EECE 5550 Mobile Robotics HW #3
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EECE 5550 Mobile Robotics HW #3
Due: Feb 23, 2023
Question 1: Optimization Methods
Please see this Colab.
Question 2: Lie algebras and left-invariant vector fields
Let G be a Lie map:
group with group operation 兴 : G x G → G. For each
Lg : G → G
Lg (x) 全 g 兴 x
g l G, the left-translation (1)
is a diffeomorphism of G. The left-translation could be used to identify1 the tangent space Te (G) of G at the identity element e l G with the Lie algebra Lie(G) (the set of left-invariant vector fields on G), as follows:
ϕ : Te (G) → Lie(G)
ϕ(ω) = Vω (2)
where Vω is the left-invariant vector field on G determined by:
Vω (x) 全 d(Lx )e (ω). (3)
In words: we associate to each element ω l Te (G) the left-invariant vector field Vω l Lie(G) whose value Vω (x) at x l G is the image of ω under the derivative of the left-translation map Lx that sends the identity e l G to x.
In this exercise, we will study the left-translation maps and left-invariant vector fields for our two favorite Lie group examples: Rn (with vector addition as the group operation) and GL(n).
(a) Given v l Rn , what is the corresponding left-translation map Lv : Rn → Rn ?
(b) What is the derivative dLv of the map Lv you found in part (a)?
(c) Given a vector ξ l T0 (Rn ) Rn in Rn ’s Lie algebra, what is the left-invariant vector field Vξ on Rn determined by ξ? Interpret this result geometrically.
(d) Given a matrix A l GL(n), what is the corresponding left-translation map LA : GL(n) → GL(n)?
(e) What is the derivative dLA of the map LA you found in part (d)?
(f) The tangent space TI (GL(n)) of GL(n) at the identity I l GL(n) is just Rn ×n , the set of all n x n matrices.2 Given a matrix Ω l TI (GL(n)), what is the left-invariant vector field VΩ on GL(n) determined by Ω?
Question 3: Exponential map of the orthogonal group
The exponential map for the general linear group GL(n) is just the usual matrix exponential:
exp: Rn ×n
exp(X) 全
→ GL(n)
2 Xk
k! .
k=0 (4)
However, this formula (4) can sometimes be significantly simplified when applied to a subgroup G S GL(n). In this exercise, we will explore what this simplification looks like for the orthogonal
group O(2).
(a) The Lie algebra Lie(O(n)) of the orthogonal group O(n) is Skew(n), the set of n-dimensional
skew-symmetric matrices:
Skew(n) 全{A l Rn ×n f AT = -A } . (5)
In particular, the Lie algebra of O(2) is:
Lie(O(2)) = Skew(2) = {╱-ω(0) 0(ω)、 l R2×2 f ω l R} . (6)
Given an element:
Ω 全 ╱ -ω(0) 0(ω)、 (7)
of Lie(O(2)), derive an expression for its kth power Ωk . (Hint: it may help to work out the first few powers of Ω . Can you spot a pattern?)
(b) Using the result of part (a), derive a simplified expression for exp(Ω). (Hint: it may help to split the series in (4) into odd and even powers. Can you recognize these series?) What is the geometric interpretation of exp(Ω)?
Question 4: Motion on Lie groups
Let G be a Lie group with group operation 兴 : GxG → G and Lie algebra Lie(G). We saw in class that each ω l Lie(G) generates a left-invariant vector field Vω on G, and that the exponential map describes the integral curves (i.e. the trajectories ) of this vector field. Specifically, the integral curve γ : R → G of the left-invariant vector field Vω that starts at the point x l G at time t = 0 is given by:
γ(t) 全 x 兴 exp(tω). (8)
Intuitively, equation (8) provides a prescription for “moving around” on the Lie group G along the “direction” determined by ω .
In this exercise, we will see how one can apply (8) to interpolate Lie group-valued data – this is an important operation for robot kinematics.
(a) If a point x l G lies in the image of G’s exponential map,3 we write “log(x)” to denote one of x’s preimages,4 so that:
x = exp(log(x)). (9)
If G’s exponential map is surjective, then there is always at least one choice of log(x) l Lie(G) that will satisfy (9).
Now suppose that x, y l G and that G’s exponential map is surjective. Using (8), derive a formula for a curve γ : [0, 1] → G such that γ(0) = x and γ(1) = y .
(b) The exponential map for Rn is just the identity map:
exp: Rn → Rn
exp(ξ) = ξ . (10)
Using (10), specialize your result from part (a) to derive a formula for a curve γ that joins x to y in Rn . Interpret this result geometrically.
(c) The Lie group SE(3) of 3D robot poses can be modeled as the product manifold M 全 R3 x SO(3),5 equipped with the following group multiplication rule:
(t1 , R1 ) 兴 (t2 , R2 ) = (R1 t2 + t1 , R1 R2 ). (11)
Given the two poses:
0-00(.)3(8)2 0-00(.)6(6)6.(、) .(、) (12)
apply the formula you derived in part (a) to calculate the “midpoint” γSE(3)(1/2) on the curve γSE(3) : [0, 1] → SE(3) from X0 to X1 . (Hint: you may find it helpful to use the homogeneous representation of SE(d).)
(d) Since R3 and SO(3) are themselves Lie groups (under vector addition and matrix multi- plication, respectively), we can construct the product Lie group P 全 R3 x SO(3): this is the group whose elements are pairs of the form (t, R) l R3 x SO(3), equipped with the multiplication law
(t1 , R1 ) 兴P (t2 , R2 ) = (t1 + t2 , R1 R2 ). (13)
That is, in the product group P , we simply apply the group operations from R3 and SO(3) separately in each component.
The Lie groups SE(3) and P thus have the same manifold structure (they are both built on the manifold R3 x SO(3)), but different group structures [compare the multiplication rules (11) and (13)].
Using the formula that you derived in part (a), compute the “midpoint” γP (1/2) of the curve γP : [0, 1] → P from X0 to X1 in P .
(e) Plot the translational components of the curves γSE(3) and γP from parts (c) and (d) over
two intervals: (i) t l [0, 1] and (ii) t l [0, 30]. Describe these curves qualitatively.
2023-02-22