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For a $k$-form $0\ne \omega^k\in \bigwedge^k(V^*)$ it is $M(\omega^k)\subset V^*$ with $$M(\omega^k)=\{\eta^1\in\bigwedge^1(V^*):\eta^1\wedge\omega^k=0\}.$$

Prove that: dim $M(\omega^k)\le k$.

I know that dim $M(\omega^k)\le n$ since dim $\bigwedge^1(V^*)=n$ and that $k$ linear forms $\sigma_1,...,\sigma_k \in V^*$ are linear independent if and only if $\sigma_1\wedge...\wedge\sigma_k\ne0$.

Anyway, I find no useful approach to this. Any help is greatly appreciated!

$\color{red}{EDIT}$:

I got the hint to first make it clear for n=4, k=2. So let $\{v_1,v_2,v_3,v_4\}$ be a basis of $V$, $\{\sigma_1,\sigma_2,\sigma_3,\sigma_4\}$ the dual basis, i.e. the basis of $V^*=\bigwedge^1(V^*)$. So we have dim($\bigwedge^k(V^*)$)$={4 \choose 2}=6$, i.e. the basis of $\bigwedge^k(V^*)$ is $\{\sigma_1\wedge\sigma_2,\sigma_1\wedge\sigma_3,\sigma_1\wedge\sigma_4,\sigma_2\wedge\sigma_3,\sigma_2\wedge\sigma_4,\sigma_3\wedge\sigma_4\}$. Now I write $\eta^1$ and $\omega^k$ as linear combination of the basis elements: $\eta^1=\sum_{i=1}^4 a_i\sigma_i$, $\omega^k=\sum_{i\ne j, i<j} b_k(\sigma_i\wedge\sigma_j)$. $$\eta^1\wedge\omega^k=0 \Leftrightarrow \sum_{i=1}^4 a_i\sigma_i \wedge \sum_{i\ne j, i<j} b_k(\sigma_i\wedge\sigma_j)=0 \Leftrightarrow$$ $$(a_1\sigma_1+a_2\sigma_2+a_2\sigma_3+a_4\sigma_4)\wedge[b_1(\sigma_1\wedge\sigma_2)+b_2(\sigma_1\wedge\sigma_3)+b_3(\sigma_1\wedge\sigma_4)+b_4(\sigma_2\wedge\sigma_3)+b_5(\sigma_2\wedge\sigma_4)+b_6(\sigma_3\wedge\sigma_4)]$$=$$a_1b_4(\sigma_1\wedge\sigma_2\wedge\sigma_3)+a_1b_5(\sigma_1\wedge\sigma_2\wedge\sigma_4)+a_1b_6(\sigma_1\wedge\sigma_3\wedge\sigma_4)+a_2b_2(\sigma_2\wedge\sigma_1\wedge\sigma_3)+a_2b_3(\sigma_2\wedge\sigma_1\wedge\sigma_4)+a_2b_6(\sigma_2\wedge\sigma_3\wedge\sigma_4)+a_3b_1(\sigma_3\wedge\sigma_1\wedge\sigma_2)+a_3b_3(\sigma_3\wedge\sigma_1\wedge\sigma_4)+a_3b_5(\sigma_3\wedge\sigma_2\wedge\sigma_4)+a_4b_1(\sigma_4\wedge\sigma_1\wedge\sigma_2)+a_4b_2(\sigma_4\wedge\sigma_1\wedge\sigma_3)+a_4b_4(\sigma_4\wedge\sigma_2\wedge\sigma_3)$$=$$(a_1b_4-a_2b_2+a_3b_1)\sigma_1\wedge\sigma_2\wedge\sigma_3+(a_1b_5-a_2b_3+a_4b_1)\sigma_1\wedge\sigma_2\wedge\sigma_4+(a_1b_6-a_3b_3+a_4b_2)\sigma_1\wedge\sigma_3\wedge\sigma_4+(a_2b_6-a_3b_5+a_4b_5)\sigma_2\wedge\sigma_3\wedge\sigma_4$$=$$0$$

But how do I see now that dim$M(\omega^k)\le 2$ here?

newbie
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2 Answers2

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Hint: take a basis $v_1,\dots,v_n$ of $V$. This gives you a basis of $\bigwedge^k(V^*)$ and a basis of $\bigwedge^1(V^*)$. Write $\eta^1$, $\omega^k$ in terms of these basis and impose the condition $\eta^1\in M(\omega^k)$. This will give you a certain number of equations on the coefficients. Try to do it for $n=4$, $k=2$ to convince yourself of the result. The general case seems more comlicated but at the end of the day it all boils down to count the number of coefficients you need to write $\eta^1$ and $\omega^k$ in terms of the basis (which actually means to find the dimension of $\bigwedge^1(V^*)$ and $\bigwedge^k(V^*)$) and to count the number of equations you obtain imposing the condition $\eta^1\wedge\omega^k=0$.

Uskebasi
  • 1,523
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As in the question, let $\omega \in \bigwedge^k V^*$ be a nonzero $k$-form, and consider the set $$ M(\omega) = \{ \eta \in V^* \mid \eta \wedge \omega = 0 \}$$ This is precisely the kernel of the linear map $$ \psi: V^* \to \bigwedge^{k+1} V^*, \quad \psi(\eta) = \eta \wedge \omega$$ Suppose that $\dim V^* = m$. Let $f_1, \ldots, f_r$ be a basis for $\ker \psi$, and extend this to a basis $f_1, \ldots, f_m$ of $V^*$. For any subset $I = \{i_1 < \ldots < i_k\} \subseteq \{1, \ldots, m\}$, with $|I| = k$, let $f_I = f_{i_1} \wedge \cdots \wedge f_{i_k}$. The $f_I$ form a basis for $\bigwedge^k V^*$, as $I$ ranges over all $k$-element subsets of $\{1, \ldots, m\}$.

Write $\omega = \sum_I a_I f_I$ in this basis. If $i \in \{1, \ldots, r\}$ (so $f_i \in \ker \psi$), then $f_i \wedge \omega = \psi(f_i) = 0$. But $$ 0 = f_i \wedge \omega = \sum_{I, i \notin I}a_I f_i \wedge f_I$$ where the sum ranges over those subsets $I$ such that $i \notin I$. All the $f_i \wedge f_I$ appearing in the sum are linearly independent in $\bigwedge^{k+1}V^*$, and hence all the corresponding $a_I = 0$. Since we can do this for any $i$ between $1$ and $r$, the only coefficients $a_I$ which are nonzero must have $I$ containing the set $\{1, \ldots, r\}$. But $r = \dim \ker \psi$, and now we have found $r \leq k$.

Joppy
  • 12,875