Understanding the Equality in Equation 2.96: Volume Element as n-Form?

[LAHLH]

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

\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}

I don't quite understand this equality. For example just taking n=2, the LHS is dx^0\otimes dx^1-dx^1\otimes dx^0 (which isn't zero because tensor products don't commute). Where on the RHS one would have \frac{1}{2}\left(dx^0\wedge dx^1-dx^1\wedge dx^0\right)=dx^0\wedge dx^1, 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.

 

[arkajad]

Formula (1.81) in Carroll's "Lecture Notes on General Relativity" (1997) tells you:

A\wedge B=A\otimes B-B\otimes A

So, what is it that causes you the problem?

 

[LAHLH]

Oh I was unaware of this formula.

Do you mean (1.80) in this edition of Carroll? namely (A\wedge B)_{\mu\nu}=A_{\mu}B_{\nu}-A_{\nu}B_{\mu}

Which I guess can be written (A\wedge B)_{\mu\nu} =(A\otimes B)_{\mu\nu}-(B\otimes A)_{\mu\nu}, leading to the equation you stated: (A\wedge B) =(A\otimes B)-(B\otimes A)

 

[arkajad]

Yes - that is another way of writing it.