Varying one factor of a list product in a ring #
In an R-algebra A, the product of a list is affine in each factor. In particular, shifting one
factor a by the scalar algebraMap R A c changes the product by c times the product of the
list with that factor deleted. This is the step that centres one letter of a word at a time.
An additive map δ of a possibly noncommutative ring obeying the Leibniz rule
δ (a * b) = δ a * b + a * δ b sends an ordered product f 0 * ⋯ * f (n - 1) to the sum, over the
positions k, of the same product with the factor f k replaced by δ (f k). The rule covers
both a derivation of a commutative ring and the inner derivation ⁅a, -⁆ of an associative ring,
which is how the commutator with a Lie generator acts on a word in a universal enveloping algebra.
Main results #
List.prod_append_sub_algebraMap_cons:(l₁ ++ (a - algebraMap R A c) :: l₂).prod = (l₁ ++ a :: l₂).prod - c • (l₁ ++ l₂).prod.List.prod_map_snd_mem_span_of_isChain: the product of the second components of a list of labelled letters with no two consecutive equal labels lies in a span determined by the first label, when each admissible letter carries the span of the rest into it.AddMonoidHom.map_prod_ofFn_of_leibniz: the Leibniz rule for an ordered product ofnfactors.
Shifting the factor a of the product of l₁ ++ a :: l₂ by the scalar algebraMap R A c
subtracts c times the product of l₁ ++ l₂.
Induction along a word with no two consecutive equal labels. Let P single out the
admissible labelled letters (k, a). If multiplying an admissible letter with label k onto
the span of T k lands in the span of V k, the span of each T k contains 1, and
V k' ⊆ T k for k ≠ k', then the product of the letters of every admissible word whose
consecutive labels differ lies in the span of V k, where k is the label of its first letter.
The Leibniz rule for an ordered product. An additive map δ with
δ (a * b) = δ a * b + a * δ b sends the product f 0 * ⋯ * f (n - 1) to the sum over k of the
same product with its k-th factor replaced by δ (f k). The ring need not be commutative.