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Hi everybody,

my question is the following: Let $(M^n,g)$ be a Riemannian manifold and $e_1,\ldots,e_n$ be an orthonomal frame in a point. Assume, that we now the sectional curvatures of all planes, spanned by these vectors, i.e. we know the components $R_{ijij}$ of the curvature tensor.

Is it then possible to calculate all other components $R_{ijkl}$?

The problem is, that if I want, e.g. to calculate $R_{ijik}$ by the polarization identity, I need to know the sectional curvature spanned by $e_1,e_j+e_k$ which I assume to be unknown.

Is it then possible to calculate, or at least to estimate the sectional curvature from above by $R_{ijij}$, $R_{jkjk}$ and $R_{ikik}$?

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  • $\begingroup$ Maybe I am missing something. But, given any 2-plane you can choose a orthonormal basis of this 2-plane as a linear combination of the vectors you are giving, and since everything is linear you can thus express the sectional curvature of this 2-plane as a combination of the data you know. But then, you know the full sectional curvature and this determines the Riemann curvature tensor... Where am I wrong? $\endgroup$
    – diverietti
    Jan 23, 2012 at 13:14
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    $\begingroup$ diverietti, I think the question assumes an initial choice of an orthonormal frame and that you know only the sectional curvatures of 2-planes spanned by pairs of vectors from the plane. The question is whether given this limited amount of information (a frame and the corresponding $n(n-1)/2$ sectional curvatures), you can compute or estimate the sectional curvature of an arbitrary 2-plane. $\endgroup$
    – Deane Yang
    Jan 23, 2012 at 13:27
  • $\begingroup$ The sectional curvatures determine the Ricci tensor. $\endgroup$
    – Ian Agol
    Jan 23, 2012 at 21:23
  • $\begingroup$ No it does'nt, by the same argument as above. You have given $n(n-1)/2$ sectional curvatures but the Ricci tensor has $n(n+1)/2$ different components. $\endgroup$
    – Klaus Kröncke
    Jan 24, 2012 at 16:54

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Not if $n > 2$. The full Riemann tensor has $n^2(n^2-1)/12$ different components. The number of sectional curvatures spanned by two basis vectors is $n(n-1)/2$. The former is always larger than the latter if $n > 2$.

As for estimating the sectional curvature, you might want to study the case $n = 3$ first, because everything reduces to studying a $3$-by-$3$ symmetric matrix. There, your question reduces to the question of whether you can estimate the eigenvalues of such a matrix from knowing its diagonal components.

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  • $\begingroup$ Wait a minute Deane, this does not prove anything. Any bilinear form has more parameters than the corresponding quadratic form. Nonetheless, the polarisation formula allows one to go from the quadratic form to the bilinear form. $\endgroup$
    – Mikhail Katz
    Jun 14, 2023 at 15:53
  • $\begingroup$ @MikhailKatz, doesn’t the polarization formula require knowing values of the quadratic form for vectors (here, pairs of vectors) other than the values for only basis vectors? $\endgroup$
    – Deane Yang
    Jun 14, 2023 at 20:25
  • $\begingroup$ Sorry, I misread the question. Arch Stanton answered the question I had in mind. $\endgroup$
    – Mikhail Katz
    Jun 15, 2023 at 7:06
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There is an explicit, but complicated, formula for going from the sectional curvature back to the curvature tensor in equation (1.10) on page 16 of Cheeger and Ebin's book.

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  • $\begingroup$ Thanks! I needed this reference badly. $\endgroup$
    – Liviu Nicolaescu
    Apr 11, 2019 at 13:28

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