PIT for depth-$4$ circuits and Sylvester-Gallai conjecture for polynomials

TL;DR

This paper explores extending Sylvester-Gallai-based techniques to develop deterministic polynomial-time blackbox identity testing algorithms for depth-4 circuits with bounded top fanin, achieving partial success and conditional results.

Contribution

It generalizes Sylvester-Gallai approach from depth-3 to depth-4 circuits, providing a conditional deterministic PIT algorithm and an unconditional algorithm for a subclass.

Findings

Conditional polynomial-time PIT algorithm for depth-4 circuits with bounded top fanin.

Unconditional polynomial-time algorithm for a specific subclass of depth-4 circuits.

Extension of Sylvester-Gallai theorem-based methods to higher circuit depths.

Abstract

This text is a development of a preprint of Ankit Gupta. We present an approach for devising a deterministic polynomial time blackbox identity testing (PIT) algorithm for depth-$4$ circuits with bounded top fanin. This approach is similar to Kayal-Shubhangi approach for depth-$3$ circuits. Kayal and Shubhangi based their algorithm on Sylvester-Gallai-type theorem about linear polynomials. We show how it is possible to generalize this approach to depth-$4$ circuits. However we failed to implement this plan completely. We succeeded to construct a polynomial time deterministic algorithm for depth-$4$ circuits with bounded top fanin and its correctness requires a hypothesis. Also we present a polynomial-time (unconditional) algorithm for some subclass of depth-$4$ circuits with bounded top fanin.

Equations

$$C(x_1,\dots ,x_n)=\sum _{i=1}^k{F}_i=\sum _{i=1}^k\prod _{j=1}^{d_i}{l}_{ij}$$

$$C(x_1,\dots ,x_n)=\text{gcd}(C)\cdot \sum _{i=1}^k{F}_i=\sum _{i=1}^k\prod _{j=1}^{d_i^{\prime }}{l}_{ij}^{\prime }$$

$$C:=\sum _{i=1}^{i=k}{F}_i=\sum _{i=1}^k\prod _{j=1}^{d_i}{l}_{ij}$$

Full text

We present an approach for devising a deterministic polynomial time blackbox identity testing (PIT) algorithm for depth-$4$ circuits with bounded top fanin. This approach is similar to Kayal-Shubhangi [15] approach for depth-$3$ circuits. Kayal and Shubhangi based their algorithm on Sylvester-Gallai-type theorem about linear polynomials. We show how it is possible to generalize this approach to depth-$4$ circuits. However we failed to implement this plan completely. We succeeded to construct a polynomial time deterministic algorithm for depth-$4$ circuits with bounded top fanin and its correctness requires a hypothesis. Also we present a polynomial-time (unconditional) algorithm for some subclass of depth-$4$ circuits with bounded top fanin.

1 Introduction

Polynomial Identity Testing : In blackbox polynomial identity testing (PIT), given only query access to a hidden circuit, one has to determine if it outputs the zero polynomial. In whitebox PIT one has to solve the same problem with possibility to see a circuit.

This problem has numerous applications and has appeared in many fundamental results in complexity theory. Although this problem exhibits a trivial randomized algorithm, designing an efficient deterministic algorithm is one of the most challenging open problems. Strong equivalence results between derandomizing PIT and proving super-polynomial circuit lower bounds for explicit polynomials are known (cf. Chapter 4 of [19]).

Depth-$4$ Circuits : In a surprising result, Agrawal-Vinay [2] showed that a complete derandomization of PIT for just depth-4 ($\Sigma\Pi\Sigma\Pi$) circuits implies an exponential lower bound for general circuits and a near complete derandomization of PIT for general circuits of poly-degree. Hence the problem of derandomizing PIT for such fanin restricted depth-$4$ circuits is equivalent to the general case.

