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If we have a Riemannian manifold $(M,g)$ with constant sectional curvature $C$ I have proven that the $(1,3)$ curvature tensor $$R(X,Y)Z = C(\langle Y,Z\rangle X - \langle X,Z\rangle Y)$$ and that the $(0,4)$ Riemannian curvature tensor is given by $$\operatorname{Rm}(X,Y,Z,W) = C(\langle X,W\rangle\langle Y,Z\rangle - \langle X,Z\rangle \langle Y,W\rangle)$$ now I would like to prove that the Ricci curvature, defined by $\operatorname{Ric}(X,Y) = \sum_{i=1}^n \operatorname{Rm}(X,e_i,e_i,Y)$ is given by $C(n-1)g(X,Y)$. I'm sure this should be a somewhat simple exercise given that we know what the Riemannian curvature is in this situation, but I cannot get any further than: $$C\sum_{i=1}^n (g(X,Y)g(e_i,e_i) - g(X,e_i)g(Y,e_i))$$

Any help is appreciated!

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    $\begingroup$ Note that proper notation is $\langle X,W\rangle$, not $<X,W>$. And if you google "latex symbols", you can find things like this. $\qquad$ $\endgroup$ – Michael Hardy Apr 20 '17 at 20:26
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    $\begingroup$ @MichaelHardy : $\langle\cdot, \cdot\rangle $ is definitely more good looking, but $<\cdot, \cdot>$ is also used a lot even in research paper. $\endgroup$ – user99914 Apr 21 '17 at 7:53
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    $\begingroup$ @JohnMa : That form is for occasions when you're limited to characters on the keyboard. It's not used in published journals whose typesetting software is not so limited, and it's not used in competently written LaTeX documents. Where LaTeX or MathJax is available, it's an incorrect usage. $\endgroup$ – Michael Hardy Apr 21 '17 at 17:02
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    $\begingroup$ @JohnMa : Notice also the lack of proper spacing: $3 - \langle a,b\rangle$ versus $3 - < a,b> \qquad$ $\endgroup$ – Michael Hardy Apr 21 '17 at 17:03
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Recall that $\{e_i\}$ is an orthonormal basis so $g(e_i, e_i) = 1$.

Writing $X = \sum_{j=1}^nX^je_j$ and $\sum_{k=1}^nY^ke_k$ we see that

$$g(X, e_i)g(Y, e_i) = g\left(\sum_{j=1}^nX^je_j, e_i\right)g\left(\sum_{k=1}^nY^ke_k, e_i\right) = \sum_{j,k=1}^nX^jY^kg(e_j, e_i)g(e_k, e_i) = X^iY^i.$$

Now note that

$$g(X, Y) = g\left(\sum_{j=1}^nX^je_j, \sum_{k=1}^nY^ke_k\right) = \sum_{j,k=1}^n X^jY^kg(e_j, e_k) = \sum_{j=1}^nX^jY^j = \sum_{i=1}^nX^iY^i.$$

Therefore,

\begin{align*} C\sum_{i=1}^n(g(X, Y)g(e_i, e_i) - g(X, e_i)g(e_i, Y)) &= C\sum_{i=1}^n(g(X, Y) - X^iY^i)\\ &= C\sum_{i=1}^ng(X, Y) - C\sum_{i=1}^nX^iY^i\\ &= Cng(X, Y) - Cg(X, Y)\\ &= C(n-1)g(X, Y). \end{align*}

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