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Let $V$ be a finite dimensional vector space and $\eta \in \Lambda^k(V^\ast)$ be decomposable and non-zero. Then there exists covectors $\omega^1,\dots,\omega^k$ such that $$\eta(v_1, \dots, v_k) = \omega_1 \wedge \dots \wedge \omega_k (v_1, \dots, v_k) = \det \left(\omega^i(v_j)\right).$$ Fix a basis for $V$, I will now treat covectors like vectors. Define the matrices \begin{align} \Omega &= (\omega_i^j) \\ V &= (v_j^i). \end{align}

Then $$\eta(v_1, \dots, v_k) = \det(\Omega V)$$ Define $S=\text{span}(\omega^1, \dots, \omega^k)$ and let $E$ be a $k\times n$ matrix whose rows form an orthonormal basis for $S$. Then $\Omega = \Omega E^T E$ so \begin{align} \eta(v_1, \dots, v_k) &= \det(\Omega E^T E V)\\ &= \det(\Omega E^T) \det(E V) \\ &= \det(E\Omega^T) \det(E V) \\ &= \text{Vol}_S(\omega^1,\dots,\omega^k) \text{Vol}_S(v_1, \dots, v_k) \end{align} Where $\text{Vol}_S(u_1, \dots, u_k)$ gives the $S$-projected volume of the parallelotope spanned by $u_1, \dots, u_k$. Assuming I didn't make any mistakes, is there an analogous viewpoint for non-decomposable $k$-covectors?

Edit:

Here is another question that probably has an obvious answer. Is every $k$-covector decomposable in the right basis?

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  • $\begingroup$ The answer to your edit is no: The covector $\omega = dx_1 \wedge dx_2 + dx_3 \wedge dx_4$ in $\mathbb{R}^4$ is not decomposable with respect to any basis. The idea is that if $\omega$ was decomposable, then $\omega^2 = \omega \wedge \omega = 0$ because it repeats some basis vector. However, $\omega\wedge \omega = 2 dx_1\wedge dx_2\wedge dx_3\wedge dx_4 \neq 0$. $\endgroup$ – Jason DeVito Apr 19 '18 at 15:43
  • $\begingroup$ @JasonDeVito, Thanks, I also found this which basically answers my whole question. So I guess my best option is to think of non-decomposable $k$-forms as linear combinations of decomposable ones. $\endgroup$ – Cairn O. Apr 19 '18 at 23:06

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