I am studying the usage of bivectors to represent the Lie algebra of the general linear group.
1) The book " 3-d transposition groups" by Michael Aschbacher states
for the general linear group. Is this a statement that says that the general linear group can be regarded as the isometry group over a null vector space?
Also in E. Chisolm, "Geometric Algebra", it is stated that any bilinear form can uniquely in geometric algebra as $f(u, v) = u \cdot (\underline{F}v) \, \forall u, v$ where $\underline{F}$ is unique to the bilinear form. Therefore, for the isometry group above, what is this unique function $\underline{F}$ (could this be the identity of the null vector space)?
The reason I ask this is because it can be shown that any element of the general linear group acting on a null vector $v$ can be written as $\underline{F}v = v \cdot F_{2}$ where $F_{2}$ is a unique bivector, and thus from Eric Chisolm, this means that $\underline{F}$ is skew-symmetric. Furthermore, the paper: C. Doran, D. Hestenes, F.Sommen and N. Van Acker, "Lie groups as spin groups", J. Math. Phys, 34 (8), 1993, shows that the Lie algebra generators of a particular Lie group are those that commute with the operator $\underline{F}$ that determines the skew-symmetric bilinear form of that group. They then state without pre-emptive motivation (it seems) that choosin $\underline{F}$ to be the identity operator of the null vector space will then generate the general linear group.
This is why I am asking if the bilinear form of the isometry group for the general linear group is simply $f(u, v) = u \cdot (\underline{F}v) \, \forall u, v \in V$ where $V$ is a null vector space and $\underline{F}$ is the identity of the null vector space.
See my comment for what does not make sense to me in the question; but the quote you quote does make sense to me, as follows: For a $K$-vector space $V$ with a bilinear form $f: V \times V \rightarrow K$, we have the isometry group $$O(V,f) := \lbrace A \in GL(V): \text{ for all } v,w: f(Av,Aw)=f(v,w) \rbrace.$$
If $V$ is $m$-dimensional and the form is given by a $m\times m$-matrix $F$, i.e. $f(v,w) = v^{tr}Fw$ (matrix products) $= v \cdot Fw$ (dot product), then the condition above is equivalent to
$$O(V,f) := \lbrace A \in GL(V): A^{tr}FA=F \rbrace.$$
For $K = \mathbb R$ these are Lie groups. Note that for example if $F = I_m$ i.e. the form $f$ is the standard inner product, this gives the standard orthogonal group. Or if $m$ is even, $m=2n$ say, and $F = \pmatrix{0 & I_n \\-I_n &0}$, this gives the standard symplectic group.
Now if one takes the trivial form $f(v,w)=0$, equivalently $F=0$, the extra condition in either of the above definitions is empty (automatically satisfied for all $A \in GL(V)$), and one just gets the full general linear group $GL(V)$ as $O(V,f)$.