Curvature and halfbrackets, part II

I continue from “Curvature and halfbrackets, part I“, with the same notations and background.

In a metric space with dilations (X,d,\delta), there are three quantities which will play a role further.

1. The first quantity is related to the “norm” function defined as

\rho_{\varepsilon}(x,u) = d^{x}_{\varepsilon}(x,u)

Notice that this is not a distance function, instead it is more like a norm of u with respect to the basepoint x, at scale \varepsilon. Together with the field of dilations, this “norm” function contains all the information about the local and infinitesimal behaviour of the distance d. We can see this from the fact that we can recover the re-scaled distance d^{x}_{\varepsilon} from this “norm”, with the help of the approximate difference (for this notion see on this blog the definition of approximate difference in terms of emergent algebras here, or go to point 3. from the post The origin of emergent algebras (part III)):

\rho_{\varepsilon}(\delta^{x}_{\varepsilon} u , \Delta^{x}_{\varepsilon}(u,v)) = d^{x}_{\varepsilon}(u,v)

(proof left to the interested reader) This identity shows that the uniform convergence of (x,u,v) \mapsto d^{x}_{\varepsilon}(u,v) to (x,u,v) \mapsto d^{x}(u,v), as \varepsilon goes to 0, is a consequence of the following pair of uniform convergences:

  • that of the function (x,u) \mapsto \rho_{\varepsilon}(x,u) which converges to (x,u) \mapsto d^{x}(x,u)
  • that of  the pair (dilation, approximate difference)  (x,u,v) \mapsto (\delta^{x}_{\varepsilon} u , \Delta^{x}_{\varepsilon}(u,v)) to (x,u,v) \mapsto (x, \Delta^{x}(u,v)), see how this pair appears from the normed groupoid formalism, for example by reading the post from the post The origin of emergent algebras (part III).

With this definition of the “norm” function, I can now introduce the first quantity of interest, which measures the difference between the “norm” function at scale \varepsilon and the “norm” function at scale 0:

A_{\varepsilon}(x,u) = \rho_{\varepsilon}(x,u) - d^{x}(x,u)

The interpretation of this quantity is easy in the particular case of a riemannian space with dilations defined by the geodesic exponentials. In this particular case

A_{\varepsilon}(x,u) = 0

because the “norm” function \rho_{\varepsilon}(x,u) equals the distance between d(x,u) (due to the definition of dilations with respect to the geodesic exponential).

In more general situations, for example in the case of a regular sub-riemannian space, we can’t define dilations in terms of geodesic exponentials (even if we may have at disposal geodesic exponentials). The reason has to do with the fact that the geodesic exponential in the case of a regular sub-riemannian manifold, is not intrinsically defined as a function from the tangent of the geodesic at it’s starting point. That is because geodesics in regular sub-riemannian manifolds (at least those which are classically, i.e. with respect to the differential manifold structure, smooth , are bound to have tangents only in the horizontal directions.

As another example, think about a sub-riemannian Lie group. Here, we may define a left-invariant dilation structure with the help of the Lie group exponential. In this case the quantity A_{\varepsilon}(x,u) is certainly not equal to 0, excepting very particular cases, as a riemannian compact Lie group, with bi-invariant distance, where the geodesic and Lie group exponentials coincide.


2.   The second quantity is the one which is most interesting for defining (sectional like) curvature, let’s call it

B_{\varepsilon}(x,u,v) = d^{x}_{\varepsilon}(u,v) - d^{x}(u,v).


3. Finally, the third quantity of interest is a kind of a measure of the convergence of (x,u,v) \mapsto (\delta^{x}_{\varepsilon} u , \Delta^{x}_{\varepsilon}(u,v)) to (x,u,v) \mapsto (x, \Delta^{x}(u,v)), but measured with the norms from the tangent spaces.  Now, a bit of notations:

dif_{\varepsilon}(x,u,v) = (\delta^{x}_{\varepsilon} u , \Delta^{x}_{\varepsilon}(u,v)) for any three points x, u, v,

dif_{0}(x,u,v) = (x, \Delta^{x}(u,v))  for any three points x, u, v  and

g(v,w) = d^{v}(v,w) for any two points v, w.

With these notations I introduce the third quantity:

C_{\varepsilon}(x,u,v) = g( dif_{\varepsilon}(x,u,v) ) - g( dif_{0}(x,u,v) ).


The relation between these three quantities is the following:

Proposition.  B_{\varepsilon}(x,u,v) = A_{\varepsilon}(dif_{\varepsilon}(x,u,v)) + C_{\varepsilon}(x,u,v).


Suppose that we know the following estimates:

A_{\varepsilon}(x,u) = \varepsilon^{\alpha} A(x,u) + higher order terms, with A(x, u) \not = 0 and \alpha > 0,

B_{\varepsilon}(x,u,v) = \varepsilon^{\beta} B(x,u,v) + higher order terms, with B(x,u,v) \not = 0 and \beta > 0,

C_{\varepsilon}(x,u) = \varepsilon^{\gamma} C(x,u,v) + higher order terms, with C(x, u) \not = 0 and \gamma > 0,

Lemma. Let  us sort in increasing order the list of the values \alpha, \beta, \gamma and denote the sorted list by a, b, c. Then a = b.

The proof is easy. The equality from the Proposition tells us that the modules of A_{\varepsilon}, B_{\varepsilon} and C_{\varepsilon} can be taken as the edges of a triangle. Suppose then that a < b < c, use the estimates from the hypothesis and divide by \varepsilon^{a} in one of the three triangle inequalities, then go with \varepsilon to 0 in order to arrive at a contradiction 0 < 0.


The moral of the lemma is that there are at most two different coefficients in the list \alpha, \beta, \gamma. The coefficient \beta is called “curvdimension”. In the next post I shall explain why, in the case of a sub-riemannian Lie group,  the coefficient \gamma is related to the halfbracket. Moreover, we shall see that in the case of sub-riemannian Lie groups all three coefficient are equal, therefore the infinitesimal behaviour of the halfbracket determines the curvdimension.


2 thoughts on “Curvature and halfbrackets, part II”

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s