Clifford Analysis, Homework 1

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1 Clifford Analysis, Homework 1 November 1, Let w v 1... v k, for vectors v j. Show that ŵ is the result of negating the vectors: ŵ ( v 1 )... ( v k ). Show that w is the result of changing the order of the vectors: w v k... v 1. Show also that the involution is an algebra automorphism, i.e. w 1 w ŵ 1 ŵ and that the reversion is an algebra anti-automorphism, i.e. w 1 w w w 1. The involution is helpful when commuting through a vector v V : Show that v w ŵ v. Solution: a) By definition ŵ ( 1) k v 1... v k and thus ŵ ( 1) k v 1... v k ( v 1 )... ( v k ). b) By definition w ( 1) k(k 1)/ v 1... v k and thus w ( 1) k(k 1)/ v 1... v k ( 1) (k 1)+(k ) v 1... v k ( 1) (k ) v v 3... v k v 1 ( 1) (k 3) v 3 v 4... v k v v 1... v k... v 1. c) Let w s n w s and v s n v s with w s, v s span{e s } s V and {e s } s n is a basis for V. Now ŵ v s,t ŵ s v t. Thus it suffices to prove ŵ v ŵ v for simple multivectors w w 1... w i and v v 1... v j where v k, w k V. We have ŵ v ( w 1 )... ( w i ) ( v 1 )... ( v j ) ŵ v 1

2 d) As above, it suffices to prove w v v w for simple multivectors w w 1... w i and v v 1... v j. We have w v w 1... w i v 1... v j v j... v 1 w i... w 1 v w i... w 1 v w. e) Write w j w j with w j j V and we get v w j v w j j ( 1) j w j v ŵ v. Show that if w k V and Θ l V then Θ w, w Θ k l V. In particular, if k l, show that Θ w w Θ Θ, w, where as usual R and 0 V are identified. Solution: Let w k V and Θ l V. Also let Φ m V. Now if l + m k, we have Φ, Θ w Θ Φ, w 0 since the above is a duality between objects with different grades. Since Φ was arbitrary, we conclude that (Θ w) m 0 for m k l. Thus Θ w k l V. The argument for w Θ is similar. Now if k l, we have 3 Θ w 1, Θ w Θ 1, w Θ, w, w Θ 1, w Recall the definition of the regressive product Θ 1 Θ, w Θ, w. x y : ((xi 1 ) (yi 1 ))I. Compute e s e t for the induced ON-basis. Also compute (( e s ) ( e t )) and compare with the previous result. Finally show that if w 1, w V are such that 0 w 1 w n V, then [w 1 w ] [w 1 ] [w ].

3 Solution: Note that since t, s {1,,..., n} : n, the set n \ s is equal to the complement s c. e s e t ((e s I 1 ) (e t I 1 ))I ((e s e n ) (e t e n ))e n ((e s en ) (e t en ))e n ((e s en ) (e t en ))e n ɛ(s, n \ s)ɛ(t, n \ t)(e n\s e n\t )e n { ɛ(s, s c )ɛ(t, t c )e s c t ce n, (n \ s) (n \ t) 0, otherwise { ɛ(s, s c )ɛ(t, t c )ɛ((s t) c, s t)e s t, s t n 0, otherwise. (( e s ) ( e t )) ((en e s ) (e n e t )) e n ɛ(n \ s, s)ɛ(n \ t, t)(en\s e n\t ) e n { ɛ(s c, s)ɛ(t c, t)ɛ(s c t c, n \ (s c t c ))e n\((n\s) (n\t)), 0, { ɛ(s c, s)ɛ(t c, t)ɛ((s t) c, s t)e s t, s t n 0, Using Propositions.1- we get [w 1 w ] [(( w 1 ) ( w )) ] [( w ) (( w 1 ) )] [( w ) w 1 ] [ w ] [w 1 ] [w ] [w 1 ] (n \ s) (n \ t) 4 The cofactor formula (.4) can be generalized to give a relation between T 1 : k V k V and T : n k V n k V for any 0 k n. Work out the details of this. Solution: Let T : V V be an invertible linear map. Denote by T : V V the induced linear map on the exterior algebra of V. Recall that if w k V then w n k V. We have T ( 1 det T (T n k V ( w)) ) T ( 1 det T (det T (T 1 w)) T (T 1 w) w, where we used the formulas T ( (T v)) det v (Corollary.9) and ( v) v. This shows that 5 (T k V ) 1 w 1 det T (T n k V ( w)). Prove Heron s formula for the euclidean area A 1 v 1 v of the triangle with edge vectors v 1, v, v 3 (so that v 1 + v + v 3 0), i.e. A 1 4 (a b + a c + b c ) a 4 b 4 c 4 p(p a)(p b)(p c) ) 3

