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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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1  
I haven't checked carefully, but you certainly want to choose $(x_0,y_0)$ and $(x_1,y_1)$ such that $F(x_0,y_0) = G(x_1,y_1)$ and at a glance $F(x,y) \neq G(x,y)$, so no wonder the things don't match up, as they live in different tangent spaces. –  t.b. Jul 3 '11 at 18:16
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Oh my god, my name is well chosen - thank you! –  Embarassed Guy Jul 3 '11 at 18:30
    
(If you make this an answer I will certainly accept it) –  Embarassed Guy Jul 3 '11 at 18:33
    
@Theo: I don't see why $\frac{\partial F}{\partial x}(x,y)=(1,0,\frac{x}{\sqrt{1+x^2+y^2}}), \ \ \frac{\partial F}{\partial y}(x,y)=(0,1,\frac{y}{\sqrt{1+x^2+y^2}})$ are a basis for the tangent space? AFAIK, the partials in the directions of the standard axes, i.e., in the direction of $e_i$ are the standard choice for basis. Is that what these are, otherwise, how do you know they are independent? –  gary Jul 3 '11 at 23:47
    
@Theo: Of course.I will rename myself "Embarrassed Guy" #2 re my last post; I need either a vacation, or a quadruple espresso ( or both); I'll walk to Starbucks after this post. –  gary Jul 4 '11 at 0:34

1 Answer 1

up vote 3 down vote accepted

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.

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