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 $X,Y\in\mathfrak{X}(\mathbb{R}^n)$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.