec{b}$ can be generated
in $\ell+1$ steps from $I'$. Since $I'\subseteq G(1)$,
this means that $\wec{b}\in G(\ell+1)$.
This completes the proof of Claim~\ref{suppk}.

\medskip

\begin{clm}\label{finish}
The set of all tuples whose support has size less than 
$m$ generates $\m a^n$.
\end{clm}

Let $A(j)$ denote the set of tuples in $A^n$ 
of support at most $j$. 
In this notation, $A(n) = A^n$.
The claim is that $A(m-1)$ is a generating set
for $\m a^n$. We prove this 
%via induction
by showing that if $r\geq m-1$, then 
$F(A(r),A(r),\ldots,A(r))\cup A(r)$ contains $A(r+1)$.

In order to prove that 
$F(A(r),A(r),\ldots,A(r))\cup A(r)$ contains $A(r+1)$, 
choose $\wec{b}\in A(r+1)\setminus A(r)$ arbitrarily.
There is no loss of generality in assuming that 
the support of $\wec{b}$ is $[r+1]=\{1,\ldots,r+1\}$. 
Given $\wec{b}$, we need to solve the equation
$F(\wec{x}_1,\ldots,\wec{x}_m) = \wec{b}$
with tuples $\wec{x}_i\in A(r)$. For this, it suffices
to find $x_{i,j}\in \{1, a_1,\ldots,a_m\}$ such that the row 
equations hold in:

\[
\small{
F\left( 
\left[
\begin{array}{c}
1\\
x_{2,1}\\
\vdots\\
x_{m-1,1}\\
\underline{\;\;\;x_{m,1}\;\;\;}\\
x_{m+1,1}\\
\vdots\\
\underline{\;\;x_{r+1,1}\;\;}\\
1\\
\vdots\\
1\\
\end{array}
\right]
\left[
\begin{array}{c}
x_{1,2}\\
1\\
\vdots\\
x_{m-1,2}\\
\underline{\;\;\;x_{m,2}\;\;\;}\\
x_{m+1,2}\\
\vdots\\
\underline{\;\;x_{r+1,2}\;\;}\\
1\\
\vdots\\
1\\
\end{array}
\right],
\ldots,
\left[
\begin{array}{c}
x_{1,m-1}\\
x_{2,m-1}\\
\vdots\\
1\\
\underline{\;\;\;x_{m,m-1}\;\;\;}\\
x_{m+1,m-1}\\
\vdots\\
\underline{\;\;x_{r+1,m-1}\;\;}\\
1\\
\vdots\\
1\\
\end{array}
\right]
\left[
\begin{array}{c}
x_{1,m}\\
x_{1,m}\\
\vdots\\
x_{m-1,m}\\
\underline{\;\;\;\;\;\;1\;\;\;\;\;\;}\\
x_{m+1,m}\\
\vdots\\
\underline{\;\;x_{r+1,m}\;\;}\\
1\\
\vdots\\
1\\
\end{array}
\right]
\right) = 
\left[
\begin{array}{c}
b_1\\
b_2\\
\vdots\\
b_{m-1}\\
\underline{\;\;\;\;b_m\;\;\;\;}\\
b_{m+1}\\
\vdots\\
\underline{\;\;\;b_{r+1}\;\;\;}\\
1\\
\vdots\\
1\\
\end{array}
\right].
}
\]

Above the first row of dividers
on the left-hand side is an $m\times m$ matrix of elements
of $A$ with $1$'s on the diagonal (and no other restriction, yet).
Below the second row of dividers is an 
$(n-r-1)\times m$ matrix of $1$'s. 
Between the two rows of dividers is
an $(r+1-m)\times m$ matrix with no restriction (yet)
on its entries. Since each column has at most $r$ 
entries that are not $1$'s,
any solution to the row equations will yield
columns $\wec{x}_i\in A(r)$ such that
$F(\wec{x}_1,\ldots,\wec{x}_m) = \wec{b}$

The first row can be filled out arbitrarily subject to the
condition that 
\[
\{1,x_{1,2}, \ldots,x_{1,m},b_1\} = 
\{1,a_1,\ldots,a_m\}.
\] 
According to the definition
of $F$, this is possible, given that $b_1\neq 1$.
This argument works for the first $m$ rows.
The next $r+1-m$ rows can be filled in arbitrarily
subject to the condition that
$\{x_{i,1}, x_{i,2}, \ldots,x_{i,m},b_i\} = 
\{1,a_0,\ldots,a_m\}$, and this is just as easy as the
earlier rows. The last $n-r-1$ rows are already filled in.
This proves that $\wec{b}$ (and hence every tuple in
$A(r+1)\setminus A(r)$) belongs to the set $F(A(r),\ldots,A(r))$.

