----- > [!proposition] Proposition. ([[2-form decomposes iff self-wedge vanishes]]) > Show that an $\omega \in \Lambda^{2}\big( (\mathbb{R}^{n})^{*} \big)$ can be written as a [[algebra of alternating multilinear forms|wedge product]] of [[dual vector space|covectors]] $\omega= \eta_{1} \wedge \eta_{2}$, $\eta_{i} \in (\mathbb{R}^{n})^{*}$, if and only if $\omega \wedge \omega =0$. > This result does not generalize to $\omega \in \Lambda^{k}( (\mathbb{R}^{n})^{*} )$ for $k>2$ and the degrees of the forms $\eta_{1}, \eta_{2}$ summing up to $k$. - [ ] provide example > [!proof]- Proof. ([[2-form decomposes iff self-wedge vanishes]]) > If $\omega=\eta_{1}\wedge \eta_{2}$, then $\omega \wedge \omega=(\eta_{1} \wedge \eta_{2}) \wedge (\eta_{1} \wedge \eta_{2})$; skew-symmetry gives $\omega \wedge \omega=-(\eta_{1} \wedge \eta_{1}) \wedge (\eta_{2} \wedge \eta_{2})=0$. > Conversely, suppose $\omega \wedge \omega=0$. Induct on $n$. For $n=0$ the result holds vacuously. For $n=1$ it holds trivially, because on the one hand$\begin{align} \omega \in \Lambda^{2}( (\mathbb{R}^{1})^{*} ) & \iff \omega(a,b)=-\omega(b,a) \text{ for all }a,b \in \mathbb{R} \\ \end{align}$ while on the other hand $\omega \in \Lambda^{2} ( (\mathbb{R}^{1})^{*}) \iff \omega(a,b)=ab \ \omega(1,1)=\omega(b, a) \text{ for all } a, b \in \mathbb{R}$ from which it follows $\Lambda^{2}\big( (\mathbb{R}^{1})^{*} \big)=(0)$. > Assume the result holds in dimension $n-1$, and consider the case of dimension $n$. > Fixing basis $dx^{1},\dots,dx^{n}$ of $(\mathbb{R}^{n})^{*}$, one has $\begin{align} \omega &= \sum_{1 \leq i < j \leq n} \omega_{ij} \ dx^{i} \wedge dx^{j} \\ & = \left( \sum_{1 \leq i < j \leq n - 1} \omega_{ij} \ dx^{i} \wedge dx^{j} \right) + \left( \sum_{1 \leq i \leq n-1} \omega_{i n } \ dx^{i} \right) \wedge dx^{n} \\ & = \widetilde{\omega} + \gamma \wedge dx^{n}. \end{align}$ Where, letting $W=\text{span}(dx^{1},\dots,dx^{n-1})$, $\widetilde{\omega } \in \Lambda^{2} (W) \cong \Lambda^{2} ( (\mathbb{R}^{n-1})^{*} )$ and $\gamma \in \Lambda^{1}( W)$. Note that the induction hypothesis implies $\widetilde{\omega}=\alpha_{1} \wedge \alpha_{2}$ for $1$-forms $\alpha_{1},\alpha_{2}$ if $\widetilde{\omega} \wedge \widetilde{\omega}=0$. Also note that $0=\omega \wedge \omega =( \widetilde{\omega} + \gamma \wedge dx^{n} ) \wedge( \widetilde{\omega} + \gamma \wedge dx^{n} ) = 2 \widetilde{\omega} \wedge \gamma \wedge dx^{n} + \widetilde{\omega} \wedge \widetilde{\omega},$ i.e., $-\widetilde{\omega} \wedge \widetilde{\omega}=2\widetilde{\omega} \wedge \gamma \wedge dx^{n}.$ Since $\widetilde{\omega}$ is based on dual vectors whose span does not contain $dx^{n}$, this equality can only happen if the left- and right-hand side are both zero, from which it follows that $\widetilde{\omega} \wedge \widetilde{\omega}=0$ and $\widetilde{\omega} \wedge \gamma=0$. > Now using the induction hypothesis we write $\widetilde{\omega}=\alpha_{1} \wedge \alpha_{2}$ and obtain $\alpha_{1} \wedge \alpha_{2} \wedge \gamma=0.