Hermite Splines in Lie Groups as Products of Geodesics

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1 Hermte Splnes n Le Groups as Products of Geodescs Ethan Eade Updated May 28, Introducton 1.1 Goal Ths document defnes a curve n the Le group G parametrzed by tme and by structural parameters n the assocated Le algebra g, and shows how to compute ts dfferentals by all parameters. The curve s expressed prmarly n the form S n : R g n G (1) S n (t) exp (a n p n ) exp (a n 1 p n 1 )... exp (a 1 p 1 ) (2) Havng establshed notaton and some useful denttes, we show how to compute dfferentals of the product of exponentals, and then descrbe how to acheve boundary condtons on the curve s values and dervatves by choosng a and p approprately. The result s a form sutable for expressng an ndvdual segment of a Hermte splne n the group, complete wth all the structural dfferentals useful for fttng the splne to data. Channg such segments wth shared boundary condtons on neghbors yelds a C 2 trajectory n the group, permttng parametrc regresson aganst observatons nvolvng the value, frst, and second tme dervatves of the trajectory. The formulaton s vald for any Le group, but as a motvatng example, consder usng such a splne to model a rgd body trajectory n SE (3), and adjustng the splne parameters to maxmze the lkelhood of nosy observatons from a gyroscope and accelerometer on the body. 1.2 Notaton Elements of the Le group G are denoted wth upper-case letters, whle elements of the assocated Le algebra g are denoted wth lower-case letters. The dentty matrx s wrtten I. The adjont representaton n the group s wrtten Ad [], whle the adjont representaton n the algebra s wrtten ad []. The exponental map s wrtten exp. It maps elements of g to G, and elements of ad [g] to Ad [G]. Its nverse s the logarthm log. Corrected left-hand-sdes of equatons from S to S. 1

2 2 Identtes Ths secton enumerates denttes useful n the rest of the document, and can be skpped f the reader s famlar wth representatons of Le groups and ther propertes. Throughout ths secton, let x, y g and X, Y G. 2.1 Basc propertes of Ad [],ad [], and exp () The adjont operator n g s ant-commutatve: The exponental map and the adjont representaton commute: ad [x] y ad [y] x (3) The adjont transforms rght tangent vectors to left tangent vectors: exp (ad [x]) Ad [exp (x)] (4) The adjont s a group homomorphsm: X exp (y) exp (Ad [X] y) X (5) Ad [X Y] Ad [X] Ad [Y] (6) 2.2 Ad [X] ad [y] ad [Ad [X] y] Ad [X] Usng the above, we can show that the adjont n the group can transform the adjont n the algebra from rght to left as well. Consder t R. Frst send all factors to the adjont representaton: X exp (t y) exp (Ad [X] t y) X (7) Ad [X] Ad [exp (t y)] Ad [exp (Ad [X] t y)] Ad [X] (8) Next re-order the applcaton of the exponental map and the adjont operator: Ad [X] exp (ad [t y]) exp (ad [Ad [X] t y]) Ad [X] (9) Then use the lnearty of the adjont operator n the algebra: Fnally, dfferentate by t at t 0: Ad [X] exp (t ad [y]) exp (t ad [Ad [X] y]) Ad [X] (10) Ad [X] ad [y] ad [Ad [X] y] Ad [X] (11) 2

3 2.3 Dfferentals n G and the adjont We defne dfferentals n the group n terms of left updates, so that the dfferental s a lnear mappng from the parameter(s) of nterest α to the algebra g: X (α) α ( log (X (α + δ) X (α) 1)) Dfferentals by left updates are then smple. Consder a lnear operator on the algebra,l: Ths makes dfferentals of products straghtforward: δ0 (12) (exp (L ɛ) X) ɛ ɛ0 L (13) (X Y) Y (X Y) X Ad [X] (14) I (15) We can also dfferentate transformatons of algebra vectors by the adjont: (Ad [exp (δ) X] y) (Ad [exp (δ)] Ad [X] y) δ0 (16) (exp (ad [δ]) Ad [X] y) (17) (ad [δ] Ad [X] y) (18) ad [Ad [X] y] δ (19) ad [Ad [X] y] (20) 3 S (t) and ts Tme Dervatves 3.1 Functon of tme We defne the product recursvely: a : R R (21) p g (22) S : R g G (23) S (t) exp (a (t) p ) S 1 (t) (24) S 1 (t) exp (a 1 (t) p 1 ) (25) S n (t) s then evaluated by left-multplyng each exponental term from 1 to n. 3

