Structure Groups and Linear Transformations

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1 Structure Groups and Linear Transformations Darien Elderfield August 20, Preliminaries Definition 1 Let V be a vector space and R : V V V V R be linear in each input R is an algebraic curvature tensor if it satisfies the following properties for all x,y,z,w V: 1 R(x, y, z, w) = R(y, x, z, w), 2 R(x, y, z, w) =R(z,w,x,y) 3 R(x, y, z, w)+r(z,x,y,w)+r(y, z, x, w) =0 The vector space of algebraic curvature tensors on V is denoted A(V) Definition 2 Let A End(V ) Theprecomposition of A, denoted A, with an algebraic curvature tensor R(x, y, z, w) is defined A R(x, y, z, w) =R(Ax, Ay, Az, Aw) IfA is invertible, we may instead define the precomposition of A with R as A R(x, y, z, w) = R(A 1 x, A 1 y, A 1 z,a 1 w) In some cases, the precompositions A and A may be used interchangably In these cases, we denote the precomposition as A,andassume the form A in any arguments given, although the corresponding arguments for A can be easily verified Definition 3 Let R be an algebraic curvature tensor on a vector space V of dimension n The structure group of R is denoted G R and defined G R = {A GL(n) A R = R} Definition 4 A Jordan block of size k corresponding to some eigenvalue λ R on R k is defined: λ λ 1 0 J(k, λ) = λ 1

2 The Jordan block corresponding to a pair of complex conjugate eigenvalues a ± b 1 is defined in the construction of the size 2k matrix: A I A I 0 J(k, a, b) = A where and a A = b I = b a Definition 5 Let {A i } be a collection of square matrices,,,n The direct sum of A i is: A n 0 A 2 0 A i = 0 0 A n Lemma 1 Let A End(V) Choosing an appropriate basis for V, A will decompose as the direct sum of Jordan blocks The unordered collection of these blocks is determined by A Definition 6 The Jordan normal form of A is the unordered collection of Jordan blocks from the preceding lemma 2 A Generalization of Previous Results Lemma 2 Let A End(V) Define T A : A(V ) A(V ) by T A (R) =A R Then T A is alineartransformationona(v ) Proof Let R 1,R 2 A(V ) and let c R Then: T A ((R 1 + R 2 )(x, y, z, w)) = A (R 1 + R 2 )(x, y, z, w) =(R 1 + R 2 )(Ax, Ay, Az, Aw) = R 1 (Ax, Ay, Az, Aw)+R 2 (Ax, Ay, Az, Aw) =A R 1 (x, y, z, w)+a R 2 (x, y, z, w) = T A (R 1 (x, y, z, w)) + T A (R 2 (x, y, z, w)) Furthermore, T A (cr 1 (x, y, z, w)) = T A (R 1 (cx, y, z, w)) = A R 1 (cx, y, z, w) =R 1 (A(cx), Ay, Az, Aw) = R 1 (cax, Ay, Az, Aw) =cr 1 (Ax, Ay, Az, Aw) =ca R 1 (x, y, z, w) = ct A (R 1 (x, y, z, w)) The following constitutes a generalization of the work done by Kaylor [1] 2

