ExampleConstruction: AI-generatedVerification: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6-luna)

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Full-lattice translation equivariance and a zero-padding boundary failure

Example

Let $f : Z^{2} \to R$ have finite support, let $K \subseteq Z^{2}$ be finite, and choose real weights $w_{δ}$ for $δ \in K$ . The full-lattice layer $C_{w} f (p) = \sum_{δ \in K} f (p + δ) w_{δ}$ commutes with every translation $L_{t} f (p) = f (p - t)$ . A finite-grid boundary can fail: take $D = {0, e_{1}}$ with $e_{1} = (1, 0)$ , let $τ$ swap the two sites and let $T f = f \circ τ^{- 1}$ , extend arrays by zero outside $D$ , and define $C_{0} f (x) = \overline{f} (x - e_{1})$ for $x \in D$ . For $f (0) = 1$ and $f (e_{1}) = 0$ , one obtains $T (C_{0} f) = (1, 0)$ but $C_{0} (T f) = (0, 0)$ , so this zero-padded layer is not equivariant for the swap action on the two-site grid.

Example references

discrete convolutional layers and boundary extensions

the action induced on feature maps

full-lattice translation equivariance

Goodfellow, Bengio, and Courville, Deep Learning, Chapter 9 §§9.2–9.5

Cohen and Welling, “Group Equivariant Convolutional Networks,” §§5–6.1

Facts & Assumptions

Given: The full-lattice layer and two-site zero-padded layer in the Example. Verify the translation identity and the finite boundary failure.

[F1]

Discrete convolutional layer with local receptive fields, channel mixing, stride, and padding: A discrete layer has finite receptive field and computes a weighted sum of shifted input values, using its chosen extension outside the input domain.

[F2]

The left action on feature maps induced by an action on sites: For a left action on sites, $(L_{u} f) (x) = f (u^{- 1} \cdot x)$ .

[F3]

Group cross-correlation and compatible layers are equivariant: The full-lattice unit-stride cross-correlation is equivariant under the translation group when its sum is finite or absolutely convergent.

Verification

1.1givenF1F3

The formula $C_{w} f (p) = \sum_{δ \in K} f (p + δ) w_{δ}$ is the first-layer translation correlation with finite filter support, so [F3] gives $C_{w} (L_{t} f) = L_{t} (C_{w} f)$ for every $t \in Z^{2}$ .

2.1givenF1F2algebra∎

Since $τ$ swaps $0$ and $e_{1}$ , $T f = (0, 1)$ . Zero extension gives $C_{0} f (0) = \overline{f} (- e_{1}) = 0$ and $C_{0} f (e_{1}) = \overline{f} (0) = 1$ , so $C_{0} f = (0, 1)$ and $T (C_{0} f) = (1, 0)$ . In turn, $C_{0} (T f) (0) = \overline{T f} (- e_{1}) = 0$ and $C_{0} (T f) (e_{1}) = T f (0) = 0$ , so $C_{0} (T f) = (0, 0)$ . These outputs differ, proving the stated failure.

Depends on

Used by

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Sources

Ian Goodfellow, Yoshua Bengio, and Aaron Courville, Deep Learning (standard reference, not scraped)
Taco S. Cohen and Max Welling, Group Equivariant Convolutional Networks (standard reference, not scraped)