Symmetroids

Computing the degree of a cubic symmetroid

We make a numerical computation to determine the degree of the cubic symmetroid

$$Q_3:= \{\det(x_0A_0 + x_1A_1 + x_2A_2 + x_3A_3)\mid A_i\in \mathbb{C}^{3\times 3}, A_i^T = A_i, 0\leq i\leq 3\}.$$

First, we compute the dimension of $Q_3$. We use coordinates by setting the first matrix to be $A_0 = \mathrm{diag}(a_{1},a_{2},a_3)$, where $a_1,a_2,a_3$ are variables. Then, we have the following situation:

$$ f:V \to W,\; (a_1,a_2,a_3,A_1,A_2,A_3) \mapsto \text{coefficients of }\det\Big(\sum_{i=0}^4x_iA_i\Big), $$

where $D = \dim_{\mathbb C} V = 21$, $N = \dim W_{\mathbb C} = 20$, and coefficients means the coefficients as a polynomial in $x$.

We set up $f$ in julia.

using HomotopyContinuation, LinearAlgebra
d = 3; n = 4
# D = dim V, N = dim W
M = binomial(d - 1 + 2, 2)
D = (n-1) * M + 3
N = binomial(n - 1 + d, d)

@var x[0:n-1] a[1:D]
# Construct symmetric matrices with variable entries
A = map(0:n-2) do
  Aᵢ  = []
  for i in 1:d
           k = * M + sum(d-j for j in 0:i-1)
           push!(Aᵢ, [zeros(i-1); a[k-(d-i):k]])
  end
  Aᵢ = hcat(Aᵢ...)
  (Aᵢ + transpose(Aᵢ)) ./ 2
end

A₀ = [a[D-2] 0 0; 0 a[D-1] 0; 0 0 a[D]]
μ = x[1] .* A₀ + sum(x[i+1] .* A[i] for i in 1:n-1)

# f is equal to the coefficients of det(μ) as a polynomial in x
f = System(coefficients(det), x))

To determine the dimension of $Q_3$ we compute the rank of the Jacobian matrix of $f$ at a random point.

J₀ = jacobian(InterpretedSystem(f), randn(ComplexF64,D))
dimQ = rank(J₀)
dim_fibers = D  - dimQ

We get

julia> dimQ, dim_fibers
(16, 5)

The degree of $Q_3$ multiplied by the degree of a general fiber $f^{-1}(h), h\in Q_3,$ is thus the number of isolated complex solutions of the following system of $16$ polynomial equations in the $16$ variables $b=(b_1,\ldots,b_{16})$:

$$ R \cdot f(a) = r \quad \text{ and } \quad a = S\cdot b + s, $$

where $R \in \mathbb{C}^{16 \times 20}$, $r \in \mathbb{C}^{16}$, $S \in \mathbb{C}^{21\times 16}$ and $s \in \mathbb{C}^{21}$ are chosen randomly.

# Compute a linear space a = S b + s
S1 = qr(randn(ComplexF64, D, D)).Q
S = S1[:, 1:dimQ]
s = S1[:, dimQ + 1]

@var b[1:dimQ]
L₁ = System(S * b + s)

We want to use monodromy to solve the intersection of f ∘ L₁ with the linear space given by $L_2 = \{R\cdot f(a) = r\}$. For this, we consider the first row of $[R, r]$ as varying. We introduce new variables k. We also introduce an initial solution b₁.

@var k[1:N+1] f₀[1:length(f)]

b₁ = randn(ComplexF64, dimQ)
f₁ = f(L₁(b₁))

R1 = qr(randn(ComplexF64, N, N)).Q
R = [transpose(k[1:N]); R1[1:(dimQ-1), :]]
r = [k[N+1]; R[2:end, :] * f₁]

L₂ = System(
           R * f₀ - r,
           variables = f₀;
           parameters = k
           );

The parameters for the initial solution b₁ are the following.

p₁ = randn(ComplexF64, N)
params = [p₁; transpose(p₁) * f₁]

Now, we can compute the zeros of L₂ ∘ f ∘ L₁ using monodromy. It is enough to compute one zero in each fiber. This is why we compare points with the distance function

$$\mathrm{dist}(b_1,b_2) = \Vert (f\circ L_1)(b_1) - (f\circ L_1)(b_2)\Vert_\infty.$$

dist(x,y) = norm(f(L₁(x)) - f(L₁(y)), Inf)

Finally, we execute the monodromy function.

julia> points = monodromy_solve(L₂ ∘ f ∘ L₁,
                   [b₁],
                   params,
                   distance = dist,
                   compile = false
                   )
MonodromyResult
===============
• return_code → :heuristic_stop
• 305 solutions
• 2435 tracked loops
• random_seed → 0xfa48c9b2

This shows that the degree of the cubic symmetroid is $305$.

(This was proven by Vainsencher in this article.)

We certify that the 305 solutions are true solutions.

julia> c = certify(L₂ ∘ f ∘ L₁,
    solutions(points);
    target_parameters = points.parameters)
CertificationResult
===================
• 305 solution candidates given
• 305 certified solution intervals (0 real, 305 complex)
• 305 distinct certified solution intervals (0 real, 305 complex)

We reuse the code for quartic symmetroids replacing d = 3 by d = 4. After three months the computation had found $849998$ solutions. At this point the computation was aborted manually, because it hadn't found any more solution in a week. This led us to state Conjecture that the degree $\delta$ of the quartic symmetroid is $850000\leq \delta \leq 851000$. Determining the exact number is an open problem.

Cite this example:
@Misc{ symmetroids2021 ,
    author =  { Paul Breiding, Christian Ikenmeyer, Mateusz Michalek, and Reuven Hodges },
    title = { Symmetroids },
    howpublished = { \url{ https://www.JuliaHomotopyContinuation.org/examples/symmetroids/ } },
    note = { Accessed: August 5, 2021 }
}

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