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Problem: I'd like to parametrize the manifold given by $\{(x,y,z)\in{\mathbb R}^{3}\,|\, x^2 + y^2 - z^2 = a\}$ for $a < 0$. The two mappings we'd use are $f(x,y) = (x,y,\sqrt{x^2 + y^2 - a})$ and a similar map for the "bottom part."

This is what it looks like for $a = -1$.

It was noted that I should prove that these are diffeomorphisms by showing that $d(f)_{(x,y)}:R^{2}\to T(M_{1})_{f(x,y)}$ with $M_{1}$ being the "top part" is nonsingular as a linear transformation and then apply the inverse function theorem. My problem is showing this map is non-singular: what exactly is the derivative here? I thought that it might be something like

$\left(\begin{array}{ccc} 1 & 0 & \frac{-x}{\sqrt{x^2 + y^2 - a}}\\ 0 & 1 &\frac{-y}{\sqrt{x^2 + y^2 - a}}\end{array}\right)$

but this isn't square and so cannot (I think!) be singular or non-singular by definition. Is there something I'm missing here?

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I thought all you needed to do was prove that the map is a bijection and both the map and the inverse are differentiable? – Muphrid Feb 12 '13 at 3:10
That's probably true; though, this method was suggested to me, and I'm not sure I understand it. I wanted to make sure there wasn't a huge hole in my understanding. – james Feb 12 '13 at 4:09
Right, I'm not sure singular is the right word to apply here. What you should be able to ensure is that the Jacobian takes a point $p$ and a vector $v \in \mathbb R^2$ and maps $v$ to some nonzero vector in $T_p M$ for all nonzero $v$. If this is the case, you should be good, and the transformation you have here is then a valid coordinate chart. – Muphrid Feb 12 '13 at 4:14
The solutions I was looking at are at, just for reference; it's the first problem in the solutions of the first problem set, I think. – james Feb 12 '13 at 4:45
@Murphid Ah, that makes sense. But the matrix above will take me from ${\mathbb R}^2$ into ${\mathbb R}^3$ --- do I need to show this linear transformation is actually going into $T_{p}M$? – james Feb 12 '13 at 4:47
up vote 1 down vote accepted

Let's go partway through the problem. Let $f(p') = p$, where $p' \in \mathbb R^2$ and $p \in M$. The Jacobian is then

$$\underline f(a) = a \cdot \nabla' f(p')$$

You can then find $\underline f(e_1)$ and $\underline f(e_2)$. These two vectors define the tangent space. We know the tangent space must be a plane, so the cross product of these vectors (both of which are in $\mathbb R^3$) gives the normal to the plane that is the tangent space at the point $p$.

Now then, to ensure that this is the case, $\underline f(a) \neq 0$ for any $a \neq 0$. Why is this so? Because then you could take $b,c$ such that $a = b-c$ and then $\underline f(b) = \underline f(c)$. This would make the Jacobian, as a linear operator, no longer bijective. You can't "invert" it in the usual matrix algebra sense anyway, as it isn't square, but the Jacobian of $f$ and the Jacobian of $f^{-1}$ have the property that, when used consecutively, they can project vectors in $\mathbb R^3$ onto the tangent space.

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Ah, that makes a lot more sense than what I was thinking. I don't think I've ever seen the notation $\nabla'$ though with the prime after the del; what does that mean? – james Feb 12 '13 at 5:24
Just that $\nabla'$ is associated with $p'$ (with $\mathbb R^2$) and not with $p$ (which is in $M$ or $\mathbb R^3$). A benefit of the notation is that you can write $\nabla$ (really, the projection of $\nabla$ onto $T^*_pM$) in terms of $\nabla'$ and the Jacobian of $f^{-1}$. – Muphrid Feb 12 '13 at 5:46

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