# Differential of Hopf's map

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.

• 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. – 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.