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To explain my question I need to discuss two notions from differential geometry (the Riemann curvature tensor and sectional curvature), but this is really a linear algebra / Euclidean geometry question, so I will provide background material.

The Riemann curvature tensor is a certain linear function $R(u,v,w,z)$ with four vector inputs and one real output: $$R:\Bbb R^n\times\Bbb R^n\times\Bbb R^n\times\Bbb R^n\to\Bbb R.$$ Such a function has, a priori, $n^4$ degrees of freedom (one for every choice of four basis vectors). However, the Riemann tensor is required to satisfy a few requirements:

  • $R(u,v,w,z)=R(w,z,u,v)$ (interchange symmetry)

  • $R(v,u,w,z)=-R(u,v,w,z)$ and $R(u,v,z,w)=-R(u,v,w,z)$ (antisymmetry)

  • $R(u,v,w,z)+R(v,w,u,z)+R(w,u,v,z)=0$ (cyclic relation)

These symmetries end up bringing the number of degrees of freedom of $R$ from $n^4$ down to $\frac{n^2(n^2-1)}{12}$.

We can also define something called the sectional curvature $K$, defined as follows. Suppose $\sigma$ is a plane through the origin, and let $u$ and $v$ be orthonormal vectors spanning $\sigma$. Then $K(\sigma)$, also denoted $K(u,v)$, is defined to be $R(u,v,u,v)$. In general, if $u$ and $v$ span $\sigma$ but are not necessarily orthonormal, then $$K(\sigma)=\frac{R(u,v,u,v)}{(\|u\|\|v\|\sin\theta)^2}=\frac{R(u,v,u,v)}{\|u\|^2\|v\|^2-(u\cdot v)^2}$$ where $\theta$ is the angle between $u$ and $v$. It is an algebraic fact that $K(\sigma)$ does not depend on the choice of two generators of the plane $\sigma$. It is also an algebraic fact that $K$ determines $R$: if you know the sectional curvature of all planes, then you know all values of the curvature tensor.

(The underlying differential geometry idea is that $R$ is used to describe the curvature at a point in curved high-dimensional space, and $K$ relates it to two-dimensional Gaussian curvature.)

This raises the following natural question. Can one choose $\frac{n^2(n^2-1)}{12}$ planes in $\Bbb R^n$ in a natural way such that the sectional curvatures $K$ of these planes determine the Riemann tensor $R$? If these planes are chosen generically or randomly, then they will almost surely determine $R$. However, is there a particularly symmetric way to do this? Note that, since the degrees of freedom match exactly, the sectional curvatures of these planes will be independent.

When $n=3$, we need $6$ planes through the origin. I believe may select the planes connecting opposite edges of the cube. When $n=4$, we need $20$ planes. I believe we may select the planes through the centers of the Boerdijk–Coxeter helix rings of the $600$-cell. (Take care that these are planes, not hyperplanes, so they each intersect the hypersphere in a circle.) Does this generalize somehow?

Akiva Weinberger
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  • So you are implicitly identifying $\mathbb R^n$ with the tangent space $T_p M$ of some Riemannian $n$-manifold $M$ at some point $p \in M$? – Lee Mosher Apr 18 '25 at 15:45
  • @LeeMosher Yes. – Akiva Weinberger Apr 18 '25 at 15:52
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    By the way I've long wondered something related to your question: thinking of $K$ as a function defined on the Grassmanian of 2-dimensional subspaces of $\mathbb R^n$ --- which is a perfectly nice manifold, even a symmetric space --- what kinds of functions does one get? – Lee Mosher Apr 18 '25 at 15:56
  • @mr_e_man Fixed. – Akiva Weinberger Apr 22 '25 at 17:23
  • @LeeMosher My suspicion is that for any multilinear function $S:(\Bbb R^n)^4\to\Bbb R$ whose values on planes are well-defined (eg for any plane $\sigma$ and any pair of orthonormal vectors $u,v$ and $u',v'$ generating it, $S(u,v,u,v)=S(u',v',u',v')$), there exists a function $R:(\Bbb R^n)^4\to\Bbb R$ satisfying these symmetries having the same values on planes. I do not have a proof, however, and this would be a good subject of another Math SE question. – Akiva Weinberger Apr 22 '25 at 17:54
  • $K$ is completely determined by a subset of $n^2$ of the coefficients of $R$, of which at most $\binom n2$ are distinct. So at most that many degrees of freedom, but it alone is enough to determine all of $R$ with its $\frac{n^4-n^2}{12}$ degrees of freedom? – Paul Sinclair Apr 23 '25 at 02:22
  • @PaulSinclair Consider $K(\langle e_1,e_2+e_3\rangle)$, $K(\langle e_1+e_2,e_3+e_4\rangle)$, etc (where $e_1,e_2,e_3,e_4$ are basis vectors). Counting planes of these types gives you $\binom n2+3\binom n3+3\binom n4$ data points, which is actually more than needed (in fact, exactly $\binom n4$ more), so they must not be independent. – Akiva Weinberger Apr 23 '25 at 02:55

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