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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

1 Answers 1

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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.

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    Thanks for the reply. The $f$ in your formulas is : $[x\;y\;z]^T$ ? – 2017-01-23
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    @minidable no, $f$ is a smooth function. Vector fields are defined as maps from the space of smooth functions into itself that satisfy the derivation property ($V(f\cdot g)=f\cdot V(g)+V(f)\cdot g$). (If you choose a coordinate system each vector field looks like a sum $\sum_i v_i\partial_i$ in this coordinate chart.) By looking at $[v,w](f)$ for an _arbitrary_ smooth function you can find what the vector field $[v,w]$ must be. – 2017-01-23
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    again thanks for your time. How do you know that $v=\sum_i v_i\partial_i$, $w=\sum_i w_i\partial_i$, how do you check that? Another thing. When you write $\partial_i$ it would be $\partial_i|_{i=1}=\frac{\partial }{\partial x}$? Thanks. – 2017-01-23
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    Before you reply I think I need to do a recap of what I studied so far about manifolds because I think I am slowly understanding. I am studying the book Nonlinear Control System by Isidori (http://www.springer.com/la/book/9783540199168) but I think I found a good presentation which explains in an easier way what I need for now. http://www.cds.caltech.edu/~marsden/wiki/uploads/cds202-08/home/Manifolds.pdf – 2017-01-23
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    With $\partial_i$ I mean $\frac{\partial}{\partial x_i}$, in my notation I have also suppressed the point in which we are. Meaning $v_i$ is a function of the coordinates $x$ and $\partial_i$ is the evaluation of the derivative at a specific point. The fact that every vector field is (in a coordinate chart) the sum of smooth functions with partial derivatives follows because vector fields can also be viewed as a function $M\to \bigcup_p T_pM$ where each point is assigned to a vector of its tangent space in a smooth way. Since $T_pM$ has $\partial_i$ as a basis (after choice of coordinates) it.. – 2017-01-24
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    .. is sort of motivated why you can write every vector field in such a form (after choice of coordinates). I cannot give full proofs here, however I recommend you look into a book about differential geometry. Personally I like the one by Jack Lee and the one by Jeffrey Lee (unrelated). Do Carmo is supposed to be another standard work but I have not used it personally. – 2017-01-24