Structure Groups and Linear Transformations

Structure Groups and Linear Transformations Darien Elderfield August 20, 2015 1 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 1 0 0 0 λ 1 0 J(k, λ) = 0 0 0 λ 1

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 0 0 0 A I 0 J(k, a, b) = 0 0 0 A where and a A = b I = b a 1 0 0 1 Definition 5 Let {A i } be a collection of square matrices,,,n The direct sum of A i is: A 1 0 0 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

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 1 + + 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

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

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 + 0 = 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 )= 1 + 1 = 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

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 0 0 0 0 η 0 0 0 0 γ 0 0 0 0 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 0 0 0 0 0 η 0 0 0 A 1 0 0 a b 0 = 0 0 b a 0 0 0 0 0 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 1331 + 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

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 0 0 0 0 η 1 0 0 0 η 0 0 0 0 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 1221 +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 0 0 0 0 a b 0 0 b a 0 0 0 0 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 1221 + 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 1 0 0 0 λ 0 0 0 0 η 0 0 0 0 7

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 1 0 0 0 0 λ 0 0 0 A 1 0 0 a b 0 = 0 0 b a 0 0 0 0 0 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 1331 + 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 1 0 0 0 λ 1 0 0 0 λ 0 0 0 0 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 1221 + 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

a b 0 0 b a 0 0 0 0 λ 0 0 0 0 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 3123 + 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 0 0 0 b a 0 0 0 A 1 0 0 c d 0 = 0 0 d c 0 0 0 0 0 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 2142 + 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 1 0 0 b a 0 1 0 A 1 0 0 a b 0 = 0 0 b a 0 0 0 0 0 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

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 2 2142 + 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-1461286 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, http://wwwmathcsusbedu/reu/lk12pdf 10