I am studying Frobenius' Theorem in Agricola & Friedrich's Global Analysis or, to be more precise, an auxiliary theorem that, roughly speaking, states that if $m-k$ $1$-forms exist on a $m$-dimensional manifold, and their derivatives can be written as: $$dw_i=\sum_{j=1}^{m-k}\theta_{ij}\wedge w_j$$ then the forms themselves can be expressed (locally) as: $$w_i=\sum_{j=1}^{m-k}h_{ij}\; df_j$$ for some functions $h_{ij},f_j$.

The proof proceeds in several steps that can be summarized by:

  1. Consider some coordinate system $(y,z)\in\mathbb{R}^k\times\mathbb{R}^{m-k}$ such that the forms take the special structure: $$w_i=dz^i-\sum_{j=1}^k A_{ij}(y,z)\,dy^j$$

  2. For each fixed $(u,v)\in\mathbb{R}^k\times\mathbb{R}^{m-k}$, build the system of ODEs: $$\left(\gamma^i\right)'=\sum_{j=1}^k A_{ij}(t\,u,\gamma)\,u^j\quad ;\qquad\qquad \gamma^i(0)=v^i$$ If $F^i(t,u,v)=\gamma^i(t)$ is the solution of such system with parameters $u$ and initial values $v$, consider the change of coordinates: $$\begin{aligned} u^i&=y^i\\ F^j(1,u,v)&=z^j \end{aligned}$$

  3. Prove that when $w_i$ is expressed in the dual basis $\{du^1,\dots,du^k,dv^1,\dots,dv^{m-k}\}$, the coefficients of the $du^i$ terms vanish, so that the functions $v^j$ are those required by the statement.

Now, all the steps are more or less involved but, in the end, mechanical. But I am intrigued by the geometric intuition of the system of ODEs that appears at step 2. Does it mean that solution trajectories of the ODE remain on the integral manifold? It is not easy to see because, strictly speaking, the ODE is time-varying thus the solutions lie on a $m+1$-dimensional space that includes time as an additional variable.

So the title of the question could be rephrased as: how could Frobenius (and the other guys in the history of the theorem) even think that the statement was true or figure out the proof procedure?

NOTE: I have tried to keep the post both brief and self-contained, but please ask for more detail if needed.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.