Understanding Riemann Curvature Tensor in Misner, Thorne and Wheeler's Gravitation

Question

I'm trying to understand section 11.4 of Misner, Thorne and Wheeler's Gravitation textbook, which explains how the output of the Riemann Curvature Tensor $Riemann(...,A,u,v)$ is a vector describing the difference between vector $A$ and a version of $A$ parallel transported around a closed loop formed using the vector fields $u$ and $v$.

They describe transporting around the loop in the image, where the Lie Bracket $[u,v]\Delta a \Delta b$ "closes" the loop. So far this makes sense to me.

To "derive" a formula for this, they add the individual parallel transport results for each of the five "legs" of the loop. Since the initial vector $A$ is only present at the starting point, they introduce $A^{(field)}$, which is a vector field defined at all points on the loop. This allows us to do a subtraction with $A^{(mobile)}$, the parallel transported version of $A$, anywhere on the loop. The resulting difference vector is:

I don't understand the leap in logic at the 2nd equal sign. The paragraph seems to indicate the difference vector would be:

$$ \nabla_v A \Delta b - \nabla_v A \Delta b - \nabla_u A \Delta a + \nabla_u A \Delta a + \nabla_{[u,v]} A \Delta a \Delta b$$

I don't understand how the first four steps become a "commutator" $ \nabla_u \nabla_v - \nabla_v \nabla_v$

Is suspect it has something to do with the fact that the legs for $-u \Delta a$ and $+u \Delta a$ are not located at the same point, but I can't figure out the exact reason.

Answer 1

Consider the quadrilateral $\mathbf{u}\Delta a$, $\mathbf{v}\Delta b$, $-\mathbf{u}\Delta a$, $-\mathbf{v}\Delta b$, plus the closer $[\mathbf{v},\mathbf{u}]\Delta a\Delta b$ in the figure. Denote its vertices by points $0, 1, 2, 3, 4$ counterclockwise with $0$ begin the starting point of the parallel transport.

Moreover, denote the vector field $\mathbf{A}$ at point $i$ by $\mathbf{A}_i^\mathrm{(field)}$ and the parallel transported vector of $\mathbf{A}_{i-1}^\mathrm{(field)}$ at $i$ by $\mathbf{A}_i^\mathrm{(mobile)}$, i.e., $\mathbf{A}_i^\mathrm{(mobile)}=\Gamma_{(i-1)\to i}(\mathbf{A}_{i-1}^\mathrm{(field)})$ with $\mathbf{A}_0^\mathrm{(mobile)}=\Gamma_{4\to 0}(\mathbf{A}_4^\mathrm{(field)})$.

For simplicity, work only to order $\Delta a\Delta b$ here. Expansion to quadratic order in $\Delta a$ and $\Delta b$ gives the same result as terms of order $(\Delta a)^2$ and $(\Delta b)^2$ cancel out separately.

Change in $\mathbf{A}_i^\mathrm{(field)}$ relative to the parallel transported $\mathbf{A}_i^\mathrm{(mobile)}$ along each leg of the quadrilateral is as follows:

1. $0\to 1$: $$\require{cancel}\mathbf{A}^\mathrm{(field)}_1-\mathbf{A}^\mathrm{(mobile)}_1=\Delta a\nabla_\mathbf{u}\mathbf{A}^\mathrm{(field)}|_0,$$
2. $1\to 2$: $$\mathbf{A}^\mathrm{(field)}_2-\mathbf{A}^\mathrm{(mobile)}_2=\Delta b\nabla_\mathbf{v}\mathbf{A}^\mathrm{(field)}|_1=\Delta b(\nabla_\mathbf{v}\mathbf{A}^\mathrm{(field)}|_0+\Delta a\nabla_\mathbf{u}\nabla_\mathbf{v}\mathbf{A}^\mathrm{(field)}|_0),$$
3. $2\to 3$: $$\mathbf{A}^\mathrm{(field)}_3-\mathbf{A}^\mathrm{(mobile)}_3=\Delta a\Delta b\nabla_{[\mathbf{v},\mathbf{u}]}\mathbf{A}^\mathrm{(field)}|_2=\Delta a\Delta b\nabla_{[\mathbf{v},\mathbf{u}]}\mathbf{A}^\mathrm{(field)}|_0,$$
4. $3\to 4$: $$\begin{align*}\mathbf{A}^\mathrm{(field)}_4-\mathbf{A}^\mathrm{(mobile)}_4=-\Delta a\nabla_\mathbf{u}\mathbf{A}^\mathrm{(field)}|_3&=-\Delta a(\nabla_\mathbf{u}\mathbf{A}^\mathrm{(field)}|_4+\cancel{\Delta a\nabla_\mathbf{u}\nabla_\mathbf{u}\mathbf{A}^\mathrm{(field)}|_4})\\&=-\Delta a(\nabla_\mathbf{u}\mathbf{A}^\mathrm{(field)}|_0+\Delta b\nabla_\mathbf{v}\nabla_\mathbf{u}\mathbf{A}^\mathrm{(field)}|_0),\end{align*}$$
5. $4\to 0$: $$\mathbf{A}^\mathrm{(field)}_0-\mathbf{A}^\mathrm{(mobile)}_0=-\Delta b\nabla_\mathbf{v}\mathbf{A}^\mathrm{(field)}|_4=-\Delta b(\nabla_\mathbf{v}\mathbf{A}^\mathrm{(field)}|_0+\cancel{\Delta b\nabla_\mathbf{v}\nabla_\mathbf{v}\mathbf{A}^\mathrm{(field)}|_0}).$$

The key point is to express covariant derivatives $\nabla_\mathbf{v}\mathbf{A}^\mathrm{(field)}|_1$, $\nabla_{[\mathbf{v},\mathbf{u}]}\mathbf{A}^\mathrm{(field)}|_2$, $\nabla_\mathbf{u}\mathbf{A}^\mathrm{(field)}|_3$ and $\nabla_\mathbf{v}\mathbf{A}^\mathrm{(field)}|_4$ in terms of that evaluated at $0$.

Using $\mathbf{A}_i^\mathrm{(mobile)}=\Gamma_{(i-1)\to i}(\mathbf{A}_{i-1}^\mathrm{(field)})$ with $\mathbf{A}_0^\mathrm{(mobile)}=\Gamma_{4\to 0}(\mathbf{A}_4^\mathrm{(field)})$ and adding all changes up give $$\begin{align*}-\delta\mathbf{A}^\mathrm{(field)}_0&=\mathbf{A}^\mathrm{(field)}_0-\Gamma_{4\to 0}\ldots\Gamma_{1\to 2}\Gamma_{0\to 1}(\mathbf{A}^\mathrm{(field)}_0)\\&=(\nabla_\mathbf{u}\nabla_\mathbf{v}\mathbf{A}^\mathrm{(field)}|_0-\nabla_\mathbf{v}\nabla_\mathbf{u}\mathbf{A}^\mathrm{(field)}|_0+\nabla_{[\mathbf{v},\mathbf{u}]}\mathbf{A}^\mathrm{(field)}|_0)\Delta a\Delta b\\ &=(\nabla_\mathbf{u}\nabla_\mathbf{v}\mathbf{A}^\mathrm{(field)}-\nabla_\mathbf{v}\nabla_\mathbf{u}\mathbf{A}^\mathrm{(field)}-\nabla_{[\mathbf{u},\mathbf{v}]}\mathbf{A}^\mathrm{(field)})|_0\Delta a\Delta b\\ &=R(\mathbf{u},\mathbf{v})\mathbf{A}^\mathrm{(field)}|_0\Delta a\Delta b. \end{align*}$$