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Let $V$ be a real $n$-dimensional vector space, and let $W \le \bigwedge^k V$ be a subspace . Suppose that $\dim W \ge 2$. Does $W$ contain a non-zero decomposable element?

If $\dim W=1$, then clearly we can take $W=\text{span} (\sigma)$ for some non-decomposable $\sigma \in \bigwedge^k V$.

Asaf Shachar
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1 Answers1

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No. Consider the $3$-dimensional subspace $W\subset\Lambda^2\mathbb R^4$ spanned by

$$B_1=e_2e_3+e_1e_4,\;B_2=e_3e_1+e_2e_4,\;B_3=e_1e_2+e_3e_4$$

where $e_i$ span $\mathbb R^4$. See that $B_iB_j=0$ for $i\neq j$, and $B_iB_i=2e_1e_2e_3e_4$.

Thus, for any $X\in W\setminus\{0\}$,

$$X=x_1B_1+x_2B_2+x_3B_3$$

$$XX=2e_1e_2e_3e_4(x_1^2+x_2^2+x_3^2)\neq0$$

which shows that $X$ is not decomposable.

These bivectors correspond to isoclinic rotations in $SO(\mathbb R^4)$.

mr_e_man
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  • Hi, I wanted to thank you for this great and very beautiful answer. I am sorry it took me so long to accept it. I am interested in your comment that these bivectors correspond to isoclinic rotations in $4D$. Can you please explain how this is so? Thanks. – Asaf Shachar Mar 19 '23 at 13:36
  • Bivectors are isomorphic to antisymmetric matrices, by $v\wedge w\cong vw^\top-wv^\top$ (where $\top$ is the matrix transpose). Then the matrix exponential maps to the special orthogonal group. -- Alternatively, bivectors are isomorphic to elements of the Clifford algebra, by $v\wedge w\cong\tfrac12(vw-wv)$, and there the exponential maps to the spin group. Given a bivector $B$, the rotation operation is $v\mapsto\exp(B/2)v\exp(-B/2)$. See How to calculate the exponential of a 4D bivector?. – mr_e_man Mar 20 '23 at 00:16