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Let the $1$-form $\theta_X$ be defined as $\theta_X(Y)=g(X,Y)$. Then my textbook says

$$(d\theta_X)(\partial_k,\partial_l)=\partial_k g(X,\partial_l)-\partial_l g(X,\partial_k)-g(X,[\partial_k,\partial_l])$$

I tried looking up exterior derivatives of $1$-forms to make sense of this formula, but I only got exterior derivatives of alternating forms. How does one get the formula given above? Is one supposed to use Cartan's formula?

  • Have you tried evaluating $\iota_Z\mathrm{L}_Y\theta_X$ two different ways, with one way using the definition of Lie derivative $\langle\mathrm{L}_Y\alpha,Z\rangle=\partial_Y\langle \alpha,Z\rangle -\langle \alpha,[Y,Z]\rangle$, and the other exploiting Cartan's formula $\mathrm{L}_Y=\iota_Y\mathrm{d}+\mathrm{d}\iota_Y$? – K B Dave Jan 03 '20 at 01:30
  • @K B Dave- Tried it. Worked. Thanks! –  Jan 03 '20 at 04:40

1 Answers1

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If $\omega$ is a $k$-form, then $d\omega$ is a $(k + 1)$-form and if $V_0, \dots, V_k$ are vector fields, then

\begin{align*} d\omega(V_0, \dots, V_k) =&\, \sum_{i=0}^k(-1)^iV_i\,\omega(V_0, \dots, V_{i-1}, V_{i+1}, \dots, V_k)\\ &+ \sum_{0 \leq i < j \leq k}(-1)^{i+j}\omega([V_i, V_j], V_0, \dots, V_{i-1}, V_{i+1}, \dots, V_{j-1}, V_{j+1}, \dots, V_k). \end{align*}

In particular, if $\omega$ is a $1$-form, then $d\omega(X, Y) = X\omega(Y) - Y\omega(X) - \omega(X, Y)$. Therefore

$$d\theta_X(\partial_k, \partial_l) = \partial_k\theta_X(\partial_l) - \partial_l\theta_X(\partial_k) - \theta_X([\partial_k, \partial_l]) = \partial_k g(X, \partial_l) - \partial_l g(X, \partial_k) - g(X, [\partial_k, \partial_l]).$$

Michael Albanese
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