5
$\begingroup$

This is exercise 7.8 b) of Nakahara's GTaP: Let $\omega\in\Omega^1(M)$ be a 1-form on a Riemannian manifold with Levi-Civita connection $\nabla$. Prove that

$$ \mathrm{d}\omega=(\nabla_\mu\omega)_\nu\, \mathrm dx^\mu\wedge\mathrm dx^\nu $$

I proved it using the fact that $\mathrm dx^\mu\wedge\mathrm dx^\nu=-\mathrm dx^\nu\wedge\mathrm dx^\mu$, so:

\begin{align} (\nabla_\mu\omega)_\nu\,\mathrm dx^\mu\wedge\mathrm dx^\nu & = (\partial_\mu\omega_\nu-{\Gamma^\lambda}_{\mu\nu}\,\omega_\lambda)\,\mathrm dx^\mu\wedge\mathrm dx^\nu\\ & = \sum_{\mu<\nu}\left(\partial_\mu\omega_\nu-\partial_\nu\omega_\mu+({\Gamma^\lambda}_{\mu\nu}-{\Gamma^\lambda}_{\nu\mu})\omega_\lambda\right)\,\mathrm dx^\mu\wedge\mathrm dx^\nu\\ & = \sum_{\mu<\nu}\left(\partial_\mu\omega_\nu-\partial_\nu\omega_\mu\right)\,\mathrm dx^\mu\wedge\mathrm dx^\nu\\ & = \mathrm d\omega \end{align}

I hope this is correct and makes sense.

I don't like my solution because starting at the second line, it "quits" Einstein summation convention and needs an explicit summation symbol.

  1. Is there a way to prove this without "quitting" Einstein summation convention?
  2. Is there maybe even a way to prove it in a coordinate-free way?
$\endgroup$

2 Answers 2

6
$\begingroup$

Here's an outline of a coordinate-free proof.

(1) For any $1$-form $\omega$ and any vector fields $X,Y$, there is the formula $$d\omega(X,Y) = X(\omega(Y)) - Y(\omega(X)) - \omega([X,Y]).$$ (2) For any affine connection $\nabla$, there is the formula $$X (\omega(Y)) = (\nabla_X\omega)(Y) + \omega(\nabla_XY).$$ By switching the roles of $X, Y$ and subtracting, this gives $$X \omega(Y) - Y \omega(X) = (\nabla_X \omega)(Y) - (\nabla_Y \omega)(X) + \omega(\nabla_XY - \nabla_YX).$$ But $\nabla$ is torsion-free, so.... (left to you)

(3) By plugging in $X = \frac{\partial}{\partial x^\mu}$ and $Y = \frac{\partial}{\partial x^\nu}$, we get..... (left to you)

Remarks: Note that the proof works for any torsion-free affine connection $\nabla$, not just the Levi-Civita connection. This formula is an instance of "Cartan's First Structure Equation." Generalizations exist to $k$-forms for any $k \geq 1$.

$\endgroup$
4
$\begingroup$

There are some wrong expressions in your solution. A possible correct solution is as follows.

\begin{align} (\nabla_\mu\omega)_\nu\,\mathrm dx^\mu\wedge\mathrm dx^\nu & = (\partial_\mu\omega_\nu-{\Gamma^\lambda}_{\mu\nu}\omega_\lambda)\,\mathrm dx^\mu\wedge\mathrm dx^\nu\\ & = \frac{1}{2}\left(\partial_\mu\omega_\nu-\partial_\nu\omega_\mu-({\Gamma^\lambda}_{\mu\nu}-{\Gamma^\lambda}_{\nu\mu})\omega_\lambda\right)\,\mathrm dx^\mu\wedge\mathrm dx^\nu\\ & = \frac{1}{2}\left(\partial_\mu\omega_\nu-\partial_\nu\omega_\mu\right)\,\mathrm dx^\mu\wedge\mathrm dx^\nu\\ & = \mathrm d\omega \end{align}

Do you agree?

$\endgroup$
1
  • 1
    $\begingroup$ Corrected my question, do you still see any errors? Your solution looks good, thanks. $\endgroup$
    – Bass
    Dec 14, 2015 at 21:09

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.