Alternating forms

A $k$ -linear form $w$ is skew-symmetric if $π w = - w$ for every odd permutation $π$ in $𝒮_{k}$ . Equivalently, $w$ is skew-symmetric if $π w = (sgn π) w$ for every permutation $π$ in $𝒮_{k}$ . (If $π w = (sgn π) w$ for all $π$ , then, in particular, $π w = - w$ whenever $π$ is odd. If, conversely, $π w = - w$ for all odd $π$ , then, given an arbitrary $π$ , factor it into transpositions, say, $π = τ_{1} \dots τ_{q}$ , observe that $sgn π = (- 1)^{q}$ , and, since $π w = (- 1)^{q} w$ , conclude that $π w = (sgn π) w$ , as asserted. This proof makes tacit use of the unproved but easily available fact that if $σ$ and $τ$ are permutations in $𝒮_{k}$ , then $(σ τ) w = (σ (τ w))$ .) The set of all skew-symmetric $k$ -linear forms is a subspace of the space of all $k$ -linear forms. To get a non-trivial example of a skew-symmetric bilinear form $w$ , let $y_{1}$ and $y_{2}$ be linear functionals and write $w (x_{1}, x_{2}) = y_{1} (x_{1}) y_{2} (x_{2}) - y_{1} (x_{2}) y_{2} (x_{1}) .$ More generally, if $w$ is an arbitrary $k$ -linear form, a skew-symmetric $k$ -linear form can be obtained from $w$ by forming $\sum (sgn π) π w$ , where the summation is extended over all permutations $π$ in $𝒮_{k}$ .

A $k$ -linear form $w$ is called alternating if $w (x_{1}, \dots, x_{k}) = 0$ whenever two of the $x$ ’s are equal. (Note that if $k = 1$ , then this condition is vacuously satisfied.) The set of all alternating $k$ -linear forms is a subspace of the space of all $k$ -linear forms. There is an important relation between alternating and skew-symmetric forms.

Theorem 1. Every alternating multilinear form is skew-symmetric.

Proof. Suppose that $w$ is an alternating $k$ -linear form, and that $i$ and $j$ are integers, $1 \leq i < j \leq k$ . If $x_{1}, \dots, x_{k}$ are vectors, we write $w_{0} (x_{i}, x_{j}) = w (x_{1}, \dots, x_{k});$ if the $x$ ’s other than $x_{i}$ and $x_{j}$ are held fixed (temporarily), then $w_{0}$ is an alternating bilinear form of its two arguments. Since, by bilinearity, $w_{0} (x_{i} + x_{j}, x_{i} + x_{j}) = w_{0} (x_{i}, x_{i}) + w_{0} (x_{i}, x_{j}) + w_{0} (x_{j}, x_{i}) + w_{0} (x_{j}, x_{j}),$ and since, by the alternating character of $w_{0}$ , the left side and the two extreme terms of the right side of this equation all vanish, we see that $w_{0} (x_{j}, x_{i}) = - w_{0} (x_{i}, x_{j})$ . This, however, says that $(i, j) w (x_{1}, \dots, x_{k}) = - w (x_{1}, \dots, x_{k}),$ or, since the $x$ ’s are arbitrary, that $(i, j) w = - w$ . Since every odd permutation $π$ is the product of an odd number of transpositions, such as $(i, j)$ , it follows that $π w = - w$ for every odd $π$ , and the proof of the theorem is complete. ◻

The connection between alternating forms and skew-symmetric ones involves one subtle point. Consider the following “proof” of the converse of Theorem 1: if $w$ is a skew-symmetric $k$ -linear form, if $1 \leq i < j \leq k$ , and if $x_{1}, \dots, x_{k}$ are vectors such that $x_{i} = x_{j}$ , then $(i, j) w (x_{1}, \dots, x_{k}) = w (x_{1}, \dots, x_{k})$ since $x_{i} = x_{j}$ , and at the same time, $(i, j) w (x_{1}, \dots, x_{k}) = - w (x_{1}, \dots, x_{k})$ since $w$ is skew-symmetric; consequently $w (x_{1}, \dots, x_{k}) = - w (x_{1}, \dots, x_{k})$ , so that $w$ is alternating. This argument is wrong; the trouble is in the inference “if $w = - w$ , then $w = 0$ .” If we examine that inference in more detail, we find that it is based on the following reasoning: if $w = - w$ , then $w + w = 0$ , so that $(1 + 1) w = 0$ . This is correct. The trouble is that in certain fields $1 + 1 = 0$ , and therefore the inference from $(1 + 1) w = 0$ to $w = 0$ is not justified; the converse of Theorem 1 is, in fact, false for vector spaces over such fields.

