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I have been going through competing dynamical systems textbooks recently, and don't understand the distinction between two definitions. One of my textbooks says that a dynamical system is defined as:

$\mathbf{x}' = f(\mathbf{x})$, where $\mathbf f(x)$ is a vector field such that $\mathbf{f}:\mathbb{R}^{n} \to \mathbb{R}^{n}$.

My other textbook says that $\mathbf{f}$ is a mapping from $M$ (The manifold) to $TM$, the tangent bundle.

Are the two definitions equivalent? Is it common in dynamical systems theory to just assume that $M = \mathbb{R}^n$, I'm not understanding the distinction.

Thanks.

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    $\begingroup$ They aren't equivalent, rather the first is a particular instance of the more general situation. Both want to say that the dynamic system is governed by a (smoothly varying) vector field that describes the infinitesimal motion of a point $x$ under the flow of $f$, only in the more abstract setting it requires more work to properly define what is meant by the vector field at a point, in both cases the proper place for the vector field to live at each point is $T_x M$, but when $M = \mathbb{R}^n$ the tangent plane at $x$ is isomorphic to $\mathbb{R}^n$ itself. $\endgroup$
    – Nadiels
    Commented Mar 4, 2017 at 15:22
  • $\begingroup$ Okay, that is helpful, thanks! Is there a situation where $M$ is not R^n? $\endgroup$ Commented Mar 4, 2017 at 15:29
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    $\begingroup$ Absolutely! Imagine any dynamic system you like on the surface of our planet, these would be better considered as dynamic systems on the sphere $S^2$ than as dynamic systems in the ambient space $\mathbb{R}^3$ $\endgroup$
    – Nadiels
    Commented Mar 4, 2017 at 15:33
  • $\begingroup$ Hi. Thanks that is very helpful. Is dimensionality a question though? If my dynamical system is on R^n, then, I suppose it is finite-dimensional, so I have a finite number of equations to work with. If my dynamical system is on $M$, the it is infinite-dimensional, and I assume I'm dealing with PDEs? Is that correct? $\endgroup$ Commented Mar 4, 2017 at 15:46
  • $\begingroup$ Maybe some confusion about manifolds - $M$ can be finite dimensional! To say that $M$ is an $n$-dimensional manifold is to say that "locally it looks like $\mathbb{R}^n$. Maybe you haven't been exposed to much differential geometry yet? In which case the reference that treats dynamic systems on $\mathbb{R}^n$ is probably the one to work through first. $\endgroup$
    – Nadiels
    Commented Mar 4, 2017 at 15:58

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First let me address what your various textbooks say.

Given a manifold $M$, to say that a vector field $f$ is a mapping from $M$ to $TM$ is incomplete. Let me use $T_p M$ to denote the fiber of $TM$ over the point $p \in M$, in other words $T_p M$ is the tangent space of $M$ at $p$. A vector field on a manifold $M$ is not just any old mapping from $M$ to $TM$. Instead, a vector field on $M$ is a mapping $f : M \to TM$ such that for each $p \in M$ we have $f(p) \in T_p M$, in other words $f(p)$ is a vector in the tangent space at $p$.

In the special case where $M = \mathbb{R}^n$, if you keep all of this notation in mind, then your two textbooks are saying essentially the same thing. The tangent bundle in this case is a product $T \mathbb{R}^n = \mathbb{R}^n \times \mathbb{R}^n$, and the tangent space at each point $p\in \mathbb{R}^n$ has the form $$T_p \mathbb{R}^n = \{(p,v) \,|\, v \in \mathbb{R}^n\} $$ where the vector operations on the vector space $T_p \mathbb{R}^n$ are defined by simply ignoring the $p$ coordinate, i.e. $(p,v) + (p,w) = (p,v+w)$ and similarly for scalar multiplication. Because of this, there is a canonical isomorphism between vector fields expressed as functions $$f : \mathbb{R}^n \to \mathbb{R}^n $$ and vector fields expressed as functions $$g : \mathbb{R}^n \to T \mathbb{R}^n $$ This canonical isomorphism is given by the formula $g(p)=(p,f(p))$.

Regarding your last sentence, perhaps there may be elementary expositions of dynamical systems that restrict attention to $M=\mathbb{R}^n$, but the full theory of dynamical systems considers manifolds in all their full and general glory, and in this theory it is not sufficient to consider $\mathbb{R}^n$. Dynamical systems on spheres, on toruses, and on all kinds of manifolds are important.

I would also point out that it is misleading to say that a dynamical system on a manifold $M$ is a vector field. What is important in dynamical systems is not the vector field in particular, but its integral curves and their behavior over long time spans.

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  • $\begingroup$ Hi. Thanks that is very helpful. Is dimensionality a question though? If my dynamical system is on R^n, then, I suppose it is finite-dimensional, so I have a finite number of equations to work with. If my dynamical system is on $M$, the it is infinite-dimensional, and I assume I'm dealing with PDEs? Is that correct? $\endgroup$ Commented Mar 4, 2017 at 15:46
  • $\begingroup$ The distinction between ODE's and PDE's is not about the dimension of the underlying manifold $M$. It is instead between the dimension of the integral spaces. To follow up on my final paragraph, given a 1st order linear ODE $x'=f(x)$ the integral curves are differentiable functions of the form $\gamma : \mathbb{R} \mapsto M$ where $\gamma'(t)=f(\gamma(t))$, i.e. they are curves and hence 1-dimensional. One can think of a 1st order linear PDE as defining "integral manifolds" of higher dimension, but that's another topic. $\endgroup$
    – Lee Mosher
    Commented Mar 4, 2017 at 16:06
  • $\begingroup$ Why aren't textbooks this explicit? $\endgroup$ Commented Jun 7, 2018 at 4:02

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