3 Definition 7 Let B = {R 1, R 2,,R N } be an ordered basis for A(V ) and let A End(V ) Define the Kaylor matrix of A with respect to B: β 11 β 21 β N1 β 12 β 22 β N2 K A = β 1N β 2N β NN where β ij is the coefficient corresponding to R j when A R i is expressed in terms of the basis B Lemma 3 K A is the matrix representation of the linear transformation T A with respect to the ordered basis B Proof Let R A(V ) Then R = N α i R i where α i R ExpressR as a column vector with respect to B: α 1 α 2 α N R = T A : A(V ) A(V ), so given R i B, T A (R i )=A R i = N β ij R j where β ij R Then T A (R) =T A ( N α i R i )= N j=1 α i T A (R i ), since T A is linear But then N α i T A (R i )= N N α i β ij R j = N N R j α i β ij Expressed as a column vector with respect to B: j=1 j=1 N α i β i1 N β 11 β 21 β N1 α 1 α i β i2 β 12 β 22 β N2 α 2 T A (R) = = = K AR N β 1N β 2N β NN α N α i β in Theorem 1 Let x = x 1 R x N R N If there exists a nonzero algebraic curvature tensor R such that A G R, then K A x = x has a non-trivial solution [1] Corollary 1 The solution space of K A x = x is the set of all algebraic curvature tensors R such that A G R [1] 3

4 3 Results Fact K I =I Proof Let R A(V ) Then: K I R(x, y, z, w) =I R(x, y, z, w) =R(I 1 x, I 1 y, I 1 z,i 1 w)=r(x, y, z, w) =IR(x, y, z, w) Fact K A = K A Proof Let R A(V ) Then: K A R(x, y, z, w) =( A) R(x, y, z, w) =R(( A) 1 x, ( A) 1 y, ( A) 1 z,( A) 1 w) = R( A 1 x, A 1 y, A 1 z, A 1 w)=( 1) 4 R(A 1 x, A 1 y, A 1 z,a 1 w) = A R(x, y, z, w) =K A R(x, y, z, w) Theorem 2 Let A,B End(V) and let K A and K B be the matrix representations of T A (R) = A R and T B (R) = B R with respect to the basis B Then the matrix representation of T AB (R) =(AB) R with respect to B is K AB = K B K A Proof Let R A(V ) Then K AB R(x, y, z, w) =(AB) R(x, y, z, w) = R(ABx, ABy, ABz, ABw) =A R(Bx,By,Bz,Bw) =B (A R(x, y, z, w)) = K B K A R(x, y, z, w) Corollary 2 Let A,B GL(n) and let K A and K B be the matrix representations of T A (R) =A R and T B (R) =B R with respect to the basis B Then the matrix representation of T AB (R) =(AB) R with respect to B is K AB = K A K B Proof Let R A(V ) Then: K AB R(x, y, z, w) =(AB) R(x, y, z, w) =R((AB) 1 x, (AB) 1 y, (AB) 1 z,(ab) 1 w) = R(B 1 A 1 