\medskip

We have shown that if $G\subseteq A^n$ is any generating set
for $\m a^n$, then $G$ contains the tuple of empty support
along with, for every subset $S\subseteq [n]$ satisfying 
$0<|S|=k<m$, an independent set $I_S$ containing $m-k$
tuples of support $S$. Since there are 
$\binom{n}{k}$ subsets $S\subseteq [n]$
for each $k$, this accounts for at least 
$1+\sum_{k=1}^{m-1} (m-k)\binom{n}{k}$ elements of $G$.
However, we have shown along the way that any set 
containing the tuple of empty support
and independent sets $I_S$ of $m-k$
tuples of support $S$ for each subset $S\subseteq [n]$
of size $0<|S|=k<m$ is a generating set for $\m a^n$.
Altogether this shows that 
$d_{\m a}(n)=1+\sum_{k=1}^{m-1} (m-k)\binom{n}{k}$. 
\end{proof}


\begin{cor}
For any $m>1$, there is a total algebra $\m a$
with $m+2$ elements which satisfies $d_{\m a}(n)\in\Theta(n^{m-1})$.
\end{cor}

\begin{proof}
Apply Theorem~\ref{partial} to Theorem~\ref{example}.
\end{proof}

\subsection{Adding pseudoprojections}\label{pseudoprojections}

Let $\Sigma$ be a set of linear identities which does not 
entail the existence of a cube term.
For each function symbol $F$ occurring in $\Sigma$
there is a variable $k$ such that $F$ is a 
$k$th variable pseudoprojection, according to 
Theorem~\ref{decide_cube}. Using this fact, we will
show that if $\m a$ is any algebra, then there is an algebra
$\m b$ with exactly the same growth function
($d_{\m a}=d_{\m b}$) which generates a variety realizing $\Sigma$.
The argument does not require $\Sigma$ or $\m a$ to be finite.

\subsection{A decision procedure for cube terms, II}\label{decision2}


\subsection{Exponential growth}\label{exp}



\section{Near Unanimity Enforces the Logarithmic Monochotomy}\label{comb}

We now begin establishing restrictions on growth rates.
In Subsection \ref{bak_pix1} we recall the Baker-Pixley
Theorem, which characterizes varieties with a
near unanimity term operation.
In Subsection \ref{solution} we solve
a combinatorial problem that enables us to 
derive in Subsection \ref{bak_pix2}
that growth rates of finite algebras are logarithmic in any 
variety with a near unanimity term operation.

\subsection{The Baker-Pixley Theorem, Part I}\label{bak_pix1}

In \cite{baker_pixley}, Kirby Baker and Alden Pixley
characterized the existence of a $(k+1)$-ary near
unanimity term operation in several ways, one of which
is the following.

\begin{thm}\label{baker_pix}
%{\rm \cite{baker_pixley}}
A variety $\mathcal V$ has a $(k+1)$-ary near unanimity
term operation if and only if, in $\mathcal V$,
whenever $\m b$ is a subalgebra of a direct product 
$\m a_1\times \cdots \times \m a_n$, $(n\geq k)$,
then $\m b$ can be uniquely determined from a knowledge 
of its $k$-fold coordinate projections.
\end{thm}

The statement that $\m b\leq \prod \m a_i$ is uniquely
determined by its $k$-fold projections means that 
if $\m c\leq \prod \m a_i$ has the same $k$-fold
projections as $\m b$, then $\m c=\m b$.
The particular case that interests us is 
the one where $\m a_1 = \cdots = \m a_n = \m a$
and $\m c = \m a^n$. In this case the theorem says
that if $\m a$ has a $(k+1)$-ary near unanimity term
and $\m b\leq \m a^n$ projects 
surjectively onto each subset of $k$ coordinates, then
$\m b = \m a^n$.

These observations prompt the following purely combinatorial
question: If $A$ is a finite set and $n\geq k$ are positive
integers, then how small can a set $G\subseteq A^n$ be
if its projection onto any subset of $k$ coordinates
is surjective? For if $A$ is the universe of an algebra
with a $(k+1)$-ary near unanimity term operation 
and $\m b = \lb G\rb$ is the subalgebra generated
by a set $G$ with the property just described,
then $\m b$ will project surjectively onto 
any subset of $k$ coordinates, hence 
$\m a^n = \m b = \lb G\rb$. That is,
a set $G$ that projects surjectively onto 
any $k$ coordinates will be a generating for $\m a^n$
set if $\m a$ has a $(k+1)$-ary near unanimity term operation.

If $k=1$, then one choice for a smallest subset $G\subseteq A^n$ 
that projects surjectively onto any $k$ coordinates is the
set of constant tuples in $A^n$. This set certainly 
projects onto each single coordinate;
on the other hand, no set of size strictly less than
$|A|$ can project onto any single coordinate.
Hence, when $k=1$, the smallest size for $G$ is $|A|$.
In the following subsection, we consider the problem for
$k>1$ and prove that the least size for such $G\subseteq A^n$
is $\Theta(\log(n))$.

