The first step is to consider the set of transformations as a group,
in addition to a topological space.^{4.8} We now derive several
important groups from sets of matrices, ultimately leading to ,
the group of
rotation matrices, which is very important
for motion planning. The set of all nonsingular
real-valued matrices is called the *general linear group*, denoted
by , with respect to matrix multiplication.
Each matrix
has an inverse
, which
when multiplied yields the identity matrix,
. The
matrices must be nonsingular for the same reason that 0 was removed
from
. The analog of division by zero for matrix algebra is the
inability to invert a singular matrix.

Many interesting groups can be formed from one group, , by
removing some elements to obtain a *subgroup*, . To be a subgroup, must be a subset of and
satisfy the group axioms. We will arrive at the set of rotation
matrices by constructing subgroups. One important subgroup of
is the *orthogonal group*, , which is the set of all
matrices
for which , in which denotes
the matrix *transpose* of . These matrices have orthogonal
columns (the inner product of any pair is zero) and the determinant is
always or . Thus, note that takes the inner product
of every pair of columns. If the columns are different, the result
must be 0; if they are the same, the result is because . The *special orthogonal group*, , is the subgroup of
in which every matrix has determinant . Another name for
is the *group of -dimensional rotation matrices*.

A chain of groups, , has been described in which denotes ``a subgroup of.'' Each group can also be considered as a topological space. The set of all matrices (which is not a group with respect to multiplication) with real-valued entries is homeomorphic to because entries in the matrix can be independently chosen. For , singular matrices are removed, but an -dimensional manifold is nevertheless obtained. For , the expression corresponds to algebraic equations that have to be satisfied. This should substantially drop the dimension. Note, however, that many of the equations are redundant (pick your favorite value for , multiply the matrices, and see what happens). There are only ways (pairwise combinations) to take the inner product of pairs of columns, and there are equations that require the magnitude of each column to be . This yields a total of independent equations. Each independent equation drops the manifold dimension by one, and the resulting dimension of is , which is easily remembered as . To obtain , the constraint is added, which eliminates exactly half of the elements of but keeps the dimension the same.

(4.10) |

which is homeomorphic to . The group is formed from the set of all nonsingular matrices, which introduces the constraint that . The set of singular matrices forms a 3D manifold with boundary in , but all other elements of are in ; therefore, is a 4D manifold.

Next, the constraint is enforced to obtain . This becomes

(4.11) |

which directly yields four algebraic equations:

Note that (4.14) is redundant. There are two kinds of equations. One equation, given by (4.13), forces the inner product of the columns to be 0. There is only one because for . Two other constraints, (4.12) and (4.15), force the rows to be unit vectors. There are two because . The resulting dimension of the manifold is because we started with and lost three dimensions from (4.12), (4.13), and (4.15). What does this manifold look like? Imagine that there are two different two-dimensional unit vectors, and . Any value can be chosen for as long as . This looks like , but the inner product of and must also be 0. Therefore, for each value of , there are two choices for and : 1) and , or 2) and . It appears that there are two circles! The manifold is , in which denotes the union of disjoint sets. Note that this manifold is not connected because no path exists from one circle to the other.

The final step is to require that , to obtain , the set of all 2D rotation matrices. Without this condition, there would be matrices that produce a rotated mirror image of the rigid body. The constraint simply forces the choice for and to be and . This throws away one of the circles from , to obtain a single circle for . We have finally obtained what you already knew: is homeomorphic to . The circle can be parameterized using polar coordinates to obtain the standard 2D rotation matrix, (3.31), given in Section 3.2.2.

Steven M LaValle 2012-04-20