x, B 1 A 1 y, B 1 A 1 z,b 1 A 1 w)=b R(A 1 x, A 1 y, A 1 z,a 1 w) = A (B R(x, y, z, w)) = K A K B R(x, y, z, w) Corollary 3 If A GL(n) then K (K A ) 1 Proof K A 1K A = K A 1 A = K I = I and K A K K A K I = I Theorem 3 If G is a subgroup of GL(n) then K G = {K A A G} is also a group Proof G is a group so I G, and so K I = I K G Let K A K G Then A G and A 1 G so K (K A ) 1 K G Let K A,K B K G Then A,B G so AB G and so K A K B = K AB K G Let K A,K B,K C K G ThenK A (K B K C )=(K A K B )K C since matrix multiplication is associative Lemma 4 Let B = {e 1,e 2,,e n } be a basis for a vector space V Then given e i,e j B, e i = e j, there exists a symmetric bilinear form φ such that the canonical algebraic curvature tensor R φ (x, y, z, w) =φ(x, w)φ(y, z) φ(x, z)φ(y, w) has R φ (e i,e j,e j,e i )=1 and, up to symmetries of algebraic curvature tensors, all other R φ (e k,e l,e m,e p )=0 4

5 Proof Let φ : V V R be defined as follows: φ(e i,e i )=φ(e j,e j ) = 1 and φ(e k,e l )=0 for all other basis vector pairs Then φ is a symmetric bilinear form, so we can form the canonical algebraic curvature tensor R φ (x, y, z, w) =φ(x, w)φ(y, z) φ(x, z)φ(y, w) Then R φ (e i,e j,e j,e i ) = 1 and, up to symmetries of algebraic curvature tensors, all other R φ (e k,e l,e m,e p ) = 0 Lemma 5 Let B = {e 1,e 2,,e n } be a basis for a vector space V Then given distinct e i,e j,e k B, there exists an algebraic curvature tensor R A(V ) such that R(e i,e j,e k,e i ) = 1 and, up to symmetries of algebraic curvature tensors, all other R(e l,e m,e p,e q )=0 Proof Let φ : V V R be defined as follows: φ(e i,e i )=φ(e j,e k )=φ(e k,e j )= 1 and φ(e l,e m ) = 0 for all other basis vector pairs Then φ is a symmetric bilinear form, so we can define the canonical algebraic curvature tensor R φ (x, y, z, w) = φ(x, w)φ(y, z) φ(x, z)φ(y, w) Then R φ (e i,e j,e k,e i ) = 1, and if E s is any other permutation of {e i,e i,e j,e k } then R φ (E s ) is predetermined by the symmetries of algebraic curvature tensors It is clear from the definition of R φ that all other R φ (e l,e m,e p,e q )=0except R φ (e j,e k,e k,e j )= 1 and its nonzero associates According to the previous lemma, there exists an algebraic curvature tensor R ψ A(V ) such that R ψ (e j,e k,e k,e j ) = 1 and all other R ψ (e l,e m,e p,e q ) = 0, up to symmetries of algebraic curvature tensors Then R = R φ + R ψ has: R(e i,e j,e k,e i )=R φ (e i,e j,e k,e i )+R ψ (e i,e j,e k,e i ) = = 1, R(e j,e k,e k,e j )=R φ (e j,e k,e k,e j )+R ψ (e j,e k,e k,e j )= = 0, and R(e l,e m,e p,e q )=R φ (e l,e m,e p,e q )+R ψ (e l,e m,e p,e q )=0+0=0 for all other R(e l,e m,e p,e q ), up to symmetries of algebraic curvature tensors Theorem 4 If A GL(n), n 3, then K A = I A = ±I Proof ( ) This follows directly from the two previously stated facts ( ) Suppose A = ±I Then ±I, so the Jordan Normal form of A 1 has a Jordan block that isn t either J(1, 1) or J(1,-1) Call this block A 1 A 1 has one of the following forms: (i) J(1,λ), λ = ±1 (ii) J(2,λ) (iii) J(m, λ), m Z, 3 m n (iv) J(1,a,b) (v) J(p, a, b), p Z, 2 p n 2 (Note that none of the Jordan blocks of A 1 can have eigenvalue zero, since A 1 is invertible, and additionally that none of the Jordan blocks with complex eigenvalues can have imaginary part zero, since they would not be complex) (i) Suppose A 1 = J(1,λ),λ = ±1 Choose an ordered basis B = {e 1,e 2,,e n } for V such that, reading down and to the right along the diagonal of the matrix representation of A 1, A 1 appears first and the remaining Jordan blocks appear in the 5

6 following order: (1) any other J(1,η), (2) any J(m, η),m Z, 3 m n 1, (3) any J(p, a, b),p Z, 1 p n 1 2 n 3, so A 1 has another Jordan block, A 2, in the second diagonal entry of A 1 A 2 will have form like that of (1),(2), or (3) above (i)(1) Suppose A 2 = J(1,η) Then A 1 has another Jordan block A 3 in the third diagonal position with form either (a) J(m, γ), 1 m n 2, or (b) J(p, a, b),p Z, 1 p n 2 2 (i)(1)(a) Suppose A 3 = J(m, γ), 1 m n 2 Then A 1 has the form: λ η γ So A 1 e 1 = λe 1, A 1 e 2 = ηe 2, and A 1 e 3 = γe 3 Then A R(e 1,e 2,e 2,e 1 )=R(A 1 e 1,A 1 e 2,A 1 e 2,A 1 e 1 )=R(λe 1,ηe 2,ηe 2,λe 1 ) = λ 2 η 2 R(e 1,e 2,e 2,e 1 ) Similarly, A R(e 1,e 3,e 3,e 1 )=R(A 1 e 1,A 1 e 3,A 1 e 3,A 1 e 1 )=R(λe 1,γe 3,γe 3,λe 1 ) = λ 2 γ 2 R(e 1,e 3,e 3,e 1 ) and A R(e 2,e 3,e 3,e 2 )=R(A 1 e 2,A 1 e 3,A 1 e 3,A 1 e 2 ) = R(ηe 2,γe 3,γe 3,ηe 2 )=η 2 γ 2 R(e 2,e 3,e 3,e 2 ) If K A = I then A R(e 1,e 2,e 2,e 1 )= R(e 1,e 2,e 2,e 1 ), but A R(e 1,e 2,e 2,e 1 )= λ 2 η 2 R(e 1,e 2,e 2,e 1 ), so this is impossible If K A = I then A R(e 1,e 2,e 2,e 1 ) = R(e 1,e 2,e 2,e 1 ),A R(e 1,e 3,e 3,e 1 )=R(e 1,e 3,e 3,e 1 ), and A R(e 2,e 3,e 3,e 2 )=R(e 2,e 3,e 3,e 2 ), so λ 2 η 2 = λ 2 γ 2 = η 2 γ 2 = 1 Thus λ 2 = γ 2,so1=λ 2 γ 2 = λ 4 But λ = ±1, so this is impossible (i)(1)(b) Suppose