Skip to content

Alternating Multilinear Forms#

Definition: Alternating Multilinear Forms

A multilinear form \(f: V^n \to F\) is alternating if for all \(\mathbf{v}_1, \dotsc, \mathbf{v}_n \in V\) we have

\[ \mathbf{v}_i = \mathbf{v}_j \text{ with } i \ne j \implies f(\mathbf{v}_1, \dotsc, \mathbf{v}_n) = 0. \]

Intuition

A multilinear form is alternating if it is zero whenever any two of its arguments are equal.

Theorem: Argument Swap \(\implies\) Sign Change

Let \(f: V^n \to F\) be a multilinear form.

If \(f\) is alternating, then

\[ f(\mathbf{v}_1, \dotsc, \mathbf{v}_i, \dotsc, \mathbf{v}_j, \dotsc, \mathbf{v}_n) = -f(\mathbf{v}_1, \dotsc, \mathbf{v}_j, \dotsc, \mathbf{v}_i, \dotsc, \mathbf{v}_n) \]

for all \(i, j \in \{1, \dotsc, n\}\) with \(i \ne j\).

Intuition

Swapping any two arguments of an alternating multilinear form switches the sign of the result.

Proof

Since \(f\) is alternating, we have

\[ f(\dotsc, \mathbf{u} + \mathbf{w}, \dotsc, \mathbf{u} + \mathbf{w}, \dotsc) = 0 \]

for all \(\mathbf{u}, \mathbf{w} \in V\). Furthermore, since \(f\) is a multilinear form, we have the following:

\[ \begin{aligned} f(\dotsc, \mathbf{u} + \mathbf{w}, \dotsc, \mathbf{u} + \mathbf{w}, \dotsc) & = f(\dotsc, \mathbf{u}, \dotsc, \mathbf{u} + \mathbf{w}, \dotsc) + f(\dotsc, \mathbf{w}, \dotsc, \mathbf{u} + \mathbf{w}, \dotsc) \\ & = f(\dotsc, \mathbf{u}, \dotsc, \mathbf{u}, \dotsc) + f(\dotsc, \mathbf{u}, \dotsc, \mathbf{w}, \dotsc) + f(\dotsc, \mathbf{w}, \dotsc, \mathbf{u}, \dotsc) + f(\dotsc, \mathbf{w}, \dotsc, \mathbf{w}, \dotsc) \\ & = 0 + f(\dotsc, \mathbf{u}, \dotsc, \mathbf{w}, \dotsc) + f(\dotsc, \mathbf{w}, \dotsc, \mathbf{u}, \dotsc) + 0\end{aligned} \]

Combining the two results, we obtain

\[ f(\dotsc, \mathbf{u}, \dotsc, \mathbf{w}, \dotsc) + f(\dotsc, \mathbf{w}, \dotsc, \mathbf{u}, \dotsc) = 0 \]

and so

\[ f(\dotsc, \mathbf{u}, \dotsc, \mathbf{w}, \dotsc) = - f(\dotsc, \mathbf{w}, \dotsc, \mathbf{u}, \dotsc) \]

for all \(\mathbf{u}, \mathbf{v} \in V\).

Theorem: Linear Dependence \(\implies\) Zero

Let \(f: V^n \to F\) be an alternating multilinear form.

If \(\mathbf{v}_1, \dotsc, \mathbf{v}_n \in V\) are linearly dependent, then \(f(\mathbf{v}_1, \dotsc, \mathbf{v}_n) = 0\).

Proof

Since \(\mathbf{v}_1, \dotsc, \mathbf{v}_n\) are linearly dependent, we know that at least one of \(\mathbf{v}_1, \dotsc, \mathbf{v}_n\) can be represented as a linear combination of the others. Assume, without loss of generality, that this is \(\mathbf{v}_n\):

\[ \mathbf{v}_n = \sum_{i = 1}^{n-1} \lambda_i \mathbf{v}_i \]

Since \(f\) is multilinear, we have the following:

\[ f\left(\mathbf{v}_1, \dotsc, \mathbf{v}_{n-1}, \sum_{i = 1}^{n-1} \lambda_i \mathbf{v}_i\right) = \sum_{i=1}^{n-1} \lambda_i f\left(\mathbf{v}_1, \dotsc, \mathbf{v}_{n-1}, \mathbf{v}_i\right) \]

For each term, \(\mathbf{v}_i\) appears twice in \(f\) - once at position \(i\) and once at position \(n\). Since \(f\) is alternating, this makes every term equal to zero.

Theorem: Basis \(\implies\) Unique Alternating Multilinear Form

Let \(f, g: V^n \to F\) be alternating multilinear forms.

If \(V\) is \(n\)-dimensional and \(\mathbf{b}_1, \dotsc, \mathbf{b}_n \in V\) are a basis for \(V\) and \(f(\mathbf{b}_1, \dotsc, \mathbf{b}_n) = g(\mathbf{b}_1, \dotsc, \mathbf{b}_n)\), then \(f = g\).

Proof

TODO