\subsection{The solution to a combinatorial problem}\label{solution}

The problem considered here is:
If $A$ is a finite set and $n\geq k>1$ are 
integers, then how small can a set $G\subseteq A^n$ be
if its projection onto any subset of $k$ coordinates
is surjective? 

If $A$ is finite, $G\subseteq A^n$ and $|G|=g$, 
then $G$ can be linearly ordered
and taken to be the sequence of 
rows of a $g\times n$ matrix of elements
of $A$, say $[a_{i,j}]$. If 
\[
\sigma\colon \quad 1 \leq j(1) < \cdots < j(k) \leq n
\]
is a selection of $k$ numbers between $1$ and $n$, then 
the projection of $G$ onto the
coordinates in $\sigma$ 
is the set of row vectors
$(a_{1, j(1)}, \ldots, a_{1, j(k)}), \ldots, 
(a_{g, j(1)}, \ldots, a_{g, j(k)})$ which 
occur as the set of rows of the $g\times k$
minor of $[a_{i,j}]$ whose column indices
are the indices in $\sigma$.
$G$ projects surjectively onto each $k$ coordinates of $A^n$ 
if and only if, for each choice $\sigma$
of $k$ column indices, the set of row vectors 
of the corresponding $g\times k$ minor of $[a_{i,j}]$ exhausts $A^k$.
Therefore, call a $g\times k$ matrix of elements of $A$
a \emph{bad minor} (or \emph{bad matrix})
if its rows fail to exhaust $A^k$.
The desired property of $G$ is that its associated
matrix has no bad minors.

\begin{thm}\label{surj_projections}
Let $A$ be a finite set of size $|A|=a > 1$.
Let $n\geq k > 1$ be natural numbers, and set
%$\displaystyle u = \frac{a^k}{a^k-1}$.
$\displaystyle u = a^k/(a^k-1)$.
If 
$
g\geq k\log_u\left(n\right) + \log_u\left(a^k/k!\right),
$
then there is a matrix in $A^{g\times n}$ with no bad minors.
\end{thm}

\begin{proof}
This is a probabilistic proof.
Our sample space is the set $A^{g\times n}$ of all $g\times n$
matrices of elements of $A$. 
Our probability distribution is the uniform one, 
so each individual matrix $M\in A^{g\times n}$ 
has probability $P(M)=|A^{g\times n}|^{-1} = a^{-gn}$.
For each matrix $M\in A^{g\times n}$ and 
each sequence of $k$ column indices, 
\[
\sigma\colon \quad 1 \leq j(1) < \cdots < j(k) \leq n,
\]
let $M_{\sigma}$ denote the $g\times k$ minor of $M$
whose column indices are those enumerated by $\sigma$
(called the \emph{$\sigma$-minor} of $M$).
Let $X_{\sigma}$ be the random variable whose value 
at the element $M\in A^{g\times n}$ is $1$ if 
$M_{\sigma}$ is a bad minor and $0$ otherwise,
i.e., $X_{\sigma}$ is the indicator variable for bad $\sigma$-minors.

\begin{clm}\label{expectation_clm}
For any $\sigma$, 
the expected value of $X_{\sigma}$ satisfies
\begin{equation}\label{first_estimate}
E(X_{\sigma})\leq a^k\left(a^k-1\right)^ga^{-gk}.
\end{equation}
\end{clm}

The expectation is computed 
\[
\begin{array}{rl}
E(X_{\sigma}) 
&=\; \sum_{M\in A^{g\times n}} (X_{\sigma}(M)\cdot P(M))\\
&=\; \sum_{M\in A^{g\times n}} (X_{\sigma}(M)\cdot a^{-gn})\\
&=\; \left(\sum_{M\in A^{g\times n}} X_{\sigma}(M)\right)a^{-gn},
\end{array}
\]
where the sum $\sum_{M\in A^{g\times n}} X_{\sigma}(M)$
on the last line represents the number matrices
in $A^{g\times n}$ whose $\sigma$-minor is bad.
By definition, a $g\times k$ matrix is bad if some
tuple $\wec{a}\in A^k$ does not appear among its rows.
So, 
for each $\wec{a}\in A^k$, let $\mathcal U_{\wec{a}}$
denote the set of all $g\times k$ matrices where
$\wec{a}$ does not appear among the rows.
$|\mathcal U_{\wec{a}}|$ can be computed by noting that
the $g$ rows of a matrix in $\mathcal U_{\wec{a}}$
may be freely chosen
from the set $A^k-\{\wec{a}\}$, which has
size $a^k-1$, so $|\mathcal U_{\wec{a}}| = (a^k-1)^g$.
The bad $g\times k$ matrices are those from
$\bigcup_{\wec{a}\in A^k} \mathcal U_{\wec{a}}$. Since
the cardinality of the union is no more than the sum of the
individual cardinalities, and these summands have the
same size,  we get that the number of bad
$g\times k$ matrices is no more than
$|A^k|\cdot |\mathcal U_{\wec{a}}| = a^k(a^k-1)^g$.
Each bad $g\times k$ matrix $N$ can be extended in 
$a^{g(n-k)}$ ways to a matrix $M\in A^{g\times n}$ whose
$\sigma$-minor satisfies $M_{\sigma}=N$, so 
the number of matrices in $A^{g\times n}$ with
a bad $\sigma$-minor is no more than 
$a^k(a^k-1)^ga^{g(n-k)}$.
Hence 
\[
E(X_{\sigma}) = 
\left(\sum_{M\in A^{g\times n}} X_{\sigma}(M)\right)a^{-gn}
\leq a^k(a^k-1)^g a^{g(n-k)} a^{-gn} = a^k(a^k-1)^g a^{-gk},
\]
as claimed.

