3
$\begingroup$

Let $$f: \mathbb{R}^{2} \times \mathbb{R}^{2} \rightarrow \mathbb{R}^{2} \quad f(z, w)=z w$$ be the map induced by the multiplication of complex numbers. Check whether it is a $C^{\infty}$-map.

By definition, $f$ is itself smooth ($C^k$) if its coordinate presentations $\hat{f} \equiv \psi \circ f \circ \varphi^{-1}$ with respect to any pair of charts happen to be smooth.

Note that $\mathcal{A} \equiv\left\{\mathcal{O}_{\alpha}, \varphi_{\alpha}\right\}$ is an atlas. It seems that we need to construct two $C^{\infty}$-maps and verify $\hat{f}$ is $C^{\infty}$-maps.

Any suggestions on how to approach this problem?

Thanks in Advance!

$\endgroup$

1 Answer 1

Reset to default
1
$\begingroup$

The identification of $\Bbb{C}$ with $\Bbb{R}^2$ is usually done by $z=x+iy\mapsto (x,y)$. Implicitly, you have used that to define a map $f:\Bbb{R}^2\times \Bbb{R}^2\to \Bbb{R}^2$ which is the map $\mu:\Bbb{C}\times \Bbb{C}\to \Bbb{C}$ given by $\mu(z,w)=zw$ in coordinates. So, use the criterion you mentioned above to check to see that the component functions are smooth. That is, if $z=x_1+iy_1$ and $w=x_2+iy_2$ then $$zw=(x_1x_2-y_1y_2)+i(x_1y_2+x_2y_1).$$

Now, using this calculation can you see why your map is smooth? What are the component functions in local coordinates?

$\endgroup$
3
  • $\begingroup$ Ahh, thanks! Just to confirm; in your notation, $\mu$ is constructed to be the coordinate presentation, i.e. $\mu = \psi \circ f \circ \varphi^{-1}$. Since it is smooth, $f: \mathbb{R}^{2} \times \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}$ is smooth. Is my understanding right? $\endgroup$
    – Z.Y.H
    Jan 12 at 6:25
  • $\begingroup$ I think of it like this : $\mu:\Bbb{C}\times \Bbb{C}\to \Bbb{C}$ is defined on an "abstract manifold" of dimension $2$. It is the map that is not yet in real coordinates. It is defined by $\mu(z,w)=zw$. Then in local coordinates we parametrize $\Bbb{C}$ as above. Then $f:\Bbb{R}^2\times \Bbb{R}^2\to \Bbb{R}^2$ is the coordinate representation of $\mu$, which is the "abstract" map of manifolds. $\endgroup$
    – Alekos Robotis
    Jan 12 at 7:08
  • $\begingroup$ As a slight clarification of the above: you should think of $\Bbb{C}$ as an abstract $2-$dimensional real manifold without a priori choice of local coordinates. The identification $z=x+iy\mapsto (x,y)$ is a canonical choice of coordinates giving an ($\Bbb{R}-$linear) identification $\Bbb{C}\cong \Bbb{R}^2$. $\endgroup$
    – Alekos Robotis
    Jan 12 at 7:16

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.