# Data-structure lower bounds without encoding arguments

I have recently posted the paper [Vio21] (download) which does something that I have been trying to do for a long time, more than ten years, on and off. Consider the basic data-structure problem of storing $m$ bits of data $x\in \{0,1\}^{m}$ into $m+r$ bits so that the prefix-sum queries

\begin{aligned} \mathbb {\text {\textsc {Rank}}}(i):=\sum _{j\le i}x_{j} \end{aligned}

can be computed by probing $q$ cells (or words) of $w$ bits each. (You can think $w=\log m$ throughout this post.) The paper [PV10] with Pǎtraşcu shows that $r\ge m/w^{O(q)}$, and this was recently shown to be tight by Yu [Yu19] (building on the breakthrough data structure [Pǎt08] which motivated the lower bound and is not far from it).

As is common in data-structure lower bounds, the proof in [PV10] is an encoding argument. In the recently posted paper, an alternative proof is presented which avoids the encoding argument and is perhaps more in line with other proofs in complexity lower bounds. Of course, everything is an encoding argument, and nothing is an encoding argument, and this post won’t draw a line.

The new proof establishes an intrinsic property of efficient data structures, whereas typical proofs including [PV10] are somewhat tailored to the problem at hand. The property is called the separator and is a main technical contribution of the work. At the high level the separator shows that in any efficient data structure you can restrict the input space a little so that many queries are nearly pairwise independent.

Also, the new proof rules out a stronger object: a sampler (see previous post here on sampling lower bounds). Specifically, the distribution Rank$(U)$ where $U$ is the uniform distribution cannot be sampled, not even slightly close, by an efficient cell-probe algorithm. This implies the data-structure result, and it can be informally interpreted as saying that the “reason” why the lower bound holds is not that the data is compressed, but rather that one can’t generate the type of dependencies occurring in Rank via an efficient cell-probe algorithm, regardless of what the input is.

Building on this machinery, one can prove several results about sampling, like showing that cell-probe samplers are strictly weaker than AC0 samplers. While doing this, it occurred to me that one gets a corollary for data structures which I had not seen in the literature. The corollary is a probe hierarchy, showing that some problem can be solved with zero redundancy ($r=0$) with $O(q)$ probes, while it requires almost linear $r$ for $q$ probes. For example I don’t know of a result yielding this for small $q$ such as $q=O(1)$; I would appreciate a reference. (As mentioned in the paper, the sampling viewpoint is not essential and just like for Rank one can prove the data-structure corollaries directly. Personally, and obviously, I find the sampling viewpoint useful.)

One of my favorite open problems in the area still is: can a uniform distribution over $[m]$ be approximately sampled by an efficient cell-probe algorithm? I can’t even rule out samplers making two probes!

### References

[Pǎt08]   Mihai Pǎtraşcu. Succincter. In 49th IEEE Symp. on Foundations of Computer Science (FOCS). IEEE, 2008.

[PV10]   Mihai Pǎtraşcu and Emanuele Viola. Cell-probe lower bounds for succinct partial sums. In 21th ACM-SIAM Symp. on Discrete Algorithms (SODA), pages 117–122, 2010.

[Vio21]   Emanuele Viola. Lower bounds for samplers and data structures via the cell-probe separator. Available at http://www.ccs.neu.edu/home/viola/, 2021.

[Yu19]    Huacheng Yu. Optimal succinct rank data structure via approximate nonnegative tensor decomposition. In Moses Charikar and Edith Cohen, editors, ACM Symp. on the Theory of Computing (STOC), pages 955–966. ACM, 2019.

# Non-abelian combinatorics and communication complexity

Below and here in pdf is a survey I am writing for SIGACT, due next week.  Comments would be very helpful.

Finite groups provide an amazing wealth of problems of interest to complexity theory. And complexity theory also provides a useful viewpoint of group-theoretic notions, such as what it means for a group to be “far from abelian.” The general problem that we consider in this survey is that of computing a group product $g=x_{1}\cdot x_{2}\cdot \cdots \cdot x_{n}$ over a finite group $G$. Several variants of this problem are considered in this survey and in the literature, including in .

Some specific, natural computational problems related to $g$ are, from hardest to easiest:

(1) Computing $g$,

(2) Deciding if $g=1_{G}$, where $1_{G}$ is the identity element of $G$, and

(3) Deciding if $g=1_{G}$ under the promise that either $g=1_{G}$ or $g=h$ for a fixed $h\ne 1_{G}$.

Problem (3) is from [MV13]. The focus of this survey is on (2) and (3).

We work in the model of communication complexity [Yao79], with which we assume familiarity. For background see [KN97RY19]. Briefly, the terms $x_{i}$ in a product $x_{1}\cdot x_{2}\cdot \cdots \cdot x_{n}$ will be partitioned among collaborating parties – in several ways – and we shall bound the number of bits that the parties need to exchange to solve the problem.

Organization.

We begin in Section 2 with two-party communication complexity. In Section 3 we give a streamlined proof, except for a step that is only sketched, of a result of Gowers and the author [GV15GVb] about interleaved group products. In particular we present an alternative proof, communicated to us by Will Sawin, of a lemma from [GVa]. We then consider two models of three-party communication. In Section 4 we consider number-in-hand protocols, and we relate the communication complexity to so-called quasirandom groups [Gow08BNP08]. In Section 6 we consider number-in-hand protocols, and specifically the problem of separating deterministic and randomized communication. In Section 7 we give an exposition of a result by Austin [Aus16], and show that it implies a separation that matches the state-of-the-art [BDPW10] but applies to a different problem.

Some of the sections follow closely a set of lectures by the author [Vio17]; related material can also be found in the blog posts [VioaViob]. One of the goals of this survey is to present this material in a more organized matter, in addition to including new material.

### 2 Two parties

Let $G$ be a group and let us start by considering the following basic communication task. Alice gets an element $x\in G$ and Bob gets an element $y\in G$ and their goal is to check if $x\cdot y=1_{G}$. How much communication do they need? Well, $x\cdot y=1_{G}$ is equivalent to $x=y^{-1}$. Because Bob can compute $y^{-1}$ without communication, this problem is just a rephrasing of the equality problem, which has a randomized protocol with constant communication. This holds for any group.

The same is true if Alice gets two elements $x_{1}$ and $x_{2}$ and they need to check if $x_{1}\cdot y\cdot x_{2}=1_{G}$. Indeed, it is just checking equality of $y$ and $x_{1}^{-1}\cdot x_{2}^{-1}$, and again Alice can compute the latter without communication.

Things get more interesting if both Alice and Bob get two elements and they need to check if the interleaved product of the elements of Alice and Bob equals $1_{G}$, that is, if

\begin{aligned} x_{1}\cdot y_{1}\cdot x_{2}\cdot y_{2}=1_{G}. \end{aligned}

Now the previous transformations don’t help anymore. In fact, the complexity depends on the group. If it is abelian then the elements can be reordered and the problem is equivalent to checking if $(x_{1}\cdot x_{2})\cdot (y_{1}\cdot y_{2})=1_{G}$. Again, Alice can compute $x_{1}\cdot x_{2}$ without communication, and Bob can compute $y_{1}\cdot y_{2}$ without communication. So this is the same problem as before and it has a constant communication protocol.

For non-abelian groups this reordering cannot be done, and the problem seems hard. This can be formalized for a class of groups that are “far from abelian” – or we can take this result as a definition of being far from abelian. One of the groups that works best in this sense is the following, first constructed by Galois in the 1830’s.

Definition 1. The special linear group $SL(2,q)$ is the group of $2\times 2$ invertible matrices over the field $\mathbb{F} _{q}$ with determinant $1$.

The following result was asked in [MV13] and was proved in [GVa].

Theorem 1. Let $G=SL(2,q)$ and let $h\ne 1_{G}$. Suppose Alice receives $x_{1},x_{2}\in G$ and Bob receives $y_{1},y_{2}\in G$. They are promised that $x_{1}\cdot y_{1}\cdot x_{2}\cdot y_{2}$ either equals $1_{G}$ or $h$. Deciding which case it is requires randomized communication $\Omega (\log |G|)$.

This bound is tight as Alice can send her input, taking $O(\log |G|)$ bits. We present the proof of this theorem in the next section.

Similar results are known for other groups as well, see [GVa] and [Sha16]. For example, one group that is “between” abelian groups and $SL(2,q)$ is the following.

Definition 2. The alternating group $A_{n}$ is the group of even permutations of $1,2,\ldots ,n$.

If we work over $A_{n}$ instead of $SL(2,q)$ in Theorem 1 then the communication complexity is $\Omega (\log \log |G|)$ [Sha16]. The latter bound is tight [MV13]: with knowledge of $h$, the parties can agree on an element $a\in {1,2,\ldots ,n}$ such that $h(a)\ne a$. Hence they only need to keep track of the image $a$. This takes communication $O(\log n)=O(\log \log |A_{n}|)$ because $|A_{n}|=n!/2.$ In more detail, the protocol is as follows. First Bob sends $y_{2}(a)$. Then Alice sends $x_{2}y_{2}(a)$. Then Bob sends $y_{1}x_{2}y_{2}(a)$ and finally Alice can check if $x_{1}y_{1}x_{2}y_{2}(a)=a$.

Interestingly, to decide if $g=1_{G}$ without the promise a stronger lower bound can be proved for many groups, including $A_{n}$, see Corollary 3 below.

In general, it seems an interesting open problem to try to understand for which groups Theorem 1 applies. For example, is the communication large for every quasirandom group [Gow08]?

Theorem 1 and the corresponding results for other groups also scale with the length of the product: for example deciding if $x_{1}\cdot y_{1}\cdot x_{2}\cdot y_{2}\cdots x_{n}\cdot y_{n}=1_{G}$ over $G=SL(2,q)$ requires communication $\Omega (n\log |G|)$ which is tight.

A strength of the above results is that they hold for any choice of $h$ in the promise. This makes them equivalent to certain $mixing$ results, discussed below in Section 5.0.1. Next we prove two other lower bounds that do not have this property and can be obtained by reduction from disjointness. First we show that for any non-abelian group $G$ there exists an element $h$ such that deciding if $g=1_{G}$ or $g=h$ requires communication linear in the length of the product. Interestingly, the proof works for any non-abelian group. The choice of $h$ is critical, as for some $G$ and $h$ the problem is easy. For example: take any group $G$ and consider $H:=G\times \mathbb {Z}_{2}$ where $\mathbb {Z}_{2}$ is the group of integers with addition modulo $2$. Distinguishing between $1_{H}=(1_{G},0)$ and $h=(1_{G},1)$ amounts to computing the parity of (the $\mathbb {Z}_{2}$ components of) the input, which takes constant communication.

Theorem 2. Let $G$ be a non-abelian group. There exists $h\in G$ such that the following holds. Suppose Alice receives $x_{1},x_{2},\ldots ,x_{n}$ and receives $y_{1},y_{2},\ldots ,y_{n}$. They are promised that $x_{1}\cdot y_{1}\cdot x_{2}\cdot y_{2}\cdot \cdots \cdot x_{n}\cdot y_{n}$ either equals $1_{G}$ or $h$. Deciding which case it is requires randomized communication $\Omega (n)$.

Proof. We reduce from unique set-disjointness, defined below. For the reduction we encode the And of two bits $s,t\in \{0,1\}$ as a group product. This encoding is similar to the famous puzzle that asks to hang a picture on a wall with two nails in such a way that the picture falls if either one of the nails is removed. Since $G$ is non-abelian, there exist $a,b\in G$ such that $a\cdot b\neq b\cdot a$, and in particular $a\cdot b\cdot a^{-1}\cdot b^{-1}=h$ with $h\neq 1$. We can use this fact to encode the And of $s$ and $t$ as

\begin{aligned} a^{s}\cdot b^{t}\cdot a^{-s}\cdot b^{-t}=\begin {cases} 1~~\text {if And\ensuremath {(s,t)=0}}\\ h~~\text {otherwise} \end {cases}. \end{aligned}

In the disjointness problem Alice and Bob get inputs $x,y\in \{0,1\}^{n}$ respectively, and they wish to check if there exists an $i\in [n]$ such that $x_{i}\land y_{i}=1$. If you think of $x,y$ as characteristic vectors of sets, this problem is asking if the sets have a common element or not. The communication of this problem is $\Omega (n)$ [KS92Raz92]. Moreover, in the “unique” variant of this problem where the number of such $i$’s is 0 or 1, the same lower bound $\Omega (n)$ still applies. This follows from [KS92Raz92] – see also Proposition 3.3 in [AMS99]. For more on disjointness see the surveys [She14CP10].

We will reduce unique disjointness to group products. For $x,y\in \{0,1\}^{n}$ we produce inputs for the group problem as follows:

\begin{aligned} x & \rightarrow (a^{x_{1}},a^{-x_{1}},\ldots ,a^{x_{n}},a^{-x_{n}})\\ y & \rightarrow (b^{y_{1}},b^{-y_{1}},\ldots ,b^{y_{n}},b^{-y_{n}}). \end{aligned}

The group product becomes

\begin{aligned} \underbrace {a^{x_{1}}\cdot b^{y_{1}}\cdot a^{-x_{1}}\cdot b^{-y_{1}}}_{\text {1 bit}}\cdots \cdots a^{x_{n}}\cdot b^{y_{n}}\cdot a^{-x_{n}}\cdot b^{-y_{n}}. \end{aligned}

If there isn’t an $i\in [n]$ such that $x_{i}\land y_{i}=1$, then for each $i$ the term $a^{x_{i}}\cdot b^{y_{i}}\cdot a^{-x_{i}}\cdot b^{-y_{i}}$ is $1_{G}$, and thus the whole product is 1.

Otherwise, there exists a unique $i$ such that $x_{i}\land y_{i}=1$ and thus the product will be $1\cdots 1\cdot h\cdot 1\cdots 1=h$, with $h$ being in the $i$-th position. If Alice and Bob can check if the above product is equal to 1, they can also solve the unique set disjointness problem, and thus the lower bound applies for the former. $\square$

We required the uniqueness property, because otherwise we might get a product $h^{c}$ that could be equal to 1 in some groups.

Next we prove a result for products of length just $4$; it applies to non-abelian groups of the form $G=H^{n}$ and not with the promise.

Theorem 3. Let $H$ be a non-abelian group and consider $G=H^{n}$. Suppose Alice receives $x_{1},x_{2}$ and Bob receives $y_{1},y_{2}$. Deciding if $x_{1}\cdot y_{1}\cdot x_{2}\cdot y_{2}=1_{G}$ requires randomized communication $\Omega (n)$.

