2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.