A 3 = J(p, a, b),p Z, 1 p n 2 2, 1 m n 2 Then A 1 has the form: λ η A a b 0 = 0 0 b a So A 1 e 1 = λe 1, A 1 e 2 = ηe 2, and A 1 e 3 = ae 3 be 4 Then A R(e 1,e 3,e 3,e 1 )= R(A 1 e 1,A 1 e 3,A 1 e 3,A 1 e 1 )=R(λe 1,ae 3 be 4,ae 3 be 4,λe 1 )=λ 2 a 2 R(e 1,e 3,e 3,e 1 ) 2λ 2 abr(e 1,e 3,e 4,e 1 )+λ 2 b 2 R(e 1,e 4,e 4,e 1 ) Let R ijji denote the algebraic curvature tensor constructed in lemma 3, R ijji = R φ for the appropriate φ, and let R ijki denote the algebraic curvature tensor constructed in lemma 4, R ijki = R for the appropriate R Let R = R a 2b R 1341 Then R(e 1,e 3,e 3,e 1 ) = 1, but A R(e 1,e 3,e 3,e 1 )=λ 2 a 2 R(e 1,e 3,e 3,e 1 ) 2λ 2 abr(e 1,e 3,e 4,e 1 )+ 6

7 λ 2 b 2 R(e 1,e 4,e 4,e 1 )=λ 2 a 2 (1) 2λ 2 ab( a 2b )+λ2 b 2 (0) = λ 2 a 2 λ 2 a 2 = 0 Thus A R = R and A R = R, sok A = ±I (i)(2) Now suppose A 2 = J(m, η), 2 m n 1 Then A 1 has the form: λ η η So A 1 e 1 = λe 1,A 1 e 2 = ηe 2, and A 1 e 3 = e 2 + ηe3 Then A R(e 1,e 3,e 3,e 1 )= R(A 1 e 1,A 1 e 3,A 1 e 3,A 1 e 1 )=R(λe 1,e 2 + ηe3,e 2 + ηe3,λe 1 )=λ 2 R(e 1,e 2,e 2,e 1 )+ λ 2 η 2 R(e 1,e 3,e 3,e 1 )+2λ 2 ηr(e 1,e 2,e 3,e 1 ) Let R = η 2 R R 1331 ThenR(e 1,e 3,e 3,e 1 )= 1, but A R(e 1,e 3,e 3,e 1 )=λ 2 R(e 1,e 2,e 2,e 1 )+λ 2 η 2 R(e 1,e 3,e 3,e 1 )+2λ 2 ηr(e 1,e 2,e 3,e 1 )= λ 2 ( η 2 )+λ 2 η 2 (1) + 2λ 2 η(0) = λ 2 η 2 + λ 2 η 2 = 0 Thus A R = R and A R = R, so K A = ±I (i)(3) Now suppose A 2 = J(p, a, b),p Z, 1 p n 1 2 ThenA 1 has the form: λ a b 0 0 b a So A 1 e 1 = λe 1 and A 1 e 2 = ae 2 be 3 ThenA R(e 1,e 2,e 2,e 1 ) = R(A 1 e 1,A 1 e 2,A 1 e 2,A 1 e 1 )=R(λe 1,ae 2 be 3,ae 2 be 3,λe 1 )=λ 2 a 2 R(e 1,e 2,e 2,e 1 )+ λ 2 b 2 R(e 1,e 3,e 3,e 1 ) 2λ 2 abr(e 1,e 2,e 3,e 1 ) Let R = R a 2b R 1231 Then R(e 1,e 2,e 2,e 1 ) = 1, but A R(e 1,e 2,e 2,e 1 ) = λ 2 a 2 R(e 1,e 2,e 2,e 1 )+λ 2 b 2 R(e 1,e 3,e 3,e 1 ) 2λ 2 abr(e 1,e 2,e 3,e 1 )=λ 2 a 2 (1) + λ 2 b 2 (0) 2λ 2 ab( a 2b )=λ2 a 2 λ 2 a 2 =0 Thus A R = R and A R = R, sok A = ±I (ii)now, when A 1 = J(2,λ), A 2 has one of the following forms: (1) J(m, η), 1 m n 2, or (2) J(p, a, b), 1 p n 2 2 (ii)(1) Suppose A 2 = J(m, η), 1 m n 2 Then A 1 has the form: λ λ η

