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Mathematics · 术语表

什么是 alternating form?

Definition 2.13 University Mathematics — Year 2 · 第 2 章 — Linear Algebra

A map f ⁣:En→Kf \colon E^n \to K is nn-linear when it is linear in each variable, and alternating when it vanishes whenever two arguments are equal. Alternating implies antisymmetric: swapping two arguments changes the sign (expand f(…,x+y,…,x+y,… )=0f(\dots, x + y, \dots, x + y, \dots) = 0); more generally, for σ∈Sn\sigma \in \mathfrak{S}_n,

f(xσ(1),…,xσ(n))=ε(σ) f(x1,…,xn),f(x_{\sigma(1)}, \dots, x_{\sigma(n)}) = \varepsilon(\sigma)\, f(x_1, \dots, x_n),

by decomposing σ\sigma into transpositions (Theorem 1.21).

示例

Example 2.15 (Sarrus, derived and demolished)

For n=3n = 3 the permutation formula has exactly 3!=63! = 6 terms. Listing S3\mathfrak{S}_3 by signature — id\mathrm{id}, (1 2 3)(1\,2\,3), (1 3 2)(1\,3\,2) even; (1 2)(1\,2), (1 3)(1\,3), (2 3)(2\,3) odd — gives

det⁡A=a11a22a33+a21a32a13+a31a12a23−a21a12a33−a31a22a13−a11a32a23:\det A = a_{11}a_{22}a_{33} + a_{21}a_{32}a_{13} + a_{31}a_{12}a_{23} - a_{21}a_{12}a_{33} - a_{31}a_{22}a_{13} - a_{11}a_{32}a_{23} :

precisely the “diagonals” rule of Sarrus taught in school — now a theorem, with the mysterious signs identified as signatures. The demolition: for n=4n = 4 there are 2424 permutations, of which only 88 are picked up by any diagonal-drawing scheme; Sarrus has no degree-44 version, and cofactor expansion (Theorem 2.17 (4)) takes over. Counting terms is also a warning: the permutation formula has n!n! summands, so it is a definition, not an algorithm — row reduction computes det⁡\det in O(n3)O(n^3) operations instead.

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Definition 21.1 University Mathematics — Year 3 · 第 21 章 — Differential Forms and Stokes’ Theorem

Let EE be a real vector space of dimension nn. A kk-linear alternating form on EE is a map α ⁣:Ek→R\alpha\colon E^k \to \R, linear in each variable, with α(v1,…,vk)=0\alpha(v_1, \dots, v_k) = 0 whenever two arguments are equal. Their space is written ΛkE∗\Lambda^k E^*; by convention Λ0E∗=R\Lambda^0E^* = \R. Alternation forces antisymmetry: exchanging two arguments changes the sign (expand α(…,v+w,…,v+w,… )=0\alpha(\dots, v + w, \dots, v + w, \dots) = 0), and more generally α(vσ(1),…,vσ(k))=ε(σ) α(v1,…,vk)\alpha(v_{\sigma(1)}, \dots, v_{\sigma(k)}) = \varepsilon(\sigma)\,\alpha(v_1, \dots, v_k) for every permutation σ\sigma.

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