\medskip

If $X:=\sum_{\sigma} X_{\sigma}$ is the sum of all
$X_{\sigma}$ as $\sigma$ ranges
over all $\binom{n}{k}$ choices of $k$ column indices
and $M\in A^{g\times n}$, then $X(M)$
equals the number of bad $g\times k$ minors of $M$. Since 
expectation is linear, and since 
$\binom{n}{k} < n^k/k!$ when $n\geq k > 1$, we get 
from (\ref{first_estimate})
that 
\[
E(X) = \sum_{\sigma} E(X_{\sigma}) \leq \binom{n}{k} a^k(a^k-1)^g a^{-gk} <
n^k(a^k/k!)(a^k-1)^g a^{-gk}.
\]
If it is the case that 
\begin{equation}\label{bound}
n^k(a^k/k!)(a^k-1)^g a^{-gk}\leq 1,
\end{equation}
then we will have $E(X) < 1$, meaning that the expected number of 
bad minors in an element of $A^{g\times n}$
is strictly less than $1$. This can only happen
if matrices without bad minors exist.
Rewriting (\ref{bound}) as
\[
n^k \leq 
\left(\frac{a^{k}}{\left(a^k-1\right)}\right)^g(a^k/k!)^{-1} = 
u^g(a^k/k!)^{-1},
\]
using the definition $u = a^k/(a^k-1)$, we can solve for $g$
to get
\begin{equation}\label{4rth}
g\geq k\log_u(n) + \log_u(a^k/k!).
\end{equation}
When this inequality holds
we get that (\ref{bound}) holds,
so a matrix with no bad minors exists. This is exactly the statement
of the theorem.
\end{proof}

\begin{cor}\label{surj_projections_cor}
Let $A$ be a finite set of size $|A|=a > 1$.
Let $n\geq k > 1$ be natural numbers, and set
$u = a^k/(a^k-1)$.
\begin{enumerate}
\item[(1)]
If 
$
g=\lceil k\log_u\left(n\right) + \log_u\left(a^k/k!\right)\rceil,
$
then there exists a subset $G\subseteq A^n$ of size $g$
whose projection onto any
$k$ coordinates of $A^n$ is surjective.
\item[(2)]
Let $G\subseteq A^n$ be a subset.
\begin{enumerate}
\item[(i)]
If the projections 
${\rm pr}_i\colon G\to A$ and 
${\rm pr}_j\colon G\to A$ 
onto distinct coordinates are different
functions for all $i$ and $j$, then
$
|G| \geq \log_a(n).
$
\item[(ii)]
If no two projections 
${\rm pr}_i\colon G\to A$ and 
${\rm pr}_j\colon G\to A$ 
differ by a permutation of $A$
(i.e., if it is not the case that 
$\pi\circ {\rm pr}_i={\rm pr}_j$
for any $i$ and $j$ or $\pi\in {\textrm{Sym}(A)}$),
then
$
|G| \geq \log_a(n) + \log_a(a!).
$
\item[(iii)]
If all projections of $G$
onto pairs of coordinates are surjective, then
$
|G| \geq \log_a(n) + \log_a(a!).
$
\end{enumerate}
\end{enumerate}
\end{cor}

When reading the statement of
this corollary one should imagine that $a$ and $k$ 
(and hence $u$) are
fixed while $n$ ranges. Item (1) implies that
there is a subset $G\subseteq A^n$ of size $O(\log(n))$
that projects surjectively onto any $k$ coordinates.
Item (2) implies that any set $G\subseteq A^n$ 
that projects surjectively onto each set of $k$ coordinates
must have size $\Omega(\log(n))$ if $k$ is at least $2$. Taken together
these statements show that, as a function of $n$, 
the least size of a set $G\subseteq A^n$ that projects onto
each $k$ coordinates is $\Theta(\log(n))$.

\begin{proof}[Proof of Corollary~\ref{surj_projections_cor}]
Part (1) is an immediate consequence of Theorem~\ref{surj_projections}.

For (2)(i), suppose that 
$G\subseteq A^n$ is a subset 
whose projections 
onto distinct coordinates are different
functions for different coordinates. 
Then the set of projections onto distinct coordinates
constitute $n$ distinct elements of the function 
space $A^G$, which has cardinality $a^g$. 
It follows that $n\leq a^g$, or $|G|=g\geq \log_a(n)$.

