E0041 proposition notation E0001

Algebraic Taylor–Möbius duality

Evaluating a polynomial on the infinitesimal cube in the Taylor algebras E0040 displays its forward differences (in \(B_k\)) or its derivatives (in \(A_k\)) as algebra coefficients.

(1) Proposition (Boolean Taylor–Möbius duality). Let \(p \in \ik[y_1, \dots, y_d]\), \(x \in \ik^d\), and \(v_1, \dots, v_k \in \ik^d\).

  • In \(B_k\):

    \[ p(x + {\textstyle\sum_i} \delta_i v_i) = \sum_{S \subseteq [k]} \Delta(p; x; v_S)\, \delta^S. \]

  • In \(A_k\):

    \[ p(x + {\textstyle\sum_i} \eps_i v_i) = \sum_{S \subseteq [k]} D(p; x; v_S)\, \eps^S. \]

Proof. Ad \(B_k\)) Expand \(p(x + \sum_i \delta_i v_i) = \sum_S c_S \delta^S\) in the free basis. The branch maps \(\rho_T: \delta_i \mapsto 1_{i \in T}\) give \(p(x + \sum_{i \in T} v_i) = \sum_{S \subseteq T} c_S\). Boolean Möbius inversion E0002 gives \(c_S = \Delta(p; x; v_S)\).

Ad \(A_k\)) The \(\eps^S\)-coefficient is \(D(p; x; v_S)\) by the multinomial theorem: \(\eps_i^2 = 0\) retains only the multilinear monomials.

(2) Proposition (Binomial Taylor–Möbius duality). In \(B_k^\nu\):

\[ p(x + {\textstyle\sum_i} \delta_i v_i) = \sum_{0 \leq \alpha \leq \nu} \frac{1}{\alpha!}\, \Delta^\alpha(p; x; v_\bullet)\, (\delta)_\alpha. \]

In \(A_k^\nu\):

\[ p(x + {\textstyle\sum_i} \eps_i v_i) = \sum_{0 \leq \alpha \leq \nu} \frac{1}{\alpha!}\, D^\alpha(p; x; v_\bullet)\, \eps^\alpha. \]

Proof. Ad \(B_k^\nu\)) The branch maps \(\rho_\beta: \delta_i \mapsto \beta_i\) satisfy \(\rho_\beta((\delta)_\alpha) = (\beta)_\alpha\), so the expansion coefficients are determined by binomial Möbius inversion E0003.

Ad \(A_k^\nu\)) The \(\eps^\alpha\)-coefficient is \(D^\alpha / \alpha!\) by the multinomial theorem.

(3) Proposition (Taylor–Möbius duality in \(C_k\)). In \(C_k\), expand \(p(x + \sum_i v_i x_i) = \sum_S c_S(t) x^S\). Then

\[ c_S(t) = \frac{1}{t^{|S|}}\, \Delta(p; x; (tv_i)_{i \in S}), \qquad c_S(0) = D(p; x; v_S). \]

The coefficients \(c_S(t)\) lie in \(\ik[t]\); in particular, \(\Delta(p; x; tv_\bullet)\) is divisible by \(t^k\).

Proof. In the generic fiber \(C_k[t^{-1}]\), set \(\delta_i = t^{-1} x_i\) so \(\delta_i^2 = \delta_i\). The branch maps give \(t^{|S|} c_S(t) = \Delta(p; x; (tv_i)_{i \in S})\) by Boolean Möbius inversion E0002. Since \(c_S \in \ik[t]\) (freeness), the divisibility follows, and \(c_S(0) = D(p; x; v_S)\).

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