There has been an incredibly large number of results for $\Sigma\Pi\Sigma\Pi$-circuits with diverse restrictions. A study for the case in which the bottom fan-in of such depth-$4$ circuits is at most $1$ ( $\Sigma\Pi\Sigma$ circuits) was initiated by Dvir-Shpilka [6] (whitebox) and Karnin-Shpilka [14] (blackbox). A different study for the case with the restriction of bounded transcendence degree was initiated by Beecken et al. [5]. Recently, Agrawal et al. [1] reproved all these diverse results using a single unified technique based on the Jacobian criterion. In allmost all these results, the fanin of the top $+$ gate is assumed to be $O(1)$. For details see the survey by Shpilka-Yehudayoff [19] or the one by Saxena[18].

The Model : In this work we consider the model of $\Sigma\Pi\Sigma\Pi(k,r)$ circuits over $\mathbb{C}$, the field of complex numbers. We first define $\Sigma\Pi\Sigma\Pi(k)$ circuits. These are circuits having four alternating layers of and gates where the fanin of the top gate is $k$. Such a circuit alternating layers of $+$ and $\times$ gates where the fanin of the top $+$ gate is $\leqslant k$. Such a circuit $C$ computes a polynomial of the form

[TABLE]

where $d_{i}$ are the fanins of the $\times$ gates at the second level. Define $\text{gcd}(C):=\text{gcd}(F_{1},\ldots,F_{k})$. A circuit is called simple if $\text{gcd}(C)=1$. The polynomial computed by a $\Sigma\Pi\Sigma\Pi(k,r)$ circuit $C$ has the same form as in (1) with added restriction that the degree of every $l_{ij}$ is at most $r$. As $l_{ij}$ can have at most $r$ irreducible factors, we can factor $l_{ij}$ while incurring a multiplicative factor of $r$ in $d_{i}$. Hence, the polynomial computed by a $\Sigma\Pi\Sigma\Pi(k,r)$ circuit $C$ is of the form

[TABLE]

where gcd($C$) is a product of polynomials of degree at most $r$ and $l^{\prime}_{ij}$ are irreducible. Such a circuit is said to be homogenous if all $F_{i}$ are homogenous of the same degree (and therefore $l^{\prime}_{ij}$ are also homogenous).

2 Results

To state the result we first need to introduce some notions from incidence geometry.

Sylvester-Gallai type problems

A well-known theorem in incidence geometry called the Sylvester-Gallai (SG) theorem states that : if there are $n$ distinct points on the real plane such that, for every pair of distinct points, the line through them also contains a third point, then they all lie on the same line. Over several decades, various variants of this result have been proved and are in general called Sylvester-Gallai type problems. Informally, in such problems, one is presented with a set of objects (points, hyperplanes, etc.) with a lot of “local” dependencies (e.g. two points are collinear with a third) and the goal is to translate these local restrictions to a global bound (usually on the dimension of the space spanned by the objects). Recently, in an impressive work by Barak et al.[3], a robust variant of the SG theorem was proved which among other things says that, even if for every point, the above stated restriction holds for a constant fraction of other points, one can still bound the dimension of the vector space spanned by the point set in $\mathbb{C}^{d}$ by a constant. Few other lines of study for the SG type problems include

• •

replacing lines by higher dimensional vector spaces (initiated by Hansen),

• •

having multiple sets of (colored) points (initiated by Motzkin-Rabin),

• •

robust/fractional versions of the above (initiated by Barak et al.).

For an introduction to the SG theorem and its variants see the survey by Borwein-Moser [4]. One interesting feature of [3] is that the robust variant of SG theorem was motivated by a problem in theoretical computer science, in particular the study of (linear) Locally Correctable Codes. A common feature of all these variants is that they only consider flats/vector spaces/linear varieties. Ankit Gupta and the author propose a new line of SG theorems for non-linear polynomials. These problems arise very naturally in our approach for devising PIT algorithms for $\Sigma\Pi\Sigma\Pi(k,r)$ circuits.

The SG theorem can be restated in terms of polynomials as follows: let $l_{1},\ldots,l_{m}$ be distinct homogenous linear polynomials in $\mathbb{R}[x_{0},\ldots,x_{n}]$ s.t. for every pair of distinct $l_{i}$ , $l_{j}$ there is a distinct $l_{k}$ s.t. $l_{k}$ belongs to the ideal $\langle l_{i},l_{j}\rangle$. Then dimension of the vector space spanned by all $l_{n}$ is at most $1$.

