# Does Frobenius theorem apply to vector-valued function?

We know Frobenius theorem handle pde systems like $\{Xf=0, Yf=0\}$ requiring Lie bracket $[X,Y]\equiv 0 \mod X, Y$ for completely integrability of the system. However, how to handle systems like $\{Xf+Yg=0\}$. For example, how to solve $\frac{\partial u_i}{\partial x_j}=\frac{\partial u_j}{\partial x_i}$ for $i,j=1..n$?

-
Your example is about the closedness of the differential 1-form $\omega=u_1dx_1+\ldots+u_ndx_n$, which has the solution $u_i=\frac{\partial F}{\partial x_i}$, with arbitrary $F$. Concerning the general question, I would think in terms of differential consequences and prolongations (see, e.g., Kuranishi's "On E. Cartan's prolongation theorem of exterior differential systems."): in this context, "integrability" means that the prolonged system projects back to the whole original one (you may check that this reduces to the Frobenius theorem in the case of a PDE given by vector fields). – Giovanni Moreno Aug 28 '13 at 9:19
Thank you for the complement. How do you get the solution? By Poincare lemma or other stuffs? And I'm only a beginner on differential geometry and feel it hard to reduce? Could it be successfully reduced or just your suggestion? Would you mind offering more explanation? – Shuchang Aug 29 '13 at 4:06

[too long for a comment] Yes, the solution follows precisely from the Poincare' lemma. To see the relationship between Frobenius theorem and the theory of prolongations, let $\pi:\mathbb{R}^2\times\mathbb{R}\to\mathbb{R}^2$ be the trivial bundle, and $x,y$ coordinates on its basis. Let $$X=a\partial_x+b\partial_y$$ $$Y=a'\partial_x+b'\partial_y$$ and regard (any) vector field $X$ as a smooth function $\psi_X$ on the first jet extension $J^1(\pi)$, i.e., the space with coordinates $x,y,u,u_x,u_y$: $$\psi_X=au_x+bu_y.$$ Recall that the function $\psi_X$ is intrinsically defined by the property $$\psi_X\circ j_1(f)=X(f)$$ for any function $f=f(x,y)$. In this "jet formalism", your system of equations corresponds to the sub manifold $E=\{\psi_X=0,\psi_Y=0\}$ of $J^1(\pi)$. Now we must introduce the first prolongation $E^{(1)}$ of $E$, which is a submanifold of $J^2(\pi)$: to define it, take the total derivatives of the functions $\psi_X$ and $\psi_Y$: $$D_x(\psi_X)=a_xu_x+au_{xx}+b_xu_y+bu_{xy}$$ $$D_y(\psi_X)=a_yu_x+au_{xy}+b_yu_y+bu_{yy}$$ $$D_x(\psi_Y)=a'_xu_x+a'u_{xx}+b'_xu_y+b'u_{xy}$$ $$D_y(\psi_Y)=a'_yu_x+a'u_{xy}+b'_yu_y+b'u_{yy}.$$ In terms of ideals, $E$ corresponds to the ideal $I=\mathrm{Span}\{\psi_X,\psi_Y\}$ in $C^\infty(J^1(\pi))$, whereas $E^{(1)}$ corresponds to the ideal $I'=\mathrm{Span}\{\psi_X,\psi_Y,D_x(\psi_X),D_y(\psi_X),D_x(\psi_Y),D_y(\psi_Y)\}$ in $C^\infty(J^2(\pi))$. You should also remember that $C^\infty(J^1(\pi))$ is embedded into $C^\infty(J^2(\pi))$: then, the "algebraic formulation" of formal integrability reads $$I'\cap C^\infty(J^1(\pi)) = I.$$ (Observe that the "$\supseteq$" inclusion always holds).
Now, you can easily check that $I'\cap C^\infty(J^1(\pi))=\mathrm{Span}\{\psi_X,\psi_Y,\psi_{[X,Y]}\}$: basically, you show that the only way (besides taking $\psi_X$ or $\psi_Y$ themselves) to get an element of $C^\infty(J^1(\pi))$ out of the generators of $I'$ is by means of the following $C^\infty(J^1(\pi))$-linear combination: $$aD_x(\psi_Y)+bD_y(\psi_Y)-a'D_x(\psi_X)-b'D_y(\psi_X),$$ which turns out to be exactly $\psi_{[X,Y]}$, i.e., the function associated to the commutator $[X,Y]$. So,"formal integrability" holds if and only if $$\mathrm{Span}\{\psi_X,\psi_Y,\psi_{[X,Y]}\}=\mathrm{Span}\{\psi_X,\psi_Y\},$$ which is just a paraphrase of Frobenius' condition.