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To paraphrase (only slightly) Guillemin and Pollack asks:

Given the function $f : R^3 \to R^1$ defined by: $$f(x,y,z) = x^2 + y^2 - z^2$$ Prove that if $a$ and $b$ are both positive or both negative then: $f^{-1}(a)$ is diffeomorphic to $f^{-1}(b)$.

My reasoning is that, $f^{-1}(a)$ is the locus of points in $R^3$ that satisfies:

$$a = x^2 + y^2 - z^2$$

In order to transform $f^{-1}(a)$ to $f^{-1}(b)$, the book hints scalar multiplication by $\sqrt{b/a}$. It doesn't seem to make sense to me why you would do that. It does seem like a good idea, however, to multiply by $b/a$. that way we get:

$$a \cdot \frac{b}{a} = \left(\frac{b}{a}\right)(x^2+y^2-z^2)$$

$$b = \left(\frac{b}{a}\right)(x^2+y^2-z^2)$$

So I am misunderstanding something, what is it?

Dair
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1 Answers1

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As pointed out by Daniel Fischer, I need to concern my self with $f(cx, cy, cz)$. Not $c \cdot f(x, y, z)$ as I have done. So:

\begin{align} f(\sqrt{b/a}x, \sqrt{b/a}y, \sqrt{b/a}z) &= (\sqrt{b/a}x^2) + (\sqrt{b/a}y)^2 - (\sqrt{b/a}z)^2 \\ &= \frac{b}{a}x^2 + \frac{b}{a}y^2 - \frac{b}{a}z^2 \\ &= \left(\frac{b}{a}\right)(x^2 + y^2 - z^2) \end{align}

And the rest is similar to the question.

Dair
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