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Let $(M, g, \nabla)$ be a Riemannian manifold (with $\nabla$ the Levi-Civita connection of $g$), and let {$E_i$} be a local orthonormal basis of vector fields ($g(E_i, E_j) = \delta_{ij} )$.

What can I say about $[E_i, E_j]$ or equivalently (using the torsion-free property of Levi-Civita connection) about $\nabla_{E_i} E_j$? Is it true that $[E_i, E_j]$ vanishes?

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  • $\begingroup$ @Travis did you delete your answer? I can't find it anymore $\endgroup$
    – Ngicco
    Commented Feb 18, 2017 at 9:31
  • $\begingroup$ Yes, but only temporarily. I've corrected and undeleted it. $\endgroup$ Commented Feb 18, 2017 at 11:37

3 Answers 3

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Given a local frame $(X_i)$, the Koszul formula gives an expression for $\nabla$ in terms of the derivatives of the metric coefficients and the commutation coefficients of the frame: $$\begin{multline}2 g(\nabla_{X_i} X_j, X_k) = X_i \cdot g(X_j, X_k) + X_j \cdot g(X_i, X_k) - X_k \cdot (X_i, X_j) \\ + g([X_i, X_j], X_k) - g([X_i, X_k], X_j) - g([X_j, X_k], X_i)\end{multline} .$$

If $(E_i)$ is orthonormal, by definition $g(E_j, E_k) = \delta_{jk}$ and so the first three terms on the right-hand side vanish, $$g(\nabla_{E_i} E_j, E_k) = \frac{1}{2}\big(g([E_i, E_j], E_k) - g([E_i, E_k], E_j) - g([E_j, E_k], E_i)\big) ,$$ and we can recover an simple explicit formula for the covariant derivative in terms of the metric and the Lie bracket alone: $$\begin{align}\nabla_{E_i} E_j &= \sum_k g(\nabla_{E_i} E_j, E_k) E_k \\ &= \frac{1}{2}\sum_k \big(g([E_i, E_j], E_k) - g([E_i, E_k], E_j) - g([E_j, E_k], E_i)\big) E_k\end{align}$$ In particular, we can see that $g(\nabla_{E_i} E_j, E_k)$ is skew in $j, k$. (On the other hand, for coordinate frames $(X_i) = (\partial_{x^i})$, it is the other three terms that automatically vanish, giving the usual coordinate formula for Christoffel symbols.)

Now, if $[E_i, E_j] = 0$ for all $i, j$, then $\nabla_{E_i} E_j = 0$ for all $i, j$, and so substituting in the definition of curvature gives that the restriction of $g$ to the domain of the local frame is actually flat. Thus, for any (locally) nonflat metric $g$ one cannot choose a commuting, orthonormal frame $(E_i)$. On the other hand, as levap's answer shows, even for flat $g$ the brackets need not vanish.

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  • $\begingroup$ Can you show explicitly the first relation? Since, if $\nabla _{E_i} E_j = \sum f_{ij}^a E_a$, one can say $$0 = \nabla _{E_k} g(E_i,E_j)= g( \nabla _{E_k} E_i, E_j) + g(E_i, \nabla _{E_k} E_j) = f_{ki}^j+ f_{kj}^i $$ which is not the skew-symmetry property. I am sure I make a mistake. $\endgroup$
    – Ngicco
    Commented Feb 17, 2017 at 12:47
  • $\begingroup$ No, it was my original answer that was mistaken. I've corrected it. $\endgroup$ Commented Feb 18, 2017 at 11:36
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The commutator doesn't have to vanish, even if the metric is flat. For example, the calculations here show that $[\hat{r}, \hat{\theta}] = \frac{1}{r} \hat{\theta}$ when

$$ \hat{r} = \cos \theta \hat{x} + \sin \theta \hat{y},\\ \hat{\theta} = -\sin \theta \hat{x} + \cos \theta \hat{y} $$

is the local orthonormal frame associated to the polar coordinate system on $\mathbb{R}^2 \setminus \{ 0 \}$ (with the standard metric).

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Actually it should work just by Schwartz's theorem on exchanging second derivatives in $\mathbb{R}^n$:

in fact, what you are doing is choosing local coordinates $x_{i}$ such that $E_i=\frac{\partial}{\partial x^i}$, and thus, around the point where you have defined your basis, you have $[E_i, E_j](f)=(\frac{\partial}{\partial x^i}\frac{\partial}{\partial x^j} - \frac{\partial}{\partial x^j}\frac{\partial}{\partial x^i})(f)=0$ for any $f\in C^{\infty}(M)$, due to Schwartz indeed.

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  • $\begingroup$ Surely using coordinate vector fields, there is no problem. However, it is always possible? Generally speaking, coordinate vector fields are not an orthonormal basis $g(\partial_i, \partial_j) = g_{ij}$ which can be not equal to $\delta_{ij}$. $\endgroup$
    – Ngicco
    Commented Feb 17, 2017 at 12:00
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    $\begingroup$ This is not true in general: In fact, one can only choose coordinates $(x^i)$ (around an arbitrary point) such that $E_i = \frac{\partial}{\partial x^i}$ if the manifold is locally flat! $\endgroup$ Commented Feb 17, 2017 at 12:05
  • $\begingroup$ If you can choose coordinates so that $\partial \over \partial x_i$ is orthonormal, your manifold is flat ie locally isometric to an Euclidian space. $\endgroup$
    – Thomas
    Commented Feb 17, 2017 at 18:39
  • $\begingroup$ @Travis. Apparently our comments are the same... $\endgroup$
    – Thomas
    Commented Feb 18, 2017 at 6:37

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