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It's in my website url, but what does it actually mean? I hope to explain what it means in this blog post. Most of this I formulated for my thesis, but I hope to keep it a little more accessible for those who do not know much about manifolds.

The vector field $v(x,y) = (-y,x)$ on $\mathbb R^2.$ Any circle centered at the origin is an integral curve of this particular $v$.

Vector fields

For the well-initiated, a vector field $v$ on a manifold $M$ is a smooth section $v\colon M\to TM$ of the tangent bundle of $M$. For those who do not understand this terminology, just read this as "it's a function which---when you hand it a point---it hands you a vector" (this vector is tangent to the manifold in question). Geometrically, this means that at each point $p$ of $M$ we can imagine an arrow called $v_p = v(p)$ based at $p$.

An integral curve to $v$ is a curve $\gamma\colon(a,b)\to M$ on $M$ such that: $$\gamma'(t) = v(\gamma(t)),\quad\quad\text{for each }t\in(a,b).$$ That is to say, the so-called "velocity vector" of the curve at time $t$ is exactly the arrow $v(\gamma(t))$ at the point $\gamma(t)$. To our vector field $v$ we may associate (locally) a system of first order ordinary differential equations (arising from the honest-to-goodness linearisation of $v$). The solutions to this system of ODEs correspond to integral curves of $v$, and hence why we call them "integral" (we integrate the system to get the solution in the form of a curve).

The Picard-Lindelof theorem guarantees the existence of solutions to a system of ODEs and so the result transfers to the manifold (at least locally). It turns out that every vector field integrates to a unique maximal "flow" (a family of integral curves). The result is easily glossed over in an ODE course, but it's actually a very special phenomenon as the moment you begin to consider higher-dimensional analogues to vector fields, it fails.

Integrability and $d$-fields

This is a $2$-field (i.e. plane field) in $\mathbb R^3$. It is translation invariant in the $z$-direction.

Before I get into the nittygritty, I will give the layman a chance. A $d$-field on a manifold $M$ is the association of a $d$-dimensional subspace of the tangent space of $M$ to each point of $M$. The small $d$ examples are the only ones you need to be concerned about: for $d=1$, a $1$-field is a vector field (more accurately, a line field); and for $d=2$, a $2$-field is a function $\xi$ which associates to each $p$ on $M$ a plane $\eta_p = \eta(p)$ tangent to $M$ at $p$. Moreover, this function is smooth, meaning that as $p$ varies, $\eta(p)$ varies smoothly with it. For $d>2$, a $d$-field is hard to imagine, so don't worry about it since we will only ever care about these two cases.

For the mathematically inclined, you might have noticed I took some artistic liberties in my explanation above. To remedy it, the actual definition of a $d$-field is a smooth subbundle of $TM$ of rank $d$. Hopefully this satisfies your craving for formalities.

The notion of integral curve then extends to the notion of integral submanifolds of $M$ to a $d$-field $\eta$. A $k$-submanifold $\Gamma\subseteq M$ is an integral submanifold to $\eta$ if for each point $p$ in $\Gamma$, the tangent space $T_p\Gamma$ of $\Gamma$ at $p$ is contained in the $d$-dimensional space $\eta(p)$. If you let $d=1$, this corresponds to integral curves for vector fields. We say that $\eta$ is integrable if each point is contained in an integral $d$-submanifold to $\eta$ (note that $k=d$ here!). For $d=1$, we have that Picard-Lindelof translates as "$1$-fields are always integrable".

If we consider a $2$-field $\eta$ (read: plane field) in $\mathbb R^3$, the plane $\eta(p)$ at the point $p$ admits a line $\ell_p$ normal to it. For any tangent vector $v$ in $T_p\mathbb R^3$, we can project $v$ onto the line $\ell_p$. This gives a projection map $\alpha_p\colon T_p\mathbb R^3\to\mathbb R$. Note that $\alpha_p$ satisfies the property: $$v\in\eta(p) \iff \alpha_p(v) = 0,$$ and so $\ker\alpha_p = \eta(p)$. The assignment $p\mapsto \alpha_p$ is a 1-form $\alpha$ on $\mathbb R^3$ such that $\ker\alpha = \eta$.