8 So A 1 e 2 = e 1 + λe 2 and A 1 e 3 = ηe 3 ThenA R(e 2,e 3,e 3,e 2 ) = R(A 1 e 2,A 1 e 3,A 1 e 3,A 1 e 2 )=R(e 1 +λe 2,ηe 3,ηe 3,e 1 +λe 2 )=λ 2 η 2 R(e 2,e 3,e 3,e 2 )+ η 2 R(e 1,e 3,e 3,e 1 )+2λη 2 R(e 3,e 1,e 2,e 3 ) Let R = R 2332 λ 2 R 1331 Then R(e 2,e 3,e 3,e 2 ) = 1, but A R(e 2,e 3,e 3,e 2 ) = λ 2 η 2 R(e 2,e 3,e 3,e 2 )+η 2 R(e 1,e 3,e 3,e 1 )+2λη 2 R(e 3,e 1,e 2,e 3 )=λ 2 η 2 (1) + η 2 ( λ 2 )+ 2λη 2 (0) = λ 2 η 2 λ 2 η 2 = 0 Thus A R = R and A R = R, sok A = ±I (ii)(2) Now suppose A 2 = J(p, a, b), 1 p n 2 2 ThenA 1 has the form: λ λ A a b 0 = 0 0 b a So A 1 e 1 = λe 1 and A 1 e 3 = ae 3 be 4 ThenA R(e 1,e 3,e 3,e 1 ) = R(A 1 e 1,A 1 e 3,A 1 e 3,A 1 e 1 )=R(λe 1,ae 3 be 4,ae 3 be 4,λe 1 )=λ 2 b 2 R(e 1,e 4,e 4,e 1 )+ λ 2 a 2 R(e 1,e 3,e 3,e 1 ) 2λ 2 abr(e 1,e 3,e 4,e 1 ) Let R = R a 2b R 1341 Then R(e 1,e 3,e 3,e 1 ) = 1, but A R(e 1,e 3,e 3,e 1 ) = λ 2 b 2 R(e 1,e 4,e 4,e 1 )+λ 2 a 2 R(e 1,e 3,e 3,e 1 ) 2λ 2 abr(e 1,e 3,e 4,e 1 )=λ 2 b 2 (0) + λ 2 a 2 (1) 2λ 2 ab( a 2b )=λ2 a 2 λ 2 a 2 = 0 Thus A R = R and A R = R, sok A = ±I (iii) Let A 1 = J(m, λ), 3 m n ThenA 1 has the form: λ λ λ So A 1 e 1 = λe 1 and A 1 e 3 = e 2 + λe 3 ThenA R(e 1,e 3,e 3,e 1 ) = R(A 1 e 1,A 1 e 3,A 1 e 3,A 1 e 1 )=R(λe 1,e 2 +λe 3,e 2 +λe 3,λe 1 )=λ 2 R(e 1,e 2,e 2,e 1 )+ 2λ 3 R(e 1,e 2,e 3,e 1 )+λ 4 R(e 1,e 3,e 3,e 1 ) Let R = λ 2 R R 1331 Then R(e 1,e 3,e 3,e 1 ) = 1, but A R(e 1,e 3,e 3,e 1 )= λ 2 R(e 1,e 2,e 2,e 1 )+2λ 3 R(e 1,e 2,e 3,e 1 )+λ 4 R(e 1,e 3,e 3,e 1 )=λ 2 ( λ 2 )+2λ 3 (0) + λ 4 (1) = λ 4 + λ 4 = 0 Thus A R = R and A R = R, sok A = ±I (iv) Let A 1 = J(1,a,b) Then A 2 has one of the following forms: (1) J(m, λ), 1 m n 2, or (2) J(q, c, d), 1 q n 2 2 (iv)(1) Let A 2 = J(m, λ), 1 m n 2 Then A 1 has the form: 8

9 a b 0 0 b a λ So A 1 e 1 = ae 1 be 2 and A 1 e 3 = λe 3 ThenA R(e 1,e 3,e 3,e 1 ) = R(A 1 e 3,A 1 e 1,A 1 e 2,A 1 e 3 )=R(ae 1 be 2,λe 3,λe 3,ae 1 be 2 )=λ 2 a 2 R(e 1,e 3,e 3,e 1 )+ λ 2 b 2 R(e 2,e 3,e 3,e 2 ) 2abλ 2 R(e 3,e 1,e 2,e 3 ) Let R = a 2b R R 1331 Then R(e 1,e 3,e 3,e 1 ) = 1, but A R(e 1,e 3,e 3,e 1 ) = λ 2 a 2 R(e 1,e 3,e 3,e 1 )+λ 2 b 2 R(e 2,e 3,e 3,e 2 ) 2abλ 2 R(e 3,e 1,e 2,e 3 )=λ 2 a 2 (1) + λ 2 b 2 (0) 2abλ 2 ( a 2b )=λ2 a 2 λ 2 a 2 = 0 Thus A R = R and A R = R, sok A = ±I (iv)(2) Let A 2 = J(q, c, d), 1 q n 2 2 ThenA 1 has the form: a b b a A c d 0 = 0 0 d c So A 1 e 1 = ae 1 be 2, A 1 e 2 = be 1 +ae 2 and A 1 e 3 = ce 3 de 4 ThenA R(e 1,e 2,e 3,e 1 )= R(A 1 e 1,A 1 e 2,A 1 e 3,A 1 e 1 )=R(ae 1 be 2,be 1 + ae 2,ce 3 de 4,ae 1 be 2 )=ac(a 2 + b 2 )R(e 1,e 2,e 3,e 1 )+bc(a 2 + b 2 )R(e 2,e 1,e 3,e 2 ) ad(a 2 + b 2 )R(e 1,e 2,e 4,e 1 ) bd(a 2 + b 2 )R(e 2,e 1,e 4,e 2 ) Let R = ac bd R R 1231 Then R(e 1,e 2,e 3,e 1 ) = 1, but A R(e 