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In Riemannian geometry, when we think of the Ricci tensor merely as $Ric(\vec a, \vec b)$, then it's hard to describe it other than saying it's the contraction of the Riemann tensor. But when we realize that, due to its symmetry, it can be written as

$$Ric(\vec a, \vec b) = \frac12(Ric(\vec a + \vec b, \vec a + \vec b) - Ric(\vec a, \vec a) - Ric(\vec b, \vec b))$$

then we see it can be described entirely in terms of its action on single vectors, rather than pairs of vectors. And for such a component, we arrive at a really neat interpretation: it is the rate (acceleration?) of convergence of geodesics along that vector.

My question is: can we apply the same kind of logic to the Riemann tensor itself to get alternate, perhaps more intuitive, descriptions of it? (The difference, of course, is that we already have a useful interpretation before applying symmetry: we know that ${R^a}_{bcd}$ is "the $a$-component picked up by the $b$ vector upon parallel transport around the $cd$ loop")

Starting from that first description, we then have to lower the first index, and then, for example, we could focus on the symmetry of exchange of index pairs. In which case we're talking about components of the form $R_{abab}$. Is there a simple geometric description of such a component only in terms of the $ab$ loop (bivector?)? And could we arrive at other descriptions by applying the other symmetries? Or even by applying multiple symmetries at once?

Adam Herbst
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  • Yes; see this answer (note that knowing $R_{abab}$ is essentially the same as knowing the sectional curvature). You can treat the $(0,4)$ Riemann tensor as a bilinear form on $\Lambda^2TM$, and the same argument applies. – Kajelad Feb 27 '22 at 00:38
  • @Kajelad Oh wow, thanks! – Adam Herbst Feb 27 '22 at 19:14
  • @Kajelad Quick follow-up: can we see the degrees of freedom from this formulation? Naively, since the space of bivectors has dimension $m = {n\choose 2}$, I would think a quadratic form on it would have $\frac12 m(m+1)$. But that is a bit higher than the true value of $\frac{1}{12}n^2(n^2 - 1)$ for the Riemann tensor... – Adam Herbst Feb 28 '22 at 14:53
  • @Kajelad Or maybe put it this way: for the Ricci tensor, there are $\frac12 n(n-1)$ degrees to choose the eigen-frame, and $n$ to choose the curvatures in that frame. Is there an analogous way of visualizing the sectional curvatures? – Adam Herbst Feb 28 '22 at 16:12
  • The fact that $R$ is a symmetric bilinear form comes from the three interchange symmetries $R_{abcd}=-R_{bacd}=-R_{abdc}=R_{cdab}$. To fully describe the symmetries at a point, you also need the first Bianchi identity $R_{a(bcd)}=0$. It might be worth noting that these aren't "degrees of freedom" in the typical sense; you can't freely prescribe the Riemann/Ricci tensors since they depend on the metric/connection. – Kajelad Feb 28 '22 at 21:33
  • @Kajelad Got it, thanks again – Adam Herbst Feb 28 '22 at 23:05

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