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feat(linear_algebra/clifford_algebra/spin_group) : Spin Group #16040

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4 changes: 4 additions & 0 deletions src/linear_algebra/clifford_algebra/basic.lean
Original file line number Diff line number Diff line change
Expand Up @@ -7,6 +7,7 @@ Authors: Eric Wieser, Utensil Song
import algebra.ring_quot
import linear_algebra.tensor_algebra.basic
import linear_algebra.quadratic_form.isometry
import tactic.equiv_rw

/-!
# Clifford Algebras
Expand Down Expand Up @@ -313,6 +314,9 @@ def invertible_ΞΉ_of_invertible (m : M) [invertible (Q m)] : invertible (ΞΉ Q m)
mul_inv_of_self := by rw [map_smul, mul_smul_comm, ΞΉ_sq_scalar, algebra.smul_def, ←map_mul,
inv_of_mul_self, map_one] }

def invertible_of_invertible_ΞΉ (m : M) [invertible (ΞΉ Q m)] [invertible (2 : R)] :
invertible (Q m) := sorry
Comment on lines +317 to +318
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Any more ideas on how to prove this or under what situations it's true?

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I already got the proof of this. I just need some time to tidy it up.

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Could you commit the messy version somewhere so that it doesn't get lost?

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Could you commit the messy version somewhere so that it doesn't get lost?

@Biiiilly Hi, I'm trying to make this PR work in Lean 4, do you still have the proof of this?


/-- For a vector with invertible quadratic form, $v^{-1} = \frac{v}{Q(v)}$ -/
lemma inv_of_ΞΉ (m : M) [invertible (Q m)] [invertible (ΞΉ Q m)] : β…Ÿ(ΞΉ Q m) = ΞΉ Q (β…Ÿ(Q m) β€’ m) :=
begin
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