4 where a v 1, b v, c v 3 and p (a + b + c)/ is half the perimeter. Solution: We use Lagrange s identity (Proposition 3.1) to obtain On the other hand, v 3 v v 3 implies v 1 v v 1 v, v 1 v v 1 v v 1, v. v 3 v 1 + v, v 1 + v v 1 + v 3 + v 1, v v 1, v v 3 v 1 v. Plugging this into our previous formula yields ( v 1 v v 1 v v3 v 1 v A 1 4 a b 1 1 (a4 + b 4 + c 4 a c b c + a b ) a b + a c + b c 8 a4 + b 4 + c 4 1 Taking the square root of this leads to the wanted formula. Work through Example 3.0 instead using the pure quaternion representation of vectors. Solution: Recall the definitions j 1 : e 1 e 3 j : e e 31 j 3 : e 3 e 1. Then the quaternion representation of v V with V Euclidean and dim V 3 is given by i j 1, j j, k j 3. We want to find the quaternion describing the rotation by angle φ : arccos(1/3) counter clockwise around e 1 + e. The rotation plane is given by w : [ (e 1 + e )] (see Proposition 3.17). We have ) w (e 1 + e ) e 3 + e 31 i j. Then, according to Proposition 3.17 our rotation is described by the map v qvq 1 where q cos(φ/) w w 3 + i + j i + j. sin(φ/) cos(arccos(1/3)/) + i + j sin(arccos(1/3)/). 4

5 Using the rules i j k 1, ijk 1, ij ji k, ki ik j, jk kj i we get and thus (a + bi + cj + dk) 1 q 1 1 i 1 j a bi cj dk a + b + c + d i j. Thus the rotation of v counter clockwise around e 1 + e by the angle arccos(1/3) is given by the quaternion product qvq 1 1 ( + i + j)(v 1i + v j + v 3 k)( i j) 1 (v 1i + v j + v 3 k v 1 + v k v 3 j v 1 k v + v 3 i)( i j) 1 3 [i(v 1 + v + v 3 ) + j(v 1 + v v 3 ) + k( v 1 + v + v 3 )] Thus the rotation matrix is The trace is w 3 e i T (e i ) iqiq 1 + jqjq 1 + kqkq 1 i1 1 (5 4i 4j) 3 Let {e 1, e, e 3 } be an ON basis such that e 3 spans the rotation axis and e 1 e spans the rotation plane with T (e 1 ) e 1 cos φ + e sin φ and T (e ) e cos φ e 1 sin φ. Also, T (e 3 ) 1 and thus w 3 e i T (e i) (cos φ + e 1 e sin φ) + 1. i1 The two expressions for w must be the same and thus cos φ φ arccos(1/3). Since i j e 3 + e 31, the rotation axis is e 1 + e because (e 1 + e ) e 3 + e 31 (see the beginning of Example 3.0). 5

6 7 Use the matrix representations [ ] x0 + x x 0 + x 1 e 1 + x e + x 1 e 1 x 1 x 1 x 1 + x 1 x 0 x for the Clifford algebra of the Euclidean plane, and x 0 + x 1 e 1 + x e + x 3 e 3 + x 1 e 1 + x 13 e 13 + x 3 e 3 + x 13 e 13 [ ] x0 + x + i(x 13 x 13 ) x 1 x 1 + i(x 3 + x 3 ) x 1 + x 1 + i(x 3 x 3 ) x 0 x + i(x 13 + x 13 ) for the Clifford algebra for Euclidean three dimensional space, to verify the calculations in Example 3.5. In both two and three dimensions, use these representations to write equation for the coordinates that determines when a multivector in the Clifford algebra is not invertible. Solution: [ ] [ Now it remains to solve the system of equations x 0 + x 8 x 1 x 1 8 x 1 + x 1 x 0 x 58 ] [ ] [ ] [ ] The solution is x 0 33 x 1 5 x 5 3 x 1 This result agrees with the computation done in Example 3.5. For the second part, we compute [ ] [ ] i( 7 5) i( + ) 3 + i( 3) i( 1 + 1) i( ) i( 7 + 5) 4 + i( 1 1) i( + 3) [ ] [ ] 1i 5 i 10 4i 4 i i 1 + 5i [ ] 18 + i i 1 + 1i 14 i

7 and then we solve x 0 + x + i(x 13 x 13 ) x 1 x 1 + i(x 3 + x 3 ) x 1 + x 1 + i(x 3 x 3 ) x 0 x + i(x 13 + x 13 ) 18 + i i 1 + 1i 14 i x 0 + x 18 x 1 x 1 18 x 1 + x 1 1 x 0 x 14 x 13 + x 13 x 3 + x x 3 + x 3 1 x 13 + x 13 x 0 x 1 1 x x 1 x 3 x x 3 57 x 13 0 A matrix [ ] [ ] a11 a A 1 x0 + x x 1 x 1 a 1 a x 1 + x 1 x 0 x is invertible iff its determinant is non-zero. Thus the conditions for invertibility in the Clifford algebra are that our matrix can be written in the above form and that det A (x 0 + x )(x 0 x ) (x 1 x 1 )(x 1 + x 1 ) x 0 x x 1 + x 1 0 and similarly for the Clifford algebra for Euclidean three dimensional space. 8 Show that C : V L + ( V ) : v v + realizes L + ( V ) as a Clifford algebra for (V,, ), whereas C : V L ( V ) : v v realizes L ( V ) as a Clifford algebra for (V,, ). Prove, through a basis calculation, that L + ( V ) V : T T (1) is an invertible linear map, and that T 1 (1) T (1) : (T 1 T )(1) gives another way to construct the Clifford product on V. Solution: We will verify that conditions (C) and (U) of Definition 3. hold. 7

