Let $$h : \mathbb{C^2} \rightarrow \mathbb{C \times R} $$ $$h(z_1, z_2) = (2z_1z_2^*, |z_1|^2-|z_2|^2)$$

How do you find the differential of $h$ and show it is onto/surjective?

I know that I can express $h$ as $\mathbb{R^4}$ instead of $\mathbb{C^2}$, but then how do I proceed? Do I just differentiate with respect to each $x_1, x_2, x_3, x_4$ instead of $z_1, z_2$ given that $x_2, x_4$ are the imaginary part?

So, let's say $h(z_1, z_2)$ becomes $$h(x_1,x_2,x_3,x_4) = (2(x_1+x_2i)(x_3-x_4i),\ x_1^2+x_2^2-x_3^2-x_4^2) \\ \rightarrow (2(x_1x_3 + x_2x_4), 2(x_2x_3-x_1x_4), x_1^2+x_2^2-x_3^2-x_4^2)$$

But if I start differentiating $h$ with respect to each variable, I am getting a Jacobian matrix $\mathbb{R^{3 \times 4}}$. I think I am doing something wrong here.

  • $\begingroup$ Welcome to Math.SE ! I recommend that you share more of your work when asking a question. This will help to clarifiy the question and understand what is the difficult you encounter. For instance you could to show what you obtain if you differentiate $h$ both ways and make the question more precise. $\endgroup$ – Tom-Tom Dec 11 '18 at 21:14

In fact you must understand $h$ as a map from $\mathbb{R}^4$ to $\mathbb{R}^3$ and the derivative of $h$ as a the derivative in the sense of real multivariable calculus. You have done this almost correctly (I corrected a typo), and you are right that the Jacobian $Jh(x)$ of $h$ at $x$ is a $3 \times 4$-matrix. With respect to the standard bases of $\mathbb{R}^4, \mathbb{R}^3$ it is the matrix representation of the differential $Dh(x)$ of $h$ at $x$ which is linear map $Dh(x) : \mathbb{R}^4 \to \mathbb{R}^3$.

You have to determine for which $x$ the map $Dh(x)$ is a surjection. This is equivalent to determining when $Jh(x)$ has maximal rank, i.e. rank $3$. You have $$Jh(x) = \left( \begin{array}{rrrr} 2x_3 & 2x_4 & 2x_1 & 2x_2 \\ -2x_4 & 2x_3 & 2x_2 & -2x_1 \\ 2x_1 & 2x_2 & -2x_3 & -2x_4 \\ \end{array}\right) = 2 \left( \begin{array}{rrrr} x_3 & x_4 & x_1 & x_2 \\ -x_4 & x_3 & x_2 & -x_1 \\ x_1 & x_2 & -x_3 & -x_4 \\ \end{array}\right) = 2 M(x) .$$ You see that $Jh(0) = 0$. i.e. $Jh(0)$ has rank $0$. Let us show that the rank is $3$ if $x \ne 0$. Denote by $M_i(x)$ the matrix obtained from $M(x)$ be deleting the $i$-th column. Easy computations show that $$\det M_1(x) = -x_2(x_1^2 + x_2^2 + x_3^2 + x_4^2)$$ $$\det M_2(x) = -x_1(x_1^2 + x_2^2 + x_3^2 + x_4^2)$$ $$\det M_3(x) = -x_4(x_1^2 + x_2^2 + x_3^2 + x_4^2)$$ $$\det M_4(x) = -x_3(x_1^2 + x_2^2 + x_3^2 + x_4^2)$$ At least one of these four expressions is $\ne 0$ which proves our claim.


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