Theorems · Theorem · linear algebra
AlternatingMap.alternatizeUncurryFin_alternatizeUncurryFinLM_comp_apply
∀ {R : Type u_1} {M : Type u_2} {N : Type u_4} [inst : CommRing R] [inst_1 : AddCommGroup M] [inst_2 : AddCommGroup N]
[inst_3 : Module R M] [inst_4 : Module R N] {n : ℕ} (f : M →ₗ[R] M →ₗ[R] M [⋀^Fin n]→ₗ[R] N) (v : Fin (n + 2) → M),
(AlternatingMap.alternatizeUncurryFin (AlternatingMap.alternatizeUncurryFinLM ∘ₗ f)) v =
∑ i,
∑ j ≥ i,
(-1) ^ (↑i + ↑j) •
(((f (v i.castSucc)) (v j.succ)) (j.removeNth (i.castSucc.removeNth v)) -
((f (v j.succ)) (v i.castSucc)) (j.removeNth (i.castSucc.removeNth v)))If f is a bilinear map taking values in the space of alternating maps,
then evaluation of the twice uncurried f on a tuple of vectors v
can be represented as a sum of
$$
f(v_i, v_j; v_0, \dots, \hat{v_i}, \dots, \hat{v_j}-) -
f(v_j, v_i; v_0, \dots, \hat{v_i}, \dots, \hat{v_j}-)
$$
over all (i j : Fin (n + 2)), i < j, taken with appropriate signs.
Here $\hat{v_i}$ and $\hat{v_j}$ mean that these vectors are removed from the tuple.
We use pairs of i j : Fin (n + 1), i ≤ j,
to encode pairs (i.castSucc : Fin (n + 2), j.succ : Fin (n + 2)),
so the power of -1 is off by one compared to the informal texts.
In particular, if f is symmetric in the first two arguments,
then the resulting alternating map is zero,
see alternatizeUncurryFin_alternatizeUncurryFinLM_comp_of_symmetric below.
- Cited by
- 2 results in Mathlib
- Foundations
- Depth 94 from the axioms · uses propext, Classical.choice, Quot.sound
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