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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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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? –  diverietti Jan 23 '12 at 13:14
2  
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. –  Deane Yang Jan 23 '12 at 13:27
    
oh, I see! Thanks! –  diverietti Jan 23 '12 at 13:39
    
The sectional curvatures determine the Ricci tensor. –  Ian Agol Jan 23 '12 at 21:23
    
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. –  Klaus Kröncke Jan 24 '12 at 16:54

2 Answers 2

up vote 7 down vote accepted

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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That sounds reasonable, thank you! –  Klaus Kröncke Jan 23 '12 at 14:40

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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Do you also have an unknown account? ;) –  M P Jan 23 '12 at 19:38

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