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How do we check exactly that a distribution is involutive?

I have the following definition in my book: A $k-$dimensional distribution $\Delta$ on a manifold $M$ is a smooth choice of a k-dimensional subspace $\Delta_{p} \subseteq T_{p}M$. The distribution is called $involutive$ if $\left [ \Delta , \Delta \right ]\subseteq \Delta$.

In my case i have the following vector fields: for $x=(x,y,\theta ,\varphi )\in \Omega :=\mathbb{R}^{2}\times S^{1}\times(- \frac{\pi}{4},\frac{\pi}{4})$, let $X=\frac{\partial }{\partial \varphi }$ and $Y=\cos\theta \frac{\partial }{\partial x}+\sin\theta \frac{\partial }{\partial y}+ \frac{\tan\varphi }{C}\frac{\partial }{\partial \theta }$, where $C$ is some constant. Find out if the distribution $\Delta = \langle X,Y \rangle$ is involutive or not.

Okay, i have computed the Lie bracket $\left [ X,Y \right ]=\frac{1}{C \cos ^{2}{\theta }}\frac{\partial }{\partial \theta }$, but i don't really know what to do now. Can anybody help me with this example, please? Thank you in advance!

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The requirement that $[\Delta, \Delta] \subseteq \Delta$ can be determined by ensuring that all pairwise Lie Brackets of the vector fields that make up the distribution are themselves in the distribution. See Distributions (differential geometry). In your scenario there is only one bracket to consider so we ask, does there exist functions $a, b$ such that $$[X, Y] = aX + bY$$ i.e. $$\frac{1}{C \cos ^{2}{\theta }}\frac{\partial }{\partial \theta } = a\frac{\partial }{\partial \varphi } + b\left(\cos\theta \frac{\partial }{\partial x}+\sin\theta \frac{\partial }{\partial y}+ \frac{\tan\varphi }{C}\frac{\partial }{\partial \theta }\right)$$

The answer is no since no linear combination of $X$ and $Y$ can produce the vector $\frac{d}{d\theta}$.

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  • $\begingroup$ You might want to point out what $a$ and $b$ are. $\endgroup$ – Michael Albanese Aug 2 '15 at 20:21
  • $\begingroup$ @Lullaby - Happy to help. $\endgroup$ – muaddib Aug 6 '15 at 20:49

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