Proof. The proof is similar to the proof of Theorem 2. We use coordinate $i$ of $G$ to encode bit $i$ of the disjointness instance. If there is no intersection in the latter, the product will be $1_{G}$. Otherwise, at least some coordinate will be $\ne 1_{G}$. $\square$

As a corollary we can prove a lower bound for $A_{n}$.

Corollary 3. Theorem 3 holds for $G=A_{n}$.

Proof. Note that $A_{n}$ contains $(A_{4})^{\lfloor n/4\rfloor }$ and that $A_{4}$ is not abelian. Apply Theorem 3. $\square$

Theorem 3 is tight for constant-size $G$. We do not know if Corollary 3 is tight. The trivial upper bound is $O(\log |A_{n}|)=O(n\log n)$.

### 3 Proof of Theorem 1

Several related proofs of this theorem exist, see [GV15GVaSha16]. As in [GVa], the proof that we present can be broken down in three steps. First we reduce the problem to a statement about conjugacy classes. Second we reduce this to a statement about trace maps. Third we prove the latter. We present the first step in a way that is similar but slightly different from the presentation in [GVa]. The second step is only sketched, but relies on classical results about $SL(2,q)$ and can be found in [GVa]. For the third we present a proof that was communicated to us by Will Sawin. We thank him for his permission to include it here.

#### 3.1 Step 1

We would like to rule out randomized protocols, but it is hard to reason about them directly. Instead, we are going to rule out deterministic protocols on random inputs. First, for any group element $g\in G$ we define the distribution on quadruples $D_{g}:=(x_{1},y_{1},x_{2},(x_{1}\cdot y_{1}\cdot x_{2})^{-1}g)$, where $x,y\in G$ are uniformly random elements. Note the product of the elements in $D_{g}$ is always $g$.

Towards a contradiction, suppose we have a randomized protocol $P$ such that

\begin{aligned} \mathbb{P} [P(D_{1})=1]\geq \mathbb{P} [P(D_{h})=1]+\frac {1}{10}. \end{aligned}

This implies a deterministic protocol with the same gap, by fixing the randomness.

We reach a contradiction by showing that for every deterministic protocol $P$ using little communication, we have

\begin{aligned} |\Pr [P(D_{1})=1]-\Pr [P(D_{h})=1]|\leq \frac {1}{100}. \end{aligned}

We start with the following standard lemma, which describes a protocol using product sets.

Lemma 4. (The set of accepted inputs of) A deterministic $c$-bit protocol for a function $f:X\times Y\to Z$ can be written as a disjoint union of $2^{c}$ rectangles, where a rectangle is a set of the form $A\times B$ with $A\subseteq X$ and $B\subseteq Y$ and where $f$ is constant.

Proof. (sketch) For every communication transcript $t$, let $S_{t}\subseteq G^{2}$ be the set of inputs giving transcript $t$. The sets $S_{t}$ are disjoint since an input gives only one transcript, and their number is $2^{c}$: one for each communication transcript of the protocol. The rectangle property can be proven by induction on the protocol tree. $\square$

Next, we show that any rectangle $A\times B$ cannot distinguish $D_{1},D_{h}$. The way we achieve this is by showing that for every $g$ the probability that $(A\times B)(D_{g})=1$ is roughly the same for every $g$, and is roughly the density of the rectangle. (Here we write $A\times B$ for the characteristic function of the set $A\times B$.) Without loss of generality we set $g=1_{G}$. Let $A$ have density $\alpha$ and $B$ have density $\beta$. We aim to bound above

\begin{aligned} \left |\mathbb{E} _{a_{1},b_{1},a_{2},b_{2}:a_{1}b_{1}a_{2}b_{2}=1}A(a_{1},a_{2})B(b_{1},b_{2})-\alpha \beta \right |, \end{aligned}

where note the distribution of $a_{1},b_{1},a_{2},b_{2}$ is the same as $D_{1}$.

Because the distribution of $(b_{1},b_{2})$ is uniform in $G^{2}$, the above can be rewritten as

\begin{aligned} & \left |\mathbb{E} _{b_{1},b_{2}}B(b_{1},b_{2})\mathbb{E} _{a_{1},a_{2}:a_{1}b_{1}a_{2}b_{2}=1}(A(a_{1},a_{2})-\alpha )\right |\\ & \le \sqrt {\mathbb{E} _{b_{1},b_{2}}B(b_{1},b_{2})^{2}}\sqrt {\mathbb{E} _{b_{1},b_{2}}\mathbb{E} _{a_{1},a_{2}:a_{1}b_{1}a_{2}b_{2}=1}^{2}(A(a_{1},a_{2})-\alpha )}.\\ & =\sqrt {\beta }\sqrt {\mathbb{E} _{b_{1},b_{2},a_{1},a_{2},a_{1}',a_{2}':a_{1}b_{1}a_{2}b_{2}=a_{1}'b_{1}a_{2}'b_{2}=1}A(a_{1},a_{2})A(a_{1}',a_{2}')-\alpha ^{2}}. \end{aligned}

The inequality is Cauchy-Schwarz, and the step after that is obtained by expanding the square and noting that $(a_{1},a_{2})$ is uniform in $G^{2}$, so that the expectation of the term $A(a_{1},a_{2})\alpha$ is $\alpha ^{2}$.

Now we do several transformations to rewrite the distribution in the last expectation in a convenient form. First, right-multiplying by $b_{2}^{-1}$ we can rewrite the distribution as the uniform distribution on tuples such that

\begin{aligned} a_{1}b_{1}a_{2}=a_{1}'b_{1}a_{2}'. \end{aligned}

The last equation is equivalent to $b_{1}^{-1}(a_{1}')^{-1}a_{1}b_{1}a_{2}=a_{2}'$.

We can now do a transformation setting $a_{1}'$ to be $a_{1}x^{-1}$ to rewrite the distribution of the four-tuple as

\begin{aligned} (a_{1},a_{2},a_{1}x^{-1},C(x)a_{2}) \end{aligned}

where we use $C(x)$ to denote a uniform element from the conjugacy class of $x$, that is $b^{-1}xb$ for a uniform $b\in G$.

Hence it is sufficient to bound

\begin{aligned} \left |\mathbb{E} A(a_{1},a_{2})A(a_{1}x^{-1},C(x)a_{2})-\alpha ^{2}\right |, \end{aligned}

where all the variables are uniform and independent.

With a similar derivation as above, this can be rewritten as

\begin{aligned} & \left |\mathbb{E} A(a_{1},a_{2})\mathbb{E} (A(a_{1}x^{-1},C(x)a_{2})-\alpha )\right |\\ & \le \sqrt {\mathbb{E} A(a_{1},a{}_{2})^{2}}\sqrt {\mathbb{E} _{a_{1},a_{2}}\mathbb{E} _{x}^{2}(A(a_{1}x^{-1},C(x)a_{2})-\alpha )}.\\ & =\sqrt {\alpha }\sqrt {\mathbb{E} A(a_{1}x^{-1},C(x)a_{2})A(a_{1}x'^{-1},C(x')a_{2})-\alpha ^{2}}. \end{aligned}

Here each occurrence of $C$ denotes a uniform and independent conjugate. Hence it is sufficient to bound

\begin{aligned} \left |\mathbb{E} A(a_{1}x^{-1},C(x)a_{2})A(a_{1}x'^{-1},C(x')a_{2})-\alpha ^{2}\right |. \end{aligned}

We can now replace $a_{2}$ with $C(x)^{-1}a_{2}.$ Because $C(x)^{-1}$ has the same distribution of $C(x^{-1})$, it is sufficient to bound

\begin{aligned} \left |\mathbb{E} A(a_{1}x^{-1},a_{2})A(a_{1}x'^{-1},C(x')C(x^{-1})a_{2})-\alpha ^{2}\right |. \end{aligned}

For this, it is enough to show that with high probability $1-1/|G|^{\Omega (1)}$ over $x'$ and $x$, the distribution of $C(x')C(x^{-1})$, over the choice of the two independent conjugates, has statistical distance $\le 1/|G|^{\Omega (1)}$ from uniform.

#### 3.2 Step 2

In this step we use information on the conjugacy classes of the group to reduce the latter task to one about the equidistribution of the trace map. Let $Tr$ be the Trace map:

\begin{aligned} Tr\begin {pmatrix}a_{1} & a_{2}\\ a_{3} & a_{4} \end {pmatrix}=a_{1}+a_{4}. \end{aligned}

We state the lemma that we want to show.

Lemma 5. Let $a:=\begin {pmatrix}0 & 1\\ 1 & w \end {pmatrix}$ and $b:=\begin {pmatrix}v & 1\\ 1 & 0 \end {pmatrix}$. For all but $O(1)$ values of $w\in \mathbb{F} _{q}$ and $v\in \mathbb{F} _{q}$, the distribution of

\begin{aligned} Tr\left (au^{-1}bu\right ) \end{aligned}

is $O(1/q)$ close to uniform over $\mathbb{F} _{q}$ in statistical distance.

To give some context, in $SL(2,q)$ the conjugacy class of an element is essentially determined by the trace. Moreover, we can think of $a$ and $b$ as generic elements in $G$. So the lemma can be interpreted as saying that for typical $a,b\in G$, taking a uniform element from the conjugacy class of $b$ and multiplying it by $a$ yields an element whose conjugacy class is uniform among the classes of $G$. Using that essentially all conjugacy classes are equal, and some of the properties of the trace map, one can show that the above lemma implies that for typical $x,x'$ the distribution of $C(x')C(x^{-1})$ is close to uniform. For more on how this fits we refer the reader to [GVa].

#### 3.3 Step 3

We now present a proof of Lemma 5. The high-level argument of the proof is the same as in [GVa] (Lemma 5.5), but the details may be more accessible and in particular the use of the Lang-Weil theorem [LW54] from algebraic geometry is replaced by a more elementary argument. For simplicity we shall only cover the case where $q$ is prime. We will show that for all but $O(1)$ values of $v,w,c\in \mathbb{F} _{q}$, the probability over $u$ that $Tr(au^{-1}bu)=c$ is within $O(1/q^{2})$ of $1/q$, and for the others it is at most $O(1/q)$. Summing over $c$ gives the result.

We shall consider elements $b$ whose trace is unique to the conjugacy class of $b$. (This holds for all but $O(1)$ conjugacy classes – see for example [GVa] for details.) This means that the distribution of $u^{-1}bu$ is that of a uniform element in $G$ conditioned on having trace $b$. Hence, we can write the probability that $Tr(au^{-1}bu)=c$ as the number of solutions in $x$ to the following three equations (divided by the size of the group, which is $q^{3}-q$):

\begin{aligned} x_{3}+x_{2}+wx_{4} & =c & \hspace {1cm}(Tr(ax)=c),\\ x_{1}+x_{4} & =v & \hspace {1cm}(Tr(x)=Tr(b)),\\ x_{1}x_{4}-x_{3}x_{3} & =1 & \hspace {1cm}(Det(x)=1). \end{aligned}

We use the second one to remove $x_{1}$ and the first one to remove $x_{2}$ from the last equation. This gives

\begin{aligned} (v-x_{4})x_{4}-(c-x_{3}-wx_{4})x_{3}=1. \end{aligned}

This is an equation in two variables. Write $x=x_{3}$ and $y=x_{4}$ and use distributivity to rewrite the equation as

\begin{aligned} -y^{2}+vy-cx+x^{2}+wxy=1. \end{aligned}

At least since Lagrange it has been known how to reduce this to a Pell equation $x^{2}+dy^{2}=e$. This is done by applying an invertible affine transformation, which does not change the number of solutions. First set $x=x-wy/2$. Then the equation becomes

\begin{aligned} -y^{2}+vy-c(x-wy/2)+(x-wy/2)^{2}+w(x-wy/2)y=1. \end{aligned}

Equivalently, the cross-term has disappeared and we have

\begin{aligned} y^{2}(-1-w^{2}/4)+y(v+cw/2)+x^{2}-cx=1. \end{aligned}

Now one can add constants to $x$ and $y$ to remove the linear terms, changing the constant term. Specifically, let $h:=(v+cw/2)/2$ and set $y=y-h$ and $x=x+c/2$. The equation becomes

\begin{aligned} (y-h)^{2}(-1-w^{2}/4)+(y-h)2h+(x+c/2)^{2}-c(x+c/2)=1. \end{aligned}

The linear terms disappear, the coefficients of $x^{2}$ and $y^{2}$ do not change and the equation can be rewritten as

\begin{aligned} y^{2}(-1-w^{2}/4)+h^{2}(-1-w^{2}/4)-2h^{2}+x^{2}+(c/2)^{2}-c^{2}/2=1. \end{aligned}

So this is now a Pell equation

\begin{aligned} x^{2}+dy^{2}=e \end{aligned}

where $d:=(-1-w^{2}/4)$ and

\begin{aligned} e:=1+h^{2}(3+w^{2}/4)+(c/2)^{2}=1+(v^{2}+(cw/2)^{2}+cvw)(1/4)(3+w^{2}/4)+(c/2)^{2}. \end{aligned}

For all but $O(1)$ values of $w$ we have that $d$ is non-zero. Moreover, for all but $O(1)$ values of $v,w$ the term $e$ is a non-zero polynomial in $c$. (Specifically, for any $v\ne 0$ and any $w$ such that $3+w^{2}/4\ne 0$.) So we only consider the values of $c$ that make it non-zero. Those where $e=0$ give $O(q)$ solutions, which is fine. We conclude with the following lemma.

Lemma 6. For $d$ and $e$ non-zero, and prime $q$, the number of solutions over $\mathbb{F} _{q}$ to the Pell equation

\begin{aligned} x^{2}+dy^{2}=e \end{aligned}

is within $O(1)$ of $q$.

This is a basic result from algebraic geometry that can be proved from first principles.

Proof. If $d=-f^{2}$ for some $f\in \mathbb{F} _{q}$, then we can replace $y$ with $fy$ and we can count instead the solutions to the equation

\begin{aligned} x^{2}-y^{2}=e. \end{aligned}

Because $x^{2}-y^{2}=(x-y)(x+y)$ we can set $x':=x-y$ and $y':=x+y$, which preserves the number of solutions, and rewrite the equation as

\begin{aligned} x'y'=e. \end{aligned}

Because $e\ne 0$, this has $q-1$ solutions: for every non-zero $y'$ we have $x'=e/y'$.

