smooth function

**rayman** · February 18th 2013, 12:39 AM

show that

$f(x,y,z)=\frac{1}{z^2+1}(x^2-y^2,2xy,2z)$ defines a smooth function $f: S^{2}\to S^{2}$ where $S^{2}=\{p\in \mathbb{R}^3, |p|=1\}$ so I am guessing it is a unit sphere.

How do we proceed with such problem?

f is a smooth function if all partial derivatives of all possible orders are defined in all points of the domain of f right?

I started with computing these partial derivatives

$\frac{\partial f}{\partial x}=\frac{1}{z^2+1}(2x,2y,0)$

$\frac{\partial f}{\partial y}=\frac{1}{z^2+1}(-2y,2x,0)$

$\frac{\partial f}{\partial z}=\frac{-2(z^2-1)}{(z^2+1)^2}$

and these partial derivatives are all defined at any point from the domain of f

so I compute second order partial derivatives

$\frac{\partial^2 f}{\partial x^2}=\frac{2}{z^2+1}$

$\frac{\partial^2 f}{\partial y^2}=\frac{-2}{z^2+1}$

$\frac{\partial^2 f}{\partial z^2}=\frac{4z(z^2-3)}{(z^2+1)^3}$

and these 3 are also defined for all the points of the domain.

if I continue differentiating

$\frac{\partial^3 f}{\partial x^3}=\frac{\partial^3 f}{\partial y^3}=0$ and

$\frac{\partial^3 f}{\partial z^3}=-\frac{12(z^4-6z^2+1)}{(z^2+1)^4}$ which is defined for all points

What should I do next? can I conclude based on the calculations that f is certainly smooth?

Thanks in advance

**xxp9** · February 18th 2013, 10:06 AM

f is obviously smooth on R^3. So what you need to show is, when f is restricted to S^2, its image is in S^2.
That is, to show that [(x^2-y^2)^2 + (2xy)^2 + (2z)^2]/(1+z^2)^2 = 1, given that x^2+y^2+z^2=1.
Which is easy to show.

**rayman** · February 18th 2013, 10:19 AM

so showing all the computations above were not necessary if I am supposed to restrict myself to $S^{2}$ right?
But I still don't get how to show $\Biggl[(x^2-y^2)^2+(2xy)^2+(2z)^2\Biggr]\frac{1}{(z^2+1)^2}=1$ how does this imply that f will be smooth on the sphere???

**xxp9** · February 18th 2013, 03:58 PM

$(x^2-y^2)^2 + (2xy)^2 + (2z)^2=(x^2+y^2)^2 + (2z)^2$
= $(1-z^2)^2+(2z)^2=(1+z^2)^2$

This shows that $f|_{S^2}$ indeed maps to $S^2$ .

Then $f$ is smooth because "the restriction of a smooth function on a smooth sub-manifold is smooth".
To prove the above statement, let $i: S^2 \rightarrow R^3$ is the smooth embedding of $S^2$ in $R^3$ ,
$f : R^3 \rightarrow S^2$ is a smooth map on $R^3$ , then $f|_{S^2} = f \circ i$ is smooth on $S^2$ .

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