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The question I would like to ask is the following one.

Consider a projective space just as a smooth manifold, e.g. $\mathbb{C}P^1$ is $S^2$. Then most maps from $S^2$ to $S^2$ even if smooth, injective and surjective do not respect the "projective space structure" (which I am not sure about how to define) and therefore are not automorphisms of $\mathbb{C}P^1$. Do all the maps which preserve the "projective space structure" arise from vector space isomorphisms?

In order for the question above to make sense two things need to be clarified first.

A projective space $P(V)$ is defined to be the set of 1-dimensional subspaces of some vector space $V$. Is it possible to give an equivalent but intrinsic definition i.e. as a set with some structure without referring to a vector space?

Any isomorphism of $V$ descends to a surjective transformation of $P(V)$ - which is usually called an automorphism of $P(V)$. Is there a meaningful way to define an automorphism of a projective space in intrinsic terms, i.e. as a map preserving the "projective space structure", without mentioning the vector space isomorphism from which it arises?

If it is possible to define projective spaces and transformations intrinsically, is it true that all the projective automorphisms come from vector space isomorphisms?

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Veblen and Young wrote a book on this. –  Mariano Suárez-Alvarez Dec 20 '13 at 20:02
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2 Answers

up vote 2 down vote accepted

First, you have to define what a "projective structure" means. Let $P^n$ be a projective space over some field $F$. In the context of classical projective geometry, a self-map $f: P^n\to P^n$ is projective (or preserves the projective structure) if it preserves projective incidences: Three points $x, y, z\in P^n$ belong to the same line if and only if their images do. (In principle, one should also make a similar requirements on configurations of $k$ points belonging to a common $k-2$-dimensional projective subspace, but this turns out to be unnecessary). If $n=1$, i.e., you are dealing with 1-dimensional projective space, then this condition, of course is meaningless and every self-map $f$ is projective. However, if $n\ge 2$ then von Staudt proved what he called the "2nd fundamental theorem of projective geometry", which states that a bijective transformation $P^n\to P^n$ is projective if and only if it is a composition of an element of $PGL(n, F)$ and an automorphism of the field $F$. If you are interested in $F={\mathbb R}$, then such an automorphism has to be the identity. If $F={\mathbb C}$, then you have an uncountable group of automorphisms (assuming Axiom of Choice). However, if you assume that your projective transformations are continuous, then the only automorphism left is the complex conjugation.

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Dear studious, how do you define a projective structure? –  GFR Jan 8 at 12:30
    
@GFR: I am using the definition that a projective structure on $P^n$ (regarded as a set) is given by the collinearity relation $R$ on triples of points in $P^n$. Of course, $R$ is not a random relation on triples. You can either describe it axiomatically (you would have a read a textbook on projective geometry to see how it is done) or simply by requiring that it comes from the geometric collinearity, where we identify $P^n$ with the projectivization of some vector space. The 1st (I think) fundamental theorem of projective geometry is that the two approaches to collinearity are equivalent. –  studiosus Jan 8 at 13:26
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I personally like the following definition of projective space, although it isn't, as you request, independent of a vector space:

Let $E$ be a vector space over a field $F$. A Projective space is a triple $\mathbb P = (\mathbb P, E, \pi)$, where $\mathbb P$ is a set called the set of points, and $\pi$ is a surjective map $${{\pi:E\setminus\{0\}\to\mathbb P} \atop {x\mapsto\pi(x) \stackrel {\text{def}}= [x] }}$$ satisfying $\pi(x) = \pi(y) \Leftrightarrow y = \lambda x $, for some $\lambda\in F\setminus 0$.

Also a characterization of projective transformations is that they are only those that preserve cross-ratios. I don't think I've answered the question but the definition is slightly more intrinsic.

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