1,e 2,e 3,e 1 ) = ac(a 2 +b 2 )R(e 1,e 2,e 3,e 1 )+bc(a 2 +b 2 )R(e 2,e 1,e 3,e 2 ) ad(a 2 +b 2 )R(e 1,e 2,e 4,e 1 ) bd(a 2 + b 2 )R(e 2,e 1,e 4,e 2 )=ac(a 2 + b 2 )(1) + bc(a 2 + b 2 )(0) ad(a 2 + b 2 )(0) bd(a 2 + b 2 )( ac bd )= ac(a 2 + b 2 ) ac(a 2 + b 2 ) = 0 Thus A R = R and A R = R, sok A = ±I (v) Let J(p, a, b), 2 p n 2 ThenA 1 has the form: a b b a A a b 0 = 0 0 b a So A 1 e 1 = ae 1 be 2, A 1 e 2 = be 1 + ae 2 and A 1 e 3 = e 1 + ae 3 be 4 Then A R(e 1,e 2,e 3,e 1 ) = R(A 1 e 1,A 1 e 2,A 1 e 3,A 1 e 1 ) = R(ae 1 be 2,be 1 + ae 2,e 1 + 9

10 ae 3 be 4,ae 1 be 2 )=b(a 2 + b 2 )R(e 1,e 2,e 2,e 1 )+a 2 (a 2 + b 2 )R(e 1,e 2,e 3,e 1 )+ab(a 2 + b 2 )R(e 2,e 1,e 3,e 2 ) ab(a 2 + b 2 )R(e 1,e 2,e 4,e 1 ) b 2 (a 2 + b 2 )R(e 2,e 1,e 4,e 2 ) Let R = a2 R b R 1231 Then R(e 1,e 2,e 3,e 1 ) = 1, but A R(e 1,e 2,e 3,e 1 ) = b(a 2 +b 2 )R(e 1,e 2,e 2,e 1 )+a 2 (a 2 +b 2 )R(e 1,e 2,e 3,e 1 )+ab(a 2 +b 2 )R(e 2,e 1,e 3,e 2 ) ab(a 2 + b 2 )R(e 1,e 2,e 4,e 1 ) b 2 (a 2 + b 2 )R(e 2,e 1,e 4,e 2 )=b(a 2 + b 2 )(0) + a 2 (a 2 + b 2 )(1) + ab(a 2 + b 2 )(0) ab(a 2 + b 2 )(0) b 2 (a 2 + b 2 )( a2 )=a 2 (a 2 + b 2 ) a 2 (a 2 + b 2 )=0 b 2 Thus A R = R and A R = R, sok A = ±I Corollary 4 Define U : G K G by U(A) =K A Then U is an onto homomorphism of groups, Ker(U) ={±I}, andsog/{±i} = K G Proof By construction, K A K G A G and U(A) =K A so U is onto Let A, B G Then U(AB) =K AB = K A K B = U(A)U(B) So U is a group homomorphism Ker(U) ={A G U(A) =K A = I} = {±I} by the previous theorem, so G/{±I} = K G by the First Isomorphism Theorem 4 Open Questions 1 Given A End(V ), A not necessarily invertible, if A G R for some R and Ker(R) = 0, must it be that Ker(A) = 0? 2 For A GL(n), does the Jordan normal form of A determine or constrain the Jordan normal form of K A? 3 Given some number of elements from a group G GL(n), can the study of Kaylor matrices determine whether or not G is the structure group for some algebraic curvature tensor R? 5 Acknowledgments I would like to thank Dr Corey Dunn for his guidance, insights, and patience throughout this project I would also like to thank Dr Rolland Trapp for the air of positivity and humor that he brings to the program This project was supported both by NSF grant DMS and California State University, San Bernardino 6 References [1] Kaylor, Lisa Is Every Invertible Linear Map in the Structure Group of some Algebraic Curvature Tensor?, CSUSB 2012 REU, 10