smooth function

Sep 2010
117
4
Germany
show that

\(\displaystyle f(x,y,z)=\frac{1}{z^2+1}(x^2-y^2,2xy,2z)\) defines a smooth function \(\displaystyle f: S^{2}\to S^{2}\) where \(\displaystyle S^{2}=\{p\in \mathbb{R}^3, |p|=1\}\) 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

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


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


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

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

so I compute second order partial derivatives

\(\displaystyle \frac{\partial^2 f}{\partial x^2}=\frac{2}{z^2+1}\)

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

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

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



if I continue differentiating

\(\displaystyle \frac{\partial^3 f}{\partial x^3}=\frac{\partial^3 f}{\partial y^3}=0\) and

\(\displaystyle \frac{\partial^3 f}{\partial z^3}=-\frac{12(z^4-6z^2+1)}{(z^2+1)^4}\) 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
 
Last edited:
Mar 2010
293
91
Beijing, China
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.
 
Sep 2010
117
4
Germany
so showing all the computations above were not necessary if I am supposed to restrict myself to \(\displaystyle S^{2}\) right?
But I still don't get how to show \(\displaystyle \Biggl[(x^2-y^2)^2+(2xy)^2+(2z)^2\Biggr]\frac{1}{(z^2+1)^2}=1\) how does this imply that f will be smooth on the sphere???
 
Last edited:
Mar 2010
293
91
Beijing, China
\(\displaystyle (x^2-y^2)^2 + (2xy)^2 + (2z)^2=(x^2+y^2)^2 + (2z)^2\)
=\(\displaystyle (1-z^2)^2+(2z)^2=(1+z^2)^2\)

This shows that \(\displaystyle f|_{S^2}\) indeed maps to \(\displaystyle S^2\).

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