Source: Pages 89 and 90 of Sean Carroll's Spacetime and Geometry

Quite a confusion in two steps of this quantity:

$$ \begin{eqnarray} \sqrt{|g|}d^n x &=& \sqrt{|g|}dx^0 \land ... \land dx^{n-1} \newline &=& \frac{1}{n!}\sqrt{|g|} \widetilde{\epsilon}_{\mu_1 ... \mu_n}dx^{\mu_1} \land ... \land dx^{\mu_n} \newline &=& \frac{1}{n!} \epsilon_{\mu_1 ... \mu_n}dx^{\mu_1} \land ... \land dx^{\mu_n} \newline &=& \epsilon_{\mu_1 ... \mu_n}dx^{\mu_1} \otimes ... \otimes dx^{\mu_n}\newline &=& \epsilon\end{eqnarray}$$

where my confusion lies in the second equality where derivatives are reindexed via the inclusion of 1 over n factorial and the Levi-Civita symbol, as well as in fourth equality where the wedge products turn to tensor products via the suppression of the 1 over $n$ factorial.

Carroll explained the former with "since the wedge product and Levi-Civita symbol are completely anti-symmetric. (The factor of $1/n!$ takes care of the overcounting introduced by summing over permutations of the indices.)"

Carroll's definition of the wedge product (page 84) : for some p-form A and some q-form B, $$ (A \land B)_{\mu_1 ... \mu_{p+q}} = \frac{(p+q)!}{p!q!}A_{[\mu_1 ... \mu_p} B_{\mu_{p+1}... \mu_{p+q}]} $$

And so for 2 vectors we have,

$$ (A \land B)_{\mu \nu} = \frac{(1+1)!}{1!1!}A_{[\mu} B_{\nu]} $$

And then Carroll's definition of antisymmetrization (page 27) is,

$$ {T_{[\mu_1... \mu_n]}}_\rho{}^\sigma = \frac{1}{n!} (T_{\mu_1... \mu_n \rho{}^\sigma} + \text{alternating sums over permutations of indices } \mu_1...\mu_n)$$

So the wedge product continues as,

$$ (A \land B)_{\mu \nu} = \frac{(1+1)!}{1!1!}A_{[\mu} B_{\nu]} = 2 \frac{1}{2!}(A_\mu B_\nu - A_\nu B_\mu) = A_\mu B_\nu - A_\nu B_\mu$$

  • 2
    $\begingroup$ Which page in Carroll? $\endgroup$
    – Qmechanic
    Aug 13, 2019 at 21:59
  • $\begingroup$ You may find this post interesting. Read comments. $\endgroup$ Aug 24, 2019 at 0:01

1 Answer 1


Looking at the two-dimensional case, it is fairly simple to see what is happening: $$\begin{eqnarray} v\land w & = & \frac{1}{2!}(v\land w-w\land v) \\ & = & \frac{1}{2!}\epsilon_{\mu\nu}v^{\mu}\land w^{\nu} \\ & = & \frac{1}{2!}\epsilon_{\mu\nu}(v^{\mu}\otimes w^{\nu}-w^{\nu}\otimes v^{\mu}) \\ & = & \epsilon_{\mu\nu}v^{\mu}\otimes w^{\nu}. \end{eqnarray}$$

The factors of $n!$ arise similarly when there are more than two one-forms involved. You can write a wedge product of $n$ factors in $n!$ different ways, but permuting the order; the overall sign for each term is then the sign of the permutation. You can also write the wedge product as a sum of all $n!$ terms, but then you need to divide by that $n!$. That is what is being done on the first line above. The conversion to the tensor products, rather than wedge products, is essentially the same thing in reverse, since the wedge product is, by definition, an antisymmetrized sum of the one-forms involved.

  • $\begingroup$ Hmmm ok I think I'm getting closer. Allow me to Tex up the definition of the wedge product that I was working with from Carroll. I had not experienced it in my undergrad GR course so it may be just that I need to go see some more definitions of it in Wald or Misner, Thorne, Wheeler. I will add this to the question in one moment. $\endgroup$
    – Lopey Tall
    Aug 14, 2019 at 17:17
  • $\begingroup$ @AntoniParellada Sorry, I don't do the Mathematics StackExchange. $\endgroup$
    – Buzz
    Aug 19, 2019 at 13:10
  • $\begingroup$ @AntoniParellada There is a slight notational confusion here. The $v^\mu$ etc. should not be the components of $v$, but rather the basis elements, i.e., $v^1$ = v, $v^2$ = $w$, something like that, since the O.P. was about basis elements, not components. That is, the O.P. wrote in terms of $x^\mu$ etc. and for simplicity, we wish here to consider 2 dimensions only, so we wish to consider the OP's $x^1$ and $x^2$, which we choose, for further simplicity, to call $v$ and $w$. But he switched back to using superscripts, which confuses things. If it weren't for this, there would be no problem $\endgroup$ Dec 9, 2020 at 4:53

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