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A complex line in $\mathbb{C}P^n$ is defined to be a copy of $\mathbb{C}P^1$. I know that this determines an embedding which can be represented in the following way: $$[u:v] \mapsto [ua_0+vb_0: \cdots :ua_n+vb_n]$$

where $[a_0:\cdots: a_n]$ and $[b_0:\cdots: b_n]$ are two points on $\mathbb{C}P^n$.

My questions are

  1. Do we have another way of representing it? Like in the real case we have the expressions like this.

  2. How many points do we need to determine a line? Like in the real case we have two distinct points in any dimension determine a unique line. How do we see that?

Evans Gambit
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    Be careful. Your description of a complex line in $\mathbb{CP}^n$ as a copy of $\mathbb{CP}^1$ is incomplete. For example, one could regard a twisted cubic as a copy of $\mathbb{CP}^1$ in $\mathbb{CP}^3$, but it is not a complex line. A complex line is a linearly embedded copy of $\mathbb{CP}^1$ in $\mathbb{CP}^n$. – Michael Albanese Sep 06 '21 at 12:57
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    Incidentally, 2 points in $\mathbb{C}P^n$ will determine a line in $\mathbb{C}P^n$. (pun absolutely intended!) – Evans Gambit Sep 06 '21 at 13:05
  • What do you mean by $u_0, v_0, \dots, u_n, v_n$? – Michael Albanese Sep 06 '21 at 13:32
  • @MichaelAlbanese Sorry that was typo. Editted. –  Sep 06 '21 at 13:44
  • @EvansGambit Thanks for the editting! It's then clear that two points determines a line. –  Sep 06 '21 at 14:15
  • @EvansGambit I just wonder how do you see that this line is unique? –  Sep 06 '21 at 14:24
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    If you think about $\mathbb{C}P^n=\mathbb{C}^{n+1}-0/\mathbb{C}^*$, then two distinct points $p, q\in \mathbb{C}P^n$, will correspond to linearly independent vectors $v_p, v_q$ in $\mathbb{C}^{n+1}$ and the line $pq$ through them corresponds to the unique 2-dimensional subspace generated by $v_p, v_q$. – Evans Gambit Sep 06 '21 at 14:48

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