Do these vector fields span an integrable distribution?

Question

Let $X_i$, $1 \le 1 \le n$, be smooth vector fields on the open set $U \subset \mathbb{R}^n$, which are linearly independent at each point. Set $[X_i, X_j] = \sum_{k=1}^n c_{ij}^kX_k$. Suppose that the coefficients $c_{ij}^k$ are all constant.

Let $\overline{X}_i$, $1 \le i \le n$, on $\overline{U} \subset \mathbb{R}^n$ be another copy (with the same coefficients). Set $Y_i = X_i + \overline{X}_i$, $1 \le i \le n$, on $U \times \overline{U}$. My question is, do the vector fields $Y_i$ necessarily span an integrable distribution $E \subseteq T(U \times \overline{U})$?

Answer 1

The vector fields $Y_i$ do necessarily span an integrable distribution $E \subseteq T(U \times \overline{U})$. $E$ is necessarily a smooth subbundle of $T(U \times \overline{U})$, so it suffices to show $E$ is closed under the Lie bracket. Note that $[X_i, \overline{X}_j] = 0$ for every $i$, $j$ because $X_i$ only differentiates with respect to the first $n$ variables and $\overline{X}_j$ only differentiates with respect to the last $n$ variables. This means that for every $i$, $j$,$$[Y_i, Y_j] = [X_i + \overline{X}_i, X_j + \overline{X}_j] = [X_i, X_j] + [\overline{X}_i, \overline{X}_j] = \sum_k c_{ij}^k X_k + \sum_k c_{ij}^k \overline{X}_k = \sum_k c_{ij}^k Y_k.$$Therefore the bracket of any linear combinations of the $Y_i$ is again a linear combination of the $Y_i$, so $[E, E] \subset E$, as required.

Answer 2

A useful fact: Let $X,Y,$ be vector fields on $U$, and define vector fields on $U\times U$ by $\overline{X}=(X,0),\quad \overline{Y}=(0,Y).$ Since $\overline{X}$ depends only on the first coordinate and $\overline{Y}$ depends only on the second, we have by definition $$\mathcal{L}_XY=0,$$in other words$$[X,Y]=0.$$And to the question: We have$$[Y_i,Y_j]=[X_i+\overline{X}_i,X_j+\overline{X}_j]=[X_i,X_j]+[\overline{X}_i,\overline{X}_j],$$where the last equality follows from the above fact. Now $$[X_i,X_j]+[\overline{X}_i,\overline{X}_j]=\sum c_{ij}^kX_k+\sum c_{ij}^k\overline{X}_k=\sum c_{ij}^kY_k,$$and hence the answer to the question is positive.