# The Algebraic Cost of a Boolean Sum

Ian Orzel, Srikanth Srinivasan, Sébastien Tavenas, Amir Yehudayoff

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Ian Orzel, Srikanth Srinivasan, Sébastien Tavenas, Amir Yehudayoff. The Algebraic Cost of a Boolean Sum. 2025. hal-05351607

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# THE ALGEBRAIC COST OF A BOOLEAN SUM

IAN ORZEL, SRIKANTH SRINIVASAN, SÉBASTIEN TAVENAS, AND AMIR YEHUDAYOFF

**ABSTRACT.** The P versus NP problem is about the computational power of an existential $\exists_{w \in \{0,1\}^n}$ quantifier. The VP versus VNP problem is about the power of a boolean sum $\sum_{w \in \{0,1\}^n}$ operation. We study the power of a single boolean sum $\sum_{w \in \{0,1\}}$, and prove that in some cases the cost of eliminating this sum is large. This identifies a fundamental difference between the permanent and the determinant. This investigation also leads to the simplest proof we are aware of that the permanent is VNP-complete.

## 1. INTRODUCTION

The P versus NP problem is about the computational power of an $\exists_{w \in \{0,1\}^n}$ quantifier or equivalently an $\bigvee_{w \in \{0,1\}^n}$ operation. The algebraic variant —the VP versus VNP problem— asks about the computational power of a $\sum_{w \in \{0,1\}^n}$ operation. The algebraic variant seems easier to approach because algebra provides rich structures to work with. In this paper $\mathbb{F}$ is a field of characteristic zero. We use standard terminology from algebraic complexity theory; see [2, 12, 17] and references within for more details and background.

The class VP captures efficient computations of polynomials. The class VNP is defined by adding a boolean sum to the class VP. This brings us to the question that guides this work:

*What is the algebraic cost of boolean summation?*

A fundamental tool in computational complexity theory is reduction between problems [6]. Reductions allow us to compare the complexities of different problems. In the algebraic setting, reductions are projections. The polynomial $f \in \mathbb{F}[\mathcal{X}]$ is a projection of $g \in \mathbb{F}[\mathcal{Y}]$ if there is a map $\phi: \mathcal{Y} \to \mathcal{X} \cup \mathbb{F}$ so that $f = \lambda(g \circ \phi)$, where $\lambda \in \mathbb{F}$. The projection is called *simple* if for each $x \in \mathcal{X}$ the cardinality of $\phi^{-1}(x)$ is at most one.

For boolean summation, we use the following notation. For $S \subseteq \mathcal{X}$, denote by $f_S$ the polynomial in the variables $\mathcal{X} \cup \{y\}$, where $y$ is a new variable, that is obtained from $f$ by the substitution $x = y$ for all $x \in S$. For example, $(x_1+x_2^2+x_3)_{\{x_1,x_2\}} = y+y^2+x_3$. Denote by $\Sigma_S f$ the polynomial

$$ \Sigma_S f = f_S|_{y=0} + f_S|_{y=1}. $$

For example, $\Sigma_{\{x_1,x_2\}}(x_1+x_2^2+x_3) = 2+2x_3$. The variables in $S$ do not appear in $f_S$ and $\Sigma_S f$.

<footnote>Srikanth Srinivasan was funded by the European Research Council (ERC) under grant agreement no. 101125652 (ALBA). Amir Yehudayoff is supported by a DNRF chair grant, and partially supported by BARC. Sébastien Tavenas is supported by ANR-22-CE48-0007.</footnote>
To measure the cost of boolean sums, we need to work with families $\mathcal{F} = \{F_n \in \mathbb{F}[\mathcal{X}_n]\}_{n \in \mathbb{N}}$ of polynomials. The two central polynomial families in algebraic complexity are the permanent and the determinant:

$$ \mathrm{perm}(X) = \mathrm{perm}_n(X) = \sum_{\pi} \prod_i x_{i, \pi(i)} $$

and

$$ \det(X) = \det_n(X) = \sum_{\pi} (-1)^{\mathrm{sign}(\pi)} \prod_i x_{i, \pi(i)}, $$