The dimension of the vector space spanned by a set of linear polynomials is a special case of the general concept of transcendence degree of a set of polynomials . Polynomials $f_{1},\ldots,f_{m}\subset\mathbb{C}[x_{1},\ldots,x_{n}]$ are called algebraically independent if there is no non-zero polynomial $F$ such that $F(f_{1},\ldots,f_{m})=0$. The transcendence degree $\text{trdeg}_{\mathbb{C}}\{f_{1},\ldots,f_{m}\}$ is the maximal number $r$ of algebraically independent polynomials in the set.

Definition 1**.**

A simple homogenous $\Sigma\Pi\Sigma\Pi(k)$ circuit $C$ such that

[TABLE]

as stated in Equation (1) is SG if for every $i\in\{1,\ldots,k\}$ and

for every $l_{1j_{1}},l_{2j_{2}}\ldots,l_{i-1,j_{i-1}},l_{i+1,j_{i+1}},\ldots,l_{k,j_{k}}$

the ideal $\langle l_{1j_{1}},l_{2j_{2}}\ldots,l_{i-1,j_{i-1}},l_{i+1,j_{i+1}},\ldots,l_{k,j_{k}}\rangle$ contains $F_{i}$.

Our motivation behind terming such circuits as SG comes from Dvir-Shpilka’s idea of using variants of the SG theorem for bounding the dimension of the vector space spanned by the linear forms occurring (at the third layer) in such circuits in the case the bottom fanin is at most $1$, i.e., it is a $\Sigma\Pi\Sigma\Pi(k,1)$ circuit. They also conjectured that, if $\mathbb{F}$ has characteristic [math] then, this dimension is bounded by a function of only $k$. Indeed later, Kayal-Saraf [15] used a colored higher-dimensional variant of the SG theorem to prove this conjecture for $\mathbb{R}$. In spirit of Dvir-Shpilka [6] we conjecture that in such SG-$\Sigma\Pi\Sigma\Pi(k,r)$ circuits the transcendence degree of the set of $l_{ij}$ is bounded by a function of $k$, $r$.

Conjecture 1**.**

Let $C$ be a $\Sigma\Pi\Sigma\Pi(k,r)$ circuit of the form (3). If $C$ is SG then $\text{trdeg}_{\mathbb{C}}\{l_{ij}\}\leqslant\lambda(k,r)$ for some function $\lambda$.

For the case $r=1$ this conjecture reduces to the case $c=1$ by the irreducibility of vector spaces and was first proved over $\mathbb{R}$ in [15]. We are now ready to state our first result for PIT.

Theorem 1**.**

Given white-box access to a $\Sigma\Pi\Sigma\Pi(k,r)\text{-circuit}$ $f\in\mathbb{C}[x_{0},\ldots,x_{n}]$ of degree $d$, the identity test for $f$ can be decided deterministically in time $\text{poly}(n,d)$ for constant $k$ and $r$ if $C$ is not SG. Moreover, if Conjecture 1 holds then the same is true even if $C$ is SG.

Remark*.*

In [11] Ankit Gupta gives another definition of SG-circuit (he uses radical ideals instead of usual ideals). Our approach is in some sense better: we obtain a similar result as in [11] under a weaker conjecture.

Our second result is a proof of Conjecture 1 in a special case—Theorem  7 in Section 5. Also we obtain a full deradomization of PIT for some subclass of $\Sigma\Pi\Sigma\Pi(3,2)$—Theorem 8 in Section 6.

3 Case $\Sigma\Pi\Sigma\Pi(3,1)$-circuits

Here we present our idea of derandomization for $\Sigma\Pi\Sigma\Pi(3,1)$-circuits, i.e., polynomials of the form $F_{1}+F_{2}+F_{3}$, where every $F_{i}$ is a product of linear homogeneous polynomials $l_{i1},l_{i2}\ldots$

