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Having tried to solve exercise 4.4-7 to have another proof of Frobenius Theorem, I would ask you a question.

This is what I have understood: In Step 2 there is to prove, for any tangent subbundle $E$ on a manifold $M,$ that if the module $\Gamma(E)$ of its sections is a Lie subalgebra of $\Gamma(TM)$ then it is locally generated by $k$ independent commuting vector fields. In Step 1 there is to prove the integrability of $E$ when it is locally generated by $k$ independent commuting vector fields.

Now this is the question: Because, in the step 1, the authors assume additionally that $\Gamma(E)$ has to be an abelian Lie subalgebra of $\Gamma(TM)$?
There is some point that I don't understand? or is it a lapsus calami?

Thank you very much for the attention.

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I'm not sure that I understand your question. I assume you are talking about what in my edition of this book is called exercise 4.4-7. If so, Step 1 is actually asking you to prove the integrability of E in the abelian case. Then in Step 2 you show the general involutive case reduces locally to the abelian case. – José Figueroa-O'Farrill Jan 30 '11 at 13:39
1)Yes. It is exactly the problem on an alternative (Hirsch, Weinstein) proof of Frobenius Theorem, and, in its statement, the distribution E is said to be (locally) abelian if (Ab): the submodule \Gamma(E) of its sections is an abelian Lie subalgebra of \Gamma(TM). 2)I just found how to solve the problem when I substitute this latter condition(Ab) with the condition(): E is locally generated by a system of independent and commuting vector fields. And I think that (Ab) is really stronger than (), while (*) is equivalent to involutivity of E. 3)So the request of (Ab) is superfluous, or not? – Giuseppe Jan 30 '11 at 16:52
It seems to me that "Lie subalgebra generated by k independent commuting vector fields" is not the same thing as "abelian Lie subalgebra". – Deane Yang Jan 30 '11 at 17:10
Please, about the exercise, should you say me if my way of proof is correct, or not? if not, where I need to use the stronger condition "\Gamma(E) abelian"? – Giuseppe Jan 30 '11 at 17:30
Does the book really use the word "abelian"? – Deane Yang Jan 30 '11 at 17:46

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