# How to get the Riemann curvature tensor from the commutator operating on a basis vector

In the following the basis vectors are assumed to be varying functions of position. This means that when a vector appears under the differentiation operator, both components and basis vectors will, in general be differentiated according the the product rule. An underline indicates a particular term is to be held constant during differentiation.

Del operating on a vector is written as

$$\nabla\left[\vec{v}\right]=\partial_{\sigma}\left[\mathfrak{e}_{\nu}v^{\nu}\right]\mathfrak{e}^{\sigma}.$$

Basis 1-forms will be treated as contravariant basis vectors. So in a coordinate basis, we have

$$\mathfrak{e}^{\sigma}=dx^{\alpha}.$$

Del followed by 'dot vector' contracts on the differentiation index. This is called the directional derivative.

$$\nabla\left[\vec{v}\right]\cdot\vec{w}=\partial_{\sigma}\left[\mathfrak{e}_{\nu}v^{\nu}\right]\mathfrak{e}^{\sigma}\cdot\mathfrak{e}_{\omega}w^{\omega}=\frac{\partial\mathfrak{e}_{\nu}v^{\nu}}{\partial x^{\omega}}w^{\omega}.$$

In particular the partial derivative with respect to the

$$\nabla\left[\varphi\right]\cdot\mathfrak{e}_{\delta}=\partial_{\delta}\varphi=\frac{\partial\varphi}{\partial x^{\delta}}$$

$$\nabla\left[\vec{v}\right]\cdot\mathfrak{e}_{\omega}=\frac{\partial\vec{v}}{\partial x^{\omega}}.$$

Del preceded by a 'vector dot' contracts on the argument of del.

$$\vec{u}\cdot\nabla\left[\vec{v}\right]=\partial_{\sigma}\left[\underline{\mathfrak{e}_{\upsilon}u^{\upsilon}}\cdot\mathfrak{e}_{\nu}v^{\nu}\right]$$

$$\vec{u}\cdot\nabla\left[\vec{v}\right]\cdot\vec{w}=\partial_{\sigma}\left[\underline{\mathfrak{e}_{\upsilon}u^{\upsilon}}\cdot\mathfrak{e}_{\nu}v^{\nu}\right]\mathfrak{e}^{\sigma}\cdot\mathfrak{e}_{\omega}w^{\omega}$$

Placing a bar under an index (or in mathjax a bar over the index) indicates a component living in the tangent plane. So the $$\beta$$ basis vector living in the manifold may be expressed on the tangent basis as

$$\mathfrak{e}_{\beta}=\mathfrak{e}_{\bar{\beta}}\frac{\partial x^{\bar{\beta}}}{\partial x^{\beta}}.$$

$$\nabla\left[\mathfrak{e}_{\beta}\right]=\partial_{\bar{\gamma}}\left[\mathfrak{e}_{\bar{\beta}}\frac{\partial x^{\bar{\beta}}}{\partial x^{\beta}}\right]\mathfrak{e}^{\bar{\gamma}}$$

$$=\mathfrak{e}_{\bar{\beta}}\partial_{\bar{\gamma}}\left[\frac{\partial x^{\bar{\beta}}}{\partial x^{\beta}}\right]\mathfrak{e}^{\bar{\gamma}}$$

$$=\mathfrak{e}_{\bar{\beta}}\frac{\partial^{2}x^{\bar{\beta}}}{\partial x^{\bar{\gamma}}\partial x^{\beta}}\mathfrak{e}^{\bar{\gamma}}$$

As unorthodox as this may seem, observe that it leads to the traditional form of the connection coefficient

$$\mathfrak{e}^{\alpha}\cdot\nabla\left[\mathfrak{e}_{\beta}\right]\cdot\mathfrak{e}_{\gamma}=\mathfrak{e}^{\alpha}\cdot\mathfrak{e}_{\bar{\beta}}\frac{\partial^{2}x^{\bar{\beta}}}{\partial x^{\bar{\gamma}}\partial x^{\beta}}\mathfrak{e}^{\bar{\gamma}}\cdot\mathfrak{e}_{\gamma}$$

$$=\frac{\partial x^{\alpha}}{\partial x^{\bar{\beta}}}\frac{\partial^{2}x^{\bar{\beta}}}{\partial x^{\bar{\gamma}}\partial x^{\beta}}\frac{\partial x^{\bar{\gamma}}}{\partial x^{\gamma}}$$

$$=\frac{\partial x^{\alpha}}{\partial x^{\bar{\beta}}}\frac{\partial^{2}x^{\bar{\beta}}}{\partial x^{\gamma}\partial x^{\beta}}=\Gamma^{\alpha}{}_{\beta\gamma}.$$

Since I use square brackets to enclose parameter lists, I use double square brackets $$\left[\![\_,\_\right]\!]$$ to indicate the commutator. As indicated above, I use the dot product notation interchangeably with contraction notation.

The above notation has proved invaluable in many circumstances. It should work to produce the Riemann curvature tensor beginning with MTW equation 8.44. Unfortunately I haven't figured out a way to translate the right-most term in the form I arrive at, into the terms involving products of Christoffel symbols.

Does anybody see a way to make this work? The first line in the following screen-scrape is a shot in the dark.

This is a more conventional derivation based on MTW Exercise 11.3 (which includes the solution).

• I believe the problem is that I need to differentiate "inside" the gradient, and treat the exposed basis vectors as constants. – Steven Thomas Hatton Aug 2 at 22:15

The lower left block rewrites $$\nabla$$ as the contracted product of manifold basis vectors and an almost legal Christoffel symbol. My only motivation for this step is that it works, and mirrors the standard derivation.
The under-bar on the $$\delta$$ is easily removed, as shown in the upper right block. A similar justification is evident for removing the under-bar on the raised $$\beta$$ and replacing it with $$\mu$$. The complete transformation of a Christoffel symbol is far more complicated, so this is a fairly dangerous slight of hand.