We can assume w.l.og. that $F_{1}$, $F_{2}$ and $F_{3}$ are pairwise coprime. Indeed, if, say, $(F_{1},F_{2})\not=1$ then either $(F_{1},F_{2})|F_{3}$ and we can devide all $F_{i}$ by $(F_{1},F_{2})$, or if $F_{3}$ does not divide $(F_{1},F_{2})$ then $F_{1}+F_{2}+F_{3}$ is not identiacally zero. If $F_{1}+F_{2}+F_{3}=0$ then $F_{3}\in\langle F_{1},F_{2}\rangle$ and hence $F_{3}\in\langle l_{11},l_{21}\rangle$. Can we verify the last belonging effectively? The answer is “yes”. First, note that $\langle l_{11},l_{21}\rangle$ contains $F_{3}=l_{31}\cdot l_{32}\ldots$ iff there exists $i$ such that $l_{3i}\in\langle l_{11},l_{21}\rangle$ because the ideal $\langle l_{11},l_{21}\rangle$ is prime. Now note that we can easily verified whether $l_{3i}\in\langle l_{11},l_{21}\rangle$ for every $i$ in polynomial time.

So, we can verify that $\langle l_{1j},l_{2j}\rangle$ contains $F_{3}$ for every $i$ and $j$. Similar for $F_{1}$ and $F_{2}$. Assume that we have verified all this and does not find contradictions with zero indentity of $F_{1}+F_{2}+F_{3}$. Does this means that this polynomial is zero? No! A conter-example is $F_{1}:=x$, $F_{2}:=y$, $F_{3}:=x+2y$.

However, the following result shows that in this case the dimension spanned by $l_{ij}$ is at most $4$. Hence, it is easy to determine the identity of the circuit by Schwartz-Zippel lemma.

Theorem 2**.**

If $\{S_{i}\}$ is a finite collection of two or more non-empty disjoint finite sets in an affine or in a projective complex space such that $\bigcup S_{i}$ spans a subspace of at least dimension $5$, then there exists a line cutting precisely two of the sets.

In fact the proof of Theorem 2 is closely follows the proof of Edelstein-Kelly theorem in [7]. We just use the following result of Kelly instead of Sylvester-Gallai theorem.

Theorem 3** ([16]).**

If a finite set of $k>2$ points in an affine or in a projective complex space is not a subset of a plane, then there exists a line in that space containing precisely two of the points.

Proof of Theorem 2.

First note that a pencil of lines in an affine or a projective $4$-space, not all in the same $3$-dimensional plane must contain a pair of lines such that the plane defined by these lines contains none of the other lines. This follows at once if we consider a section of the pencil by a $3$-dimensional plane and appeal to Theorem 3 in the $3$-dimensional plane of the section. We call this fact Motzkin’s observation since he observed it for $\mathbb{R}$ in [17].

We now choose a pair of points $p_{1}$ and $p_{2}$ of $\bigcup S_{i}$ where $p_{1}$ and $p_{2}$ are from different $S_{i}$. The points of $\bigcup S_{i}\setminus\{p_{1},p_{2}\}$ define a pencil of $2$-dimensional planes with line $p_{1}p_{2}$ as axis. A section of this pencil by a properly chosen $4$-space defines a pencil of lines in that $4$-space not all in a plane. (Indeed, since points of $\bigcup S_{i}$ do not belongs to any $4$-space there exist points $A,B,C,D$ such that the vectors $p_{2}A$, $p_{2}B$, $p_{2}C$, $p_{2}D$, $p_{2}p_{1}$ are linearly independent. The $4$-space $p_{2}A,p_{2}B,p_{2}C,p_{2}D$ is suitable for us.) By the Motzkin’s observation, two of the lines of this pencil define a $2$-plane free of any of the other lines of the pencil. This plane together with the points $p_{1}$ and $p_{2}$ spans a $3$-space $\Gamma$ such that the points of $\bigcup S_{i}$ in this $3$-space are on precisely two $2$-planes of the original pencil of $2$-planes. Each of these planes contains at least one point of $\bigcup S_{i}\setminus\{p_{1},p_{2}\}$.