where $X = (x_{i,j})$ is an $n \times n$ matrix of variables, the sum is over permutations $\pi$ of $[n]$, and the product is over $i \in [n]$. The permanent is VNP-complete, so that $\mathrm{VP} \neq \mathrm{VNP}$ if and only if $\mathrm{perm}$ is not in $\mathrm{VP}$ [14]. It also appears in many places in computer science because it encapsulates many counting problems (see e.g. [15, 13, 1] and references within); by definition, it counts the number of perfect matchings in a bipartite graph. The $\mathrm{VP}$ versus $\mathrm{VNP}$ problem is (essentially) about the relationship between the permanent and the determinant (see [14, 11, 9] and references within). The determinant defines a complexity measure; the determinantal complexity $\mathrm{dc}(f)$ of a polynomial $f$ is the minimum $k$ so that $f$ is a projection of $\det_k$. It is known that $\mathrm{dc}(f) < \infty$ for all $f$. The $\mathrm{VP}$ versus $\mathrm{VNP}$ problem is roughly captured by what is $\mathrm{dc}(\mathrm{perm}_n)$? Proving that $\mathrm{dc}(\mathrm{perm}_n)$ is super-polynomial in $n$ is one of the major open problems in computer science. The best lower bound we know is $\mathrm{dc}(\mathrm{perm}_n) \ge \frac{n^2}{2}$; see [10, 3].

The aim of this paper is to try and characterize the properties of the permanent that make it VNP-complete and distinguish it from the determinant. We start by outlining the classic proof of the VNP-completeness of the permanent [2]. It is known that any polynomial $f_n$ in VNP can be written as

$$ f_n(X) = \sum_{y_1 \in \{0,1\}} \sum_{y_2 \in \{0,1\}} \cdots \sum_{y_m \in \{0,1\}} g_{m'}(X, Y) $$

where $m, m'$ are polynomial in $n$, and $g_{m'}$ can be computed by an arithmetic formula of size $s \le \mathrm{poly}(n)$. The permanent can efficiently simulate formulas [14]; that is, there is an $s+1 \times s+1$ matrix $D'$ whose entries are elements of $X \cup Y \cup \mathbb{F}$ such that $g_{m'}(X, Y) = \mathrm{perm}(D')$. To reduce $f_n$ to the permanent, one just needs to find a polynomial size matrix $E$ such that

$$ \mathrm{perm}(E) = \sum_{y_1 \in \{0,1\}} \sum_{y_2 \in \{0,1\}} \cdots \sum_{y_m \in \{0,1\}} \mathrm{perm}(D'). $$

The VNP-completeness proof is thus really based on a reduction of a boolean sum of permanents to a single permanent.

The standard proof goes from $D'$ to $E$ in “one big step”. To better understand the process and identify what fails for the determinant, we will break down this transformation into smaller steps. Instead of a sum of $m$ boolean sums, let us consider a single boolean sum. In this case, the standard proof constructs for each matrix $\tilde{D}$, a matrix $\tilde{E}$ so that $\mathrm{perm}(\tilde{E}) = \sum_{y \in \{0,1\}} \mathrm{perm}(\tilde{D})$. The size of $\tilde{E}$ is the size of $\tilde{D}$ plus $O$ (the number of occurrences of the variables $y$ in the matrix $\tilde{D}$).

In this sense, it would be “nice” to assume that a $y$ variable does not occur many times in the matrix $\tilde{D}$. Returning to the case of $m$ variables, it is easier to work with a matrix $D$ where each $y_i$ appears once (we call such matrices read-once). We
replace each variable $y_i$ by a set of distinct variables $S_i$, and the goal is to construct $E$ so that

$$ \mathrm{perm}(E) = \sum_{S_1} \sum_{S_2} \cdots \sum_{S_m} \mathrm{perm}(D). $$

We also work with simple reductions, since it allows us to compose them without increasing the number of occurrences of the variables. We can even refine this decomposition (this step is not explicit in the standard proof). Lemma 2 below shows that for a read-once matrix $D_0$ of size $d_0$ and a set $S$ of cardinal $s$, there is a read-once matrix $D_1$ of size $d_1 = d_0 + O(s)$ such that

$$ \sum_{S} \mathrm{perm}(D_0) = \sum_{T_1} \sum_{T_2} \cdots \sum_{T_s} \mathrm{perm}(D_1) $$

where each $T_i$ has size only two. In Observation 1, we remark that we can not hope to get a similar decomposition where sums have size only one. Such results allow us to safely assume that a $y$ variable occurs at most twice in the matrix we sum over.

