Let $(M,g)$ be a Riemannian manifold. From $g$ and a fixed vector field $V$ we can derive the following two differential forms:

  • A $1$- form $\alpha(X) = g(V,X)$, i.e. $\alpha = \iota_Vg$.
  • A $2$-form $\beta(X,Y) = g(V,[X,Y]) = g(V,\mathcal{L}_XY)$.

Notice that we obviously have $\beta(X,Y) = \alpha([X,Y]).$

Is there a nice expression for the exterior derivative $d\alpha$ and $d\beta$ of these forms? In particular, when are they zero?

I can get expressions in local coordinates, but they are kinda ugly and I cannot relate them to other known quantities.

  • $\begingroup$ The first is just the usual musical isomorphism $TM \to T^\ast M$ defined by the Riemannian metric $g$ (en.wikipedia.org/wiki/Musical_isomorphism). The second looks familiar, but I can't place it for the life of me. $\endgroup$ – Branimir Ćaćić Feb 21 '15 at 19:36
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    $\begingroup$ @BranimirĆaćić I know, that's how I obtained it :) However I really need to understand its exterior derivative, and damn but I don't seem to be able to find anything nice... $\endgroup$ – Daniel Robert-Nicoud Feb 22 '15 at 12:56
  • $\begingroup$ O gosh, I'm stupid. I didn't notice you were looking for the exterior derivatives... $\endgroup$ – Branimir Ćaćić Feb 22 '15 at 17:02

A literal back-of-the envelope calculation leads me to $$d\alpha(X,Y)=g(\nabla_XV,Y)-g(\nabla_YV,X).$$ Does that help?

EDIT: Here is what I presume, 2 1/2 years later, was on the back of the envelope (with a few extra words).

Since $\nabla$ is $g$-compatible, we have $\nabla g = 0$, and so $$X(g(V,Y)) = \nabla_X(g(V,Y)) = g(\nabla_X V,Y) + g(X,\nabla_X Y).$$ Also, since $\nabla$ is torsion-free, we have $\nabla_X Y - \nabla_Y X = [X,Y]$, so $g(V,[X,Y]) = g(V,\nabla_X Y)-g(V,\nabla_Y X)$.

Using the standard formula for the derivative of a $1$-form, we have \begin{align*} d\alpha(X,Y) &= X(g(V,Y)) - Y(g(V,X))- g(V,[X,Y]) \\ &= g(\nabla_X V,Y) + g(V,\nabla_X Y) - g(\nabla_V, X) - g(V,\nabla_Y X) - \big(g(V,\nabla_X Y)-g(V,\nabla_Y X)\big)\\ &= g(\nabla_X V,Y) - g(\nabla_Y V,X), \end{align*} as required.

  • $\begingroup$ Nope, not really. Thanks for the effort anyway :) $\endgroup$ – Daniel Robert-Nicoud Feb 24 '15 at 17:27
  • $\begingroup$ Would the downvoter care to contribute a superior answer? $\endgroup$ – Ted Shifrin Mar 9 '15 at 15:01
  • $\begingroup$ I don't understand why this answer was downvoted. $\endgroup$ – calculusbyparts Mar 10 '15 at 7:00
  • $\begingroup$ @TedShifrin Is it possible if you can provide the calculation here? $\endgroup$ – Cookie Oct 6 '17 at 20:02
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    $\begingroup$ @Cookie. I've edited. $\endgroup$ – Ted Shifrin Oct 6 '17 at 20:51

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