Now it is easy to check that if a collection of two or more finite non-empty and disjoint sets in a $3$-dimensional space lie on two planes and not on one, then there is a line intersecting precisely two of the sets. Indeed, denote these planes as $\alpha$ and $\beta$. If there exist two points from $\alpha\bigcup\beta\setminus\{p_{1},p_{2}\}$ from different sets then the lines that connect these points is what we want. Else we consider any line that connects some point from $\alpha\bigcup\beta\setminus\{p_{1},p_{2}\}$ and $p_{1}$ or $p_{2}$. ∎

4 General case $\Sigma\Pi\Sigma\Pi(k,r)$-circuits

Now we will try to use the same idea for general $\Sigma\Pi\Sigma\Pi(k,r)$-circuits. To simplify notation we consider $\Sigma\Pi\Sigma\Pi(3,2)$-circuits. So, we consider the circuits of the form $F_{1}+F_{2}+F_{3}$, where $F_{i}$ is a product of linear or quadratic (irredicuble) homogenuos polynomials $l_{i1}l_{i2}\ldots$. We can assume that $F_{1}$, $F_{2}$ and $F_{3}$ are pairwise coprime by the same reasons as before. Again, we want to verify whether $F_{3}\in\langle l_{11},l_{21}\rangle$. However, it is not as simple as in the previous section. Membership of $F_{3}\in\langle l_{11},l_{21}\rangle$ does not mean that there exists $l_{3i}$ such that $l_{3i}\in\langle l_{11},l_{21}\rangle$. We use the following analogue of this statement.

Theorem 4** ([12]).**

Let $P_{1},\ldots,P_{d},Q_{1},\ldots,Q_{k}\in\mathbb{C}[x_{0},\ldots,x_{n}]$ be homogenous polynomials of degree at most $r$.

Assume that $P_{1}\cdot P_{2}\cdots P_{d-1}\cdot P_{d}\in\langle Q_{1},\ldots,Q_{k}\rangle$.

Then there exist $\{i_{1},\ldots,i_{f}\}\subseteq\{1,\ldots,d\}$, where $f=f(k,r)$ such that the polynomial $P_{i_{1}}\cdots P_{i_{f}}\in I:=\langle Q_{1},\ldots,Q_{k}\rangle$.

The proof of this theorem was given by Hailong Dao at MathOverflow [12]. We present it here for the convenience of the reader.

Proof.

The point is that many invariants of the ideal $I=(Q_{1},\dots,Q_{k})$ can be bounded depending only on $k$ and $r$:

Theorem 5** ([8, Proposition 4.6], [10], [9], [13]).**

There exists a primary decomposition of $I=I_{1}\cap\dots\cap I_{l}$ such that each of the $I_{i}$ is $\mathfrak{p}_{i}$-primary and the number of generators of $I_{i}$ as well as degrees and $l$ itself are bounded by some function of $k$ and $r$. 2. 2.

If $I$ is $\mathfrak{p}$-primary then the minimal $B$ such that $\mathfrak{p}^{B}\subseteq I$ is upper bounded by some function from $k$ and $r$.

By the first item of this theorem the problem reduces to the case when $I$ is $\mathfrak{p}$-primary. By the second item there is $B$ such that $\mathfrak{p}^{B}\subseteq I$, and this number is also bounded by the degrees and number of generators of $I$. Remove all the $P_{i}$ that is not in $\mathfrak{p}$. The product of the rest is still in $I$ because $I$ is a primary ideal. If there are at most $B$ elements remaining, we are done. If not, then choose $B$ of them, the product is in $\mathfrak{p}^{B}\subseteq I$. ∎

By this theorem it is simple to recognize membership of $F_{3}$ in $\langle l_{11},l_{21}\rangle$ by a polynomial time-bounded algorithm that proves the first part of Theorem 1. The second part of this theorem follows from the following result.

Theorem 6** ([5]).**

Let $C$ be an $m$-variate circuit. Let $f_{1},\ldots,f_{m}$ be $l$-sparse, $\delta$-degree, $n$-variate polynomials with trdeg $r$. Suppose we have oracle access to the $n$-variate $d$-degree circuit $C′:=C(f_{1},\ldots,f_{m}).$ There is a blackbox $\text{poly}(size(C′)\cdot dl\delta)^{r}$ time test to check $C′=0$ over $\mathbb{C}$.

Proof of Theorem 1.

