Installation¶

The following instructions were prepared using

In [1]:
versioninfo()
Julia Version 1.8.5
Commit 17cfb8e65ea (2023-01-08 06:45 UTC)
Platform Info:
  OS: Linux (x86_64-linux-gnu)
  CPU: 16 × AMD Ryzen 7 PRO 4750U with Radeon Graphics
  WORD_SIZE: 64
  LIBM: libopenlibm
  LLVM: libLLVM-13.0.1 (ORCJIT, znver2)
  Threads: 8 on 16 virtual cores
Environment:
  JULIA_NUM_THREADS = 8
In [2]:
using Pkg
Pkg.activate(".")
using Dates
now()
  Activating project at `~/Mathematics/Teaching/2022/Computational Group Theory/2022 WS Karlsruhe/Symbolic SOS`
Out[2]:
2023-02-17T12:04:31.847
In [3]:
using Plots
using LinearAlgebra
using Groups
using PropertyT
using PropertyT.StarAlgebras

The computation¶

We will show that $$\Delta^2 - 0.2\Delta \in \operatorname{SL}(3, \mathbb{Z})$$ is very close to being a sum of squares.

So far we only made the needed packages available in the notebook. In the next cell we define G to be the set of all $3\times 3$ matrices over $\mathbb Z$.

In [4]:
N = 3
G = Groups.MatrixGroups.SpecialLinearGroup{N}(Int8)
Out[4]:
special linear group of 3×3 matrices over Int8

Generating set¶

Now we create the elementary matrices $E_{i,j}$. The set of all such matrices and their inverses is denoted by S.

In [5]:
S = gens(G)
S = union!(S, inv.(S))
Out[5]:
12-element Vector{FPGroupElement{Groups.MatrixGroups.SpecialLinearGroup{3, Int8, DataType, Alphabet{Groups.MatrixGroups.ElementaryMatrix{3, Int8}}, Vector{Groups.MatrixGroups.ElementaryMatrix{3, Int8}}}, …}}:
 E₁₂
 E₁₃
 E₂₁
 E₂₃
 E₃₁
 E₃₂
 E₁₂^-1
 E₁₃^-1
 E₂₁^-1
 E₂₃^-1
 E₃₁^-1
 E₃₂^-1

Group Ring and Laplacians¶

Now we will generate the ball E_R of radius $R=4$ in $\operatorname{SL}(N,\mathbb{Z})$ and use this as a (partial) basis in a group ring (denoted by RG below). Such group ring also needs a multiplication table (pm, which is actually a division table) which is created as follows: when $x,y$ reside at positions i-th and j-th in E_R, then pm[i,j] = k, where k is the position of $x^{-1}y$ in E_R.

In [6]:
RG, S, sizes = PropertyT.group_algebra(G, S; halfradius=2, twisted=true)
[ Info: generating wl-metric ball of radius 4
  0.347782 seconds (871.81 k allocations: 30.018 MiB, 92.16% compilation time)
[ Info: sizes = [13, 121, 883, 5455]
[ Info: computing the *-algebra structure for G
  0.053713 seconds (158.85 k allocations: 7.217 MiB, 93.01% compilation time)
Out[6]:
(*-algebra of special linear group of 3×3 matrices over Int8, FPGroupElement{Groups.MatrixGroups.SpecialLinearGroup{3, Int8, DataType, Alphabet{Groups.MatrixGroups.ElementaryMatrix{3, Int8}}, Vector{Groups.MatrixGroups.ElementaryMatrix{3, Int8}}}, …}[E₁₂, E₁₃, E₂₁, E₂₃, E₃₁, E₃₂, E₁₂^-1, E₁₃^-1, E₂₁^-1, E₂₃^-1, E₃₁^-1, E₃₂^-1], [13, 121, 883, 5455])
In [7]:
sizes
Out[7]:
4-element Vector{Int64}:
   13
  121
  883
 5455
In [8]:
Δ = length(S)*one(RG) - sum(RG(s) for s in S)
# Δ.coeffs
Out[8]:
12·(id) -1·E₁₂ -1·E₁₃ -1·E₂₁ -1·E₂₃ -1·E₃₁ -1·E₃₂ -1·E₁₂^-1 -1·E₁₃^-1 -1·E₂₁^-1 -1·E₂₃^-1 -1·E₃₁^-1 -1·E₃₂^-1
In [9]:
using JuMP
using SCS
In [10]:
UB = 0.2
SDP_problem = PropertyT.sos_problem_primal(Δ^2, Δ, upper_bound=UB, augmented=false);
SDP_problem
Out[10]:
A JuMP Model
Maximization problem with:
Variables: 7382
Objective function type: VariableRef
`AffExpr`-in-`MathOptInterface.EqualTo{Float64}`: 5455 constraints
`AffExpr`-in-`MathOptInterface.LessThan{Float64}`: 1 constraint
`Vector{VariableRef}`-in-`MathOptInterface.PositiveSemidefiniteConeTriangle`: 1 constraint
Model mode: AUTOMATIC
CachingOptimizer state: NO_OPTIMIZER
Solver name: No optimizer attached.
Names registered in the model: P, psd, λ
In [11]:
function scs_optimizer(;
    accel = 50,
    alpha = 1.95,
    eps = 1e-10,
    max_iters = 10_000,
    verbose = true,
    linear_solver = SCS.DirectSolver,
)
    return JuMP.optimizer_with_attributes(
        SCS.Optimizer,
        "acceleration_lookback" => accel,
        "acceleration_interval" => 10,
        "alpha" => alpha,
        "eps_abs" => eps,
        "eps_rel" => eps,
        "linear_solver" => linear_solver,
        "max_iters" => max_iters,
        "warm_start" => true,
        "verbose" => verbose,
    )
end