4 3.2 Frst dervatve by tme Consder the exponental factor A (t) exp (a (t) p ): A (t + ɛ) exp ((a (t) + ɛ a (ɛ (t) + O 2)) ) p (26) exp ((ɛ a (ɛ (t) + O 2)) ) p exp (a (t) p ) (27) exp ((ɛ a (ɛ (t) + O 2)) ) p A (t) (28) A (t) A (t + ɛ) ɛ0 (29) ɛ a (t) p (30) The frst tme dervatve S n (t) of S n (t) s then straghtforward to express recursvely usng the product and chan rules: S (t) A (t) S 1 (t) (31) S (t) A (t) S 1 (t) + Ad [A (t)] S 1 (t) (32) a (t) p + Ad [A (t)] S 1 (t) (33) 3.3 Second dervatve by tme The second dervatve s the partal dervatve by t of Eq.33: S (t) S (t) t a (t) p + Ad [A (t)] S 1 (t) ad [ Ad [A (t)] S 1 (t) ] A (34) (t) (35) a (t) p + Ad [A (t)] S 1 (t) + ad [ A (t)] Ad [A (t)] S 1 (t) (36) 4 Structural Dfferentals for Regresson In order to perform regresson of the model aganst observatons, we ll need dfferentals of S, S, and S by the parameters {p }. To ths end, we ll requre the dfferental of the exponental map, descrbed n detal n another document ( Dfferental of the Exponental Map ): D exp [x] exp (x) x log (exp (x + δ) exp ( x)) (37) For many Le groups of nterest, D exp [x] can be computed n closed form. For convenence, we label the dfferental of the exponental factor A of S by ts argument: The tme parameter t s omtted n most of the equatons below to avod clutter. D D exp [a (t) p ] (38) 4

5 4.1 Structural dfferentals of S (t) Eq.24 s dfferentated usng the chan rule and Eq.14: S A p p (39) D exp [a p ] a (40) D a (41) S p j { Ad [A ] S 1 p j j < 0 j > (42) 4.2 Structural dfferentals of S (t) Eq.33 s dfferentated usng the product rule and Eq.20: S ( A ) ad [ Ad [A p p ] S 1] A p a I ad [ Ad [A ] S 1] A p S S 1 Ad [A ] p j < j p j 0 j > 4.3 Structural dfferentals of S (t) Eq.36 s dfferentated usng the product rule and Eq.20: S p a I ad [ Ad [A ] S 1] S p j a I ( ad [ Ad [A ] S A ad [ A [ p ] ad Ad [A ] S 1] A p ( [ ] [ 1] + ad A ad Ad [A ] S 1]) A p S 1 Ad [A ] p + ad [ A S j ] Ad [A 1 ] p j j < 0 j > a ad [ Ad [A ] S 1 ] (43) ) a ad [ Ad [A ] S 1 ] (44) (45) 4.4 Effcent computaton of structural dfferentals Drectly applyng the recursve defntons above requres O ( n 2) tme to compute structural dfferentals S n, S n, and S n. By reorderng the recurson from left to rght and accumulatng products and sums, these dfferentals can all be computed n O (n) tme: 5

6 Note that Eq. 50 makes use of the dentty gven n Eq.11. L n I (46) l n 0 (47) S n S Ad [L p ] (48) p S n ( S ) Ad [L p ] (49) p S n ( S ) ( S ) Ad [L p ] + ad [l p ] n (50) p L 1 L A (51) l 1 l + Ad [L ] A (52) 5 Constructng a Splne Segment wth Boundary Condtons By choosng a and p approprately, we can construct a C curve segment B (t) wth specfed boundary values and dervatves (sometmes called a Hermte splne). Such segments can then be concatenated to form a C 2 splne n the group by matchng the boundary condtons (value, frst, and second tme dervatves) between neghborng segments. The tme nterval n each segment s, wthout loss of generalty, assumed to be the unt nterval [0, 1]. Any other nterval can be remapped to ths by an affne transformaton (as can the dfferentals). The desred boundary condtons of the segment are specfed as group elements for the functon values at the boundares, and algebra elements for the tme dervatves: B (0) B 0 G (53) B (1) B 1 G (54) B (0) d 0 g (55) B (1) d 1 g (56) B (0) h 0 g (57) B (1) h 1 g (58) 5.1 Segment defnton In order to have enough degrees of freedom to acheve the boundary condtons, the Hermte splne segment s defned as a product of sx terms, one of whch s constant over tme: B (t) S 5 (t) B 0 (59) We choose the {a } to be quntc polynomals wth values and dervatves at 0 and 1 that make satsfyng the boundary condtons of B (t) trval. 6

