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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 $R(x,y)x$?

Zev Chonoles
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Lonely Penguin
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1 Answers1

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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.

  • How do I know that the choosen vector fields is well defined? – Lonely Penguin Jun 02 '12 at 16:45
  • Yes, that is what I wrote. What do you mean well-defined Jr? That they exist? – M.B. Jun 02 '12 at 16:46
  • I mean if $A(p)=x, B(p)=y$ is it true that $R(A,B)A=R(X,Y)X$? – Lonely Penguin Jun 02 '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. –  Jun 02 '12 at 16:50
  • (Existence: use local coordinates, that choice the choice of vector field does not matter: see Thomas' comment.) edit: oh, a second too slow :-) – M.B. Jun 02 '12 at 16:50
  • I still don't understand why $R(A,B)A=R(X,Y)X$... – Lonely Penguin Jun 02 '12 at 16:54
  • This equation is the definition of $R(X,Y)X$. The point is that it is well defined, that is, if you choose $C,D$ which coincide with $A, B$ in $p$, you will get the same result (in $p$). –  Jun 02 '12 at 16:55
  • yes, but why I'll get the same result, if $R(X,Y)X(p)$ depends not only of $X(p),Y(p)$ but of $R^{l}_{ijk}(p)$ – Lonely Penguin Jun 02 '12 at 17:01
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    $R^l_{ijk}(p)$ does not depend on $A, B, C, D$. It depends on $p$ alone in the given coordinate system. –  Jun 02 '12 at 17:02
  • that's right, thanks :) – Lonely Penguin Jun 02 '12 at 17:09