@time status, _ = PropertyT.solve(SDP_problem, scs_optimizer());
@info status
 25.111573 seconds (25.65 M allocations: 1.333 GiB, 1.55% gc time, 62.95% compilation time: 0% of which was recompilation)
------------------------------------------------------------------
	       SCS v3.2.1 - Splitting Conic Solver
	(c) Brendan O'Donoghue, Stanford University, 2012
------------------------------------------------------------------
problem:  variables n: 7382, constraints m: 12837
cones: 	  z: primal zero / dual free vars: 5455
	  l: linear vars: 1
	  s: psd vars: 7381, ssize: 1
settings: eps_abs: 1.0e-10, eps_rel: 1.0e-10, eps_infeas: 1.0e-07
	  alpha: 1.95, scale: 1.00e-01, adaptive_scale: 1
	  max_iters: 10000, normalize: 1, rho_x: 1.00e-06
	  acceleration_lookback: 50, acceleration_interval: 10
lin-sys:  sparse-direct-amd-qdldl
	  nnz(A): 22036, nnz(P): 0
------------------------------------------------------------------
 iter | pri res | dua res |   gap   |   obj   |  scale  | time (s)
------------------------------------------------------------------
     0| 1.56e+02  1.00e+00  5.29e+01  2.20e+01  1.00e-01  2.38e-02 
   250| 1.96e-04  2.35e-05  1.69e-03 -1.99e-01  1.00e-01  8.39e-01 
   500| 1.83e-04  8.40e-06  3.12e-04 -2.00e-01  5.42e-03  1.67e+00 
   750| 1.27e-04  4.40e-06  1.99e-04 -2.00e-01  5.42e-03  2.54e+00 
  1000| 7.03e-05  2.86e-06  1.29e-04 -2.00e-01  5.42e-03  3.43e+00 
  1250| 6.36e-05  2.20e-06  8.56e-05 -2.00e-01  5.42e-03  4.27e+00 
  1500| 5.24e-05  1.80e-06  6.80e-05 -2.00e-01  5.42e-03  5.10e+00 
  1750| 5.35e-05  1.38e-06  5.63e-05 -2.00e-01  5.42e-03  5.92e+00 
  2000| 2.78e-03  2.23e-05  4.87e-04 -2.00e-01  5.42e-03  6.77e+00 
  2250| 1.87e-04  8.19e-07  1.07e-06 -2.00e-01  5.42e-03  7.64e+00 
  2500| 1.59e-07  8.62e-10  3.97e-08 -2.00e-01  5.42e-03  8.40e+00 
  2725| 9.50e-10  4.10e-12  8.27e-11 -2.00e-01  5.42e-03  9.08e+00 
------------------------------------------------------------------
status:  solved
timings: total: 9.08e+00s = setup: 1.62e-02s + solve: 9.06e+00s
	 lin-sys: 7.26e-01s, cones: 7.68e+00s, accel: 1.63e-01s
------------------------------------------------------------------
objective = -0.200000
------------------------------------------------------------------
[ Info: OPTIMAL
In [12]:
λ = value(SDP_problem[:λ])
Out[12]:
0.19999999990216422
In [13]:
P = value.(SDP_problem[:P])
Q = real.(sqrt(P))
heatmap(Q, clim=(-0.4,0.4), color=:bluesreds, yflip=true, aspect_ratio=:equal, size=(500,500), legend=false)
Out[13]:
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In [14]:
sos = PropertyT.compute_sos(RG, Q, augmented=false)
residual = (Δ^2 - λ*Δ) - sos
coeffs(residual)
Out[14]:
5455-element Vector{Float64}:
 -8.526512829121202e-13
  1.9184653865522705e-13
  2.2737367544323206e-13
  2.6290081223123707e-13
  1.9184653865522705e-13
  1.8474111129762605e-13
  1.9895196601282805e-13
  1.9184653865522705e-13
  2.2737367544323206e-13
  2.5579538487363607e-13
  1.8118839761882555e-13
  1.9184653865522705e-13
  2.0250467969162855e-13
  ⋮
  9.159550025439237e-14
  9.206124132886958e-14
  1.9525493143152618e-13
  9.287367375738445e-14
 -2.16886519241217e-13
  9.212404629560732e-14
  9.307885538200859e-14
  1.3393404144724133e-13
  9.173053169614453e-14
  9.447255742562001e-14
 -2.174385935898179e-13
  1.9500200065802886e-13
In [15]:
norm(residual, 1)
Out[15]:
8.013365126577617e-10

