# Do differential equations correspond to vector fields, or differential forms, or both?

Suppose we have two populations, one of predators, the other of prey. Let $$x$$ denote the number of prey and $$y$$ denote the number of predators. Suppose we've chosen to model the rise and fall in population numbers by the Lotka-Voltera equations: $$\frac{dx}{dt} = \alpha x - \beta xy, \qquad \frac{dy}{dt} = -\gamma y + \delta xy$$

It seems to me that there's both a vector field interpretation of these equations, and also a differential forms interpretation.

For the differential forms viewpoint, we work on a three dimensional manifold with coordinate axes labelled $$x$$, $$y$$ and $$t$$, and we look for submanifolds whose cotangent bundles satisfy the following equations: $$dx = (\alpha x -\beta xy)dt, \qquad dy = (-\gamma y + \delta xy)dt.$$

Another viewpoint is that we work on a two-dimensional manifold with coordinate axes labelled $$x$$ and $$y$$ and we look for integral curves of the vector field $$(\alpha x - \beta xy)\frac{\partial}{\partial x} + (-\gamma y + \delta xy) \frac{\partial}{\partial y}$$

Question. Are differential equations vector fields or differential forms, or both? More generally, how does this all work?

1. On Cartesian space, are both viewpoints equally valid? Why are they equivalent?

2. When we work on smooth manifolds, are both viewpoints equally valid? Is there a sense in which they're equivalent, but not canonically equivalent, kind of like how a finite-dimensional vector space is isomorphic to its dual, but not in a canonical way?

• Once the manifold has been given a Riemannian metric $g$ (this is always possible on a $\sigma$-compact manifold), there is a canonical $1:1$ correspondence between vector fields and $1$-form given by $X\mapsto g(X,\cdot)$. Jan 11 '19 at 20:13

Good questions! As far as I see, an ODE system could find its interpretation by either a vector field or a differential form. Let me try to provide a view using fiber bundle. I hope this may partly provide some answer to your questions, and I also very much look forward to further answers from geniuses in our community.

In the following paragraphs, let us focus on the following ODE system $$\dot{x}^{\mu}(t)=f^{\mu}(t,x^1(t),x^2(t),...,x^n(t))$$ for $$\mu=1,2,...,n$$ (I would like to thank @MoisheCohen for suggesting this clarification).

Let $$M$$ be an $$n$$-dimensional differentiable manifold, with $$\left\{x^{\mu}\right\}_{\mu=1}^n$$ being its local coordinates. Let $$P=\mathbb{R}\times M$$. Consider the following trivial fiber bundle $$\pi:P\to\mathbb{R}$$, where $$\mathbb{R}$$ serves as the base space with $$t$$ being its local coordinate, $$M$$ plays the role of the fiber, and $$P$$ becomes the total space.

Consider the tangent bundle of $$P$$, denoted by $$TP$$. Obviously, $$\left\{\frac{\partial}{\partial t},\frac{\partial}{\partial x^{\mu}}:\mu=1,2,...,n\right\}$$ forms a basis of $$TP$$. Now, we hope to assign a connection on $$TP$$ by seeking for a direct-sum decomposition $$TP=HP\oplus VP$$, where $$VP$$ denotes the vertical space of $$TP$$, i.e., $$\forall\,p=\left(t,x\right)\in\mathbb{R}\times M=P$$, $$V_pP=T_x(\pi^{-1}(\pi(p))$$, while $$HP$$ is called the horizontal space of $$TP$$.

Such an assignment could be done in two identical ways. Let us start from an intuitive one. It is obvious that a basis for $$VP$$ could be $$\left\{\frac{\partial}{\partial x^{\mu}}:\mu=1,2,...,n\right\},$$ since each of these vectors is tangential to the fiber. Therefore, it suffices to construct a basis for $$HP$$. Since $$HP$$ is a subspace of $$TP$$, its basis is no more than a linear combination of all base vectors of $$TP$$. Since $$TP=HP\oplus VP$$ and the basis for $$VP$$ does not include $$\partial/\partial t$$, the basis for $$HP$$ must include a non-degenerating $$\partial/\partial t$$ term. Thus a trivial choice of the base vector is $$\frac{\partial}{\partial t}+f^{\mu}\frac{\partial}{\partial x^{\mu}},$$ where each $$f^{\mu}$$ is a function of $$t$$ and all $$x^{\nu}$$'s, and the Einstein notation has been employed. To sum up, we have \begin{align} HP&=\text{span}\left\{\frac{\partial}{\partial t}+f^{\mu}\frac{\partial}{\partial x^{\mu}}\right\},\\ VP&=\text{span}\left\{\frac{\partial}{\partial x^{\mu}}:\mu=1,2,...,n\right\}. \end{align} This assignment defines a connection on $$P$$.