All the steps we described so far were done for the permanent, but they can also be achieved by the determinant. Consequently, the difference between the permanent and the determinant must boil down to eliminating a boolean sum over a single variable that appears twice. Indeed, in Theorem 3 we show that the cost for the permanent is constant and for the determinant is linear in the size of the matrix (see Sections 4 and 5). If we compose these costs a polynomial number of times, the cost for the permanent stays polynomial and, for the determinant, it becomes exponential.

This approach implies a new proof of the VNP-hardness of the permanent. Even though it continues to be based on the construction of so-called “gadgets”, the proof is more modular and exposes a fundamental difference between permanent and determinant. Our proof uses one gadget for decomposing the full summation to summations of size two (going from $D_0$ to $D_1$ above). One can compare this with the “iff-coupling” gadget in [2]. Then, we need another gadget to simulate a sum of size two. In the original proof, the “rosette” gadget takes care of the sum. The “rosette” gadget should properly interact with the “iff-coupling” gadget, making the overall description more complicated.

We are ready for a central definition. The algebraic additive cost of boolean summation over a family $\mathcal{F}$ is measured via the map $\alpha_{\mathcal{F}} : \mathbb{N} \to \mathbb{N} \cup \{\infty\}$, defined as follows: for each $s \in \mathbb{N}$, let $\alpha_{\mathcal{F}}(s)$ be the minimum integer$^1 k$ so that for every $n$ and $S \subset \mathcal{X}_n$ of size $|S| \le s$, the polynomial $\sum_S F_n$ is a simple projection of $F_{n+k}$. In other words, $\alpha_{\mathcal{F}}(s)$ is the additive $\mathcal{F}$-cost of performing a boolean sum over $s$ variables.

As explained before, both the permanent and the determinant are known to simulate formulas; the proofs in both cases are basically identical.

**Theorem ([14]).** *Both perm and det simulate formulas.*

The key difference between them must, therefore, reside in the cost of boolean summation. We first observe that for singletons, a boolean summation is “free”.

**Observation 1.** $\alpha_{\mathrm{perm}}(1) = \alpha_{\mathrm{det}}(1) = 0$.

<footnote>$^1$If there is no such integer, then it is infinite.</footnote>
Next, we observe that it suffices to sum over “variables that appear twice”. The following lemma is an algebraic analog of the well-known fact that MAX 2-SAT is NP-complete.

**Lemma 2.** For $\mathcal{F} \in \{\text{perm}, \text{det}\}$, if $\alpha_{\mathcal{F}}(2) < \infty$ then $\mathcal{F}$ is VNP-complete.

The main result of our work, however, is the following.

**Theorem 3.** $\alpha_{\text{perm}}(2) \le 3$ and $\alpha_{\text{det}}(2) = \infty$.

The theorem highlights a key difference between perm and det. It shows, in a strong sense, that the known proofs that perm is VNP-complete fail for det. In addition, it leads to the simplest proof of the VNP-completeness of the permanent that we know of. The standard proofs that perm supports efficient boolean summation construct various “gadgets”, which are directed graphs with certain properties. The above theorem shows that there are no such constructions for det.

The proof of the theorem is based on the following bound

$$dc(\Sigma_{\{x_{1,1}, x_{2,2}\}} \det(X)) \ge 2dc(\det(X)) - 2).$$

It shows that a single boolean summation doubles the computational complexity. We are led to the following direct-sum-type problem: *how is the cost of* $\Sigma_{S_1} \Sigma_{S_2} \dots \Sigma_{S_m}$ *related to the costs of each* $\Sigma_{S_i}$?

As a final remark, we mention that a similar investigation can be performed in the boolean setting, where instead of $\Sigma$ we have $\exists$.

**1.1. Related Work.** At a high level, our result shows a difference between the computational power of the permanent polynomial and the determinant polynomial. Other works in a similar spirit have been conducted, all of which have the ultimate goal of helping us understand why VP and VNP are likely different from each other.

The results of von zur Gathen [16] and Mignon and Ressayre [10] (see also [3]) show a difference in the “geometry of the underlying varieties” of the two polynomials. This difference implies a quadratic lower bound on the determinantal complexity of the permanent (we shall also use techniques from [10]).

Recent work of Hruběs and Joglekar [5] points to a distinction in reductions to read-once determinants and permanents. This work also gives a lower bound on the cost of certain reductions which are incomparable to ours.

Our work is different from the two lines of works above in that we seek to justify why one might expect an exponential blow-up is required in reductions from the permanent to the determinant.