So now we're in position where $$\Delta^2 - \lambda \Delta = \sum \xi_i^*\xi_i + residual$$

and we would like to dominate $residual$ by $R\Delta$ for some $R$ depending on $|residual|_1$. The problem is that we can do it only if $residual$ belongs to the augmentation ideal, however...

In [16]:
StarAlgebras.aug(residual)
Out[16]:
-1.928179000086013e-10
In [17]:
StarAlgebras.aug(sos)
Out[17]:
1.9280101511025753e-10

There are two ways to solve this:

  • Project $Q$ to "close enough" rational numbers and then align all columns so that $\xi_i \in I[G]$
  • use interval arithmetic!
In [18]:
0.1 + 0.2 - 0.3
Out[18]:
5.551115123125783e-17
In [19]:
using PropertyT.IntervalArithmetic
In [20]:
u = @interval(0.1)
Out[20]:
[0.0999999, 0.100001]
In [21]:
r = @interval 0.1 + @interval 0.2 - @interval 0.3
Out[21]:
[-6.9389e-17, 2.77556e-17]
In [22]:
0 ∈ r
Out[22]:
true
In [23]:
PropertyT.augment_columns!(Q)
Q_aug = Interval.(Q)
Out[23]:
121×121 Matrix{Interval{Float64}}:
  [2.00364, 2.00365]        …  [-0.00705892, -0.00705891]
 [-0.309393, -0.309392]         [0.0593636, 0.0593637]
 [-0.309393, -0.309392]         [0.0552353, 0.0552354]
 [-0.309393, -0.309392]         [0.0552353, 0.0552354]
 [-0.309393, -0.309392]        [-0.0250792, -0.0250791]
 [-0.309393, -0.309392]     …   [0.0593636, 0.0593637]
 [-0.309393, -0.309392]        [-0.0237578, -0.0237577]
 [-0.309393, -0.309392]         [0.0593636, 0.0593637]
 [-0.309393, -0.309392]         [0.0552353, 0.0552354]
 [-0.309393, -0.309392]         [0.0552353, 0.0552354]
 [-0.309393, -0.309392]     …   [0.030087, 0.0300871]
 [-0.309393, -0.309392]         [0.0593636, 0.0593637]
 [-0.309393, -0.309392]        [-0.218148, -0.218147]
   ⋮                        ⋱    ⋮
  [0.0226735, 0.0226736]        [0.0399426, 0.0399427]
 [-0.0078798, -0.00787979]  …   [0.0079027, 0.00790271]
  [0.0226735, 0.0226736]       [-0.0333089, -0.0333088]
 [-0.00705891, -0.0070589]     [-0.0585376, -0.0585375]
  [0.00680343, 0.00680344]     [-0.132314, -0.132313]
  [0.0226735, 0.0226736]        [0.0248844, 0.0248845]
  [0.0226735, 0.0226736]    …   [0.0248844, 0.0248845]
  [0.0530596, 0.0530597]        [0.116407, 0.116408]
  [0.0226735, 0.0226736]        [0.0248844, 0.0248845]
  [0.0226735, 0.0226736]        [0.0248844, 0.0248845]
 [-0.0078798, -0.00787979]      [0.105081, 0.105082]
 [-0.00705891, -0.0070589]  …   [0.339944, 0.339945]
In [24]:
0 in sum(Q_aug[:, 1])
Out[24]:
true
In [25]:
sos_int = PropertyT.compute_sos(RG, Q_aug, augmented=true);
In [26]:
residual_int = Δ^2 - @interval(λ)*Δ - sos_int;
In [27]:
aug(residual_int)
Out[27]:
[-1.75794e-10, 1.72987e-10]
In [28]:
0 in aug(residual_int)
Out[28]:
true

Recall: if $x \in I[G]$, then there exists $R\in \mathbb{R}$ such that $x + R\Delta$ is a sum of squares.

