Then, if $k>n$, can it be shown that $\psi$ is the trivial one which sends every element to $0 $?

Question

Let $\psi:V\times \cdots \times V\to \mathbb R$ be an antisymetric $k$ tensor on $V$, which is $n$ dimensional.

Then, if $k>n$, can it be shown that $\psi$ is the trivial one which sends every element to $0 $?

Answer 1

$\psi$ is antisymmetric iff $\psi(v_1,\ \cdots,\ v_k)=0$ where $v_i=v_j$ and $i<j$.

Let $\{ e_i\}_{i=1}^n$ be a basis for $V$. Since $\psi$ is $k$-linear, $\psi( e_{i_1},\cdots ,e_{i_k})$ determines $\psi$.

Note that $k> n$ so that there exists $j<l$ s.t. $e_{i_j}=e_{i_l}$ So $ \psi( e_{i_1},\cdots ,e_{i_k})=0$.

Answer 2

Fix a basis of $V$. By multilinearity it is sufficient to prove every input consisting of basis vectors gets sent to zero. But inputting $k>n$ vectors from a basis of size $n$ results in a redundant entry, which by antisymmetry must map to zero.