I've been dealing with the following problem for a while and meanwhile I have no idea how to move on. Maybe one of you can help me? :)

Let $M = \mathbb{R}^2\subset \mathbb{R}^3$ a manifold. I need to find two tangent-vectorfields $\xi, \eta$ such that

(i) $[\xi,\eta]=0$, but $\nabla_{\xi}\eta \neq 0$ at at least one point P

(ii)$[\xi,\eta]\neq0$, but $\nabla_{\xi}\eta = 0$ at at least one point P

First I calculated (for an arbitrary function $\phi$ and $\xi := \xi_1 \frac{\partial}{\partial x}+\xi_2 \frac{\partial}{\partial y}$, $\eta := \eta_1 \frac{\partial}{\partial x}+\eta_2 \frac{\partial}{\partial y}$):

$[\xi,\eta] = ... = \frac{\partial \phi}{\partial x} \left[ \xi_1 \frac{\partial \eta_1}{\partial x} + \xi_2 \frac{\partial \eta_1}{\partial y}- \eta_1 \frac{\partial \xi_1}{\partial x} - \eta_2 \frac{\partial \xi_1}{\partial y}\right] + \frac{\partial \phi}{\partial y} \left[ \xi_1 \frac{\partial \eta_2}{\partial x} + \xi_2 \frac{\partial \eta_2}{\partial y}- \eta_1 \frac{\partial \xi_2}{\partial x} - \eta_2 \frac{\partial \xi_2}{\partial y}\right]$

For (ii) I get by using $\nabla_{\xi} = \nabla_{ \xi_1 \frac{\partial}{\partial x}+\xi_2 \frac{\partial}{\partial y}} = \xi_1 \nabla_{\frac{\partial}{\partial x}}+\xi_2 \nabla_{\frac{\partial}{\partial y}}$ for continous $\xi_i$ and $\eta_i$:

$\nabla_{\xi}\eta = 0 \Leftrightarrow \xi_1(\frac{\partial \eta}{\partial x} <\frac{\partial \eta}{\partial x},n> + \xi_2(\frac{\partial \eta}{\partial y} <\frac{\partial \eta}{\partial y},n> = 0 $ Due to the normal vector $n$ has only a z direction because $M=\mathbb{R}^2$, the scalar product is zero and we get:

$\xi_1 \frac{\partial \eta}{\partial x} + \xi_2 \frac{\partial \eta}{\partial y} = 0$

So the equation above simplifies to:

$\frac{\partial \phi}{\partial x} \left[- \eta_1 \frac{\partial \xi_1}{\partial x} - \eta_2 \frac{\partial \xi_1}{\partial y}\right] + \frac{\partial \phi}{\partial y} \left[ - \eta_1 \frac{\partial \xi_2}{\partial x} - \eta_2 \frac{\partial \xi_2}{\partial y}\right] \neq 0$

Both of the brackets have to be nonzero, so the solution of these two equations gives the solution.

Is this right so far?

How, beside try and error, can I find out the solution of this equations?

How does it work for (i)? In this case I can't eliminate anything from the first equation, because the second condition then is not equal Zero.

Thanks a lot for your help!

  • $\begingroup$ Are $\xi$ and $\eta$ tangent to $\mathbb{R}^2$ or $\mathbb{R}^3$? $\endgroup$ – Amitai Yuval Jun 28 '17 at 1:19


Notice that in $\mathbb{R}^n$ the Levi-Civita connection $\nabla$ is simply the directional derivate. In other words, if you take two tangent vectors $X=\sum_i x^i\dfrac{\partial }{\partial r^i}$ and $Y=\sum_iy^i\dfrac{\partial}{\partial r^i}$ in $\mathfrak{X}(\mathbb{R}^n)$, where the $r^i$ are the standard coordinates, we have $$\nabla_XY=\sum_{i=1}^nX(y^i)\dfrac{\partial}{\partial r^i}.$$ In particular you can check that $\nabla_XY-\nabla_YX=[X,Y]$, for all $X,Y\in\mathfrak{X}(\mathbb{R}^n)$.

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