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I would like to know how the measure changes in $CP^1$ mapping from Riemann sphere (2-sphere) to $\mathbb{C}^2$-plane.

Let a point on the 2-sphere is given by the vector $\hat{n}(\mathbf{x})=(n^x(\mathbf{x}),n^y(\mathbf{x}),n^z(\mathbf{x}))$ with constraint $|\hat{n}|^2=1$. Now the $CP^1$ mapping is defined as follows \begin{equation} n^\alpha(\mathbf{x}) = \mathbf{z}^\dagger (\mathbf{x})\sigma^\alpha \mathbf{z}(\mathbf{x}), \quad \alpha=x,y,z \end{equation} where $\mathbf{z}(\mathbf{x})=[z_1(\mathbf{x}),z_2(\mathbf{x})]$ is a two component spinor, $z_1, z_2\in \mathbb{C}$ are two complex numbers, and $\sigma$ are Pauli matrices. The $CP^1$ mapping leads \begin{eqnarray} n^x &=& z_1^\ast z_2 + z_2^\ast z_1= 2\Re \left( z_1^\ast z_2 \right)\\ n^y &=& -iz_1^\ast z_2 + i z_2^\ast z_1=2\Im \left( z_1^\ast z_2 \right) \\ n^z &=&z_1^\ast z_1 - z_2^\ast z_2 \end{eqnarray} and \begin{equation} |\hat{n}|= |\mathbf{z}|^2 = |z_1|^2 + |z_2|^2=1. \end{equation}

Now my question is how to find measure $dn^xdn^ydn^z$ in terms of complex numbers $z_1$ and $z_2$ with appropriate normalization constant.

maxr
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  • $\mathbb CP_1 = S^2$ has 2 freedom degrees, while $\mathbb C^2$ has 4 freedom degrees, so you have no map from $\mathbb CP_1$ to $\mathbb C^2$, where $\mathbb C^2$ is the image of this map. You may think to stereographic projection, which is a map from $S^2$ to $\mathbb C$. – Trimok Sep 29 '13 at 13:55
  • @Trimok: Actually $\mathbf{z}\in\mathbb{ℂ}^2$ have 3 degrees of freedom [See constraint: $|z_1|^2+|z_2|^2=1$]. Additionally, the $\mathbf{z}$ fields have gauge symmetry [See Fradkin's book]. So degrees of freedom have no issue here. In literature, the $CP^1$ mapping, mapping from $S^2$ to $\mathbb{C}^2$, has been done for example in the case of nonlinear sigma model. – maxr Sep 29 '13 at 21:49
  • $dn^xdn^ydn^z$ (without constraint) is not a measure of $CP_1$ because $CP_1$ has only 2 degrees of freedom, so you had better to take $sin \theta ~d\theta ~d\phi$. Any equivalent measure should have only $2$ degrees of freedom too. So you have to express the constraints between $z_1$ and $z_2$ to eliminate the unnecessary $4-2=2$ degrees of freedom – Trimok Sep 30 '13 at 08:59
  • Thanks Trimok! May be I'm missing something. Could you please tell me what kind of constraint on $\hat{n}$ fields you are talking about as we already have constraint $|\hat{n}|^2=1$. I know there is mismatch in degrees of freedom. But as I said in Fradkin's book it is said that the gauge degree of freedom on $\mathbf{z}$ does not influence the physical results. – maxr Sep 30 '13 at 11:20

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