The argument for (2)(ii) is essentially the same as for 
(2)(i). The assumptions imply that the
set of all functions of the form $\pi\circ {\rm pr}_i\colon G\to A$,
$\pi\in {\textrm{Sym}(A)}$, are distinct. 
This yields $|{\textrm{Sym}(A)}|\cdot n = a!\cdot n$ functions
from the set $A^G$, which has size $a^g$,
hence $a!\cdot n\leq a^g$. Solving for $g$
yields $|G|=g\geq \log_a(n) + \log_a(a!)$.

For (2)(iii), if two projections differ by a permutation,
say $\pi\circ {\rm pr}_i = {\rm pr}_j$, then the projection
of $G$ onto the $i$th and $j$ coordinates maps $G$ into the
the set of pairs of the form $(x,\pi(x))$. All such pairs
lie on the graph of $\pi$, which is a proper
subset of $A^2$ when $|A|>1$. Thus, if all projections
of $G$ onto pairs of coordinates are surjective, 
no two single coordinate projections
can differ by a permutation. This shows that 
item (2)(iii) follows from (2)(ii).
\end{proof}

\begin{rem}
It is possible to refine the estimate
in Corollary~\ref{surj_projections_cor}~(1)
when we are dealing with algebras instead of just sets.
Suppose that there is a subset $B\subseteq A$ such
that $B^k$ generates $\m a^k$. Then we can apply
the arguments above to obtain a set $G\subseteq B^n$
whose projection onto any $k$ coordinates
is $B^k$. According to the arguments, we can find such a $G$
of size 
\[
|G| = \lceil k\log_v\left(n\right) + \log_v\left(b^k/k!\right)\rceil,
\]
where $b = |B|$ and $v = b^k/(b^k-1)$. This is a larger
base for the logarithm and a smaller constant on the right,
so an overall smaller estimate for $G$.
The subalgebra of $\m a^n$ generated
by $G$ will project surjectively onto each $k$
coordinates, since its projection contains the generating
set $B^k$. Note here that there is no longer any
need to assume that $\m a$ is finite, only that $B$
is finite.

Another refinement can be made on top of the preceding one.
Suppose that there is a subset $P\subseteq B^k$
of size $|P|=p$
that generates $\m a^k$. (Here we do not assume that $P$
is the $k$th power of a subset of $A$.)
The argument we have given does not generalize to produce
a subset $G\subseteq A^n$ whose projection onto any $k$ coordinates
is exactly $P$, but it does generalize to produce a subset
$G\subseteq A^n$ whose projection onto any $k$ coordinates will be
contained in $B^k$ and will contain $P$. This enough to ensure
that the subalgebra of $\m a^n$ that
is generated by $G$ projects surjectively
onto any $k$ coordinates. How much of an improvement do
we get in our estimate of $|G|$?

The assumption that $P\;(\subseteq B^k)$
generates $\m a^k$ affects the estimate 
in Claim~\ref{expectation_clm} as follows.
For each $\wec{b}\in P$, let $\mathcal U_{\wec{b}}$
denote the set of all matrices in $B^{g\times k}$
where $\wec{b}$ does not appear among the rows.
Then $|\mathcal U_{\wec{b}}| = (b^k-1)^g$, so 
$|\bigcup_{\wec{b}\in P}\mathcal U_{\wec{b}}| \leq |P|(b^k-1)^g=p(b^k-1)$.
Each of these $g\times k$ matrices can be extended in 
$b^{g(n-k)}$ ways to a matrix in $B^{g\times n}$, so 
we obtain the estimate $E(X_{\sigma})\leq p(b^k-1)b^{-gk}$.
This allows us to choose $G$ so that it
is slightly smaller, namely 
\[
|G|= \lceil k \log_v(n) + \log_v(p/k!) \rceil.
\]

Let's summarize the two refinements.
Let $\m a$ be a (possibly infinite) algebra.
Suppose that $P\subseteq A^k$ is a finite
generating set for $\m a^k$. Let $B$ be the smallest
subset of $A$ for which $P\subseteq B^k$, i.e., let
$B\subseteq A$
be the set of elements that appear in the coordinates
of the tuples in $P$. 
Let $b = |B|$,  $p = |P|$,
and $v = b^k/(b^k-1)$.
Assertion: There is a set $G\subseteq A^n$ of size 
$\lceil k \log_v(n) + \log_v(p/k!) \rceil$ that generates
a subalgebra of $\m a^n$ which projects surjectively onto any 
$k$ coordinates.
\end{rem}

\subsection{The Baker-Pixley Theorem, Part II}\label{bak_pix2}

The combinatorial arguments given in the preceding
subsection 
may be combined with Theorem~\ref{baker_pix} to produce 
a result about growth sequences.

\begin{thm}
{\rm (Near unanimity terms
enforce the logarithmic monochotomy.)}
If $\m a$ is a finite algebra with a near
unanimity term operation, then  $d_{\m a}\in \Theta(log(n))$.
\end{thm}

This theorem will be generalized in Theorem~\ref{????}.

