# How to choose basis vectors for a given metric to calculate curl of a vector in General Relativity?

The space-time interval in General Relativity is $$ds^2=g_{\mu\nu}dx^\mu dx^\nu$$, where $$g_{\mu\nu}$$ is the metric tensor. If the space-time is stationary and axisymmetric, the corresponding geometry is described by the Kerr metric, where the non-zero metric coefficients are $$g_{tt}$$, $$g_{rr}$$, $$g_{\theta\theta}$$, $$g_{\phi\phi}$$ and $$g_{t\phi}$$. The off-diagonal term $$g_{t\phi}\ne0$$ implies the axisymmetry.

In the textbook by Landau & Lifshitz, the authors define the 3-dimensional metric tensor $$\gamma_{ij}=\left(-g_{ij}+\frac{g_{0i}g_{0j}}{g_{00}}\right),$$ such that the spatial distance is expressed as $$dl^2=\gamma_{ij}dx^idx^j$$. Then, the authors define the 3-dimensional vector $$\mathbf{g}$$ as $$g_i=-\frac{g_{0i}}{g_{00}}.$$ I am trying to calculate the curl of the vector $$\mathbf{g}$$, i.e., $$\mathbf{\nabla}\times\mathbf{g}$$, and for this I require to choose a suitable set of basis vectors. However, I am not able to understand how to find the basis vectors.

Could someone please suggest me any hints on how to proceed with the calculation?

• The basis vectors are obtainable from the metric $\gamma$. Commented Jun 10, 2022 at 14:15
• @K.defaoite Okay. I had tried but having trouble with the off-diagonal term $g_{t\phi}$ that appears in the metric coefficient $\gamma_{\phi\phi}$. Commented Jun 10, 2022 at 15:47
• My only suggestion would be to remember that a vector field (here $\mathbf g$) corresponds (using the metric) to a $1$-form $\omega$. Curl is found by taking $d\omega$, applying the Hodge star operator, and then converting back to a vector field. The Hodge star is easiest to compute when the metric is diagonal, but can be calculated in any eventuality. Commented Jun 10, 2022 at 18:41

This might not be exactly, what you are looking for, but I am unsure, what you mean exactly by "finding basis vectors". If the task is to compute the curl, I suggest to proceed exactly as suggested by Ted Shifrin in the comments. Let me try to outline the concrete computation steps you need to do:

In the given situation, you are working on a 3-dimensional Riemannian manifold with metric $$\gamma$$. So the components of $$\gamma$$ can be expressed as $$\gamma_{ij}$$ with $$i,j\in\{1,2,3\}$$ in any given chart/coordinate system. So you can choose any coordinate expression for the Kerr metric you like and compute $$\gamma$$ using the formula from Landau/Lifshitz.

In the same way you can use the other formula from Landau/Lifshitz you gave to compute the components of $$\mathbf{g}$$. Note that although Landau/Lifshitz call this a vector, it is really a 1-form or a co-vector field, because you compute $$g_i$$ with a lower index. The components of the actual vector field (contra-variant components) can be computed by $$g^i = \gamma^{ij}g_j$$, where you need the contra-variant metric components $$\gamma^{ij}$$. Compute them, e.g., by computing the matrix inverse of $$(\gamma_{ij})$$.

On the Riemannian manifold with metric $$\gamma$$, we can define the curl of a vector field $$X$$ with components $$X^i$$ as $$\nabla\times X := (\star_{\gamma}(\mathrm{d}X^\flat))^\sharp.$$ Here the musical isomorphisms $$\flat$$ and $$\sharp$$, just correspond to raising and lowering indices, i.e., $$(X^\flat)_i = \gamma_{ij}X^j$$ for a vector field $$X$$ and $$(u^\sharp)^i = \gamma^{ij}u_j$$ for a co-vector field or 1-form $$u$$. The operator $$\mathrm{d}$$ is the exterior derivative. For a 1-form $$u$$ the component expression for the resulting 2-form $$\mathrm{d}u$$ is just $$(\mathrm{d}u)_{ij} = \partial_iu_j - \partial_ju_i.$$ $$\star_\gamma$$ is the Hodge star operator for the metric $$\gamma$$. In the 3-dimensional case here, it converts a 2-form into a 1-form. You can compute it for the metric $$\gamma$$, e.g., using the formula in this answer. For the 2-form $$\mathrm{d}u$$ in the 3-dimensional case here, it breaks down to $$(\star_\gamma(\mathrm{d}u))_k = \sqrt{\det(\gamma)}\gamma^{il}\gamma^{jm}\varepsilon_{klm}(\mathrm{d}u)_{ij}.$$

Putting everything together, what you might want to compute is: $$(\nabla\times\mathbf{g})^k = \sqrt{\det(\gamma)}\gamma^{kr}\gamma^{il}\gamma^{jm}\varepsilon_{rlm}(\partial_ig_j - \partial_jg_i).$$

On a side note: I think what Landau/Lifshitz is doing there is called 3+1-splitting or lapse/shift-splitting in general relativity (cf. for example $$\S$$ 21.4 in Misner, Thorne, Wheeler: Gravitation). The Riemannian metric $$\gamma$$ will then only be independent of the time $$t$$ you split away if the spacetime is stationary, which of course is true for the Kerr metric.

• Thank you for the detailed answer. Though I was looking for an answer based on basis vectors corresponding to the metric, your approach appears to be more elegant. Commented Jun 21, 2022 at 7:47