This question is essentially a tweak of under what conditions can orthogonal vector fields make curvilinear coordinate system? .

Suppose we have a frame of vector fields $\nu_i$ on $\mathbb{R}^n$. As discussed in the linked question, these $\nu_i$ will (at least locally) form the $\frac{\partial}{\partial x^i}$ of a coordinate system on $\mathbb{R}^n$ iff they commute pairwise.

Suppose that the $\nu_i$ do not commute. When is it possible (say, locally) to find $n$ non-vanishing functions $f_i$ such that


This collection of equations can be regarded as an overdetermined system of $\frac{1}{2}n^2(n-1)$ nonlinear PDE. What is the solubility condition for this system?


1 Answer 1


Without loss of of generality, we consider strictly positive functions $ f_i := \exp {g^i} $ and slove the problem for functions $g^i$. After calculation, the equation $[\exp g^i \cdot v_i, \exp g^j \cdot v_j] = 0$ is the same as $$ [v_i, v_j] = v_j g^i \cdot v_i - v_i g^j \cdot v_j .$$

Therefore, when the given involutive frame has dimension 2, i.e., $$ [v_1, v_2] = \alpha \cdot v_1 - \beta \cdot v_2 $$ for two smooth functions $\alpha$ and $\beta$, we only need to solve the following two independent differential equations: $$ v_2 g^1 = \alpha,\quad v_1 g^2 = \beta.$$ Locally, this is always possible.

For involutive frame of higher dimension, we should necessarily have that each two vector fields are involutive from our rewriting of the equations. In this case, we denote by $\Gamma^i_j$ the corresponding coefficients (they are smooth functions): $$ [v_i, v_j] = \Gamma^i_j \cdot v_i - \Gamma_i^j \cdot v_j, \quad i \neq j. $$ And we then need to solve the overdetermined system for each $g^i$ : $$ v_j g^i = \Gamma^i_j, \quad j \neq i.$$ The following sufficient condition for the existence of $g^i$ comes from the Frobenius theorem, stated as Theorem 19.27 in Lee's Introduction to Smooth Manifolds, $$ v_k \Gamma^i_j - v_j \Gamma^i_k = \Gamma^k_j \cdot \Gamma^i_k - \Gamma^j_k \cdot \Gamma^i_j , \quad \forall k \neq i, \, j \neq i,$$ which is eq.(19.10) in the book. It is obvious that the above condition is also necessary as $ v_k(v_j g^i) - v_j(v_k g^i) = [v_k, v_j] g^i $.


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