\input fontmac
\input mathmac
\def\FF{{\bf F}}
\def\T{{\rm T}}
\def\bias{\op{\rm bias}}
\def\prk{\op{\rm prk}}
\def\ark{\op{\rm ark}}
\def\bar{\overline}
\def\hat{\widehat}
\def\norm#1{|\!|#1|\!|}
\def\bignorm#1{\big|\!\big|#1\big|\!\big|}
\def\Norm#1{\Big|\!\Big|#1\Big|\!\Big|}
\def\normm#1{\bigg|\!\bigg|#1\bigg|\!\bigg|}
\def\x{{\bf x}}
\def\y{{\bf y}}
\def\b{{\bf b}}
\widemargins
\bookheader{THE ANALYTIC RANK OF A TENSOR}{MARCEL K. GOH}
\maketitle{The analytic rank of a tensor}{by}{Marcel K. Goh}{22 April 2022}
\floattext4.5 \ninebf Note. \ninepoint These notes are more or less a retelling of
some results in a 2019 paper entitled ``The analytic rank of tensors and its applications'' by
S.~Lovett. I rearranged results in an order that makes more sense to me.
I also go more deeply into definitions (for my own sake) and skip fewer steps,
so for other students this may be easier to
follow than the original paper.
\bigskip
\advsect Introduction
\noindent
There are various compatible definitions of the {\it rank} of a matrix. The one that
extends most easily to the context of tensors, which we define later, is the following.
An $m\times n$ matrix $A$ is said to be {\it rank one} if there exist vectors $u\in \FF^m$
and $v\in \FF^n$ such that $A = uv^\T$. The {\it rank} of a general matrix $A$ is the minimum
number $k$ such that we can write $A = A_1 + \cdots + A_k$, where $A_i$ is a rank one matrix
for all $1\le i\le k$.
In a first course on linear algebra, one usually learns that the rank of a matrix is the
dimension of its column (or row) space. To see that the above definition of rank is equivalent,
let $A$ be a rank-$k$ matrix and let $B = \{b_1, \ldots, b_k\}$
be a basis of its column space. Since every column of $A$
can be written as a linear combination of vectors in $B$, so there is a $k\times n$ matrix $C$
such that $A = BC$. Now letting $c_1, \ldots, c_k$ be the rows of $C$, we have
$$A = b_1{c_1}^\T + \cdots + b_k{c_k}^\T,$$
so the rank of $A$ is at most $k$. On the other hand, if
$$A = u_1{v_1}^\T + \cdots + u_{k'}{v_{k'}}^\T,$$
for some $k'< k$, then for all $x \in \FF^n$, we have
$$Ax = u_1{v_1}^\T x + \cdots + u_{k'}{v_{k'}}^\T x.$$
Since ${v_i}^\T x$ is a scalar for all $1\le i\le k'$, we conclude that $u_1, \ldots, u_{k'}$
span the image of $A$ and the column space of $A$ is at most $k'0$
there exists $n$ such that every subset $A\subseteq \FF_3^n$ with $|A|\ge \delta 3^n$ contains three pairwise
distinct elements $x$, $y$, and $z$ with $x+y+z=0$.
A later paper by R.~Meshulam gave better quantitative
bounds on $n$ with respect to $\delta$; namely, it was shown that we need only take $n > 2/\delta$.
This means that if $A$ is a cap set in $\FF_3^n$, then $|A| \le 2\cdot 3^n/n$. However, it was long
suspected that this bound could be improved to $|A| \le O(c^n)$ for some $c<3$. This was finally proved
in 2017 by J.~S.~Ellenberg and D.~Gijswijt, and T.~Tao showed in a blog post (dated 18 May 2016)
that thir proof can be restated in terms of the partition rank of a function in $3$ variables. This
can actually be modified so the function is a $3$-tensor, but to just get the general idea, let
us extend our definition of partition rank to general functions of three variables temporarily.
