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Me again, still learning my lesson of "don't drink and derive":

I have got two parametrizations of the surface $H :=\{ (x,y,z) \in \mathbb{R}^3 \, | \, z^2 = 1+x^2+y^2, \, z > 0\}$,

$F:\mathbb{R}^2 \rightarrow H, \ (x,y) \mapsto (x,y,\sqrt{1+x^2+y^2})$

$G:D^2 \rightarrow H, \ (x,y) \mapsto (\frac{x}{\sqrt{1-x^2-y^2}}, \frac{y}{\sqrt{1-x^2-y^2}}, \frac{1}{\sqrt{1-x^2-y^2}})$

where $D^2$ denotes the open unit disk in $\mathbb{R}^2$. I take the first to give me a base of the tangent space at a point $F(x,y)$:

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

Now I triple checked the fact that $G$ really goes to $H$ and that

$\frac{\partial G}{\partial x}(x,y) = (\frac{1-y^2}{\sqrt{1-x^2-y^2}^3}, \frac{xy}{\sqrt{1-x^2-y^2}^3}, \frac{x}{\sqrt{1-x^2-y^2}^3})$

I even got the latter confirmed from a book. But then $\frac{\partial G}{\partial x}(x,y)$ should lie in the tangent space, hence be expressible as linear combination of the above basis vectors. Due to the $1$s and $0$s in our basis vectors the coefficients are easy to read off, we must have

$ \frac{\partial G}{\partial x}(x,y) = \frac{1-y^2}{\sqrt{1-x^2-y^2}^3}\frac{\partial F}{\partial x}(x,y) + \frac{xy}{\sqrt{1-x^2-y^2}^3}\frac{\partial F}{\partial y}(x,y)$

But then the third coordinate doesn't match:

$\frac{x}{\sqrt{1-x^2-y^2}^3} \neq \frac{1-y^2}{\sqrt{1-x^2-y^2}^3} \cdot \frac{x}{\sqrt{1+x^2+y^2}} + \frac{xy}{\sqrt{1-x^2-y^2}^3} \cdot \frac{y}{\sqrt{1+x^2+y^2}}$

The difference is exactly the factor of $\frac{1}{\sqrt{1+x^2+y^2}}$ which I can't get rid of. Where is my mistake (apart from drinking too much yesterday)? Thank you!

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    @gary: no worries, I'll delete my two comments, the previous one and this one soon, and if you delete yours, (almost) no one will know :) Such things happen to all of us... By the way, enjoy your coffee and hopefully a vacation, soon. Best wishes,2011-07-04

1 Answers 1

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Let $\psi: D^2 \to \mathbb{R}^2$ be given by $\psi(u,v) = \frac{1}{\sqrt{1-u^2-v^2}}(u,v)$. Then it is easy to check that $G(u,v) = F(\psi(u,v))$. Given this, I suggest to simply apply the chain rule in order to express $\frac{\partial G}{\partial u}$ as a linear combination of $\frac{\partial F}{\partial x}(\psi(u,v))$ and $\frac{\partial F}{\partial y}(\psi(u,v))$.

The reason your expressions didn't work out the way you expected is that $G(x,y) \neq F(x,y)$ (unless $x = y = 0$), so your tangent vectors live in different tangent spaces.