So now we can assume that $d\ne -f^{2}$ for any $f\in \mathbb{F} _{q}$. Because the number of squares is $(q+1)/2$, the range of $x^{2}$ has size $(q+1)/2$. Similarly, the range of $e-dy^{2}$ also has size $(q+1)/2$. Hence these two ranges intersect, and there is a solution $(a,b)$.

We take a line passing through $(a,b)$: for parameters $s,t\in \mathbb{F}$ we consider pairs $(a+t,b+st)$. There is a bijection between such pairs with $t\ne 0$ and the points $(x,y)$ with $x\ne a$. Because the number of solutions with $x=a$ is $O(1)$, using that $d\ne 0$, it suffices to count the solutions with $t\ne 0$.

The intuition is that this line has two intersections with the curve $x^{2}+dy^{2}=e$. Because one of them, $(a,b)$, lies in $\mathbb{F} _{q}$, the other has to lie as well there. Algebraically, we can plug the pair in the expression to obtain the equivalent equation

\begin{aligned} a^{2}+t^{2}+2at+d(b^{2}+s^{2}t^{2}+2bst)=e. \end{aligned}

Using that $(a,b)$ is a solution this becomes

\begin{aligned} t^{2}+2at+ds^{2}t^{2}+2dbst=0 \end{aligned}

We can divide by $t\ne 0$. Obtaining

\begin{aligned} t(1+ds^{2})+2a+2dbs=0. \end{aligned}

We can now divide by $1+ds^{2}$ which is non-zero by the assumption $d\ne -f^{2}$. This yields

\begin{aligned} t=(-2a-2dbs)/(1+ds^{2}). \end{aligned}

Hence for every value of $s$ there is a unique $t$ giving a solution. This gives $q$ solutions. $\square$

### 4 Three parties, number-in-hand

In this section we consider the following three-party number-in-hand problem: Alice gets $x$, Bob gets $y$, Charlie gets $z$, and they want to know if $x\cdot y\cdot z=1_{G}$. The communication depends on the group $G$. We present next two efficient protocols for abelian groups, and then a communication lower bound for other groups.

#### 4.1 A randomized protocol for the hypercube

We begin with the simplest setting. Let $G=(\mathbb {Z}_{2})^{n}$, that is $n$-bit strings with bit-wise addition modulo 2. The parties want to check if $x+y+z=0^{n}$. They can do so as follows. First, they pick a hash function $h$ that is linear: $h(x+y)=h(x)+h(y)$. Specifically, for a uniformly random $a\in \{0,1\}^{n}$ define $h_{a}(x):=\sum a_{i}x_{i}\mod 2$. Then, the protocol is as follows.

• Alice sends $h_{a}(x)$,
• Bob send $h_{a}(y)$,
• Charlie accepts if and only if $h_{a}(x)+h_{a}(y)+h_{a}(z)=0s$.

The hash function outputs 1 bit, so the communication is constant. By linearity, the protocol accepts iff $h_{a}(x+y+z)=0$. If $x+y+z=0$ this is always the case, otherwise it happens with probability $1/2$.

#### 4.2 A randomized protocol for $\mathbb {Z}_{N}$

This protocol is from [Vio14]. For simplicity we only consider the case $N=2^{n}$ here – the protocol for general $N$ is in [Vio14]. Again, the parties want to check if $x+y+z=0\mod N$. For this group, there is no 100% linear hash function but there are almost linear hash functions $h:\mathbb {Z}_{N}\rightarrow \mathbb {Z}_{2^{\ell }}$ that satisfy the following properties. Note that the inputs to $h$ are interpreted modulo $N$ and the outputs modulo $2^{\ell }$.

1. for all $a,x,y$ there is $c\in \{0,1\}$ such that $h_{a}(x+y)=h_{a}(x)+h_{a}(y)+c$,
2. for all $x\neq 0$ we have $\mathbb{P} _{a}[h_{a}(x)\in \{-2,-1,0,1,2\}]\leq O(1/2^{\ell })$,
3. $h_{a}(0)=0$.

Assuming some random hash function $h$ that satisfies the above properties the protocol works similarly to the previous one:

• Alice sends $h_{a}(x)$,
• Bob sends $h_{a}(y)$,
• Charlie accepts if and only if $h_{a}(x)+h_{a}(y)+h_{a}(z)\in \{-2,-1,0\}$.

We can set $\ell =O(1)$ to achieve constant communication and constant error.

To prove correctness of the protocol, first note that $h_{a}(x)+h_{a}(y)+h_{a}(z)=h_{a}(x+y+z)-c$ for some $c\in \{0,1,2\}$. Then consider the following two cases:

• if $x+y+z=0$ then $h_{a}(x+y+z)-c=h_{a}(0)-c=-c,$ and the protocol is always correct.
• if $x+y+z\neq 0$ then the probability that $h_{a}(x+y+z)-c\in \{-2,-1,0\}$ for some $c\in \{0,1,2\}$ is at most the probability that $h_{a}(x+y+z)\in \{-2,-1,0,1,2\}$ which is $\leq 2^{-\Omega (\ell )}$; so the protocol is correct with high probability.

The hash function..

For the hash function we can use a function analyzed in [DHKP97]. Let $a$ be a random odd number modulo $2^{n}$. Define

\begin{aligned} h_{a}(x):=(a\cdot x\gg n-\ell )\mod 2^{\ell } \end{aligned}

where the product $a\cdot x$ is integer multiplication, and $\gg$ is bit-shift. In other words we output the bits $n-\ell +1,n-\ell +2,\ldots ,n$ of the integer product $a\cdot x$.

We now verify that the above hash function family satisfies the three properties we required above.

Property (3) is trivially satisfied.

For property (1) we have the following. Let $s=a\cdot x$ and $t=a\cdot y$ and $u=n-\ell$. To recap, by definition we have:

• $h_{a}(x+y)=((s+t)\gg u)\mod 2^{\ell },$
• $h_{a}(x)=(s\gg u)\mod 2^{\ell }$,
• $h_{a}(x)=(t\gg u)\mod 2^{\ell }$.

Notice that if in the addition $s+t$ the carry into the $u+1$ bit is $0$, then

\begin{aligned} (s\gg u)+(t\gg u)=(s+t)\gg u \end{aligned}

otherwise

\begin{aligned} (s\gg u)+(t\gg u)+1=(s+t)\gg u \end{aligned}

which concludes the proof for property (1).

Finally, we prove property (2). We start by writing $x=s\cdot 2^{c}$ where $s$ is odd. So the binary representation of $x$ looks like

\begin{aligned} (\cdots \cdots 1\underbrace {0\cdots 0}_{c~\textrm {bits}}). \end{aligned}

The binary representation of the product $a\cdot x$ for a uniformly random $a$ looks like

\begin{aligned} (\textit {uniform}~1\underbrace {0\cdots 0}_{c~\textrm {bits}}). \end{aligned}

We consider the two following cases for the product $a\cdot x$:

1. If $a\cdot x=(\underbrace {\textit {uniform}~1\overbrace {00}^{2~bits}}_{\ell ~bits}\cdots 0)$, or equivalently $c\geq n-\ell +2$, the output never lands in the bad set $\{-2,-1,0,1,2\}$;
2. Otherwise, the hash function output has $\ell -O(1)$ uniform bits. For any set $B$, the probability that the output lands in $B$ is at most $|B|\cdot 2^{-\ell +O(1)}$.

#### 4.3 Quasirandom groups

What happens in other groups? The hash function used in the previous result was fairly non-trivial. Do we have an almost linear hash function for $2\times 2$ matrices? The answer is negative. For $SL_{2}(q)$ and $A_{n}$ the problem is hard, even under the promise. For a group $G$ the complexity can be expressed in terms of a parameter $d$ which comes from representation theory. We will not formally define this parameter here, but several qualitatively equivalent formulations can be found in [Gow08]. Instead the following table shows the $d$’s for the groups we’ve introduced.

 $G$ : abelian $A_{n}$ $SL_{2}(q)$ $d$ : $1$ $\Omega (\frac {\log |G|}{\log \log |G|})$ $|G|^{\Omega (1)}$

.

Theorem 1. Let $G$ be a group, and let $h\in G$. Let $d$ be the minimum dimension of any irreducible representation of $G$. Suppose Alice, Bob, and Charlie receive $x$, y, and $z$ respectively. They are promised that $x\cdot y\cdot z$ either equals $1_{G}$ or $h$. Deciding which case it is requires randomized communication complexity $\Omega (\log d)$.

This result is tight for the groups we have discussed so far. The arguments are the same as before. Specifically, for $SL_{2}(q)$ the communication is $\Omega (\log |G|)$. This is tight up to constants, because Alice and Bob can send their elements. For $A_{n}$ the communication is $\Omega (\log \log |G|)$. This is tight as well, as the parties can again just communicate the images of an element $a$ such that $h(a)\ne a$, as discussed in Section 1. This also gives a computational proof that $d$ cannot be too large for $A_{n}$, i.e., it is at most $(\log |G|)^{O(1)}$. For abelian groups we get nothing, matching the efficient protocols given above.

### 5 Proof of Theorem 1

First we discuss several “mixing” lemmas for groups, then we come back to protocols and see how to apply one of them there.

##### 5.0.1 $XY$ mixing

We want to consider “high entropy” distributions over $G$, and state a fact showing that the multiplication of two such distributions “mixes” or in other words increases the entropy. To define entropy we use the norms $\lVert A\rVert _{c}=\left (\sum _{x}A(x)^{c}\right )^{\frac {1}{c}}$. Our notion of (non-)entropy will be $\lVert A\rVert _{2}$. Note that $\lVert A\rVert _{2}^{2}$ is exactly the collision probability $\mathbb{P} [A=A']$ where $A'$ is independent and identically distributed to $A$. The smaller this quantity, the higher the entropy of $A$. For the uniform distribution $U$ we have $\lVert U\rVert _{2}^{2}=\frac {1}{|G|}$ and so we can think of $1/|G|$ as maximum entropy. If $A$ is uniform over $\Omega (|G|)$ elements, we have $\lVert A\rVert _{2}^{2}=O(1/|G|)$ and we think of $A$ as having “high” entropy.

Because the entropy of $U$ is small, we can think of the distance between $A$ and $U$ in the 2-norm as being essentially the entropy of $A$:

\begin{aligned} \lVert A-U\rVert _{2}^{2} & =\sum _{x\in G}\left (A(x)-\frac {1}{|G|}\right )^{2}\\ & =\sum _{x\in G}A(x)^{2}-2A(x)\frac {1}{|G|}+\frac {1}{|G|^{2}}\\ & =\lVert A\rVert _{2}^{2}-\frac {1}{|G|}\\ & =\lVert A\rVert _{2}^{2}-\lVert U\rVert _{2}^{2}\\ & \approx \lVert A\rVert _{2}^{2}. \end{aligned}

Lemma 7. [Gow08BNP08] If $X,Y$ are independent over $G$, then

\begin{aligned} \lVert X\cdot Y-U\rVert _{2}\leq \lVert X\rVert _{2}\lVert Y\rVert _{2}\sqrt {\frac {|G|}{d}}, \end{aligned}

where $d$ is the minimum dimension of an irreducible representation of $G$.

By this lemma, for high entropy distributions $X$ and $Y$, we get $\lVert X\cdot Y-U\rVert _{2}\leq \frac {O(1)}{\sqrt {|G|d}}$. The factor $1/\sqrt {|G|}$ allows us to pass to statistical distance $\lVert .\rVert _{1}$ using Cauchy-Schwarz:

\begin{aligned} \lVert X\cdot Y-U\rVert _{1}\leq \sqrt {|G|}\lVert X\cdot Y-U\rVert _{2}\leq \frac {O(1)}{\sqrt {d}}.~~~~(1) \end{aligned}

This is the way in which we will use the lemma.

Another useful consequence of this lemma, which however we will not use directly, is this. Suppose now you have $three$ independent, high-entropy variables $X,Y,Z$. Then for every $g\in G$ we have

\begin{aligned} |\mathbb{P} [X\cdot Y\cdot Z=g]-1/|G||\le \lVert X\rVert _{2}\lVert Y\rVert _{2}\lVert Z\rVert _{2}\sqrt {\frac {|G|}{d}}.~~~~(2) \end{aligned}

To show this, set $g=1_{G}$ without loss of generality and rewrite the left-hand-side as

\begin{aligned} |\sum _{h\in G}\mathbb{P} [X=h](\mathbb{P} [YZ=h^{-1}]-1/|G|)|. \end{aligned}

By Cauchy-Schwarz this is at most

\begin{aligned} \sqrt {\sum _{h}\mathbb{P} ^{2}[X=h]}\sqrt {\sum _{h}(\mathbb{P} [YZ=h^{-1}]-1/|G|)^{2}}=\lVert X\lVert _{2}\lVert YZ-U\lVert _{2} \end{aligned}

and we can conclude by Lemma 7. Hence the product of three high-entropy distributions is close to uniform in a point-wise sense: each group element is obtained with roughly probability $1/|G|$.

At least over $SL(2,q)$, there exists an alternative proof of this fact that does not mention representation theory (see [GVa] and [VioaViob]).

With this notation in hand, we conclude by stating a “mixing” version of Theorem 2. For more on this perspective we refer the reader to [GVa].

Theorem 1. Let $G=SL(2,q)$. Let $X=(X_{1},X_{2})$ and $Y=(Y_{1},Y_{2})$ be two distributions over $G^{2}$. Suppose $X$ is independent from $Y$. Let $g\in G$. We have

\begin{aligned} |\mathbb{P} [X_{1}Y_{1}X_{2}Y_{2}=g]-1/|G||\le |G|^{1-\Omega (1)}\lVert X\rVert _{2}\lVert Y\rVert _{2}. \end{aligned}

For example, when $X$ and $Y$ have high entropy over $G^{2}$ (that is, are uniform over $\Omega (|G|^{2})$ pairs), we have $\lVert X\rVert _{2}\le \sqrt {O(1)/|G|^{2}}$, and so $|G|^{1-\Omega (1)}\lVert X\rVert _{2}\lVert Y\rVert _{2}\le 1/|G|^{1+\Omega (1)}$. In particular, $X_{1}Y_{1}X_{2}Y_{2}$ is $1/|G|^{\Omega (1)}$ close to uniform over $G$ in statistical distance.

##### 5.0.2 Back to protocols

As in the beginning of Section 3, for any group element $g\in G$ we define the distribution on triples $D_{g}:=(x,y,(x\cdot y)^{-1}g)$, where $x,y\in G$ are uniform and independent. Note the product of the elements in $D_{g}$ is always $g$. Again as in Section 3, it suffices to show that for every deterministic protocols $P$ using little communication we have

\begin{aligned} |\Pr [P(D_{1})=1]-\Pr [P(D_{h})=1]|\leq \frac {1}{100}. \end{aligned}

Analogously to Lemma 4, the following lemma describes a protocol using rectangles. The proof is nearly identical and is omitted.

