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By von Oniscik A.L., Sulanke R.

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Example text

4 for the special cases T = J(k, a) and T = J(k, a, b) described in Eqs. b). , ek } be the standard basis for Rk . The Jordan block J(k, a) defines the linear transformation: J(k, a)ei := aei + ei−1 if i > 1, aei if i = 1 . We define a non-degenerate inner product ·, · on Rk by setting: ei , ej := δi+j,k+1 . February 7, 2007 9:33 WSPC/Book Trim Size for 9in x 6in aGilkeyCurvHomogenBook-v21e The Geometry of the Riemann Curvature Tensor 35 Since J(k, a) = a · Id +J(k, 0), and since Id is symmetric with respect to any inner product, we may take a = 0.

2) J (π) is complex linear for every π in CP± (V, ·, · , J). (3) A(π) is complex linear for every π in CP± (V, ·, · , J). Proof. Suppose J ∗ A = A. Then for all x, y, z, we have that: A(y, x, x, z) + A(y, Jx, Jx, z) = A(Jy, x, x, Jz) + A(Jy, Jx, Jx, Jz) . Replacing z by Jz yields A(y, x, x, Jz) + A(y, Jx, Jx, Jz) = −A(Jy, x, x, z) − A(Jy, Jx, Jx, z) . This implies that J (πx )y, Jz = − J (πx )Jy, z . This shows that JJ (πx ) = J (πx )J as desired. Thus Assertion (1) implies Assertion (2). Similarly, we may compute that for all x, y, z we have: JA(πx )y, z = − A(πx )y, Jz = −A(x, Jx, y, Jz) = −A(Jx, JJx, Jy, JJz) = A(x, Jx, Jy, z) = A(πx )Jy, z .

The map Id +α splits the projection πs . Proof. If A ∈ ker(πs ), then A satisfies: Aijkl = −Ajikl , Aijkl + Ajkil + Akijl = 0, Aijkl = −Aijlk . 1, Aijkl = Aklij and thus A is an algebraic curvature tensor. Conversely, of course, if A ∈ Alg0 , then Aijkl + Aijlk = 0. Thus ker(πs ) ∩ Fg = Alg0 . If S ∈ Λ2 (V ∗ ) ⊗ S02 (V ∗ , ·, · ), let T = α(S). We have Tijkl = 21 {Skjil + Sikjl − Sljik − Siljk }, Tjikl = 21 {Skijl + Sjkil − Slijk − Sjlik } = −Tjikl , Tklij = 21 {Silkj + Skilj − Sjlki − Skjli } = −Tijkl .

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Algebra und Geometrie 2. Moduln und Algebren by von Oniscik A.L., Sulanke R.


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