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Gravitation By Charles Misner, Kip Thorene, John Wheeler page 129.

The components of the exterior product of $p$ vectors

$\displaystyle (\cal u_1\wedge u_2\wedge...\wedge u_p)^{\alpha_1...\alpha_p} =... =p!u_1^{[\alpha_1}u_2^{\alpha_2}...u_p^{\alpha_p]} =\delta^{\alpha_1\alpha_2...\alpha_p}_{1,2...p}\det[(u_\mu)^\lambda]$

Some related information could be found here. Determinant and Levi-Civita symbol But still, I'm a bit confused of why exterior product could be so closely connected to determinant.

Is it because Levi-Civita symbol could be computed from the determinate of Kronecker delta? Does this carry any further implications?

(A related reference could be found here: Product of Levi-Civita symbol is determinant? where guestDiego's answer provided some insights.)

ShoutOutAndCalculate
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    If $V$ is an $n$-dimensional vector space, then its $n$-th exterior product $\bigwedge^n V$ is a one-dimensional vector space. A linear map $T:V\to V$ induces a linear map $\bigwedge^n T:\bigwedge^n V\to\bigwedge^n V$. Then $\bigwedge^n T$ is just multiplication by $\det T$. – Angina Seng Oct 22 '19 at 06:24
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    @LordSharktheUnknown That looks pretty much like an answer to me. – amd Oct 22 '19 at 07:04

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