Intuition for Interchange Symmetry of the Riemann Tensor

Question

Is there an intuitive/geometric picture for the interchange symmetry of the Riemann tensor? I have seen plenty of algebraic derivations, but would like to understand if the symmetry expresses something intuitive or obvious (in hindsight at least).

Here is what I have worked out so far (please let me know if I've gotten it wrong!)

I don't know if this notation is standard (I can't quite keep track of the variety of conventions) so I apologize if it's off. Work with a torsion free, metric connection, namely the Levi-Civita Connection.

Define the Reimann Curvature Tensor as the infinitesimal rotation matrix resulting from parallel transport of a vector $Z$ around a parallelogram-shaped loop defined by $X$ and $Y$. $$R(X,Y)Z = R_{\mu\nu\lambda}^\sigma X^\mu Y^\nu Z^\lambda$$ (1) Anti-symmetry in the first two indices. Reversing the direction of the loop makes the effect opposite. Therefore $(R_{\mu\nu})^\sigma_\lambda = -(R_{\nu\mu})^\sigma_\lambda$. (These parentheses makes the matrix explicit). In other words $R(X,Y) = - R(Y,X)$. $$R_{\mu\nu\lambda}^\sigma = -R_{\nu\mu\lambda}^\sigma$$

(2) Anti-symmetry in the second two indices. An infinitesimal rotation matrix (like a Lorentz transform) is anti-symmetric when its indices are lowered. Putting the upper index in the last position, $$g_{\sigma\tau} R^\tau_{\mu\nu\lambda} = R_{\mu\nu\lambda\sigma} = -R_{\nu\mu\sigma\lambda}$$

(3) First Bianchi Identity (via this question on Math Stack Exchange). The fact that the connection is Torsion-free forces the lateral faces of a cube to close. The Bianchi identity expresses that they form a triangle. $$R_{[\mu\nu\lambda]}^\sigma = 0$$

(4) Interchange symmetry. From these three properties, we can derive the interchange symmetry. For example in these notes. You can also get it, I have seen, by expanding out the Levi-Civita connection in terms of the metric.

I am looking for something other than pure index juggling! For example, some sort of picture similar to that in the top answer on that Math Stack Exchange post.

(5) Bonus: After this I'm going to get to the Second Bianchi identity, I imagine. Is there any good intuition for that? Wikipedia gives it as the following, though I'm willing to bet that the indices are in a different order than I put them. $$R_{ab[cd;e]}^{}= R_{abcd;e}^{}+R_{abde;c}^{}+R_{abec;d}^{}=0$$

Answer 1

I can elucidate a different perspective on the Riemann tensor. It is helpful to write it not as $R^\sigma_{\mu\nu\lambda}$ but rather as $R^i_{j\mu\nu}$ to start off with to distinguish the two types of indices, despite them running over the same range.

This is because if we consider the frame bundle which is in fact a $GL(d,\mathbb R)$ principal bundle, the connection one form (pulled back by a section) belongs to $\Omega^1(M) \otimes T_eG$, i.e. it is Lie algebra-valued form.

The Lie algebra is matrix valued, so when we take the curvature which is a Lie algebra valued two form, we have two indices $ij$ for this, and a further two $\mu\nu$ indices as it is a two form on the base space.

As such, clearly it must be anti-symmetric in the $\mu\nu$ indices, by what we know about forms. But why also anti-symmetric in $ij$? Recall with the existence of a metric we can restrict to orthogonal frames, and then $ij$ are anti-symmetric precisely because of the anti-symmetric matrices representing the Lie algebra of the orthogonal group.

Answer 2

Naturally the identities will not mean much unless either you are of Riemann sort, or, you start comprehending where it makes sense; put in contexts of say, space-time in GR, and further restricting to only acting upon vectors with physical meaning. For example start with Bel decomposition. Then trace it back to merely the index juggling.

Another obvious first move is the equation $R_{uv}=0$ holding in empty space.

Another one is the Einstein tensor equals Stress energy tensor being the one and only one way to put the principle of general covariance into an equation.