# Show that the multiplication of two complex numbers is a $C^{\infty}$ map

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?

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 $$zw=(x_1x_2-y_1y_2)+i(x_1y_2+x_2y_1).$$
• 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? Jan 12 at 6:25
• 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. Jan 12 at 7:08
• 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 $\Bbb{C}\cong \Bbb{R}^2$. Jan 12 at 7:16