How does this give a vector bundle?!?

Question

My text says that the map $p:S^2\to \text{Idem}( \text{Mat}_3(\Bbb R))$ given by $p(x)v=v-\langle x,v\rangle x$ defines a vector bundle $E(p)$ on $S^2$.

So I assume that $E(p)=\bigcup_xp(x)\subset \text{Idem}( \text{Mat}_3(\Bbb R))$ and the vector bundle projection is $\pi:E(p)\to S^2, p(x)\mapsto x$. But this map isn't continuous since $p(x_n)\to p(x)$ only implies $x_n\to \pm x$.

What is going on here?

Answer 1

Is there a $\langle x, x \rangle $ term missing? Notice then $\langle x, p(x)v \rangle = \langle x, v \rangle - \langle x , v \rangle = 0$. So what you have is the Normal bundle.