Can I determine the projectivity $ h: \mathbb {P}_{1}(\mathbb {R}) \rightarrow \mathbb {P}_{1}(\mathbb {R})$ , where $ (1: 1) $ is the fix point and swaps the points $ (3: 0) $ and $ (0: 2) $?
I first thought of finding some matrix $H \in Mat_2(\mathbb{R})$ but ended up trying with double ratio which we defined as:
Definition: For four distinct points $a=\left[a_{0}: a_{1}\right], b=\left[b_{0}: b_{1}\right]$, $c=\left[c_{0}, c_{1}\right]$ and $ d=\left[d_{0}: d_{1}\right]$ we define
$ \operatorname{DR}(a, b, c, d):=\frac{\operatorname{det}\left(\begin{array}{ll} a_{0} & c_{0} \\ a_{1} & c_{1} \end{array}\right)}{\operatorname{det}\left(\begin{array}{ll} b_{0} & c_{0} \\ b_{1} & c_{1} \end{array}\right)}: \frac{\operatorname{det}\left(\begin{array}{ll} a_{0} & d_{0} \\ a_{1} & d_{1} \end{array}\right)}{\operatorname{det}\left(\begin{array}{ll} b_{0} & d_{0} \\ b_{1} & d_{1} \end{array}\right)} $ the double ratio of these points.
But I'm not sure, since I have to find a new point $d$ which is different to the points above. Also I'm not sure how I can make $(1:1)$ a fix point.
I could find some related problems here but none with swapping points and fixed points
And there is another one I'm stuck with:
Let $f: \mathbb {P}_{n}(\mathbb {F}) \rightarrow \mathbb {P}_{n}(\mathbb {F}) $ be a projectivity. First let $\mathbb {F}=\mathbb {C} $. Is there a $n \in \mathbb{N} $ for which one can choose $f$ such that $f $ has no fixed point? And I'm also interested in the case when $\mathbb {F}=\mathbb {R}$.
Thanks!
For your first question, it is a standard fact in projective geometry that given any two pairs of three distinct points in $P_1(\mathbb{R})$ there is a unique projective map taking the first three points to the second three points.
In your case this can be made quite explicit. You can just take the matrix $$ \begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix} $$
As for your second question, every projective map $f:P_n(\mathbb{C}) \to P_n(\mathbb{C})$ has a fixed point. If you think about the definition of $P_n(\mathbb{C})$ as the space of lines through the origin in $\mathbb{C}^{n+1}$ then you quickly see that fixed points of $f$ correspond to eigenvectors of any of the linear maps that projects to $f$. It is a standard fact that every linear map from $\mathbb{C}^{n+1}$ to itself has an eigenvalue and so every projective map has a fixed point
The case is slightly more subtle over the reals, but the idea is the same. Fixed points of maps from $P_n(\mathbb{R})$ to itself correspond to real eigenvectors of corresponding linear automorphisms of $\mathbb{R}^{n+1}$. While not every linear map of a real vector space has a real eigenvector, it is the case that if $n$ is even then every linear automorphism of $\mathbb{R}^{n+1}$ has a real eigenvector. This just comes from the fact that the characteristic polynomial has odd degree and hence has a real root. Thus if $n$ is even then every projective map from $P_n(\mathbb{R})$ to itself has a fixed point.
This is false if $n$ is odd. For instance the map defined by the matrix
$$ \begin{pmatrix} 0 & -1\\ 1 & 0 \end{pmatrix} $$
has no fixed points in $P_1(\mathbb{R})$. Taking block diagonal versions of this example gives fixed point free automorphisms of $P_n(\mathbb{R})$ for all odd $n$.