$n$-linear alternating form with $\dim{V}<n$ $\overset{?}{\text{is}}$ the $0$-form

Question

Prove that every $n$-linear alternating form on a vector space of dimension less than $n$ is the zero form.

Answer 1

Let $k$ be a field, and let $V$ be a vector space over $k$, with $\dim V <n$. Let $\omega$ be an alternating $n$-form on $V$. Then $\omega : V^n \to k$ is an alternating, multilinear map. For general input $v_1,\ldots,v_n$, there exists a linear dependence $\sum c_i v_i=0$ among the $v_i$, by dimension of $V$. If $c_n \neq 0$ (e.g.), we can scale $c_n=1$. Then solving for $v_n$, we may write $$\omega(v_1,\ldots,v_n)=\omega\left(v_1,v_2,\ldots,v_{n-1},-\sum_{i<n} c_iv_i\right)=-\sum_{i<n}c_i \omega\left(v_1,v_2,\ldots,v_{n-1},v_i\right),$$ in which the last step uses multilinearity. Since $\omega$ is alternating, switching the $i$th and $n$th inputs negates the value of $\omega$. Yet here $\omega(v_1,\ldots,v_{n-1},v_i)$ is clearly invariant under such a permutation, hence $$\omega(v_1,\ldots,v_{n-1},v_i)=0$$ for all $i$. Thus $\omega$ is the zero map, as desired.

In general, we can show (by constructing a basis) that the dimension of the space of alternating $n$-forms on a $d$-dimensional space is given by $\binom{d}{n}$, which vanishes for $n > d$.

Answer 2

Just use the definition of alternating, along with the fact that multilinearity says that the function is totally determined by what happens when you put basis vectors in each slot.

Answer 3

Let $\omega$ be an $n$-multilinear, alternating form on $V$, where $\dim(V)=m<n$. Let $\{e_1,\ldots,e_m\}$ be any basis of $V$. Because $\omega$ is multilinear, $\omega$ is determined entirely by its values on $n$-tuples of elements from $\{e_1,\ldots,e_m\}$ (apply $\omega$ to an $n$-tuple of arbitrary vectors; write each of them as a linear combination of the $e_i$'s, and use the linearity of $\omega$ in each entry). But $n>m$, so every such $n$-tuple has a repeat...