Given $A\subseteq \FF_3^n$, let $T : V^3 \to \FF_3$, where $V = {\FF_3}^{\FF_3^n}$, be given by
$$T(e_a,e_b,e_c) = \cases{1, & if $a+b+c=0$;\cr 0, & otherwise,}$$
for basis vectors $e_v$ and extended to all other vectors by linearity. (The function $e_v$ has
$e_v(v) = 1$ and $e_v(x)=0$ for all $x\ne v$.)
Now for a tensor $T : (\FF^X)^d \to \FF$, we say that a set $A\subseteq X$ is an {\it independent
set in $T$} if for all $i_1,\ldots,i_d\in A$, the condition that the coefficient $T_{i_1,\ldots,i_d}$
be nonzero is equivalent to $i_1=\cdots=i_d$.
We then give an upper bound on the size of a cap set by proving that
\medskip
\item{i)} if $A$ contains no nontrivial solutions to $x+y+z=0$, then $A$ is an independent set in $T$;
\smallskip
\item{ii)} if $A$ is an independent set in $T$ then $\prk(T)\ge |A|$; and
\smallskip
\item{iii)} the partition rank of $T$ is low.
\medskip
In these notes, we aim to show that this general strategy may be performed with the partition rank
replaced by something called the analytic rank.
\medskip\boldlabel The analytic rank.
In a 2011 paper, W.~T.~Gowers and J.~Wolf introduced another definition of rank that is Fourier-analytic
in nature. Now we require the field $\FF$ to be finite, and let $\chi:\FF\to\CC$ be any nontrivial
additive character. Recall that for such a character, $\ex_{a\in \FF} \chi(a) = 0$.
The {\it bias} of a tensor $T : V^d\to \FF$ is the average
$$\bias(T) = \ex_{x\in V^d} \chi\bigl(T(x)\bigr).$$
Note that if $T$ is a linear form (i.e., an order-$1$ tensor) that is not identically zero,
then $\bias(T) = 0$, since we can bring the sum inside all three functions and the sum over all elements
a vector space over a finite field is zero. If $T$ is identically zero then $\bias(T) = 1$.
Now to see that the bias of a tensor is always in $(0,1]$, note that if we fix any $(x_2, \ldots, x_d)\in V^{d-1}$,
then $T(x_1,x_2,\ldots,x_d)$ becomes a linear form (order-$1$ tensor) in $x_1$ and
$$\eqalign{
\bias(T)
&= \ex_{x_2,\ldots,x_d\in V} \ex_{x_1\in V} \chi\bigl(T(x_1,\ldots,x_d)\bigr)\cr
&=\pr_{x_2,\ldots,x_d\in V} \bigl\{ T(x_1,\ldots,x_d) \equiv 0\bigr\},\cr
}$$
from our earlier observation about order-$1$ tensors.
The {\it analytic rank} is defined to be the quantity
$$\ark(T) = -\log_{|\FF|} \bias(T);$$
since $\bias(T) \in (0,1]$ we have $\ark(T)\ge 0$. In the case of order-$2$ tensors, the analytic rank
is once again equivalent to ordinary matrix rank. To see this, suppose that $T : (\FF^n)^2\to \FF$
is defined as $T(x,y) = \sum_{i=1}^r x_iy_i$. Then $\bias(T)$ is the probability that, fixing
$y$, the linear form $T(x,y)$ is identically zero. This is equivalent to every coordinate of $y$ being
zero, which happens with probability $1/|\FF|^r$, and hence we see that $\ark(T) = r$.
\advsect Subadditivity of analytic rank
The goal of this section is to prove that if $T$ and $S$ are tensors, then $\ark(T+S)\le \ark(T)+\ark(S)$.
Our first small lemma is the following.