The work of Landsberg and Ressayre [7] gives an exponential lower bound in a restricted setting. They showed that any reductions from the permanent to the determinant that preserve the underlying symmetries must have exponential cost. Our results imply a similar bound within the general framework of Valiant’s reduction, but without assuming symmetry. Our bounds are restricted in a different way; they hold just for a particular type of reductions that are built by “iteratively applying a single building block”.
## 2. SINGLE VARIABLE SUMMATION

The explanation of Observation 1 is that

$$ \Sigma_{\{x_{1,1}\}} \det(X) $$

$$ = \det \begin{bmatrix} 0 & x_{1,2} & x_{1,3} & \dots & x_{1,n} \\ x_{2,1} & x_{2,2} & x_{2,3} & \dots & x_{2,n} \\ x_{3,1} & x_{3,2} & x_{3,3} & \dots & x_{3,n} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ x_{n,1} & x_{n,2} & x_{n,3} & \dots & x_{n,n} \end{bmatrix} + \det \begin{bmatrix} 1 & x_{1,2} & x_{1,3} & \dots & x_{1,n} \\ x_{2,1} & x_{2,2} & x_{2,3} & \dots & x_{2,n} \\ x_{3,1} & x_{3,2} & x_{3,3} & \dots & x_{3,n} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ x_{n,1} & x_{n,2} & x_{n,3} & \dots & x_{n,n} \end{bmatrix} $$

$$ = 2\det \begin{bmatrix} 1/2 & x_{1,2} & x_{1,3} & \dots & x_{1,n} \\ x_{2,1} & x_{2,2} & x_{2,3} & \dots & x_{2,n} \\ x_{3,1} & x_{3,2} & x_{3,3} & \dots & x_{3,n} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ x_{n,1} & x_{n,2} & x_{n,3} & \dots & x_{n,n} \end{bmatrix}. $$

A similar calculation holds for the permanent.

## 3. TWO VARIABLES SUFFICE FOR VNP-COMPLETNESS

The aim of this section is to prove Lemma 2 that shows that summation over two variables suffices for VNP-completeness. We prove it for the determinant (the proof for the permanent is similar). The following construction shows how to replace a boolean summation over $m$ variables by $m$ boolean summations over two variables.

**Observation 4.** Let $f \in \mathbb{F}[X]$ and $S = \{x_1, \dots, x_m\} \subseteq X$. Let $y_1, \dots, y_m$ be $m$ new variables, and define

$$ g(y) = \left( \prod_{i \in [m]} y_i \right) + \left( \prod_{i \in [m]} (1 - y_i) \right). $$

Then

$$ \Sigma_{\{x_1, y_1\}} \Sigma_{\{x_2, y_2\}} \dots \Sigma_{\{x_m, y_m\}} g f = \Sigma_S f. $$

Indeed, on the l.h.s. there is a sum over $2^m$ terms, but $g$ picks exactly two of them that perfectly agree with the r.h.s.

The observation allows to replace a boolean sum over many variables by several sums, each over just two.

**Observation 5.** Let $g$ be the polynomial defined above. Then, there is a $2m \times 2m$ matrix $A$ with entries in $\{y_1, \dots, y_m, 0, 1, -1\}$ so that each $y_i$ appears once in $A$ and

$$ g = \det(A). $$

The matrix $A$ is the $m$-variate version of the following matrix:

$$ A = \begin{bmatrix} -1 & 1 & -1 \\ -1 & y_1 & \\ & -1 & 1 \\ & -1 & y_2 \\ & & -1 & 1 \\ & & -1 & y_3 \\ & & & -1 & 1 \\ & & & -1 & y_4 \end{bmatrix}. $$
<figure><img src="image_2.png" /><figcaption>FIGURE 1. The graph $G_A$</figcaption></figure>

To see that $\det(A)$ is the polynomial $g$, it is easier to see the determinant as a signed sum of the weights of the cycles covers of the graph $G_A$ whose $A$ is the adjacency matrix (see Figure 1)

$$ \det(A) = \sum_{C \text{ cycles cover}} \left[ \prod_{\sigma \text{ cycle of } C} (-1)^{1+\text{length of } C} \text{weight}(C) \right]. $$

The reader could find more details about this characterization by cycles covers in [8]. The only cycle which contains a dashed edge is exactly the union of the dashed edges. So the only cover that contains this cycle also contains the $m$ $y_i$-loops. It corresponds to one cover of signed weight $((-1)^{m+1}(-1)^{m-1}) \prod_{i=1}^{m} y_i = \prod_{i=1}^{m} y_i$. Otherwise, the covers contain no dashed edges and are given by $m$ disjoint graphs. The signed weight is the product of them: $\prod_i((-1)(-1) + (1)(-y_i)) = \prod_i(1 - y_i)$. The sum of the covers gives exactly the polynomial $g$.

