3 added 688 characters in body edited Aug 29 '16 at 20:54 H. H. Rugh 23.8k11 gold badge1212 silver badges3636 bronze badges 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 $$w\in T_v \mapsto w'\in T_{v'}$$ 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. 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 $$w\in T_v \mapsto w'\in T_{v'}$$ 2 added 688 characters in body edited Aug 29 '16 at 20:49 H. H. Rugh 23.8k11 gold badge1212 silver badges3636 bronze badges It is a bit indirect, but there is also a notion of representative onGiven non-zero $$T_[v]P(V)$$ with respect$$v$$ we want to describe local coordinates of $$P(V)$$ in a neighborhood of the baseprojective point $$[v]={\Bbb C}v$$. If you scaleWe do so through the orthogonal complement: $$T_v=\{z: \langle a,v \rangle=0\}$$. $$v$$ by 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 $$\lambda$$ then you also scaleand $$a$$$$w\in T_v$$ 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 $$b$$. Note that$$w$$: $$\phi_v(x):= w = \frac{\langle v,v\rangle}{\langle x,v\rangle}x-v.$$ Now, for any non-zero $$|\lambda|^2$$ is real$$\mu$$ we have $$\phi_v(\mu x)=\phi_v(x)$$ so it disappears inthis 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 ratiowanted coordinate map. It is a bit indirect, but there is also a notion of representative on $$T_[v]P(V)$$ with respect to the base point. If you scale $$v$$ by $$\lambda$$ then you also scale $$a$$ and $$b$$. Note that $$|\lambda|^2$$ is real so it disappears in the ratio. 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. 1 answered Aug 28 '16 at 19:58 H. H. Rugh 23.8k11 gold badge1212 silver badges3636 bronze badges It is a bit indirect, but there is also a notion of representative on $$T_[v]P(V)$$ with respect to the base point. If you scale $$v$$ by $$\lambda$$ then you also scale $$a$$ and $$b$$. Note that $$|\lambda|^2$$ is real so it disappears in the ratio.