Return to Question

Let $V$ be a vector space over the finite field $\mathbb{F}_q$ of dimension $n\geq 6$. Let $\omega\in\bigwedge^{n-3}V$ be a nonzero element. Define the anihalator subspace of $\omega$ by $V_\omega=\{z\in V: \omega\wedge z= 0\}$. It is well known that $0\leq \dim V_\omega \leq n-3$. Suppose for this given $\omega$ we have $\dim V_\omega= n-k$ where $k\geq 5$. I want to prove that $$|\{x\in V: \omega\wedge x\in \bigwedge^{n-2}V\text{ is decomposable}\}|\leq q^{n-k}(q^{k-3}-1)(q^2+1). $$

By a decomposable element of $\bigwedge^{d}V$, we mean a nonzero element that can be written as $v_1\wedge\ldots\wedge v_k$ for some linearly independent elements $\{v_1,\ldots,v_k\}$ of $V$. I tried that and could prove this when $k=6$ but I am not able to prove this in general.

Let $V$ be a vector space of dimension $n\geq 6$ over the finite field $\mathbb{F}_q$. Let $\omega\in\bigwedge^{n-3}V$ be a nonzero element. Define the annihilator subspace of $\omega$ by $V_\omega=\{z\in V: \omega\wedge z= 0\}$. It is well known that $0\leq \dim V_\omega \leq n-3$. Suppose for this given $\omega$ that we have $\dim V_\omega= n-k$ where $k\geq 5$. I want to prove that $$\left|\{x\in V: \omega\wedge x\in \bigwedge^{n-2}V\text{ is decomposable}\}\right|\leq q^{n-k}(q^{k-3}-1)(q^2+1). $$

By a decomposable element of $\bigwedge^{d}V$, we mean a nonzero element that can be written as $v_1\wedge\ldots\wedge v_d$ for some linearly independent elements $\{v_1,\ldots,v_d\}$ of $V$. I can prove this when $k=6$ but I am not able to prove this in general.

Let $V$ be a vector space over the finite field $\mathbb{F}_q$ of dimension $n\geq 6$. Let $\omega\in\bigwedge^{n-3}V$ be a nonzero element. Define the anihalator subspace of $\omega$ by $V_\omega=\{z\in V: \omega\wedge z\neq 0\}$. It is well known that $0\leq \dim V_\omega \leq n-3$. Suppose for this given $\omega$ we have $\dim V_\omega= n-k$ where $k\geq 5$. I want to prove that $$|\{x\in V: \omega\wedge x\in \bigwedge^{n-2}V\text{ is decomposable}\}|\leq q^{n-k}(q^{k-3}-1)(q^2+1). $$

By a decomposable element of $\bigwedge^{d}V$, we mean a nonzero element that can be written as $v_1\wedge\ldots\wedge v_k$ for some linearly independent elements $\{v_1,\ldots,v_k\}$ of $V$. I tried that and could prove this when $k=6$ but I am not able to prove this in general.

Let $V$ be a vector space over the finite field $\mathbb{F}_q$ of dimension $n\geq 6$. Let $\omega\in\bigwedge^{n-3}V$ be a nonzero element. Define the anihalator subspace of $\omega$ by $V_\omega=\{z\in V: \omega\wedge z= 0\}$. It is well known that $0\leq \dim V_\omega \leq n-3$. Suppose for this given $\omega$ we have $\dim V_\omega= n-k$ where $k\geq 5$. I want to prove that $$|\{x\in V: \omega\wedge x\in \bigwedge^{n-2}V\text{ is decomposable}\}|\leq q^{n-k}(q^{k-3}-1)(q^2+1). $$

By a decomposable element of $\bigwedge^{d}V$, we mean a nonzero element that can be written as $v_1\wedge\ldots\wedge v_k$ for some linearly independent elements $\{v_1,\ldots,v_k\}$ of $V$. I tried that and could prove this when $k=6$ but I am not able to prove this in general.