**Remark.** For the permanent, every $-1$ above the diagonal of the matrix $A$ should be changed into a $1$, and the $2 \times 2$ blocks into $\begin{bmatrix} -1 & 1 \\ 1 & y_1 \end{bmatrix}$.

*Proof of Lemma 2.* Assume that $\alpha_{\det}(2) = C < \infty$. Because $\det$ simulates formulas, we can write every $n$-variate $f$ in VNP (see [2, 12])

$$ f = \sum_{S_1} \dots \sum_{S_{\ell}} \det(D) $$

where $D$ is a $d \times d$ matrix with $d \le \text{poly}(n)$ so that each entry in $D$ is a variable or a field element, $S_1, \dots, S_\ell$ are pairwise disjoint, and each variable in $S_i$ appears at most once in $D$. Let $g, A$ be as above for $m = |S_\ell|$. Consider the $b_\ell \times b_\ell$ matrix with $b_\ell = d + 2|S_\ell|$:

$$ B_{\ell} = \begin{bmatrix} A & \\ & D \end{bmatrix}. $$

By the above, $g \cdot \det(D) = \det(B_\ell)$ and

$$ \sum_{S_\ell} \det(D) = \sum_{T_1} \sum_{T_2} \dots \sum_{T_{|S_\ell|}} \det(B_\ell) $$

where the size of each $T_i$ is two (and each variable not in $S_\ell$ appears once in $B_\ell$). Because $C < \infty$,

$$ \sum_{S_\ell} \det(D) = \det(D_\ell) $$
where $D_\ell$ is a $d_\ell \times d_\ell$ matrix with $d_\ell \le d+(C+2) \cdot |S_\ell|$. Continuing in this manner, we can eliminate all of $\Sigma_{S_1} \dots \Sigma_{S_\ell}$. Here we used the fact that we are dealing with simple projections and that $\sum_i |S_i| \le d^2$. We conclude that $\mathrm{dc}(f) \le d+(C+2)d^2 \le \mathrm{poly}(n)$ so that $\mathrm{det}$ is VNP-complete. $\square$

### 4. THE PERMANENT IS VNP-COMPLETE

We now describe a proof of the well-known fact that $\mathrm{perm}$ is VNP-complete. By Lemma 2, the VNP-completeness of $\mathrm{perm}$ reduces to the following construction.<fnref n="2" />

**Claim.** If $X = (x_{i,j})$ is an $n \times n$ matrix of variables then

$$ \Sigma_{\{x_{1,1}, x_{2,2}\}} \mathrm{perm}(X) = \mathrm{perm}(E) $$

where

$$ E = \left[ \begin{array}{ccc|cccc} 1 & 1 & 1 & & & & \\ 1 & 0 & -1 & & & & \\ -\frac{1}{2} & \frac{1}{2} & \frac{3}{2} & 1 & & & \\ \cline{2-7} & & & & 1 & & \\ & 1 & & 0 & x_{1,2} & x_{1,3} & \cdots \\ & & 1 & x_{2,1} & 0 & x_{2,3} & \\ & & & x_{3,1} & x_{3,2} & x_{3,3} & \\ & & & & \vdots & & \vdots \\ & & & & & & \end{array} \right] $$

*Proof.* To see this, denote by $X[b]$ the matrix $X$ after the substitution $x_{1,1} = x_{2,2} = b$ for $b \in \{0, 1\}$. For $S \subseteq \{1, 2\}$, denote by $X[b]_{-S}$ the matrix $X[b]$ after deleting the rows and columns in $S$. We have

$$ \begin{aligned} \mathrm{perm}(X[1]) &= \mathrm{perm} \left( \left[ \begin{array}{ccc} 1 & x_{1,2} & x_{1,3} \\ x_{2,1} & 1 & x_{2,3} \\ x_{3,1} & x_{3,2} & x_{3,3} \end{array} \right] \right) \\ &= \mathrm{perm}(X[0]) + \mathrm{perm}(X[0]_{-\{1\}}) + \mathrm{perm}(X[0]_{-\{2\}}) + \mathrm{perm}(X[0]_{-\{1,2\}}). \end{aligned} $$

