Conjugacy classes of involutions in the Lorentz group \(\Omega(V)\) and in \(\text{SO}(V)\)
The author tackles the problem of describing the conjugacy classes of involutions in the orthogonal group \(O(V)\) and in its commutator subgroup \(\Omega(V)\), and thus, in particular, in the Lorentz group. Let \((V, f)\) be a nonsingular vector space over a finite or ordered field \(K\) with \(\text{char\,}K\neq 2\) and \(\dim V= n\), and let \(f\) be a bilinear form. For an element \(\pi\) in the orthogonal group \(O(V)\) define \(B(\pi)= \{v^\pi- v\mid v\in V\}\) and \(F(\pi)= \{v\in V\mid v^\pi= v\}\). Then the following results hold: 1. First let \(V\) have dimension 4 and be of index 1. Assume that all two-dimensional nonsingular subspaces of \(V\) of index \(0\) are isometric and that \(f: U^\perp\to K: t\to f(t, t)\) is universal for every nonsingular two-dimensional subspace \(U\) of \(V\) of index \(0\). Then all involutions in \(\Omega(V)\) are conjugate in \(\Omega(V)\) and \(\Omega(V)\) is bireflectional. 2. Now let \(\dim V= n\) and index \(V= m\geq 1\). Assume that for given \(r\) and \(s\) all nonsingular subspaces of \(V\) of even dimension \(r\) and index \(s\) are isometric and \(f\) is universal on every nonsingular at least two-dimensional subspace of \(V\). (E.g., this is the case if \(K\) is finite and \(r\) is even.) Then two involutions \(\rho\) and \(\tau\) are conjugate in \(\text{SO}(V)\) if and only if \((r_\rho, s_\rho)= (r_\tau, s_\tau)\), where \(r_\tau= \dim B(\rho)\) and \(s_\rho= \text{index\,} B(\rho)\). If \(\tau= \kappa\rho\kappa^{-1}\), then \(\kappa\) can be chosen in \(\Omega(V)\). Thus, if \(C\) is a conjugacy class of involutions in \(\text{SO}(V)\), then \(C\subset\Omega(V)\) or \(C\cap\Omega(V)= \varnothing\). Let the field \(K\) be Euclidean and therefore ordered, let \(V\) be a nonsingular vector space over \(K\), and let \(U\) be a nonsingular subspace of \(V\). Denote the dimension of any maximal positive and maximal negative definite subspace of \(U\) by \(m\) and \(n\), respectively. Define \(t(U)= \langle m, n\rangle\). Then \(\dim U= m+ n\) and signature \(\text{sgn\,} U= m- n\). In this case one has the following result: 3. Assume \(V\) is a vector space over a Euclidean field \(K\) with \(t(V)= \langle p,q\rangle\) and suppose that \(p,q> 0\). Then two involutions \(p\) and \(r\) in \(\text{SO}(V)\) are conjugate if and only if \(t(B(\rho))= t(B(\tau))\). Let \(t(B(\rho))= t(B(\tau))= \langle k,\ell\rangle\) and \(t(F(\rho))= t(F(\tau))= \langle k',\ell'\rangle\). If \(k',\ell'> 0\) or if \(k,\ell> 0\), then \(\tau= \kappa\rho\kappa^{-1}\) where \(\kappa\) may be chosen in \(\Omega(V)\). Therefore \(C_\rho\subset\Omega(V)\) or \(C_\rho\cap\Omega(V)= 0\), where \(C_\rho\) is the conjugacy class of \(\rho\) in \(\text{SO}(V)\).
scientific article; zbMATH DE number 2081922
scientific article