Complex lines in C^2

Problem: (a) Suppose {l_1} and {l_2} are two complex lines in {\mathbb{C}^2} which are not orthogonal. Prove that the projection of circles {\Gamma\subset l_1} to {l_2} are also circles.

(b) Suppose {l_1} is a 2 dimensional real plane in {\mathbb{C}^2} and {l_2} is a complex line. If for any circle {\Gamma\subset l_1}, its projection to {l_2} is a circle on {l_2}, then {l_1} must be a complex or anticomplex line. Proof: (a) Under some translation and rotation which preserves the projection and circles, we can assume

\displaystyle l_1=\{(z,wz)|z\in \mathbb{C}\}

\displaystyle l_2=\{(\xi,0)|\xi\in \mathbb{C}\}

where {w} is a fixed constant in {\mathbb{C}}. Here we have used the property that {l_2} is not orthogonal to {l_1}. Suppose {\Gamma} is circle in {l_1} with center {(z_0, wz_0)} with radius {r}, then points on {\Gamma} satisfy

\displaystyle r=|(z-z_0,w(z-z_0))|=|z-z_0|\sqrt{1+|w|^2}

which means {|z-z_0|=const}. This exactly means the projection on {l_2} is a circle.

(b) Any 2 dimensional real space through origin can be respresented by

\displaystyle l_1=\{\alpha\zeta+\beta\bar{\zeta}|\zeta\in \mathbb{C}\}

where {\alpha=(\alpha_1,\alpha_2)\in\mathbb{C}^2,\beta=(\beta_1,\beta_2)\in\mathbb{C}^2}. WLOG, assume {l_2} is the same as (a).

If {|\alpha_1|\neq |\beta_1|}, then let {z=\alpha_1\zeta+\beta_1\bar{\zeta}},

\displaystyle l_1=\left\{\left(z,\frac{(\bar{\alpha}_1\alpha_2-\bar{\beta}_1\beta_2)z+(\alpha_1\beta_2-\alpha_2\beta_1)\bar{z}}{|\alpha_1|^2-|\beta_1|^2}\right)\right\}

By projecting to {l_2}, we get {|z|=const} must imply {(\bar{\alpha}_1\alpha_2-\bar{\beta}_1\beta_2)z+(\alpha_1\beta_2-\alpha_2\beta_1)\bar{z}} has constant norm also. One can verify that this is only true when {\bar{\alpha}_1\alpha_2-\bar{\beta}_1\beta_2=0} or {\alpha_1\beta_2-\alpha_2\beta_1=0}, which means {l_1} must be an anti-complex line or complex line.

If {|\alpha_2|\neq |\beta_2|}, we can porject circles on {l_1} to {l_3=\{(0,\zeta)\}}, which are also circles by assumption. So we can apply all the above proof.

Next suppose {|\alpha_1|=|\beta_1|\neq 0} and {|\alpha_2|= |\beta_2|\neq 0}, first note that

\displaystyle \frac{|\alpha_1\zeta+\beta_1\bar{\zeta}|}{|\alpha_2\zeta+\beta_2\bar{\zeta}|}\equiv const

In fact, for any {\zeta}, there exists {c_1>0} such that

\displaystyle |\alpha_1\zeta+\beta_1\bar{\zeta}|+|\alpha_1\zeta+\beta_1\bar{\zeta}|=c_1

then {\zeta/c_1} will be in a circle in {l_1} whose center is origin and radius 1. So {|\alpha_1\zeta+\beta_1\bar{\zeta}|=c_1c_0} for some constant {c_0}. Then

\displaystyle \frac{|\alpha_1\zeta+\beta_1\bar{\zeta}|}{|\alpha_2\zeta+\beta_2\bar{\zeta}|}=\frac{c_1c_0}{c_1-c_1-c_0}=const

Choosing {\zeta} in particular, this constant must be {|\alpha_1|/|\alpha_2|=|\beta_1|/|\beta_2|}. From

\displaystyle \frac{|\alpha_1\zeta+\beta_1\bar{\zeta}|}{|\alpha_2\zeta+\beta_2\bar{\zeta}|}=\frac{|\alpha_1|}{|\alpha_2|}

Squaring it and write it as equality, we get

\displaystyle [|\alpha_2|^2\alpha_1\bar{\beta}_1-|\alpha_1|^2\alpha_2\bar{\beta}_2]\zeta^2=[|\alpha_1|^2\bar{\alpha_2}\beta_2-|\alpha_2|^2\bar{\alpha_1}\beta_1]\bar{\zeta}^2

Then

\displaystyle |\alpha_2|^2\alpha_1\bar{\beta}_1=|\alpha_1|^2\alpha_2\bar{\beta}_2

\displaystyle |\alpha_1|^2\bar{\alpha_2}\beta_2=|\alpha_2|^2\bar{\alpha_1}\beta_1

This two imply {\alpha_1/\beta_1=\alpha_2/\beta_2=\eta^{-1}}. Hence

\displaystyle l_1=\{(\alpha_1,\alpha_2)(\zeta+\eta\bar{\zeta})|\zeta\in\mathbb{C}\}

However, {|\zeta|=|\eta\bar{\eta}|} implies that

\displaystyle \text{Arg}\,(\zeta+\eta\bar{\zeta})=\frac{1}{2}\text{Arg}\,\eta

then {l_1} is in fact only one dimesional in this case. Contradiction. \Box

Remark: Shabat, Introduction to complex analysis Part II. Functions of several variables.

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