Theorem 2. If $x_{1}, \dots, x_{k}$ are linearly dependent vectors and if $w$ is an alternating $k$ -linear form, then $w (x_{1}, \dots, x_{k}) = 0$ .

Proof. If $x_{i} = 0$ for some $i$ , the conclusion is trivial. If all the $x_{i}$ are different from $0$ , we apply the theorem of Section: Linear combinations to find an $x_{h}$ , $2 \leq h \leq k$ , that is a linear combination of the preceding ones. If, say, $x_{h} = \sum_{i = 0}^{h - 1} α_{i} x_{i}$ , we replace $x_{h}$ in $w (x_{1}, \dots, x_{k})$ by this expansion, and use the linearity of $w$ in its $h$ -th argument, and draw the desired conclusion by an argument of the same type. ◻

In one extreme case (namely, when $k = n$ ) a sort of converse of Theorem 2 is true.

Theorem 3. If $w$ is a non-zero alternating $n$ -linear form, and if $x_{1}, \dots, x_{n}$ are linearly independent vectors, then $w (x_{1}, \dots, x_{n}) \neq 0$ .

Proof. Since ( Section: Dimension , Theorem 2) the vectors $x_{1}, \dots, x_{n}$ form a basis, we may, given an arbitrary set of $n$ vectors $y_{1}, \dots, y_{n}$ , write each $y$ as a linear combination of the $x$ ’s. If we replace each $y$ in $w (y_{1}, \dots, y_{n})$ by the corresponding linear combination of $x$ ’s and expand the result by multilinearity, we obtain a long linear combination of terms such as $w (z_{1}, \dots, z_{n})$ , where each $z$ is one of the $x$ ’s. If, in such a term, two of the $z$ ’s coincide, then, since $w$ is alternating, that term must vanish. If, on the other hand, all the $z$ ’s are distinct, then $w (z_{1}, \dots, z_{n}) = π w (x_{1}, \dots, x_{n})$ for some permutation $π$ . Since (Theorem 1) $w$ is skew-symmetric, it follows that $w (z_{1}, \dots, z_{n}) = (sgn π) w (x_{1}, \dots, x_{n})$ . If $w (x_{1}, \dots, x_{n}) = 0$ , it would follow that $w (z_{1}, \dots, z_{n}) = 0$ , and hence that $w (y_{1}, \dots, y_{n}) = 0$ for all $y_{1}, \dots, y_{n}$ , contradicting the assumption that $w \neq 0$ . ◻

The proof (not the statement) of this result yields a valuable corollary.

Theorem 4. Any two alternating $n$ -linear forms are linearly dependent.

Proof. Suppose that $w_{1}$ and $w_{2}$ are alternating $n$ -linear forms and that ${x_{1}, \dots, x_{n}}$ is a basis. Given any $n$ vectors $y_{1}, \dots, y_{n}$ , write each of them as a linear combination of the $x$ ’s, and, just as above, replace each of them, in both $w_{1} (y_{1}, \dots, y_{n})$ and $w_{2} (y_{1}, \dots, y_{n})$ , by the corresponding linear combination. It follows that each of $w_{1} (y_{1}, \dots, y_{n})$ and $w_{2} (y_{1}, \dots, y_{n})$ is a linear combination (the same linear combination) of terms such as $w_{1} (z_{1}, \dots, z_{n})$ and $w_{2} (z_{1}, \dots, z_{n})$ , where each $z$ is one of the $x$ ’s. Since $w_{1} (x_{1}, \dots, x_{n})$ and $w_{2} (x_{1}, \dots, x_{n})$ are scalars, they are linearly dependent, so that there exist scalars $α_{1}$ and $α_{2}$ not both zero, such that $α_{1} w_{1} (x_{1}, \dots, x_{n}) + α_{2} w_{2} (x_{1}, \dots, x_{n}) = 0$ ; from these facts we may infer that $α_{1} w_{1} + α_{2} w_{2} = 0$ , as asserted. ◻

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