\proclaim Lemma \advthm. Let $W_0, W_1,\ldots,W_n : \FF^m\to \FF$ be functions. Let
functions $A,B : \FF^n\times \FF^m\to \FF$ be given by
$$A(x,y) = \sum_{i=1}^n x_i W_i(y)\qquad\hbox{and}\qquad B(x,y) = A(x,y) + W_0(y).$$
Then
$$\bigl| \bias(B)\bigr| \le \bias(A).$$
\proof We expand
$$\bias(B) = \ex_{x\in \FF^n, y\in \FF^m} \chi\bigl(B(x,y)\bigr)
= \ex_{y\in \FF^m} \indic{W_1(y) = \cdots = W_n(y) = 0} \cdot \chi\bigl(W_0(y)\bigr)$$
and by the triangle inequality,
$$\bigl|\bias(B)\bigr| \le \ex_y \indic{W_1(y) = \cdots = W_n(y) = 0} = \bias(A).\noskipslug$$
This lemma is used to prove a bound on the bias of a certain sum of tensors.
First, we introduce some notation. With some $d$ fixed, we let $\x = (x_1,\ldots,x_d)$ and
similarly for $\y = (y_1,\ldots,y_d)$. Then for
$I\subseteq [d]$, define $I^c = [d]\setminus I$ and let $\x_I = (x_i : i\in I)$.
\edef\normbound{\the\thmcount}
\proclaim Lemma \advthm. Let $d\ge 1$ and for each $I\subseteq [d]$, let $R_I : V^I\to \FF$
be an order-$|I|$ tensor. Consider the function
$$R(\x) = \sum_{I\subseteq [d]} R_I(x_I).$$
Then
$$\bigl|\bias(R)\bigr| \le \bias(R_{[d]}).$$
\proof Fix some $i\in [d]$ and write $R(\x)$ as
$$R(\x) = \sum_{I \ni i} R_I(x_I) + \sum_{I\not\ni i} R_I(x_I).$$
Setting $x = x_i$ and $y = \x_{[d]\setminus\{i\}}$, the first sum has the form
$$\sum_{i=1}^n x_iW_i(y)$$ and the second sum does not depend on $x$ at all, so we can set it
to be $W_0(y)$. The previous lemma now tells us that
$$\bigl|\bias(R)\bigr| \le \bias\Bigl(\sum_{I\ni i} R_I(\x_I)\Bigr).$$
Now iterate this with $i=d$ all the way down to $i=1$ (replacing $d$ by $d-1$ each time)
to get the statement of the lemma.\slug
Before proving subadditivity, we introduce another bit of notation. For $I\subseteq [d]$,
let
$$T_I(\x,\y) = T(\x_I, \y_{I^c}) = T(z_1, \ldots, z_d),$$
where $z_i = x_i$ if $i\in I$ and $z_i = y_i$ if $i\notin I$. After expanding out the definition
of multilinearity, we see that $T(\x+\y)$ decomposes as
$$ T(\x+\y) = \sum_{I\subseteq[d]} T_I(\x,\y).$$
\edef\subadditivity{\the\thmcount}
\proclaim Theorem \advthm. Let $T,S: V^d\to \FF$ be order-$d$ tensors. Then
$$\ark(T+S) \le \ark(T) + \ark(S).$$
\proof It is enough to show that
$$\bias(T+S) \ge \bias(T)\bias(S).$$
We express
$$\eqalign{
\bias(T)\bias(S) &= \Bigl(\ex_{\x\in V^d} \chi\bigl(T(\x)\bigr)\Bigr)
\Bigl(\ex_{\y\in V^d} \chi\bigl(T(\y)\bigr)\Bigr)\cr
&= \ex_{\x\in V^d} \ex_{\y\in V^d} \chi\bigl( T(\x) + S(\y)\bigr) \cr
&= \ex_{\x\in V^d} \ex_{\y\in V^d} \chi\bigl( T(\x) + S(\x+\y)\bigr) \cr
&= \bias\bigl( T(\x) + S(\x+\y)\bigr), \cr
}$$
and by our earlier decomposition the tensor product of a sum, this gives us
$$\bias(T)\bias(S)
= \bias \Bigl( T(\x) + \sum_{I\subseteq [d]} S_I(\x,\y) \Bigr)
= \bias \Bigl( (T+S)(\x) + \sum_{I\ne [d]} S_I(\x,\y) \Bigr), $$
Setting $\y = \b\in V^d$ to maximise the norm of this right-hand side, we have
$$\bias(T)\bias(S) \le \Bigl| \bias\Bigl( (T+S)(\x) + \sum_{I\ne [d]} S_I(\x, \b)\Bigr)\Bigr|.$$
Here $S_I(\x,\b)$ is an order-$|I|$ tensor in the variables $\x_I$. Thus we may apply
the previous lemma with
$$R_{[d]}(\x) = (T+S)(\x)\qquad\hbox{and}\qquad R_I(\x_I) = S_I(\x,\b)\ \hbox{for}\ I\ne [d]$$
to get $\bias(T)\bias(S) \le \bias(T+S)$.\slug
As an application of this theorem, we show that common roots of tensors on a common input are
positively correlated.
