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I have the following two sets

$\mathcal{S}= \left\lbrace (x,y,z,w) \in \mathbb{R}^4 \mid x^2+y^2- z^2-w^2 = 1 \right\rbrace$

and $\mathcal{S}' = \left\lbrace (x,y,z,w) \in \mathbb{R}^4 \mid x^2+y^2- z^2-w^2 = r \right\rbrace$

for some non-zero real number $r$.

I need to show that these two sets are diffeomorphic.

I considered taking two cases for $r$ ($r>0$ and $r<0$);

then considering the map $f :\mathcal{S} \rightarrow \mathcal{S}' $ where $f( x,y,z,w) = \sqrt{|r|} ( x,y,z,w) $ when $r>0$.

Is this correct?

Thank you.

Kerry H
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1 Answers1

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As you wrote yourself, your argument works only for $r\gt 0$.

If $r\lt 0$ the trick is to notice that the change of variables $x=Z,y=W,z=X, w=Y$ shows that $S$ is diffeomorphic to $Z^2+W^2-X^2-Y^2=1$ and thus to $x^2+y^2-z^2-w^2=-1$.
You can then apply your argument to show that you can replace the $-1$ on the right hand side of the equation $x^2+y^2-z^2-w^2=-1$ by any negative number $r$ to prove that $S$ is also diffeomorphic to the manifold $S'$ given by $x^2+y^2-z^2-w^2=r$ for any $r\lt 0$.

Georges Elencwajg
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  • Thanks alot... On a different note if I may. I know that $\mathcal{S}'$ is a 3-dimensional manifold when $r \neq 0$ (by the level set theorem). How about when $r=0$? – Kerry H Jan 08 '14 at 10:56
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    Dear Kerry, if $r=0$, then $S$ is not a manifold. It is a cone (= union of lines ) through the origin $O=(0,0,0,0)$ and is a manifold except at $O$, where one says that $S$ has a singularity . $S$ is an example of an algebraic variety, a more general structure in that an algebraic variety is allowed to have singular points. Singular points of an algebraic variety are points at which it is not a manifold but however at which it is not too wild, due to to the variety being described by polynomials rather than by arbitrary smooth functions. – Georges Elencwajg Jan 08 '14 at 11:09