4
$\begingroup$

How do you show that every $2$-covector $\omega\in\Lambda^2((\mathbb{R}^3)^\ast)$ is decomposable i.e. that $$\omega=u\wedge v$$ for some $u,v\in(\mathbb{R}^3)^\ast$?

In general we have $$\omega=\omega_{12}e^1\wedge e^2+\omega_{13}e^1\wedge e^3+\omega_{23}e^2\wedge e^3,$$ so we can set $$\omega_{12}e^1\wedge e^2+\omega_{13}e^1\wedge e^3+\omega_{23}e^2\wedge e^3=(a_1e^1+a_2e^2+a_3e^3)\wedge(b_1e^1+b_2e^2+b_3e^3),$$ and try to solve for $a_i,b_i$. But there doesn't seem to have a well-defined solution to these equations unless we separate it in several different cases ($\omega_{12} \neq 0$, etc). What's a better way of proving it?

$\endgroup$
  • $\begingroup$ FWIW, this follows from the occasionally useful fact that a $2$-covector $\omega$ (on any f.d. vector space) is decomposable iff $\omega \wedge \omega = 0$, but presumably you don't have access to it here. $\endgroup$ – Travis Mar 19 '15 at 3:23
3
$\begingroup$

Here is one way to do this without the case splitting you mention, but it uses the important Hodge star operator, which, given the question, you may or may not have available:

Pick an inner product $\langle \, , \, \rangle$ and orientation on $\mathbb{R}^3$ (the usual ones will do), which defines a volume form $\Omega \in \Lambda^3 (\mathbb{R}^3)^*$; this defines a Hodge star operator $$\ast: \Lambda^2 (\mathbb{R}^3)^* \to (\mathbb{R}^3)^*$$ characterized by $$\alpha \wedge \ast \beta = \langle \alpha, \beta \rangle \Omega,$$ where $\langle \, , \, \rangle$ here denotes the induced inner product on $\Lambda^2 (\mathbb{R}^3)^*$.

Henceforth assume that $\omega \neq 0$ (if $\omega = 0$ it is clearly decomposable, as in that case $\omega = 0 \wedge 0$). Now, $\ast \omega$ is a $1$-form, and we can extend it to some orthogonal basis $(\ast \omega, \zeta, \eta)$ of $(\mathbb{R}^3)^*$. Then, $(\omega, \ast \zeta, \ast \eta)$ is an orthogonal basis of $\Lambda^2 (\mathbb{R}^3)^*$. Computing gives that \begin{align} \langle \zeta \wedge \eta, \ast \zeta \rangle \Omega &= \zeta \wedge \eta \wedge \zeta = 0 \\ \langle \zeta \wedge \eta, \ast \eta \rangle \Omega &= \zeta \wedge \eta \wedge \eta = 0 ,\\ \end{align} so by orthogonality $\omega$ is a multiple of $\zeta \wedge \eta$, and in particular $\omega$ is decomposable.

$\endgroup$
1
$\begingroup$

Alternatively, the components of the wedge product with respect to a basis $(e_1, e_2, e_3)$ and the induced basis $(e_2 \wedge e_3, e_3 \wedge e_1, e_1 \wedge e_2)$ are given by the familiar cross product. More precisely, $$\omega_{23}e^2\wedge e^3 + \omega_{31}e^3\wedge e^1 + \omega_{12}e^1\wedge e^2= (a_1e^1+a_2e^2+a_3e^3)\wedge(b_1e^1+b_2e^2+b_3e^3)$$ is equivalent to $$(\omega_{23}, \omega_{31}, \omega_{12}) = (a_1, a_2, a_3) \times (b_1, b_2, b_3).$$ So, the statement that every $2$-covector $\omega \in \Lambda^2 (\mathbb{R}^3)^*$ is decomposable is equivalent to the statement that every vector in $\mathbb{R}^3$ can be written as the cross product of two other vectors.

But this latter claim is easy to prove: Pick any two linearly independent vectors $X, Y$ orthogonal to $\ast \omega := (\omega_{23}, \omega_{31}, \omega_{12})$. Then, by construction $X \times Y$ is some multiple of $\ast \omega$, and by rescaling, say, $X$, we can ensure that $\ast \omega = X \times Y$. (In fact, the relationship between $\wedge$ and $\times$ can be written in terms of the Hodge star operator and the isomorphism $\mathbb{R}^3 \cong (\mathbb{R}^3)^*$ determined by the standard inner product.)

$\endgroup$
1
$\begingroup$

This can be done using clifford algebra. Call a product of $k$ vectors under the clifford product a $k$-versor. Versors are closed under multiplication: the product of two versors is another versor. Note that a $k$-vector that is also a versor is necessarily simple.

Now, consider a bivector $B$ that might not be simple. Multiply it by $\epsilon = e_1 e_2 e_3$, the unit pseudoscalar, which is also a versor (this is the operation of Hodge duality Travis mentions). The result is necessarily some vector. All vectors are simple, and all vectors are versors. $B\epsilon$ is some vector.

Now, consider $(B \epsilon) \epsilon^{-1}$. The result is necessarily a bivector, but it also must be a versor since $B\epsilon$ is a vector. But $B \epsilon \epsilon^{-1} = B$, and since the product operation is associative, this implies $B$ is a versor, and therefore a simple $2$-vector.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.