\section{Few subpowers Enforces the Logarithmic/Linear Dichotomy}\label{comb}

Now we move outside the realm of varieties with near unanimity
term operation. Our main goal is to prove that if
a variety has a cube term, then finite perfect algebras
have logarithmic growth while finite imperfect algebras
have linear growth.

\subsection{Maximal subuniverses}\label{max_sec}

%\subsection{Critical relations}\label{critical}


When $\m a^n$ is finitely generated, then a 
subset $G\subseteq \m a^n$ is a generating subset if and only 
if $G$ is contained in no maximal subuniverse of $A^n$.
For this reason, the problem understanding the growth rates of 
finitely generated algebras is 
related to the problem of understanding 
the maximal subuniverses of powers.

Let $\varphi\colon \m b\to \m c$ be a surjective homomorphism.
If $M\subseteq B$ is a maximal subuniverse, then either
$\varphi(M) = C$ or $\varphi(M)$ is a maximal subuniverse
of $C$. In the latter case, $M = \varphi^{-1}(\varphi(M))$
is the full preimage of $\varphi(M)$, and we
say that $M$ is \emph{induced} by $\varphi$. Thus, some information
about the maximal subuniverses of $\m b$ can be derived
from information about the maximal subuniverses of quotients of $\m b$.
We begin by studying the maximal subuniverses of 
$\m b = \m a^n$ in the case where the homomorphism $\varphi$
is a projection onto a subset of coordinates.

We will need a concept from \cite{parallelogram}.

\begin{df}
An $n$-ary \emph{critical} compatible relation on $\m a$
is a subuniverse of $\m a^n$ that is completely meet--irreducible 
in the lattice of subuniverses of $\m a^n$
and is directly indecomposable
as a relation on $A$, i.e., it is not of the form 
$R\times S$ for subsets $R\subseteq A^U$
and $S\subseteq A^V$, where $\{U,V\}$ is a partition
of $[n]$ into two cells.
\end{df}

Since any maximal subuniverse $M$ of $\m a^n$ is 
completely meet--irreducible, a
critical maximal subuniverse of $\m a^n$
is just a directly indecomposable 
$n$-ary compatible relation on $\m a$.
These can be described in terms of the following
notion.

\begin{df}
If $M$ is a subuniverse of $\m a^n$, then a 
\emph{support} of $M$ is a subset $S\subseteq [n]$
that is minimal under inclusion for the property
that ${\rm pr}_U(M)\neq A^S$,
where ${\rm pr}_U\colon \m a^n\to \m a^S$ is the projection map.
\end{df}


\begin{lm}
If $M$ is a maximal subuniverse of $\m a^n$, then $M$
has a unique support. If $S$ is the support of $M$,
then ${\rm pr}_S(M)$ is a critical maximal subuniverse
of $\m a^S$, and $M$ is induced by the projection
${\rm pr}_S$.
In particular, $M\leq \m a^n$ 
is critical if and only if its support is $[n]$.
\end{lm}

\begin{proof}
Consider the following property of an index $i\in [n]$ relative
to a maximal subuniverse $M$: 
***tag this ***
if tuples $\wec{a}, \wec{b}\in A^n$ 
agree at every coordinate except possibly at $i$, 
then both or neither belong to $M$.
We shall argue that if $i$ lies outside of at least
one support of $M$, then $i$ has this property relative to $M$.
Conversely, we shall argue that if $j$ lies in any support,
then $j$ does not have this property relative to $M$.
This will show that the unique support of $M$ is the set
of indices that do not have this property.

Suppose that $S\subseteq [n]$ is a support and that 
$i\in [n]-S$. 
If $\wec{a}, \wec{b}\in \m a^n$ agree in all coordinates
except $i$, then ${\rm pr}_S(\wec{a}) = {\rm pr}_S(\wec{b})$.
The fact that $S$ is a support implies that
$M = {\rm pr}_S^{-1}({\rm pr}_S(M))$; hence, if $\wec{a}\in M$, then
${\rm pr}_S(\wec{b}) = {\rm pr}_S(\wec{a})\in {\rm pr}_S(M)$,
so $\wec{b}\in {\rm pr}_S^{-1}({\rm pr}_S(M)) = M$.
Similarly $\wec{b}\in M$ implies $\wec{a}\in M$.

