I am currently reading the book Mathematics in physics by Michael Stone and Paul Goldbart. In chapter 11, page 421, the authors say that

"Except in pathological cases, the configuration space M of a mechanical system is a manifold."

Are the authors correct? What examples are there of these pathological cases? Are there any physically relevant cases where the configuration space in classical mechanics is not a manifold? And why does configuration space have to be a manifold - what about variables taking on discrete values?

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    $\begingroup$ If the system has a configuration space of discrete values then anything which requires continuity won't apply. Like, for instance, Newton's laws. $\endgroup$ – user107153 Jan 8 '18 at 18:10
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    $\begingroup$ Don't understand either.... What's the author's definition of manifolds? Strictly speaking manifolds should always be finite dimension, right (of course you could generalize to infinite dimension but that's the generalization...)? So even "manifolds" on Hilbert space are not manifolds... $\endgroup$ – Kite.Y Jan 8 '18 at 18:36
  • $\begingroup$ This post (v1) seems to be a list question. $\endgroup$ – Qmechanic Jan 8 '18 at 18:37

Consider a pair of points $P$ and $Q$, say of mass $m>0$, and suppose that they are constrained as follows.

  1. $Q$ stays on the $z$ axis and can freely move along it up to restrictions said below.

  2. $P$ can freely rotate around $z$ and is connected to $Q$ by means of an ideal rod (zero mass) of length $\ell$, and it is also connected to the origin $O$ by means of another ideal rod of length $\ell$.

This is a quite standard idealized mechanical system (I could use it as a starting point for an exercise of my undergraduate course of Analytic Mechanics).

A suitable coordinate system to describe the system is apparently given by the signed distance $Z\in (-2\ell, 2\ell)$ of $Q$ from $O$ along $z$ and an angular coordinate $\theta\in (-\pi,\pi]$ describing the position of $P$ around $z$ in the plane $x,y$.

Well you see that, discarding the extreme points $Z= \pm 2\ell$ (they could be included with a more precise discussion, see the ADDENDUM) a disaster shows up when $Z=0$.

Concerning the set $Z\in (-2\ell, 0)$ and $Z\in (0, +2\ell)$, the space of configurations is diffeomorphic to $\mathbb R \times \mathbb S^1$. When $Z=0$ the configuration space (at fixed $Z$) instead of being $\mathbb S^1$, it becomes $\mathbb S^2$ and another set of coordinates should be used.

There are two possibilities: either declaring that the configuration space is diffeomorphic to $\mathbb R \times \mathbb S^1$ deliberately ignoring the problem at $Z=0$, or declaring that it is not a ($2$-dimensional) manifold (because each point in the subset at $Z=0$ has a neighborhood that is not diffeomorphic to $\mathbb R^2$), but is made of the union of three manifolds respectively diffeomorphic to $\mathbb R \times \mathbb S^1$, $\mathbb S^2$, and $\mathbb R \times \mathbb S^1$. In practice, with an imprecise but pictorical description (see the ADDENDUM for a precise description) it is the union of a cylinder and a sphere inside the cylinder, tangent to the cylinder at the equator. This is not a manifold because it is not locally homeomorphic to $\mathbb R^n$ for some fixed $n$ ($2$ in our case).

Generally speaking the space of configuration is almost always a manifold because it is obtained by imposing constraints on a set of $N$ matter points initially described in $\mathbb R^{3N}$. Constrains are determined by a family of $c< 3n$ real-valued functions $f_k = f_k(t,\vec{x}_1,\cdots, \vec{x}_N)$ by imposing that every admissible configuration $\vec{x}_1,\cdots, \vec{x}_N$ at any time $t$ must satisfy $$f_k(t,\vec{x}_1,\cdots, \vec{x}_N) =0\:, \quad k=1,\ldots, c\:. \tag{1}$$ If the functions $f_k$ are smooth and functionally independent, the theorem of regular values proves that (1) defines an embedded submanifold of $\mathbb R \times \mathbb R^{3N}$ of dimension $1+ 3N-c$. Fixing time $t \in \mathbb R$, we have an embedded submanifold of $\mathbb R^{3N}$, of dimension $3N-c$, called space of configurations.

The two conditions of the constraints may be false for some points and it sometimes happens in particular when dealing with constraints like rigidity in some geometrically involved way.

ADDENDUM. If $X,Y, Z$ denote the coordinate set of $Q$ and $x,y,z$ the coordinate set of $P$, both in the whole $\mathbb R^3$ space, the four constraints, corresponding to the set of conditions 1 and 2 above, read

$$f(x,y,z, X,Y,Z)=0\:, \quad g(x,y,z,X,Y, Z)=0\:, \quad h(x,y,z, X,Y,Z)=0, \quad i(x,y,z, X,Y,Z)=0\tag{2}$$ where $$f(x,y,z, X,Y, Z) := x^2+y^2+z^2 -\ell^2\:, \quad g(x,y,z, X,Y,Z) := x^2+y^2+(z-Z)^2 -\ell^2\:, \quad g(x,y,z, X,Y,Z)= X\:, \quad i(x,y,z, X,Y,Z)= Y\:.$$ These constraints are functionally independent if, by definition, their differentials are linearly independent on the set of points $(x,y,z,X,Y, Z)$ where all conditions (2) are valid.

In this case theorem of regular values implies that this set is an embedded submanifold of $\mathbb R^3 \times \mathbb R^3$ with dimension $6-4 =2$. It is clear by direct computations that the four differentials are not linearly independent (because $df=dg$) when $Z=0$. This is the problem with the said system of constraints.

A precise description of the configuration space when including also the points $Z= \pm 2\ell$ is the union of two $2$-spheres in $\mathbb R^4$ that have an equator in common.

  • $\begingroup$ Sorry, I do not know, advanced geometric formulations of classical mechanics are not part of my research field. What I wrote arises from my undergraduate lectures (I teach also Lagrangian and Hamiltonian Mechanics though my research field is QM and QFT). $\endgroup$ – Valter Moretti Feb 27 '18 at 10:29

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