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Let $V$ be a real $n$-dimensional oriented inner product space, and let $1 \le k < n$. I am trying to find different simple proofs for the following claim:

There is no non-zero $\omega \in \bigwedge^k (V^*) \cong (\bigwedge^k V)^*$ which is $\text{SO}$-invariant. That is

$$ Q^*\omega=\omega \,\text{ for every }\, Q \in \text{SO}(V) \Rightarrow \omega=0.$$

I present two different proofs below, but I think that there should be a simpler (or more conceptual) argument. I Would like to see more proofs.

First proof:

Consider $\ker \omega$; it can be shown that it contains a non-zero decomposable element $v_1 \wedge \cdots \wedge v_k$. (here we use $k<n$). W.L.O.G we can assume that the $v_i$ form an orthonormal set of vectors. (Indeed, we can apply the Gram–Schmidt process on them, thus replacing them with orthonormal vectors having the same span).

Now, we know that $$0=\omega(v_1 \wedge \cdots \wedge v_k)=Q^*\omega(v_1 \wedge \cdots \wedge v_k)=\omega(Qv_1 \wedge \cdots \wedge Qv_k). \tag{1}$$

Complete $v_1, \dots ,v_k$ into an orthonormal basis $v_1, \dots ,v_n$ for $V$. Since $v_1, \dots ,v_k$ can be mapped into any orthonormal $k$-tuple of the form $v_{i_1}, \dots ,v_{i_k}$ via an element of $\text{SO}(V) $, equation $(1)$ implies that $\omega$ vanishes on a basis of $\bigwedge^k V$, so it must be zero.


Second proof:

Let $e_i$ be an orthonormal basis for $V$, and let $e^i$ be its dual basis. Write $\omega=a_{i_1i_2\dots i_k}e^{i_1} \wedge e^{i_2} \dots \wedge e^{i_k}$. Then $\omega(e_{i_1} \wedge \dots \wedge e_{i_k})=a_{i_1i_2\dots i_k}$, which we rewrite as $\omega(e_I)=a_I$ where $I=(i_1i_2\dots i_k)$ is the corresponding multi-index.

Now define $Q \in \text{SO}(V)$ by setting $Q(e_1)=-e_1,Q(e_2)=-e_2,Q(e_i)=e_i$ for $i \ge 3$. We then have $$ a_I=\omega(e_I)=Q^*\omega(e_I)= \begin{cases} a_I, & \text{if $1,2 \in I$ or $1,2 \notin I$} \\ -a_I, & \text{if ($1 \in I$ and $2 \notin I$) or ($2 \in I$ and $1 \notin I$)} \end{cases}.$$

In particular, whenever $1 \in I$ and $2 \notin I$, $a_I=0$. Since the choice of the index "2" was arbitrary, we conclude that whenever $1 \in I$, we have $a_I=0$. Since the choice of the index "1" was arbitrary, this forces $\omega$ to be zero.


Comment:

When $k$ is odd and $d$ is even, one can take $Q(v)=-v \in \text{SO}(V)$. Then $\omega=Q^*\omega=-\omega$ so this gives a "one-line proof" that $\omega=0$.

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