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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

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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.

Hello
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  • Several things here I don't understand. Via the correspondence in your third paragraph, the trivial form $f(u,v)=0$ corresponds to $\underline{F}=0$, the zero matrix, not "the identity of the null space", whatever that is supposed to be. Then, in your fourth paragraph, several things seem strange. Suddenly $\underline{F}$ is supposed to be invertible? And then, are you claiming that every invertible matrix is skew-symmetric? And then, for a general Lie group, what is "the skew-symmetric bilinear form of that group" -- are you sure such a thing is there for all Lie groups? – Torsten Schoeneberg Jan 15 '20 at 07:15
  • I think Chisolm is using the standard positive-definite dot product (or at least a non-degenerate dot product) to represent other arbitrary bilinear forms. – mr_e_man Jan 20 '20 at 04:46
  • As for $\underline F(v)=v\cdot F_2$, this can be an arbitrary linear transformation on $V$ (the null space), and it's skew-symmetric (in a generalized sense) on $V\oplus V^*$. See my answer on geometric product between matrices. – mr_e_man Jan 20 '20 at 04:46
  • There are many interesting isomorphisms between different spaces at play here; the problem is that it's hard to keep track of which space or representation we're talking about. – mr_e_man Jan 20 '20 at 04:54

1 Answers1

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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 $O(V,f)$.

Torsten Schoeneberg
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  • Hello, thank you very much for your reply. So this means that $\underline{F}$ used to defined the bilinear form does not have to be invertible even if the bilinear form is that describing the general linear group (because the $\underline{F}$ defined for the orthogonal and symplectic group is)? – Hello Jan 15 '20 at 09:55
  • The reason I ask this question is because in the paper "Lie groups as spin groups" referred to above, they define their bilinear form for the general linear group as $a \cdot (\underline{F}b)$ where $a, b \in V$ where $V$ is a null space and $\underline{F}$ is the identity in the sense that $\underline{F}b = b$. Could $\underline{F}$ also be 0 here but can define a different $\underline{F}$ in this case because of the choice of vector space? – Hello Jan 15 '20 at 09:59
  • Where in the paper do they do what you claim? As far as I can tell, they rather define (for an $n$-dimensional $V$) a form on the $2n$-dimensional space $V \otimes V^$ which has maximal (namely $n$-dimensional) totally isotropic subspace, probably to make things basis-free / invariant. That's interesting but, as far as I can tell, significantly different from what you write. I mean, if $F$ is the identity and $\cdot$ is the dot product, then $v \cdot Fw$ obviously does not* define the identically zero form on $V$, unless $V$ is literally zero-dimensional $=0$. – Torsten Schoeneberg Jan 16 '20 at 05:05
  • @TorstenSchoeneberg - $F$ is the identity on $V$, but a different operation on $V\oplus V^$. And I think you're interpreting "dot product" too strictly; it is precisely the symmetric bilinear form that makes $V$ "isotropic" or "null", so indeed $v\cdot w=0$ identically on $V$ (but not on $V\oplus V^$). – mr_e_man Jan 20 '20 at 05:15
  • @mr_e_man: First of all good catch that $\oplus$ makes more sense than the $\otimes$ I wrote; then, ok on what you write and that might make sense of the final paragraphs of the OP, but then that is a very different usage of $F$ than the one in the OP's third paragraph quoting Chisolm. One should make clear that the symbols $F$ and $\cdot$ have very different meanings in those different sources. Given that my answer takes the meaning of Chisolm, maybe you want to turn your comments into an answer that makes sense of that other usage? – Torsten Schoeneberg Jan 20 '20 at 21:03