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I have a follow-up question to this one.

We define $\mathbb{P}^n\mathbb{C}$ as $(\mathbb{C}^{n+1}-\{0\})/(a\sim ta)$. So, if we include $0$ in our definition, then we will have that all points are equivalent to $0$ as $$ta\sim a\sim 0\cdot a\sim 0$$ i.e. we will have one equivalent class. Also, more geometric explanation: projective space is the space of lines, but $0$ is not a line.

Is it correct?

user89898989
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    Yes, the projective space is the space of lines and $0$ is not a line. You can include $0$ as a point without using it as a scalar but then e.g. the result will fail to be a manifold. – Qiaochu Yuan Sep 18 '20 at 06:02

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No, the relation is $v \sim w \iff \exists \hspace{0.1cm} \lambda \in \mathbb{K}^{*} = \mathbb{K}-\left\lbrace 0 \right\rbrace : w = \lambda w$. So $v \sim 0 \iff \exists \hspace{0.1cm} \lambda \in \mathbb{K}^{*} : 0 = \lambda v$, but $\lambda \ne 0$, do you see what happens ?

jacopoburelli
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