\proclaim Corollary \advthm. Let $T_1,\ldots,T_m, S_1,\ldots,S_n : V^d\to \FF$ be order-$d$ tensors.
Then
$$\eqalign{
\pr_{x\in V^d}&\bigl\{ T_1(x) = \cdots = T_m(x) = S_1(x) = \cdots = S_n(x) = 0\bigr\}\ge \cr
&\pr_{x\in V^d}\bigl\{ T_1(x) = \cdots = T_m(x) = 0\}\cdot
\pr_{x\in V^d}\bigl\{ S_1(x) = \cdots = S_n(x) = 0\}
}$$
\proof Define order-$(d+1)$ tensors $T$ and $S$ by setting
$$T(x_0,x_1,\ldots,x_d) = \sum_{i=1}^m x_{0,i} T_i(x_1,\ldots, x_d)$$
and
$$S(x_0,x_1,\ldots,x_d) = \sum_{i=m+1}^{m+n} x_{0,i} S_{i-m}(x_1,\ldots,x_d).$$
The left-hand side of the desired inequality is $\bias(T+S)$ and the right-hand side is
$\bias(T)\bias(S)$.\slug
\advsect Analytic rank and partition rank
In this section, we shall discuss properties of the analytic rank that allow it to be used as
the partition rank was used above in the cap-set problem. First, we show that the analytic rank
is bounded above by the partition rank.
\proclaim Theorem \advthm. Let $T:V^d\to\FF$ be an order-$d$ tensor. Then $\ark(T)\le \prk(T)$.
\proof By Theorem~\subadditivity, we only need to consider the case where $T$ has partition rank one.
So we assume that $T:V^d\to \FF$ factors as
$$T(\x) = T_1(\x_A)T_2(\x_B),$$
where $A\cup B = [d]$ is a nontrivial partition of $[d]$. We shall show that $\ark(T)\le 1 = \prk(T)$ by
showing that $\bias(T) \ge 1/|\FF|$. For $a,b\in \FF$ define the function
$$ F_{a,b}(\x) = \bigl(T_1(\x_A) + a\bigr) \bigl( T_2(\x_B) + b\bigr).$$
Letting $R_{[d]} = T$, $R_{\emptyset} = ab$, $R_A = bT_1(\x_A)$, $R_B = aT_2(\x_B)$, and $R_I$ be the
zero tensor (of the corresponding order)
for all other $I\subseteq [d]$, we see that $R = F_{a,b}$ and
by Lemma~\normbound, $\bigl|\bias(F_{a,b})\bigr| \le \bias(T)$.
On the other hand, we have
$$\eqalign{
\ex_{a,b\in \FF} \bias(F_{a,b}) &= \ex_{a,b\in \FF} \ex_{\x\in V^d}
\chi\bigl( (T_1(\x_A) + a)(T_2(\x_B) + b) \bigr) \cr
&= \ex_{a,b\in F} \chi(ab) \cr
&= \pr_{b\in \FF} \{b=0\} \cr
&= {1\over |\FF|}. \cr
}$$
So we may find $a,b$ such that $\bias(T) \ge \bigl| \bias(F_{a,b})\bigr| \ge 1/|\FF|$, which
is what we wanted.\slug
Since we are often interested in finding some upper bound on the partition rank in applications,
this theorem suggests that it should actually be easier to upper bound the analytic
rank, though {\it ad hoc} methods to do so have not currently been developed.
