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I have the following exercise that I have not been able to solve:

Consider three simple bivectors $B_i=\alpha_i\wedge\beta_i, i=1,2,3$ in $\mathbb{R}^{4}$. The bivectors $B_i$ are linearly independent in the space $\bigwedge^2\mathbb{R}^{4}$. Let's define $V_{i}=$span$\{\alpha_i,\beta_i\}$. Also consider that dim$(V_i\cap V_j)=1$

a) Let $x_1$ be any nonzero vector in $V_2\cap V_3$ and cyclically (that is to say $x_2\in V_1\cap V_3$ and $x_3\in V_1\cap V_2$). Prove that the space span$\{x_1,x_2,x_3\}$ is three dimensional or one dimensional, but not two dimensional.

b)Now suppose that there exists $n\in\mathbb{R}^4$ such that $n_{I}B_{i}^{IJ}=0$ for $i=1,2,3$. Prove that span$\{x_1,x_2,x_3\}$ is actually three dimensional in this case.

I have tried to solve it by seeing $\{x_1,x_2,x_3\}$ as $\{x_1\}\cup\{x_2\}\cup\{x_3\}$, considering that span$\{x_1,x_2,x_3\}=$span$\{x_1\}+$span$\{x_2\}$+span$\{x_3\}$ and using the formula dim$(W_1+W_2+W_3)=$dim$(V_1)$+dim$(V_2)$+dim$(V_3)$-dim$(V_1\cap(V_2+V_3))$-dim$(V_2\cap V_3)$, but the formula doesn't help to prove that dim(span$\{x_1,x_2,x_3\})\neq2$

The inequality dim$(W_1+W_2+W_3)\leq$ dim$(W_1)$+ dim$(W_2)$+ dim$(W_3)$- dim$(W_1\cap W_2)$- dim$(W_1\cap W_3)$- dim$(W_2\cap W_3)$+ dim$(W_1\cap W_2\cap W_3)$ doesn't work either.

And I have found that trying to prove the second item is even more difficult to me.

Could anybody help me??

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1 Answer 1

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Hint

(a) It's straightforward to find examples for which $\dim \operatorname{span}\{x_1, x_2, x_3\}$ is $1$, $3$.

So, suppose $\dim \operatorname{span} \{x_1, x_2, x_3\} < 3$. Then (by relabeling if necessary) we may assume that $x_3 \in S := \operatorname{span}\{x_1, x_2\}$. What can we conclude about $V_1, V_2, V_3$ if $\dim S = 2$?

(b) Suppose $\dim \operatorname{span} \{x_1, x_2, x_3\} = 1$. Then, we may take $\alpha_1 = \alpha_2 = \alpha_3 = x_1$ and thus $B_i = x_1 \wedge \beta_i$ for $i=1,2,3$. What can we then conclude from the linear independence of $\{B_1, B_2, B_3\}$?

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  • $\begingroup$ In b) how do you use the assertion $n_I B^{IJ}=0$ ? $\endgroup$ Apr 15, 2021 at 23:51
  • $\begingroup$ If $\{\beta_1, \beta_2, \beta_3\}$ is linearly dependent, then so is $\{B_1, B_2, B_3\}$, which contradicts our hypothesis, i.e., $\{\beta_1, \beta_2, \beta_3\}$ is linearly independent. Since we're proving the contrapositive of the original statement, it's enough to show that that the common kernel of the contraction maps $x_I \to x_I B_i^{IJ}$ is trivial, i.e., that the only $n_I$ satisfying the conditions $n_I B_i^{IJ}$ is the zero vector. $\endgroup$ Apr 16, 2021 at 4:01

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