I am studying distributions and when they are involutive and the Froebenius's Theorem. I cam across an example but I don't understand how to solve it.
I have a distribution $\Delta$ defined as:
$\Delta=span\{v,w\}$
where
$v= \left( \begin{matrix} -y \\ x\\0 \end{matrix} \right)$
$w= \left( \begin{matrix} 2zx \\ 2yz\\ z^2+1-x^2-y^2 \end{matrix} \right)$
In the slides I found on the internet it says:
$\left[ v,w\right]\equiv0$
But I tried to compute it (with $s=[x\;y\;z]^T$):
$\left[ v,w\right] \frac{\partial w}{\partial s}v-\frac{\partial v}{\partial s}w = -[2z\;2z\;2z][-y\;x\;0]^T=2zy-2zx \neq0$
What I do wrong?
Thanks for the help
If in local coordinates you have for two vector fields $v=\sum_i v_i\partial_i$, $w=\sum_i w_i\partial_i$, then: $$[v,w](f)=\sum_iv_i\partial_i\left(\sum_jw_i\partial_j(f)\right)-\sum_iw_i\partial_i\left(\sum_jv_i\partial_j(f)\right)$$ After applying the product rule, doing some substitutions and using $\partial_i\partial_j(f)=\partial_j\partial_i(f)$, you get:
$$[v,w]=\sum_i\left(\sum_j v_j\partial_jw_i-w_j\partial_jv_i\right) \partial_i$$ So this is the formula you need to use. Apply it on your example to see that the first component is: \begin{align} \sum_j v_j\partial_j w_1-w_j\partial_jv_1&=-y\partial_x (2xz)+x\partial_y(2xz)+0\partial_z(z^2+1-x^2-y^2)\\ &\qquad -2zx\partial_x(-y)-2yz\partial_y(-y)-(z^2+..)\partial_z(-y)\\ &=-2yz+2yz=0 \end{align} The other componenets must be done in the same way.