8 First we observe that (Cv) w v + v + (w) v + (v w + v w) v (v w) + v (v w) + v (v w) + v v w 0 + v (v w) + v (v w) + 0 v, v w where in the last equality we used Theorem.75. Thus (Cv) v, v. To verify (U) of Definition 3., we use Automatic universality (Proposition 3.4). Since dim L + (V ) n where n dim V, we have automatic universality and thus (U) holds. The proof for v v is almost identical. Let F : L + ( V ) V be the map F (T ) T (1). Clearly F is linear and for s n we have F (e + s ) e + s (1) e s 1 + es e s e s. Thus the inverse F 1 : V L + ( V ) is given by F 1 ( s n v s e s ) s n v s e + s. Now e + s (1) e + t (1) : (e+ s e + t )(1) e s e t 1 e s e t Solution: b(v) n e + i ( e i v + e i v) i1 n e i ( ei v) + ei (ei v) + e i ( e i v) + ei e i v i1 n 0 + e i (ei v) e i (e i v) + 0. i1 Now set v e j1... e jk k V. Then e i v 0 if i {j 1,..., j k } and e i (e i v) 0 if i / {j 1,..., j k }. Recalling the anticommutation relation (Theorem.75) e i (ei v) + e i (e i v) ei, e i v v 8

9 we get b(v) n e i (ei v) e i (e i v) i1 i {j 1,...,j k } i {j 1,...,j k } ( e i (e i v)) + v + kv + (n k)v (n k)v. i/ {j 1,...,j k } v i/ {j 1,...,j k } The reversion w w in dimensions is (recall that b i e + i e i ) e i (ei v) 1 Re ( (1 + i) 1 (1 ib 1 )(1 ib ) ) 1 Re ((1 + i)(1 ib 1 ib b 1 b )) 1 (1 + b 1 + b b 1 b ) Since e n w e 1 e... e n w e + 1 e+... e+ n w, it clearly holds that w e n w belongs to (L + ( V )) n. Since n dim V, an explicit computation with w w 0 + w 1 e 1 + w e + w 1 e 1 yields we (e e 1 ) e + (w e 1 ) e + (w e 1 ) e + w e 1 e w 1 + w 1 e w e 1 + w 0 e 1 e 1 e w e 1 (e w) e1 (e w) e1 (e w) + e 1 e w w 1 + w 1 e w e 1 + w 0 e 1 i.e. w we e 1 e w and thus the map belongs to (L ( V )) n. Corollary 3.13 tells us that w e n w and w we n and thus 10 w e + 1 e+ 1 (1 + e+ 1 e 1 + e+ e e+ 1 e 1 e+ e )w 1 (e+ 1 e+ e+ e 1 + e+ 1 e e 1 e )w w e 1 e 1 (1 + e+ 1 e 1 + e+ e e+ 1 e 1 e+ e )w 1 (e 1 e + e+ 1 e e+ e 1 e+ 1 e+ ) Show the following. In two dimensional euclidean space, the group Spin(V ) is the unit circle in C ev V. In three dimensional euclidean space Spin(V ) is the unit three sphere in H ev V. Solution: By definition Spin(V ) {q ev V : q 1, q V }. 9

10 Since we have the isomorphism between C and ev V, the Spin group consists of complex numbers z C satisfying z 1 and z z 1 z for some z 1, z C. For any complex number z we can write z z z. Since every element of C has a square root, we have ev V V and thus Spin(V ) {q ev V : q 1} i.e. the unit circle in C. More explicit proof would be to write q q 0 + q 1 e 1 and then notice that e 1 (q 0 e 1 + q 1 e ) q 0 + q 1 e 1 q, implying that ev V V. For quaternions we can use the same proof (square root) by observing that if q H, q u + qv where u, v R, q q i + q j j + q k k and q 1, then q 1 and hence computing the square root of this works identically to computing the square root of the complex number u + iv. 11 Define V : {w V : Sw w} where S(v 1 v ) v v 1. The map B(v w)x w, x v defines a mapping from V to the space of skew symmetric linear maps so(v ). Also, the map A : V V, A(v w) v w is bijective and Lemma 4.14 yields a bijective map C : op(v ) V defined by Mv (CM) v. Show that A B 1 C. Solution: For v 1 v V we have which implies that Then B(v 1 v ) 1 B(v 1 v ) 1 B(v 1 v + v 1 v ) 1 B(v 1 v ) 1 B(v v 1 ) B(v 1 v )w 1 v, w v 1 1 v 1, w v. u 1, CB(v 1 v ) u u 1, B(v 1 v )u 1 v, u u 1, v 1 1 v 1, u u 1, v 1 u 1 u, v 1 v u 1, (v 1 v ) Thus CB(v 1 v ) v 1 v A(v 1 v ) which means that CB A, which is equivalent with AB 1 C. u. 10