It can be shown that for our $residual_{int}$ it is enough to take $R >= 2|residual_{int}|_1$.

Therefore

\begin{align*} \Delta^2 - \lambda \Delta & = \sum_i\xi_i^*\xi_i + residual\\ \Delta^2 - \lambda \Delta + 2|residual|_1\Delta & = \sum_i\xi_i^*\xi_i + \underbrace{residual + 2|residual|_1\Delta}_{\in \Sigma^2 I[G]} \end{align*}

i.e. $\Delta^2 - (\lambda - 2|residual|_1)\Delta \in \Sigma^2I[G]$

In [29]:
λ_certified = @interval(λ) - 2norm(residual_int, 1)
Out[29]:
[0.199995, 0.199996]
In [ ]:

In [30]:
round(λ_certified.lo, digits=7, RoundDown)
Out[30]:
0.1999958
In [31]:
κ = sqrt(2λ_certified/length(S))
Out[31]:
[0.182572, 0.182573]
In [32]:
0.1825740
Out[32]:
0.182574

Previously the best known bound was $0.00017...$ by M. Kassabov.

Wedderburn Decomposition¶

Now something happens: in the next cell we split the subspace of $\mathbb{R} \operatorname{SL}(N, \mathbb{Z})$ supported on E_R into irreducible representations of the wreath product $\mathbb Z / 2 \mathbb Z \wr \operatorname{Sym}_N$. The action of wreath product on the elements of the matrix space is by conjugation, i.e. permutation of rows and columns. We also compute projections on the invariant subspaces to later speed up the optimisation step.

In [33]:
using Groups
using PermutationGroups
using SymbolicWedderburn
In [34]:
wd = let RG = RG, N = N
    G = StarAlgebras.object(RG)
    P = PermGroup(perm"(1,2)", Perm(circshift(1:N, -1)))
    Σ = Groups.Constructions.WreathProduct(PermGroup(perm"(1,2)"), P)
    act = PropertyT.action_by_conjugation(G, Σ)

    wdfl = @time SymbolicWedderburn.WedderburnDecomposition(
        Float64,
        Σ,
        act,
        basis(RG),
        StarAlgebras.Basis{UInt16}(@view basis(RG)[1:sizes[2]]),
    )
end
 17.944732 seconds (43.99 M allocations: 2.208 GiB, 5.26% gc time, 99.84% compilation time)
Out[34]:
Wedderburn Decomposition into 247 orbits and 5 simple summands of sizes
[14, 9, 6, 14, 12]
In [35]:
elt = Δ^2
order_unit = Δ
UB = 0.200001