Now suppose that $j\in T$ and
that $T$ is a support. If $T = \{j\}$, then 
$M'={\rm pr}_T(M)$
is a maximal subuniverse of $\m a$ and 
$M = {\rm pr}_T^{-1}({\rm pr}_T(M))$ consists of all tuples
whose $j$th coordinate lies in $M'$. If $a_j\in M'$ and $b_j\notin M'$,
then we may extend these to $n$-tuples that agree everywhere
except in coordinate $j$, and they witness the fact that
$j$ does not have the property we are considering.
Now suppose that $T$ properly contains $\{j\}$. 
Since $T-\{j\}$ is a proper subset of a support
we get ${\rm pr}_{T-\{j\}}(M) = A^{T-\{j\}}$.
Since ${\rm pr}_{T}(M)$ is a proper subuniverse of $\m a^T$
there is a tuple $\wec{a}\in A^n$ such that 
${\rm pr}_{T}(\wec{a}) \notin {\rm pr}_{T}(M)$.
Since ${\rm pr}_{T-\{j\}}(\wec{a})\in A^{T-\{j\}}={\rm pr}_{T-\{j\}}(M)$
there is a tuple $\wec{b}\in M$ such that 
${\rm pr}_{T-\{j\}}(\wec{a}) = {\rm pr}_{T-\{j\}}(\wec{b})$,
so $\wec{a}$ and $\wec{b}$ agree on $T-\{j\}$.
They do not agree at $j$, since
${\rm pr}_{T}(\wec{a}) \notin {\rm pr}_{T}(M)$ while
${\rm pr}_{T}(\wec{b}) \in {\rm pr}_{T}(M)$.
Using the result established in the preceding paragraph
we may change the coordinate values of $\wec{b}\in M$
at every index outside of the support $T$, 
one by one, to obtain a new tuple $\wec{b}'\in M$ that agrees
with $\wec{a}$ in every coordinate except $j$.
The tuples $\wec{a}\notin M$ and $\wec{b}'\in M$ witness
that the property fails for $j$.

We have established that $M$ has a unique support, say $S$.
To show that ${\rm pr}_S(M)$ is critical we must show that 
it is directly indecomposable. If not, then there is a partition
$\{U,V\}$ of $S$ such that  
\[
{\rm pr}_S(M) = {\rm pr}_U(M)\times {\rm pr}_V(M).
\]
The minimality of $S$ implies that ${\rm pr}_U(M) = A^U$
and ${\rm pr}_V(M) = A^V$, so the above direct decomposition
would imply 
${\rm pr}_S(M) = A^U\times A^V = A^S$, contradicting the definition
of a support.


****

\end{proof}







\subsection{The parallelogram property}\label{parallel}


\subsection{Growth rates for algebras with a cube term}\label{growth_cube}


\subsection{Enumerating maximal subuniverses}\label{enum}



Let $\max(n)$ (or $\max_{\m a}(n))$ denote the number
of maximal subuniverses of $\m a^n$, and let 
$\crit(n)$ (or $\crit_{\m a}(n)$) denote the number
of critical maximal subuniverses of $\m a^n$.

\begin{cor}
(Need CP, or at least that maximal subuniverses 
have the parallelogram property,
so that subuniverses induced by critical maximals
are maximal.)
If $\max(n)$ is a finite-valued function, then it is related to 
$\crit(n)$ by the equations
\[
\max(n) = \sum_{k=1}^n \binom{n}{k} \crit(k)
\quad {\textrm and}\quad 
\crit(n) = \sum_{k=1}^n (-1)^k\binom{n}{k} \max(k).
\]
\end{cor}


\section{Discussion and problems}\label{discussion}


\subsection{}

linear maltsev conditions do not need the finiteness condition


\begin{prb}
Is there a finite algebra $\m a$ where
$d_{\m a}(n)\notin O(\log(n))$ and 
$d_{\m a}(n)\notin \Omega(n)$?
\end{prb}


\subsection{Basic identities}
Kelly's Completeness Theorem was originally proved for
basic identities instead of linear identities.
A term is \emph{basic} if it is a variable, a constant,
or a function symbol applied to variables or constants.
An identity $s=t$ is basic if both terms involved are basic.

Let $C$ denote the set of constant symbols of an
algebraic language $\mathcal L$, let $\Sigma$
be a set of basic $\mathcal L$-identities, and let
$X$ denote a set of variables containing all variables
occurring in $\Sigma$.
The Kelly closure of $\Sigma$ 
relative to $X$ is
the smallest set $\overline{\Sigma}$ 
of basic identities obtained from $\Sigma$ by the following rules:
\begin{enumerate}
\item[(i)]
$\Sigma\subseteq \overline{\Sigma}$;
\item[(ii)] $(t\approx t)\in \overline{\Sigma}$ for all basic
$\mathcal L$-terms $t$ in the variables $X$;
\item[(iii)] if $(s\approx t)\in \overline{\Sigma}$, then 
$(t\approx s)\in \overline{\Sigma}$;
\item[(iv)] if $(r\approx s)\in \overline{\Sigma}$ and
$(s\approx t)\in \overline{\Sigma}$, then 
$(r\approx t)\in \overline{\Sigma}$; and
\item[(v)] if $(s\approx t)\in \overline{\Sigma}$
and $f\colon X\to X\cup C$ is a function, then 
$(s[f]\approx t[f])\in \overline{\Sigma}$, where 
$s[f]$ denotes the basic term obtained from 
$s$ by replacing each variable $x\in X$ with $f(x)\in X\cup C$.
\item[(vi)] if $(s\approx t)\in \overline{\Sigma}$
and $(c\approx d)\in \overline{\Sigma}$ for $c, d\in C$,
then $(s\approx s')\in \overline{\Sigma}$, where
$s'$ is the basic term obtained from $s$
by replacing one occurrence of $c$ with $d$.
\end{enumerate}

