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I have seen two different expositions of the meaning of the indices of the Riemann tensor metric:

On the first one, the indices $\color{red}{\mu\;\nu}$ would indicate the tangent vectors to the coordinates defining the sides of the infinitesimally small loop around a point, through which parallel transport takes place, and they are the last two indices:

$$R^{\alpha}{}_{\color{blue}\beta\,\color{red}{\mu\,\nu}}$$

while $\color{blue}\beta$ is the basis vector being transported. Further, both $\alpha$ and $\beta$ would come from the Christoffel symbols.

On the second one, it is the very last index that corresponds to the vector in the basis vector that it is undergoing parallel transport, whereas $\beta \, \mu$ would define the loop.

In either case, after "feeding" the curvature tensor $2$ vectors, say $\color{red}{\vec \mu}, \,\color{red}{\vec \nu}$ the result will be a matrix. Hence, this matrix can be fed a third vector $\color{blue} {\vec \beta}$. So you feed that matrix a third vector, and I suppose it would yield a vector as the output as follows: The physical interpretation of the output vector (after feeding $3$ vectors) is the effect that parallely transporting that third vector $\color{blue} {\vec \beta}$ along the differentially small parallelogram formed by the $2$ first vectors fed to the tensor, $\color{red}{\vec \mu}, \,\color{red}{\vec \nu}$, would have on the third vector $\color{blue} {\vec \beta}$.

The question is two-pronged:

  1. Which one of the two explanations above is correct, and does it make a difference? How do you make sure you get the order correct if it's not standardized?

  2. Is my interpretation of the meaning of the rank $2$ tensor or matrix resultant after feeding two vectors to the curvature tensor correct?

Antoni Parellada
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  • Sounds like they're just using different conventions for the ordering of the indices. – Nicholas Todoroff Sep 01 '24 at 19:46
  • @NicholasTodoroff Absolutely, but how much freedom is there in the reordering / convention, and what is determined by the derivation of the Riemann tensor from the Christoffel symbols and the metric tensor? Also, what happens when two vectors are fed to the tensor? – Antoni Parellada Sep 01 '24 at 19:48
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    The proper way to think of the Riemann curvature is as an $\text{End}(TM)$-valued $2$-form (see here for a brief explanation). It is an accident of the tangent bundle that these indices run over the same range. The question of how you want to notate these various indices is completely arbitrary; there are only so many ways to do so, and even then based on the (anti) symmetries of the curvature these are all related in some simple fashion. – peek-a-boo Sep 01 '24 at 20:14
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    So as long as you’re consistent, neither is more correct than the other (you can even see I made a little comment in my link above regarding this placement). Also the entries of the matrix are irrelevant; it is the endomorphism which matters; that’s the endomorphism which tells you (at a second order level) what parallel transport around a loop does. The entries of the matrix have the same interpretation as in basic linear algebra: they’re just numbers which represent the endomorphism relative to a given basis. – peek-a-boo Sep 01 '24 at 20:17
  • @peek-a-boo When you say endomorphism between the tangent bundle and itself, what do you mean? I can see how the vector expressing the difference between the beginning and end of the parallel transport is an endomorphism, but it would be a vector, not a matrix. So what is the information in the output of feeding two vectors into the curvature tensor? – Antoni Parellada Sep 01 '24 at 21:02
  • Look at the link. Saying that the curvature is an “$\text{End}(E)$-valued $2$-form on $M$” means that for each point $p\in M$, I have a linear map $R_p:\bigwedge^2(T_pM)\to \text{End}(E_p)$, or what amounts to the same thing, a bilinear alternating map $R_p:T_pM\times T_pM\to\text{End}(E_p)$. So, if we’re further given two tangent vectors $v,w\in T_pM$, then we get an element $R_p(v,w)\in \text{End}(E_p)$, i.e we have a linear map $R_p(v,w):E_p\to E_p$. If you evaluate this linear map on a vector in the fiber, i.e if you take $\xi\in E_p$, then you get an output $R_p(v,w)(\xi)\in E_p$. – peek-a-boo Sep 01 '24 at 21:24
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    Yes, the parallel transport around the loop is an endomorphism $P_{\gamma_{ij},\epsilon,\delta}:E_p\to E_p$, but what I’m now saying is that the endomorphism $R_p\left(\frac{\partial}{\partial x^i}(p),\frac{\partial}{\partial x^j}(p)\right):E_p\to E_p$ is related to this parallel transport map by virtue of being the (negative of the) coefficient of $\epsilon\delta$. – peek-a-boo Sep 01 '24 at 21:27

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