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.

  • $\begingroup$ 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


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

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.