In applications, it is also important that the partition rank of a tensor $T$
is bounded below by the size of an
independent set in $T$. Towards showing a version of this for the analytic rank, we first show
that the analytic rank does not increase when the tensor is restricted to a subspace.
\proclaim Lemma \advthm. Let $T : V^d \to \FF$ be an order-$d$ tensor and let $U\subseteq V$ be a
subspace. If $T|_U : U^d\to \FF$ is the restriction of $T$ to $U$, then $\ark(T|_U) \le \ark(T)$.
\proof Let $W\subseteq V$ be a subspace so that $V = U\oplus W$. Then any $v\in V$ has a unique
expression as $v = u+w$ for $u\in U$ and $w\in W$, so
$$\bias(T) = \ex_{u_1,\ldots,u_d\in U} \ex_{w_1,\ldots,w_d\in W}\chi\bigl(T(u_1+w_1,\ldots,u_d+w_d)\bigr).$$
Fix an arbitrary choice of $w_1,\ldots,w_d\in W$. Then
$$T(u_1+w_1,\ldots, u_d+w_d) = \sum_{I\subseteq [d]} T_I(u_I, w_{I^c}),$$
where $T_I(u_I, w_{I^c}) = T(z_1,\ldots,z_d)$ where $z_i = u_i$ if $i\in I$ and $z_i = w_i$ if $i\notin I$.
By Lemma~\normbound, we have
$$\eqalign{
\bigl| \ex_{u_1,\ldots,u_d\in U} \chi\bigl( T(u_1+w_1,\ldots,u_d+w_d) \bigr) \bigr|
&\le \ex_{u_1,\ldots,u_d\in U} \chi\bigl( T(u_1,\ldots,u_d)\bigr) \cr
&= \bias(T|_U) \cr
}$$
Since this is for all $w_1,\ldots, w_d$, averaging the left-hand side over all choices and
applying the triangle inequality gives us what we want.\slug
We are now able to show that for any tensor $T$ and independent set $A$ in $T$, we have
$\ark(T) \ge c|A|$ for $c$ depending on $d$ and $|\FF|$.
\proclaim Theorem \advthm. Let $T : (\FF^n)^d\to \FF$ be an order-$d$ tensor. Assume that
$A\subseteq [n]$ is an independent set in $T$. Then $\ark(T)\ge c|A|$ where
$$c = -\log_{|\FF|} \biggl( 1-\biggl( 1-{1\over |\FF|}\biggr)^{d-1}\biggr),$$
so that $c\ge 2^{-d}$ and $c\ge 1-\log(d-1)/\log |\FF|$.
\proof Let $S : (F^A)^d\to \FF$ be the restriction of $T$ to $\FF^A$.
By the previous lemma, $\ark(S)$ is a lower bound on $\ark(T)$, so we shall prove the theorem
by expressing $\ark(S)$ as a function of $|A|$. We have
$$\bias(S) = \ex_{x_1,\ldots,x_d\in \FF^A} \chi\Bigl( \sum_{i\in A} c_i \prod_{j\in [d]} x_{i,j}\Bigr),$$
where, since $A$ is an independent set, $c_i\ne 0$ for all $i\in A$. But we also know that
$$\bias(S) =
\pr_{x_2,\ldots,x_d\in \FF^A} \Bigl\{ \bigcap_{i\in A} \{x_{i,2}\cdots x_{i,d} = 0\}\Bigr\},$$
and this is easily seen to be
$$\bias(S) = \biggl( 1-\biggl( 1-{1\over |\FF|}\biggr)^{d-1} \biggr)^{|A|}.$$
Taking the logarithm of both sides, we see that $\ark(S) = c|A|$, where $c$ is as in the theorem
statement.
By convexity,
$$c \ge -\log_2 (1-2^{-(d-1)}) \ge 2^{-(d-1)}.$$
Also, as long as $|\FF|\ge d$, we have
$$c \ge -\log_{|\FF|}\bigl((d-1)/|\FF|\bigr) = 1-\log_{|\FF|} (d-1).\noskipslug$$
\section References
\bye