Making sense of ‘Floppy’ Structures on Manifolds

differential-geometrymathematical physicssmooth-manifoldssoft-questionsymplectic-geometry

Roger Penrose in his book "The Road to Reality" (Section 14.8 – Symplectic Manifolds) loosely defines a "floppy" structure to be one which if we apply two variants of it on two copies of the same manifold, the two manifolds are locally isomorphic (and thus locally indistinguishable from one another). The example he brings is that of two symplectic manifolds with the same dimension and signature. I think what he is referring to (although he does not make it explicit) is Darboux's Theorem which again asserts that two symplectic manifolds of the same dimension are locally isomorphic (more accurately: symplectomorphic).

In his exact words and I quote:

The local structure of a symplectic manifold is an example of what
might be called a ‘Floppy’ structure. There is, for example, no notion
of curvature for a symplectic manifold, which might serve to
distinguish one symplectic manifold from another, locally. If we have
two real symplectic manifolds of the same dimension (and the same
‘signature’, cf. §13.10), then they are locally completely identical
(in the sense that for any point p in one manifold and any point q in
the other, there are open sets of p and q that are identical). This
is in stark contrast with the case of (pseudo-) Riemannian manifolds,
or manifolds in which merely a connection is specified. In those
cases, the curvature tensor (and, for example, its various covariant
derivatives) defines some distinguishing local structure which is
likely to be different for different such manifolds.

This makes sense to me. Then he goes on to describe two more examples of manifolds one of which is floppy while the other isn't. In particular he makes the following two claims:

  1. Let $M_1$ be a real manifold with a nowhere vanishing vector field $F$ on it. Then $M_1$ is a floppy manifold.
  2. In contrast, let $M_2$ be a real manifold with two general vector fields. Then $M_2$ is NOT a floppy manifold

Question 1: For the 1st part, I think the family of tori would be a good place to start since assigning a (smooth) nowhere vanishing vector field is always possible. I can also see how we can use $F$ to define a global frame on the entire $M_1$ starting by setting $\partial_1 := F \neq 0$ etc. However, in what sense are all tori (with $F$) locally indistinguishable? For example, shouldn't we able to (locally) distinguish a torus with circular cross sections versus one with elliptical just by looking at the difference in curvatures between a circle and an ellipse?

Question 2: Now to his second claim. First of all, what do you think he means by "general" vector fields? Am I right to assume that they are linearly independent and by extension neither one of these can be anywhere vanishing (as this would violate independence)? And how does the existence of the second field makes them distinguishable?

Best Answer

Question 1: I think Penrose is thinking of manifolds (which are floppy) as opposed to Riemannian manifolds (which are not). Specifically, the idea of a torus having a cross sectional circle vs a cross sectional ellipse is implicitly viewing tori as Riemannian manifolds, not as plain manifolds.

The sense in which a manifold with vector field is floppy is the following:

Suppose $M$ is a smooth manifold and $V$ is a smooth vector field on $M$ with $V(p)\neq 0$ for some $p\in M$. Then there is a chart $U$ containing $p$ for which $V$ is a coordinate vector field.

(See, e.g., Theorem 2.1 of these notes)

Question 2: I'm not exactly sure what Penrose means by "general", but his comment certainly applies to linearly independent vector fields. Specifically, given linearly independent vector fields $V_1$ and $V_2$ on (a neighborhood in) a manifold, one can ask whether or not there is a single chart where $V_1$ and $V_2$ are both coordinate vector fields.

In general, the Lie bracket $[V_1,V_2]$ measures the obstruction to this. That is, if $[V_1,V_2] = 0$, then the answer is yes, while if the bracket is non-zero, the answer is no.