Lemma 8. (The set of accepted inputs of) A deterministic $c$-bit number-in-hand protocol with three parties can be written as a disjoint union of $2^{c}$ “rectangles,” that is sets of the form $A\times B\times C$.

Next we show that these product sets cannot distinguish these two distributions $D_{1},D_{h}$, via a straightforward application of lemma 7.

Lemma 9. For all $A,B,C\subseteq G$ we have $|\mathbb{P} (A\times B\times C)(D_{1})=1]-\mathbb{P} [(A\times B\times C)(D_{h})=1]|\leq 1/d^{\Omega (1)}.$

Proof. Pick any $h\in G$ and let $x,y,z$ be the inputs of Alice, Bob, and Charlie respectively. Then

\begin{aligned} \mathbb{P} [(A\times B\times C)(D_{h})=1]=\mathbb{P} [(x,y)\in A\times B]\cdot \mathbb{P} [(x\cdot y)^{-1}\cdot h\in C|(x,y)\in A\times B],~~~~(3) \end{aligned}

where $(x,y)$ is uniform in $G^{2}$. If either $A$ or $B$ is small, that is $\mathbb{P} [x\in A]\leq \epsilon$ or $\mathbb{P} [y\in B]\leq \epsilon$, then also $\mathbb{P} [(x,y)\in A\times B]\le \epsilon$ and hence (??) is at most $\epsilon$ as well. This holds for every $h$, so we also have $|\mathbb{P} (A\times B\times C)(D_{1})=1]-\mathbb{P} [(A\times B\times C)(D_{h})=1]|\leq \epsilon .$ We will choose $\epsilon$ later.

Otherwise, $A$ and $B$ are large: $\mathbb{P} [x\in A]>\epsilon$ and $\mathbb{P} [y\in B]>\epsilon$. Let $(x',y')$ be the distribution of $(x,y)$ conditioned on $(x,y)\in A\times B$. We have that $x'$ and $y'$ are independent and each is uniform over at least $\epsilon |G|$ elements. By Lemma 7 this implies $\lVert x'\cdot y'-U\rVert _{2}\leq \lVert x'\rVert _{2}\cdot \lVert y'\rVert _{2}\cdot \sqrt {\frac {|G|}{d}}$, where $U$ is the uniform distribution. As mentioned after the lemma, by Cauchy–Schwarz we obtain

\begin{aligned} \lVert x'\cdot y'-U\rVert _{1}\leq |G|\cdot \lVert x'\rVert _{2}\cdot \lVert y'\rVert _{2}\cdot \sqrt {\frac {1}{d}}\leq \frac {1}{\epsilon }\cdot \frac {1}{\sqrt {d}}, \end{aligned}

where the last inequality follows from the fact that $\lVert x\rVert _{2},\lVert y\rVert _{2}\leq \sqrt {\frac {1}{\epsilon |G|}}$.

This implies that $\lVert (x'\cdot y')^{-1}-U\rVert _{1}\leq \frac {1}{\epsilon }\cdot \frac {1}{\sqrt {d}}$ and $\lVert (x'\cdot y')^{-1}\cdot h-U\rVert _{1}\leq \frac {1}{\epsilon }\cdot \frac {1}{\sqrt {d}}$, because taking inverses and multiplying by $h$ does not change the distance to uniform. These two last inequalities imply that

\begin{aligned} |\mathbb{P} [(x'\cdot y')^{-1}\in C]-\mathbb{P} [(x'\cdot y')^{-1}\cdot h\in C]|\le O(\frac {1}{\epsilon \sqrt {d}}); \end{aligned}

and thus we get that

\begin{aligned} |\mathbb{P} [(A\times B\times C)(D_{1})=1]-\mathbb{P} [(A\times B\times C)(D_{h})=1]|\le O(\frac {1}{\epsilon \sqrt {d}}). \end{aligned}

Picking $\epsilon =1/d^{1/4}$ completes the proof. $\square$

Returning to arbitrary deterministic protocols $P$ (as opposed to rectangles), write $P$ as a union of $2^{c}$ disjoint rectangles by Lemma 8. Applying Lemma 9 and summing over all rectangles we get that the distinguishing advantage of $P$ is at most $2^{c}/d^{1/4}$. For $c\leq (1/100)\log d$ the advantage is at most $1/100$, concluding the proof.

### 6 Three parties, number-on-forehead

In number-on-forehead (NOH) communication complexity [CFL83] with $k$ parties, the input is a $k$-tuple $(x_{1},\dotsc ,x_{k})$ and each party $i$ sees all of it except $x_{i}$. For background, it is not known how to prove negative results for $k\ge \log n$ parties.

We mention that Theorem 1 can be extended to the multiparty setting, see [GVa]. Several questions arise here, such as whether this problem remains hard for $k\ge \log n$, and what is the minimum length of an interleaved product that is hard for $k=3$ parties (the proof in 1 gives a large constant).

However in this survey we shall instead focus on the problem of separating deterministic and randomized communication. For $k=2$, we know the optimal separation: The equality function requires $\Omega (n)$ communication for deterministic protocols, but can be solved using $O(1)$ communication if we allow the protocols to use public coins. For $k=3$, the best known separation between deterministic and randomized protocol is $\Omega (\log n)$ vs $O(1)$ [BDPW10]. In the following we give a new proof of this result, for a different function: $f(x,y,z)=1_{G}$ if and only if $x\cdot y\cdot z=1$ for $x,y,z\in SL(2,q)$. As is true for some functions in [BDPW10], a stronger separation could hold for $f$. For context, let us state and prove the upper bound for randomized communication.

Claim 10. $f$ has randomized communication complexity $O(1)$.

Proof. In the number-on-forehead model, computing $f$ reduces to two-party equality with no additional communication: Alice computes $y\cdot z=:w$ privately, then Alice and Bob check if $x=w^{-1}$. $\square$

To prove the lower bound for deterministic protocols we reduce the communication problem to a combinatorial problem.

Definition 11. A corner in a group $G$ is a set $\{(x,y),(xz,y),(x,zy)\}\subseteq G^{2}$, where $x,y$ are arbitrary group elements and $z\neq 1_{G}$.

For intuition, if $G$ is the abelian group of real numbers with addition, a corner becomes $\{(x,y),(x+z,y),(x,y+z)\}$ for $z\neq 0$, which are the coordinates of an isosceles triangle. We now state the theorem that connects corners and lower bounds.

Lemma 12. Let $G$ be a group and $\delta$ a real number. Suppose that every subset $A\subseteq G^{2}$ with $|A|/|G^{2}|\ge \delta$ contains a corner. Then the deterministic communication complexity of $f$ (defined as $f(x,y,z)=1\iff x\cdot y\cdot z=1_{G}$) is $\Omega (\log (1/\delta ))$.

It is known that $\delta \ge 1/\mathrm {polyloglog}|G|$ implies a corner for certain abelian groups $G$, see [LM07] for the best bound and pointers to the history of the problem. For $G=SL(2,q)$ a stronger result is known: $\delta \ge 1/\mathrm {polylog}|G|$ implies a corner [Aus16]. This in turn implies communication $\Omega (\log \log |G|)=\Omega (\log n)$.

Proof. We saw already twice that a number-in-hand $c$-bit protocol can be written as a disjoint union of $2^{c}$ rectangles (Lemmas 4, 8). Likewise, a number-on-forehead $c$-bit protocol $P$ can be written as a disjoint union of $2^{c}$ cylinder intersections $C_{i}:=\{(x,y,z):f_{i}(y,z)g_{i}(x,z)h_{i}(x,y)=1\}$ for some $f_{i},g_{i},h_{i}\colon G^{2}\to \{0,1\}$:

\begin{aligned} P(x,y,z)=\sum _{i=1}^{2^{c}}f_{i}(y,z)g_{i}(x,z)h_{i}(x,y). \end{aligned}

The proof idea of the above fact is to consider the $2^{c}$ transcripts of $P$, then one can see that the inputs giving a fixed transcript are a cylinder intersection.

Let $P$ be a $c$-bit protocol. Consider the inputs $\{(x,y,(xy)^{-1})\}$ on which $P$ accepts. Note that at least $2^{-c}$ fraction of them are accepted by some cylinder intersection $C=f\cdot g\cdot h$. Let $A:=\{(x,y):(x,y,(xy)^{-1})\in C\}\subseteq G^{2}$. Since the first two elements in the tuple determine the last, we have $|A|/|G^{2}|\ge 2^{-c}$.

Now suppose $A$ contains a corner $\{(x,y),(xz,y),(x,zy)\}$. Then

\begin{aligned} (x,y)\in A & \implies (x,y,(xy)^{-1})\in C & & \implies h(x,y)=1,\\ (xz,y)\in A & \implies (xz,y,(xzy)^{-1})\in C & & \implies f(y,(xyz)^{-1})=1,\\ (x,zy)\in A & \implies (x,zy,(xzy)^{-1})\in C & & \implies g(x,(xyz)^{-1})=1. \end{aligned}

This implies $(x,y,(xzy)^{-1})\in C$, which is a contradiction because $z\neq 1$ and so $x\cdot y\cdot (xzy)^{-1}\neq 1_{G}$. $\square$

### 7 The corners theorem for quasirandom groups

In this section we prove the corners theorem for quasirandom groups, following Austin [Aus16]. Our exposition has several minor differences with that in [Aus16], which may make it more computer-science friendly. Possibly a proof can also be obtained via certain local modifications and simplifications of Green’s exposition [Gre05bGre05a] of an earlier proof for the abelian case. We focus on the case $G=\textit {SL}(2,q)$ for simplicity, but the proof immediately extends to other quasirandom groups (with corresponding parameters).

Theorem 1. Let $G=\textit {SL}(2,q)$. Every subset $A\subseteq G^{2}$ of density $|A|/|G|^{2}\geq 1/\log ^{a}|G|$ contains a corner $\{(x,y),(xz,y),(x,zy)~|~z\neq 1\}$.

#### 7.1 Proof idea

For intuition, suppose $A$ is a product set, i.e., $A=B\times C$ for $B,C\subseteq G$. Let’s look at the quantity

\begin{aligned} \mathbb {E}_{x,y,z\leftarrow G}[A(x,y)A(xz,y)A(x,zy)] \end{aligned}

where $A(x,y)=1$ iff $(x,y)\in A$. Note that the random variable in the expectation is equal to $1$ exactly when $x,y,z$ form a corner in $A$. We’ll show that this quantity is greater than $1/|G|$, which implies that $A$ contains a corner (where $z\neq 1$). Since we are taking $A=B\times C$, we can rewrite the above quantity as

\begin{aligned} \mathbb {E}_{x,y,z\leftarrow G}[B(x)C(y)B(xz)C(y)B(x)C(zy)] & =\mathbb {E}_{x,y,z\leftarrow G}[B(x)C(y)B(xz)C(zy)]\\ & =\mathbb {E}_{x,y,z\leftarrow G}[B(x)C(y)B(z)C(x^{-1}zy)] \end{aligned}

where the last line follows by replacing $z$ with $x^{-1}z$ in the uniform distribution. If $|A|/|G|^{2}\ge \delta$, then both |B|/|G|$\ge \delta$ and $|B|/|G|\ge \delta$. Condition on $x\in B$, $y\in C$, $z\in B$. Then the distribution $x^{-1}zy$ is a product of three independent distributions, each uniform on a set of density $\ge \delta$. (In fact, two distributions would suffice for this.) By Lemma 7, $x^{-1}zy$ is $\delta ^{-1}/|G|^{\Omega (1)}$ close to uniform in statistical distance. This implies that the above expectation equals

\begin{aligned} \frac {|A|}{|G|^{2}}\cdot \frac {|B|}{|G|}\cdot \left (\frac {|C|}{|G|}\pm \frac {\delta ^{-1}}{|G|^{\Omega (1)}}\right ) & \geq \delta ^{2}\left (\delta -\frac {1}{|G|^{\Omega (1)}}\right )\geq \delta ^{3}/2>1/|G|, \end{aligned}

for $\delta >1/|G|^{c}$ for a small enough constant $c$. Hence, product sets of density polynomial in $1/|G|$ contain corners.

Given the above, it is natural to try to decompose an arbitrary set $A$ into product sets. We will make use of a more general result.

#### 7.2 Weak Regularity Lemma

Let $U$ be some universe (we will take $U=G^{2}$) and let $f:U\rightarrow [-1,1]$ be a function (for us, $f=1_{A}$). Let $D\subseteq \{d:U\rightarrow [-1,1]\}$ be some set of functions, which can be thought of as “easy functions” or “distinguishers” (these will be rectangles or closely related to them). The next theorem shows how to decompose $f$ into a linear combination $g$ of the $d_{i}$ up to an error which is polynomial in the length of the combination. More specifically, $f$ will be indistinguishable from $g$ by the $d_{i}$.

Lemma 13. Let $f:U\rightarrow [-1,1]$ be a function and $D\subseteq \{d:U\rightarrow [-1,1]\}$ a set of functions. For all $\epsilon >0$, there exists a function $g:=\sum _{i\le s}c_{i}\cdot d_{i}$ where $d_{i}\in D$, $c_{i}\in \mathbb {R}$ and $s=1/\epsilon ^{2}$ such that for all $d\in D$

\begin{aligned} \left |\mathbb {E}_{x\leftarrow U}[f(x)\cdot d(x)]-\mathbb {E}_{x\leftarrow U}[g(x)\cdot d(x)]\right |\le \epsilon . \end{aligned}

A different way to state the conclusion, which we will use, is to say that we can write $f=g+h$ so that $\mathbb{E} [h(x)\cdot d(x)]$ is small.

The lemma is due to Frieze and Kannan [FK96]. It is called “weak” because it came after Szemerédi’s regularity lemma, which has a stronger distinguishing conclusion. However, the lemma is also “strong” in the sense that Szemerédi’s regularity lemma has $s$ as a tower of $1/\epsilon$ whereas here we have $s$ polynomial in $1/\epsilon$. The weak regularity lemma is also simpler. There also exists a proof [Tao17] of Szemerédi’s theorem (on arithmetic progressions), which uses weak regularity as opposed to the full regularity lemma used initially.

Proof. We will construct the approximation $g$ through an iterative process producing functions $g_{0},g_{1},\dots ,g$. We will show that $||f-g_{i}||_{2}^{2}$ decreases by $\ge \epsilon ^{2}$ each iteration.

