Let $(M,g)$ be a Riemannian manifold and let $\omega$ be a $1$-form on $M$. I want to rewrite $d^2\omega=0$ in terms of the Levi-Civita connection.

I can show the following:

$$d\omega(X,Y) = (\nabla_X\omega)(Y) - (\nabla_Y\omega)(X), $$ which in index notation reads $$(d\omega)_{ab} =2 \nabla_{[a}\omega_{b]}.$$

Similiarly for a $2$-form, $\mu$, we have: $$d\mu(X,Y,Z) = (\nabla_X\mu)(Y,Z) - (\nabla_Y\mu)(X,Z) + (\nabla_Z\mu)(X,Y) ,$$

which in index notation reads $$(d\mu)_{abc} = 3\nabla_{[a}\phi_{bc]}.$$

Now plugging in $d\omega$ for $\mu$ we get $$0 = d^2\omega = (d(d\omega))_{abc} = \nabla_{[a}(d\omega)_{bc]}.$$

I want to plug in the above expression (in index notation) for $d\omega$ but I'm not really sure how to handle the indices. Do I just get $$3\nabla_{[a} 2\nabla_{[b}\omega_{c]]} = 6\nabla_{[a}\nabla_b\omega_{c]}?$$

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    $\begingroup$ The formulae for the exterior derivative that you come up with hold for any torsion-free connection, not necessarily for the Levi-Civita connection. $\endgroup$ – Yuri Vyatkin Mar 23 '14 at 0:44

This is correct. See M.Spivak, Calculus on manifolds, 1965, Theorem 4-4 (2), p.80.

Of course, this can be also verified directly by writing out the expansion: $$ \begin{align} 3\nabla_{[a} 2\nabla_{[b}\omega_{c]]} & = \nabla_{a} 2\nabla_{[b}\omega_{c]} + \nabla_{b} 2\nabla_{[c}\omega_{a]} +\nabla_{c} 2\nabla_{[a}\omega_{b]} \\ & = \nabla_{a} \nabla_{b}\omega_{c} - \nabla_{a} \nabla_{c}\omega_{b} + \nabla_{b} \nabla_{c}\omega_{a} - \nabla_{b} \nabla_{a}\omega_{c} + \nabla_{c} \nabla_{a}\omega_{b} - \nabla_{c} \nabla_{b}\omega_{a} \\ & = 6 \nabla_{[a} \nabla_b \omega_{c]} \end{align} $$ where we have used that for a tensor $t_{a b c}$ with a symmetry $$ t_{a b c} = t_{a [b c]} = \tfrac{1}{2} \left( t_{a b c} - t_{a c b} \right) $$ the alternation is expressed by $$ t_{[a b c]} = \tfrac{1}{3} \left( t_{a b c} + t_{b c a} + t_{c a b} \right) $$ which is again easy to prove: $$ \begin{align} t_{[a b c]} & = \tfrac{1}{6} \left( t_{a b c} - t_{a c b} + t_{b c a} - t_{b a c} + t_{c a b} - t_{c b a} \right) \\ & = \tfrac{1}{6} \left( 2 t_{a b c} + 2 t_{b c a} + 2 t_{c a b} \right) \end{align} $$

To prove the Poincare lemma $d^2 \omega = 0 $ locally one can use the Euclidean connection on a chosen coordinate system, and since the exterior derivative is independent of a choice of torsion-free connection, the lemma follows from the fact that partial derivatives commute. The details see in R. Wald, General relativity, 1984, p. 429.

  • $\begingroup$ Thanks for the answer. This helps a lot. $\endgroup$ – Bohring Mar 23 '14 at 4:22

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