# Inner product on tangent space of projective space

I just got stuck at a certain part of my book which concern projective spaces. Really need a help to understand this.

First, their definition.

They define an inner product in $V$ in order to induce an inner product $\mathbb{P}(V)$.

Finally, the inner product is defined.

What I don't get is how this is independent of the choice of the representative? For instance, we can choose $\lambda v$ to be representative, for some $\lambda\in\mathbb{C}$ such that $|\lambda|=2$. Then

$$\langle a,b \rangle_{\lambda v} = \frac{\langle a,b \rangle_\mathbb{R}}{\|\lambda v\|^2} = \frac{1}{4}\frac{\langle a,b \rangle_\mathbb{R}}{\| v\|^2} \neq \frac{\langle a,b \rangle_\mathbb{R}}{\|v\|^2}.$$

EDIT: Just to show all details, here is how they defined the charts and how they are working the tangent spaces.

Given non-zero $v$ we want to describe local coordinates of $P(V)$ in a neighborhood of the projective point $[v]={\Bbb C}v$. We do so through the orthogonal complement: $T_v=\{z: \langle a,v \rangle=0\}$.
Except for projective points in $P(T_v)$ we will expres any $[x]$ in the form $[v+w]$ for a suitable $w\in T_v$. To get this map let $x\in V\setminus T_v$ and let us seek $\lambda$ and $w\in T_v$ for which $$x = \lambda (v+w)$$ Taking scalar product with $v$ we get: $\langle x,v\rangle = \lambda \langle v,v\rangle$ from which we first isolate $\lambda$ and insert in the first to obtain $w$: $$\phi_v(x):= w = \frac{\langle v,v\rangle}{\langle x,v\rangle}x-v.$$ Now, for any non-zero $\mu$ we have $\phi_v(\mu x)=\phi_v(x)$ so this lifts to a well-defined map: $\hat{\phi}_v([x]) = \phi_v(x)$ and $\Psi_v^{-1}=\hat{\phi}_v: P(V)\setminus P(T_v) \rightarrow T_v$ is the wanted coordinate map.
Exercise: Given $v,v'$ and corresponding $[x]\in A_v\cap A_{v'}$ find the change of coordinate map taking $w\in T_v \mapsto w'\in T_{v'}$
• I'm very glad for your answer. Could you give me more details of how is this notion of representatives in $T_{[v]}P(V)$ ? Thanks. – Integral Aug 28 '16 at 20:02
• Ok, but the idea is that you treat the triple $v,a,b$ as projective (so they all scale with the same const $\lambda$) and then construct the scalar product. I think it is convenient to project away the $v$ direction in the scalar product (which does not seem to be the case in (14.13) above?) – H. H. Rugh Aug 28 '16 at 21:38
• My book is identifying $T_{[v]}P(V)$ with the orthogonal complement of $v$, which they denoted by $T_v$. This identification is made through the differential $D\phi_v(0):T_v\to T_{[v]}P(V)$, where $\phi_v$ is a local chart they defined. I'm getting confuse with this identification, I just can't make the calculations work out. They defined $\phi_v([x]) = \frac{\|v\|^2}{<x,v>}x - v.$ – Integral Aug 29 '16 at 1:40