What is alternating form? — Mathematics

Definition 2.13 University Mathematics — Year 2 · Chapter 2 — Linear Algebra

A map $f \colon E^n \to K$ is $n$ -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 $f(\dots, x + y, \dots, x + y, \dots) = 0$ ); more generally, for $\sigma \in \mathfrak{S}_n$ ,

f(x_{\sigma(1)}, \dots, x_{\sigma(n)}) = \varepsilon(\sigma)\, f(x_1, \dots, x_n),

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

Examples

Example 2.15 (Sarrus, derived and demolished)

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

\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 $n = 4$ there are $24$ permutations, of which only $8$ are picked up by any diagonal-drawing scheme; Sarrus has no degree- $4$ version, and cofactor expansion (Theorem 2.17 (4)) takes over. Counting terms is also a warning: the permutation formula has $n!$ summands, so it is a definition, not an algorithm — row reduction computes $\det$ in $O(n^3)$ operations instead.

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

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

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Examples

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