7 a 5 (t) 1 2 t5 t t3 (60) a 4 (t) 1 2 t t4 3 2 t t2 (61) a 3 (t) 3t 5 + 7t 4 4t 3 (62) a 2 (t) 3t 5 + 8t 4 6t 3 + t (63) a 1 (t) 6t 5 15t t 3 (64) or equvalently: a 1 (t) a 2 (t) a 3 (t) a 4 (t) a 5 (t) t t 2 t 3 t 4 t 5 (65) These polynomals are constructed to satsfy the followng boundary condtons: a 5 (0) a 4 (0) a 3 (0) a 2 (0) a 1 (0) 0 (66) a 5 (1) a 4 (1) a 3 (1) a 2 (1) 0 (67) a 1 (1) 1 (68) a 2 (0) 1 (69) a 5 (0) a 4 (0) a 3 (0) a 1 (0) 0 (70) a 5 (1) a 4 (1) a 2 (1) a 1 (1) 0 (71) a 3 (1) 1 (72) a 5 (0) a 3 (0) a 2 (0) a (0) 0 (73) a 4 (1) a 3 (1) a 1 a 4 (0) 1 (74) 2 (1) a 1 (1) 0 (75) a 5 (1) 1 (76) Thus each polynomal selects for the value, frst tme dervatve, or second tme dervatve at one of the boundares. The parameters of S 5 (t) are then straghtforward to assgn: p 5 h 1 (77) p 4 h 0 (78) p 3 d 1 (79) p 2 d 0 (80) p 1 log B 1 B0 1 (81) By Eq.66 all of the factors of S 5 (0) are the dentty, so we satsfy the frst boundary condton trvally: B (0) B 0 (82) 7

8 By Eq.67 and 68 the second boundary condton s satsfed: Smlar nspecton of S 5 are satsfed. B (1) S 5 (1) B 0 (83) B 1 B0 1 B 0 (84) B 1 (85) (t) and S 5 (t) shows that all four of the boundary condtons on the dervatves {d 0, d 1, h 0, h 1 } 5.2 Tme dervatves of B (t) Because the B 0 factor s constant over tme, the tme dervatves of B (t) are taken drectly from those of S 5 (t), descrbed n detal above: B (t) S 5 (t) (86) B (t) S 5 (t) (87) 5.3 Structural dfferentals of B (t) The dfferentals of B (t) by the boundary parameters {B 0, B 1, d 0, d 1, h 0, h 1 } follow from the dfferentals of S 5 (t) by p, along wth the chan rule. Frst we make explct the parametrc perturbatons of B 0 and B 1 : B 0 exp (ɛ 0 ) B 0 (88) B 1 exp (ɛ 1 ) B 1 (89) p 1 log B 1 B 0 1 (90) ( ) log exp (ɛ 1 ) B 1 B0 1 exp ( ɛ 0 ) (91) B (t) exp (a 5 (t) p 5 ) exp (a 4 (t) p 4 ) exp (a 3 (t) p 3 ) exp (a 2 (t) p 2 ) exp (a 1 (t) p 1 ) B 0 (92) Note that changes to the ntal value parameter B 0 (va ɛ 0 ) affect B(t) both drectly and through the resultng modfcaton of p 1, whle changes to B 1 (va ɛ 1 ) affect B(t) only through the modfcaton of p 1. To dfferentate p 1 by ts arguments, we employ the property that the dfferental of the nverse s the nverse of the dfferental (see document Dfferental of the Exponental Map ) : p 1 p 1 ɛ 0 ( D exp [p B 1 ɛ 1 ] ) 1 1 p 1 p 1 ɛ 0 ( D exp [p B 0 ɛ 1 ] ) [ ] 1 Ad B 1 B0 1 0 (93) (94) Fnally, the structural dfferentals of B (t) follow by the chan rule: 8

9 B (t) S5 (t) p1 + Ad [S 5 (t)] (95) C 0 p 1 B 0 B (t) S5 (t) p1 (96) C 1 p 1 B 1 B (t) S 5 (t) (97) d 0 p 2 B (t) S 5 (t) d 1 p 3 (98) B (t) S 5 (t) h 0 p 4 (99) B (t) S 5 (t) h 1 p 5 (100) 9

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