Split A.4.35 into two separate @test_broken assertions - #20
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The original test chained three different things in one equality:
refl(Vr, B) == B∧Vr*(Vr)⁻¹ == (B∧Vr)⨽(Vr)⁻¹
But refl() in galgebra is a grade-by-grade sandwich (Vr*B[r]*Vr⁻¹ with
sign (-1)^{s(r+1)}), which is not the same as the rejection (B∧Vr)*Vr⁻¹.
The chained equality was conflating two unrelated operations.
Split into:
- (B∧Vr)*(Vr)⁻¹ == (B∧Vr)⨽(Vr)⁻¹ (rejection = right-contraction form)
- refl(Vr, B) == (B∧Vr)*(Vr)⁻¹ (sandwich vs rejection)
CI will now show which half passes and which is genuinely broken.
Refs #15
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Refs #15
The original test had a chained
==mixing three unrelated things:After reading galgebra's
reflect_in_bladesource:refl(Vr, B)is a grade-by-grade sandwich:(-1)^{s(r+1)} * Vr * B[r] * Vr⁻¹for each grader, wheres = grade(Vr). For a grade-2 blade (likeVr = u∧vin Cl3), the sign is always+1, so it reduces toVr * B * Vr⁻¹.(B ∧ Vr) * Vr⁻¹is the rejection (outer projection onto the complement), a completely different operation.Split into two
@test_brokenso CI shows which part actually fails:(B ∧ Vr) * (Vr)⁻¹ == (B ∧ Vr) ⨽ (Vr)⁻¹— pure rejection formula (A.4.35 as written in the paper)refl(Vr, B) == (B ∧ Vr) * (Vr)⁻¹— whether the sandwich equals rejection (likely false in general)