Now, consider a permutation $\pi$ such that the associated value of the permutation on $E$ is nonzero. We can then notice that such a permutation falls into one of four categories:

* $\pi(\{1, 2, 3\}) = \{1, 2, 3\}$,

* $\pi(2) = 4$ and $\pi(\{1, 3\}) = \{1, 3\}$,

* $\pi(3) = 5$ and $\pi(\{1, 2\}) = \{1, 2\}$, or

* $\pi(2) = 4$, $\pi(3) = 5$, and $\pi(1) = 1$.

Let $U$ be the $3 \times 3$ upper-left block

$$ U = \left[ \begin{array}{ccc} 1 & 1 & 1 \\ 1 & 0 & -1 \\ -\frac{1}{2} & \frac{1}{2} & \frac{3}{2} \end{array} \right]. $$

<footnote n="2">²It suffices to consider two variables that are not in the same row or column by the construction of the matrices $B_\ell$ in the previous subsection.</footnote>
For $S, T \subseteq \{1, 2, 3\}$ with $|S| = |T|$, we denote by $U_S^T$ the sub-matrix of $U$ obtained by keeping the rows of $S$ and the columns of $T$. The matrix $U$ is chosen such that

$$ \mathrm{perm}(U_{\{1,2\}}^{\{1,3\}}) = \mathrm{perm}(U_{\{1,3\}}^{\{1,2\}}) = 0, $$

$$ \mathrm{perm}(U_{\{1\}}^{\{1\}}) = \mathrm{perm}(U_{\{1,2\}}^{\{1,2\}}) = \mathrm{perm}(U_{\{1,3\}}^{\{1,3\}}) = 1, $$

$$ \mathrm{perm}(U) = 2. $$

In the first of the above 4 cases, it must be that $\pi(\{4, 5, \dots, m\}) = \{4, 5, \dots, m\}$. Therefore, the sum of the weights of these permutations corresponds to

$$ \mathrm{perm}\left(\left[\begin{array}{c|c} U & \\ \hline & X[0] \end{array}\right]\right) = 2\mathrm{perm}(X[0]). $$

In the second case, it must be that $\pi(4) = 2$ and $\pi(\{5, 6, \dots, m\}) = \{5, 6, \dots, m\}$. Thus, this case corresponds to

$$ \mathrm{perm}\left(\left[\begin{array}{c|c} U_{\{1,3\}}^{\{1,3\}} & \\ \hline & X[0]_{-\{1\}} \end{array}\right]\right) = \mathrm{perm}(X[0]_{-\{1\}}). $$

In the third case, we have that $\pi(5) = 3$ and $\pi(\{4, 6, 7, \dots, m\}) = \{4, 6, 7, \dots, m\}$. Thus, this case corresponds to

$$ \mathrm{perm}\left(\left[\begin{array}{c|c} U_{\{1,2\}}^{\{1,2\}} & \\ \hline & X[0]_{-\{2\}} \end{array}\right]\right) = \mathrm{perm}(X[0]_{-\{2\}}). $$

Finally, in the fourth case, it must be that $\pi(4), \pi(5) \in \{2, 3\}$ and $\pi(\{6, 7, \dots, m\}) = \{6, 7, \dots, m\}$. This case then corresponds to

$$ \mathrm{perm}\left(\left[\begin{array}{c|c} U_{\{1\}}^{\{1\}} & \\ \hline & X[0]_{-\{1,2\}} \end{array}\right]\right) = \mathrm{perm}(X[0]_{-\{1,2\}}). $$

Summing it together,

$$ \mathrm{perm}(E) = \mathrm{perm}(X[0]) + \mathrm{perm}(X[1]). \quad \square $$

