On the structure group of a decomposable model space

Abstract

We study the structure group of a canonical algebraic curvature tensor built from a symmetric bilinear form, and show that in most cases it coincides with the isometry group of the symmetric form from which it is built. Our main result is that the structure group of the direct sum of such canonical algebraic curvature tensors on a decomposable model space must permute the subspaces V i on which they are defined. For such an algebraic curvature tensor, we show that if the vector space V is a direct sum of subspaces V 1 and V 2 , the corresponding structure group decomposes as well if V 1 and V 2 are invariant of the action of the structure group on V . We determine the freedom one has in permuting these subspaces, and show these subspaces are invariant if dim V 1 = dim V 2 or if the corresponding symmetric forms defined on those subspaces have different (but not reversed) signatures, so that in this situation, only the trivial permutation is allowable. We exhibit a model space that realizes the full permutation group, and, with exception to the balanced signature case, show the corresponding structure group is isomorphic to the wreath product of the structure group of a given symmetric bilinear form by the symmetric group. Using these results, we conclude that the structure group of any member of this family is isomorphic to a direct product of wreath products of pseudo-orthogonal groups by certain subgroups of the symmetric group. Finally, we apply our results to two families of manifolds to generate new isometry invariants that are not of Weyl type.

Kernels of Algebraic Curvature Tensors of Symmetric and Skew-Symmetric Builds

“…The defining properties of algebraic curvature tensors can be used to show (see [4]) that the definition of the kernel of an algebraic curvature tensor is not biased towards the first entry, that is, ker(R) = {y ∈ V | R(x, y, z, w) = 0 for all x, z, w ∈ V } = {z ∈ V | R(x, y, z, w) = 0 for all x, y, w ∈ V } = {w ∈ V | R(x, y, z, w) = 0 for all x, y, z ∈ V }.…”

An Examination of Linear Combinations of Skew-Adjoint Type Algebraic Curvature Tensors

“…We can use Properties 2 and 3 of algebraic curvature tensors to show that for any x ∈ Ker(R) and y, z, w ∈ V , R(x, y, z, w) = R(y, x, z, w) = R(y, z, x, w) = R(y, z, w, x) = 0 [2].…”

“…2 ik − 3b 2 ik Hence a ik = ±b ik for all i ≤ n and k > n. Now let's consider R(e i , f k , f l , e i ):R(e i , f k , f l , e i ) = 3a il a ik − 3b il b ik = 0Since a il = ±b il and a ik = ±b ik , a il a ik = b il b ik if and only if a ih = δb ih for all i ≤ n and h > n where δ is either 1 or −1. This proves 1).…”