I have the following question: in Carroll's book we're asked to show that $$ \text{d}(\omega \wedge \eta) = (\text{d}\omega)\wedge \eta +(-1)^q\omega\wedge (\text{d}\eta) $$ For a $p$-form $\omega$ and $q$-form $\eta$. Where we have the following definitions: $$(\omega\wedge \eta)_{\mu_1...\mu_{p+q}}\equiv\frac{(p+q)!}{p!q!}\omega_{[\mu_1...\mu_p}\eta_{\mu_{p+1}...\mu_{p+q}]}$$ $$ (\text{d}\omega)_{\mu_1...\mu_{p+1}}\equiv (p+1)\partial_{[\mu_1}\omega_{\mu_2...\mu_{p+1}]} $$

I tried to prove the identity just from the definitions, but arrived at the following issue (which I'll just illustrate with two 2-forms for simplicity).

If I expand the left side of the identity, I get: $$ \text{d}(\omega\wedge \eta) = 5\cdot \frac{4!}{2!2!} \partial_{[\mu_1}\omega_{[\mu_2\,\mu_3}\eta_{\mu_4\mu_5]]} $$

Maybe it would've been more convenient to try a different method from this, but upon seeing it I got curious... What exactly is the meaning of that double antisymmetrization going on down there? Or have I done something wrong to reach that expression?


1 Answer 1


There's no such thing, really. As a reminder, antisymmetrizing $n$ indices meaning taking the average over all permutations, with alternating signs:

$$X_{[i_1 \dotsm i_n]} = \frac{1}{n!} \sum_{\sigma} (-1)^\sigma \, X_{i_{\sigma(1)} \dotsm i_{\sigma(n)}}$$ with the sum running over all permutations $\sigma$ of $\{1,2,\ldots,n\}$, and $(-1)^\sigma$ is $\pm 1$ if $\sigma$ is even/odd.

That's the definition. You can now easily see that antisymmetrizing twice works trivially, for instance

$$X_{[i_1} Y_{[i_2 i_3]]} = X_{[i_1} Y_{i_2 i_3]}.$$

There's a formal way to prove this (using group theory), but there's a simple heuristic method too. The idea is to expand $Y_{i_2 i_3}$ into symmetric and antisymmetric parts: $$Y_{i_2 i_3} = Y_{[i_2 i_3]} + Y_{(i_2 i_3)}.$$ In order to prove the above equation, it suffices to show that

$$X_{[i_1} Y_{(i_2 i_3)]} = 0.$$

But this is a simple exercise:

$$X_{[i_1} Y_{(i_2 i_3)]} = \frac12 \big\{ X_{[i_1} Y_{i_2 i_3]}+X_{[i_1} Y_{i_3 i_2]} \big\} = \frac12 \big\{ X_{[i_1} Y_{i_2 i_3]}-X_{[i_1} Y_{i_2 i_3]} \big\} = 0.$$ (The first equality is the definition of $Y_{(i_2 i_3)}$, the second one from the fact that $X_{[i_1} Y_{i_3 i_2]}$ is by construction an antisymmetric tensor.)

The same generalizes to higher rank tensor: the brackets $[\ldots]$ kill any symmetric parts of a tensor and leave only the fully antisymmetric part.


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