We claim that there exists an algorithm verifying that a given $\Sigma\Pi\Sigma\Pi(k,r)$-circuit is SG. Indeed, let $C$ be a circuit of the form (3). We need to verify that $F_{i}$ belongs to ideal $I:=\langle l_{1j_{1}},l_{2j_{2}}\ldots,l_{i-1,j_{i-1}},l_{i+1,j_{i+1}},\ldots,l_{k,j_{k}}\rangle$ for every $i$ and for every $l_{1j_{1}},l_{2j_{2}}\ldots,l_{i-1,j_{i-1}},l_{i+1,j_{i+1}},\ldots,l_{k,j_{k}}$ (note that there are only $\text{poly}(n,d)$ such conditions for constant $k$). By Theorem 4 $F_{i}=l_{i,1}\cdot\ldots\cdot l_{i,d_{i}}$ belongs to $I$ iff there exists $\{i_{1},\ldots,i_{f}\}\subseteq\{1,\ldots,d\}$ such that $l_{i_{1}}\cdots l_{i_{f}}\in I$ where $f=f(k,r)$. So there are only $\text{poly}(n,d)$ such conditions for constant $k$ and $d$. One such condition can be verifiyed in polynomial time (here it is crucial that all these polynomials are homogeneous). Indeed, a homogeneous polynomial $A$ of degree $a$ belongs to $\langle B_{1},\ldots,B_{s}\rangle$, where $B_{i}$ are homogeneous polynomial of degree $b$ if and only if there exists homogeneous polynomials $D_{1},\ldots,D_{s}$ of degree $a-b$ such that $A=B_{1}\cdot D_{1}+\ldots B_{s}\cdot D_{s}$. Hence, to verify that $l_{i_{1}}\cdots l_{i_{f}}\in I$ we need to solve a system of $\text{poly}(n,d)$ linear equations.

Now assume we are given a SG $\Sigma\Pi\Sigma\Pi(k,r)$-circuit (if a circuit is not SG then it is not identically zero). If Conjecture 1 holds then the trdeg of this circuit is constant. Then by Theorem 6 there exists a polynomial (in $n$ and $d$) algorithm solving PIT for this circuit. ∎

5 Proof of Conjecture 1 in a special case

We do not know the correctness of Conjecture 1 even for $\Sigma\Pi\Sigma\Pi(3,2)$ circuits. For this reason we consider a simple subclass of such circuits. Namely, we consider circuits with the following property: all ideals $\langle l_{ik},l_{jt}\rangle$ for different $i$ and $j$ and for quadratic $l_{ik}$, $l_{jt}$ are prime. Also, we need that not all quadratic polynomials $l_{ij}$ have the same index $i$.

Theorem 7**.**

Conjecture 1 holds for such circuits.

Proof.

Denote the set of quadratic polynomials $l_{ij}$ as $Q$. First, we prove that even the dimension of $\text{span}(Q)$ is bounded by a constant. Indeed, a quadratic polynomial $l$ belongs to $\langle l_{1},l_{2}\rangle$ where $l_{1}$, $l_{2}\in Q$ iff $l$ is a linear combination of $l_{1}$ and $l_{2}$. Besides, every such ideal $\langle l_{ij},l_{ks}\rangle$ where $l_{ij},l_{ks}\in Q$ and $i\not=k$ must contain a quadratic polynomial $l_{tu}$ where $t\not=i,k$ since the ideal $\langle l_{ij},l_{ks}\rangle$ is prime. Hence $\dim(\text{span}(Q))$ is at most $2$ by Theorem 2. Here, it is important that $\langle l_{1},l_{2}\rangle$ is prime and not all quadratic polynomials $l_{ij}$ have the same index $i$.

Consider an ideal of the form $\langle l_{1},l_{2}\rangle$ where $l_{1}$ and $l_{2}$ are linear. Recall, that this ideal is prime. Denote by $L$ the set of all $l_{ij}$ such that there exists $l_{kt}$ with $k\not=i$ such that the ideal $\langle l_{ij},l_{kt}\rangle$ contains some quadratic $l_{fu}$ with $f\not=i,k$.

Lemma 1**.**

The dimension of $\text{span}(L)$ is at most $6$.

