- #1
LAHLH
- 409
- 1
Hi,
I'm reading Sean Carroll's text, ch2, and believe I understood most of the discussion on Integration in section 2.10. However in equation 2.96, he states that
[tex] \epsilon\equiv\epsilon_{\mu_1\mu_2...\mu_n}dx^{\mu_1}\otimes dx^{\mu_2}\otimes...\otimes dx^{\mu_n} =\frac{1}{n!}\epsilon_{\mu_1\mu_2...\mu_n}}dx^{\mu_1}\wedge dx^{\mu_2}\wedge...\wedge dx^{\mu_n}[/tex]
I don't quite understand this equality. For example just taking n=2, the LHS is [tex] dx^0\otimes dx^1-dx^1\otimes dx^0[/tex] (which isn't zero because tensor products don't commute). Where on the RHS one would have [tex] \frac{1}{2}\left(dx^0\wedge dx^1-dx^1\wedge dx^0\right)=dx^0\wedge dx^1[/tex], by the antisymmetry of the wedge product.
So I'm at a loss to understand this part of 2.96 despite understand the following lines.
Thanks a lot for any replies.
I'm reading Sean Carroll's text, ch2, and believe I understood most of the discussion on Integration in section 2.10. However in equation 2.96, he states that
[tex] \epsilon\equiv\epsilon_{\mu_1\mu_2...\mu_n}dx^{\mu_1}\otimes dx^{\mu_2}\otimes...\otimes dx^{\mu_n} =\frac{1}{n!}\epsilon_{\mu_1\mu_2...\mu_n}}dx^{\mu_1}\wedge dx^{\mu_2}\wedge...\wedge dx^{\mu_n}[/tex]
I don't quite understand this equality. For example just taking n=2, the LHS is [tex] dx^0\otimes dx^1-dx^1\otimes dx^0[/tex] (which isn't zero because tensor products don't commute). Where on the RHS one would have [tex] \frac{1}{2}\left(dx^0\wedge dx^1-dx^1\wedge dx^0\right)=dx^0\wedge dx^1[/tex], by the antisymmetry of the wedge product.
So I'm at a loss to understand this part of 2.96 despite understand the following lines.
Thanks a lot for any replies.