### 5. COST OF A BOOLEAN SUMMATION FOR THE DETERMINANT

We now prove that $\alpha_{\det}(2) = \infty$. We shall in fact prove the following:

$$ \mathrm{dc}(\Sigma_{\{x_{1,1}, x_{2,2}\}} \det_n) \geq 2(n-2). $$

The proof relies on a mechanism developed by Mignon and Ressayre to lower bound the determinantal complexity [10]. For a polynomial $f \in \mathbb{F}[\mathcal{X}]$, denote by $H_f$ the $|\mathcal{X}| \times |\mathcal{X}|$ Hessian matrix of $f$ defined by

$$ (H_f)_{x,x'} = \frac{\partial^2}{\partial x \partial x'} f. $$

**Lemma 6** ([10]; see also Lemma 13.3 in [4]). Let $f$ be a polynomial in the variables $x = (x_1, \dots, x_n)$ and let $a \in \mathbb{F}^n$ be so that $f(a) = 0$ then

$$ \mathrm{dc}(f) \geq \frac{\mathrm{rank}(H_f|_{x=a})}{2}. $$
The theorem follows by considering the polynomial

$$f = \sum_{\{x_{1,1}, x_{2,2}\}} \det_n$$

and the $n \times n$ matrix

$$A = \begin{bmatrix} 1 & \frac{1}{2} \\ & I \end{bmatrix},$$

where $I$ is the $(n-2) \times (n-2)$ identity matrix. In the two claims below (which complete the proof), let $H = H_f$.

**Claim 7.** $f(A) = 0$.

*Proof of Claim 7.*

$$ \begin{align*} f(A) &= \det \left( \begin{bmatrix} 1 & \frac{1}{2} \\ & I \end{bmatrix} \right) + \det \left( \begin{bmatrix} 1 & \frac{1}{2} \\ & 1 \end{bmatrix} \right) \\ &= -\frac{1}{2} + 1 - \frac{1}{2} = 0. \tag*{\hspace{2em} \square} \end{align*} $$

**Claim 8.** $\mathrm{rank}(H|_{X=A}) \ge 4(n-2)$.

*Proof of Claim 8.* Focus on the $4(n-2)$ variables $V_i = \{x_{1,i}, x_{2,i}, x_{i,1}, x_{i,2}\}$ for $i > 2$. Furthermore, we consider $G$ the $4(n-2) \times 4(n-2)$ sub-matrix of $H$ whose rows and columns are labeled by variables in $\bigcup_{i>2} V_i$. Let $f_0 = \det_n|_{x_{1,1}=x_{2,2}=0}$ and $f_1 = \det_n|_{x_{1,1}=x_{2,2}=1}$ so that $f = f_0 + f_1$.

An entry in the Hessian of $\det$ is plus/minus the determinant of the $(n-2)$-minor obtained by deleting the corresponding rows and columns from $X$ (or zero if the two variables share a row or a column).

For convenience, denote by $H_{k\ell m p}$ the $(x_{k,\ell}, x_{m,p})$-entry in $H$ (and in $G$).

We first claim that after substituting $A$, the sub-matrix $G$ of $H$ has $n-2$ blocks, each of size $4 \times 4$:

$$ \begin{bmatrix} G_3 & & & & \\ & G_4 & & & \\ & & \ddots & & \\ & & & G_n & \end{bmatrix} $$

where for $i > 2$

$$ G_i = \begin{bmatrix} H_{1i1i} & H_{1i2i} & H_{1ii1} & H_{1ii2} \\ H_{2i1i} & H_{2i2i} & H_{2ii1} & H_{2ii2} \\ H_{i11i} & H_{i12i} & H_{i1i1} & H_{i1i2} \\ H_{i21i} & H_{i22i} & H_{i2i1} & H_{i2i2} \end{bmatrix}. $$

Indeed, for $i \ne j$, the matrix obtained from $A$ by deleting row $i$ and column $j$ has a zero row so its determinant is zero. It follows for $k, \ell \in \{1, 2\}$ that

$$(H_f)_{ik\ell j} |_{X=A} = (H_{f_0})_{ik\ell j} |_{X=A} + (H_{f_1})_{ik\ell j} |_{X=A} = 0 + 0 = 0.$$

A similar argument holds for the other entries $H_{ikj\ell}$, $H_{ki\ell j}$, and $H_{kij\ell}$.

