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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.

enter image description here

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:

enter image description here

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.

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  • $\begingroup$ It really is a bit of a slick derivation. I think the key thing to keep in mind is that they are throwing terms away, but the error is all higher-order so it's alright. $\endgroup$
    – knzhou
    Jun 10, 2019 at 11:53
  • $\begingroup$ @eigenchris Your video on the topic was good. but I really don't prefer this method as the covariant derivative $\nabla_XY$ of a vector field $Y$ is taken along the flow curves of $X$, working with vectors directly can be confusing here. a much simpler "IMO" derivation of the riemann curvature tensor is by writing the parallel transport map along the flow of $X$ : $\Pi_{tX}$ in terms of the covariant derivative along $X$ : $\nabla_X$. see this post: mathoverflow.net/questions/272253/… $\endgroup$ Jul 26, 2021 at 16:27
  • $\begingroup$ this is a post of mine too were I asked about a similar method before seeing the post I linked above: mathoverflow.net/questions/398269/… . this part of the wiki page about parallel transport is helpful : en.wikipedia.org/wiki/… . $\endgroup$ Jul 26, 2021 at 16:36
  • $\begingroup$ "iirc" you didn't introduce in your videos on connections the parallel transport "map" itself. you discussed it in the context of connections and connection coefficients but I think it would be clearer to write the connection in terms of the parallel transport map to clearly see the relation. $\endgroup$ Jul 26, 2021 at 16:38
  • $\begingroup$ At last I just want to say that your two series's on tensors were really beautiful. I remember that I didn't really understand tensors until I saw your series and Justin C. Feng's "Poor man's Introduction to tensors". $\endgroup$ Jul 26, 2021 at 18:46

1 Answer 1

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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 $$\begin{align*}-\delta\mathbf{A}^\mathrm{(field)}_0&=\mathbf{A}^\mathrm{(field)}_0-\Gamma_{4\to 0}\ldots\Gamma_{1\to 2}\Gamma_{0\to 1}(\mathbf{A}^\mathrm{(field)}_0)\\&=(\nabla_\mathbf{u}\nabla_\mathbf{v}\mathbf{A}^\mathrm{(field)}|_0-\nabla_\mathbf{v}\nabla_\mathbf{u}\mathbf{A}^\mathrm{(field)}|_0+\nabla_{[\mathbf{v},\mathbf{u}]}\mathbf{A}^\mathrm{(field)}|_0)\Delta a\Delta b\\ &=(\nabla_\mathbf{u}\nabla_\mathbf{v}\mathbf{A}^\mathrm{(field)}-\nabla_\mathbf{v}\nabla_\mathbf{u}\mathbf{A}^\mathrm{(field)}-\nabla_{[\mathbf{u},\mathbf{v}]}\mathbf{A}^\mathrm{(field)})|_0\Delta a\Delta b\\ &=R(\mathbf{u},\mathbf{v})\mathbf{A}^\mathrm{(field)}|_0\Delta a\Delta b. \end{align*}$$

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  • $\begingroup$ In the first step. how can you just equate the term in LHS that lives in the tangent space of $1$ directly to RHS that lives at $0$? don't u need a parallel transport map for that? $\endgroup$ Jul 24, 2021 at 20:52
  • $\begingroup$ Exactly, they are related by a parallel transport, which to lowest order in $\Delta a$ is given by the RHS. $\endgroup$
    – shangyung
    Jul 26, 2021 at 15:43
  • $\begingroup$ Do you mean something like $\Gamma_{0\to 1}(\Delta a \nabla_{\mathbf{u}}\mathbf{A}|_0) \approx \Delta a \nabla_{\mathbf{u}}\mathbf{A}|_1$? $\endgroup$ Jul 26, 2021 at 16:17
  • $\begingroup$ No, I mean $\mathbf{A}_1^{(\mathrm{mobile})}=\Gamma_{0\to 1}(\mathbf{A}_0^{(\mathrm{field})})\approx\mathbf{A}_1^{(\mathrm{field})}-\Delta a\nabla_\mathbf{u}\mathbf{A}^{(\mathrm{field})}|_0$. $\endgroup$
    – shangyung
    Jul 27, 2021 at 7:34

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