Start: Define $g_{0}=0$ (which can be realized setting $c_{0}=0$).

Iterate: If not done, there exists $d\in D$ such that $|\mathbb {E}[(f-g)\cdot d]|>\epsilon$. Assume without loss of generality $\mathbb {E}[(f-g)\cdot d]>\epsilon$.

Update: $g':=g+\lambda d$ where $\lambda \in \mathbb {R}$ shall be picked later.

Let us analyze the progress made by the algorithm.

\begin{aligned} ||f-g'||_{2}^{2} & =\mathbb {E}_{x}[(f-g')^{2}(x)]\\ & =\mathbb {E}_{x}[(f-g-\lambda d)^{2}(x)]\\ & =\mathbb {E}_{x}[(f-g)^{2}]+\mathbb {E}_{x}[\lambda ^{2}d^{2}(x)]-2\mathbb {E}_{x}[(f-g)\cdot \lambda d(x)]\\ & \leq ||f-g||_{2}^{2}+\lambda ^{2}-2\lambda \mathbb {E}_{x}[(f-g)d(x)]\\ & \leq ||f-g||_{2}^{2}+\lambda ^{2}-2\lambda \epsilon \\ & \leq ||f-g||_{2}^{2}-\epsilon ^{2} \end{aligned}

where the last line follows by taking $\lambda =\epsilon$. Therefore, there can only be $1/\epsilon ^{2}$ iterations because $||f-g_{0}||_{2}^{2}=||f||_{2}^{2}\leq 1$. $\square$

#### 7.3 Getting more for rectangles

Returning to the main proof, we will use the weak regularity lemma to approximate the indicator function for arbitrary $A$ by rectangles. That is, we take $D$ to be the collection of indicator functions for all sets of the form $S\times T$ for $S,T\subseteq G$. The weak regularity lemma shows how to decompose $A$ into a linear combination of rectangles. These rectangles may overlap. However, we ideally want $A$ to be a linear combination of non-overlapping rectangles. In other words, we want a partition of rectangles. It is possible to achieve this at the price of exponentiating the number of rectangles. Note that an exponential loss is necessary even if $S=G$ in every $S\times T$ rectangle; or in other words in the uni-dimensional setting. This is one step where the terminology “rectangle” may be misleading – the set $T$ is not necessarily an interval. If it was, a polynomial rather than exponential blow-up would have sufficed to remove overlaps.

Claim 14. Given a decomposition of $A$ into rectangles from the weak regularity lemma with $s$ functions, there exists a decomposition with $2^{O(s)}$ rectangles which don’t overlap.

Proof. Exercise. $\square$

In the above decomposition, note that it is natural to take the coefficients of rectangles to be the density of points in $A$ that are in the rectangle. This gives rise to the following claim.

Claim 15. The weights of the rectangles in the above claim can be the average of $f$ in the rectangle, at the cost of doubling the error.

Consequently, we have that $f=g+h$, where $g$ is the sum of $2^{O(s)}$ non-overlapping rectangles $S\times T$ with coefficients $\mathbb{P} _{(x,y)\in S\times T}[f(x,y)=1]$.

Proof. Let $g$ be a partition decomposition with arbitrary weights. Let $g'$ be a partition decomposition with weights being the average of $f$. It is enough to show that for all rectangle distinguishers $d\in D$

\begin{aligned} |\mathbb {E}[(f-g')d]|\leq |\mathbb {E}[(f-g)d]|. \end{aligned}

By the triangle inequality, we have that

\begin{aligned} |\mathbb {E}[(f-g')d]|\leq |\mathbb {E}[(f-g)d]|+|\mathbb {E}[(g-g')d]|. \end{aligned}

To bound $\mathbb {E}[(g-g')d]|$, note that the error is maximized for a $d$ that respects the decomposition in non-overlapping rectangles, i.e., $d$ is the union of some non-overlapping rectangles from the decomposition. This can be argued using that, unlike $f$, the value of $g$ and $g'$ on a rectangle $S\times T$ from the decomposition is fixed. But, from the point of “view” of such $d$, $g'=f$! More formally, $\mathbb {E}[(g-g')d]=\mathbb {E}[(g-f)d]$. This gives

\begin{aligned} |\mathbb {E}[(f-g')d]|\leq 2|\mathbb {E}[(f-g)d]| \end{aligned}

and concludes the proof. $\square$

We need to get still a little more from this decomposition. In our application of the weak regularity lemma above, we took the set of distinguishers to be characteristic functions of rectangles. That is, distinguishers that can be written as $U(x)\cdot V(y)$ where $U$ and $V$ map $G\to \{0,1\}$. We will use that the same guarantee holds for $U$ and $V$ with range $[-1,1]$, up to a constant factor loss in the error. Indeed, let $U$ and $V$ have range $[-1,1]$. Write $U=U_{+}-U_{-}$ where $U_{+}$ and $U_{-}$ have range $[0,1]$, and the same for $V$. The error for distinguisher $U\cdot V$ is at most the sum of the errors for distinguishers $U_{+}\cdot V_{+}$, $U_{+}\cdot V_{-}$, $U_{-}\cdot V_{+}$, and $U_{-}\cdot V_{-}$. So we can restrict our attention to distinguishers $U(x)\cdot V(y)$ where $U$ and $V$ have range $[0,1]$. In turn, a function $U(x)$ with range $[0,1]$ can be written as an expectation $\mathbb{E} _{a}U_{a}(x)$ for functions $U_{a}$ with range $\{0,1\}$, and the same for $V$. We conclude by observing that

\begin{aligned} \mathbb{E} _{x,y}[(f-g)(x,y)\mathbb{E} _{a}U_{a}(x)\cdot \mathbb{E} _{b}V_{b}(y)]\le \max _{a,b}\mathbb{E} _{x,y}[(f-g)(x,y)U_{a}(x)\cdot V_{b}(y)]. \end{aligned}

#### 7.4 Proof

Let us now finish the proof by showing a corner exists for sufficiently dense sets $A\subseteq G^{2}$. We’ll use three types of decompositions for $f:G^{2}\rightarrow \{0,1\}$, with respect to the following three types of distinguishers, where $U_{i}$ and $V_{i}$ have range $\{0,1\}$:

1. $U_{1}(x)\cdot V_{1}(y)$,
2. $U_{2}(xy)\cdot V_{2}(y)$,
3. $U_{3}(x)\cdot V_{3}(xy)$.

The first type is just rectangles, what we have been discussing until now. The distinguishers in the last two classes can be visualized over $\mathbb {R}^{2}$ as parallelograms with a 45-degree angle. The same extra properties we discussed for rectangles can be verified hold for them too.

Recall that we want to show

\begin{aligned} \mathbb {E}_{x,y,g}[f(x,y)f(xg,y)f(x,gy)]>\frac {1}{|G|}. \end{aligned}

We’ll decompose the $i$-th occurrence of $f$ via the $i$-th decomposition listed above. We’ll write this decomposition as $f=g_{i}+h_{i}$. We apply this in a certain order to produce sums of products of three functions. The inputs to the functions don’t change, so to avoid clutter we do not write them, and it is understood that in each product of three functions the inputs are, in order $(x,y),(xg,y),(x,gy)$. The decomposition is:

\begin{aligned} & fff\\ = & ffg_{3}+ffh_{3}\\ = & fg_{2}g_{3}+fh_{2}g_{3}+ffh_{3}\\ = & g_{1}g_{2}g_{3}+h_{1}g_{2}g_{3}+fh_{2}g_{3}+ffh_{3}. \end{aligned}

We first show that the expectation of the first term is big. This takes the next two claims. Then we show that the expectations of the other terms are small.

Claim 16. For all $g\in G$, the expectations $\mathbb {E}_{x,y}[g_{1}(x,y)g_{2}(xg,y)g_{3}(x,gy)]$ are the same up to an error of $2^{O(s)}/|G|^{\Omega (1)}$.

Proof. We just need to get error $1/|G|^{\Omega (1)}$ for any product of three functions for the three decomposition types. We have:

\begin{aligned} & \mathbb {E}_{x,y}[c_{1}U_{1}(x)V_{1}(y)\cdot c_{2}U_{2}(xgy)V_{2}(y)\cdot c_{3}U_{3}(x)V_{3}(xgy)]\\ = & c_{1}c_{2}c_{3}\mathbb {E}_{x,y}[(U_{1}\cdot U_{3})(x)(V_{1}\cdot V_{2})(y)(U_{2}\cdot V_{3})(xgy)]\\ = & c_{1}c_{2}c_{3}\cdot \mathbb {E}_{x}[(U_{1}\cdot U_{3})(x)]\cdot \mathbb {E}_{y}[(V_{1}\cdot V_{2})(y)]\cdot \mathbb {E}_{z}[(U_{2}\cdot V_{3})(z)]\pm \frac {1}{|G|^{\Omega (1)}}. \end{aligned}

This is similar to what we discussed in the overview, and is where we use mixing. Specifically, if $\mathbb {E}_{x}[(U_{1}\cdot U_{3})(x)]$ or $\mathbb {E}_{y}[(V_{1}\cdot V_{2})(y)]$ are at most $1/|G|^{c}$ for a small enough constant $c$ than we are done. Otherwise, conditioned on $(U_{1}\cdot U_{3})(x)=1$, the distribution on $x$ is uniform over a set of density $1/|G|^{c}$, and the same holds for $y$, and the result follows by Lemma 7. $\square$

Recall that we start with a set of density $\ge 1/\log ^{a}|G|$.

Claim 17. $\mathbb {E}_{x,y}[g_{1}(x,y)g_{2}(x,y)g_{3}(x,y)]>1/\log ^{4a}|G|$.

Proof. We will relate the expectation over $x,y$ to $f$ using the Hölder inequality: For random variables $X_{1},X_{2},\ldots ,X_{k}$,

\begin{aligned} \mathbb {E}[X_{1}\dots X_{k}]\leq \prod _{i=1}^{k}\mathbb {E}[X_{i}^{c_{i}}]^{1/c_{i}}\text { such that }\sum 1/c_{i}=1. \end{aligned}

To apply this inequality in our setting, write

\begin{aligned} f=(f\cdot g_{1}g_{2}g_{3})^{1/4}\cdot \left (\frac {f}{g_{1}}\right )^{1/4}\cdot \left (\frac {f}{g_{2}}\right )^{1/4}\cdot \left (\frac {f}{g_{3}}\right )^{1/4}. \end{aligned}

By the Hölder inequality the expectation of the right-hand side is

\begin{aligned} \leq \mathbb {E}[f\cdot g_{1}g_{2}g_{3}]^{1/4}\mathbb {E}\left [\frac {f}{g_{1}}\right ]^{1/4}\mathbb {E}\left [\frac {f}{g_{2}}\right ]^{1/4}\mathbb {E}\left [\frac {f}{g_{3}}\right ]^{1/4}. \end{aligned}

The last three terms equal to $1$ because

\begin{aligned} \mathbb {E}_{x,y}\frac {f(x,y)}{g_{i}(x,y)} & =\mathbb {E}_{x,y}\frac {f(x,y)}{\mathbb {E}_{x',y'\in \textit {Cell}(x,y)}[f(x',y')]}=\mathbb {E}_{x,y}\frac {\mathbb {E}_{x',y'\in \textit {Cell}(x,y)}[f(x',y')]}{\mathbb {E}_{x',y'\in \textit {Cell}(x,y)}[f(x',y')]}=1. \end{aligned}

where $\textit {Cell}(x,y)$ is the set in the partition that contains $(x,y)$. Putting the above together we obtain

\begin{aligned} \mathbb {E}[f]\leq \mathbb {E}[f\cdot g_{1}g_{2}g_{3}]^{1/4}. \end{aligned}

Finally, because the functions are positive, we have that $\mathbb {E}[f\cdot g_{1}g_{2}g_{3}]^{1/4}\leq \mathbb {E}[g_{1}g_{2}g_{3}]^{1/4}$. This concludes the proof. $\square$

It remains to show the other terms are small. Let $\epsilon$ be the error in the weak regularity lemma with respect to distinguishers with range $\{0,1\}$. Recall that this implies error $O(\epsilon )$ with respect to distinguishers with range $[-1,1]$. We give the proof for one of the terms and then we say little about the other two.

Claim 18. $|\mathbb {E}[f(x,y)f(xg,y)h_{3}(x,gy)]|\leq O(\epsilon )^{1/4}$.

The proof involves changing names of variables and doing Cauchy-Schwarz to remove the terms with $f$ and bound the expectation above by $\mathbb {E}[h_{3}(x,g)U(x)V(xg)]$, which is small by the regularity lemma.

Proof. Replace $g$ with $gy^{-1}$ in the uniform distribution to get

\begin{aligned} & \mathbb {E}_{x,y,g}^{4}[f(x,y)f(xg,y)h_{3}(x,gy)]\\ & =\mathbb {E}_{x,y,g}^{4}[f(x,y)f(xgy^{-1},y)h_{3}(x,g)]\\ & =\mathbb {E}_{x,y}^{4}[f(x,y)\mathbb {E}_{g}[f(xgy^{-1},y)h_{3}(x,g)]]\\ & \leq \mathbb {E}_{x,y}^{2}[f^{2}(x,y)]\mathbb {E}_{x,y}^{2}\mathbb {E}_{g}^{2}[f(xgy^{-1},y)h_{3}(x,g)]\\ & \leq \mathbb {E}_{x,y}^{2}\mathbb {E}_{g}^{2}[f(xgy^{-1},y)h_{3}(x,g)]\\ & =\mathbb {E}_{x,y,g,g'}^{2}[f(xgy^{-1},y)h_{3}(x,g)f(xg'y^{-1},y)h_{3}(x,g')], \end{aligned}

where the first inequality is by Cauchy-Schwarz.