An important example is the pictured plane field above is the kernel of the 1-form $\alpha = dz - y\,dx$. That is, the plane at $p = (p_1,p_2,p_3)$ is given by the span: $$\ker\alpha_p = \text{span}\left\lbrace\begin{bmatrix}0\\1\\0\end{bmatrix},\begin{bmatrix}1\\0\\p_2\end{bmatrix}\right\rbrace.$$ Let's call this plane field $\xi = \ker(dz-y\,dx)$. Note that if $\Gamma$ is an integral surface of $\xi$ containing the origin, then since $\Gamma$ is tangent to the plane $$\xi(0) = \text{span}\left\lbrace\begin{bmatrix}0\\1\\0\end{bmatrix},\begin{bmatrix}1\\0\\0\end{bmatrix}\right\rbrace,$$ it ought to have a small neighbourhood containing an integral curve with tangent vector $\begin{bmatrix}1\\0\\0\end{bmatrix}$ through 0 (i.e. it contains a small segment of the $x$-axis), as well as an integral curve with the tangent vector $\begin{bmatrix}0\\1\\0\end{bmatrix}$ through 0 (i.e. it contains a small segment of the $y$-axis). Thus $\Gamma$ contains an open set of the $x,y$-plane!

But alas! that is not possible! Any little open set of $x,y$-plane about the origin intersects $\xi(p)$ transversely for $p$ not the origin. Thus $\Gamma$ cannot exist, and $\xi$ is non-integrable. The 2-dimensional analogue to Picard-Lindelof doesn't exist! Because integrability in the $d=2$ case is not as simple as the $d=1$ case, we will seek to understand it a bit better.


For the uninitiated, note that $\frac{\partial}{\partial x}$ is the vector field (in $\mathbb R^3$) which we identify with the constant unit vector field $p\mapsto(1,0,0)$ in the $x$-direction. With this notation, it's apparent that we can view the vector field $\frac{\partial}{\partial x}$ as an operator which takes in a smooth function $f$ on $M$ (the set of which we denote by $C^\infty(M)$) and spits out a smooth function $\frac{\partial f}{\partial x}$ (as well, it is linear with regards to this association, and it satisfies the Leibniz rule---a.k.a. the product rule). In a similar way, any vector field can be viewed as a map $C^\infty(M)\to C^\infty(M)$.

Given two vector fields $v$ and $w$ we might initially wonder if we can then compose them to yield another vector field $w\circ v$. The disappointing answer is that this is not the case, but we can remedy this by taking the commutator of the vector fields. Indeed, $w\circ v$ may not be a vector field but $$[v,w] := v\circ w - w\circ v,$$ always is a vector field. The commutator $[v,w]$ measures the extent that the flows of $v$ and $w$ fail to commute.

A vector field on a manifold $M$ whose given vectors only reside in a $d$-field $\eta$ is called a section of $\eta$. If $\Gamma$ is an integral submanifold to $\eta$, it turns out that the commutator of any two sections of $\eta$ is again a section of $\eta$. This is technically due to the nice embedding of $e\colon\Gamma\to M$ which has the property: $$e_{*,p}(T_p\Gamma) = \eta_{e(p)},\quad\quad\text{for }p\in\Gamma,$$ due to the integrability condition. Thus we have the following fact:

Proposition: Every integrable $d$-field is involutive.

This gives us a speedy proof that $\xi$ from before is not integrable, as we have: $$\left[\frac{\partial}{\partial y},\frac{\partial}{\partial x} + y\frac{\partial}{\partial z}\right] = \frac{\partial}{\partial z},$$ which is never contained in $\xi$.

Our observation is one direction of a result known as Frobenius's theorem:

Frobenius's Theorem: A $d$-field is integrable if and only if it is involutive.

The reason why is a little bit mysterious. For the layman, you might be satisfied to just hear that it relies on Picard-Lindelof, used inductively. Someone more familiar with geometry might better appreciate the following explanation:

If a $d$-field is involutive, it is locally represented by a Lie subalgebra of the algebra of vector fields on $M$. Using the integrability of line-fields, one can use the flow of vector fields as a Lie group which acts (as $\mathbb R^d$) on $M$, inducing a foliation on $M$, the leaves being the orbits of this action. The action (and hence the foliation into leaves) is not well-defined unless the flows commute, which is the case for involutive $d$-fields. Thus we have a hunch that involutive implies integrable.
That being said, I am not happy with this explanation myself so if anyone has a good reason to guess that involutivity implies integrability, then please send me an email.

