The question is: given a Riemannian manifold with a non-degenerate metric g and an inner product $u_{\beta}u^{\beta}=1$, is $\nabla_{\mu} (u_{\alpha}u_{\beta})=0$ without demanding the trivial solution $\nabla_{\mu}u_{\alpha}=0$? There are two ways to support this.

1) Direct calculation. $\nabla_{\mu} (u_{\alpha}u_{\beta})=\partial_{\mu}(u_{\alpha}u_{\beta})-\Gamma^{\lambda}_{\mu\alpha}u_{\lambda}u_{\beta}-\Gamma^{\lambda}_{\mu\beta}u_{\lambda}u_{\alpha}$=$\partial_{\mu}(u_{\alpha}u_{\beta})-\frac{1}{2}u^{\rho}u_{\beta}(\partial_{\mu}g_{\varrho\alpha}+\partial_{\alpha}g_{\varrho\mu}-\partial_{\varrho}g_{\mu\alpha})-\frac{1}{2}u^{\rho}u_{\alpha}(\partial_{\mu}g_{\varrho\beta}+\partial_{\beta}g_{\varrho\mu}-\partial_{\varrho}g_{\mu\beta})$.

If we now evaluate this in an orthonormal basis $(e_{\alpha})$ at a point $p\in M$ with $ e_{0}=u. $ Then, $ u^{0}u_{0}=1 $, $ u^{i}u_{i}=0 $, $g_{\alpha\beta}=\delta_{\alpha\beta}$ and it follows that $\nabla_{\mu} (u_{\alpha}u_{\beta})=0$. Since it is a tensor, if it vanishes in one frame, it vanishes in all frames.

2)$\nabla_{\mu}(u_{\beta}u^{\beta})=0$ which means $ g^{\alpha\beta}\nabla_{\mu}(u_{\alpha}u_{\beta})=0 $. Since g is non-degenerate, $ \mid \nabla_{\mu}(u_{\alpha}u_{\beta})\mid=0 $ and $\nabla_{\mu}(u_{\alpha}u_{\beta})=0 $ is a solution that does not require the trivial solution $\nabla_{\mu}u_{\alpha}=0 $. The trivial solution is sufficient but not necessary. In fact, $ u_{\beta}\nabla_{\mu}u_{\alpha}+u_{\alpha}\nabla_{\mu}u_{\beta}=0 $ means $ u^{\beta}\nabla_{\mu}u_{\beta}=0 $ so in general, the vector is orthogonal to the covariant derivative of its covector.

Am I missing something? Please advise in detail if so.


There is no reason why this should be true, just think about what this maens for the flat connection on an open subset $U\subset\mathbb R^n$: In this case $u$ is just a function $u:U\to\mathbb R^n$ and the condition you impose is that $\sum_i(u_i(x))^2=1$ for all $x\in U$, where $u_i$ are the components of $u$. So this just means that $u$ has values in the unit sphere. In contrast, the equation $\nabla_\mu(u_\alpha u_\beta)=0$ would say that for each choice of $i$ and $j$ all partial derivatives of the $x\mapsto u_i(x)u_j(x)$ vanish, so all these funcitons have to be constant. So already the function $x\mapsto x/|x|$ on $\mathbb R^2\setminus\{0\}$ gives a counter example.

Both your agruments are flawed for different reason. In 1) you have to compute in local coordinates, so you cannot require $g_{\alpha\beta}=\delta_{\alpha\beta}$. And you not only need this at a point, since you have to take partial derivatives. You could compute in an orthonormal frame, but there the description of the covariant derivative takes a completely different form.

In 2) there is no interpretation of the contraction of two indices of a tensor with three indices as a norm. To get a norm, you would need two copies of the tensor and theree copies of the metric, so this would be $g^{\mu\nu}g^{\alpha\gamma}g^{\beta\rho}(\nabla_\mu (u_\alpha u_\beta))( \nabla_\nu (u_\gamma u_\rho))$.

  • $\begingroup$ Andreas, since each of the products of unit vectors are constants, $ u_{\beta}\nabla_{\mu}u_{\alpha}+u_{\alpha}\nabla_{\mu}u_{\beta}=0 $ and by contacting with the inverse metric, $ u^{\beta}\nabla_{\mu}u_{\beta}=0 $. This means the vector $ u^{\beta} $ is orthogonal to the covariant derivative of its covector. It is not true that $ u^{\beta} $ must be constant; that is a sufficient but not a necessary condition for $\nabla_{\mu}(u_{\alpha}u_{\beta})=0 $. $\endgroup$ – Kolten Apr 4 '20 at 14:03
  • $\begingroup$ Furthermore, in the particular orthonormal frame as stated, $ e_{0} $ in a coordinate system and $ \hat{e}_{0} $ in a non-coordinate frame are both 1. All of the non-holonomic terms in the extended connection vanish, so there are no corrections to the Christoffel connection coefficients. In local coordinates, the calculation holds. Secondly, $\nabla_{\mu}(u_{\alpha}u_{\beta})=0 $ is to be interpreted as its equivalent: the vector $ u^{\beta} $ is orthogonal to the covariant derivative of its covector. Again, that does not require the vector to be constant. $\endgroup$ – Kolten Apr 4 '20 at 14:07
  • $\begingroup$ The covariant derivative of the covector is a bilinear form, the fact that the vector inserts trivially into this, does not say the the bilinear form vanishes identically. Concerning the other "argument", where do you get information on the (non-holomorphic version of) the Christoffels (even in the point you are looking at)? Finally, a simple counter example should be a good argument to show that computations are worng. Just think about $u(x,y)=\tfrac{xdx+ydy}{x^2+y^2}$ on $\mathbb R^2$ as indicated in the answer. $\endgroup$ – Andreas Cap Apr 5 '20 at 7:55
  • $\begingroup$ $ u_{\alpha}u_{\beta} $ is a product of two vectors, neither of which are constants. Your "counter example" is not an example of this product. In non-holonomic coordinates, $ g_{\mu\nu}=e^{\alpha}_{\mu}e^{\beta}_{\nu}\delta_{\mu\nu} $ but in the orthonormal system described, $\delta_{\mu\nu}=g_{\mu\nu} $ so the vierbiens are all 1 and the Lie bracket vanishes. $\endgroup$ – Kolten Apr 5 '20 at 15:05
  • $\begingroup$ My example is that $u_\alpha$ itself is $\tfrac{xdx+ydy}{\sqrt{x^2+y^2}}$. Then $u_\alpha u_\beta$ is $\tfrac{x^2}{x^2+y^2}dx\otimes dx+\tfrac{xy}{x^2+y^2}(dx\otimes dy+dy\otimes dx)+\tfrac{y^2}{x^2+y^2}(dy\otimes dy)$ which evidently is not parallel.But if you prefer not to believe it, feel free to do so. $\endgroup$ – Andreas Cap Apr 5 '20 at 17:52

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