# Is every rotationally-invariant $k$-covector zero?

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 $$k). W.L.O.G we can assume that the $$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 $$\omega=0$$.