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A symmetric determinantal lower bound for diagonal power sums via polar degree

This paper establishes a new lower bound for the symmetric determinantal complexity of diagonal power sums over complex numbers by proving that the symmetric determinantal complexity of $\sum_{i=1}^n x_i^n$ is at least $(1/(2e) - o(1))n^2$ through a self-contained geometric analysis of polar degrees and symmetric matrix kernels.

Original authors: Karthik Sheshadri

Published 2026-06-10

📖 6 min read🧠 Deep dive

CC BY 4.0

Original authors: Karthik Sheshadri

Original paper licensed under CC BY 4.0 (http://creativecommons.org/licenses/by/4.0/). ✨ This is an AI-generated explanation of the paper below. It is not written or endorsed by the authors. For technical accuracy, refer to the original paper. Read full disclaimer

The Big Picture: Building a House with a Specific Blueprint

Imagine you have a complex mathematical shape (a polynomial) and you want to build it using a specific type of machine: a determinant machine.

In math, a "determinant" is a special calculation you do with a grid of numbers (a matrix). If you can arrange your numbers in a grid so that when you run the determinant calculation, it spits out your original shape, you have found a "determinantal representation."

This paper asks a very specific question: How big does this grid need to be?

The Grid: Think of the grid as a square frame. The size of the frame is $m \times m$ .
The Constraint: The paper insists the grid must be symmetric. This means the grid looks the same if you fold it along the diagonal (like a butterfly). If you swap the top-left number with the bottom-right, it's the same.
The Goal: The author wants to find the smallest possible frame size ( $m$ ) needed to build a specific type of shape called a "diagonal power sum" (basically, adding up numbers raised to a power, like $x_1^n + x_2^n + \dots$ ).

The Main Discovery: A Tighter Squeeze

The author proves that for these specific shapes, the grid cannot be small. It has to be quite large.

The Old Way: Previous research looked at grids that didn't have to be symmetric. They found the grid had to be at least a certain size.
The New Way: This paper looks at the symmetric grids. Because the symmetry forces the "left" and "right" sides of the machine to be identical, it actually makes the machine more efficient at hiding complexity, but the author proves that this efficiency has a limit.
The Result: The author shows that the minimum size of the grid grows roughly as the square of the number of variables ( $n^2$ ). Specifically, the size is at least about $\frac{1}{2e}$ times $n^2$ (where $e$ is a famous math constant, roughly 2.718).

In simple terms: If you try to build a complex shape using a symmetric grid, you can't cheat by making the grid tiny. The grid must be big, and the author calculated exactly how big it has to be.

The Detective Work: How They Proved It

To prove the grid must be big, the author acts like a detective looking for "loopholes" where the grid could be smaller than expected.

The "Kernel" (The Hidden Key):
Imagine the grid has a "secret key" (a vector) that makes the whole thing collapse to zero. In a non-symmetric grid, there are two different keys (a left key and a right key). In a symmetric grid, there is only one key.
- Analogy: Think of a door. A non-symmetric door needs two people to push it open from different sides. A symmetric door only needs one person pushing from the middle. The author uses this "single person" fact to tighten the math.
The "Polar Degree" (Counting the Intersections):
The author uses a geometric trick called "polar degree." Imagine shining a light on your shape and counting how many shadows it casts on a wall.
- The author sets up a scenario where they count how many times the "secret key" intersects with the shape's surface.
- They use a rule called Bezout's Theorem (a fancy way of saying "counting intersections"). It's like saying: "If I have 3 lines crossing a circle, I can count exactly how many times they touch."
The "Local Normal Form" (The Microscope):
The hardest part of the proof was checking if the "single key" (the symmetric kernel) behaves nicely.
- The Fear: What if the key gets "stuck" or creates a weird, messy blob instead of a clean point?
- The Fix: The author used a mathematical microscope (a "Schur complement") to zoom in on the grid. They proved that even though the grid is symmetric, the key behaves perfectly cleanly. It doesn't create messy blobs; it stays a single, sharp point. This allowed them to trust their "intersection count."

The "Magic Number" ( $1/2e$ )

The paper concludes with a specific number: $1/(2e)$ .

In the non-symmetric version of this problem, the number was $1/(4e)$ .
Because the symmetric version is more constrained (the left and right sides are glued together), the "intersection count" is twice as efficient.
This doubles the lower bound, changing the constant from $1/4$ to $1/2$ .

Why This Matters (According to the Paper)

It's a "Lower Bound": The paper doesn't say "You can build it with a grid of size $X$ ." It says "You cannot build it with a grid smaller than $X$ ."
It's Exact: The proof is for exact calculations, not approximations.
It's Not a "Border" Case: The paper admits it doesn't prove anything about grids that are almost symmetric or grids that get bigger as you get closer to a limit. It's strictly about perfect, exact symmetric grids.
It's Not for All Math Worlds: The proof only works in "characteristic zero" (basically, the standard world of real and complex numbers we use in school). It breaks down in other weird mathematical universes (like those using modular arithmetic).

Summary Analogy

Imagine you are trying to pack a very specific, bulky suitcase (the polynomial) into a box (the matrix).

The Rule: The box must be perfectly symmetrical (the left side mirrors the right).
The Question: How small can the box be?
The Answer: The author proves that no matter how cleverly you fold the clothes, the box must be at least a certain size. If you try to use a smaller box, the suitcase simply won't fit, and the math proves it impossible. The author calculated that the minimum size is roughly proportional to the square of the number of items you are packing.

The paper is a rigorous mathematical proof that symmetry, while helpful in some ways, actually forces the "box" to be larger than previously thought for these specific types of shapes.

