let $H$ be a real subspace of $C^3\setminus\{0\}$ of codimension $2$. $\pi (H)$ defined like a subvarietie of CP^2,?

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let $H$ be a real subspace of $C^3\setminus \{0\}$ of codimension $2$. is $\pi (H)$ defined like a subvarietie of $CP^2$, where Quotient map $$\pi:C^3\setminus \{0\}→CP^2$$ is the projection.

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I believe that in general you will get a codimension-one linear real subvariety of $\Bbb CP^2$. I suggest you try the example of the real codimension-two plane given by $y_1=y_2=0$, when we write $z_j = x_j+\sqrt{-1}y_j$, $j=0,1,2$. I believe that in that case you get the subvariety of $\Bbb CP^2$ given by the closure of $\{[1,Z_1,Z_2]: Z_2 \text{ is a real multiple of } Z_1\}$, which you can see easily enough is a real $3$-dimensional submanifold.

EDIT: That is, in fact, the general case. Suppose we start with the two real linear equations $$a_0x_0+a_1x_1+a_2x_2+b_0y_0+b_1y_1+b_2y_2 = a'_0x_0+a'_1x_1+a'_2x_2+b'_0y_0+b'_1y_1+b'_2y_2 = 0.$$ You can form the two complex vectors $c=a+ib, c'=a'+ib'\in\Bbb C^3$. Then this real $4$-dimensional subspace of $\Bbb C^3$ is a complex $2$-dimensional subspace if and only if $c$ and $c'$ are linearly dependent over $\Bbb C$. When they are linearly independent, we can find an element of $GL(3,\Bbb C)$ (which, of course, induces a linear automorphism of $\Bbb CP^2$) carrying $c,c'$ to $(1,0,0)$ and $(0,1,0)$, respectively, and this is (equivalent to) the case we already discussed.

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I'm now looking on this quetion : can we indendify any real subspace of codimension 2 (of $\Bbb{C}^3$) To $\lbrace a_1z_1+a_2z_2+a_3z_3+2\varepsilon R(a_1z_1)=0\rbrace ?~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$ let $a_1,a_2,a_3$ in $\Bbb{C}\setminus\lbrace0\rbrace$ and R is The real part . I use $\varepsilon$, $( 2\varepsilon R(a_1z_1))$, to get small real perturbation of $\lbrace a_1z_1+a_2z_2+a_3z_3=0\rbrace$ such that it become real subspace of $\Bbb{C}^3$. I want to know if any real subspace of $\Bbb{C}^3$ of real codimension 2 can be writen H= $\lbrace a_1z_1+a_2z_2+a_3z_3+2\varepsilon R(a_1z_1)=0\rbrace$ (maybe by some tronsformation). we can even take $a_1=1,a_2=2,a_3=3.$