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Let me first quote the definition of Ricci curvature from Wikipedia. Let $M$ be an n-dimensional Riemannian manifold equipped with its Levi-Civita connection $\nabla$. The Riemannian curvature tensor $R$ of $M$ is the $(1,3)$ tensor defined by $$ R(X,Y)Z=\nabla_X\nabla_YZ-\nabla_Y\nabla_XZ-\nabla_{[X,Y]}Z $$ on vector fields $X,Y,Z$. Let $T_pM$ denote the tangent space of $M$ at a point $p$. For any pair $(\xi,\eta)$ of tangent vectors at $p$, the Ricci tensor $R(\xi,\eta)$ evaluated at $(\xi,\eta)$ is defined to be the trace of the linear map $T_pM\rightarrow T_pM$ given by $$ \zeta \mapsto R(\zeta,\eta)\xi. $$

My question is why is this a reasonable definition? It seems to me that there is more natural linear map $T_pM\rightarrow T_pM$ given by $$ \zeta \mapsto R(\xi,\eta)\zeta $$ and we may take the trace of this map. In this case what shall we get?

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Isn't this an antisymmetrical map (with respect to the scalar product on the tangent space given by the metric)? Its trace ought to be $0$. –  Olivier Bégassat Oct 7 '12 at 5:48
    
You are right. This obvious map doesn't work. But I still wonder where the first linear map comes from. –  M. K. Oct 7 '12 at 6:23
    
Explaining why a definition is "reasonable" turns often out to be quite tricky. "Why should the definition not be reasonable?" However, as far as geometric intuition can justify a definition this document www.yann-ollivier.org/rech/publs/visualcurvature.pdf and the question math.stackexchange.com/questions/468651/… might be helpful. –  gofvonx Aug 20 '13 at 14:14

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