Now replace $g\rightarrow x^{-1}g,g'\rightarrow x^{-1}g$ and reason in the same way:

\begin{aligned} & =\mathbb {E}_{x,y,g,g'}^{2}[f(gy^{-1},y)h_{3}(x,x^{-1}g)f(g'y^{-1},y)h_{3}(x,x^{-1}g')]\\ & =\mathbb {E}_{g,g',y}^{2}[f(gy^{-1},y)\cdot f(g'y^{-1},y)\mathbb {E}_{x}[h_{3}(x,x^{-1}g)\cdot h_{3}(x,x^{-1}g')]]\\ & \leq \mathbb {E}_{x,x',g,g'}[h_{3}(x,x^{-1}g)h_{3}(x,x^{-1}g')h_{3}(x',x'^{-1}g)h_{3}(x',x'^{-1}g')]. \end{aligned}

Replace $g\rightarrow xg$ to rewrite the expectation as

\begin{aligned} \mathbb {E}[h_{3}(x,g)h_{3}(x,x^{-1}g')h_{3}(x',x'^{-1}xg)h_{3}(x',x'^{-1}g')]. \end{aligned}

We want to view the last three terms as a distinguisher $U(x)\cdot V(xg)$. First, note that $h_{3}$ has range $[-1,1]$. This is because $h_{3}(x,y)=f(x,y)-\mathbb{E} _{x',y'\in \textit {Cell}(x,y)}f(x',y')$ and $f$ has range $\{0,1\}$, where recall that $Cell(x,y)$ is the set in the partition that contains $(x,y)$. Fix $x',g'$. The last term in the expectation becomes a constant $c\in [-1,1]$. The second term only depends on $x$, and the third only on $xg$. Hence for appropriate functions $U$ and $V$ with range $[-1,1]$ this expectation can be rewritten as

\begin{aligned} \mathbb {E}[h_{3}(x,g)U(x)V(xg)], \end{aligned}

which concludes the proof. $\square$

There are similar proofs to show the remaining terms are small. For $fh_{2}g_{3}$, we can perform simple manipulations and then reduce to the above case. For $h_{1}g_{2}g_{3}$, we have a slightly easier proof than above.

##### 7.4.1 Parameters

Suppose our set has density $\delta \ge 1/\log ^{a}|G|$, and the error in the regularity lemma is $\epsilon$. By the above results we can bound

\begin{aligned} \mathbb {E}_{x,y,g}[f(x,y)f(xg,y)f(x,gy)]\ge 1/\log ^{4a}|G|-2^{O(1/\epsilon ^{2})}/|G|^{\Omega (1)}-\epsilon ^{\Omega (1)}, \end{aligned}

where the terms in the right-hand size come, left-to-right from Claim 17, 16, and 18. Picking $\epsilon =1/\log ^{1/3}|G|$ the proof is completed for sufficiently small $a$.

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# bounded independence plus noise fools space

There are many classes of functions on $n$ bits that we know are fooled by bounded independence, including small-depth circuits, halfspaces, etc. (See this previous post.)

On the other hand the simple parity function is not fooled. It’s easy to see that you require independence at least $n-1$. However, if you just perturb the bits with a little noise $N$, then parity will be fooled. You can find other examples of functions that are not fooled by bounded independence alone, but are if you just perturb the bits a little.

In [3] we proved that any distribution with independence about $n^{2/3}$ fools space-bounded algorithms, if you perturb it with noise. We asked, both in the paper and many people, if the independence could be lowered. Forbes and Kelley have recently proved [2] that the independence can be lowered all the way to $O(\log n)$, which is tight [1]. Shockingly, their proof is nearly identical to [3]!

This exciting result has several interesting consequences. First, we now have almost the same generators for space-bounded computation in a fixed order as we do for any order. Moreover, the proof greatly simplifies a number of works in the literature. And finally, an approach in [4] to prove limitations for the sum of small-bias generators won’t work for space (possibly justifying some optimism in the power of the sum of small-bias generators).

My understanding of all this area is inseparable from the collaboration I have had with Chin Ho Lee, with whom I co-authored all the papers I have on this topic.

### The proof

Let $f:\{0,1\}^{n}\to \{0,1\}$ be a function. We want to show that it is fooled by $D+E$, where $D$ has independence $k$, $E$ is the noise vector of i.i.d. bits coming up $1$ with probability say $1/4$, and $+$ is bit-wise XOR.

The approach in [3] is to decompose $f$ as the sum of a function $L$ with Fourier degree $k$, and a sum of $t$ functions $H_{i}=h_{i}\cdot g_{i}$ where $h_{i}$ has no Fourier coefficient of degree less than $k$, and $h_{i}$ and $g_{i}$ are bounded. The function $L$ is immediately fooled by $D$, and it is shown in [3] that each $H_{i}$ is fooled as well.

To explain the decomposition it is best to think of $f$ as the product of $\ell :=n/k$ functions $f_{i}$ on $k$ bits, on disjoint inputs. The decomposition in [3] is as follows: repeatedly decompose each $f_{i}$ in low-degree $f_{L}$ and high-degree $f_{H}$. To illustrate:

\begin{aligned} f_{1}f_{2}f_{3} & =f_{1}f_{2}(f_{3H}+f_{3L})=f_{1}f_{2}f_{3H}+f_{1}(f_{2H}+f_{2L})f_{3L}=\ldots \\ = & f_{1H}f_{2L}f_{3L}+f_{1}f_{2H}f_{3L}+f_{1}f_{2}f_{3H}+f_{1L}f_{2L}f_{3L}\\ = & H_{1}+H_{2}+H_{3}+L. \end{aligned}

This works, but the problem is that even if each time $f_{iL}$ has degree $1$, the function $L$ increases the degree by at least $1$ per decomposition; and so we can afford at most $k$ decompositions.

The decomposition in [2] is instead: pick $L$ to be the degree $k$ part of $f$, and $H_{i}$ are all the Fourier coefficients which are non-zero in the inputs to $f_{i}$ and whose degree in the inputs of $f_{1},\ldots ,f_{i}$ is $\ge k$. The functions $H_{i}$ can be written as $h_{i}\cdot g_{i}$, where $h_{i}$ is the high-degree part of $f_{1}\cdots f_{i}$ and $h_{i}$ is $f_{i+1}\cdots f_{\ell }$.

Once you have this decomposition you can apply the same lemmas in [3] to get improved bounds. To handle space-bounded computation they extend this argument to matrix-valued functions.

### What’s next

In [3] we asked for tight “bounded independence plus noise” results for any model, and the question remains. In particular, what about high-degree polynomials modulo $2$?

### References

[1]   Ravi Boppana, Johan Håstad, Chin Ho Lee, and Emanuele Viola. Bounded independence vs. moduli. In Workshop on Randomization and Computation (RANDOM), 2016.

[2]   Michael A. Forbes and Zander Kelley. Pseudorandom generators for read-once branching programs, in any order. In IEEE Symp. on Foundations of Computer Science (FOCS), 2018.

[3]   Elad Haramaty, Chin Ho Lee, and Emanuele Viola. Bounded independence plus noise fools products. SIAM J. on Computing, 47(2):295–615, 2018.

[4]   Chin Ho Lee and Emanuele Viola. Some limitations of the sum of small-bias distributions. Theory of Computing, 13, 2017.

# Nonclassical polynomials and exact computation of Boolean functions

Guest post by Abhishek Bhrushundi.

I would like to thank Emanuele for giving me the opportunity to write a guest post here. I recently stumbled upon an old post on this blog which discussed two papers: Nonclassical polynomials as a barrier to polynomial lower bounds by Bhowmick and Lovett, and Anti-concentration for random polynomials by Nguyen and Vu. Towards the end of the post, Emanuele writes:

“Having discussed these two papers in a sequence, a natural question is whether non-classical polynomials help for exact computation as considered in the second paper. In fact, this question is asked in the paper by Bhowmick and Lovett, who conjecture that the answer is negative: for exact computation, non-classical polynomials should not do better than classical.”

In a joint work with Prahladh Harsha and Srikanth Srinivasan from last year, On polynomial approximations over $\mathbb {Z}/2^k\mathbb {Z}$, we study exact computation of Boolean functions by nonclassical polynomials. In particular, one of our results disproves the aforementioned conjecture of Bhowmick and Lovett by giving an example of a Boolean function for which low degree nonclassical polynomials end up doing better than classical polynomials of the same degree in the case of exact computation.

The counterexample we propose is the elementary symmetric polynomial of degree $16$ in $\mathbb {F}_2[x_1, \ldots , x_n]$. (Such elementary symmetric polynomials also serve as counterexamples to the inverse conjecture for the Gowers norm [LMS11GT07], and this was indeed the reason why we picked these functions as candidate counterexamples),

\begin{aligned}S_{16}(x_1, \ldots , x_n) = \left (\sum _{S\subseteq [n],|S| = 16} \prod _{i \in S}x_i\right )\textrm { mod 2} = {|x| \choose 16} \textrm { mod 2},\end{aligned}

where $|x| = \sum _{i=1}^n x_i$ is the Hamming weight of $x$. One can verify (using, for example, Lucas’s theorem) that $S_{16}(x_1, \ldots , x_n) = 1$ if and only if the $5^{th}$ least significant bit of $|x|$ is $1$.

We use that no polynomial of degree less than or equal to $15$ can compute $S_{16}(x)$ correctly on more than half of the points in $\{0,1\}^n$.

Theorem 1. Let $P$ be a polynomial of degree at most $15$ in $\mathbb {F}_2[x_1, \ldots , x_n]$. Then

\begin{aligned}\Pr _{x \sim \{0,1\}^n}[P(x) = S_{16}(x)] \le \frac {1}{2} + o(1).\end{aligned}

[Emanuele’s note. Let me take advantage of this for a historical remark. Green and Tao first claimed this fact and sent me and several others a complicated proof. Then I pointed out the paper by Alon and Beigel [AB01]. Soon after they and I independently discovered the short proof reported in [GT07].]

The constant functions (degree $0$ polynomials) can compute any Boolean function on half of the points in $\{0,1\}^n$ and this result shows that even polynomials of higher degree don’t do any better as far as $S_{16}(x_1, \ldots , x_n)$ is concerned. What we prove is that there is a nonclassical polynomial of degree $14$ that computes $S_{16}(x_1, \ldots , x_n)$ on $9/16 \ge 1/2 + \Omega (1)$ of the points in $\{0,1\}^n$.

Theorem 2. There is a nonclassical polynomial $P$ of degree $14$ such that

\begin{aligned}\Pr _{x \sim \{0,1\}^n}[P(x) = S_{16}(x)] = \frac {9}{16} - o(1).\end{aligned}

A nonclassical polynomial takes values on the torus $\mathbb {T} = \mathbb {R}/\mathbb {Z}$ and in order to compare the output of a Boolean function (i.e., a classical polynomial) to that of a nonclassical polynomial it is convenient to think of the range of Boolean functions to be $\{0,1/2\} \subset \mathbb {T}$. So, for example, $S_{16}(x_1, \ldots , x_n) = \frac {1}{2}$ if $|x|_4 = 1$, and $S_{16}(x_1, \ldots , x_n) = 0$ otherwise. Here $|x|_4$ denotes the $5^{th}$ least significant bit of $|x|$.

We show that the nonclassical polynomial that computes $S_{16}(x)$ on $9/16$ of the points in $\{0,1\}^n$ is

\begin{aligned}P(x_1, \ldots , x_n) = \frac {\sum _{S \subseteq [n], |S|=12} \prod _{i \in S}x_i}{8} \textrm { mod 1}= \frac {{|x| \choose 12}}{8} \textrm { mod 1} .\end{aligned}

The degree of this nonclassical polynomial is $14$ but I wouldn’t get into much detail as to why this is case (See [BL15] for a primer on the notion of degree in the nonclassical world).

Understanding how $P(x)$ behaves comes down to figuring out the largest power of two that divides $|x| \choose 12$ for a given $x$: if the largest power of two that divides $|x| \choose 12$ is $2$ then $P(x) = 1/2$, otherwise if the largest power is at least $3$ then $P(x) = 0$. Fortunately, there is a generalization of Lucas’s theorem, known as Kummer’s theorem, that helps characterize this:

Theorem 3.[Kummer’s theorem] The largest power of $2$ dividing $a \choose b$ for $a,b \in \mathbb {N}$, $a \ge b$, is equal to the number of borrows required when subtracting $b$ from $a$ in base $2$.
Equipped with Kummer’s theorem, it doesn’t take much work to arrive at the following conclusion.

Lemma 4. $P(x) = S_{16}(x)$ if either $|x|_{2} = 0$ or $(|x|_2, |x|_3, |x|_4, |x|_5) = (1,0,0,0)$, where $|x|_i$ denotes the $(i+1)^{th}$ least significant bit of $|x|$.

If $x = (x_1, \ldots , x_n)$ is uniformly distributed in $\{0,1\}^n$ then it’s not hard to verify that the bits $|x|_0, \ldots , |x|_5$ are almost uniformly and independently distributed in $\{0,1\}$, and so the above lemma proves that $P(x)$ computes $S_{16}(x)$ on $9/16$ of the points in $\{0,1\}^n$. It turns out that one can easily generalize the above argument to show that $S_{2^\ell }(x)$ is a counterexample to Bhowmick and Lovett’s conjecture for every $\ell \ge 4$.

We also show in our paper that it is not the case that nonclassical polynomials always do better than classical polynomials in the case of exact computation — for the majority function, nonclassical polynomials do as badly as their classical counterparts (this was also conjectured by Bhowmick and Lovett in the same work), and the Razborov-Smolensky bound for classical polynomials extends to nonclassical polynomials.

We started out trying to prove that $S_4(x_1, \ldots , x_n)$ is a counterexample but couldn’t. It would be interesting to check if it is one.

### References

[AB01]    N. Alon and R. Beigel. Lower bounds for approximations by low degree polynomials over z m. In Proceedings 16th Annual IEEE Conference on Computational Complexity, pages 184–187, 2001.

[BL15]    Abhishek Bhowmick and Shachar Lovett. Nonclassical polynomials as a barrier to polynomial lower bounds. In Proceedings of the 30th Conference on Computational Complexity, pages 72–87, 2015.

[GT07]    B. Green and T. Tao. The distribution of polynomials over finite fields, with applications to the Gowers norms. ArXiv e-prints, November 2007.

[LMS11]   Shachar Lovett, Roy Meshulam, and Alex Samorodnitsky. Inverse conjecture for the gowers norm is false. Theory of Computing, 7(9):131–145, 2011.

# Entropy polarization

Sometimes you see quantum popping up everywhere. I just did the opposite and gave a classical talk at a quantum workshop, part of an AMS meeting held at Northeastern University, which poured yet another avalanche of talks onto the Boston area. I spoke about the complexity of distributions, also featured in an earlier post, including a result I posted two weeks ago which gives a boolean function $f:\{0,1\}^{n}\to \{0,1\}$ such that the output distribution of any AC$^{0}$ circuit has statistical distance $1/2-1/n^{\omega (1)}$ from $(Y,f(Y))$ for uniform $Y\in \{0,1\}^{n}$. In particular, no AC$^{0}$ circuit can compute $f$ much better than guessing at random even if the circuit is allowed to sample the input itself. The slides for the talk are here.

