Vrftsjtryk

first merry Christmas to the team. Now let be $M$ a manifold of dimension $n in mathbb{N}$. Let be $alpha_{1}, alpha_{2}, ..., alpha_{n-k}$ $n-k$ $1$-forms linearly independent in each points of $M$. I suppose those forms are smooth. For all $x in M$, let be $P_{x} = bigcap_{i = 1}^{n-k}Ker(alpha_{i}(x))$ and $omega = alpha_{1} wedge alpha_{2} wedge ... wedge alpha_{n-k}$.
I shown $P$ is a subbundle of $TM$.

I would like to show that [for all $x in M$ it exists $f in C^{infty}(M)$ such that $f(x) neq 0$ and such that $d(fomega) = 0$ on neighbourhood of $x$] if I suppose $P$ is integrable.

Here is the reason I made :
Let be $x in M$. Let be $U$ a small neighbourhood of $x$ and $(a_{i})_{i in <n-k+1, n>}$ 1-differentials forms and $(X_{i})_{i in <1, n>}$ smooth vectorial fields construct the following way : lets complete $(alpha_{i}(x))_{i in <1, n-k>}$ in a base $(alpha_{i}(x))_{i in <1, n>}$ of $T^{*}_{x}M$. We can suppose $U$ small enough such that it's the domain of a map $phi$. For all $i in <n-k+1, n>$, $alpha_{i}(x) = sum_{j = 1}^{n}a_{j}^{i}d(phi_{j})(x)$. As the determinant is a continuous map we can suppose $U$ small enough such that $alpha_{i}(.) := dbeta_{i}(.)$(where $beta_{i}(.) = sum_{j = 1}^{n}a_{j}^{i}phi_{j}(.)$) is such that for all $y in U$, $(alpha_{i}(x))_{i in <1, n>}$ is a base of $T_{y}^{*}M$. Now we can construct $(X_{i})_{i in <1, n>}$ smotth vectorial fields on $U$ such that $alpha_{i}(X_{j}) = delta_{ij}(1)$ on $U$.

Now, I wrote $[X_{i}, X_{j}] = sum_{l = 1}^{n}c_{ij}^{l}X_{l}$ and using $(1)$ I show $d alpha_{l} = -sum_{i < j} c_{ij}^{l} alpha_{i} wedge alpha_{j} (3)$.

As $P$ is integrable, $c_{ij}^{h} equiv 0$ on $U$ if $i, j in <n-k+1, n>$ and $h in <1, n-k>(*)$. So using $(*)$$ d omega = sum_{l = 1}^{n-k}epsilon_{l}' omega wedge sum_{i in <n-k +1, n>}c_{il}^{l} alpha_{i}$ where $epsilon_{l}' in {pm 1}(2)$. So in $(2)$, $d omega = omega wedge sum_{l = 1}^{n-k}epsilon_{l}' sum_{i in <n-k +1, n>}c_{il}^{l} dbeta_{i}$. Let's considering $f = e^{(-1)^{n-k} times (sum_{l = 1}^{n-k}epsilon_{l}' sum_{i in <n-k +1, n>}c_{il}^{l} beta_{i})}$ which is smooth and positive on $U$. So $d(fomega) = (-1)^{n-k}f times ((sum_{l = 1}^{n-k}epsilon_{l}' sum_{i in <n-k +1, n>}c_{il}^{l} dbeta_{i}) wedge omega - (sum_{l = 1}^{n-k}epsilon_{l}' sum_{i in <n-k +1, n>}dc_{il}^{l} beta_{i} ) wedge omega) - f domega = (sum_{l = 1}^{n-k}epsilon_{l}' sum_{i in <n-k +1, n>}dc_{il}^{l} beta_{i} ) wedge omega + f domega - f domega = (sum_{l = 1}^{n-k}epsilon_{l}' sum_{i in <n-k +1, n>}dc_{il}^{l} beta_{i} ) wedge omega$

on $U$.

If I show $(sum_{l = 1}^{n-k}epsilon_{l}' sum_{i in <n-k +1, n>}dc_{il}^{l} beta_{i} ) wedge omega = 0$ on $U$ I'll won.

I try to differentiate the expression $(3)$ but it doesn't give me informations. If the $c_{lj}^{l}$ could be constant. But I don't have freedom to choose them in constant in my construction.

Do you have ideas?

Thanks for all the answers and merry Christmas again.

thank you for your answer. The criterion you gave to me is the Froebenius theorem I use to get (2).

And at the end I get $d omega = 0 mod (<omega>)$ : more precisely I get $ d omega = omega wedge sum_{l = 1}^{n-k}epsilon_{l}' sum_{i in <n-k +1, n>}c_{il}^{l} dbeta_{i} $ but then I can't conclude at the end.

If the $c_{il}^{l}$ where constants on $U$ I could conclude with the function $f$ I wrote.

Perhaps you have got an another idea?

I wish you a good day.