# What is the mapping from $\Bbb R^n$ to the (real) projective space?

I'm particularly interested in the case where $n=2$. The definition of $\mathbb RP^n$ as a manifold maps the point $(x_1,\ldots,x_n)$ in $\mathbb{R}^n$ by means of the function $\frac{x_i}{x_j}$ (where $i\ne j$) though it isn't specific about the range of indexes for the denominator.

So in the case of a point $p=(x,y)$ in $\mathbb R^2$ such that $(x,y)\ne (0,0)$, using the mapping we'd have $(x/y)$ and(/or?) $(y/x)$. The trouble here is that points that lie on either the x-axis or the y-axis result in division by zero.

Would a better mapping be $(x/r)$, $(y/r)$ where $r = \sqrt{x^2+y^2}$. As long as $(x,y)\ne(0,0)$, this avoids division by zero. And, for any non-zero real number $a$, it maps $(ax,ay)$ to $\frac{ax}{ar}$, and $\frac{ay}{ar} = (\frac{x}{r},\frac{y}{r})$. So am I missing something here?

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Two points $(x_1,y_1)$ and $(x_2,y_2)$ are in the same projective equivalence class if $x_1y_2 = y_1x_2$. This avoids the problem of dividing by zero. – orlandpm Mar 9 '13 at 0:14
I've edited your question to use $\LaTeX$. Please make sure it still represents your original intent. Specifically, I wasn't sure what you meant by $RP^n$, so I left it basically as-is. Also, the last paragraph was a little difficult to edit, so see if I did that right. :) For help with formatting in the future, please see this meta question. – apnorton Mar 9 '13 at 0:32
First of all, $\mathbb RP^n$ is the image of $\mathbb R^{n+1}\setminus 0$, not of $\mathbb R^n$. In particular, you are defining $\mathbb RP^1$, not $\mathbb RP^2$ – Thomas Andrews Mar 9 '13 at 0:33
anorton, your editing is correct. Thanks! – Mr X Mar 9 '13 at 0:33
More generally, it is easier to define $\mathbb RP^n$ in terms of equivalence classes, not fractions (where division by zero is an issue.) $(x_0,\dots,x_n)\sim (y_0,\dots,y_n)$ if there is a $\lambda\in\mathbb R\setminus\{0\}$ such that $x_i=\lambda y_i$ for all $i$. – Thomas Andrews Mar 9 '13 at 0:37