$ If $\alpha_{1},\alpha_{2},\gamma$ were linearly independent as vectors in $(\mathbb{R}^{n})^{*}$, then they would extend to a basis of $(\mathbb{R}^{n})^{*}$, inducing a basis of $\Lambda^{3}( (\mathbb{R}^{n})^{*} )$ which contains $\alpha_{1} \wedge \alpha_{2} \wedge \gamma$ as an element. In particular, we would have $\alpha_{1} \wedge \alpha_{2} \wedge \gamma \neq 0$. So it must be the case that there exists a nontrivial relation $a_{1}\alpha_{1}+a_{2}\alpha_{2}+\lambda \gamma = 0$ among $\alpha_{1},\alpha_{2},\gamma$. If $\lambda=0$ then $\widetilde{\omega}=\alpha_{1} \wedge \alpha_{2}=\alpha_{1} \wedge\left( -\frac{a_{1}}{a_{2}}\alpha_{1} \right)=0$ and therefore $\omega=\gamma \wedge dx^{n}$ is a wedge of 1-forms. Else $\lambda \neq 0$ and we have $\gamma=-(\frac{a_{1}}{\lambda}\alpha_{1} + \frac{a_{2}}{\lambda}\alpha_{2})=b_{1} \alpha_{1} + b_{2}\alpha_{2}$. In this case, $\omega= \widetilde{\omega} + \gamma \wedge dx^{n}= \alpha_{1} \wedge \alpha_{2} + b_{1} \alpha_{1} \wedge dx^{n} + b_{2} \alpha_{2} \wedge dx^{n}$ and we may identify $\omega$ with a $3$-form. The map $\begin{align} (\mathbb{R}^{3})^{*} &\xrightarrow{\phi} \Lambda^{3}((\mathbb{R}^{3})^{*}) \\ f & \mapsto \omega \wedge f \end{align}$ has 2-dimensional kernel. Fix a basis $u^{1}, u^{2}$ for $\text{ker }\phi$, and extend to a basis $u^{1},u^{2},u^{3}$ for $(\mathbb{R}^{3})^{*}$. In this new basis, one has $\begin{align} \omega = \omega_{12} \ u^{1} \wedge u^{2} + \omega_{13} \ u^{1} \wedge u^{3} + \omega_{23} \ u^{2} \wedge u^{3} \end{align}$ for some scalars $\omega_{12},\omega_{13}, \omega_{23}$. It is apparent that $0=\omega \wedge u^{1}=\omega_{23}\ u^{2} \wedge u^{3} \text{ and } 0=\omega \wedge u^{2}=\omega_{13} \ u^{1} \wedge u^{3}.$ Hence $\omega_{23}=\omega_{13}=0$ and we are left with $\omega=\omega_{12} \ u^{1} \wedge u ^{2}$, which proves the result. ^344be0 ---- Also tried a proof without induction, but didn't succeed: Assume $\omega \wedge \omega=0$. We have $\begin{align} 0=\omega \wedge \omega &= \left( \sum_{1 \leq i < j \leq n} \omega_{ij} \ dx^{i} \wedge dx^{j} \right) \wedge \left( \sum_{1 \leq k <\ell \leq n} \omega_{k \ell} \ dx^{k} \wedge dx^{\ell} \right) \\ &= \sum_{i < j} \sum_{k < \ell} \omega_{ij} \omega_{k \ell} \ dx^{i} \wedge dx^{j} \wedge dx^{k} \wedge dx^{\ell} \\ &= \sum_{i<j<k<\ell} (\omega_{ij}\omega_{k\ell} - \omega_{ik}\omega_{j\ell} + \omega_{i\ell}\omega_{jk}) \ dx^i \wedge dx^j \wedge dx^k \wedge dx^\ell \end{align}$ the last summation is a [[linear combination]] of [[linearly independent]] elements of $\Lambda^{4}( (\mathbb{R}^{n})^{*})$, thus $\omega_{ij} \omega_{k \ell}= \omega_{ik} \omega_{j \ell}- \omega_{i \ell} \omega_{jk}$ and we may write $\omega \wedge \omega = \sum_{i < j} \sum_{k < \ell} (\omega_{ik}\omega_{j \ell} - \omega_{i \ell} \omega_{jk}) \ dx^{i} \wedge dx^{j} \wedge dx^{k} \wedge dx^{\ell} \ $ . We want to show $\omega=\eta_{1} \wedge \eta_{2}$ for some $\eta_{1}=\sum_{q}f_{q} \ dx^{q}$, $\eta_{2}=\sum_{r}g_{r}\ dx^{r}$, i.e., we want to show $\omega$ has the form $\omega=\sum_{q < r} (f_{q}g_{r} - f_{r}g_{q}) \ dx^{q} \wedge dx^{r},$ so that $\omega \wedge \omega = \sum_{q < r} \sum_{s < t} $ ----- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```