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Consider an ordered set of five points $\{p_1, p_2, \dots, p_5\}$ in linear general position in $\mathbb{CP}^3$ and another ordered set of five points $\{q_1, q_2, \dots, q_5\}$, also in linear general position in $\mathbb{CP}^3$. These two sets are of course projectively equivalent, but is there a unique projective transformation that will map $p_i$ to $q_i$?

I seem to have an argument for it by translating the problem to mapping ordered sets of 1-dimensional subspaces in $\mathbb{C}^4$. However, I wish to confirm the result, just in case I may be making some mistake.

Thank you.

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  • $\begingroup$ I think the correct word is "uniqueness"? $\endgroup$ – Nishant May 2 '15 at 14:22
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Indeed there is a theorem which say:

Let $V,W$ be $(n+1)$-dimensional vector spaces further on let $P(V),P(W)$ be two $n$-dimensional projective spaces and assume that $$ A_1, \dots,A_{n+2} \in P(V) \quad \text{and} \quad B_1, \dots,B_{n+2} \in P(W) $$ are in general position. Then there exists a unique projective transformation $f \colon P(V) \to P(W)$ with $f(A_i) = B_i$ for $i = 1, \dots, n+2$.

So for $p_1, \dots, p_5 \in \mathbb{CP}^3$ and $q_1, \dots, q_5 \in \mathbb{CP}^3$ which are in general positions, resp. There exists a unique projective transformation $f \colon \mathbb{CP}^3 \to \mathbb{CP}^3$ such that $f(p_i) = q_i$ for $i= 1,\dots,5$.

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