Intuition for the indices of the Riemann tensor I have some trivial questions about the Riemann tensor.

*

*$R_{\alpha\beta\mu\nu}$, is the first two indices reserved for the components of the geometric object and the last two for the path it travels in?


*What is a good physical intuition for $\alpha\beta$ in the tensor? Are they the components of the Geometric object that is being transported?


*If its a 4D space that the geometric object is being transported in why is there only two indices for the path ($\mu\nu$)?
 A: One may determine the answer to this question by re-tracing the derivation of the Riemann curvature tensor. You can try to check things using the derivation via the lack of commutativity of covariant derivatives, below I will give it via the integral/parallel-transport derivation of the curvature tensor.
A vector field is sometimes called homogeneous (or sometimes homogeneous and isotropic) if it's magnitude and direction are constant at all points. In 3D for example this means
$$\mathbf{F}(\mathbf{r}) = \mathbf{F}(\mathbf{r}+d\mathbf{r})$$
should hold, i.e. $\mathbf{F}(\mathbf{r}+d\mathbf{r})$ is the result of displacing the vector $\mathbf{F}$ parallel to itself, so that
$$\mathbf{F}(\mathbf{r}) = \mathbf{F}(\mathbf{r}+d \mathbf{r}) = \mathbf{F}(\mathbf{r}) + d \mathbf{F}(\mathbf{r})$$
implies the requirement that $d \mathbf{F}(\mathbf{r}) = 0$ which in in curvilinear coordinates $(x^i)$ reads as
$$d \mathbf{F}(\mathbf{r}) = dx^i \partial_i \mathbf{F}(\mathbf{r}) = 0.$$
For a general vector field in a curved space, we can always infinitesimally shift a vector at a point in a direction parallel (i.e. not changing the direction) to the initial vector without changing it's magnitude, and so we can take the above as the condition for such a 'parallel displacement'.
If we expand $\mathbf{F}$ in a (position-dependent) contravariant basis
$$\mathbf{F}(\mathbf{r}) = F_j(\mathbf{r}) \mathbf{e}^j(\mathbf{r})$$
then even if $\mathbf{F}$ if homogeneous and isotropic, although the overall vector (i.e. the sum of the components contracted against the basis) should not change, the components and basis can themselves separately change as long as the whole thing does not change, and more generally doing a parallel transport of an arbitrary vector field the same behaviour can occur. Thus $d \mathbf{F} = dx^i \partial_i \mathbf{F}(\mathbf{r}) = 0$ reduces to requiring
$$dx^i \partial_i \mathbf{F}(\mathbf{r})  = dx^i \partial_i [F_j(\mathbf{r}) \mathbf{e}^j(\mathbf{r})] = dx^i (\partial_i F_j) \mathbf{e}^j + dx^i  F_j (\partial_i \mathbf{e}^j) = 0$$
and on defining $\partial_i \mathbf{e}^j(\mathbf{r}) = - \Gamma_{ik}^j \mathbf{e}^k(\mathbf{r})$ i.e. the change in the $\mathbf{e}^j$ basis vector is a new vector which is just linear combination of the basis we started with using some new coefficients $- \Gamma_{ik}^j$ called connection coefficients, this becomes
$$(\nabla_i F_j) \mathbf{e}^j(\mathbf{r}) = 0$$
where I defined
$$\nabla_i F_j = F_{j;i} = \partial_i F_j - \Gamma^k_{ji} F_k = F_{j,i} - \Gamma_{ji}^k F_k.$$
This is just the definition of the covariant derivative of a covariant vector field. For a homogeneous vector field we have the extra condition $\nabla_i F_j = 0$ and so this can naturally be taken as the condition on the covariant components of a general vector field for 'parallel displacement' to occur. The reason for the minus sign is because in the contravariant case, $\mathbf{F} = F^i \mathbf{e}_i$, the definition $\partial_i \mathbf{e}_j$ is taken by convention with a plus sign, and you can derive the minus sign as a consequence from differentiating a scalar which is the dot product of two vector fields of the above form.
If we sum up the changes from moving around through a homogeneous and isotropic vector field along a closed path $\partial D$ we should have $\mathbf{F}(\mathbf{r}) = \mathbf{F}(\mathbf{r}) + \oint d \mathbf{F}(\mathbf{r})$, so that $\oint d \mathbf{F}(\mathbf{r}) = 0$, which reduces to requiring
$$\oint dx^l \partial_l F_j - \oint \Gamma_{jl}^i F_i dx^l = 0.$$
In other words, the direct change in the components over $\partial D$, i.e. $\Delta F_j = \oint dx^l \partial_l A_j$, is counteracted by the second term which expresses how the basis vector change, so that:
$$\Delta F_j = \oint_{\partial D} \Gamma_{jl}^i F_i dx^l$$
Using Stokes' theorem this reads as
\begin{align}
\Delta F_j &= \oint_{\partial D} \Gamma_{jl}^i F_i dx^l \\
&= \int_{D} d( \Gamma_{jl}^i F_i dx^l) \\
&= \iint_{D} \partial_k (\Gamma_{jl}^i F_i) dx^k \wedge dx^l \\
&= \frac{1}{2} \iint_{D} [ \partial_k (\Gamma_{jl}^i F_i) - \partial_l (\Gamma_{jk}^i F_i)] dx^k \wedge dx^l.
\end{align}
We interpret this as follows:

