1

I need help in this:

Let $F$ is $\mathbb{R},\mathbb{C},H$ ($H$ means the quaternions). Prove that the projective space $PF^n$ $$ PF^n= \{ F^{n+1}\backslash \{0\})/\sim | x{\sim}\lambda x , \lambda \neq 0\in{F} \}$$ is homeomorphic to the quoitent space $SF^{n}/{\sim}$ under the proportion equivalence $x\sim\lambda x$ for $\lambda\in{S^1_F}$. Where

$S^n_{F}={x\in{F^{n+1}}: ||x||=1}$

user652838
  • 762
  • 1
    Each element of $FP^n$ is a line without $0$, which intersects $S_F^n$ at two points. Show that these two points are equivalent wrt the equivalence relation defined on the sphere. Then identify the lines with the equivalence classes of the intersection points with the sphere. – Vercassivelaunos Oct 01 '20 at 09:00

1 Answers1

1

In this forum you will find several similar questions, but usually only for $F = \mathbb R$.

So let $p : F^{n+1} \setminus \{0\} \to PF^n$ denote the quotient map. Define $r : F^{n+1} \setminus \{0\} \to S^n_F, r(x) = \frac{x}{\lVert x \rVert}$. This map is a retraction.

Consider the restriction $q : S^n_F \to PF^n$.

  1. We have $q \circ r = p$. In fact, for each $x \in F^{n+1}$ we have $x \sim r(x) \in S^n_F$. Thus $q(r(x)) = p(r(x)) = p(x)$.

  2. $q$ is surjective. This follows from 1. since each $\xi \in PF^n$ has the form $\xi = p(x)$ which implies $\xi = q(r(x))$.

  3. We have $q(x) = q(y)$ if and only if $x = \lambda y$ for some $\lambda \in S^1_F$. The "if" part is trivial. So let $q(x) = q(y)$ which is the same as $p(x) = p(y)$, Hence there exists $\lambda \in F \setminus \{0\}$ such that $x = \lambda y$. This implies $1 = \lVert x \rVert = \lVert \lambda y \rVert = \lvert \lambda \rvert \lVert x \rVert = \lvert \lambda \rvert$, i.e. $\lambda \in S^1_F$.

It remains to show that $q$ is a quotient map, i.e. that $U \subset PF^n$ is open if and only if $q^{-1}(U) \subset S^n_F$ is open. Since $q$ is continuous, the "only if" part is trivial. So let $q^{-1}(U) \subset S^n_F$ be open. Hence $r^{-1}(q^{-1}(U)) \subset F^{n+1} \setminus \{0\}$ is open. But $r^{-1}(q^{-1}(U)) = (q \circ r)^{-1}(U) = p^{-1}(U)$, thus $p^{-1}(U)$ is open in $F^{n+1} \setminus \{0\}$. This implies that $U$ is open in $PF^n$.

Paul Frost
  • 87,968
  • Thank you for the deep explanation. just a notification, in 3 in the first rows it is $S^1_F$ right? And in the endit is $FP^n$. Showing that q is surjective and a quoitent map is enough for saying that it is homeomorphism, right? – user652838 Oct 02 '20 at 09:50
  • 1
    Yes, $S^n_F$ was a typo. I corrected it. If you have a quotient map (which includes surjective) $f : X \to Y$, then there is a unique homeomorphism $h : X/\sim \phantom{.} \to Y$ such that $h \circ \pi = f$. Here the definition of $\sim $ is $x \sim x'$ iff $f(x) = f(x')$, and $\pi : X \to X/\sim$ is the quotient map. See https://math.stackexchange.com/q/3064037. – Paul Frost Oct 02 '20 at 11:27