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Let $P = \{(x, y, z) \in R^3\;|\; x = y\}$ (a plane) and let $X: U \subset R^2\to R^3$ be given by $X(u, v) = (u + v, u + v, uv)$, where $U = \{(u, v)\in R^2\; |\; u> v\}$. P. Is $X$ a parametrisation of $P$?

I know that $P$ is a plane and so it is a regular surface. So, I need to check if $X$ is a parametrisation by verifying only the following details:

Note that $X(U) \subset P$

  1. $X$ is smooth. This is clearly true as the components are smooth on $U$.
  2. $dX $ is one-one. This is true because $dX$ which is a $3\times 2$ matrix has rank $2$/
  3. $X$ is one-one. This is the only detail left to check. However, I am struggling in being able to decide if this is an injective map.

if $X(a,b) = X(x,y) \rightarrow (a+b,a+b,ab) = (x+y, x+y, xy)$ with $a>b$ and $x>y$.

How can I prove or disprove that this is injective?

chesslad
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1 Answers1

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Every points on $P$ can be written as $(y, y, z)$, where $y, z\in \mathbb R$.

To check if $X:U\to P$ is bijective, let $p = (y, y, z)\in P$. Then we need to find $(u, v)\in U$ so that

\begin{align} u+v=y ,\ \ \ uv = z. \end{align}

Or put it another way, $(u, v)$ are the roots of the quadratic polynomial

$$ t^2 - y t + z = 0,$$

which has solutions

$$ t = \frac{y}{2} \pm \frac{\sqrt{y^2 - 4z}}{2}.$$

This implies that if $X(u, v) = (y, y, z)$, $(u, v)$ must be (since $u>v$)

$$ u = \frac{y}{2} + \frac{\sqrt{y^2 - 4z}}{2}, v = \frac{y}{2} - \frac{\sqrt{y^2 - 4z}}{2}.$$

Hence $X$ is one-to-one.

Thus $X$ is a local parametrization: from the above it is also clear that $X$ is not surjective: the image of $X$ are precisely $(y, y, z)$ so that $y^2 - 4z >0$.

Arctic Char
  • 16,007