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Recently I have been using a video guide/explanation type thing on YouTube to gain a better understanding of the Riemann tensor and it's links to the Ricci tensor: https://youtu.be/sBDGPIuJ-HA . I understand the formulation of the tensor using a parallelogram he uses however and just in general how the tensor is formed. However I'm unsure as to what indices mean, how the dimensions are managed and how contraction works- specifically.

Is it like the stress energy tensor where certain information is kept in certain indices, such that information concerning volume deviation is contained in certain indices, meaning those indices survive the contraction and remain in the Ricci tensor?

I guess my real question is where do the indices come from and what do they represent?

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    $\begingroup$ I've read the Wikipedia page many times and it doesn't cover specifically what the indices represent. Additionally I explain in the question that I'm comfortable with how it's formulated; Christoffel symbols etc.. and it's also evident from the question that I understand the act of contraction. I'm looking specifically for an intuition behind the indices. Thank you for your suggestions it's just not quite what I was looking for. $\endgroup$ – Oliver Gregory Jul 19 '17 at 22:43
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You should familiarize yourself with a geometric interpretation of the Riemann-tensor:

The Riemann-tensor expresses the path-dependence of parallel transport.

Now, best way to ascertain path-dependence is to parallel transport a vector around a closed loop. Because the curvature tensor depends on only one point (wherever it is evaulated), the curvature tensor cannot provide a finite description of parallel transport around a closed loop, only infinitesimal.

Let us define the index structure of the Riemann tensor now as $$ R_{ab\ \ d}^{\ \ \ \ c}=\partial_a\Gamma^c_{bd}-\partial_b\Gamma^c_{ad}+\Gamma^c_{ae}\Gamma^e_{bd}-\Gamma^c_{be}\Gamma^e_{ad}$$ or as $$ R_{ab\ \ d}^{\ \ \ \ c}X^d=(\nabla_a\nabla_b-\nabla_b\nabla_a)X^c. $$

This index structure is precisely what is necessary to express the concept of infinitesimal parallel transport around a loop, it is because we need

1) a loop,

2) a linear transformation that describes the change in the vector.

  • On a manifold, vectors do not describe finite displacements, unlike in euclidean geometry, but they do describe infinitesimal displacements, so we can take two vectors, $v^a$ and $w^a$ to span an infinitesimal parallelogram. But as you know (cross products, determinants), generalized, signed area measures are antisymmetric, so $v^aw^b-w^av^b$ is what describes the "infinitesimal parallelogram" spanned by $v$ and $w$. $$ $$ For this reason, the first two indices of the Riemann-tensor are antisymmetric, because they "accept" a parallelogram: $$ R_{ab\ \ d}^{\ \ \ \ c}v^aw^b. $$ Now we get an object that has two indices remaining: $$ \mathfrak{O}^c_{\ d}=R_{ab\ \ d}^{\ \ \ \ c}v^aw^b. $$

    • Now that we have a loop, we need a linear transformation. A lin. transformation is provided by a tensor with one up and one down index, so $\mathfrak{O}^c_{\ d}$ is our transformation. The expression $$ \mathfrak{O}^c_{\ d}X^d $$ describes the transformation the vector $X$ undergoes, as it is parallel transported along the closed loop spanned by $v$ and $w$. However the Levi-Civita connection (eg. the covariant derivative described by Christoffel-symbols) preserves the lengths of vectors during parallel transport, so the vector $X$ can only change by rotation. $$ $$ This gives us the reason why the Riemann-tensor is antisymmetric in the $c,d$ indices as well (provided they are simultaneously raised or lowered). If you have a one-parameter family of rotations $O_\epsilon$ such that $O_\epsilon=1$ (the rotation at $\epsilon=0$ is the identity), then to first order in $\epsilon$ we have $$ O_\epsilon=1+\epsilon \mathfrak{O}, $$ where $\mathfrak{O}$ is antisymmetric. If instead of euclidean rotations, we consider rotations of a metric of signature (-+++) (these are essentially Lorentz-transforms), then the "antisymmetry" property applies only if this matrix has its indics raised or lowered.

In conclusion, $$ R_{ab\ \ d}^{\ \ \ \ c}v^aw^bX^d $$ describes the infinitesimal change the $X$ vector picks up, when it is parallel transported along the infinitesimal parallelogram spanned by $v$ and $w$. We need two antisymmetric indices to give a parallelogram ($a$ and $b$) and one index to act on a the vector to be transported ($d$). That's 3 indices, but the end result of the parallel transport is a vector itself, so that's the fourth index ($c$).

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  • $\begingroup$ Is this tensor like the Ricci and Stress-energy tensors in the sense that each index can take on one of 4 values: 0, 1, 2 or 3? $\endgroup$ – Oliver Gregory Jul 20 '17 at 13:36
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    $\begingroup$ @OliverGregory Yes. On a four-dimensional spacetime, tensor indices take on the values $0,1,2,3$, and the curvature tensor is no exception. $\endgroup$ – Bence Racskó Jul 20 '17 at 13:39
  • $\begingroup$ +1 but if you have a second, could tell me if my probably naive intuition as to the physical meaniing of the indices is true, that is the comments at the bottom of my answer . Thanks very much $\endgroup$ – user163104 Jul 21 '17 at 7:38
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This is my intuitive understanding of the origin of the indices.

The indices arise from the fact that a change in curvature of one dimension impacts all of the curvatures inherent in the other 3 direction.

Locally, we are allowed a transform to a flat space on any sufficiently limited region of a 4 D manifold, but globally this type of transformation is forbidden due to the presence of the second derivatives of the underlying metric tensor, upon which the Riemann Tensor components are derived.

In other words, a wrinkle in space time cannot be removed, intrinsic curvature is implicit in the manifold.

So this is where the indices arise. 4 is the minimum number required "to keep track of" the effects of curvature in one direction on the curvature of the other directions.

That's why they are needed, and that is the physical property they represent.

I admit to be slightly worried that by asking how the indices arise, you are looking for a deeper meaning than I have listed, but in that case I can put hope you get an answer from someone who understands what you after. I self study, so please also bear that in mind.

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    $\begingroup$ "The indices arise from the fact that a change in curvature of one dimension impacts all of the curvatures inherent in the other 3 direction." That simple little sentence is almost exactly what I was looking for. Thank you. $\endgroup$ – Oliver Gregory Jul 20 '17 at 8:36
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    $\begingroup$ I intend to, that's why I'm holding out on accepting. Thank you still. $\endgroup$ – Oliver Gregory Jul 20 '17 at 9:31
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    $\begingroup$ @BobBee Hi Bob, I am CountTo10 from a while back. If you have a second could you just look at the comment from my answer The indices arise from the fact that a change in curvature of one dimension impacts all of the curvatures inherent in the other 3 direction and tell me if you agree, just yes or no. The rest of the answer is telling Oliver stuff already known to him, but it might help someone else. But mainly another opinion on the significance of the indices is all I ask. Thank you very much. $\endgroup$ – user163104 Jul 20 '17 at 10:21

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