# Relationship between Frobenius theorem, curvature, and integrability

In this answer to References for "modern" proof of Newlander-Nirenberg Theorem John Hubbard alluded to something called the "Frobenius integrability form" $$\phi\mapsto\bar\partial\phi-\frac12[\phi\wedge\phi]$$. To me this looks just like curvature, but I don't understand how it's related to Frobenius. The Frobenius theorem for a Pfaffian system generated by a 1-form $$\theta$$ states that the system is integrable iff $$\mathrm{d}\theta\wedge\theta=0$$. Does this condition have any relation to the curvature form?

Generalizing Pedro's comment somewhat, one can think about the relationship between the Frobenius theorem and curvature as follows.

Let $$M$$ be a smooth real $$m$$-manifold, and $$H\subseteq TM$$ a smooth $$k$$-dimensional ($$0) distribution on $$M$$. The Frobenius theorem then states that $$H$$ is (completely) integrable if and only if for each pair $$X,Y\in\Gamma(H)$$ of smooth vector fields belonging to $$H$$, we also have $$[X,Y]\in\Gamma(H)$$, i.e. $$H$$ is involutive.

So we would like a way to measure how much the distribution $$H$$ fails to be involutive. But without some extra structure we cannot assign any "tensorial" measure of involutivity.

Let this extra structure be a distribution $$V\subseteq TM$$ which is complementary to $$H$$, i.e. we have $$TM=V\oplus H$$. Then we have projection operators $$P_V:TM\rightarrow V$$ and $$P_H:TM\rightarrow H$$ obeying $$\mathrm {id}_{TM}=P_V+P_H$$.

Let's say that the complementary pair $$(V,H)$$ is a connection on $$M$$ with $$H$$ the horizontal distribution and $$V$$ the vertical distribution. In this general setting these two distributions are on completely equal footing, so the distinction of "horizontal" and "vertical" is completely arbitrary.

But now we can construct a tensor field which measures how much $$H$$ fails to be involutive as $$R:\mathcal D\times\mathcal D\rightarrow\Gamma(V),\quad R(X,Y)=P_V[P_HX,P_HY],$$ where $$\mathcal D=\Gamma(TM)$$ is the module of smooth vector fields. This is indeed a tensor because if $$f\in C^\infty(M)$$ is a smooth function, we have $$R(X,fY)=P_V[P_HX,fP_HY]=fP_V[P_HX,P_HY]+(P_HX)fP_VP_HY=fP_V[P_HX,P_HY],$$and analogously for the first slot in $$R$$. We can then call $$R$$ the curvature of the connection $$(V,H)$$.

Note that since $$V$$ is on equal footing with $$H$$, we can also define the cocurvature $$Q(X,Y)=P_H[P_VX,P_VY],$$which then measures how much $$V$$ fails to be involutive.

We find that $$H$$ is involutive (and thus Frobenius-integrable) if and only if its curvature vanishes. The value of the curvature depends on the choice of $$V$$ but whether it is nonzero or not does not.

The situation with fibered manifolds is a special case of this one. If $$\pi:N\rightarrow M$$ is a fibered manifold, then the total space $$N$$ already has an involutive distribution $$V\subseteq TN$$ given by $$V=\ker(T\pi)$$. A connection on $$(N,\pi,M)$$ is then a smooth distribution $$H\subseteq TN$$ which is complementary to $$V$$. The curvature of $$H$$ is defined the same way it was above in the more general situation and once again $$H$$ is integrable if and only if the curvature vanishes. The cocurvature trivially vanishes because $$V$$ is always involutive.

• This is perfect, thank you! The connection (no pun intended) back to the fibered case is beautiful. Commented Mar 20 at 4:26

There is a formal analogy between the Frobenius theorem and the Newlander-Nirenberg theorem: the bracket closure of the vector fields tangent to a real subbundle of the tangent bundle is analogous to the bracket closure of a complex subbundle of the complexified tangent bundle. In case of an almost complex structure $$J$$, we consider the subbundle of complex tangent vectors $$u+iv$$ for which $$v=Ju$$. This is bracket closed just when the Nijenhuis tensor vanishes, which the Newlander-Nirenberg theorem tells us is just when the almost complex structure is a complex structure. If we try to make this analogy more precise by writing out what you called curvature forms, we get the expressions in the reference you mentioned. In the real analytic category, we can make these analogies precise by complexifying our manifold, perhaps only locally, and then the Frobenius theorem gives the real analytic Newlander-Nirenberg theorem.

• This makes it sound like the appearance of a curvature-like expression is accidental. However I'm under the impression that vanishing curvature is often treated as synonymous with integrability, so I'm really wondering whether there's a deeper connection here. Commented Mar 18 at 19:27
• Indeed vanishing of the curvature of a(n Ehresmann) connection (on a fiber bundle $\pi:E\rightarrow M$ = fiberwise linear projection $\Phi$ of $TE$ onto the vertical bundle $VE = \ker T\pi$) is synonymous with integrability - to wit, that of the horizontal bundle $HE = \ker\Phi$. More precisely, the curvature of the connection $\Phi$ is given by $$R(X,Y)=\Phi[(1-\Phi)X,(1-\Phi)Y]\ ,X,Y\in\mathfrak{X}(E)$$ so it measures how involutivity of $HE$ fails. Commented Mar 18 at 20:03
• The similarity of $R$ with the Nijenhuis tensor of an almost complex structure $J$ is due to the fact that the latter is half the Frölicher-Nijenhuis bracket of the vector-valued 1-form $J$ with itself, whereas $R$ is half the Frölicher-Nijenhuis bracket of the connection $\Phi$ with itself. Commented Mar 18 at 20:03
• @PedroLauridsenRibeiro this is very clarifying, thanks! Commented Mar 19 at 18:21