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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?

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I am finding this question a bit hard to pin down: what is your definition of "a Frobenius morphism for $G$"? What do you mean by $F$-stability? Also, it sounds like you are asking questions based on some specific text(s) you are reading. Telling us what these are could only help us help you... – Pete L. Clark Dec 6 '11 at 12:57
Among other texts, I've been working through the first chapter of Carter's Finite Groups of Lie Type. I don't have the book at hand at the moment, but I believe that Frobenius morphisms are introduced in I.17. I'm still quite confused with the concept, hence the confusion in my question. What I'm trying to ask/understand: If $F$ is the Frobenius map raising the entries of a matrix to the $q$-th power, why is $F(G) \subseteq G$? I don't see why the map should respect quadratic forms. It's probably quite easy to see, I'm just not sure how to deal with the concept.. Thanks in advance! – F_q Dec 6 '11 at 14:50

First, the Frobenius morphism is an automorphism of your field (since it's a finite field).

Next, saying a matrix is orthogonal is equivalent to saying it satisfies a set of algebraic relations: think of a matrix $X$ where each entry is an indeterminate $x_{ij}$. Then the equation $XX^T=I$ is equivalent to $n^2$ polynomial relations in $x_{ij}$'s.

If $A$'s entries satisfy these relations (i.e. $AA^T=I$), then so will $A$'s image under the Frobenius morphism since the Frobenius morphism preserves the algebraic relations defined by $XX^T=I$.

Why are the relations preserved? Basically because they are homogeneous polynomials relations whose coefficients ($1$'s) are not changed by the Frobenius morphism.

$$ XX^T = \begin{pmatrix} x_{11} & x_{12} \\ x_{21} & x_{22} \end{pmatrix} \begin{pmatrix} x_{11} & x_{21} \\ x_{12} & x_{22} \end{pmatrix} = \begin{pmatrix} x_{11}^2+x_{12}^2 & x_{11}x_{21}+x_{12}x_{22} \\ x_{11}x_{12}+x_{12}x_{22} & x_{21}^2+x_{22}^2 \end{pmatrix} $$ $$= \begin{pmatrix} 1 & 0 \\\\ 0 & 1 \end{pmatrix} = I $$

So a $2\times 2$ matrix is orthogonal if its entries satisfy the equations: $x_{11}^2+x_{12}^2=1$, $x_{11}x_{21}+x_{12}x_{22}=0$, $x_{11}x_{12}+x_{12}x_{22}=0$, and $x_{21}^2+x_{22}^2=1$.

If the $a_{ij}$'s satisfy these relations, then so do the $(a_{ij})^q$'s.

For example: $((a_{11})^q)^2+((a_{22})^q)^2=((a_{11})^2)^q+((a_{22})^2)^q=(a_{11}^2+a_{22}^2)^q=1^q=1$.

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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 $A$.

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