The new technique that enables this result I’ve called entropy polarization. Basically, for every AC$^{0}$ circuit mapping any number $L$ of bits into $n$ bits, there exists a small set $S$ of restrictions such that:

(1) the restrictions preserve the output distribution, and

(2) for every restriction $r\in S$, the output distribution of the circuit restricted to $r$ either has min-entropy $0$ or $n^{0.9}$. Whence polarization: the entropy will become either very small or very large.

Such a result is useless and trivial to prove with $|S|=2^{n}$; the critical feature is that one can obtain a much smaller $S$ of size $2^{n-n^{\Omega (1)}}$.

Entropy polarization can be used in conjunction with a previous technique of mine that works for high min-entropy distributions to obtain the said sampling lower bound.

It would be interesting to see if any of this machinery can yield a separation between quantum and classical sampling for constant-depth circuits, which is probably a reason why I was invited to give this talk.

# Hardness amplification proofs require majority… and 15 years

Aryeh Grinberg, Ronen Shaltiel, and myself have just posted a paper which proves conjectures I made 15 years ago (the historians want to consult the last paragraph of [2] and my Ph.D. thesis).

At that time, I was studying hardness amplification, a cool technique to take a function $f:\{0,1\}^{k}\to \{0,1\}$ that is somewhat hard on average, and transform it into another function $f':\{0,1\}^{n}\to \{0,1\}$ that is much harder on average. If you call a function $\delta$-hard if it cannot be computed on a $\delta$ fraction of the inputs, you can start e.g. with $f$ that is $0.1$-hard and obtain $f'$ that is $1/2-1/n^{100}$ hard, or more. This is very important because functions with the latter hardness imply pseudorandom generators with Nisan’s design technique, and also “additional” lower bounds using the “discriminator lemma.”

The simplest and most famous technique is Yao’s XOR lemma, where

\begin{aligned} f'(x_{1},x_{2},\ldots ,x_{t}):=f(x_{1})\oplus f(x_{2})\oplus \ldots \oplus f(x_{t}) \end{aligned}

and the hardness of $f'$ decays exponentially with $t$. (So to achieve the parameters above it suffices to take $t=O(\log k)$.)

At the same time I was also interested in circuit lower bounds, so it was natural to try to use this technique for classes for which we do have lower bounds. So I tried, and… oops, it does not work! In all known techniques, the reduction circuit cannot be implemented in a class smaller than TC$^{0}$ – a class for which we don’t have lower bounds and for which we think it will be hard to get them, also because of the Natural proofs barrier.

Eventually, I conjectured that this is inherent, namely that you can take any hardness amplification reduction, or proof, and use it to compute majority. To be clear, this conjecture applied to black-box proofs: decoding arguments which take anything that computes $f'$ too well and turn it into something which computes $f$ too well. There were several partial results, but they all had to restrict the proof further, and did not capture all available techniques.

Should you have had any hope that black-box proofs might do the job, in this paper we prove the full conjecture (improving on a number of incomparable works in the literature, including a 10-year-anniversary work by Shaltiel and myself which proved the conjecture for non-adaptive proofs).

### Indistinguishability

One thing that comes up in the proof is the following basic problem. You have a distribution $X$ on $n$ bits that has large entropy, very close to $n$. A classic result shows that most bits of $X$ are close to uniform. We needed an adaptive version of this, showing that a decision tree making few queries cannot distinguish $X$ from uniform, as long as the tree does not query a certain small forbidden set of variables. This also follows from recent and independent work of Or Meir and Avi Wigderson.

Turns out this natural extension is not enough for us. In a nutshell, it is difficult to understand what queries an arbitrary reduction is making, and so it is hard to guarantee that the reduction does not query the forbidden set. So we prove a variant, where the variables are not forbidden, but are fixed. Basically, you condition on some fixing $X_{B}=v$ of few variables, and then the resulting distribution $X|X_{B}=v$ is indistinguishable from the distribution $U|U_{B}=v$ where $U$ is uniform. Now the queries are not forbidden but have a fixed answer, and this makes things much easier. (Incidentally, you can’t get this simply by fixing the forbidden set.)

### Fine, so what?

One great question remains. Can you think of a counter-example to the XOR lemma for a class such as constant-depth circuits with parity gates?

But there is something more why I am interested in this. Proving $1/2-1/n$ average-case hardness results for restricted classes “just” beyond AC$^{0}$ is more than a long-standing open question in lower bounds: It is necessary even for worst-case lower bounds, both in circuit and communication complexity, as we discussed earlier. And here’s hardness amplification, which intuitively should provide such hardness results. It was given many different proofs, see e.g. [1]. However, none can be applied as we just saw. I don’t know, someone taking results at face value may even start thinking that such average-case hardness results are actually false.

### References

[1]   Oded Goldreich, Noam Nisan, and Avi Wigderson. On Yao’s XOR lemma. Technical Report TR95–050, Electronic Colloquium on Computational Complexity, March 1995. http://www.eccc.uni-trier.de/.

[2]   Emanuele Viola. The complexity of constructing pseudorandom generators from hard functions. Computational Complexity, 13(3-4):147–188, 2004.

# Matrix rigidity, and all that

The rigidity challenge asks to exhibit an n × n matrix M that cannot be written as M = A + B where A is “sparse” and B is “low-rank.” This challenge was raised by Valiant who showed in [Val77] that if it is met for any A with at most n1+ϵ non-zero entries and any B with rank O(n∕ log log n) then computing the linear transformation M requires either logarithmic depth or superlinear size for linear circuits. This connection relies on the following lemma.

Lemma 1. Let C : {0, 1}n →{0, 1}n be a circuit made of XOR gates. If you can remove e edges and reduce the depth to d then the linear transformation computed by C equals A + B where A has ≤ 2d non-zero entries per row (and so a total of ≤ n2d non-zero entries), and B has rank ≤ e.

Proof: After you remove the edges, each output bit is a linear combination of the removed edges and at most 2d input variables. The former can be done by B, the latter by A. QED

Valiant shows that in a log-depth, linear-size circuit one can remove O(n∕ log log n) edges to reduce the depth to nϵ – a proof can be found in [Vio09] – and this gives the above connection to lower bounds.

However, the best available tradeoff for explicit matrices give sparsity n2∕r log(n∕r) and rank r, for any parameter r; and this is not sufficient for application to lower bounds.

#### Error-correcting codes

It was asked whether generator matrixes of good linear codes are rigid. (A code is good if it has constant rate and constant relative distance. The dimensions of the corresponding matrixes are off by only a constant factor, and so we can treat them as identical.) Spielman [Spi95] shows that there exist good codes that can be encoded by linear-size logarithmic depth circuits. This immediately rules out the possibility of proving a lower bound, and it gives a non-trivial rigidity upper bound via the above connections.

Still, one can ask if these matrices at least are more rigid than the available tradeoffs. Goldreich reports a negative answer by Dvir, showing that there exist good codes whose generating matrix C equals A + B where A has at most O(n2∕d) non-zero entries and B has rank O(d log n∕d), for any d.

A similar negative answer follows by the paper [GHK+13]. There we show that there exist good linear codes whose generating matrix can be written as the product of few sparse matrixes. The corresponding circuits are very structured, and so perhaps it is not surprising that they give good rigidity upper bounds. More precisely, the paper shows that we can encode an n-bit message by a circuit made of XOR gates and with say n log *n wires and depth O(1) – with unbounded fan-in. Each gate in the circuit computes the XOR of some t gates, which can be written as a binary tree of depth log 2t + O(1). Such trees have poor rigidity:

Lemma 2.[Trees are not rigid] Let C be a binary tree of depth d. You can remove an O(1∕2b) fraction of edges to reduce the depth to b, for any b.

Proof: It suffices to remove all edges at depths d – b, d – 2b, …. The number of such edges is O(2d-b + 2d-2b + …) = O(2d-b). Note this includes the case d ≤ b, where we can remove 0 edges. QED

Applying Lemma 2 to a gate in our circuit, we reduce the depth of the binary tree computed at that gate to b. Applying this to every gate we obtain a circuit of depth O(b). In total we have removed an O(1∕2b) fraction of the n log *n edges.

Writing 2b = n∕d, by Lemma 1 we can write the generating matrixes of our code as C = A + B where A has at most O(n∕d) non-zero entries per row, and B has rank O(d log *n). These parameters are the same as in Dvir’s result, up to lower-order terms. The lower-order terms appear incomparable.

#### Walsh-Fourier transform

Another matrix that was considered is the n×n Inner Product matrix H, aka the Walsh-Hadamard matrix, where the x,y entry is the inner product of x and y modulo 2. Alman and Williams [AW16] recently give an interesting rigidity upper bound which prevents this machinery to establish a circuit lower bound. Specifically they show that H can be written as H = A + B where A has at most n1+ϵ non-zero entries, and B has rank n1-ϵ′, for any ϵ and an ϵ′ which goes to 0 when ϵ does.

Their upper bound works as follows. Let h = log 2n. Start with the univariate, real polynomial p(z1,z2,…,zh) which computes parity exactly on inputs of Hamming weight between 2ϵn and (1∕2 + ϵ)n. By interpolation such a polynomial exists with degree (1∕2 – ϵ)n. Replacing zi with xiyi you obtain a polynomial of degree n – ϵn which computes IP correctly on inputs x,y whose inner product is between 2ϵn and (1∕2 + ϵ)n.

This polynomial has 2(1-ϵ′)n monomials, where ϵ′ = Ω(ϵ2). The truth-table of a polynomial with m monomials is a matrix with rank m, and this gives a low-rank matrix B′.

The fact that sparse polynomials yield low-rank matrixes also appeared in the paper [SV12], which suggested to study the rigidity challenge for matrixes arising from polynomials.

Returning to the proof in [AW16], it remains to deal with inputs whose inner product does not lie in that range. The number of x whose weight is not between (1∕2 – ϵ)n and (1∕2 + ϵ)n is 2(1-ϵ′)n. For each such input x we modify a row of the matrix B′. Repeating the process for the y we obtain the matrix B, and the rank bound 2(1-ϵ′)n hasn’t changed.

Now a calculation shows that B differs from H in few entries. That is, there are few x and y with Hamming weight between (1∕2 – ϵ)n and (1∕2 + ϵ)n, but with inner product less than 2ϵn.

### Boolean complexity

There exists a corresponding framework for boolean circuits (as opposed to circuits with XOR gates only). Rigid matrixes informally correspond to depth-3 Or-And-Or circuits. If this circuit has fan-in fo at the output gate and fan-in fi at each input gate, then the correspondence in parameters is

 rank = log fo sparsity = 2fi .

More precisely, we have the following lemma.

Lemma 3. Let C : {0, 1}n →{0, 1}n be a boolean circuit. If you can remove e edges and reduce the depth to d then you can write C as an Or-And-Or circuit with output fan-in 2e and input fan-in 2d.

Proof: After you remove the edges, each output bit and each removed edge depends on at most 2d input bits or removed edges. The output Or gate of the depth-3 circuit is a big Or over all 2e assignments of values for the removed edges. Then we need to check consistency. Each consistency check just depends on 2d inputs and so can be written as a depth-2 circuit with fan-in 2d. QED

The available bounds are of the form log fo = n∕fi. For example, for input fan-in fi = nα we have lower bounds exponential in n1-α but not more. Again it can be shown that breaking this tradeoff in certain regimes (namely, log 2fo = O(n∕ log log n)) yields lower bounds against linear-size log-depth circuits. (A proof appears in [Vio09].) It was also pointed out in [Vio13] that breaking this tradeoff in any regime yields lower bounds for branching programs. See also the previous post.

One may ask how pairwise independent hash functions relate to this challenge. Ishai, Kushilevitz, Ostrovsky, and Sahai showed [IKOS08] that they can be computed by linear-size log-depth circuits. Again this gives a non-trivial upper bound for depth-3 circuits via these connections, and one can ask for more. In [GHK+13] we give constructions of such circuits which in combination with Lemma 3 can again be used to almost match the available trade-offs.

The bottom line of this post is that we can’t prove lower bounds because they are false, and it is a puzzle to me why some people appear confident that P is different from NP.

### References

[AW16]    Josh Alman and Ryan Williams. Probabilistic rank and matrix rigidity, 2016. https://arxiv.org/abs/1611.05558.

[GHK+13]   Anna Gál, Kristoffer Arnsfelt Hansen, Michal Koucký, Pavel Pudlák, and Emanuele Viola. Tight bounds on computing error-correcting codes by bounded-depth circuits with arbitrary gates. IEEE Transactions on Information Theory, 59(10):6611–6627, 2013.

[IKOS08]    Yuval Ishai, Eyal Kushilevitz, Rafail Ostrovsky, and Amit Sahai. Cryptography with constant computational overhead. In 40th ACM Symp. on the Theory of Computing (STOC), pages 433–442, 2008.

[Spi95]    Daniel Spielman. Computationally Efficient Error-Correcting Codes and Holographic Proofs. PhD thesis, Massachusetts Institute of Technology, 1995.

[SV12]    Rocco A. Servedio and Emanuele Viola. On a special case of rigidity. Available at http://www.ccs.neu.edu/home/viola/, 2012.

[Val77]    Leslie G. Valiant. Graph-theoretic arguments in low-level complexity. In 6th Symposium on Mathematical Foundations of Computer Science, volume 53 of Lecture Notes in Computer Science, pages 162–176. Springer, 1977.

[Vio09]    Emanuele Viola. On the power of small-depth computation. Foundations and Trends in Theoretical Computer Science, 5(1):1–72, 2009.

[Vio13]    Emanuele Viola. Challenges in computational lower bounds. Available at http://www.ccs.neu.edu/home/viola/, 2013.

# Mixing in groups, II

In the previous post we have reduced the “three-step mixing” over SL(2,q), the group of 2×2 matrices over the field with q elements with determinant 1, to the following statement about mixing of conjugacy classes.

Theorem 1.[Mixing of conjugacy classes of SL(2,q)] Let G = SL(2,q). With probability ≥ 1 -|G|-Ω(1) over uniform a,b in G, the distribution C(a)C(b) is |G|-Ω(1) close in statistical distance to uniform.

Here and throughout this post, C(g) denotes a uniform element from the conjugacy class of g, and every occurrence of C corresponds to an independent draw.

In this post we sketch a proof of Theorem 1, following [GV15]. Similar theorems were proved already. For example Shalev [Sha08] proves a version of Theorem 1 without a quantitative bound on the statistical distance. It is possible to plug some more representation-theoretic information into Shalev’s proof and get the same quantitative bound as in Theorem 1, though I don’t have a good reference for this extra information. However the proof in [Sha08] builds on a number of other things, which also means that if I have to prove a similar but not identical statement, as we had to do in [GV15], it would be hard for me.