Alternatively, the assignment could also be done by defining a $$VP$$-valued differential $$1$$-form on $$P$$, denoted by $$\omega$$, such that $$HP$$ is its kernel and $$VP$$ is its image. This $$\omega$$ is easy to be determined. Compatible to the basis for $$HP$$ and $$VP$$ from above, it is clear that $$\omega=\left({\rm d}x^{\mu}-f^{\mu}{\rm d}t\right)\otimes\frac{\partial}{\partial x^{\mu}}.$$ One may check that its kernel is exactly $$HP$$ defined from above.

Before moving to our ODE system, let us recap the definition of horizontal lift. Let $$\gamma:\mathbb{R}\to\mathbb{R}$$ be a curve in the base space $$\mathbb{R}$$. A curve $$\sigma:\mathbb{R}\to P$$ is called a horizontal lift of $$\gamma$$ into the total space $$P$$ if $$\pi(\sigma)=\gamma$$ and its tangent vector $$\dot{\sigma}\in HP$$.

Now, we may relate all these arguments to our ODE system. In local coordinates, without loss of generality, take $$\gamma(t)=t$$. Let its horizontal lift be $$\sigma(t)=\left(t,x^1(t),x^2(t),...,x^n(t)\right)$$, where each $$x^{\mu}=x^{\mu}(t)$$ is to be determined. Note that $$\dot{\sigma}$$ can be taken as a tangent vector by $$\dot{\sigma}=\frac{\rm d}{{\rm d}t}=\frac{\partial}{\partial t}+\dot{x}^{\mu}\frac{\partial}{\partial x^{\mu}}.$$ Thus the form of base vectors of $$HP$$, as well as the condition $$\dot{\sigma}\in HP$$, implies $$\dot{x}^{\mu}(t)=f^{\mu}(t,x^1(t),x^2(t),...,x^n(t))$$ for $$\mu=1,2,...,n$$, which is exactly our ODE system. Alternatively, provided that $$\dot{\sigma}\in HP$$ is equivalent to $$\omega(\dot{\sigma})=0$$, we have $$\left({\rm d}x^{\nu}-f^{\nu}{\rm d}t\right)\biggl(\frac{\partial}{\partial t}+\dot{x}^{\mu}\frac{\partial}{\partial x^{\mu}}\biggr)=0$$ for $$\nu=1,2,...,n$$, which also leads to our ODE system $$\dot{x}^{\nu}(t)=f^{\nu}(t,x^1(t),x^2(t),...,x^n(t)).$$

To sum up, the solution to an ODE system $$\dot{x}=f$$ can be taken as the horizontal lift of $$\mathbb{R}$$ (which is the same as the curve in $$\mathbb{R}$$) into the fiber bundle $$P=\mathbb{R}\times M$$ whose connection is assigned by $$f$$, where $$x$$ is the local coordinate of the fiber $$M$$. In constructing the connection, $$f$$ appears to be either the coefficients of the linear combination of tangent vectors, or the coefficients of a differential $$1$$-form.

That's it! I hope this partly answers your question. And I am looking forward to further answers from geniuses in our community.

• This looks interesting. I'll check it out when I have more time. Jan 10 '19 at 9:07
• @goblin: Sure! Take your time :-) Hope this could be somewhat helpful for you. Jan 10 '19 at 9:26
• You should add that this works only for ODEs which are linear with respect to the derivative term. Jan 11 '19 at 1:37
• @MoisheCohen: Thanks for your suggestion. I added a clarification in the second paragraph, credited to you. And I would like to learn more about its necessity. My thought, if an ODE system observes nonlinearity in its derivative terms, most likely we can solve some algebraic equation(s) and re-express the system with linear derivatives (this works merely locally; but the existence/uniqueness of the solution to an ODE system is also merely locally guaranteed). Hence I would really appreciate it if you could provide some examples where the linearity requirement is a must. Jan 11 '19 at 2:18