Now fix $i$ and focus on $G_i$. Entries of the form $1i1i$ and $1i2i$ are also zero. So, we are left with

$$ G_i = \begin{bmatrix} & & & H_{1ii1} & H_{1ii2} \\ H_{i11i} & H_{i12i} & & H_{2ii1} & H_{2ii2} \\ & & & H_{i21i} & H_{i22i} & \end{bmatrix}. $$
The matrix $G_i$ is symmetric so it is enough to understand

$$ \begin{bmatrix} H_{1ii1} & H_{1ii2} \\ H_{2ii1} & H_{2ii2} \end{bmatrix}. $$

This matrix is a sum of two matrices (one from $f_0$ and one from $f_1$). For $f_0$, this matrix is

$$ \begin{bmatrix} 1 \\ \frac{1}{2} \end{bmatrix}. $$

For $f_1$, this matrix is

$$ \begin{bmatrix} -1 & 1 \\ \frac{1}{2} & -1 \end{bmatrix}. $$

The sum of the two matrices is

$$ \begin{bmatrix} -1 & 2 \\ 1 & -1 \end{bmatrix}, $$

which always has rank two. Rolling back

$$ \text{rank} \left( \begin{bmatrix} G_3 & & & \\ & G_4 & & \\ & & \ddots & \\ & & & G_n \end{bmatrix} \right) = 4(n-2). \quad \square $$

REFERENCES

1. Scott Aaronson, A linear-optical proof that the permanent is #P-hard, Proceedings of the Royal Society A: Mathematical, Physical and Engineering Sciences 467 (2011), no. 2136, 3393–3405.

2. Peter Bürgisser, Completeness and reduction in algebraic complexity theory, vol. 7, Springer Science & Business Media, 2000.

3. Jin-Yi Cai, Xi Chen, and Dong Li, Quadratic lower bound for permanent vs. determinant in any characteristic, computational complexity 19 (2010), no. 1, 37–56.

4. Xi Chen, Neeraj Kayal, and Avi Wigderson, Partial derivatives in arithmetic complexity and beyond, Foundations and Trends in Theoretical Computer Science 6 (2011), no. 1–2, 1–138.

5. Pavel Hrubes and Pushkar S. Joglekar, On read-k projections of the determinant, Electron. Colloquium Comput. Complex. TR24-125 (2024).

6. Richard M. Karp, On the computational complexity of combinatorial problems, Networks 5 (1975), no. 1, 45–68.

7. Joseph M. Landsberg and Nicolas Ressayre, Permanent v. determinant: An exponential lower bound assuming symmetry, Proceedings of the 2016 ACM Conference on Innovations in Theoretical Computer Science, Cambridge, MA, USA, January 14-16, 2016 (Madhu Sudan, ed.), ACM, 2016, pp. 29–35.

8. Meena Mahajan and V Vinay, Determinant: Old algorithms, new insights, SIAM journal on Discrete Mathematics 12 (1999), no. 4, 474–490.

9. Guillaume Malod and Natacha Portier, Characterizing Valiant's algebraic complexity classes, Journal of complexity 24 (2008), no. 1, 16–38.

10. Thierry Mignon and Nicolas Ressayre, A quadratic bound for the determinant and permanent problem, International Mathematics Research Notices (2004), no. 79, 4241–4253.

11. Ketan D Mulmuley, On P vs. NP and geometric complexity theory, Journal of the ACM 58 (2011), no. 2, 1–26.

12. Amir Shpilka and Amir Yehudayoff, Arithmetic circuits: A survey of recent results and open questions, Foundations and Trends in Theoretical Computer Science 5 (2010), no. 3–4, 207–388.

13. Seinosuke Toda, PP is as hard as the polynomial-time hierarchy, SIAM Journal on Computing 20 (1991), no. 5, 865–877.

14. Leslie G. Valiant, Completeness classes in algebra, STOC, 1979, pp. 249–261.
15. ________, *The complexity of computing the permanent*, Theoretical computer science **8** (1979), no. 2, 189-201.

16. Joachim von zur Gathen, *Permanent and determinant*, 27th Annual Symposium on Foundations of Computer Science, Toronto, Canada, 27-29 October 1986, IEEE Computer Society, 1986, pp. 398-401.

17. Avi Wigderson, *P, NP and mathematics—a computational complexity perspective*, Proceedings of the ICM, vol. 6, 2006, pp. 665–712.

DEPARTMENT OF COMPUTER SCIENCE, THE UNIVERSITY OF COPENHAGEN

DEPARTMENT OF COMPUTER SCIENCE, THE UNIVERSITY OF COPENHAGEN

CHARGÉ DE RECHERCHE CNRS AU LABORATOIRE LAMA DE L'UNIVERSITÉ SAVOIE MONT BLANC

DEPARTMENT OF COMPUTER SCIENCE, THE UNIVERSITY OF COPENHAGEN, AND DEPARTMENT OF MATHEMATICS, TECHNION-IIT