If one or more of these matrices have low rank, then one could try to employ low-rank factorizations. Of course, not all matrices have low rank. I just try to point our that low rank can be exploited nicely if present.
In order to demonstrate the effect, here is an implementation of Adaptive Cross Approximation (ACA) (see this paper by Mario Bebendorf for more details on the algorithm). Because this implementation uses LinearAlgebra`BLAS`IAMAX which was introduced with version 11.2, this won't run on older versions of Mathematica.
Needs["LinearAlgebra`BLAS`"]
ClearAll[ACACompression];
Options[ACACompression] = {"MaxRank" -> 50, "Tolerance" -> 1. 10^-4,
"StartingIndex" -> 1,
"MaxPivotTrials" -> 100};
ACACompression::maxrank = "Warning: Computed factorization has maximal rank.";
ACACompression[A_?MatrixQ, opts : OptionsPattern[]] := ACACompression[i [Function] A[[i]], j [Function] A[[All, j]], opts];
ACACompression[row_, column_, OptionsPattern[]] :=
Module[{maxrank, ϵ, u, v, k, ik, jk, uk, vk, test, norm2, δ, w, iter, uQ, uR, vQ, vR, rank, σ, m, n, eps, maxiter}, eps = $MachineEpsilon;
maxiter = OptionValue["MaxPivotTrials"];
maxrank = OptionValue["MaxRank"];
ϵ = OptionValue["Tolerance"];
k = 1;
ik = OptionValue["StartingIndex"];
vk = row[ik];
m = Length[vk];
v = ConstantArray[0., {maxrank, m}];
jk = IAMAX[vk];
v[[k]] = vk = vk/vk[[jk]];
uk = column[jk];
n = Length[uk];
u = ConstantArray[0., {maxrank, n}];
u[[k]] = uk;
test = uk.uk vk.vk;
norm2 = test;
maxrank = Min[m, n, maxrank];
While[test > ϵ^2 norm2 && k < maxrank,
k++;
ik = IAMAX[uk];
vk = Subtract[row[ik], u[[;; k - 1, ik]].v[[;; k - 1]]];
jk = IAMAX[vk];
δ = vk[[jk]];
If[Abs[δ] < eps, w = uk;
δ = 0.;
iter = 0;
While[Abs[δ] < eps && iter < maxiter,
iter++;
w[[ik]] = 0.;
ik = IAMAX[w];
vk = Subtract[row[ik], u[[;; k - 1, ik]].v[[;; k - 1]]];
jk = IAMAX[vk];
δ = vk[[jk]];
];
];
v[[k]] = vk = vk/δ;
uk = Subtract[column[jk], v[[;; k - 1, jk]].u[[;; k - 1]]];
u[[k]] = uk;
test = uk.uk vk.vk;
norm2 += test + Dot[u[[1 ;; k - 1]].uk, v[[1 ;; k - 1]].vk]
];
If[k == OptionValue["MaxRank"], Message[ACACompression::maxrank];];
{u[[1 ;; k]], v[[1 ;; k]]}]
Now let's consider the following setup. I assume that the matrix G has low rank; I ensure that by taking the squared distance matrix of a set of homogeneously distributed points. (I am well aware that $G$ in OP's example is probably thought of as a Gram matrix of an inner product and, as such, positive definite so that it may be not of low rank. Still, ΓR, ΓL may be of low rank.)
n = 2000;
ΓR = RandomReal[{-1, 1}, {n, n}];
ΓL = RandomReal[{-1, 1}, {n, n}];
G = DistanceMatrix[RandomReal[{-1, 1}, n]]^2;
Now, let's compute the trace in the straight-forward way and compare the result and its timing to an ACA followed by a careful evaluation of the trace. Notice that the order in which the Dot-operations are performed is crucial here.
a = Tr[ΓR.G.ΓL.Transpose[G]]; // RepeatedTiming // First
First@RepeatedTiming[
{u, v} = ACACompression[G, "Tolerance" -> 1. 10^-6];
b = Tr[(u.ΓR.Transpose[u]).(v.ΓL.Transpose[v])];
]
Abs[a - b]/Abs[a]
0.33
0.0033
4.71289*10^-16
Let's also check that the relative accuracy of ACA:
Max[Abs[Transpose[v].u - G]]/Max[Abs[G]]
3.33565*10^-16
Edit
Squared distance matrices in Euclidean space can actually factored explicitly:
FactoredSquaredDistanceMatrix[x_?VectorQ, y_?VectorQ] := {
{x^2, x, ConstantArray[1., {Length[x]}]},
{ConstantArray[1., {Length[y]}], -2. y, y^2}
}
FactoredSquaredDistanceMatrix[x_?MatrixQ, y_?MatrixQ] := {
Join[Partition[Total[x^2, {2}], 1], x,
ConstantArray[1., {Length[x], 1}], 2],
Join[ConstantArray[1., {Length[y], 1}], -2. y,
Partition[Total[y^2, {2}], 1], 2]
}
So the following can do the job without ever forming the squared distance matrix.
First@RepeatedTiming[
{u, v} = FactoredSquaredDistanceMatrix[x, x];
c = Tr[(u.\[CapitalGamma]R.Transpose[
u]).(v.\[CapitalGamma]L.Transpose[v])];
]
Abs[a - c]/Abs[a]
