Definition of Tangent Bundle on $S^2$

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 $S^2$.

How exactly does the map $p$ specify a vector bundle $E(p)\to S^2$? What is $E(p)$?

Answer 1

The map given is not the usual form for a tangent bundle, but note that $\operatorname{Idem}(M_3(\mathbb{R}))$ is just the space of projections $\pi_V:\mathbb{R}^3 \to V$ onto a subspace $V\subset \mathbb{R}^3$. Thus we have a continuous map $x \to V_x$ (defining the topology here is a bit tricky, which may be one reason why the text does so in this way). We then have a vector bundle $E(p)\subset \mathbb{R}^3$ with fibers $V_x$ (skipping over the verification that that is in fact a well-defined, locally trivial vector bundle.) In fact, $V_x = \langle{x}\rangle^\perp\subset \mathbb{R}^3$, which agrees with the usual construction of the tangent bundle over $S^2$.