there is a short remark in the textbook about GR I have been reading, which has been bugging me for the last few hours: enter image description here

My questions are:

(1): In multivariable calculus, curl can be intuitively described as the rotation vector of a vector field. Why is it here a (0,2)-tensor field?

(2): How can one derive the second equation from the first equation?

Best regards.


The second equation only holds for vector fields $Y,Z \perp X$.


1 Answer 1

  1. If $\Gamma^k_{ij} = 0$ the equation for the curl in local coordinates is $$ \text{curl}(X) = \left(\frac{\partial X_j}{\partial{x^i}}-\frac{\partial X_i}{\partial{x^j}}\right) dx^n\wedge dx^m. $$

In $\mathbb R^3$ this expression becomes the usual curl if you apply hodge star (identify $dx\wedge dy$ with $dz$ etc.) and lower the index (identify $\partial_x$ with $dx$ etc.).

  1. You can start with the last formula and plug-in the formula for $dX^\flat$ and the definition of $\wedge$. You will find a bunch of zeroes because $X^\flat(Y) = X^\flat(Z) = 0$, and what remains is $g(X,[Y,Z])$. Using that $\nabla$ is torsion-free you find that this is equal to $g(X,\nabla_YZ)-g(X,\nabla_ZY)$. You complete this by moving the connection across brackets - because $\nabla g=0$ and the fields are perpendicular.

Try these out and let me know if you need more detail.


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