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!


1 Answer 1


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?

  • $\begingroup$ 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. $\endgroup$ 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 $\Bbb{C}\cong \Bbb{R}^2$. $\endgroup$ Jan 12 at 7:16

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