Let $R$ be the curvature of a Riemannian manifold $M$ defined by $$ R(X,Y)Z=\nabla_Y \nabla_X Z - \nabla_X \nabla_Y Z + \nabla_{[X,Y]}Z, $$ where $X,Y,Z$ are vector fields and $\nabla$ is the Riemannian connection of $M$.

Let us indicate as usual $\frac{\partial}{\partial x_i}=X_i$. We put $$ R(X_i,X_j)X_k = \sum_\ell R_{ijk}^\ell X_\ell. $$ Thus, $R_{ijk}^\ell$ are the components of the curvature $R$ in $(U,\mathbf x)$. If $$ X=\sum_i u^i X_i, \qquad Y=\sum_j v^j X_j, \qquad Z=\sum_k w^k X_k, $$ we obtain, from the linearity of $R$, $$ R(X,Y)Z=\sum_{i,j,k,\ell} R_{ijk}^\ell u^i v^j w^k X_\ell. $$ To express $R_{ijk}^\ell$ in terms of the coefficients $\Gamma_{ij}^k$ of the Riemannian connection, we write \begin{align} R(X_i,X_j)X_k &= \nabla_{X_j} \nabla_{X_i} X_k - \nabla_{X_i} \nabla_{X_j} X_k \\ &= \nabla_{X_j} \left(\sum_{\ell} \Gamma_{ik}^\ell X_\ell \right) - \nabla_{X_i} \left(\sum_{\ell} \Gamma_{jk}^\ell X_\ell \right), \end{align} which by a direct calculation yields $$ R_{ijk}^s = \sum_\ell \Gamma_{ik}^\ell \Gamma_{j\ell}^s - \sum_\ell \Gamma_{jk}^\ell \Gamma_{i\ell}^s + \frac{\partial}{\partial x_j} \Gamma_{ik}^s - \frac{\partial}{\partial x_i} \Gamma_{jk}^s. $$

This is taken from Riemannian Geometry by Manfredo do Carmo, pp. 93–94.

My question is how do we justify this direction calculation?

There are two expressions of $R(X_i,X_j)X_k$. Equating those two expressions and taking the $s$-th index of the summation $\ell=1,\ldots,n$, $$ R_{ijk}^s X_s = \nabla_{X_j} \Gamma_{ik}^s X_s - \nabla_{X_i} \Gamma_{jk}^s X_s. $$ Using product rule, I think we obtain that \begin{align} R_{ijk}^s X_s &= \left(\frac{\partial}{\partial x_j} \Gamma_{ik}^s X_s + \Gamma_{ik}^s \nabla_{X_j} X_s \right)- \left(\frac{\partial}{\partial x_i} \Gamma_{jk}^s X_s + \Gamma_{jk}^s \nabla_{X_i} X_s \right) \\ &= \Gamma_{ik}^s \nabla_{X_j} X_s - \Gamma_{jk}^s \nabla_{X_i} X_s + \frac{\partial}{\partial x_j} \Gamma_{ik}^s X_s - \frac{\partial}{\partial x_i} \Gamma_{jk}^s X_s. \end{align} Recalling the definition $\nabla_{X_i} X_k = \sum_\ell \Gamma_{ik}^\ell X_\ell$, \begin{align} R_{ijk}^s X_s &= \sum_\ell \Gamma_{ik}^s \Gamma_{js}^\ell X_\ell - \sum_\ell \Gamma_{jk}^s \Gamma_{is}^\ell X_\ell + \frac{\partial}{\partial x_j} \Gamma_{ik}^s X_s - \frac{\partial}{\partial x_i} \Gamma_{jk}^s X_s \\ &= \sum_s \Gamma_{ik}^\ell \Gamma_{j\ell}^s X_s - \sum_s \Gamma_{jk}^\ell \Gamma_{i\ell}^s X_s + \frac{\partial}{\partial x_j} \Gamma_{ik}^s X_s - \frac{\partial}{\partial x_i} \Gamma_{jk}^s X_s. \end{align} (I achieved the last step by switching the dummy indices: interchanging the roles of $\ell$ and $s$.)

Then I would cancel $X_s$ from both sides to obtain: $$ R_{ijk}^s = \sum_s \Gamma_{ik}^\ell \Gamma_{j\ell}^s - \sum_s \Gamma_{jk}^\ell \Gamma_{i\ell}^s + \frac{\partial}{\partial x_j} \Gamma_{ik}^s - \frac{\partial}{\partial x_i} \Gamma_{jk}^s. $$

However, the summations in my "equality" are over $s$, not over $\ell$ which is what the textbook printed. I would like to know where I went wrong in my alleged justification above.


1 Answer 1


I think by $R_{ijk}^s X_s$ you mean $\sum_sR_{ijk}^s X_s$. The way you change summation indices suggests this. You should either use the Einstein summation convention always or not at all.

Adding the sums over $s$ everywhere, your calculations lead to $$ \sum_sR_{ijk}^s X_s = \sum_{s,\ell} \Gamma_{ik}^s \Gamma_{js}^\ell X_\ell - \sum_{s,\ell} \Gamma_{jk}^s \Gamma_{is}^\ell X_\ell +\sum_s \frac{\partial}{\partial x_j} \Gamma_{ik}^s X_s -\sum_s \frac{\partial}{\partial x_i} \Gamma_{jk}^s X_s. $$ If you swap the dummy indices $s$ and $\ell$ in the first two sums, you get $$ \sum_sR_{ijk}^s X_s = \sum_{s,\ell} \Gamma_{ik}^\ell \Gamma_{j\ell}^s X_s - \sum_{s,\ell} \Gamma_{jk}^\ell \Gamma_{i\ell}^s X_s +\sum_s \frac{\partial}{\partial x_j} \Gamma_{ik}^s X_s -\sum_s \frac{\partial}{\partial x_i} \Gamma_{jk}^s X_s. $$ If this holds for any vector $X$, then $$ R_{ijk}^s = \sum_{\ell} \Gamma_{ik}^\ell \Gamma_{j\ell}^s - \sum_{\ell} \Gamma_{jk}^\ell \Gamma_{i\ell}^s +\frac{\partial}{\partial x_j} \Gamma_{ik}^s -\frac{\partial}{\partial x_i} \Gamma_{jk}^s . $$ Compare: If $u,v\in\mathbb R^n$ are two vectors so that $\sum_iu_ix_i=\sum_iv_ix_i$ for all $x\in\mathbb R^n$, then $u_i=v_i$ for all $i$.

The only problem was sloppiness with indices and summation. I hope you continue learning Riemannian geometry; it's both fun and useful!

  • 1
    $\begingroup$ Your answer had me realize that my mistake occurred when I wrote $\sum_\ell R_{ijk}^\ell X_\ell$ ... I should have written $\sum_s R_{ijk}^s X_s$ like you did since $\ell$ and $s$ are treated here as dummy variables, after all. Then I see that everything falls into place. Thanks for your answer. $\endgroup$ Feb 19, 2017 at 1:38

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