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Is the following idea something that is known?

I call a metric group[1] $G$ "approximately commutative near the identity" if there exists a $K$ such that for small enough $\epsilon$, when $d(g,id) < \epsilon$ and $d(h, id) < \epsilon$ then $d(gh, hg) < K \epsilon^2$.

Any commutative group is an example of this and I believe it should be straightforward to show that every Lie group satisfies this condition.

Which other groups do? Perhaps if a group is locally compact and "approximately commutative near the identity" then it must be a Lie group, but I don't know anything about any sort of topological groups.

I feel that groups of operators on Banach spaces ought to satisfy this condition too, or at the very least groups of operators on Hilbert space.

Which groups don't? Well if the metric and the group operation are both somehow "scale invariant" then $d(gh, hg)$ will be of first order, but I have no idea of a more general characterisation.

I'd also be interested in a discussion of this idea carried over to semigroups too.

Edit: [1] a metric group being a group whose operations are continuous with respect to the metric

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Can you state precisely what do you mean by a "metric group"? –  Łukasz Grabowski Jan 2 '11 at 22:14
    
@Lukasz: Yes, that’s a good question: there would be several different reasonable notions of “metric group”, corresponding to the several reasonable notions of morphisms of metric spaces. The first possibilities that spring to my mind are (from lax to strict): continuous; locally Lipschitz; Lipschitz; contraction; isometry. (These latter two give the same notion of metric group, since an invertible contraction is an isometry.) –  Peter LeFanu Lumsdaine Jan 2 '11 at 22:27
    
Can you write the condition on the group more precisely? In particular - how do you quantify $\epsilon$? –  Vít Tuček Jan 2 '11 at 22:28
    
How can the multiplication $G \times G \to G$ possibly be an isometry? $G \times G$ looks like having "dimension" the double of the "dimension" of $G$. –  Qfwfq Jan 2 '11 at 22:47
    
It is elementary that the multiplicative group of a unital Banach algebra satisfies the condition with $c=2$: write $g=I+a$, $h=I+b$ and multiply out. –  Bill Johnson Jan 2 '11 at 23:15

1 Answer 1

Yes, it's an easy and well-known fact that for a Lie group with a smooth Riemannian metric (which we may assume left-invariant) $$d([g,h], 1) = d(g*h, h*g) = O(d(1,g)*d(1,h)) . $$ This follows from differentiability, and from the observation that the commutator map $[,]: G \times G \rightarrow G$ maps the submanifolds $1 \times G$ and $G \times 1$ both to $1$, so the first derivative of the commutator is 0. The second derivative is given by the Lie bracket.

However, any Lie group whose identity component is not abelian admits left-invariant path-metrics where this inequality fails, in particular, its Carnot-Caratheodory metrics. Consider, for instance, the group of isometries of the plane from the point of view of driving a car in a flat area. Positions of the car are in 1-1 correspondence with the isometries of the plane that take it from its initial parked position to a given position. Define a driving metric that is the minimum total number of turns of the drive shaft (~mean arclength of drive wheels) to get from one position to another. As every driver knows from parallel parking, the driving distance to move sideways by a small Euclidean distance $x$, annoyingly, does not decrease to 0 linearly with x: it is proportional to $\sqrt x$, because the front wheels need (approximately) to enclose an area proportional to the sideways distance. In the driving metric, the distance from $g*h*$ to $h*g$, for example if $g = $ cramp the steering wheel to the left and drive forward one unit while $h = $ cramp the steering wheel to the right and drive forward 1 unit is sometimes as much as $2( d(1,g)+d(1,h))$: in general, there may not be a more efficient way to get from $g*h$ to $h*g$ than to retrace: first do the inverse of $g*h$, then do $h*g*$.

More generally, for any Lie group and any subspace $V$ of its Lie algebra (the tangent space at $1$) that generates the Lie algebra, there is a Carnot-Caratheodory metric that measures path lengths with along the left-invariant plane field which agrees with $V$, with respect to a left-invariant Riemannian metric defined on this plane field. These metrics are important in many real situations. Any non-abelian Lie group has many such subspaces. Just as in driving, where cars follow paths that have bounded curvature, one can further restrict to paths whose tangent vectors are in the left-invariant cone field that extends any cone in the Lie algebra that generates the Lie algebra, although the Lipschitz equivalence class of the resulting metric depends only on the linear span of the cone (assuming a finite dimensional Lie group).

The question has at least as much to do with the metric as with the group.

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