A simple example of a set of basic identities is the set
\[
\Sigma: \qquad 1\cdot x \approx x\qquad\textrm{and}\qquad
x\cdot 1 \approx x,
\]
which expresses that the binary term operation, $x\cdot y$,
has a $2$-sided unit. Since the variety
of groups realize this $\Sigma$, there exist
finite algebras with logarithmic or linear growth
rates in varieties realizing this $\Sigma$.
On the other hand, it is easy to see that no finite algebra
in a variety realizing this $\Sigma$ can have larger
than linear growth, since, associating products to the left,
\[
\left[\begin{array}{c}
a_1\\
1\\
\vdots\\
1\\
1\\
\end{array}\right]
\cdot \left[\begin{array}{c}
1\\
a_2\\
\vdots\\
1\\
1\\
\end{array}\right]
\cdots
\left[\begin{array}{c}
1\\
1\\
\vdots\\
a_{n-1}\\
1\\
\end{array}\right]
\cdot
\left[\begin{array}{c}
1\\
1\\
\vdots\\
1\\
a_n\\
\end{array}\right]
=
\left[\begin{array}{c}
a_1\\
a_2\\
\vdots\\
a_{n-1}\\
a_n\\
\end{array}\right]
\]
shows that the set of tuples with at most one entry
different from $1$ is a generating set.
There are $n|A|$ such tuples in $\m A^n$, so 
$d_{\m a}(n)\in O(n)$ when $\m a$ generates a variety
realizing $\Sigma$. We do not know whether there is
a finite algebra generating a variety realizing this $\Sigma$
that has any growth rate other than 
logarithmic or linear, so we ask:

\begin{prb}
Is it true that a finite algebra with a $2$-sided
unit for some binary term has logarithmic or linear growth?
\end{prb}

Here is another example that shows how the addition of constants
affects the results of this paper. Let $C$ be a set of constant symbols,
and for each pair of distinct elements $c, d\in C$
introduce a binary function symbol 
$m_{c,d}(x,y)$. Let $\Sigma$ consist of all identities
of the form
\[
\Sigma: \qquad m_{c,d}(c,x)\approx x\qquad\textrm{and}\qquad
m_{c,d}(d,x) \approx d.
\]
These basic identities express that $c$ is a left unit and $d$
is a left zero for $m_{c,d}$. If $\m a$ is an algebra
in a variety realizing this $\Sigma$
in which two constants interpret as the same element,
say $c^{\m a}=d^{\m a}$, then $m^{\m a}(c^{\m a},x)$ is
both the identity function and a constant function on $A$.
Thus, if $\m a$ is in a variety realizing this $\Sigma$,
then $|A|=1$ when distinct constants
have the same interpretation and  $|A|\geq |C|$ otherwise. 
In particular, if $C$
is infinite, then no variety realizing $\Sigma$
contains a nontrivial finite algebra.



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\end{document}

\section{Generic Interpretability Filters}\label{dichotomy_sec}

Show that polynomial growth is possible for an algebra
satisfying all linear(?) Maltsev conditions where
every end term is a pseudoprojection.

\subsection{Wiegold dichotomy holds for few subpowers}\label{fewsub}

\subsection{Wiegold dichotomy fails for many subpowers}\label{manysub}


There are two principal results in this section.
The first result is the construction, for any 
positive $m$, of a finite algebra whose growth is a polynomial
of degree $m$. According to the recent literature
(cf. \cite{?,?}), this is the first example of 
a finite algebra whose growth rate
is not logarithmic, linear nor exponential.
This particular example has tame congruence
theoretic type-set equal to $\{\styp\}$,
and the variety it generates contains type $\utyp$.
(This information is a measure
of the pathology of this example.)

The second result of this section is rather astonishing.
Let $\m a$ be any finite algebra, and let
$\mathcal L$ be any set of linear Maltsev conditions.
We show that, if $\mathcal L$ does not 
force few subpowers, then 
there is a finite algebra $\m b$ whose growth
is exactly the same as the growth of $\m a$
such that $\m b$ generates
a variety that satisfies all the Maltsev conditions
in $\mathcal L$. 
For example, if one starts with with the algebra
described in the previous paragraph, with growth
$d_{\m a}(n)\in \Theta(n^m)$, one can construct another
finite algebra $\m b$ with $d_{\m b}(n) = d_{\m a}(n)$ such that
$\m b$ generates a variety that is, say,
congruence distributive and congruence
$3$-permutable variety and has any other
prescribed property that is definable by linear Maltsev conditions.