Proof of Lemma 1.

A quadratic homogeneous polynomial $f$ over $\mathbb{C}$ is irreducible iff $\text{rk}(f)\geqslant 3$. Here, $\text{rk}(f)$ is the rank of $f$ as a quadratic form. Indeed, if $\text{rk}(f)<3$ then it is obvious that $f$ is not irreducible. To prove that in other cases $f$ is irreducible it is enough to show that the polynomial $x^{2}+y^{2}+z^{2}$ is irreducible and this is folklor. So, all elements of $Q$ have rank at least $3$. 2. 2.

Denote by $Q^{\prime}$ the subset of all elements of $q\in Q$ such that there exist $i$, $j$, $k$ and $t$ with $k\not=i$ s. t. $\langle l_{ij},l_{kt}\rangle\in q$. Of course the dimension of $\text{span}(Q^{\prime})$ is at most $2$ as the dimension of $\text{span}(Q)$. 3. 3.

Consider some $l$, $m\in L$ and $q\in Q^{\prime}$ such that $\langle l,m\rangle$ contains $q$. This means that the intersection of quadric $Q^{\prime}$ with line $l$ is a quadric with rank at most $2$. Therefore, $\text{rk}(q)\leqslant 3$. Combining this result with the first item we conclude that $\text{rk}(q)=3$ for every $q\in Q^{\prime}$. 4. 4.

Consider the largest linear independent subset in $L$. Denote this set as $\{l_{1},\ldots,l_{t}\}$ . We will show that $t\leqslant 3\cdot\text{dim}(\text{span}(Q^{\prime}))\leqslant 6$. This give us what we want. 5. 5.

Add new $l^{\prime}_{t+1},\ldots,l^{\prime}_{n}$ such that $\{l_{1},\ldots,l_{t},l^{\prime}_{t+1},\ldots,l^{\prime}_{n}\}$ is a basis of the linear form from $x_{1},\ldots,x_{n}$. Consider the (symmetric) matrices $A_{1},\ldots,A_{s}$ of all quadratics from $Q^{\prime}$ in the dual basis of $\{l_{1},\ldots,l_{t},l^{\prime}_{t+1},\ldots,l^{\prime}_{n}\}$. 6. 6.

The rank of every $A_{i}$ is equal to $3$. Hence, there exist $3\cdot\text{dim}(\text{span}(Q^{\prime}))$ numbers of rows such that other rows are linearly depend from these in every matrix $A_{i}$. The same is true for columns since these matrices are symmetric. 7. 7.

For every $l_{j}$ the exists $q\in Q^{\prime}$ such that $l_{i}\cap q$ is a quadratic form of rank $2$. Hence, for every $i=1,\ldots,t$ there exists a matrix $A_{j}$ such that matrix $A_{j,i}$ obtained from $A$ by deleting the $i$th row and the $i$th column has rank $2$. But from 6, it follows that there are at most $3\cdot\text{dim}(\text{span}(Q^{\prime}))$ such numbers $i$. Therefore $t\leqslant 3\cdot\text{dim}(\text{span}(Q^{\prime}))$.

∎

Add to the set $L$ the polynomials $l_{ij}$ that are linear combinations of $L$. Lemma 1 shows that the dimension of $L$ is not greater than $6$. We need to prove that the dimension of the span of the remaning linear polynomials is also bounded by a constant. Denote the set of such polynomials by $T$. The elements of $T$ have the following property. If $l_{ij}\in T$ and $l_{ts}\in T\cup L$ with $i\not=t$ then there exist $l_{pu}\in T\cup L$ such that $p\not=i,t$ and $l_{pu}$ is a linear combination of $l_{ij}$ and $l_{ts}$. Note that we can not say that if $l_{ij}\in T$ and $l_{ts}\in T$ then there exists $l_{pu}\in T$ that is a linear combination of $l_{ij}$ and $l_{ts}$, so we can not apply Theorem 2 directly. However the idea of the proof of Theorem 2 works.

We claim that dim$(\text{span}(T\cup L))<\text{dim }(\text{span}(L))+4=10$. Together with dim $(\text{span}(Q))\leqslant 2$ this implis that trdeg of all polynomials $l_{ij}$ is less than $12$ (this proves the theorem).