SDP_problem, varPs = PropertyT.sos_problem_primal(elt, order_unit, wd, upper_bound=UB, augmented=false);
SDP_problem
Out[35]:
A JuMP Model
Maximization problem with:
Variables: 355
Objective function type: VariableRef
`AffExpr`-in-`MathOptInterface.EqualTo{Float64}`: 247 constraints
`AffExpr`-in-`MathOptInterface.LessThan{Float64}`: 1 constraint
`Vector{VariableRef}`-in-`MathOptInterface.PositiveSemidefiniteConeTriangle`: 5 constraints
Model mode: AUTOMATIC
CachingOptimizer state: NO_OPTIMIZER
Solver name: No optimizer attached.
Names registered in the model: λ
In [36]:
@time status, _ = PropertyT.solve(SDP_problem, scs_optimizer());
@info status
  0.528419 seconds (19.09 k allocations: 1.431 MiB)
------------------------------------------------------------------
	       SCS v3.2.1 - Splitting Conic Solver
	(c) Brendan O'Donoghue, Stanford University, 2012
------------------------------------------------------------------
problem:  variables n: 355, constraints m: 602
cones: 	  z: primal zero / dual free vars: 247
	  l: linear vars: 1
	  s: psd vars: 354, ssize: 5
settings: eps_abs: 1.0e-10, eps_rel: 1.0e-10, eps_infeas: 1.0e-07
	  alpha: 1.95, scale: 1.00e-01, adaptive_scale: 1
	  max_iters: 10000, normalize: 1, rho_x: 1.00e-06
	  acceleration_lookback: 50, acceleration_interval: 10
lin-sys:  sparse-direct-amd-qdldl
	  nnz(A): 4132, nnz(P): 0
------------------------------------------------------------------
 iter | pri res | dua res |   gap   |   obj   |  scale  | time (s)
------------------------------------------------------------------
     0| 1.56e+02  1.00e+00  2.57e+01  9.66e+00  1.00e-01  2.89e-03 
   250| 2.28e-04  4.35e-05  7.65e-04 -2.00e-01  1.00e-01  4.01e-02 
   500| 1.24e-04  2.01e-05  3.00e-04 -2.00e-01  3.08e-02  7.77e-02 
   750| 4.08e-04  8.17e-06  1.39e-04 -2.00e-01  5.76e-03  1.16e-01 
  1000| 3.27e-04  5.15e-06  7.08e-05 -2.00e-01  5.76e-03  1.56e-01 
  1250| 2.98e-04  3.91e-06  5.14e-05 -2.00e-01  5.76e-03  1.95e-01 
  1500| 3.03e-04  2.93e-06  3.72e-05 -2.00e-01  5.76e-03  2.34e-01 
  1750| 3.16e-04  2.10e-06  2.55e-05 -2.00e-01  5.76e-03  2.73e-01 
  2000| 3.30e-04  1.43e-06  1.73e-05 -2.00e-01  5.76e-03  3.12e-01 
  2250| 3.39e-04  8.61e-07  1.12e-05 -2.00e-01  5.76e-03  3.51e-01 
  2500| 3.42e-04  3.29e-07  6.55e-06 -2.00e-01  5.76e-03  3.90e-01 
  2750| 1.65e-03  9.81e-06  4.10e-05 -2.00e-01  5.76e-03  4.31e-01 
  3000| 2.60e-04  4.02e-06  2.21e-05 -2.00e-01  5.76e-03  4.73e-01 
  3250| 2.43e-08  3.61e-10  3.34e-09 -2.00e-01  5.76e-03  5.14e-01 
  3325| 4.54e-09  1.19e-11  1.19e-10 -2.00e-01  5.76e-03  5.25e-01 
------------------------------------------------------------------
status:  solved
timings: total: 5.25e-01s = setup: 2.52e-03s + solve: 5.23e-01s
	 lin-sys: 7.01e-02s, cones: 4.03e-01s, accel: 1.49e-02s
------------------------------------------------------------------
objective = -0.200001
------------------------------------------------------------------
[ Info: OPTIMAL
In [37]:
λ = value(SDP_problem[:λ])
Out[37]:
0.2000009999714222
In [38]:
using SparseArrays
let Ps = [value.(P) for P in varPs]
    Qs = real.(sqrt.(Ps));
    heatmap(Matrix(blockdiag(sparse.(Qs)...)), clim=(-0.4,0.4), color=:bluesreds, yflip=true, aspect_ratio=:equal, size=(500,500), legend=false)
end
Out[38]:
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In [39]:
Q_new = let Ps = [value.(P) for P in varPs]
    Qs = real.(sqrt.(Ps));
    PropertyT.reconstruct(Qs, wd)
end
# p1 = heatmap(Q_new, clim=(-0.4,0.4), color=:bluesreds, yflip=true, aspect_ratio=:equal, size=(500,500), legend=false)
# p2 = heatmap(Q, clim=(-0.4,0.4), color=:bluesreds, yflip=true, aspect_ratio=:equal, size=(500,500), legend=false)
# plot(p1, p1, layout=(1,2), legend = false, size=(1000, 500))
heatmap(Q-Q_new, clim=(-0.4,0.4), color=:bluesreds, yflip=true, aspect_ratio=:equal, size=(500,500), legend=false)
Out[39]:
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In [40]:
# project to the augmentation ideal
Q_aug = Interval.(PropertyT.augment_columns!(Q_new));
In [41]:
# compute sum of squares in RG defined by projected Q
sos_int = PropertyT.compute_sos(RG, Q_aug, augmented=false);
In [42]:
# compute the residual
residual_int = Δ^2 - @interval(λ)*Δ - sos_int;
# @assert 0 in StarAlgebras.aug(residual_int)
norm(residual_int, 1)
Out[42]:
[1.71435e-08, 1.71651e-08]
In [43]:
λ_certified = (@interval(λ) - 2norm(residual_int, 1)).lo
Out[43]:
0.20000096564128686
In [44]:
round(λ_certified, digits=6, RoundDown)
Out[44]:
0.2
In [ ]: