Complex projective line (or $\mathbb{CP}^1$) is the Hilbert space where pure qubits live in. It is defined as the quotient space of $\mathbb{C}^2$ and $\mathbb{C}$, both of which are punctured in their respective zero elements.

$$\mathbb{CP}^1 \stackrel{\text{def}}{=} (\mathbb{C}^2 \setminus\{0\})/\mathbb{C}^\times\qquad \mathbb{C}^\times\stackrel{\text{def}}{=}\mathbb{C} \setminus \{0\}$$

To do quantum computing, we usually use the Hilbert space $\mathbb{C}^2$. That's the Hilbert space in which all qubit state vectors (or qubit “kets”) $\lvert\psi\rangle$ live in. However, many elements in this space correspond to something unphysical. For example, it contains state vectors whose probabilities that don't add up to 100%. There are also elements which we would consider to be duplicates of each other – up to a global phase.

Here are the 3 major issues with the space $\mathbb{C}^2$

1) The zero vector is in $\mathbb{C}^2$ but it's unphysical. Applying the “Born rule” yields 0% + 0% = 0%. Meaning, the qubit doesn't even exist. This violates the second axiom of probability which states that probabilities must add up to 100%. $$\lvert\psi\rangle = \begin{pmatrix}0\\ 0 \end{pmatrix}\qquad \lvert\psi\rangle\in\mathbb{C}^2 \qquad |0|^2 + |0|^2 = 0$$

2) There are many other states which violate the “Born rule” and are thus considered unphysical. For example, the following is a state which is in state $\lvert 0\rangle$ with 300% probability and in state $\lvert 1\rangle$ with 200% yielding the total probability of 500%. But according to the second axiom of probabilitiy, the probabilities must always add up to exactly 100%. $$\lvert\psi\rangle = \begin{pmatrix}\sqrt{3} \\ \sqrt{2}\end{pmatrix}\qquad \lvert\psi\rangle\in\mathbb{C}^2\qquad |\sqrt{3}|^2 + |\sqrt{2}|^2 = 3 + 2 = 5$$

3) Two states can differ by some phase $e^{i\phi}$ called the global phase. This phase does not have any effect on – in fact it is completely unobservable. Therefore, we $$\lvert\psi_1\rangle = \begin{pmatrix}1\\ 0\end{pmatrix}\qquad \lvert\psi_2\rangle = \begin{pmatrix}i \\ 0\end{pmatrix}\qquad \lvert\psi_1\rangle = e^{i\pi/2}\lvert\psi_2\rangle\qquad \lvert\psi\rangle \sim e^{i\phi}\lvert\psi\rangle$$.

We construct $\mathbb{CP}^1$ by resolving each of these issues individually.

1) We request that the zero state vector is not in $\mathbb{CP}^1$ $$\lvert\psi\rangle = \begin{pmatrix}0\\0\end{pmatrix}\qquad \lvert\psi\rangle\notin\mathbb{CP}^1$$

2) We can always normalize $$\lvert 0\rangle = \left\{\begin{pmatrix}1\\0\end{pmatrix}, \begin{pmatrix}2\\0\end{pmatrix}, \begin{pmatrix}-1\\ 0\end{pmatrix}\right\}\qquad \lvert \psi\rangle\sim A\lvert\psi\rangle, A\in\mathbb{R},A\neq 0$$

3)