Assume that dim$(\text{span}(T\cup L))\geqslant\text{dim }(\text{span}(L))+4$. Devide $T\cup L$ in three sets $S_{1}$, $S_{2}$ and $S_{3}$ in a natural way (in accordance with indexes $i$ of $l_{ij}$). As in the proof of Theorem 2, take $p_{1}\in S_{1}$ and $p_{2}\in S_{2}$. Again we consider the pencil of $2$-dimensional planes with line $p_{1}p_{2}$ as axis. Since dim$(\text{span}(T\cup L))<\text{dim }(\text{span}(L))+4$, there exist $t_{1},t_{2},t_{3},t_{4}\in T$ such that $p_{2}t_{1}$, $p_{2}t_{2}$, $p_{2}t_{3}$ and $p_{2}t_{4}$ are linear independent and there are no points from $L$ in subspace $p_{2}t_{1}t_{2}t_{3}t_{4}$. As in the proof of Theorem 2 we can conclude that there exist points $T_{1}$, $T_{2}\in T$ such that in the 3-space plane generated by $p_{1}$, $p_{2}$, $T_{1}$ and $T_{2}$ all points from $T\cup L$ belong to two $2$-spaces $p_{1}p_{2}T_{1}$ and $p_{1}p_{2}T_{2}$.

If $T_{1}$ and $T_{2}$ are from different $S_{i}$ then we get a contradict (there are no another points from $L\cup T$ at line $T_{1}T_{2}$). Otherwise, all points in 3-space $p_{1}p_{2}T_{1}T_{2}\setminus\{p,q\}$ belong to one $S_{i}$. Then we get a contradiction considering line $T_{1}p_{1}$ or $T_{1}p_{2}$. ∎

6 Derandomization of PIT for some subclass of $\Sigma\Pi\Sigma\Pi(3,2)$ circuits

In Theorem 7 we have the strange condition that not all quadratic polynomials $l_{ij}$ have the same index $i$. To cover this case we present an algorithm solving PIT for such circuits.

More precisely, we consider $\Sigma\Pi\Sigma\Pi(3,2)$-circuits of the form $F_{1}+F_{2}+F_{3}$, where $F_{1}$ and $F_{2}$ are products of homogenous linear polynomials and $F_{3}$ is a product of homogeneous quadratic and linear polynomials.

Theorem 8**.**

There exists a polynomial-time algorithm solving PIT for such circuits.

Proof.

Let $F_{i}=l_{i1}\cdot\ldots$ for $i=1,2$ (here $l_{ij}$ are linear polynomials ) and $F_{3}=q_{31}\cdot\ldots q_{3s}\cdot l_{31}\cdot\ldots l_{3r}$ (here $q_{3j}$ are irredicable quadratic and $l_{3j}$ are linear polynomials). We assume that $s\geqslant 1$(otherwise we can just use results of Section 3).

We can assume that $l_{11}=x$. Then $F_{1}+F_{2}+F_{3}=0$ implies $F_{2}|_{x=0}+F_{3}|_{x=0}=0$. Since $F_{2}|_{x=0}$ is a product of linear polynomials, $q_{31}|_{x=0}$ must be factorized. Hence (see the proof of Lemma 1), the rank of $q_{31}=3$. The polynomial $q_{31}|_{l_{ij}=0}$ must be factorized for all linear polynomials from $F_{1}$ and $F_{2}$ (otherwise $F_{1}+F_{2}+F_{3}\not=0$). This implies that the dimension of linear forms spanned by polynomials from $F_{1}$ and $F_{2}$ is at most $3$.

In other words $F_{1}$ and $F_{2}$ depends only on $3$ variables (after linear changing or variables). If $F_{3}$ depends another variables, then a given circuit is not identically zero. Otherwise, trdeg of all $l_{ij}$ and $q_{kt}$ is not greater than $3$. Hence, by Theorem 6 there exists a polynomial-time algorithm for such circuits.

∎

Acknowledgments

I would like to thank Hailong Dao for useful discussions and Bruno Bauwens for help in writing this paper.