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do Carmo gives a definition of sectional curvature as follows:

$$K(x,y) = \frac{\langle R(x,y)x,y\rangle}{|x\times y|^2}$$

where $x,y \in T_pM$ are linearly independent vectors.

My question: The curvature of a riemannian manifold is a correspondence that associates, for each vector fields $X,Y$, a linear map $R(X,Y)$, which takes vector fields in vector fields. In the definition above $x,y \in T_pM$ which means they are not vector fields, so how could one interpret $R(x,y)x$?

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one upvote for being perseverant ;-) – user20266 Jun 2 '12 at 17:23
up vote 5 down vote accepted

One of the main points about tensor fields is that they can be evaluated pointwise. That is, you don't need vector fields, but only tangent vectors. This applies to all three arguments of the curvature tensor (your $x, y$).

Edit after reading M.B.s comment: the point about his remark is, that the result will not depend on the choice of vector fields. Too bad, he removed the comment. It said (wording may differ): choose vector fields $X, Y$ such that $X(p)=x, Y(p)= y$ and evaluate along these.

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How do I know that the choosen vector fields is well defined? – Jr. Jun 2 '12 at 16:45
Yes, that is what I wrote. What do you mean well-defined Jr? That they exist? – M.B. Jun 2 '12 at 16:46
I mean if $A(p)=x, B(p)=y$ is it true that $R(A,B)A=R(X,Y)X$? – Jr. Jun 2 '12 at 16:49
just choose a coordinate neighbourhood around $p$. In this, e.g, $x$ will have local representations $x=(x_1,\ldots,x_n)$. Extend that as a constant vector field locally. That the result does not depend on the extension is a result you should find in do Carmo's book. – user20266 Jun 2 '12 at 16:50
$R^l_{ijk}(p)$ does not depend on $A, B, C, D$. It depends on $p$ alone in the given coordinate system. – user20266 Jun 2 '12 at 17:02

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