66 Quantifying and Learning Linear Symmetry-Based Disentanglement (LSBD) metric equals DLSBD = inf ρ∈P(G,Z) 1 2 ZGZX∥ ρ(g)−1 ◦h(g · x) −h(x)∥2 ρ,h,µdm(g)dµ(x). (4.20) 4.4.3 Practical Computation of DLSBD There are two main challenges for computing the DLSBD metric from Definition 4.3. First, to calculate the integrals in the formula, all possible data points that can be expressed as g · x0 with g ∈G=G1 ×· · ·×GK must be available. Second, the infimum of the integrals over all possible linearly disentangled representations must be estimated. This requires finding the possible invariant subspaces Z =Z1 ⊕· · ·⊕ZK induced by the encoding function h over which the group representations are disentangled. We present a practical implementation of an upper bound to DLSBD, for a given encoding functionhand dataset Xgenerated by some known group transformations. This approximation of DLSBD is designed for a group decomposition G = G1 × · · · ×GK where each Gk = SO(Dk) with k ∈ {1, . . . ,K}, i.e. the group of rotations in Dk dimensions. This implementation approximates the integrals of Definition 4.3 by using the empirical distribution of X. The invariant subspaces of Z to the subgroup actions are found by applying a suitable change of basis. In the new basis, the disentangled group representations are expressed in a parametric form whose parameters are optimised to find the tightest bound to DLSBD. See Figure 4.2 for an intuitive description of the process. Assume there is a dataset X that can be modelled in terms of the group decompositionG=G1 ×· · · Gk. For each subgroupGk there is a set of known group elements Gk ⊆Gk uniformly sampled such that the dataset is described in terms of all elements in G = G1 × · · · ×GK and a base point x0 as X = {(g1, . . . ,gK) · x0|gk ∈Gk, k ∈{1, . . . ,K}}. For each subgroup Gk we construct a set of encoded data Zk ⊆Z whose variability should only depend on the action of Gk. The set Zk is given by Zk ={zk(g1, . . . ,gK)|gj ∈Gj ,j ∈{1, . . . ,K}}, where zk(g1, . . . ,gK) :=h((g1, . . . ,gK) · x0) − 1 |Gk| X g′∈Gk h((g1, . . . ,gk−1,g′,gk+1, . . . ,gK) · x0). (4.21)
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