my aim : Why $GL(V)/Z(GL(V))$ is termed as projective general linear group?

The reason seen in books says This group acts on the projective set/space faithfully.

But, there can be many groups acting faithfully on projective space, can we call them also as projective general linear group?

I confused, and not understood the reasonings. I faced many questions from it:

Let $V$ be $n$-dimensional vector space over $F$ and $V^*$ denote the collection of one dimensional subspaces of $V$, it is called projective space.

For $V^*$, we can first consider its symmetric group $Symm(V^*)$- the set of all the bijections from $V^*$ to itself. If we put some structure on $V^*$, and if we look bijections preserving the structure on $V^*$, then we obtain a subgroup of $Symm(V^*)$

Question 1. What stuctural properties on $V^*$ the group $GL(V)/Z(GL(V))$ preserves?

question 2. Can we say that, $GL(V)/Z(GL(V))$ is precisely the set of all bijections from $V^*$ to itself which preserve certain structure on $V^*$?

Question 3. Is there some notion like Projective group associated to $V^*$ instead of Projective Linear Group? If yes, how that group differs from $GL(V)/Z(GL(V))$?

Question 4. Since passing from $V$ to $V^*$, we have lost linearity: $V^*$ is not a vector (linear) space. Then what the term linear refers to in Projective Linear Group?

[Please point out, if questions are not clear.]


$V^*$ has been traditionally used to denote the dual vector space. Let us use P$(V)$ for the projective space of 1-dimenaional subspaces of $V$.

$GL(V)$ definitely acts on P$(V)$. The subgroup of scalars, its centre, acts trivially. That is, it fixes every element of P$(V)$. (If you work with a basis the whole diagonal subgroup of $GL(n)$ acts trivially, but that subgroup is not normal).

Therefore the quotient group $PGL(V)$ acts on P$(V)$.

  • $\begingroup$ A subgroup of $GL(V)$ modulo scalars also acts on $P(V)$; why that subgroup is not called as Projective general linear group? $\endgroup$ – Beginner Dec 9 '15 at 4:52
  • 1
    $\begingroup$ For the same reason the set of cyclic permutations of a set is not called the symmetric group on the set: we want to include all . $\endgroup$ – P Vanchinathan Dec 9 '15 at 4:54

I will refer to the projective space as $\mathrm{PG}(V)$ and not $V^{*}$, since $V^{*}$ usually has a different meaning.

The projective space $\mathrm{PG}(V)$ has an underlying vector space structure, so $\mathrm{GL}(V)$ acts naturally on it; however this action is not faithful, since scalar multiples of the identity matrix all induce the identity map on $\mathrm{PG}(V)$. So by modding out by the center we make a group that acts faithfully. It is called the projective general linear group because it (faithfully) represents the action of the general linear group on the projective space.

The group preserving the structure of $\mathrm{PG}(V)$ is called the collineation group. For a projective space of the form $\mathrm{PG}(V)$ (where $V$ is a vector space over a field $\mathbb{F}$), the full collineation group is $\mathrm{PGL}(V) \rtimes \mathrm{Gal}(\mathbb{F})$. Notice that $\mathrm{Gal}(\mathbb{F})$ cannot be represented as an $\mathbb{F}$-linear map on $V$. This is sometimes called the projective semilinear group.

All elements of the collineation group preserve all subspaces of $V$. There are some differences between linear collineations and semilinear collineations. The linear maps are what are called homographies. These are typically defined in terms of maps known as central collineations. In particular, every homography is a product of a finite number of central collineations.

Central collineation: A map $\mu : \mathrm{PG}(V) \to \mathrm{PG}(V)$ is called a central collineation if there is some hyperplane $H$ fixed by $\mu$ (that is, $\mu$ restricted to the hyperplane acts as the identity map) called the axis, and a point $O$ (a point in a projective geometry is a one-dimensional subspace of the vector space) which is fixed linewise by $\mu$ (that is, each line through $O$ is stabilized by $\mu$).


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