Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Starting with a system of partial differential equations for functions on $\mathbb{R}^n$, Frobenius' theorem gives a bunch of integrability conditions (on e.g. functions which are 'data' for the system of PDEs, e.g. sourcing terms). Now let's say I'm working on a curved manifold, and have a system of PDEs that arise from covariant derivatives. I can always convert this problem into the previous one by choosing local coordinates; then my functions will be components of tensors, and my derivatives turn into ordinary partial derivatives corrected with components of a connection $\Gamma^i{}_{jk}$ (the components of this connection can now be thought of as 'data' for the system of PDEs). If I then apply the Frobenius theorem, the integrability conditions will contain $\Gamma$ and $\partial\Gamma$ and derivatives of other data, e.g. $\partial_i v^j$ for some vector $v$.

Is there any guarantee that the integrability conditions, or various linear combinations thereof, are covariant? That is, do all instances of $\partial\Gamma$ sum with other terms to give components of a curvature tensor?

I expect this to be true, since after all I could have picked any coordinate system, and the existence of a local solution should not depend on the coordinate system.

Is there a way to find the "covariant integrability conditions", if such a thing exists?

share|improve this question

1 Answer 1

up vote 12 down vote accepted

Your question is a bit vague, but let me try the following statement, which might be the kind of answer you are looking for: If $M$ is a manifold and $S\to M$ is a vector bundle over $M$ endowed with a connection $\nabla$, then a 'total differential equation' for sections $s$ of $S$ is an equation of the form $\nabla_X s = 0$ for all vector fields $X$. Then the Frobenius integrability conditions for this system are simply $\Omega^\nabla=0$, where $\Omega^\nabla$ is the curvature of $\nabla$, i.e., it is the $2$-form on $M$ with values in $\mathrm{End}(S)$ that is defined by $$ \Omega^\nabla(X,Y)s = \nabla_X\nabla_Ys - \nabla_Y\nabla_Xs - \nabla_{[X,Y]}s $$ for all vector fields $X$ and $Y$ on $M$. So, yes, the Frobenius integrability conditions are 'covariant', i.e., they can be formulated independently of coordinates.

Now, one may wonder, "But suppose I start with some linear total differential system $$ \frac{\partial u^a}{\partial x^i} = \Gamma^a_{bi}(x) u^b \tag{1} $$ in local coordinates. Does this apply?" The answer is 'yes': One regards the $u^a$ as the components of a section $s = (u^a)$ of a trivialized vector bundle $S$ and the matrices $\Gamma_i = \bigl(\Gamma^a_{bi}(x)\bigr)$ as sections of $\mathrm{End}(S)$ and one defines a connection $\nabla$ on this trivial bundle by the rule $$ \nabla_{\frac{\partial}{\partial x^i}}s = \frac{\partial s}{\partial x^i} - \Gamma_i\ s. $$ Then the curvature of $\nabla$ vanishes if and only if the Frobenius conditions are satisfied for the system (1).

If one has an inhomogeneous linear system $$ \frac{\partial u^a}{\partial x^i} = \Gamma^a_{bi}(x) u^b + \Lambda^a_i(x), \tag{2} $$ one can convert it to a homogeneous system by replacing $\Lambda^a_i(x)$ by $\Lambda^a_i(x)u^0$ in the above equation and adjoining the equations $$ \frac{\partial u^0}{\partial x^i} = 0, $$ making it a homogeneous linear system in one more variable. (Effectively, one allows the $a$ and $b$ indices (which run from, say, $1$ to $s$) to attain the value $0$, and sets $\Gamma^0_{bi}= \Gamma^0_{0i}=0$ and $\Gamma^a_{0i}=\Lambda^a_i$.) Again, the Frobenius conditions are equivalent to the vanishing of the curvature of the connection associated to the augmented system (now in $s{+}1$ unknowns rather than $s$ unknowns).

If one goes all the way to a fully nonlinear total differential system, say, $$ \frac{\partial u^a}{\partial x^i} = \Gamma^a_{i}(x,u), \tag{3} $$ then one can't (usually) reformulate this as the vanishing of the covariant derivatives of a section of something. However, there is still a 'covariant' coordinate-free interpretation of the Frobenius condition in terms of a tensor, but the tensor is defined on the total space $S$ (which has the $x$'s and $u$'s as local coordinates). The point is then, that, on $S$, one has a plane field $D$ (sometimes called a 'distribution') defined by the Pfaffian equations $$ du^a - \Gamma^a_{i}(x,u)\ dx^i = 0 $$ (summation on $i$ intended). Associated to this plane field $D\subset TS$, there is a natural skew-symmetric linear operator $$ \Phi: D\times D \to TS{\bigl/}D $$ i.e., a section of the bundle $\bigl(TS{\bigl/}D\bigr)\otimes \Lambda^2(D^\ast)$ over $S$ that vanishes if and only if the system (3) satisfies the Frobenius conditions. It is this operator that is the 'covariant' interpretation of the Frobenius integrability conditions. Of course, this operator is defined via the Lie bracket. On sections $X$ and $Y$ of $D$, it is just $$ \Phi(X,Y) \equiv [X,Y]\ \mathrm{mod}\ D, $$ so one gets the usual statement that the Frobenius integrability conditions are equivalent to the condition that the sections of $D$ be closed under Lie bracket.

Finally, to get back to when one can interpret the Frobenius conditions as the vanishing of the curvature of a connection, there is a rather general formulation of this that goes back all the way to Lie: Suppose that the system (3) can be written in the form $$ \frac{\partial u^a}{\partial x^i} = \Gamma^a_{i}(x,u) = \gamma_{i\sigma}(x) F^{a\sigma}(u), \tag{4} $$ for some functions $\gamma_{i\sigma}$ of $x$ and $F^{a\sigma}$ of $u$ (where summation over a new index $\sigma$ is assumed), and, further, assume that the vector fields $$ U^\sigma = F^{a\sigma}(u)\ \frac{\partial\ \ }{\partial u^a} $$ form a Lie algebra $L$, i.e., that there exist constants $c_\sigma^{\rho\tau}=-c_\sigma^{\tau\rho}$ such that $[U^\rho,U^\tau] = c_\sigma^{\rho\tau}U^\sigma$. Then the system (4) is what is known as a system of Lie type.

To avoid degenerate situations, one can assume, without loss of generality, that the $U^\sigma$ are a basis of $L$. In this case, one has that the Frobenius integrability condition for (4) is equivalent to the vanishing of the curvature of the $L$-valued connection $1$-form $$ \omega = U^\sigma\ \gamma_{i\sigma}(x)\ dx^i. $$ The use of this form of the equations belongs in the theory of symmetry analysis of differential equations, about which there is a very large literature.

Of course, recognizing when (3) can be written in the form (4) is not a trivial matter; there are methods for doing this involving Cartan's Method of Equivalence, but that would take a while to explain.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.