Complete non-integrability

The point of all this build up is to take what we know and turn it all upside down. We now want a $d$-field $\eta$ to not have a snowball's chance in hell of being integrable. In this case, we will call it completely non-integrable, and that's the namesake of this website. The geometric picture you should take home is that our plane fields (much like $\xi$ from before) twist far too much to ever integrate to a submanifold at any point. How we formulate this exactly is a little bit technical, and the layman who doesn't care can skip to the next section.

If you are still reading this, then you are apparently going to attempt to read the technical details. Recall that we had some plane fields given by the kernel of a 1-form $\alpha$. A more general statement about hyperplanes goes as follows:

Proposition: Any hyperplane field on a manifold $M$ is given locally as the kernel of a 1-form.

The existence of a globally defined 1-form is a little too much to always ask for, so we only require that this phenomenon occurs locally. Let $\eta = \ker\alpha$ be an integrable hyperplane field on $M$. For sections $v,w$ of $\eta$ we have the identity: $$d\alpha(v,w) = v(\alpha(w)) - w(\alpha(v)) - \alpha([v,w]),$$ where the right-hand side is zero as involutivity (thanks, Frobenius) suggests $[v,w]$ also takes values in $\eta$. Hence $d\alpha$ annihilates $\eta$. If $\eta$ were not integrable, but was annihilated by $d\alpha$, then $\alpha([v,w])=0$ for any sections of $\eta$, and hence we have:

Proposition: If a hyperplane field $\eta$ is given locally as the kernel of $\alpha$, then $\eta$ is involutive (and hence integrable) if and only if $d\alpha$ annihilates $\eta$.

It's not hard then to show that in the special case of 3-manifolds:

Proposition: If $M$ is a 3-manifold with a (hyper)plane field $\eta$, then if locally $\eta = \ker\alpha$ for a 1-form $\alpha$, we have: $$\eta\text{ is integrable} \iff \alpha\wedge d\alpha = 0.$$

Now here is where we turn it on its head: if we wanted $\eta$ to not have a snowball's chance in hell of being integrable, then the obvious choice is to guarantee that $d\alpha$ is non-degenerate on $\eta$. Since then $d\alpha|_\eta$ is a non-degenerate, closed 2-form (i.e. a symplectic form on $\eta$!), $\eta$ must have even dimension (a little bit of linear algebra magic), and thus forces $M$ to be of odd dimension!

A little bit more linear algebra gives that a closed 2-form $\omega$ is symplectic (read: non-degenerate, since we assumed closedness) is equivalent to $\omega^n$ being a volume form, and hence the non-degeneracy of $d\alpha|_\eta$ is equivalent to $(d\alpha)^n|_\eta\neq 0$ being a volume form on $\eta$. Even more linear algebra magic yields that this is equivalent to requiring $\alpha\wedge(d\alpha)^n\neq 0$! It is when this condition is sattisfied that we say that $\eta$ is completely non-integrable. Now go check that $\xi$ from before is completely non-integrable!

It turns out that when $M$ is a 3-manifold, the notion of complete non-integrability coincides with the property that for each $p\in M$, there is no integral surface containing $p$. This is not true in general, however. It's an immediate corollary from symplectic geometry that any integral submanifold to a completely non-integrable hyperplane field on a $(2n+1)$-manifold has dimension of at most $n$. If $n=2$, then a complete non-integrability does not preclude the possibility that your manifold is non-integrable "at each point", but yet still admits an inegral submanifold of dimension 3. Thus completely non-integrable is the other extreme of a spectrum to "integrable", the spectrum being the shortest possible when $n=1$.

Contact manifolds

In closing, I guess it's a good thing to mention the point of studying completely non-integrable hyperplane fields. A contact manifold is a manifold endowed with a completely non-integrable hyperplane field. For example, $\mathbb R^3$ with the completely non-integrable plane field $\xi = \ker(dz-y\,dx)$. We call the hyperplane field the contact structure of the contact manifold. Contact manifolds are the major object of study in contact topology, a sister field of symplectic topology. The brief answer as to why we study contact manifolds is because we study symplectic manifolds (a subject that we often can view from the lens of contact topology). Symplectic manifolds (and thus contact manifolds) appear naturally in many physical contexts (e.g. Hamiltonian mechanics) and thus have long been proven to be an important object.