Now consider defining tangent spaces on manifolds. Intuitively, the tangent space at a point on an -dimensional manifold is an -dimensional hyperplane in that best approximates around , when the hyperplane origin is translated to . This is depicted in Figure 8.8. The notion of a tangent was actually used in Section 7.4.1 to describe local motions for motion planning of closed kinematic chains (see Figure 7.22).
To define a tangent space on a manifold, we first consider a more complicated definition of the tangent space at a point in , in comparison to what was given in Section 8.3.1. Suppose that , and consider taking directional derivatives of a smooth function at a point . For some (unnormalized) direction vector, , the directional derivative of at can be defined as
(8.31) |
(8.32) |
Now consider taking (unnormalized) directional derivatives of a smooth function, , on a manifold. For an -dimensional manifold, the tangent space at a point can be considered once again as the set of all unnormalized directions. These directions must intuitively be tangent to the manifold, as depicted in Figure 8.8. There exists a clever way to define them without even referring to specific coordinate neighborhoods. This leads to a definition of that is intrinsic to the manifold.
At this point, you may accept that is an -dimensional vector space that is affixed to at and oriented as shown in Figure 8.8. For the sake of completeness, however, a technical definition of from differential geometry will be given; more details appear in [133,872]. The construction is based on characterizing the set of all possible directional derivative operators. Let denote the set of all smooth functions that have domains that include . Now make the following identification. Any two functions are defined to be equivalent if there exists an open set such that for any , . There is no need to distinguish equivalent functions because their derivatives must be the same at . Let denote under this identification. A directional derivative operator at can be considered as a function that maps from to for some direction. In the case of , the operator appears as for each direction . Think about the set of all directional derivative operators that can be made. Each one must assign a real value to every function in , and it must obey two axioms from calculus regarding directional derivatives. Let denote a directional derivative operator at some (be careful, however, because here is not explicitly represented since there are no coordinates). The directional derivative operator must satisfy two axioms:
You may recall these axioms from standard vector calculus as properties of the directional derivative. It can be shown that the set of all possible operators that satisfy these axioms forms an -dimensional vector space [133]. This vector space is called the tangent space, , at . This completes the definition of the tangent space without referring to coordinates.It is helpful, however, to have an explicit way to express vectors in . A basis for the tangent space can be obtained by using coordinate neighborhoods. An important theorem from differential geometry states that if is a diffeomorphism onto an open set , then the tangent space, , is isomorphic to . This means that by using a parameterization (the inverse of a coordinate neighborhood), there is a bijection between velocity vectors in and velocity vectors in . Small perturbations in the parameters cause motions in the tangent directions on the manifold . Imagine, for example, making a small perturbation to three quaternion parameters that are used to represent . If the perturbation is small enough, motions that are tangent to occur. In other words, the perturbed matrices will lie very close to (they will not lie in because is defined by nonlinear constraints on , as discussed in Section 4.1.2).
Now consider different ways to express the tangent space at some point , other than the poles (a change of coordinates is needed to cover these). Using the coordinates , velocities can be defined as vectors in . We can imagine moving in the plane defined by and , provided that the limits and are respected.
We can also use the parameterization to derive basis vectors for the tangent space as vectors in . Since the tangent space has only two dimensions, we must obtain a plane that is ``tangent'' to the sphere at . These can be found by taking derivatives. Let be denoted as , , and . Two basis vectors for the tangent plane at are
(8.35) |
(8.36) |
The tangent vectors can now be imagined as lying in a plane that is tangent to the surface, as shown in Figure 8.8. The normal vector to a surface specified as is , which yields after normalizing. This could alternatively be obtained by taking the cross product of the two vectors above and using the parameterization to express it in terms of , , and . For a point , the plane equation is
(8.37) |
Steven M LaValle 2020-08-14