11
$\begingroup$

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 $$3\nabla_{[a} 2\nabla_{[b}\omega_{c]]} = 6\nabla_{[a}\nabla_b\omega_{c]}?$$

$\endgroup$
  • 1
    $\begingroup$ 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
4
$\begingroup$

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.