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Take $M=\mathbb{R}^3$ be a smooth manifold. Consider a distribution

$\Delta_{(x,y,z)} = Span\{y\frac{\partial}{\partial x}-x\frac{\partial}{\partial y}, z\frac{\partial}{\partial x} - x\frac{\partial}{\partial z} \}$.

1) Show that the distribution is integrable. 2) Describe the maximal integral submanifolds.

Here is some of my drafts:

Take $V = y\frac{\partial}{\partial x}-x\frac{\partial}{\partial y}$ and $W = z\frac{\partial}{\partial x} - x\frac{\partial}{\partial z}$, then the Lie bracket $[V,W]=VW-WV=-y\frac{\partial}{\partial z}+z\frac{\partial}{\partial y}$.

If $[V_p, W_p]\in \Delta_p$, then the distribution is involutive and further integrable. However, for $p=(0,y,z)$, $V_p=y\frac{\partial}{\partial x}$, $W_p = z\frac{\partial}{\partial x}$, and $[V_p, W_p]=-y\frac{\partial}{\partial z}+z\frac{\partial}{\partial y}$. Easy to see the Lie bracket is not in the span. If this is true, then the distribution is not integrable. Could anyone remind me what is wrong?

As to the second part, I did not have a clear clue of how to complete. A concrete computation would be appreciated.

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Excuse me, are you sure to have correctly written $V$? Because your following calculation is wrong. By the way, if $V$ were $y\partial_x-x\partial_y$ then the result of your calculation should be correct. – Giuseppe Aug 22 '12 at 19:28
If you find a function $f:\mathbb{R}^3\to\mathbb{R}$ with the property that $V(f)=0$ and $W(f)=0$, then since the tangent space to a level set is the kernel of the differential of $f$ you will get integral submanifolds as the level sets of $f$ if these are maximal will depend on what $f$ turns out to be if it is even possible to write down such an $f$ explicitly. – Matt Aug 22 '12 at 23:15
@GiuseppeTortorella, thank you pointing the error out. I have rewritten $V$, the expression follows kind of symmetry there. In this case, do you mind explain the question above? – Honghao Aug 24 '12 at 3:26
up vote 1 down vote accepted

The tangent distribution $\Delta$ on $\mathbb R^3$ is spanned by $E_2=-x\partial_z+z\partial_x$ and $E_3=-y\partial_x+x\partial_y,$ which are the infinitesimal generators of the rotations of $\mathbb R^3$ respectively around the $y$-axis and the $z$-axis.

Computing the minors of $\begin{pmatrix}z & 0 & -x\\ -y & x & 0\end{pmatrix},$ we observe that $\Delta$ is a singular tangent distribution on $\mathbb R^3,$ because $$\textrm{rank}\Delta_{(x,yz)}=\begin{cases} 0 & \textrm{ if } x=y=z=0,\\ 1 & \textrm{ if } x=0 \textrm{ and } y^2+z^2\neq 0,\\ 2 & \textrm{ if } x\neq 0. \end{cases}$$ As you have computed $[E_2,E_3]=-y\partial_z+z\partial_y\equiv-E_1,$ where $E_1$ is the infinitesimal generator of the rotations around the $x$-axis.

Observe that $xE_1=yE_2+zE_3,$ so $\Delta$ is involutive on $\{(x,y,z)\mid x\neq 0\}.$

The Lie algebra generated by $E_2$ and $E_3$ is $\mathfrak{so}(3)=\mathrm{span}\{E_1,E_2,E_3\},$ the Lie algebra of the infinitesimal generators of the natural action $\mathrm{SO}(3)$ on $\mathbb R^3:$ $$(A,p)\in\mathrm{SO}(3)\times\mathbb R^3\to A.p\in\mathbb R^3.$$
Therefore the maximal integral manifolds of $\mathrm{span}\{E_1,E_2,E_3\}$ are the $\mathrm{SO}(3)$-orbits, i.e. the spheres centered at the origin.

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How to prove that spheres at O are "maximal" integral manifolds? Are you using any nontrivial theorems concerning about maximal integral manifolds in the last paragraph? – Ti Wen Mar 6 at 22:05

Let $f\in C^\infty(\Bbb R^3).$ We have \begin{align*} [V,W]f &= V(Wf)-W(Vf) \\ &= (y\frac{\partial}{\partial x}-x\frac{\partial}{\partial z})(z\frac{\partial f}{\partial x}-x\frac{\partial f}{\partial z}) - (z\frac{\partial}{\partial x}-x\frac{\partial}{\partial z})(y\frac{\partial f}{\partial x}-x\frac{\partial f}{\partial z}) \\ &= y\left(0\frac{\partial f}{\partial x} + z\frac{\partial^2 f}{\partial x^2} - 1\frac{\partial f}{\partial z} - x\frac{\partial^2 f}{\partial x\partial z}\right) - x\left(1\frac{\partial f}{\partial x} + z\frac{\partial^2 f}{\partial z\partial x} - 0\frac{\partial f}{\partial z} - x\frac{\partial^2 f}{\partial z^2}\right) \\ &~~~- z\left(0+y\frac{\partial^2 f}{\partial x^2} -1\frac{\partial f}{\partial z} -x\frac{\partial^2 f}{\partial x\partial z}\right) + x\left(0+y\frac{\partial^2 f}{\partial z\partial x} - 0 - x\frac{\partial^2 f}{\partial z^2}\right) \\ &= (yz-zy)\frac{\partial^2 f}{\partial x^2} + (-yx-xz+zx+xy)\frac{\partial^2 f}{\partial z\partial x} + (x^2-x^2)\frac{\partial^2 f}{\partial z^2} \\ &~~~+ \left(-y\frac{\partial f}{\partial z} -x\frac{\partial f}{\partial x} +z\frac{\partial f}{\partial z} \right) \\ &= (z-y)\frac{\partial f}{\partial z} -x\frac{\partial f}{\partial x} \end{align*}

which shows that your calculation for the Lie bracket is incorrect. The Lie bracket should be $[V,W]=(z-y)\frac{\partial}{\partial z}-x\frac{\partial}{\partial x}.$

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thanks for the explicit computation. I am a little bit confused about why there are second order partial derivatives in the calculation? – Honghao Aug 24 '12 at 3:27
Dear @Honghao, you're welcome. The second order partials come from applying the product formula for differentiation. – Andrew Aug 25 '12 at 5:38

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