Fixed point spaces in $\mathbb{RP}^1$

Question

The fixed point space of a self-homomorphism of $\mathbb{RP}^1$, $f : \mathbb{R}^2 \rightarrow \mathbb{R}^2$, is given by:

$$Fix([f]) = \{[x,y] \in \mathbb{RP}^1 | [f(x,y)] = [x,y]\} \subseteq \mathbb{RP}^1$$

then let:

$$f_1: \mathbb{R}^2 \rightarrow \mathbb{R}^2; (x,y) \rightarrow (y,x)$$

$$f_2: \mathbb{R}^2 \rightarrow \mathbb{R}^2; (x,y) \rightarrow (-y,x)$$

The claim in my notes say that $Fix([f_1]) = \{[1,1],[1,-1]\}$ and $Fix([f_2]) = \emptyset$, however I don't quite get why this is the case. How does one work this out?

Answer 1

Observe that not every map $f : \mathbb{R}^2 \rightarrow \mathbb{R}^2$ induces a map $[f ]: \mathbb{RP}^1 \to \mathbb{RP}^1$. We need to know that $f$ maps lines through the origin to lines through the origin. This is certainly the case if $f$ is a linear isomorphism. So your maps $f_1, f_2$ are okay.

Note that each line through the origin has the form $[a,b] = \{ t (a,b) \mid t \in \mathbb R\}$ with $(a,b) \ne (0,0)$. We have $[a,b] = [c,d]$ iff $(a,b) = \lambda(c,d)$ for some $\lambda \in \mathbb R$.

1. We have $[f_1]([a,b]) = [b,a]$. Thus $[a,b]$ is a fixed point iff $(a,b) = \lambda(b,a)$. One of $a, b$ must be non-zero. If $a \ne 0$, we get $a = \lambda b = \lambda^2 a$, thus $\lambda = \pm 1$. The same is true if $b \ne 0$. For $\lambda = 1$ we get $a = b$, thus $[a,b] = [a,a] = [1,1]$. For $\lambda = 1$ we get $a = -b$, thus $[a,b] = [a,-a] = [1,-1]$.

2. We have $[f_2]([a,b]) = [-b,a]$. Thus $[a,b]$ is a fixed point iff $(a,b) = \lambda(-b,a)$. One of $a, b$ must be non-zero. If $a \ne 0$, we get $a = -\lambda b = -\lambda^2 a$, which is impossible. The same is true if $b \ne 0$. Thus there are no fixed points.

Answer 2

By the way, you need a condition like $f(\lambda x)=\lambda f(x)$ for $\lambda \in \Bbb{R}$ to ensure that your map $f$ induces a map on projective space. Happily, both of these transformations of $\Bbb{R}^2$ are linear and take lines to lines. $f_2$ is rotation by $\frac{\pi}{2}$ radians counterclockwise and hence it fixes no line through the origin. $f_1$ is a reflection through the line $y=x$. Hence, it preserves the line $y=x$ which corresponds to the point $[1:1]$ in $\Bbb{RP}^1$. It fixes also the line orthogonal to $x=y$, namely $y=-x$.

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If you want to see this another way, you can argue algebraically by manipulating the coordinates $[x:y]$. (Edit: I'd actually missed one until I checked the algebraic way.) For $f_1$, $[x:y]=[y:x]$ if and only if there exists $\lambda \in \Bbb{R}^\times$ such that $\lambda x=y$ and $\lambda y=x$. This implies $\lambda^2y=y$. Hence, $\lambda =\pm 1$. This corresponds to solutions (hence fixed points) $[1:1]$ and $[1:-1]$.

For $f_2$, $[x:y]=[-y:x]$ if and only if we can find $\lambda \in \Bbb{R}^\times$ such that $\lambda x=-y$ and $\lambda y=x$. Then $\lambda^2 y=-y$ and hence $y=0$. But then our only option is $[1:0]$ which we can see is not fixed.