On the existence of an invariant non-degenerate bilinear form under a linear map

Details On the existence of an invariant non-degenerate bilinear form under a linear map On the existence of an invariant non-degenerate bilinear form under a linear map

2009-03-04

arxiv.org/abs/0903.0826v3

Mathematics

Commutative Algebra

completely revised version

Scientific paper

Let $\V$ be a vector space over a field $\F$. Assume that the characteristic of $\F$ is \emph{large}, i.e. $char(\F)>\dim \V$. Let $T: \V \to \V$ be an invertible linear map. We answer the following question in this paper: When does $\V$ admit a $T$-invariant non-degenerate symmetric (resp. skew-symmetric) bilinear form? We also answer the infinitesimal version of this question. Following Feit-Zuckerman \cite{fz}, an element $g$ in a group $G$ is called real if it is conjugate in $G$ to its own inverse. So it is important to characterize real elements in $\G(\V, \F)$. As a consequence of the answers to the above question, we offer a characterization of the real elements in $\G(V, \F)$. Suppose $\V$ is equipped with a non-degenerate symmetric (resp. skew-symmetric) bilinear form $B$. Let $S$ be an element in the isometry group $I(\V, B)$. A non-degenerate $S$-invariant subspace $\W$ of $(\V, B)$ is called orthogonally indecomposable with respect to $S$ if it is not an orthogonal sum of proper $S$-invariant subspaces. We classify the orthogonally indecomposable subspaces. This problem is nontrivial for the unipotent elements in $I(\V, B)$. The level of a unipotent $T$ is the least integer $k$ such that $(T-I)^k=0$. We also classify the levels of unipotents in $I(\V, B)$.

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