Technical Summary: A Symmetric Determinantal Lower Bound for Diagonal Power Sums via Polar Degree

Problem Statement
The paper addresses the problem of determining the symmetric determinantal complexity, denoted $sdc(f)$, of a polynomial $f$ . This is defined as the smallest integer $m$ such that $f$ can be expressed as the determinant of an $m \times m$ symmetric matrix $M$ whose entries are affine-linear forms. While lower bounds for the standard (non-symmetric) determinantal complexity $dc(f)$ are a central topic in algebraic complexity theory (specifically Valiant's program), the symmetric case presents distinct geometric constraints. The paper focuses on diagonal power sums, specifically $F_n = \sum_{i=1}^n x_i^n$ , aiming to establish a superlinear lower bound for $sdc(F_n)$ over the complex numbers $\mathbb{C}$ .

Methodology
The proof strategy parallels the author's previous non-symmetric work but adapts the geometric machinery to the symmetric setting. The core methodology relies on polar degree and multihomogeneous Bezout theory.

Geometric Setup: The authors consider a smooth hypersurface $X = V(f) \subset \mathbb{P}^{N-1}$ of degree $d$ . If $f$ admits a symmetric determinantal representation of size $m$ , the paper analyzes the incidence scheme between the hypersurface and the kernel of the representing matrix $M(z, x) = zA_0 + \sum x_i A_i$ .
Symmetric Kernel Analysis: Unlike the non-symmetric case, which involves distinct left and right kernel lines, the symmetric model has a single kernel line at rank $m-1$ . The paper utilizes a symmetric Schur-complement normal form to analyze the local structure of the incidence scheme at a "genuine polar point" (a point where the matrix has rank $m-1$ ).
Local Isolatedness: A critical technical hurdle is ensuring that the lifted polar equations do not create positive-dimensional components (spurious solutions) near genuine points. The author proves that, scheme-theoretically, the kernel incidence is a graph over the ordinary polar slice. Specifically, the quadratic conormal forms $u^T A_i u$ are shown to be unit multiples of the ordinary partial derivatives $\partial_i f$ on this local graph. This ensures that the lifted solutions are zero-dimensional and isolated.
Multihomogeneous Counting: The proof applies multihomogeneous Bezout's theorem on the product space $\mathbb{P}^N_{[z:x]} \times \mathbb{P}^{m-1}_{[u]}$ $P_{[z : x]}^{N} \times P_{[u]}^{m - 1}$ . The classes of the defining equations are determined:
- The $m$ kernel equations $M(z,x)u=0$ have class $H+U$ .
- The $N-2$ polar equations (derived from the gradient) have class $2U$ (due to the quadratic nature of $u^T A_i u$ ).
- The slicing hyperplane has class $H$ .
  The intersection number is computed as $[H^N U^{m-1}] H(H+U)^m (2U)^{N-2}$ .

Key Results
The paper establishes the following main theorem and corollaries:

Theorem 1 (Symmetric Polar-Degree Bound): For a smooth degree- $d$ hypersurface $X = V(f) \subset \mathbb{P}^{N-1}$ ( $N \ge 3$ ) with a symmetric determinantal representation of size $m$ , the topological polar degree $\delta_{top}(X) = d(d-1)^{N-2}$ satisfies:
$d(d-1)^{N-2} \le 2^{N-2} \binom{m}{N-1}$
Corollary 1 (Diagonal Power Sums): For the polynomial $F_n = \sum_{i=1}^n x_i^n$ , the symmetric determinantal complexity satisfies:
$sdc(F_n) \ge \left( \frac{1}{2e} - o(1) \right) n^2$
This improves upon the constant derived from non-symmetric bounds (which yield $1/4e$ ) by a factor of 2, specifically due to the reduction in kernel variables in the symmetric model.
Corollary 2 (General Diagonal Sums): For $F_{N,d} = \sum_{i=1}^N x_i^d$ , as $N \to \infty$ uniformly for $d \ge 2$ :
$sdc(F_{N,d}) \ge \left( \frac{1}{2e} - o_N(1) \right) N(d-1)$
Non-Vacuity: The paper provides an explicit symmetric determinantal representation for $F_{N,d}$ of size $2N(d+1)+1$ , demonstrating that the lower bounds are non-vacuous and tight up to a constant factor.

Scope and Limitations
The results are strictly valid over fields of characteristic zero (specifically $\mathbb{C}$ ). The paper explicitly notes that the theorem fails if stated uniformly over all fields of characteristic $p > 2$ . In positive characteristic, Frobenius identities can reduce the complexity of diagonal power sums (e.g., $\sum x_i^p = (\sum x_i)^p$ ), and the associated hypersurfaces may lose smoothness, invalidating the polar degree argument. Furthermore, the result applies to exact symmetric determinantal complexity, not border complexity; the isolatedness of solutions is not preserved under arbitrary degenerations.

Significance and Claims
The paper claims to provide a self-contained proof of a symmetric determinantal lower bound that does not rely on importing lemmas from the non-symmetric case. The primary contribution is the rigorous verification of the symmetric isolatedness step via the Schur-complement normal form, addressing a specific "adversarial" concern regarding whether the coincidence of left and right kernels in the symmetric model introduces new failure modes (such as spurious positive-dimensional components).

The improvement of the asymptotic constant from $1/4e$ (non-symmetric) to $1/2e$ (symmetric) is presented as a direct consequence of the symmetric geometry reducing the number of kernel variables from two to one, thereby altering the multidegree of the polar equations from $U+V$ to $2U$ . The work is positioned as a symmetric companion to the author's previous non-symmetric preprint, refining the understanding of determinantal complexity within the specific constraints of symmetric matrices.