Why it is a question
THEORY-176, promoted on 2026-10-19, holds that Bansal, Nakkiran and Barak’s stitching penalty (LIT-850) compares representations in one decision problem, through a fixed top network and a restricted map class, and so is not Blackwell’s order. Its clause that representations which are functions of each other can stitch differently is an existence claim. A two-position, one-channel example, in which a 1×1 stitcher cannot undo a swap of positions, now shows it. Until that date THEORY-176’s promote_when asked for that possibility measured in trained networks. At the owner’s direction, from the promotion audit’s batch 027 (ADR-047), the THEORY was promoted on the example, and the condition moved here unchanged.
It is open because the example shows the possibility, not its size in practice. The paper argues that a pixel shuffle would stitch worse, and does not run the experiment (NOTE-653).
What would answer it
The condition formerly on THEORY-176, unchanged. Two representations, each an invertible function of the other (the paper’s own case is a spatial permutation of a layer’s activations), stitched into the same top network under the paper’s protocol (1×1 convolutional stitcher, trained on the task’s training set, penalty on its test set), with the difference in penalty larger than the spread across stitching seeds and across two top networks. Or an independent published treatment that relates the stitching penalty to the comparison of experiments, or to a usable-information order, and reaches the same relation. The relation is refuted by a step of its derivation shown wrong, or by the stitching order shown, for the paper’s map class, to be invariant under every invertible transformation of the representation. More experiments in which a better-trained representation lowers a weaker network’s error cannot settle it, since they are compatible with both readings.