Instead, here is how you can proceed to prove the theorem. First, we remark that the distribution of C(a)C(b) is the same as that of

C(C(a)C(b)),

because for uniform x, y, and z in Fq we have the following equalities of distributions:

C(C(a)C(b)) = x-1(y-1ayz-1bz)x = x-1(y-1ayxx-1z-1bz)x = C(a)C(b)

where the last equality follows by replacing y with yx and z with zx.

That means that we get one conjugation “for free” and we just have to show that C(a)C(b) falls into various conjugacy classes with the right probability.

Now the great thing about SL(2,q) is that you can essentially think of it as made up of q conjugacy classes each of size q2 (the whole group has size q3 – q). This is of course not exactly correct, in particular the identity element obviously gives a conjugacy class of size 1. But except for a constant number of conjugacy classes, every conjugacy class has size q2 up to lower-order terms. This means that what we have to show is simply that the conjugacy class of C(a)C(b) is essentially uniform over conjugacy classes.

Next, the trace map Tr  : SL(2,q) → Fq is essentially a bijection between conjugacy classes and the field Fq. To see this recall that the trace map satisfies the cyclic property:

Tr xyz = Tr yzx.

This implies that

Tr u-1au = Tr auu-1 = Tr a,

and so conjugate elements have the same trace. On the other hand, the q matrixes

x  1

1  0

for x in Fq all have different traces, and by what we said above their conjugacy classes make up essentially all the group.

Putting altogether, what we are trying to show is that

Tr C(a)C(b)

is |G|-Ω(1) close to uniform over F q in statistical distance.

Furthermore, again by the cyclic property we can consider without loss of generality

Tr aC(b)

instead, and moreover we can let a have the form

0  1

1  w

and b have the form

v  1

1  0

(there is no important reason why w is at the bottom rather than at the top).

Writing a generic g in SL(2,q) as the matrix

u1   u2

u3   u4

you can now with some patience work out the expression

Tr au-1bu = vu 3u4 – u32 + u 42 – vu 1u2 + u12 – vwu 2u3 + wu1u3 – u22 – wu 2u4.

What we want to show is that for typical choices of w and v, the value of this polynomial is q-Ω(1) close to uniform over F q for a uniform choice of u subject to the determinant of u being 1, i.e, u1u4 – u2u3 = 1.

Maybe there is some machinery that immediately does that. Lacking the machinery, you can use the equation u1u4 – u2u3 = 1 to remove u4 by dividing by u1 (the cases where u1 = 0 are few and do not affect the final answer). Now you end up with a polynomial p in three variables, which we can rename x, y, and z. You want to show that p(x,y,z) is close to uniform, for uniform choices for x,y,z. The benefit of this substitution is that we removed the annoying condition that the determinant is one.

To argue about p(x,y,z), the DeMillo–Lipton–Schwartz-Zippel lemma comes to mind, but is not sufficient for our purposes. It is consistent with that lemma that the polynomial doesn’t take a constant fraction of the values of the field, which would give a constant statistical distance. One has to use more powerful results known as the Lang-Weil theorem. This theorem provides under suitable conditions on p a sharp bound on the probability that p(x,y,z) = a for a fixed a in Fq. The probability is 1∕q plus lower-order terms, and then by summing over all a in Fq one obtains the desired bound on the statistical distance.

I am curious if there is a way to get the statistical distance bound without first proving a point-wise bound.

To apply the Lang-Weil theorem you have to show that the polynomial is “absolutely irreducible,” i.e., irreducible over any algebraic extension of the field. This can be proven from first principles by a somewhat lengthy case analysis.

### References

[GV15]   W. T. Gowers and Emanuele Viola. The communication complexity of interleaved group products. In ACM Symp. on the Theory of Computing (STOC), 2015.

[Sha08]   Aner Shalev. Mixing and generation in simple groups. J. Algebra, 319(7):3075–3086, 2008.

# Mixing in groups

Non-abelian groups behave in ways that are useful in computer science. Barrington’s famous result [Bar89] shows that we can write efficiently an arbitrary low-depth computation as a group product over any non-solvable group. (Being non-solvable is a certain strengthening of being non-abelian which is not important now.) His result, which is false for abelian groups, has found myriad applications in computer science. And it is amusing to note that actually results about representing computation as group products were obtained twenty years before Barrington, see [KMR66]; but the time was not yet ripe.

This post is about a different property that certain non-abelian groups have and that is also useful. Basically, these groups ”mix” in the sense that if you have several distributions over the group, and the distributions have high entropy, then the product distribution (i.e., sample from each distribution and output the product) is very close to uniform.

First, let us quickly remark that this is completely false for abelian groups. To make an example that is familiar to computer scientists, consider the group of n-bit strings with bit-wise xor. Now let A be the uniform distribution over this group where the first bit is always 0. Then no matter how many independent copies of A you multiply together, the product is always A.

Remarkably, over other groups it is possible to show that the product distribution will become closer and closer to uniform. A group that works very well in this respect is SL(2,q), the group of 2×2 matrices over the field with q elements with determinant 1. This is a group that in some sense is very far from abelian. In particular, one can prove the following result.

Theorem 1.[Three-step mixing [Gow08BNP08]] Let G = SL(2,q), and let A, B, and C be three subsets of G of constant density. Let a, b, and c be picked independently and uniformly from A, B, and C respectively. Then for any g in G we have

| Pr[abc = g] – 1∕|G|| < 1∕|G|1+Ω(1).

Note that the conclusion of the theorem in particular implies that abc is supported over the entire group. This is remarkable, since the starting distributions are supported over only a small fraction of the group. Moreover, by summing over all elements g in the group we obtain that abc is polynomially close to uniform in statistical distance.

Theorem 1 can be proved using representation theory. This must be a great tool, but for some reason I always found it a little difficult to digest the barrage of definitions that usually anticipate the interesting stuff.

Luckily, there is another way to prove Theorem 1. I wouldn’t be surprised if this is in some sense the same way, and moreover this other way is not sometimes I would call elementary. But it is true that I will be able to sketch a proof of the theorem without using the word ”representation”. In this post we will prove some preliminary results that are valid for all groups, and the most complicated thing used is the Cauchy-Schwarz inequality. In the next post we will work specifically with the group SL(2,q), and use more machinery. This is all taken from this paper with Gowers [GV15] (whose main focus is the study of mixing in the presence of dependencies).

First, for convenience let us identify a set A with its characteristic function. So we write A(a) for a belongs to the set A. It is convenient to work with a slightly different statement:

Theorem 2. Let G = SL(2,q) and let A,B,C be three subsets of G of densities α,β,γ respectively. For any g in G,

|Eabc=gA(a)B(b)C(c) – αβγ|≤|G|-Ω(1)

where the expectation is over uniform elements a, b, and c from the group G such that their product is equal to g.

This Theorem 2 is equivalent to Theorem 1, because

 Eabc=gA(a)B(b)C(c) = Pr[A(a),B(b),C(c)|abc = g] = Pr[abc = g|A(a),B(b),C(c)]|G|αβγ

by Bayes’ rule. So we can get Theorem 1 by dividing by |G|αβγ.

Now we observe that to prove this ”mixing in three steps” it actually suffices to prove mixing in four steps.

Theorem 3.[Mixing in four steps] Let G = SL(2,q) and let A,B,C,D be four subsets of G of densities α,β,γ,δ respectively. For any g in G,

Eabcd=gA(a)B(b)C(c)D(d) – αβγδ ≤|G|-Ω(1),

where the expectation is over uniform elements a, b, c, and d from the group G such that their product is equal to g.

Lemma 4. Mixing in four steps implies mixing in three.

Proof: Rewrite

|Eabc=gA(a)B(b)C(c) – αβγ| = |Eabc=gf(a)B(b)C(c)|

where f(a) := A(a) – α.

In these proofs we will apply Cauchy-Schwarz several times. Each application ”loses a square,” but since we are aiming for an upper bound of the form 1∕|G|Ω(1) we can afford any constant number of applications. Our first one is now:

 (Eabc=gf(a)B(b)C(c))2 ≤ (E cC(c)2)(E c(Eab=gc-1f(a)B(b))2) = γEcEab=a′b′=gc-1f(a)B(b)f(a′)B(b′) = γEab=a′b′(A(a) – α)B(b)B(b′)(A(a′) – α).

There are four terms that make up the expectation. The terms that involve at least one α sum to -α2β2. The remaining term is the expectation of A(a)B(b)B(b′)A(a′). Note that ab = a′b′ is equivalent to ab(1∕b′)(1∕a′) = 1G. Hence by Theorem 3 this expectation is at most |G|-Ω(1). QED

So what remains to see is how to prove mixing in four steps. We shall reduce the mixing problem to the following statement about the mixing of conjugacy classes of our group.

Definition 5. We denote by C(g) the conjugacy class {h-1gh : h in G} of an element g in G. We also denote by C(g) the uniform distribution over C(g) for a uniformly selected g in G.

Theorem 6.[Mixing of conjugacy classes of SL(2,q)] Let G = SL(2,q). With probability ≥ 1 -|G|-Ω(1) over uniform a,b in G, the distribution C(a)C(b) is |G|-Ω(1) close in statistical distance to uniform.

Theorem 6 is proved in the next blog post. Here we just show that is suffices for our needs.

Lemma 7. Theorem 6 implies Theorem 3.

Proof: We rewrite the quantity to bound as

Ea,c(A(a)C(c)Eb,d:abcd=gf(b,d))

for f(b,d) = B(b)D(d) – βδ.

Now by Cauchy-Schwarz we bound this above by

Ea,c,b,d,b′,d′f(b,d)f(b′,d′)

where the expectation is over variables such that abcd = g and ab′cd′ = g. As in the proof that mixing in four steps implies mixing in three, we can rewrite the last two equations as the single equation bcd = b′cd′.

The fact that the same variable c occurs on both sides of the equation is what gives rise to conjugacy classes. Indeed, this equation can be rewritten as

c-1(1∕b)b′c = d(1∕d′).

Performing the following substitutions: b = x,b′ = xh,d′ = y we can rewrite our equation as

d = c-1hcy.

Hence we have reduced our task to that of bounding

Ef(x,C(h)y)f(xh,y)

for uniform x,y,h.

We can further replace y with C(h)-1y, and rewrite the expression as

Ex,y(f(x,y)Ehf(xh,C(h-1)y)).

This is at most

(Ex,yf2(x,y))E x,y,h,h′f(xh,C(h-1)y)f(xh′,C(h′-1)y).

Recalling that f(b,d) = B(b)D(d) – βδ, and that E[f] = βδ, the first factor is at most 1. The second can be rewritten as

Ex,y,h,h′f(x,y)f(xh-1h′,C(h′-1)C(h)y)

replacing x with xh-1 and y with C(h-1)-1y = C(h)y.

Again using the definition of f this equals

Ex,y,h,h′B(x)D(y)B(xh-1h′)D(C(h′-1)C(h)y) – β2δ2.

Now Lemma 6 guarantees that the distribution (x,y,xh-1h′,C(h′-1)C(h)y) is 1∕|G|Ω(1)-close in statistical distance to the uniform distribution over G4, and this concludes the proof. QED

### References

[Bar89]    David A. Mix Barrington. Bounded-width polynomial-size branching programs recognize exactly those languages in NC1. J. of Computer and System Sciences, 38(1):150–164, 1989.

[BNP08]    László Babai, Nikolay Nikolov, and László Pyber. Product growth and mixing in finite groups. In ACM-SIAM Symp. on Discrete Algorithms (SODA), pages 248–257, 2008.

[Gow08]    W. T. Gowers. Quasirandom groups. Combinatorics, Probability & Computing, 17(3):363–387, 2008.

[GV15]    W. T. Gowers and Emanuele Viola. The communication complexity of interleaved group products. In ACM Symp. on the Theory of Computing (STOC), 2015.

[KMR66]   Kenneth Krohn, W. D. Maurer, and John Rhodes. Realizing complex Boolean functions with simple groups. Information and Control, 9:190–195, 1966.

# Bounded indistinguishability

Countless papers study the properties of k-wise independent distributions, which are distributions where any k bits are uniform and independent. One property of interest is which computational models are fooled by such distributions, in the sense that they cannot distinguish any such distribution from a uniformly random one. Recently, Bazzi’s breakthrough, discussed earlier on this blog, shows that k = polylog(n) independence fools any polynomial-size DNF on n bits.

Let us change the question. Let us say that instead of one distribution we have two, and we know that any k bits are distributed identically, but not necessarily uniformly. We call such distributions k-wise indistinguishable. (Bounded independence is the special case when one distribution is uniform.) Can a DNF distinguish the two distributions? In fact, what about a single Or gate?

This is the question that we address in a paper with Bogdanov, Ishai, and Williamson. A big thank you goes to my student Chin Ho Lee for connecting researchers who were working on the same problems on different continents. Here at NEU the question was asked to me by my neighbor Daniel Wichs.

The question turns out to be equivalent to threshold/approximate degree, an influential complexity measure that goes back to the works by Minsky and Papert and by Nisan and Szegedy. The equivalence is a good example of the usefulness of duality theory, and is as follows. For any boolean function f on n bits the following two are equivalent:

1. There exist two k-wise indistinguishable distributions that f tells apart with advantage e;

2. No degree-k real polynomial can approximate f to pointwise error at most e/2.

I have always liked this equivalence, but at times I felt slightly worried that could be considered too “simple.” But hey, I hope my co-authors don’t mind if I disclose that it’s been four different conferences, and not one reviewer filed a complaint about that.

From the body of works on approximate degree one readily sees that bounded indistinguishability behaves very differently from bounded independence. For example, one needs k = Ω(√ n) to fool an Or gate, and that is tight. Yes, to spell this out, there exist two distributions which are 0.001 √ n indistinguishable but Or tells them apart with probability 0.999. But obviously even constant independence fools Or.

The biggest gap is achieved by the Majority function: constant independence suffices, by this, while linear indistinguishability is required by Paturi’s lower bound.

In the paper we apply this equivalence in various settings, and here I am just going to mention the design of secret-sharing schemes. Previous schemes like Shamir’s required the computation of things like parity, while the new schemes use different types of functions, for example of constant depth. Here we also rely on the amazing ability of constant-depth circuits to sample distributions, also pointed out earlier on this blog, and apply some expander tricks to trade alphabet size for other parameters.