*

*The first equality is telling us that the change in the $j$'th component of $\mathbf{F}$ around a closed loop is just due to changes in the basis of the vector field $\mathbf{F}$ as we move around the loop: at each point we take the vector field and slightly perturb it, re-express the changed basis as a linear combination of the basis at the point we perturbed about, using the vector field components at that point, and take the limit, then take a limit (to be clear I'm just describing what you do when you set up a usual (line) integral here, but adapted to the case at hand).

*In the second equality we just used Stokes' theorem to say these changes
over the boundary are due to changes inside the region bounded by
that boundary.

*In the third equality I just wrote it in components
and the fourth equality just makes explicit the anti-symmetry
implicit in the third equality. Note the integrand of the final
equality is just the change in the change of the basis about an
infinitesimal closed loop inside the region bounded by $\partial D$,
(where I am of course again really just interpreting the integrand of
the area integral in terms of the line integral it is equivalent to).
You can expect that the change in how the basis changes is due to
both to the change in the connection coefficients (which express how
the basis at a new nearby point is just a linear combination of the
basis at the initial point) and due to where the initial vector field
is even evaluated, expressed by the change in the $F_i$ components
themselves. We will see this below.

*Already we can expect that the area integrand can be re-expressed the way it was with the line integral integrand, as something contracted against the components of the vector field, so that we can interpret the integral as summing up the changes in this new 'something' as we move over the area (we were changing the basis as we moved around the loop, we're now going to be summing up changes in the 'curvature' of the area), because the changes in the $F_i$ components themselves can be re-expressed in terms of more connection coefficients, so the whole thing will be 'something' contacted against the $F_i$ components...

