TITLE: Frobenius maps and orthogonal groups
QUESTION [6 upvotes]: I have recently encountered Frobenius maps in the context of orthogonal and symplectic groups. However, some aspects still strike me as odd. Maybe I'm making some stupid mistakes here. 
Let's just assume that $G$ is an orthogonal group of dimension $n$ over the algebraic closure of a finite field with $q$ elements. Let's view $G$ as a matrix group and say that $F$ is the standard Frobenius morphism (of the general linear group) raising each entry of a given matrix to the $q$-th power. Why is $F$ then a Frobenius morphism for $G$? I understand that the basic concept involved is $F$-stability, but... Is it obvious?

REPLY [2 votes]: In general, if $A$ is the matrix of the bilinear or quadratic form fixed by $G$, then the fixed form of $F(G)$ will be $F(A)$. So if the entries of $A$ lie in the fixed field of $F$, which in this case is the subfield of order $q$, then $F(G)$ will also fix $A$. So if $G$ is, for example, the full symplectic group fixing an alternating form, or the full special or general orthogonal group fixing a bilinear/quadratic form, and the matrix $A$ of the form has its entries in the prime field, then $F(G) = G$.
For most of classical groups, including the unitary, symplectic and orthogonal groups of plus-type (Witt defect 0), we can choose the form matrix to have all of entries $\pm 1$, and hence get $F(G)=G$. (But this still allows some flexibility in the choice of form, and the outer automorphism of the associated simple group defined in this way can depend on your choice of form in some cases.)
The exception is the orthogonal groups of minus-type (Witt defect 1), when it is not always possible to choose the fixed form matrix to have entries in the prime field. In that case, to define the so-called Frobenius automorphism of $G$, you need to raise matrix entries to the $q$-th power and then conjugate by a suitable matrix to change the form back to $A$.