Show that $P(V^{*})$ may be interpreted as the set of all hyperplanes in $V$ ($V^{*}=\operatorname{Hom}(V,\mathbb{R}))$

Question

I am starting to study cohomology of manifolds using the Bott To book. Trying to solve some exercises in the book I have run into a problem and I little bite lost on it. Let $V$ be a real vector space and $V^{*}=\operatorname{Hom}(V,\mathbb{R})$ I need to prove that :

1. $P(V^{*})$ can be seen as the set of all hyperplanes in $V$.(Solved)

2. Let $Y\subset P(V)\times P(V^{*})$ be defined by:

$Y=\bigl\{([v]),[H])\mid H(v)=0,v\in V, H\in V^{*}\bigr\}$

How can I compute $H^{*}(Y)$?

My attempt: I think that $Y$ is the projectivization of the tautological bundle $S$ but I only know how to compute cohomology of the projectivization in complex vector bundles and this one is real.

Any help will be appreciated!!

Answer 1

1. Any two non zero functionals are scalar multiples of each other if and only if they share the same kernel. The kernel of a non zero functional is a hyperplane and every hyperplane can be written as such a kernel. Hence, $P(V^*)$ can be seen as the set of all hyperplanes by associating the projectivisation of any non zero functional to its kernal.

2. What is your question regarding this?