Expanding out the derivatives in the area integrand we find
$$\Delta F_j = \frac{1}{2} \iint_{D} [ (\partial_k \Gamma_{jl}^i) F_i + \Gamma_{jl}^i (\partial_k F_i) - (\partial_l \Gamma_{jk}^i) F_i -  \Gamma_{jk}^i (\partial_l F_i)] dx^k \wedge dx^l  $$
The meaning of the first and third term is that they are just the change in the connection coefficients. The meaning of the second and fourth term is that they express a change due to where we even evaluate the initial vector field, so these changes themselves can also be re-expressed in terms of more connection coefficients
$$\Delta F_j = \frac{1}{2} \iint_{D} [ (\partial_k \Gamma_{jl}^i) F_i + \Gamma_{jl}^i \Gamma_{ik}^m F_m  - (\partial_l \Gamma_{jk}^i) F_i -  \Gamma_{jk}^i \Gamma_{il}^m F_m ] dx^k \wedge dx^l $$
We now just pull out an overall $F_i$
\begin{align}
\Delta F_j &= \frac{1}{2} \iint_{D} [ (\partial_k \Gamma_{jl}^i) + \Gamma_{jl}^m \Gamma_{mk}^i - (\partial_l \Gamma_{jk}^i) -  \Gamma_{jk}^m \Gamma_{ml}^i  ] F_i dx^k \wedge dx^l \\
&= \frac{1}{2} \iint_{D} R^i_{\ jkl} F_i dx^k \wedge dx^l .
\end{align}
The quantity
$$R^i_{\ jkl} = \partial_k \Gamma_{jl}^i -  \partial_l \Gamma_{jk}^i  + \Gamma_{jl}^m \Gamma_{mk}^i -  \Gamma_{jk}^m \Gamma_{ml}^i $$
represents the change in the $j$'th covariant component of $\mathbf{F}$ around a closed path, projected onto the $i$'th covariant component of $\mathbf{F}$, caused by the change in the basis around a loop bounding a region with area element $dx^k \wedge dx^l$. The physical reason for why the basis changed over the closed loop giving a non-zero value in the closed loop integal (and so the surface area integral) is that the region was 'curved' allowing the basis to accumulate non-trivial changes, thus a natural name for the tensor $R^i_{\ jkl}$ expressing this is the curvature tensor.
If we used the non-conventional notation
$$ (R_j^{ \ \ i})_{kl} = \partial_k (\Gamma_{j}^{\ \ i})_{l} -  \partial_l (\Gamma_{j}^{\ \ i})_{k}  + (\Gamma_{j}^{\ \ m})_{l} (\Gamma_{m}^{\ \ \ i})_{k} -  (\Gamma_{j}^{\ \ m})_{k} (\Gamma_{m}^{\ \ \ i})_{l} $$
it would look as close to matrix multiplication as possible, where the $_j^{\ \ \ i}$ indices act on the covariant components $F_i$ of $\mathbf{F}$, while the $kl$ indices act on the $dx^i$ vector components.
The quantity $R_{ijkl} = g_{im} R^m_{\ jkl}$ is then just expressing the change in the $j$'th covariant component of $\mathbf{F}$ in terms of the $i$'th contravariant component of $\mathbf{F}$ via
$$\Delta F_j = \frac{1}{2} \iint R_{ijkl} F^i dx^k \wedge dx^l.$$
This or reference 2 has a good reminder of the difference between contravariance and covariance.
So with this understanding, the answer to the first question is: yes, the first two indices act on the `geometric object' i.e. the vector field we are parallel transporting, the last two indices represent the components of the infinitesimal area we integrated against. The answer to the second question is that $R^i_{\ jkl}$ is just expressing the change of the $j$'th covariant component $\Delta F_j$ as a multiple of the $i$'th covariant component $F_i$, clearly it shouldn't matter that we began with covariant components and so $R^i_{\ jkl}$ should also express the change of the $j$'th contravariant component $\Delta F^i$ as a multiple of the $i$'th contravariant component $F^j$. Since a scalar should not change under parallel transport, and since there is no basis whose change can cancel the change in the 'components' of the scalar field, a scalar field should satisfy $\Delta \phi = 0$. By evaluating $\Delta (F^i G_i) = 0$, noting $F^i G_i$ is a scalar, we can derive
$$\Delta F^i = - \frac{1}{2} \iint R^i_{\ jkl} F^j dx^k \wedge dx^l.$$
The answer to the third question should now be clear, there are only two indices $\mu \nu$ ($kl$ in my notation) because of the meaning in terms of projecting the line integral over an infinitesimal loop onto an area in the $\mu \nu$ direction, it doesn't matter what dimension $\geq 2$ we do all this in, I set things up in 3D above, there's no change if you do it in 4D etc...
References:

*

*Landau and Lifshitz, Classical Theory of Fields;

*Borisenko and Tarapov, Vector and Tensor Analysis With Applications;

*Relativity 102b: Keys to Relativity - Covariance and Contravariance.

